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PROBLEMS & SOLUTIONS IN GROUP TH€ORY FOR PHYSICISTS

GROUP TH€ORY FOR PHYSICISTS

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PROBLEMS & SOLUTIONS IN GROUP TH€ORY FOR PHYSICISTS

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PROBLEMS & SOLUTIONS IN GROUP THEORY FOR PHYSICISTS

Zhong-Qi Xicro-Yan

MQ Gu

Institute of High Energy Physics China

vp World Scientific N E W JERSEY - LONDON * SINGAPORE B E l J l N G * S H A N G H A I - HONG KONG TAIPEI * CHENNAI

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Published by

World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA oflce: Suite 202, 1060 Main Street, River Edge, NJ 07661 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Ma, Zhongqi, 1940 -

p. cm. Problems and solutions in group theory for physicists / by Z.Q. Ma and X.Y. Gu.

Includes bibliographical references and index. ISBN 98 1-238-832-X (alk. paper) -- ISBN 98 1-238-833-8 (pbk.: alk. paper)

1. Group theory. 2. Mathematical physics. I. Gu, X.Y. (Xiao-Yan). 11. Title.

QC20.7.G76 M3 2004 530.15’22--d~22 2004041980

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright 0 2004 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereox may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any informution storage und retrievul system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

Printed in Singapore by World Scientific Printers (S) Pte Ltd

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Preface

Group theory is a powerful tool for studying the symmetry of a physical system, especially the symmetry of a quantum system. Since the exact solution of the dynamic equation in the quantum theory is generally difficult to obtain, one has to find other methods to analyze the property of the system. Group theory provides an effective method by analyzing symmetry of the system to obtain some precise information of the system verifiable with observations. Now, Group Theory is a required course for graduate students major in physics and theoretical chemistry.

The course of Group Theory for the students major in physics is very different from the same course for those major in mathematics. A graduate student in physics needs to know the theoretical framework of group theory and more importantly to master the techniques in application of group theory to various fields of physics, which is actually his main objective for taking the class. However, no course or textbook on group theory can be expected to include all explicit solutions to every problem of group theory in his research field of physics. A student of physics has to know the fundamental theory of group theory, otherwise he may not be able to apply the techniques creatively. On the other hand, physics students are not expected to completely grasp all the mathematics behind group theory due to the breadth of the knowledge required.

One of the authors (Ma) first taught the group theory course in 1962. Since 1986, he has been teaching Group Theory to graduate students mainly major in physics a t the Graduate School of Chinese Academy of Sciences. In addition, most of his research work has been related to applications of group theory to physics. In 1996 the Chinese Academy of Sciences decided to publish a series of textbooks for graduate students. He was invited to write a textbook on group theory for the series. In his book, based on his

v

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vi Problems and Solutions in Group Theory

experience in teaching and research work, he explained the fundamental concepts and techniques of group theory progressively and systematically using the language familiar to physicists, and also emphasized the ways with which group theory is applied to physics. The textbook (in Chinese) has been widely used for Group Theory classes in China since it was published by Science Press in Beijing six years ago. He is honored and flattered by the tremendous reception the book has received.

By the request of the readers, an exercise book on group theory by the same author was published in 2002 by Science Press to form a complete set of textbooks on group theory. In order to make the exercise book self-contained, a brief review of the main concepts and techniques is given before the problems in each section. The reviews can be used as a concise textbook on group theory. The present book is the new edition of that book. A great deal of new materials drawn from teaching and research experiences since the publication of the previous edition are included. The reviews of each chapter has been extensively revised. Last four chapters are essentially new.

This book consists of ten chapters. Chapter 1 is a short review on lin- ear algebra. The reader is required not only to be familiar with its basic concepts but also to master its applications, especially the similarity trans- formation method. In Chapter 2, the concepts of a group and its subsets are studied through examples of some finite groups, where the importance of the multiplication table of a finite group is emphasized. Readers should pay special attention to Problem 17 of Chapter 2 and Problem 14 of Chap- ter 3 which demonstrate a systematic method for analyzing a finite group. The theory of representations of a group is studied in Chapter 3. The trans- formation operator PR for the scalar functions bridges the gap between the representation theory and the physical application. The subduced and in- duced representations of groups are used to construct the character tables of finite groups in Chapter 3, and to study the outer product of representa- tions of the permutation groups in Chapter 6. The symmetry groups T for a tetrahedron, 0 for a cube and I for an icosahedron are studied in Chap- ters 3 and 4. The Clebsch-Gordan series and coefficients are introduced in Chapter 3 and are calculated for various situations in the subsequent chapters. The calculated results of the Clebsch-Gordan coefficients and the Clebsch-Gordan series for the group I and for the permutation groups listed in Problem 27 of Chapter 3 and Problem 31 of Chapter 6, due to its complexity, are only for reference.

The classification and representations of semisimple Lie algebras are

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Preface vi i

introduced in Chapter 7 and partly in Chapter 4 by the language familiar to physicists. The methods of block weight diagrams and dominant weight diagrams are recommended for calculating the representation matrices of the generators and the Clebsch-Gordan series in a simple Lie algebra. The readers who are interested in the strict mathematical definitions and proofs in the theory of semisimple Lie algebras are recommended to read the more mathematically oriented books, e.g. [Bourbaki (1989)l.

The remaining part of the book is devoted to the properties of some important symmetry groups of physical systems. In Chapter 4 the symme- try group SO(3) of a spherically symmetric system in three dimensions is studied. The unitary representations with infinite dimensions of the non- compact group are discussed with the simplest example SO(2,l) in Problem 25 of Chapter 4. In Chapter 5 we introduce the symmetry of the crystals. More attention should be paid to the analysis method for the symmetry of a crystal from its International Notation (Problem 6). The commonly used matrix groups are studied in the last three chapters, while the Lorentz group is briefly discussed in Chapter 9.

The systematic examination of Young operators is an important char- acteristic of this book. We calculate the characters, the representation ma- trices, and the outer product of the irreducible representations of the per- mutation groups using the Young operators. The method of block weight diagrams can only symbolically give the basis states in the representation space of a Lie algebra. However, for the matrix groups SU(N), SO(N) and Sp(24, which are related to four classical Lie algebras, the basis states can be explicitly calculated using the Young operators. The relationship between two methods for the irreducible representations of the classical Lie algebras is demonstrated in the last three chapters in detail. The di- mensions of the irreducible representations of the permutation groups, the SU(N) groups, the SO(N) groups, and the Sp(2l) groups are all calculated with the hook rule, a method based on the Young diagram.

In summary, this book is written mainly for physics students and young physicists. Great, care has been taken to make the book as self-contained as possible. However, this book reflects mainly the experiences of the authors. We sincerely welcome any suggestions and comments from the readers. This book was supported by the National Natural Science Foundation of China.

Institute of High Energy Physics Beijing, China December, 2003

Zhong-Qi Ma Xiao-Yan Gu

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Contents

Preface V

1 . REVIEW ON LINEAR ALGEBRAS 1 1.1 Eigenvalues and Eigenvectors of a Matrix . . . . . . . . . . 1 1.2 Some Special Matrices . . . . . . . . . . . . . . . . . . . . . 1.3 Similarity Transformation . . . . . . . . . . . . . . . . . . . 7

4

2 . GROUP AND ITS SUBSETS 27

29 33

2.1 Definition of a Group . . . . . . . . . . . . . . . . . . . . . . 27 Subsets in a Group . . . . . . . . . . . . . . . . . . . . . . .

2.3 Homomorphism of Groups . . . . . . . . . . . . . . . . . . . 2.2

3 . THEORY OF REPRESENTATIONS 43 3.1 Transformation Operators for a Scalar.Function . . . . . . . 43 3.2 Inequivalent and Irreducible Representations . . . . . . . . 47 3.3 Subduced and Induced Representations . . . . . . . . . . . 65 3.4 The Clebsch-Gordan Coefficients . . . . . . . . . . . . . . . 79

4 . THREE-DIMENSIONAL ROTATION GROUP 115 4.1 SO(3) Group and Its Covering Group SU(2) . . . . . . . . . 115 4.2 Inequivalent and Irreducible Representations . . . . . . . . 123

Lie Groups and Lie Theorems . . . . . . . . . . . . . . . . . 140 4.4 Irreducible Tensor Operators . . . . . . . . . . . . . . . . . 146 4.5 Unitary Representations with Infinite Dimensions . . . . . . 166

4.3

5 . SYMMETRY OF CRYSTALS 173 5.1 Symmetric Operations and Space Groups . . . . . . . . . . 173

ix

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X Problems and Solutions in Group Theory

5.2 Symmetric Elements . . . . . . . . . . . . . . . . . . . . . . 5.3 International Notations for Space Groups . . . . . . . . . .

6 . PERMUTATION GROUPS 6.1 Multiplication of Permutations . . . . . . . . . . . . . . . .

6.4 Irreducible Representations and Characters

6.2 Young Patterns, Young Tableaux and Young Operators . . 6.3 Primitive Idempotents in the Group Algebra . . . . . . . .

6.5 The Inner and Outer Products of Representations . . . . . . . . . . . . . .

7 . LIE GROUPS AND LIE ALGEBRAS

7.2 Irreducible Representations and the Chevalley Bases . . . . 7.1 Classification of Semisimple Lie Algebras . . . . . . . . . .

7.3 Reduction of the Direct Product of Representations . . . .

8 . UNITARY GROUPS 8.1 8.2 Irreducible Tensor Representations of SU(N) . . . . . . . . 8.3 Orthonormal Bases for Irreducible Representations . . . . .

The SU(N) Group and Its Lie Algebra . . . . . . . . . . . .

8.4 Subduced Representations . . . . . . . . . . . . . . . . . . . 8.5 Casimir Invariants of SU(N) . . . . . . . . . . . . . . .. . .

9 . REAL ORTHOGONAL GROUPS 9.1

9.3

Tensor Represent ations of SO ( N ) . . . . . . . . . . . . . . . 9.2 Spinor Representations of SO(N) . . . . . . . . . . . . . . .

SO(4) Group and the Lorentz Group . . . . . . . . . . . . .

10 . THE SYMPLECTIC GROUPS 10.1 The Groups Sp(2l. R) and USp(2l) . . . . . . . . . . . . . . 10.2 Irreducible Representations of Sp( 2t) . . . . . . . . . . . . .

177 186

193 193 197 205 211 237

269 269 279 299

317 317 321 336 362 369

375 375 403 415

433 433 440

Bibliography 457

Index 461

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PROBKMS &I SOLUTIONS IN GROUP THEORY FOR PHYSICISTS

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Chapter 1

REVIEW ON LINEAR ALGEBRAS

1.1 Eigenvalues and Eigenvectors of a Matrix

* The eigenequation of a matrix R is

Ra = Xa, (1.1)

where X is the eigenvalue and a is the eigenvector for the eigenvalue. An eigenvector is determined up to a constant factor. A null vector is a trivial eigenvector of any matrix. We only discuss nontrivial eigenvectors.

Equation (1.1) is a set of linear homogeneous equations with respect to the components ap. The necessary and sufficient condition for the existence of a nontrivial solution to Eq. (1.1) is the vanishing of the coefficient determinant :

= + (-X)”-’TrR + , . . + det R = 0,

where rn is the dimension of the matrix R, TrR is its trace (the sum of the diagonal elements), and detR denotes its determinant. Equation (1.2) is called the secular equation of R. Evidently, this is an algebraic equation of order m with respect to A, and there are rn complex roots including multiple roots. Each root is an eigenvalue of the matrix. For a given eigenvalue A, there is at least one eigenvector a obtained from solving Eq. (1.1). However, for a root X with multiplicity n, it is not certain to obtain n linearly independent eigenvectors from Eq. (1.1).

1

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2 Problems and Solutions in Group Theory

The rank of a matrix R is said to be r if only r vectors among rn row-vectors (or column-vectors) of R are linearly independent. If the rank of ( R - X l ) is r , then (rn - r ) linearly independent eigenvectors can be obtained from Eq. (1.1). Any linear combination of eigenvectors for a given eigenvalue is still an eigenvector for this eigenvalue.

For a lower-dimensional matrix or for a sparse matrix (with many zero matrix entries), its eigenvalues and eigenvectors can be obtained by guess from experience or from some known results. So long as the eigenvalue and the eigenvector satisfy the eigenequation (l.l), they are the correct results. On the other hand, even if the results are calculated, they should also be checked whether Eq. (1.1) is satisfied.

* A matrix R is said to be hermitian if Rt = R. R is said to be unitary if Rt = R-l. A real and hermitian matrix is a real symmetric matrix. A real and unitary matrix is a real orthogonal matrix. R is said to be positive definite if its eigenvalues are all positive. R is said to be positive semi- definite if its eigenvalues are non-negative. A negative definite or negative semidefinite matrix can be defined similarly.

1. Prove that the sum of the eigenvalues of a matrix is equal to the trace of the matrix, and the product of eigenvalues is equal to the determinant of the matrix.

Solution. The secular equation of an m-dimensional matrix R is an alge- braic equation of order rn. Its rn complex roots, including multiple roots, are the eigenvalues of the matrix R. So, the secular equation can also be expressed in the following way:

m

det(R- X l ) = n ( X j - A ) j=l

m m

j= 1 j=1

= 0.

Comparing it with Eq. (1.2), we obtain that the sum of the eigenvalues is the trace of the matrix and the product of the eigenvalues is equal to the determinant of the matrix.

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Review on Linear Algebras 3

2. Calculate the eigenvalues and eigenvectors of the Pauli matrices 01 and u2 :

a , = ( ; ;), 0 2 = ( i 0 -i 0 ) .

Solution. The action of 01 on a vector is to interchange two components of the vector. A vector is an eigenvector of 01 for the eigenvalue 1 if its two components are equal. A vector is an eigenvector of 01 for the eigenvalue -1 if its two components are different by sign.

.1(:)=(:>. Q(!l)= - (1'). Similarly, if two components of a vector are different by a factor fi, then it is an eigenvector of 0 2 :

. 2 ( ] ) = ( ] ) , ",( -i '>= - ( ' ) . -i

Sometimes a matrix R may contain a submatrix 01 or 0 2 in the form of direct sum or direct product. Thus, some eigenvalues and eigenvectors of R can be obtained in terms of the above results. A matrix R is said to contain a two-dimensional submatrix in the form of direct sum if for given a and b, R,, = R,, = Rbc = Rcb = 0, where c is not equal to a and b.

3. Calculate the eigenvalues and eigenvectors of the matrix R

Solution. R can be regarded as the direct sum of two submatrices 01, one lies in the first and fourth rows (columns), the other in the second and third rows (columns). F'rom the result of Problem 2, two eigenvalues of R are 1, the remaining two are -1. The relative eigenvectors are as follows.

1 : (i) , (i). - l : ( i ) , -1 (;l).

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4 Problems and Solutions an Group Theory

4. Calculate the eigenvalues and eigenvectors of the matrix R

Solution. This is the generalization of 01. The action of R on a vector is to make a cycle on the three components of the vector in order. R3 = 1, so the eigenvalues of the matrix R is 1, w and w 2 , where w = exp{-i2~/3}. The ratio of two adjacent components is for each eigenvalue is:

the eigenvalue. The eigenvector

5. If det R # 0, prove that both RtR and tian matrices.

RRt are positive definite hermi-

Solution. It is obvious that both RtR and RRt are hermitian. Letting X and a be the eigenvalue and the eigenvector of RtR, where the eigenvector is written in the form of a column matrix:

(RtR)g = Xa, at, > 0,

we have

A {ata} = at(RtR)a = ( R G ) ~ (Ra) 2 0,

namely, X 2 0. Since detR # 0, det(RtR) # 0 and X # 0. Thus, (RtR) is positive definite. Since detRt # 0, one can similarly prove that RRt is positive definite.

1.2 Some Special Matrices

* The inner product of two column matrices a and b is defined as

Two column matrices are called orthogonal to each other if their inner product is vanishing. Q~CJ must be a non-negative real number, called the

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Review on Linear Algebras 5

square of the module of a. at, = 0 only when g = 0.

* R is called a unitary matrix if RtR = 1. The column (or row) matrices of R are normalized and orthogonal to each other:

D D

P P

A unitary transformation R does not change the inner product of any two vectors:

(1.5) t (Rg) (Rb) = gtRtRb = gtb.

The module of any eigenvalue of a unitary matrix is one, and its two eigen- vectors for different eigenvalues are orthogonal to each other. The module of the determinant of a unitary matrix is one, I det RI = 1. Letting Ra = Ag and Rb = d, we have

The product of two unitary matrices is a unitary matrix. There are rn lin- early independent orthonormal eigenvectors for an rn-dimensional unitary matrix.

A real unitary matrix R is called a real orthogonal matrix. The product of two real orthogonal matrices is a real orthogonal matrix. The complex conjugate vector a* of any eigenvector g of a real orthogonal matrix R for the eigenvalue X is an eigenvector of R for the eigenvalue A*. The module of the determinant of an orthogonal matrix R is det R = fl .

* R is called a hermitian matrix if Rt = R, or REu = Rup. A hermitian matrix R satisfies

Any eigenvalue of a hermitian matrix R is real, and two eigenvectors of R for different eigenvalues are orthogonal to each other. The determinant of a hermitian matrix is real. Letting Rg = Xa and Rb = rb, we have

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6 Problems and Solutions in Group Theory

The sum of two hermitian matrices is a hermitian matrix. There are rn linearly independent orthonormal eigenvectors for an rn-dimensional her- mitian matrix.

A real hermitian matrix R is called the real symmetric matrix. The sum of two real symmetric matrices is a real symmetric matrix. There are rn linearly independent or thonormal real eigenvectors for an rn-dimensional real symmetric matrix.

6. Prove: (I) if RtR = 1, then RRt = 1;

(2) if R-lR = 1, then RR-l = 1;

(3) if RTR = 1, then RRT = 1.

Solution. The method for proving these three equalities are similar. We prove the first equality as example.

Since RtR = 1, Rt are nonsingular. Let S be the inverse matrix of Rt, SRt = 1. So RRt = (SRt) RRt = S (RtR) Rt = SRt = 1. Replacing Rt with R-' or RT, we can prove the remaining two equalities.

7. Find the independent real parameters in a 2 x 2 unitary matrix, a real orthogonal matrix and a hermitian matrix, and give their general ex- pressions.

Solution. A 2 x 2 complex matrix contains four complex parameters, i.e., eight real parameters. For a unitary matrix, the normalization conditions of two columns give two real constraints for the parameters, and the or- thogonal condition of two column matrices gives a complex constraint, i.e., two real constraints. Altogether, there are four real constraints. A 2 x 2 unitary matrix contains four independent real parameters. For a hermitian matrix, its diagonal elements are real, that is, their imaginary parts are zero. It leads to two real constraints. Two non-diagonal matrix entries are complex conjugate to each other. It leads to a complex constraint. Thus, a 2 x 2 hermitian matrix also contains four independent real parameters.

A 2 x 2 real matrix contains four real parameters. For a real orthog- onal matrix, the normalization conditions of two columns give two real constraints for the parameters. The orthogonal condition of two column matrices gives a real constraint. There are three real constraints altogether. Thus, a 2 x 2 real orthogonal matrix contains only one independent real parameter .

Letting u be an arbitrary 2 x 2 unitary matrix, we have detu = exp{icp}. Attracting the factor exp{icp/2}, we obtain

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Review on Linear Algebras 7

From the definition,

aa* + bb* = cc* + dd* = 1, ac* + bd* = 0 , ad - bc = 1,

we have

a = a ( c c * + d d * ) = d * ( - b c + a d ) = d * , b = b(cc* + dd*) = c* (bc - ad) = -c*.

So, the general expression for a 2 x 2 unitary matrix is

U = e ip/2 ( ;!*) aa* + bb* = 1,

where two complex parameters a and b with a constraint contain three real parameters. Together with cp, there are four independent real parameters.

A real orthogonal matrix is also a unitary matrix. Its determinant may be fl. When its determinant is 1, cp = 0 and both a and b are real numbers, satisfying a2 + b2 = 1. Usually, we take a = cose and b = sine. When the determinant is -1, one may change the sign of the matrix elements in the first row. Thus, we obtain the general expression of a 2 x 2 real orthogonal matrix

sin8 c o d -cos8 sin8 , or R' = cos8 -sin8

sine C O S ~ R = (

The general expression of a 2 x 2 hermitian matrix is

where a, b, c and d all are independent real parameters.

1.3 Similarity Transformation

Ir In an m-dimensional space L, any vector a can be expanded in the chosen bases e,,

m

a = C e,aiL. ,=l

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8 Problems and Solutions an Group Theory

a, is called the component of the vector a in the basis e,. Arrange these components in one column matrix a with rn rows, which is called the column matrix form of a vector a with respect to the bases e,,

* An rn-dimensional space L is said to be invariant for an operator R if any vector in C is transformed by R to a vector belonging to C. Specially, the basis e, is transformed by R to a linear combination of the bases,

(1.10)

One may arrange the combination coefficients D,,(R) as an rn x rn matrix D ( R ) which is called the matrix form of an operator R in the bases e,.

* Choose a new set of the bases e i . Their column matrix forms S., in the original bases e, can be arranged as a matrix S:

m

e i = C e,s,,. (1.11) ,=I

This matrix S is called the transformation matrix between two sets of bases. S is an rn-dimensional nonsingular matrix. In the new bases e:, the matrix forms of a vector a and an operator R are

m m

(1.12) ,=l p= 1

The relations between two matrix forms in two sets of bases are -

- a' = S-'a, D ( R ) = S-lD(R)S. (1.13)

The matrix D ( R ) is transformed into the matrix D(R) through the similar- ity transformation S. The matrices related by a similarity transformation have the same eigenvalues. They are the matrix forms of one operator in two sets of bases. It is also said in literature that B(R) is equivalent to D(R) .

Write the similarity transformation in the form of matrix entries: m m m

S., is the column matrix form of the new basis ei in the original bases, while B,,(R) in Eq. (1.14) plays the role of the combination coefficients.

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Review o n Linear Algebras 9

So Eq. (1.14) is the matrix form of Eq. (1.12) in the original bases e,. This form is very useful in later calculation.

* If the first column of S is an eigenvector for XI of the matrix D ( R ) , then all the components in the first column of D(R) are vanishing except for the first one, which is equal to XI. If all columns of S are the linearly independent eigenvectors of D ( R ) , then D(R) is a diagonal matrix where the diagonal elements are the eigenvalues of the corresponding eigenvec- tors. This is the essence of diagonalization of a matrix through a similarity transformation.

The necessary and sufficient condition for diagonalization of an m- dimensional matrix D ( R ) is that there must exist m linearly independent eigenvectors for D ( R ) . Each column of the similarity transformation ma- trix S which diagonalizes D ( R ) is the eigenvector of D ( R ) . Such a matrix S is not unique because it may be right-multiplied by a matrix X which commutes with the diagonal matrix D(R). If the m eigenvalues of D ( R ) are different from each other, then X is a diagonal matrix. This means that each eigenvector may include an arbitrary constant factor. If the multiplic- ity of an eigenvalue X j of D ( R ) is nj, C j nj = m, then X is a block matrix, containing n; parameters. These parameters reflect the arbitrary linear combination between eigenvectors for the same eigenvalue.

A unitary matrix or a hermitian matrix can be diagonalized by a unitary similarity transformation. A real symmetric matrix can be diagonalized by a real orthogonal similarity transformation. A real orthogonal matrix can be diagonalized by a unitary similarity transformation, but generally not by a real orthogonal similarity transformation, A unitary matrix remains unitary in any unitary similarity transformation. A hermitian matrix re- mains hermitian in any unitary similarity transformation. A real symmetric matrix remains real symmetric in any real orthogonal similarity transfor- mation.

8. Find the similarity transformation to diagonalize the following matrices:

Solution. We will denote the given matrix by D ( R ) .

*

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10 Problems and Solutions in Group Theory

(1) The secular equation of D ( R ) is

-A3 + 2X2 - 4X + 8 = (2 - X)(X2 + 4) = 0.

We obtain the eigenvalues 2, 2i and -2i. Noticing that the matrix D ( R ) / 2 is a real orthogonal matrix, so its eigenvectors are orthogonal to each other. If the second component of an eigenvector is zero, we see from the second row of D ( R ) that the first component of the eigenvector has to be equal to its third component. Then, from the first or third row of D ( R ) we know that (1, 0, l)T is the eigenvector for the eigenvalue 2. Due to orthogonality, the remaining eigenvectors take the form of (1, a, -l)T. Substituting it into the eigenequation

we obtain a = ria. After normalization, we have

(2) The secular equation of D ( R ) is

The eigenvalues are exp( M ) . Substituting them into the eigenequation,

we obtain a = ri. After normalization, we have

9. Find a similarity transformation matrix M which satisfies

0 - cos 8 sin 8 sin cp -1 0 M - l ( case 0 -sinecosy) 0 M = (! 0 0 0 0 0 ) I

- sin 8 sin cp sin 8 cos cp

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Review on Linear Algebras 11

Solution. Denote by D ( R ) and D(R) the matrices before and after the similarity transformation M, respectively. D(R) is the direct sum of -ia2 and 0, so it is commutable with matrix X:

If M satisfies M - l D ( R ) M = D(R) , M X also satisfies this equation. Namely, the similarity transformation M is determined up to the three parameters .

From a(R) we see that the eigenvalues of D ( R ) are 0 and f i . The third column of M is the eigenvector for the zero eigenvalue, and the first two columns are the linear combinations of eigenvectors for the eigenvalues f i . We first calculate the eigenvector for the zero eigenvalue. Take the third component of the eigenvector to be cos 8, which is a choice for the parameter y. From the eigenequation, the first two components of the eigenvectors are calculated to be sin B cos cp and sin 8 sin cp, respectively. Let

al a2 sin8coscp M = bl b2 sinesincp . ( c1 c2 case )

Substituting M into

we obtain

a2 = -bl cos 8 + c1 sin 8 sin cp, b2 = a1 cos8 - c1 sinOcoscp, c2 = -a1 sin 8 sin cp + bl sin 8 cos cp, -a1 = 4 2 cos 8 + c2 sin 8 sin cp, -bl = a2 cos 8 - c2 sin 0 cos cp, -c1 = -a2 sin 8 sin cp + b2 sin 8 cos cp.

Substituting the first three equations into the last three, or vice versa, we have

a1 sinecoscp + bl sin8sincp + c1 cos8 = 0, a2 sin8coscp + bz sinosincp + c2 c o d = 0.

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12 Problems and Solutions an Group Theory

These two relations reflect the fact that the first two column vectors of M are orthogonal to the third one. The reason is that D ( R ) is antisymmetric. In fact, denoting by b the eigenvector for the zero eigenvalue and D(R)a = Xa # 0, we have bTD(R) = 0, and

0 = bTD(R)a = XbTa.

Now, choosing c1 = - sine, we obtain a solution from the orthogonality: a1 = cos8coscp and bl = cos8sincp. It is a choice for the parameters (u and p. Substituting the results into the preceding formulas, we have a2 = - sincp, bz = coscp and c2 = 0. Finally, we obtain the real orthogonal matrix M :

cos 8 cos cp - sin cp sin 8 cos cp cos8sincp coscp sin8sincp - sin8 0 cos e

10. Find a similarity transformation matrix M which satisfies the following three equations simultaneously

1 0 0

0 0 0 0 0 -1

M - l ( : y 0 0 ; i ) i M = A 0 (i 0 1 0 y A), 0 O i -1 0

-i 0 0 0 1

Solution. The first equation is the diagonalization of a matrix by a simi- larity transformation M , whose three column matrices are eigenvectors of the matrix. In order to satisfy the last two equations, we have to keep the multiplied factors in the eigenvectors temporarily, i.e.,

M = ia 0 -ic . (:: :) Choose the common factor of A4 such that b = 1. Substituting Jd into the second equation, where M-' is moved to the right-hand side of the

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Review on Linear Algebras 13

equation to avoid the calculation of M - l , we obtain

0 0 0 0 a + c 0 (-", ;i 8) = 5 (; i (a;c) ;).

Ad=+ -1 0 0 -i). 1

fi 0 4 0

After checking, it does satisfy the third equation.

11. Let

Find the common similarity transformation matrix X satisfying

1 0 0 0

0 0 0 !l), X - l ( R x R ) X = (: 0' 0

o o f i -1

Solution. According to the definition of the direct product of matrices,

Since the matrices before and after the similarity transformation X are both real orthogonal, we can also choose X to be real orthogonal. R x R is a diagonal matrix. It is easy to see that its eigenvectors for the eigenvalues

Thus,

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14 Problems and Solutions an Group Theory

1 and -1 respectively are

Thus, X is in the form of

Since the first two columns of X have to be the eigenvectors of S x S for the eigenvalue 1, we obtain a = b and c = -d. Because X is a real orthogonal matrix, we have a' = -b' and c' = d'. Choosing a = c = c' and calculating the application of S x S to the fourth column of X , we have

Thus, a' = -a. After normalization we obtain

1 x=- Jz

1 0 - 1 0 0 1 0 1 0 -1 0 1 1 0 1 0

12. Find the similarity transformation matrix X to diagonalize the follow- ing three matrices simultaneously,

' 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0

\ 0 0 1 0 0 0

' 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0

, 0 1 0 0 0 0

/ 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0

\ l l l O O O

Solution. Both the first two matrices are the direct sum of three 01, but different in the rows (columns). In the first matrix, the submatrices relate to the rows (columns) (1 ,4) , (2,5) and (3,6), while in the second matrix

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Review on Linear Algebras 15

they relate to (1,4), (2,6) and (3,5). So, three eigenvalues for each of two matrices are 1, and the other three are -1. The corresponding eigenvectors are as follows:

d e

d

e

f

f

1 -1

The eigenequations for the third matrix are

~ 0 0 0 1 1 1 ' 0 0 0 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0

\ I 1 1 0 0 0

d' e' f'

-f' - d'

-el -1

P s + t + u s + t + u s + t + u :;=I S t p + q + r p + q + r

U p + q + r

,

S

t ':I* U

Except for X = 0, the first three components must be equal to each other, so do the last three components. Namely, the eigenvalues are f3, and the eigenvectors are

1 + 3 : - 4

1 7 - 3 : - 4

1 1 1

-1 -1 -1

These two vectors are also the eigenvectors of the first two matrices for the eigenvalues 1 and -1, respectively. In the remaining subspace, all the eigenvalues of the third matrix are zero, but the eigenvalues of the first two matrices are f 1, respectively. Arranging the six eigenvectors, we obtain

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16 Problems and Solutions in Group Theory

the similarity transformation matrix X

which diagonalize all three matrices simultaneously:

diag(1, - 1, 1, 1, - 1, - l},

diag{l, - 1, 1, - 1, 1, - l},

diag (3, - 3, 0, 0, 0, 0 ) .

13. Show the general form of an m x m matrix, both unitary and hermitian.

Solution. Both a unitary matrix and a hermitian matrix can be diago- nalized by a unitary similarity transformation. After diagonalization, the module of each diagonal element of the unitary matrix is 1, while each diag- onal element of the hermitian matrix is real. So, each diagonal element in the diagonalized matrix, both unitary and hermitian, must be 1 or -1. We denoted by rn the diagonalized matrix where the first n diagonal elements are 1, and the last m - n diagonal elements are -1. In general, a matrix both unitary and hermitian is in the form of UI',U-l, where U is a unitary matrix with determinant one.

14. Prove that any unitary matrix R can be diagonalized by a unitary simi- larity transformation, and any hermitian matrix R can be diagonalized by a unitary similarity transformation.

Solution. We prove this proposition by induction. Obviously, the proposi- tion holds when the dimension of R is one. Assuming that the proposition holds when the dimension of R is (m - l), we are going to prove that it holds when the dimension of R is rn. Let A1 be an eigenvalue of a unitary matrix R with dimension m and S(t) be the normalized eigenvector of it. Arbitrarily find a complete set of orthonormal column matrices S!;), where S!:) is the first one in the set. Thus, we obtain a unitary matrix S(l). After the similarity transformation S( l ) , (S('))-' RS(l) keeps unitary, where all its matrix entries in the first column are vanishing except for the first one which is X I . Since [ A l l 2 = 1 and (S(l))- ' RS(l) is a unitary matrix, all its

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Review on Linear Algebras 17

matrix entries in the first row vanish except the first one.

where the submatrix R(l) is unitary and (m - 1)-dimensional. Therefore, R(l) as well as R can be diagonalized by a unitary similarity transformation. It means that there exist m orthonormal eigenvectors for an m-dimensional unitary matrix.

If R is a hermitian matrix, we introduce the similarity transforma- tion S( l ) in the same way. After the similarity transformation S(l), ($I))- RS(l) keeps hermitian, where all its matrix entries in the first col- umn are vanishing except for the first one which is A1. Since ( S ( l ) ) -' RS(l) is hermitian, all its matrix entries in the first row vanish except the first one. The remaining proof is the same as that for a unitary matrix. It means that there exist m orthonormal eigenvectors for an m-dimensional hermitian matrix.

1

15. Prove that R and Rt can be diagonalized by a common unitary similar- ity transformation if Rt is commutable with R. Further prove that the necessary and sufficient condition for a matrix R which can be diago- nalized by a unitary similarity transformation is that Rt is commutable with R.

Solution. The first part of this problem is similar to the proof given in the preceding problem. We prove the first part of the present problem by induction. Obviously, it holds when the dimension of R is one. Assuming that it holds when the dimension of R is (m - 1), we are going to prove that it holds when the dimension of R is m. We introduce the similarity transformation S( l ) in the same wa . After the similarity transformation S( l ) , (S('))-' Bt (S(l)) and ( S ( l ) ) RS(l) are still conjugate to each other and commutable with each other. Because

-7

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18 Problems and Solutions in Group Theory

we have

Thus, all rn vanish. The submatrix R(l) is still commutable with ( I t ( ' ) ) + and is (m - 1)-dimensional. Therefore, R and Rt can be diagonalized by a unitary similarity transformat ion simultaneously.

Conversely, if R can be diagonalized by a unitary similarity transfor- t mation S, then S-lRtS = (S- lRS) is also diagonal. Since two diagonal

matrices are commutable, R and Rt are also commutable.

16. Prove that any matrix can be transformed into a direct sum of the standard Jordan forms, each of which is in the form

when a = b

the remaining cases. or 1 when a + 1 = b

Solution. As is well known, for each eigenvalue, there exists at least one corresponding eigenvector. But when the eigenvalue is multiple, i t is not certain if the same number of linearly independent eigenvectors as the mul- tiplicity of the eigenvalue can be found. If there are m linear independent eigenvectors of an rn-dimensional matrix R, the similarity transformation matrix X by arranging the eigenvectors as its column matrices can diag- onalize R. Now, we only need to deal with the case where the number of linearly independent eigenvectors is less than, at least for one eigenvalue, the multiplicity of the eigenvalue. We will divide this problem into three propositions. First, any matrix R can be transformed to an upper trian- gular matrix, where the diagonal elements are the eigenvalues of R and the same eigenvalues are arranged together. Second, by a further similar- ity transformation, R can be transformed into a block matrix, where each submatrix relates to only one eigenvalue. In other words, any nondiago- nal matrix entry related to two different eigenvalues is vanishing. Third, each submatrix related to a given eigenvalue can be transformed into the standard Jordan form by a similarity transformation.

We will prove the first proposition by induction. The proposition ob- viously holds for a one-dimensional matrix. If the proposition holds for any n-dimensional matrix where n < m. We are going to show that the proposition also holds for any rn-dimensional matrix R.

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Review on Linear Algebras 19

Let R have d different eigenvalues X j with multiplicity nj, respectively. For X I , say, we can find one eigenvector S.1. Arbitrary choose a complete set of vector bases S., where S.l is the first one in the set. Arrange them into an rn-dimensional non-singular matrix S. After the similarity transformation S, R becomes R' = S-lRS, where all matrix entries in the first column are vanishing except for the first one which is X I . Removing the first row and the first column in the matrix R', we obtain an (rn - 1)-dimensional submatrix A of R'. In comparison with the set of eigenvalues of R, the multiplicity of A1 in the set of eigenvalues of A decreases by one. Being an ( rn - 1)-dimensional matrix, as we supposed, A can be transformed into an upper triangular matrix A' = X-lAX by an ( rn - 1)-dimensional similarity transformation X . On the diagonal line the matrix entries of A' are arranged in the order: X I , . .., X I , Xz, ..., Xa, . .., Ad. Let an rn-dimensional matrix T be the direct sum of the digit one and the (rn - 1)- dimensional matrix X. Then, R" = T-lR'T = (ST)- lR(ST) satisfies the first proposition: R" is an upper triangular matrix, where the diagonal elements are the eigenvalues of R and the same eigenvalues are arranged toget her.

For convenience, we remove the double primes on R". Its matrix entry is denoted by Rja,kb, where the row (column) subscript is designated by two indices ( j a ) , 1 _< j _< d, 1 5 a 5 nj:

when j > k when j = k and a > b when j = k and a = b Rja,kb =

Rja,kb the remaining cases.

The second proposition says that there exists a similarity transformation X such that after the transformation all Ria,$, with j # k become vanishing. Let S(ab) be a matrix, where all the diagonal matrix entries are one, and the nondiagonal matrix entries are vanishing except for sj::ib = w , where j < k. Denote by T(ab) the inverse of S(ab) for convenience. In fact, the difference between S(ab) and T(ab) is that w is replaced with -w. By this similarity transformation, R' = T(ab) RS("'), only the following matrix entries are changed:

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20 Problems and Solutions in Group Theory

when k < k' or when k = k' and b < b',

Rj la l ,kb - - Rj'aJ,kb -k RjJa',jaSja,kb (ab) = Rj'aJ ,kb -k wRjlaJ,ja,

when j ' < j or when j ' = j and a' < a .

Choosing

we have R i a , k b = 0. Now, we annihilate the nondiagonal matrix entries R j a , k b with j < k by a series of the above similarity transformations S(ab) one by one in the following order. First, beginning with j = 1, k = 2 and b = 1, we annihilate Rla,21 one by one in the decreasing order of a from n1 to 1. Second, increasing 13 by one from 1 to nk, we repeat the above process to annihilate Rla,2b in the decreasing order of a. Third, increasing k by one from 2 to d, for a given k we decrease j by one from k - 1 to 1, and for the given j and k we repeat the above two processes to annihilate R j a , k b .

Thus, after a series of similarity transformations, R is transformed into a block matrix, where each submatrix is an upper

when j # k when j = k when j = k R j a , k b =

R j a , j b when j = k

triangular matrix, namely,

and a > b and a = b and a < b.

The third proposition says that each submatrix related to a given eigen- value can be transformed into the standard Jordan form by a similarity transformation. Let a submatrix, denoted still by R for convenience, be n-dimensional and have the same eigenvalue A. Assume that the rank of the matrix ( R - X l ) is (n - t ) , namely, there are t linearly independent eigenvectors of R for the eigenvalue A. According to the first proposition, we can find a similarity transformation matrix X , where the first t columns are the eigenvectors of R, such that after the similarity transformation X , R becomes an upper triangle matrix, where the diagonal matrix entries are X and the nondiagonal matrix entries in the first t columns are vanishing. Thus, the problem becomes how to transform an n-dimensional echelon matrix R into the standard Jordan form:

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Review o n Linear Algebras 21

where A is a t x t constant matrix, A = X1, S is an (n - t ) x (n - t ) upper triangular matrix with the diagonal elements A, and T is a t x (n-t ) matrix. We denote the indices of R by a = 1, 2, ..., n. When c11 takes the first t values, we denote a by the first lowercase Latin letter, say, a = 1 ,2 , . . . , t . When a takes the last (n - t ) values, we denote a by the middle lowercase Latin letter, say, j = ( t + I), ( t + 2)) . . ., n.

X when j = k

Sj,k when j < k . Aa,b = XS,b, Tu,j no limit, Sj,k = 0 when j > k (1.15)

In the following we will introduce a series of similarity transformations, each of which keeps the form (1.15) invariant, but simplifies the nondiagonal matrix entries of R. Our aim is to transform the nondiagonal matrix entries of R such that in each row and in each column of R there is at most only one nonzero nondiagonal matrix entry, which is equal to 1. Then, by a simple similarity transformation, which only changes the order of the rows and the columns, R becomes the standard Jordan form.

Define a similarity transformation X, where the similarity transforma- tion matrix X ( M , b ) is the direct sum of a b-dimensional unit matrix, a ( t - b)-dimensional non-singular matrix M-' and an (n - t)-dimensional unit matrix, whose action is to transform each column from the (b + 1)th row to the t th row of T by M , but to keep A and S invariant. We can choose M to annihilate all the matrix entries in one column from the (b+ 1)th row to the t th row of T except for one, which is equal to 1.

We study the nondiagonal matrix entries in the ( t + 1) column of T . They are not all vanishing, otherwise R have (t + 1) linearly independent eigenvectors. We choose M in the similarity transformation X ( M , 0) to annihilate all matrix entries in the ( t+l) th column of T except for Tl,(t+l) = 1.

Define a similarity transformation Y , where the similarity transforma- tion matrix Y, ( P I , a1 ; p2, a2; . . .) is a unit matrix plus a few nonzero non- diagonal matrix entries. Those matrix entries are all equal to w and are located in the a1 column of the row, the a2 column of the ,!?2 row, and so on. All ap and Py are different from each other. Its inverse matrix can be obtained from it just by replacing w with --w.

We can make a similarity transformation Y, (t + 1, k) with w = -Tl,k to annihilate T 1 , k where k > t+ l . From the viewpoint of basis transformation, the original basis is the unit vector e , in the rectangular coordinate system

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22 Problems and Solutions in Group Theory

and the application of R on the basis is

The similarity transformation Yw(t + 1, k ) will transform the kth basis e k

into a new basis e l = e k - Tl,ket+l, which just cancels the term related to el in Rek:

After a series of similarity transformations Y, ( t + 1, k ) , w = -Tl,k, where k runs from t + 2 to n one by one, the matrix entries in the first row of T will all be annihilated except for T1,(t+l) = 1. This method for simplifying the matrix entries in the (t + 1)th column and in the first row of T is called the method of the first kind.

Now, we investigate the ( t + 2)th column of R. There are two cases. The first case is S(t+1),(t+2) = 0. In this case, the method of the first kind can be used to simplify the matrix entries in the (t + 2)th column and in the second row of T . We choose M in the similarity transformation X ( M , 1) to annihilate all matrix entries in the ( t + 2)th column of T except for T2,(t+2) = 1. In the transformation the matrix entries in the ( t + 1)th column and in the first row of T are kept invariant. Then, by a series of similarity transformations Yw(t + 2, k ) , where w = - T 2 , k , t + 3 5 k 5 n, we can annihilate all matrix entries T2,k except for T2,(t+2) = 1.

The second case is S(t+l),(t+2) # 0. Define a similarity transformation 2, where the similarity transformation matrix Zc(cy) is a diagonal matrix with all diagonal matrix entries to be 1 except for the a t h entry to be <. This similarity transformation enlarges the matrix entries in the a t h column of R by C times and reduces the matrix entries in the a t h row by C times, but keeps the diagonal matrix entries invariant. We first choose the similarity transformation &(t + 2) with C = { S(t+l),(t+a)) to change S(t+l),(t+2) to be 1. Second, we make a series of similarity transformations Yw(a, t + 1) with w = Tu,(t+2), 2 5 a 5 t , to annihilate the matrix entries in the ( t + 2)th column of T , where the matrix entries in the ( t + 1)th column

-1

of T are kept application of

invariant. From the viewpoint of base transformation, the R on the original basis is

t

a=2

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Review on Linear Algebras 23

A series of similarity transformations Yw(a, t + 1) with LJ = Ta,(t+2), 2 5 a 5 t , define the vector in the bracket of the above formula as a new basis et$, to annihilate all Ta,(t+2).

Third, we annihilate the matrix entries S(t+l),k, k > t+2 , in the ( t+ l ) th row of S by a series of similarity transformations Yw (t + 2, k ) with w = -S(t+l),k, where k runs from t + 3 to n one by one. The action of this series of similarity transformations is the same as that to annihilate T1,k in the method of the first kind. This method for simplifying the matrix entries in the ( t + 2)th column of T and in the (t + 1)th row of S in the second case is called the method of the second kind. In the later application some revision of this method will be made.

Based on the above simplification, we are going to prove by induction that R can be transformed into the standard Jordan form by a series of similarity transformations. We define the d-simplified R-matrix as the R matrix in the form (1.15) satisfying that in each column of R from the ( t + 1)th column to the (t + d)th column there is only one nonzero non- diagonal matrix entry which is 1 and those nonzero nondiagonal matrix entries are all located in different rows. We will prove that the d-simplified R-matrix can be transformed into the (d + 1)-simplified R-matrix by a se- ries of similarity transformations. We have proved that the l-simplified R-matrix can be transformed into the 2-simplified R-matrix by a series of similarity transformations.

For definiteness, we assume that from the ( t+l) th column to the (t+d)th column of a d-simplified R-matrix, there are g nonzero nondiagonal matrix entries located in the first g rows of T and (d - g) nonzero nondiagonal matrix entries located in S. There are two cases for the positions of the nonzero nondiagonal matrix entries in the (t + d + 1)th column of R. In the first case, all nonzero nondiagonal matrix entries in the (t + d + 1)th column of R appear in the submatrix T . In the second case, a t least one nonzero nondiagonal matrix entry in the ( t + d + 1)th column of R appears in the submatrix S. We will use the method of the first kind to simplify R in the first case, but the method of the second kind in the second case.

In the first case, we choose M in the similarity transformation X ( M , d) to annihilate all matrix entries in the (t + d + 1)th column of T except for T(g+l),( t+d+l) = 1, where the matrix entries in the first g rows of T are kept invariant. Then, we annihilate all T(g+llk, where k runs from t + d + 2 to n one by one, by the similarity transformations Yw(t + d + 1, k) with w = -T (g+l )k . The result is nothing but the (d + 1)-simplified R-matrix.

In the second case, there is a t least one nonzero nondiagonal matrix

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24 Problems and Solutions in Group Theory

entry in the ( t + d + 1)th column of S. We have known that there is only one nonzero nondiagonal matrix entry (which is 1) in each column from the ( t + 1)th column to the ( t + d)th column of the d-simplified R-matrix. For a nonzero nondiagonal matrix entry S j , , ( t + d + l ) in the ( t + d + 1)th column of S , if the nonzero nondiagonal matrix entry in the j1 th column of R lies in the a2th row, the nonzero nondiagonal matrix entry in the a2th column lies in the a3th row, and so on; till a, 5 t , then we define the degree of Sjl , ( t+d+l) to be u and the degree indices of Sjl , ( t + d + l ) to be (jl, a 2 , a3, . . . , a,). Recall that in the a,th column of R, all nondiagonal matrix entries are vanishing. Among nonzero nondiagonal matrix entries in the ( t + d + 1)th column of S , we choose the one with the highest degree, say Sj, , ( t+d+l ) with the degree u and the degree indices (jl , a 2 , a 3 , . . . , a,). We first choose a similarity transformation Zc(t+d+l) with C = (S j l , ( t+d+l ) )

to make Sjl,(t+d+l) = 1. This similarity transformation does not change the positions of the nonzero nondiagonal matrix entries in the ( t + d + 1)th column of S.

The degree v of any other nonzero nondiagonal matrix entry in the ( t + d + 1)th column of S is not larger than u, say Sj , , ( t+d+l) = a with the degree v and the degree indices ( j 2 , p2, . . . , p,), v 5 u. In the following we will show that the similarity transformation Y, ( j 2 , j1; p2, a 2 ; . . . ; p,, a,) with LJ = CT can annihilate the matrix entry Sjz,(t+d+l). The reason for this can be understood by the viewpoint of the basis transformation. The new basis induced by this similarity transformation is

-1

ej, + oej,

e,, + aeg,

when r = j1,

when r = a 2 ,

I er the remaining cases.

Before this transformation we have

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Review on Linear Algebras 25

After the similarity transformation we have

When v = u, the terms of e,,,+, and e,lu+l in the last formula disappear. Thus, the action of R on the new basis edp is the same as that on the original basis eap except for that on the basis et'+d+l, where the term of aej, disappears. The action of R on the bases e i , k > t + d + 1, may also be changed, namely, the nondiagonal matrix entries in the kth column of R may be changed. But we do not care about them. Through a series of similarity transformations, the matrix entries in the (t + d + 1)th column of S become vanishing except for Sj,,(t+d+l) = 1.

If a matrix entry in the ( t + d + 1)th column of T is nonzero, say T a ( t + d + l ) # 0, we can use the similarity transformation Y w ( a , j l ) with w = Ta(t+d+l) to annihilate Ta(,+dfl). Finally, we choose a series of simi- larity transformations Yw(t + d + 1, k) with w = - S j l , k , t + d + 2 5 k 5 n, to annihilate all S j l , k in the j l t h row of S . Thus, we have found a se- ries of similarity transformations to change a d-simplified R-matrix into a (d + 1)-simplified R-matrix, namely, we have proved the third proposition. Therefore, we have proved that any matrix R can be transformed into a direct sum of the standard Jordan forms.

We will give a typical and instructive example to show how to transform a d-simplified R-matrix into a (d + 1)-simplified R-matrix by a series of similarity transformations. Assume a 7-simplified R-matrix with t = 6, n = 15, g = 4, and d = 7,

T =

' 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 ~ 0 0 1 0 0 0 0 0 0 0 0 0 0 T5,14 T5,15

, O 0 0 0 0 0 0 T6,14 T6,15

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26 Problems and Solutions in Group Theory

S =

/ x 0 1 0 0 0 0 0 0 ' 0 0 0 0 0 0 s 8 , 1 4 s 8 , 1 5

0 0 x 1 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 s 1 0 , 1 4 s 1 0 , 1 5

0 0 s 1 1 , 1 4 s 1 1 , 1 5

0 0 0 0 0 x 1 0 0 0 0 0 0 0 0 s 1 3 , 1 4 s 1 3 , 1 5

0 0 0 0 0 0 0 A S14 ,15

\ o o o o o o o 0 x

According to the matrix entries in the 14th column (in general, the (t + d + 1)th column) of S, there are two cases. One is the case where the nondiagonal matrix entries in the 14th column of S are all zero. In this case the method of first kind can be used to simplify R. We choose a two- dimensional matrix M in the similarity transformation X(M,4) to make T5,14 = 1 and T6,14 = 0, but to keep the first four rows of T invariant. Then, we use the similarity transformation Yw(14, 15) with w = -T5,15 to make T5,15 = 0. This has been the 8-simplified R-matrix. The other is the case where the nondiagonal matrix entries in the (t + d + 1)th column of S are not all zero. In the example we assume that the nondiagonal matrix entries given in S are all nonzero. First to determine the degrees and the degree indices for these matrix entries. s 8 , 1 4 has the degree two and the degree indices (8,2). S10 ,14 has the highest degree four and the degree indices (10,9,7,1). S11 ,14 has the degree two and the degree indices (11,3). s 1 3 , 1 4 has the degree three and the degree indices (13,12,4). We use the similarity transformation Zc(14), with C = ( s lO,14)- ' to make s 1 0 , 1 4 = 1. The remaining nondiagonal matrix entries in the 14th column of S have changed their values, but are still denoted by the original symbols. Now, we use the following similarity transformations one by one to obtain the 8-simplified R-matrix:

y, (8,10; 219) with w = s 8 , 1 4 transforms S 8 , 1 4 to be zero,

Yw(l l , 10; 3,9)

Yw(13, 10; 12,9; 4,7)

Yw(5,W with w = T5,14 transforms T5,14 to be zero,

yu ( 6 , W with w = T 6 , ~ 4 transforms T6,14 to be zero,

y, (14,151

with w = S11 ,14

with w =

transforms S11 ,14 to be zero,

transforms S13 ,14 to be zero,

with w = -S10 ,15 transforms S10,15 to be zero.

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Chapter 2

GROUP AND ITS SUBSETS

2.1 Definition of a Group

* A group is a set G of elements R satisfying four axioms with respect to the given multiplication rule of elements. The axioms are: a) The set is closed to this multiplication; b) The multiplication between elements satisfy the associative law; c) There is an identity E E G satisfying E R = R ; d) The set contains the inverse R-' of any element R E G satisfying R-lR = E . The multiplication rule of elements completely describes the structure and property of a group. A group G is called a finite group if it contains finite number g of elements, and g is called the order of G. Otherwise, the group is called an infinite group. For a finite group, the multiplication rule can be given by the multiplication table, or called the group table. A group is called the Abelian group if the product of its elements is commutable. A few elements in G are called the generators of G if any element in G can be expressed as their product. The rearrangement theorem says RG = G R = G, namely, there are no duplicate elements in each row and in each column of the multiplication table.

* Two groups are called isomorphic, G M G', if there is a one-to-one correspondence between elements of two groups in such a way products correspond to products. From the viewpoint of group theory, two isomor- phic groups are the same as each other.

1. Let E be the identity of a group G, R and S be any two elements in the group G, R-' and S-' be the inverses of R and S, respectively. Try to show from the definition of a group: (a) RR-l = E ; (b) R E = R; (c) if T R = R, then T = E ; (d) if T R = E , then T = R-'; (e) The inverse of ( R S ) is S-l R-'.

27

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28 Problems and Solutions an Group Theory

Solution. The key to the proof is that each element in a group has its inverse. Recall that only the definition of a group and the proved conclusion can be used in the later proof. (a) We prove the conclusion by two methods. In the first method, since R-' is an element in the group, there exists its inverse in the group, denoted by S , SR-l = E. So, from the definition of a group we have

Another method is to denote by W-' the inverse of RR-l = W , W-lW = E . Then, we have:

WW = R (R-lR) R-' = RER-l = W, RR-l= W = EW = (W-lW) W = W - ' ( W W ) = W-lW = E.

(b) RE = R (R-IR) = (RR-l) R = ER = R.

(c) If T R = R, then

T = T (RR-l) = (TR) R-' = RR-l = E.

The conclusion says that the identity in a group is unique.

(d) If T R = E, then

T = T E = T ( R R - ~ ) = (TR) R - ~ = E R - ~ = R - ~

The conclusion says that the inverse of any element in a group is unique.

(e) Since (S-lR-') (RS) = S-' (R-lR) S = S-lES = S-lS = E and the conclusion (d), S-lI2-l is the inverse of RS.

2. Show that a group composed of all positive real numbers where the multiplication rule of elements is defined with the product of digits, is isomorphic onto a group composed of all real numbers where the multiplication rule of elements is defined with the addition of digits.

Solution. Denote by H the set of all positive real numbers R. The multi- plication of elements is defined to be the product of digits. The set is closed for the multiplication. The product of digits satisfies the associative law. The positive real number 1 is the identity in the set. The reciprocal digit 1/R is still a positive real number, which is the inverse of R. Therefore, the set of H is a group, called the multiplication group of positive real numbers.

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Group and Its Subsets 29

Denote by G the set of all real numbers S. The multiplication of el- ements is defined to be the addition of digits. The set is closed for the multiplication. The addition of digits satisfies the associative law. The real number 0 is the identity in the set. -S is still a real number, which is the inverse of S. Therefore, the set of G is a group, called the addition group of real numbers.

The exponential relation R = es gives a one-to-one correspondence between the positive real number R in H and the real number S in G, and it is invariant for the multiplication of elements:

Therefore, the group H is isomorphic onto the group G.

2.2 Subsets in a Group

* A subset of a group is called a subgroup of the group if in the multiplica- tion rule of the group elements the subset satisfies four axioms in the group definition. The set composed of the powers of any element in a group forms a cyclic subgroup, also called the period of the element. For a finite group, a subset of a group is its subgroup if and only if the subset is closed with respect to the multiplication rule because the subset contains the period of any element in it. In the same reasoning, a finite set of elements is a group if and only if the set is closed with respect to an associative multiplication rule. It is easy to judge by the multiplication table of the group whether or not a subset is a subgroup. The order of the cyclic subgroup generated by the element is called the order of the element. Only the order of the identity is 1. The identity and the whole group are two trivial subgroups. We will only consider the nontrivial subgroups.

* A subset in a group G obtained by left-multiplying (or right-multiplying) an element R E G to a subgroup H of G, where R does not belong to H , is called a left coset RH (or a right coset H R ) of the subgroup H . For a finite group, the number of the elements in a coset is equal to the order of the subgroup. Since a coset does not contain any element in the subgroup, any coset must not be a subgroup due to lack of the identity. Different left cosets (or right cosets) do not contain any common element. The necessary and sufficient condition for two elements R and S belonging to the same left coset is R-lS E H , and that for them belonging to the same right coset

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30 Problems and Solutions i n Group Theory

is RS-l E H . Hence, the order h of a subgroup H must be a whole number divisor of the order g of the group G. The integer n = g/h is called the index of the subgroup H in G. The number of the different left cosets (or right cosets) of a subgroup is (n - 1). In the columns of the multiplication table, related to the elements of a subgroup, the set of elements in each row is the subgroup itself or the left coset of the subgroup. In the rows of the multiplication table, related to the elements of a subgroup, the set of elements in each column is the subgroup itself or the right coset of the subgroup.

* The element R' = SRS-' is said to be an element conjugate to R in a group G, where R, R' and S all belong to G. The conjugate relation of two elements is mutual. If two elements are both conjugate to a third element, they are also conjugate to each other. The subset of all mutually conjugate elements in a group G is called a class C, in G. Any element in G belongs and only belongs to one class. No class forms a subgroup of G except for the class C1 composed of the identity. The set of all the inverse elements R-' of R E C, is also a class, denoted by C;'. These two classes are called the reciprocal classes mutually. It is called the self- reciprocal class if C, = C;'. RS is conjugate to SR. Conversely, two conjugate elements can be expressed as two products RS and S R where I? and S are two elements in G. For a finite group, the elements in one class have the same order. However, two elements with the same order are not necessary conjugate to each other. For a multiplication table of a finite group where the arrangement for the rows is the same as that for the columns, two conjugate elements appear and must appear at least once in two symmetrical positions with respect to the diagonal line of the multiplication table. This is the main method to check by the multiplication table whether two elements with the same order are conjugate to each other.

A subgroup H is called an invariant subgroup (or normal subgroup) of the group G if for any element R E G the left coset is equal to the right coset ,

R H = H R , RHR-I = H .

A subgroup with index 2 must be an invariant subgroup. An invariant subgroup is composed of a few whole classes and satisfies the criterion for a subgroup. It is the main method to find an invariant subgroup of a finite group G that one first gathers a few classes including the identity to see whether the number of elements in the set is a whole number divisor of the

*

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Group and I ts Subsets 31

order of G and whether the set contains the whole period of any element in the set, then, to check whether the set is closed for the multiplication rule of the group elements.

* The multiplication of two subsets is defined to be a set composed of all the products of any two elements respectively belonging to two subsets. Note that in a set only one element is taken for the duplicated element. Denote the invariant subgroup H of G and its cosets uniformly by R j H , where R1 = E. The aggregate of the subsets R j H satisfies the four axioms with respect to the multiplication rule of subsets. In fact,

( R j H ) ( R k H ) = R j R k H H = (RjRk) Ha

The multiplication rule satisfies the associate law. R1 H = H is the identity in the aggregate. Ryl H is a coset of H and is the inverse of Rj H in the aggregate. Therefore, the aggregate of the subsets R j H forms a group, called the quotient group G / H of the invariant subgroup H .

3. If H1 and HZ are two subgroups of a group G , prove that the common

Solution. Denote the set of the common elements of the subgroups H1 and H2 by H3 = (R1, R2, + - .}, where Rj E H1 and Rj E H2. The set of H3 is also a subset of G , and the product of elements in H3 satisfies the multiplication rule of G . So the associative law is satisfied for H3. Since H1 and H2 are two subgroups of G, the product RiRj of any two elements in H3 must belong to both subgroups H1 and H2 such that it belongs to the set H3. For the same reason, the identity E in G and the inverse Rj' of any element Rj in H3 belong to both subgroups H1 and H2, so that they also belong to the set H3. Therefore, the set H3 is a subgroup of G .

elements in HI and H2 also form a subgroup of G.

4. Prove that a group whose order g is a prime number must be a cyclic

Solution. The order of an element is nothing but the order of the cyclic subgroup generated by the element. As we have known, the order h of the subgroup H c G is a whole number divisor of the order g of the group G. When g is a prime number, except for the identity, the order of any element in G must be the order g of the group, so the cyclic subgroup is the group G itself. From it we come to the conclusion that any two groups whose orders are the same prime number must be isomorphic onto each other.

group C,.

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32 Problems and Solutions in Group Theory

5 . Show that up to isomorphism, there are only two different fourth-order groups: The cyclic group Cq and the fourth-order inversion group V4.

Solution. Since the order of any element in a fourth-order group, except for the identity, has to be 2 or 4. If there is at least one fourth-order element R, the group is the cyclic group Cq = { E , R, R2, R3}. If the order of any element in the group, except for the identity, is 2, denote the group by V4 = {e, o, r, p } , where e is the identity and o2 = r2 = p2 = e. From Problem 1, the product of any two elements, say or, cannot be equal to e, o and r , so or = p and the group is the fourth-order inversion group. This completes the proof.

6. Show that up to isomorphism, there are only two different sixth-order groups: The cyclic group C6 and the symmetry group D3 for the regular triangle.

Solution. Since the period of any element is a cyclic subgroup, in a sixth- order group, except for the identity, the order of any element has to be 2, 3 or 6. There are three cases. In the first case where there is at least one sixth-order element, the group is a cyclic group CS. In the second case the group contains no sixth-order element, but contains a t least one third-order element, denoted by R. The cyclic subgroup H = { E , R, R2} generated by R has the index two so that it is an invariant subgroup of the group. Denote its coset by HSo = { S O , S ~ , S ; ? } , where RmSo = Sm, R"Sj = Sj+m, and Sj+3 = Sj. S; cannot be equal to R or R2, otherwise the order of Sj is 6. From the rearrangement theorem, S; cannot be equal to Sk. Thus, S; = E. It can be deduced that Rm = Sj+mSj and SjRm = Sj-m. This group is nothing but D3. In the third case, the order of any element in the group is 2 except for the identity. Arbitrarily take two elements R and S in the group, and let RS = T . From Problem 1 we conclude that T is not equal to E , R or S. The subset composed of E , R, S and T forms a subgroup. It is isomorphic onto the inversion group V4 of order 4. Since 6/4 is not an integer, the third case cannot exist. This completes the proof.

7. Show that a group must be an Abelian group if the order of any element

Solution. What we need to show is whether RS = S R for any two different elements R and S in the group G. Letting RS = T , we have R2 = S2 = T 2 = E . From Problem 1, T S = RS2 = R, T R = T2S = S, and SR = TR2 = T = RS. Therefore, the group is an Abelian group.

in the group, except for the identity, is 2.

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Group and I ts Subsets 33

2.3 Homomorphism of Groups

* A group G is said to be homomorphic onto another group G', G' N G, if one and only one element of G' corresponds to any element of G, if at least one element of G corresponds to any element of G', and if the correspon- dence is such that the product of two elements of G is in the same way to map onto the product of two corresponding elements of G'. In other words, if G' - G, there is a one-to-many correspondence between elements of G' and G and the correspondence is invariant in the multiplication of elements. The set H of elements in G which corresponds to the identity in G' forms an invariant subgroup of G, which is called the kernel of homomorphism. The quotient group G / H is isomorphic onto G'. G' describes only part of properties of G, namely, G' describes the property of the quotient group G / H , but does not describe the difference among elements in the kernel of homomorphism.

8 . The Pauli matrices aa are defined as follows:

;) , u 2 = ( 5 ) , u 3 = ( ; " 1 ) 7

where Eabd is the totally antisymmetrical tensor of rank 3. Show that all possible products generated by 01 and 02 form a group. List the multiplication table of this group. Point out the order of this group, the order of each element, the classes, the invariant subgroups and their quotient groups. Prove that this group is isomorphic onto the symmetry group Dq for the square.

Solution. According to the multiplication rule of the Pauli matrices, there are 8 matrices generated by 01 and 02. The multiplication table is given in the Table. From the multiplication table, the set of these 8 elements is closed for the multiplication of elements. The multiplication of matrices satisfies the associative law. 1 is the identity E in the set. Any element in the set is self-inverse except for fia3. -ia3 is the inverse of ia3. Therefore, this set forms a group G with order 8. The order of the identity E is 1. The orders of -1, fa1 and f a 2 all are 2. The orders of f ig3 are both 4. {fal}, ( f a 2 ) and {fia3} form 1 and -1 form two classes.

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34 Problems and Solutions in Group Theory

the classes, respectively. There are five classes altogether. The invariant subgroups in the group are (1, -l}, (1, -1,01, -01}, (1, - 1 , 0 2 , -02} , and { 1, -1, i03,-i03}. The cosets of (1, -1) are {kal}, {&02} and {fias}. The square of each coset is the invariant subgroup {fl}, so the quotient group is isomorphic onto the fourth-order inversion group Vq. The quotient groups of the next three invariant subgroups are all isomorphic onto the second-order inversion group V2. G is homomorphic onto V4 with a 2 : 1 correspondence and onto V2 with the 4 : 1 correspondence where there are three different kernels of homomorphism.

I 1 (71 a2 ia3 -1 -a1 -a2 -ia3

1 cJ1

a2 ia3

-1 -a1 - a 2 -ia3

1 cJl

0 2 ia3

a1 1

-ia3 - 0 2

0 2

i u3 1 a1

ia3 0 2

-01 -1

-1 -01

- 0 2

-iu3

- 0 1 -1 ia3 u2

- 0 2 -ia3 -1 -a1

-ia3 - a2 a1 1

-1 - 0 1

- a2 -ia3

-81 -1 icJ3

cJ2

- cJ2

- ia3 -1 -01

-ia3 - a2 cJ1 1

1 0 1

0 2 ia3

a1 1

-ia3 -a2

(72 ia3 1 0 1

ia3 u2

-01 -1

Let the fourfold rotation C4 around the 2 axis in the group D4 corre- spond to io3, and the twofold rotation C; around the X axis correspond to 01. The correspondence between the remaining elements can be obtained by the products of elements:

Thus, the multiplications of elements also satisfy the one-to-one correspon- dence, so two groups are isomorphic.

9. Show that the set of all possible products generated by ial and i 0 2

forms a group. List the multiplication table of this group. Point out the order of this group, the order of each element, the classes, the invariant subgroups and their quotient groups in the group, respectively. Show that this group is not isomorphic onto the group D4.

Solution. There are 8 matrices generated by the products of iol and ia2. The multiplication table is as follows.

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Group and I t s Subsets

1 ial iu2 ia3

-1 -a01

-ia2 -ia3

35

1 ial a02 ia3 -1 -2u1 -aa2 -ia3 ial -1 -iu3 iaz -ia1 1 ia3 -iff2

ia2 iu3 -1 -a01 -iaz -ia3 1 ial ia3 -a02 ial -1 -au3 iaz -ial 1

-1 -ial -iaz -ia3 1 iu1 ia2 ia3 -ial 1 a03 -auz ial -1 -ia3 iaz -iu2 -ia3 1 ial ia2 ia3 -1 - a 0 1

-a03 iaz -ia1 1 ia3 -aaz ia1 -1

From the multiplication table, the set of these 8 elements is closed for the multiplication of elements. The multiplication of matrices satisfies the associative law. Therefore, this set forms a group G with order 8. The order of the identity 1 is 1. The order of -1 is 2. The orders of fia, all are 4. 1 and -1 form two classes. (fial}, (fiaz} and (&ia3} form the classes, respectively. There are five classes altogether. The invariant subgroups in the group are (1, -1}, (1, -l,iol, -ia1}, (1, -1,ia2, -ia2},

and (1, -1,203, -ia3}. Thecosetsof (&1} are (fial}, (fiaz} and {fia3}. The square of each coset is the invariant subgroup (kl}, so the quotient group is isomorphic onto the fourth-order inversion group V4. The quotient groups of the next three invariant subgroups all are isomorphic onto the second-order inversion group V2. G is homomorphic onto V4 with a 2 : 1 correspondence and onto V2 with the 4 : 1 correspondences where there are three different kernels of homomorphism. Because this group contains six fourth-order elements, this group is isomorphic neither onto the group D4, nor onto the group C4h. In literature, this group is usually called the quaternion group Q8.

10. Up to isomorphism, prove that there are only five different eighth-order group: The cyclic group c8, C4h =C4 xV2, the symmetry group D4 of a square, the quaternion group Q8 and D2h =D2 x V ~ .

Solution. The order of an element in an eighth-order group G, except for the identity, has to be 2, 4 or 8. If there is at least one eighth-order element, G is the cyclic group c8. If the orders of all elements in G , except for the identity E , are 2, we obtain from Problem 7 that G is an Abelian group, which is isomorphic onto D2h =D2 xV2.

If there is no eighth-order element G, but there is at least one fourth- order element, denoted by R, the cyclic subgroup H = { E , R, R2, R3} gen- erated by R is an invariant subgroup of G because its index is 2. Denote its coset by HSo = {So,SlrS2,S3), where RmSo = Sm, RmSj = Sj+m, and

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36 Problems and Solutions in Group Theory

S j + 4 = Sj. Sj cannot be equal to R or R3, otherwise the order of Sj is 8. From the rearrangement theorem, S; cannot be equal to Sk. If Sj" = R2, then Sj is a fourth-order element. Thus, SJrl = Sj" = R2Sj = Sj+2, and Sj"+2 = SJT2 = R2. Now, there are three cases. The first case is that all four Sj satisfy 5'; = R2. Then, 5'15'0 = RS; = R3, SoS1 = R3S? = R, and this group is isomorphic onto the quaternion group Qs. The correspondence is

The second case is that only a pair of Sj satisfy Sj" = R2, say S: = S$ = R2. Namely, S1 and S3 are the fourth-order elements, and SO and S2 are the second-order elements, So2 = Sz = E. Thus, S1So = RS; = R, SoS1 =

and this group is isomorphic onto the Abelian group C4h =Cq xV2. The correspondence is

R3S: = R, SoR = SiS1 = S1 = RSo, SIR = SfSo = R2So = S2 = RS1,

where a is the space inversion. The third case is that the orders of all Sj are 2, Sj" = E , then from R"Sj = Sj+" we have R" = Sj+"Sj and Sj R" = Sj-m. Thus, the group is isomorphic onto D4. The correspondence is

The orders of elements and the classes in these five groups are listed as follows.

Group c8

Q8

D4 C4h

D2h

No. of elements 1st-order 2nd-order 4th-order 8th-order

1 1 2 4 1 1 6 0 1 3 4 0 1 5 2 0 1 7 0 0

No. of classes

11. Investigate all ninth-order group which are not isomorphic onto each

Solution. The order of any element in a ninth-order group G, except for the identity, has to be 3 or 9. If there is at least one ninth-order element in the group, then this group is a cyclic group Cg.

If there is no ninth-order element, all elements in G, except for the identity, are the third-order elements. Arbitrary take a third-order element,

other.

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Group and I ts Subsets 37

denoted by A. The cyclic subgroup generated by A is H = { E , A , A 2 } . Denote one right coset by H B = { B , C, D } where A B = C , AC = D , and A D = B. Since B , C and D are all third-order elements, their square cannot be equal to E , A or A 2 . From the rearrangement theorem, their square cannot be equal to B, C or D. They have to be different from each other due to the uniqueness of the inverse element. Therefore, the remaining three elements in the group are B2, C2 and D2, which form another right coset of H . From the rearrangement theorem, AB2 = C B cannot be equal to C2 and B2, so it has to be equal to D2. Similarly, we can calculate the remaining products and give the following multiplication table of the group. From the multiplication table, we know that this group is an Abelian group with nine classes. There are only two different ninth-order groups.

D B2 C2 1 D2

E A A2 B C D E A A2 B C D B2 C2 D2

A A 2 E C D B

D2 B2 C2

A2 E A D B C c2 D2 B2

B C D B2 0 2

C2 E A2 A

C D B

0 2

C2 B2

A E A2

D B c C2 B2 D2 A2 A E

B2 B2 D2 C2 E A

A2 B D C

C2 c2 B2 D2 A2 E A D C B

D2 0 2

C2 B2 A

A2 E c B D

12. Investigate all tenth-order groups which are not isomorphic.

Solution. The order of any element in a tenth-order group G, except for the identity, has to be 2, 5 and 10. If there is at least one tenth-order element in the group, then this group is a cyclic group Clo.

If all elements in a tenth-order group, except for the identity, are second- order elements, arbitrary taking two different elements R and s, we have R2 = S2 = E , RS = T . T has to be a new element as we have known in Problem 1. Thus, we obtain a fourth-order subgroup { E , R, S , T } iso- morphic onto the fourth-order inversion group V4. Since the order of the subgroup is not a whole number divisor of ten, a contradiction.

If there is no tenth-order element in G, but there is at least one fifth- order element, denoted as R, then the cyclic subgroup { E , R, R2, R3, R4} generated by R is an invariant subgroup because its index is 2. Denote its coset by {SO,S~,S~,S~,S~}, where R"S0 = S,, R"Sj = Sj+", and Sj+5 = Sj. From the rearrangement theorem, S; cannot be equal to Sk.

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38 Problems and Solutions in Group Theory

If it is equal to Rj, where j is not a multiple of 5, then Sj is a tenth-order element, thus a contradiction. So Sj" = E , Sj all are second-order elements. Thus, we have R" = Sj+"Sj and SjR" = Sj-m. This group is the group D5. Up to isomorphism, there are only two tenth-order group, C ~ O and D5. From the investigation on the sixth-order group (Problem 6) and the tenth-order group, we conclude that when the order g of a group G is the double of a prime number n > 1, the group has to be a cyclic group C2n or the symmetry group D, for a regular n-sided polygon.

13. Give an example to show that the invariant subgroup of an invariant subgroup of a group G is not necessary an invariant subgroup of the group G. Conversely, show that if an invariant subgroup of a group G completely belongs to a subgroup H of G, then i t is also an invariant subgroup of H .

Solution. Let H be a subgroup of G, and let H1 be a subgroup of H . For an arbitrary element R in H I , an element S R S - l conjugate to R for G may not be an element conjugate to R for H , because S E G may not belong to H . Conversely, an element SRS-l conjugate to R for H must be an element conjugate to R for G, because S E H must belong to G. Therefore, the invariant subgroup of an invariant subgroup H of a group G is not necessary an invariant subgroup of G, but the invariant subgroup HI of a group G must be an invariant subgroup of the subgroup H c G if

An example is as follows. The group T contains three twofold axes and four threefold axes. T contains an invariant subgroup D2 composed of the identity E and three twofold rotations. However, three second-order subgroups in D2 are all the invariant subgroups of Dz, but not the invariant subgroup of T.

Hi C H .

14. In a finite group G of order g, let C, = { S1, S2, . . . , } be a class of G containing n(a) elements. For any two elements Si and Sj in the class C, (may be different or the same), show that the number m(a) of elements P E G satisfying Si = PSjP-l is g/n(a).

Solution. For a given element Sj E C,, let the number of elements R E G, commutable with Sj , be rn(a). It is easy to show that m(a) does not depend on Sj. In fact, if Si = PSjP-l, PRP-l commutes with Si.

Under the multiplication rule of G, the set H of R E G, commutable with Sj, forms a subgroup of the finite group G, because if both R and

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Group and Its Subsets 39

R’ can commute with Sj , RR’ also commutes with Sj. The order of H is m(a). Note that H is not necessary an invariant subgroup of G. Any element T E G, which does not belong to the subgroup H , cannot commute with Sj. Let TSjT-l = Si E Ca and Sj # Si. Any element T R in the left coset T H of the subgroup H satisfies TRSjR-lT-l = Si. The number of elements in the left coset TH is still rn(.>. On the other hand, for any element P which satisfies PSjP-l = Si, we have

Thus, P-lT E H , and P belongs to the left coset TH. Therefore, we obtain a one-to-one correspondence between the element Si E C, and the left coset T H by the relation

TRSjR-lT-l = Si.

When Si = Sj , T E H and T H = H . The index g/m(a) of the subgroup H is equal to the number n(a) of the elements Sj in the class C,. Namely, g = m(cr)n(a). Both the order m(a) of a subgroup H and the number n(a) of elements in a class are the divisors of the order g of the group G.

15. Prove that being the product of two subsets, the product of two classes in a group G must be a sum aggregate of a few whole classes. Namely, the sum aggregate contains all elements conjugate to any product of two elements belonging to the two classes, respectively.

Solution. Let C, and C p be two classes in a group G. For given R E C, and S E C p , we have TRT-l E C, and TST-l E C p , where T is an arbitrary element in G. Therefore,

T(I2S)T-l = TRT-lTST-l E c,co. The conclusion is proved. It is easy to prove that the multiplicity of RS in the product space of C,Cp is the same as that of the element conjugate to RS. Therefore, considering the multiplicity f ( a , p, r), we have

16. Calculate the multiplication table of the group T by extending the multiplication table of the subgroup C3 = { E , R1, R:} of T.

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40 Problems and Solutions in Group Theory

Solution. The symmetry group T for a tetrahedron contains three twofold axes orthogonal to each other and four threefold axes. Denote the twofold rotations by TZ, Ty” and TZ, respectively. TZ = T:Ti. The threefold axes in T are respectively along the directions e , f ey f e,, and -ex ey =t e,. Denoting by R1 the rotation through 2 ~ / 3 around the direction ez+ey+e,, we have

R;T:RI = T,”, R;T:R1 = T:, R;T;R1 = Ty”, R1T:R: = Ty”, R1Ty”Rq = T:, RlTzR? = T,”.

The remaining threefold rotations can be defined as

Substituting them into the above formulas, we obtain

R1Ti = R3, R1Ty” = R2,

Taking the inverse of the formulas, we have

In terms of TzTy” = T: we obtain

R1T: = R4.

T: Rq = R1, Ty”R4 = R2, T;R4 = R3, T,2Ri = Ri, Ty“Ri = Rz, T,”R,” = R;.

The above equations give the formulas for the right cosets of the subgroup C3 and the formulas for left-multiplying T:, T; and T,” on the elements in T. From them we can obtain the multiplication table of T.

Arrange the rows in the multiplication table (the first column of the table) in the order of the subgroup C3 and its left cosets T;C3, Tic3 and T;C3, and the columns (the first row of the table) in the order of C3 and its right cosets C3T:, C3Ti and C3TZ. Thus, the multiplication table is divided into 16 subtables. Each subtable contains 3 x 3 multiplication elements. The first subtable in the first row is just the multiplication table

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Group and I ts Subsets 41

of the subgroup Cs. The remaining three subtables are obtained by right- multiplying the first subtable with Tz, T i and T j , respectively. The next three rows are obtained by left-multiplying the first row with T:, T i and T’:, respectively. The relative arrangement of elements in each subtable is the same as the multiplication table of C3, where the only difference is the contained elements, which are given in the above formulas. Finally, we obtain multiplication table of the symmetry group T for the regular te t r ahedr a1 .

17. The multiplication table of the finite group G is as follows.

I E A B C D F I J K L M N

E A B C D F I J K L M N

E A B C A E F I B F A K C I L A D J K L F B E M I C N E J D M N K M J F L N I J M K D B N L C D

D F I J J B C D L E M N K N E M A M N E N A K L M L A K E K L A I D B C F C D B C J F I B I J F

K L M N M N K L I J C D J F D B F I B C C D I J D B J F B C F I N E L A E M A K L A N E A K E M

a) Find the inverse of each element in G; b) Point out which elements can commute with any element in G; c) List the period and order of each element; d) Find the elements in each class of G; e) Find all invariant subgroups in G. For each invariant subgroup, list

its cosets and point out onto which group its quotient group is iso-

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42 Problems and Solutions in Group Theory

morp hic; f ) Make a judgment whether G is isomorphic onto the tetrahedral sym-

metric group T, or isomorphic onto the regular six-sided polygon symmetric group DG.

Solution. a) The pairs of elements which are inverse to each other are given as follows:

E-' = E , A-' = A , B-' = F , C-' = I D-' = J , K-' = L, M-' = N .

b) E and A can commute with any element in the group. c) The element whose order is 1 is the identity E.

The element of order 2 is A. Its period is { E , A } . The elements of order 3 are M and N . They belong to a common period

The elements of order 4 are B, C , D , F , I and J . They belong to the

The sixth-order elements are K and L. They belong to a common period

d) E and A form two classes, { E } and { A } . The elements of order 3 and 6 respectively form two classes, { M , N } and { K , L} . The elements of order 4 are divided into two classes, { B , C, D} and { F , I , J } . e) The invariant subgroup consists of several whole classes, and contains the period of each element. Its order has to be a divisor of 12, the order of G, namely, it has to be 2, 3, 4 or 6 .

{ E , A } is an invariant subgroup of G. Its cosets are { B , F } , {C, I } , { D , J } , { K , M}, and { L , N } . Because B2 = C2 = D2 = A, the squares of the first three cosets all are the invariant subgroup. Thus, its quotient group is isomorphic onto the regular triangle symmetric group D3.

{ E , M , N } is an invariant subgroup of G. Its cosets are { A , K , L } , {B ,C , D } and { F , I , J } . Due to B2 = F 2 = A, the squares of the last two cosets are not the invariant subgroup. Thus, its quotient group is isomorphic onto fourth-order cyclic group Cq.

{ E , K , N , A , M, L } is an invariant subgroup with index 2. Its coset is { B , I , D , F, C, J } . Its quotient group is isomorphic onto two order inversion group V2.

f ) Since the group T does not contain any sixth-order element, the group G is not isomorphic onto T. Since D6 does not contain any fourth-order element, the group G is not isomorphic onto Dg.

{ E , M, N } .

periods { E , B , A , F } , { E , C, A , I } , and { E , D , A, J } , respectively.

{ E , K , N , A, M, L}.

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Chapter 3

THEORY OF REPRESENTATIONS

3.1 Transformation Operators for a Scalar Function

* An rn-dimensional matrix group D(G) is called a representation of the given group G, if G is isomorphic or homomorphic onto D(G) . The element D (R) in D (G) , which is nonsingular , is called the representation matrix of the group element R E G in the representation D(G) . The trace TrD(R) x (R) is called the character of R in the representation D(G) . The representation matrix D ( E ) of the identity E is a unit matrix, and the representation matrices of two elements R-' and R are mutually inverse matrices, D ( R - l ) = D(R)- ' . The representation D(G) is said to be faithful if G is isomorphic onto D(G) . If all matrices D ( R ) are unitary, D(G) is called a unitary representation. If all matrices D ( R ) are real orthogonal, D(G) is called a real orthogonal representation. D(G)* composed of the complex conjugate D(R)* of the representation matrices D ( R ) is the conjugate representation of G with respect to D(G) . The representation is called a self-conjugate representation if all characters in the representation are real. The representation D(G) is called the identical representation if D ( R ) = 1 for any element R in G. The dimension m of the representation is assumed to be finite in this book if without special notification.

* Denote simply by x all coordinates of degrees of freedom in a quantum system, and by $(x) the scalar wave function. Under a transformation R, 12: is transformed into x' = Rx, the wave function $(x) is transformed into $'(x) = PR$(z). Being a scalar wave function, the value of the transformed wave function PR$ at the point Rx must be equal to that of the wave

43

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44 Problems and Solutions an Group Theory

function $ before transformation a t the point z, namely

The transformed wave function PR$(~) can be obtained from the original wave function $(z) according to Eq. (3.1), namely, first to replace x with R - l x in the wave function $(x), then to regard it as the function of x, which is nothing but the transformed function PR$(~). PR is a linear operator and has a one-to-one correspondence with the transformation R. This correspondence is invariant in the product of transformations, namely, PRPS corresponds to RS in the same rule. If the set of transformations R forms a group G, then the set PG of PR also forms a group and is isomorphic onto G. A linear operator L(z ) for the scalar wave functions is transformed into L'(z) in the transformation R:

* Assume that a quantum system with the hamiltonian H ( z ) is described by a scalar wave function. R is said to be a symmetry transformation if PR can commute with H ( z ) . The set of symmetry transformations R forms the symmetry group G of the system. If the energy level E is m degeneracy, we denote by Qp(z) the basis in the eigenfunction space. Then, the m- dimensional eigenfunction space for the energy E keeps invariant under the application of PR. The matrix form of PR in this space with respect to the basis qp(z) is D ( R ) :

The set of D ( R ) forms a representation of the symmetry group.

1. Let G be a non-Abelian group, D(G) be a faithful representation of the group G, and D ( R ) be the representation matrix of element R. If the element R in G corresponds to the following matrix in the set, please decide whether the following set forms a representation of the group G. For example, in a), if R H D ( R ) t , please decide whether the set D(G)t composed of D ( R ) t forms a representation of G. a) D ( R ) t ; b) D ( R ) T ; c) D(I2-l); d) D(R)*; e) D ( R - l ) t ; f ) detD(R); g) Tr D ( R ) .

Solution. Each subproblem gives a one-to-one correspondence between the element in G and the matrix in the given set of matrices. We should

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Theory of Representations 45

judge whether their products satisfy the same one-to-one correspondence in the given rule. If yes, it is a representation of G. Otherwise, it is not. Notice that D ( R S ) = D(R)D(S) .

a) Since D ( R ) t D ( S ) t # D ( R S ) t , the set of D ( R ) t is not a representa- tion of G.

b) Since D(R)TD(S)T # D(RS)T, the set of D(R)T is not a represen- tation of G.

c) Since D(R- '>D(S- ' ) # D [(RS)- '] , the set of D ( R - l ) is not a representation of G.

d) Since D(R)*D(S)* = D(RS)*, the set of D(R)* is a representation of G, which is called the conjugate representation of D(G).

e) Since D(R- ' ) tD(S- ' ) t = D [(RS)-'It, the set of D(R- l ) t is a representation of G.

f ) Since det D ( R ) det D ( S ) = det D ( R S ) , the set of det D ( R ) is a one- dimensional representation of G.

g) Since " r D ( R ) TrD(S) # TrD(RS), the set of TrD(R) is not a rep- resentation of G.

2. The homogeneous function space of degree 2 spanned by the basis func- tions $ l ( z , y ) = x2, $2(z,y) = zy, and +3(z,y) = y2 is invariant in the following rotations R in the two-dimensional coordinate space. Calcu- late the matrix form D ( R ) of the corresponding transformation operator PR for the scalar function in the three-dimensional function space:

a ) R = ( I " ) 0 - 1 ' cos a - s ina s ina cosa

13) .=A( -1 -a ) , c) R = 2 fi -1

If replacing the basis function $2(z,y) = zy with fizy, how will the matrix form of the operator PR change?

Solution. Three transformation matrices in the problem are all real or- thogonal. The inverse matrix of R is equal to its transpose.

a)

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46 Problems and Solutions in Group Theory

2 c) P R $ ~ (z, y) = (z cos a + y sin a )

= cos2a$1(z,y) + sin(2a)$a(z,y) + sin2a$3(x,y),

= - sin a cos a$1 (z, y) + cos(2a)$2(x, y) + sin a cos a$3 (z, y) ,

= sin2 a$l(z ,y) - sin(2a)$+,y) + cos2 a!$3(z,y),

PR$2(z,y) = ( z c o s a + y s i n a ) ( -zs ina + ycosa)

2 P ~ $ 3 ( z , y ) = (-zsina! + ycosa)

2 cos2 a - sin(2a) 2 sin2 a 2sin(2a) 2cos(2a) -2sin(2a) 2 sin2 a sin(2a) 2 cos2 a

Multiplying the basis function $2(z,y) by a, we obtain the matrix form of PR by a diagonal similarity transformation M .

- M = diag (1, h, I}, D ( R ) = M - ' D ( R ) M .

Each D(R) becomes a real orthogonal matrix.

b)

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Theory of Representations 47

3.2 Inequivalent and Irreducible Representations

Ir Two representations D(G) and D ( G ) with the same dimension are called equivalent to each other if there is a similarity transformation X relating two representation matrices for each element R in the group G, D ( R ) = X - ' D ( R ) X . The characters of each element R in two equivalent representations must be the same. For a finite group and a compact Lie group, any representation is equivalent to a unitary representation, two equivalent unitary representations can be related by a unitary similarity transformation, and two representations are equivalent if and only if the corresponding characters of each element in two representations are equal to each other.

-

j, A representation D(G) of the group G is said to be reducible if the representation matrix D ( R ) of any element R of G in D(G) can be trans- formed into the same form of the echelon matrix by a common similarity transformation X ,

Otherwise, it is called an irreducible representation. The necessary and sufficient condition for a reducible representation is that there is a nontrivial invariant subspace with respect to D(G) in its representation space. If both complementary subspaces are invariant with respect to D(G) , the reducible representation is called the completely reducible one where there exists a common similarity transformation X such that the T ( R ) in Eq. (3.3) for each element R of G is vanishing:

This form of reducible representation, where the representation matrix D ( R ) of each element R of G is a direct sum of two submatrices D ( l ) ( R ) and (R) , is called the reduced representation. The representation D(G) is said to be the direct sum of two representations D ( l ) ( G ) and D(2) (G) . For a finite group and a compact Lie group, any reducible representation is complete reducible, and can be transformed into the direct sum of two representations by a similarity transformation.

j , The Schur theorem says that X must be a null matrix if D ( l ) ( R ) X = X D ( 2 ) ( R ) holds for each element R of G where D ( l ) ( G ) and D(2) (G) are

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48 Problems and Solutions i n Group Theory

two inequivalent and irreducible representations of G, and X must be a constant matrix if D ( R ) X = X D ( R ) holds for each element R of G where D(G) is an irreducible representations of G.

* The necessary and sufficient condition for two inequivalent and irre- ducible representations Di(G) and Dj(G) of a finite group G is

where n(a) denotes the number of elements in the class Ca, and g is the order of G. Equation (3.5) says that the characters of two inequivalent and irreducible representations are orthogonal to each other in the class space with the weight n(a). When i = j , Eq. (3.5) is the necessary and sufficient condition for an irreducible representation of G. The characters in all inequivalent and irreducible representations of G form a complete bases in the class space, namely the characters satisfy

The number of all inequivalent and irreducible representations of a finite group G is equal to the number gc of the classes in G. The sum of dimension squares m; of all inequivalent and irreducible representations is equal to the order g of G,

Equations (3.5-7) are the necessary conditions satisfied by the charac- ters in any finite group G. However, they are not enough to determine all characters of G. Considering some other methods, such as the representa- tions of the quotient group of the invariant subgroups of G, the subduced and induced representations, the representations of the direct product of two subgroups etc., we can determine the characters in all inequivalent and irreducible representations of G, and obtain the character table of G. Based on the character table, we can find out the convenient forms of the unitary representation matrices of G. Usually, we choose as many diagonal repre- sentation matrices of generators as possible. For the complicated groups, the special method are needed.

* For a given quantum system with the Hamiltonian H ( z ) , we first find its symmetry group G, where the transformation operator PR for each element

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Theory of Representations 49

R in G can commute with the Hamiltonian H ( x ) ,

PR is called the symmetry operator of the system. Second, we study the symmetry group G of the system. We want to

find out its character table and the convenient forms of the representation matrices of all generators R in G. The representation matrices of other elements can be calculated from those of the generators.

Third, if the energy level E is rn degeneracy, we can find arbitrarily rn linearly independent eigenfunctions $p (x) for the energy E :

Due to Eq. (3.8), PR$~(Z) must be an eigenfunction of H ( z ) with the same energy E , namely, the rn-dimensional space spanned by $p ( 2 ) is invariant in the action of the symmetry operator PR. We can calculate the matrix form D ( R ) of PR in the basis function $,(x):

The set of D ( R ) forms a representation D(G) of the symmetry group G of the system, called the representation corresponding to the energy level E . The character x(R) =TrD(R) of R is easy to calculate. Generally speaking, the representation D(G) is reducible and not in the convenient form. In terms of the following method, the representation D(G) can be reduced into the direct sum of the irreducible representations and the eigenfunctions +,(x) can be combined into the basis functions transformed according to the irreducible representation.

Make a similarity transformation X,

X - ' D ( R ) X = @ U j D j ( R ) , xff = c ajx:. (3.10)

The rnultiplicity aj of the irreducible representation D j (G) in the reducible representation D(G) can be calculated from the orthogonal relation (3.5):

(3.11) ' REG

Substituting it into Eq. (3.10), one is able to calculate the similarity trans- formation matrix X . Let R be the generator A whose representation matrix

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50 Problems and Solutions in Group Theory

D j ( A ) is diagonal, where the problem becomes that of finding a similarity transformation X to diagonalize a matrix D ( A ) . An important point is to keep all the undetermined parameters in X waiting for the subsequent calculation. Substituting the matrix X into Eq. (3.10) for the remaining generator B whose representation matrix D J ( B ) may not be diagonal. In the calculation, X-' in Eq. (3.10) should be moved to the right-hand side of the equation for avoiding the calculation for X - l . After the calculation of Eq. (3.10) for all generators of G, if there are still some parameters un- determined, they should be chosen by any feasible way. Do not leave them as the undetermined parameters. Make sure that detX should not be van- ishing. The row index p of X is the same as that of D ( R ) , and the column index of X is a combined set of the index j for the irreducible representa- tion and its row index p. When aj > 1, an additional index r is needed to distinguish different Dj in the reduction (3.10). The new eigenfunctions are the combinations of $CL by X :

(3.12)

(ai,(x) is called a function belonging to the pth row of the irreducible representation Dj of G. When aj > 1, any linear combination of the functions (a{,(.) with the same j and p, where the combination coefficients are independent of j and p, is the eigenfunction belonging to the pth row of D j . This combination reflects the arbitrary choice of the undetermined parameters. The number of the undetermined parameters is Cj us .

* For the inner product of two wave functions used in quantum mechan- ics, the symmetry operator PR is usually unitary (there is exceptional case). Under this condition, the functions belonging to two inequivalent and ir- reducible representations of the symmetry group G are orthogonal to each other.

(3.13)

where ($i(I!Pj) is called the reduced matrix elements, which is a parameter independent of the subscripts u and p. Equation (3.13) is called the Wigner-

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Theory of Representations 51

Eckart theorem.

* A group G is called the direct product of two subgroups, G = H I 8 H2, if each element in G can be expressed as a product of R E H I and S E H2, RS = SR, and there is no common element in two subgroups HI and H2 except for the identity E. A pure rotation in the three-dimensional space is called a proper rotation. A rotation together with a space inversion is called an improper rotation. A proper point group consists of the proper rotations, and improper point group consists of the proper rotations and the improper rotations. The subset of the proper rotations in an improper point group G forms an invariant subgroup H in G with index 2. An improper point group G is said to be I-type if G = H 8 V 2 , where V2 = {e, a } is the two-order inversion group. An improper point group G is said to be P-type if G does not contain the space inversion a. Multiplying a on each improper rotation in a P-type improper point group G, we obtain a proper point group G'. G' is isomorphic onto G and contains an invariant subgroup with index 2.

3. Prove that the module of any representation matrix in a one-dimensional

Solution. In fact, the representation matrix in a one-dimensional repre- sentation is a complex number. For a finite group, any representation is equivalent to the unitary one, while a one-dimensional representation is in- variant under any similarity transformation. Therefore, it is a unitary one, where the module of any representation matrix is 1.

There are many other methods to show this conclusion. In a finite group G, any element R in G with the order n satisfies R" = E , and the representation matrix of the identity E in any representation is the unit matrix. Therefore, in a one-dimensional representation D ( G ) of G, D ( R ) n = D ( E ) = 1, namely, its module is 1.

4. Prove that any irreducible representation of an infinite Abelian group is

Solution. In an Abelian group G, the product of elements are commutable, so the representation matrix D ( R ) of any element R in an irreducible rep- resentation is commutable with the representation matrix of any other el- ement in G. From the Schur theorem, D ( R ) has to be a constant matrix. Since R is an arbitrary element in G, and the representation is irreducible, the dimension of the representation has to be one.

representation of a finite group is equal to 1.

one-dimensional.

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52 Problems and Solutions in Group Theory

5 . Prove that the similarity transformation matrix between two equivalent irreducible unitary representations of a finite group, if restricting its determinant to be 1, has to be unitary.

Solution. Let D(G) and D(G) be two equivalent irreducible unitary repre- sentations of a finite group. There exists a unitary similarity transformation M , M t M = 1 and D(R) = M - l D ( R ) M . If they can be related by another similarity transformation X , D(R) = X - l D ( R ) X , then

D ( R ) = ( X M y D ( R ) ( X M - 1 ) .

From the Schur theorem, X M - ' = c l and X = cM, where c is a constant. Since detX = 1, IcI = 1. Thus, X is a unitary matrix.

6. Show that for a finite group G, the sum of the characters of all ele- ments in any irreducible representation of G, except for the identical representation, is equal to zero.

Solution. For a finite group G, the characters of two inequivalent and irreducible representation satisfy Eq. (3.5). If the representation Di (R) is the identical representation, D i ( R ) = xi(R) = 1, Eq. (3.5) shows that the sum of characters xj(R) of any irreducible representation @(G) of G, except for the identical representation, is equal to zero.

7. The linear space spanned by the elements of a finite group G is called the group space, where the group element is the basis and the vector is a linear combination of the group elements. The group space is invariant for left- or right-multiplication by any element of G. The set of the matrix forms D ( S ) of left-multiplying an element S in the group space with respect to the basis R forms the regular representation D(G) of G. The set of the matrix forms D(S) of right-multiplication forms an equivalent representation D(G) of G.

1 when S R = T DTR(s) { 0 when S R # T ,

1 when R S = T 0 when RS # T ,

- DRT(s) = {

g when S = E 0 when S # E .

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Theory of Representations 53

Calculate the similarity transformation matrix X between two equiva- lent regular representations in the group space of the D3 group. How to generalize the result to any other finite group?

Solution. Let X be the similarity transformation relating two equivalent regular representations:

PEG PEG

Substituting the values of the representation matrix entries into the equa- tion, we have X ( T S ) R = XT(SR) . Noticing that the rows and the columns are designated by the group elements. From the associative law which is satisfied by the multiplication of elements, the matrix entries of X are equal if the products of their row index and the column index as the group el- ements are equal. Namely, we may choose the matrix entries of X such that the entries are one when the product of the row index and the column index as the group elements is equal to a given element, say E , and the remaining entries are zero. In fact, the similarity transformation matrix X can be calculated from the multiplication table of the group, where the ar- rangement of rows and columns is the same as those in X , in the following way: The matrix entry in X is one if its position is the same as that of a given element, say E , in the multiplication table, otherwise it is zero. For the group D3, we may choose X by the positions of the identity E in the multiplication table. The reader is encouraged to write the matrix form of the generators of D3 in two equivalent regular representations according to the multiplication table of D3, and to check whether they are related by the similarity transformation X.

The multiplication table of D3

I E D F A B C

E D F A B C , D F E B C A

F E D C A B A C B E F D B A C D E F C B A F D E

X =

'1 0 0 0 0 0 ' 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1

For the different choice of the given element, we can find g linearly independent similarity transformation matrices X , where g is the order of the group. In fact, the reduced form of the regular representation is the direct sum of all irreducible representations, where the multiplicity of each

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54 Problems and Solutions in Group Theory

irreducible representation is equal to its dimension mj. The matrix, which can commute with all representation matrices in the reduced form of the regular representation, contains Cj m3 = g parameters.

8 . C, is a class in a finite group G, and W is a vector in the group space composed of the sum of all elements in C,. Prove that the representation matrix of W in an irreducible representation is a constant matrix, and calculate this constant.

Solution. Let C, = (R1, R2, . . . , &(,I} be a class in G, where n(a) is the number of elements in the class. Denote by x: the character of RI, in the mi-dimensional irreducible representation DJ (R). For any element S in G , we have SRI,S-' E C,. Obviously, if R k # Ri, then SRkS-' # SRiS-'. Therefore, the set of SRI,S-' is the same as the class C,, so that sws-' = w.

Since Dj(S)Di(W)Dj(S)- l = Dj(W) , according to the Schur theorem, Dj (W) = c l . Taking its trace, we obtain n(a)xi = mjc. Then,

9. Prove that the number of the self-reciprocal classes in a finite group G is equal to the number of the inequivalent and irreducible self-conjugate representations of G. In other words, the number of pairs of the re- ciprocal classes is equal to the number of pairs of the inequivalent and irreducible non-self-conjugate representations.

Solution. Let n be the number of the self-reciprocal classes Ci in G, and m be the number of pairs of the reciprocal classes C, and CL'. The number of classes in G is gc = n + 2m. In any representation, the character xi of a self-reciprocal class is real, and the characters of two reciprocal classes are complex conjugate to each other, x: = x,-I. On the other hand, any character in a self-conjugate representation is real, but two characters of a class in the pair of two non-self-conjugate irreducible representations are complex conjugate to each other. The complex characters only occur for the non-self-reciprocal class in a non-self-conjugate representation. The sum of two characters of a class in two irreducible representations which are com- plex conjugate to each other is real, and the difference of two characters is

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Theory of Representations 55

pure imaginary for a non-self-reciprocal class and is zero for a self-reciprocal class.

Being a function of the classes, the characters in the inequivalent and irreducible representations of G are linearly independent, and form a com- plete set of bases in the class space. Any class function can be expressed as a linear combination of characters of the inequivalent and irreducible rep- resentations of G. On the one hand, since the differences of two characters in the different pairs of non-self-conjugate irreducible representations, be- ing a function of the classes, is linear independent, the number of pairs of non-self-conjugate irreducible representations cannot be greater than the number m of pairs of the reciprocal classes Ccy and (7;'. On the other hand, define m functions of the classes FB,

Being a function of the classes, Fp can be expressed as a linear combination of characters of all inequivalent irreducible representations of G. Due to the explicit form of F', the characters of the self-conjugate representations will not appear in the combination, and the characters of the non-self- conjugate representations will appear only in the form of difference of two characters in a pair of representations. Therefore, the numbers of pairs of non-self-conjugate irreducible representations cannot be smaller than rn. In summary, the number of pairs of non-self-conjugate irreducible representa- tions of a finite group G is equal to the number m of pairs of the reciprocal classes in G. As a result, the number of the self-conjugate irreducible rep- resentations of G is equal to the number n of the self-reciprocal classes in G. This completes the proof.

10. If the group G is a direct product H1 @H2 of two subgroups, show that the direct product of two irreducible representations of two subgroups is an irreducible representation of G.

Solution. Suppose that R E H I , S E H2, RS = S R E G = H1 8 H2, D j ( H I ) is an irreducible representation of H I with the representation space L1, and D"(H2) is an irreducible representation of H2 with the represen- tation space L 2 . ,C1 and L2 have no non-trivial invariant subspace for the groups H I and H2, respectively.

First, we will show that the direct product of two irreducible represen- tations of subgroups is a representation of the direct product group. In fact, if there is a correspondence from the elements RS and R'S in G to the matrices D(RS) = @ ( R ) x D " ( S ) and D(R'S') = Dj(R ' ) x D " ( S ) ,

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56 Problems and Solutions in Group Theory

respectively, we have the correspondence from RSR'S' = RR'SS' to

D(RS)D(R'S') = [Dj (R) x D"S)] [Dj(R ' ) x Dk(S')]

= [Dj(R)Dj(R')] x [D"S)D"S')]

Therefore, the set of D(RS) forms a representation of the group G, denoted by D(G) = @ ( H I ) x Dk(H2). Its representation space, denoted by L, is the direct product of the spaces L1 and L2. Let a and b respectively be the arbitrary vectors in the spaces C1 and L2. Then, their direct product a x b is a vector in the space C.

Second, we will demonstrate that this representation is irreducible. Ac- cording to the definition of irreducible representation, we are going to show by reduction to absurdity that there is no non-trivial invariant subspace in its representation space. If there is a nonzero invariant subspace L' in the space L with respect to D(G), where a x b E C', C' must con- tain all vectors in the subspace L1 x b because DZ(H1) x D j ( E ) c D(G) . Further, L' must contain all vectors in the space L1 x L2 = L because DZ(E) x Dj(H2) c D(G) . This leads to C = L'.

If D j ( R ) or D k ( S ) is replaced with another inequivalent irreducible representation, it is easy to prove by reduction to absurdity that the rep- resentation by the direct product is also changed into another inequivalent irreducible representation. If G is a finite group, the number of the inequiv- alent irreducible representations D J ( H 1 ) is equal to the number of classes in the subgroup H I , and the number of the inequivalent irreducible repre- sentations D"(H2) is equal to the number of classes in the subgroup H2.

While the number of classes in the direct product group G is equal to the product of the numbers of classes in two subgroups, which is just equal to the number of the inequivalent irreducible representations of G obtained from the direct product of representations. Therefore, each irreducible rep- resentation of the direct product group can all be expressed as the direct product of the irreducible representations of two subgroups.

11. Let each element in D3 be the coordinate transformation in two- dimensional space:

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Theory of Representations 57

where R is equal to the representation matrix in the two-dimensional representation DE(D3). For the generators D and A in D3, we have

The four-dimensional function space spanned by the following basis functions is invariant in the group D3:

$l(XC,Y> = x3, $2(X,Y> = X2Y, +3(X,Y) = XY2, 744(X,Y> = Y3.

Calculate the representation matrices of the generators D and A of D3

in this representation. Then, reduce this representation into the direct sum of the irreducible representations of D3, and construct new basis function which is the linear combination of the original basis functions and belongs to the irreducible representation.

Solution. First, according to the formula

P R $ p ( X ) = $p(R-lx) = c + u ( x ) D u p ( R ) , U

one is able to calculate the representation matrices of the generators D and A of D3 in the four-dimensional space spanned by the given basis functions, where R-lx means

For the element D , we have

2" = (-x + &y) /2, yt/ = (-&x - y) /2,

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58

Hence,

Problems and Solutions in Group Theory

3& -3 &I For the element A, we have

Hence,

1 0 0 0

D ( A ) = (O-" 0 0 1 0 O ) *

0 0 0 - 1

The characters are

x(D) = 1 = XA1(D) + XA"D) + XE(D),

x(A) = O = xA1(A) + xA'(A) + xE(A).

The similarity transformation matrix X satisfies

2 0 0 0 1 0 0 0 x 0 2 0 0 0 - 1 0 0

D ( D ) X = - 2 ( 0 0 -1-& ) , D ( A ) X = X ( 0 0 1 0 0 0 & - 1 0 0 0 - 1

The second equality gives

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Theory of Representations 59

Substituting it into the first equality, we have

The solutions are a2 = - 3 ~ 1 , bl = -3b2 and c1 = c2 = dl = d2. Arbitrary choose three constants, we obtain

1 0 1 0

x=( -3 O 0 0 1 ) 1 0

0 - 1 0 1

The new basis qf,, which belongs to the pth row of the irreducible repre- sentation I?, can be calculated by X ,

12. Calculate the characters and the representation matrices of the proper symmetry group 0 of a cube with the method of the quotient group and the method of coordinate transformations.

Solution. The group 0 is the proper symmetry group of a cube, composed of three fourfold axes, four threefold axes and six twofold axes. Its order is g = 24. The generators of three fourfold axes are the rotations around the directions of three coordinate axes through 7r angle, denoted by Tz, Ty and T,, respectively. Four threefold axes are along the following four directions with the generators Rj:

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60 Problems and Solutions in Group Theory

right-multiply by SI

left-multiply by SI

The twofold axes are along the following six directions with the generators Sk:

s1 : (ex + e y > / f i , s2 : (ez - e y > / d ,

s 3 : (ey f e J J 2 , s4 : (ey - e z > / J 2 , s5 : (ez + e z > / f i , s6 : (ez - e z ) / f i -

T is an invariant subgroup of 0 with the index 2. Through the direct calculation we obtain the coset formulas of T:

E TZ T,” T: R1 Rz R3 R4 R: R i R i R i

SI T: TZ SZ T: S5 Ty S6 Tx S4 TZ S3

SI Tz T2 S2 TZ S4 TX S3 Ty 5 5 Ty” S6

Now, right-multiplying S1 on the multiplication table of T given in Problem 16 of Chapter 2, we obtain the right-upper part of the multiplication table of 0. Then, left-multiplying S1 on the upper part of the multiplication table of 0, we obtain its lower part. In the calculation, we only make the replacement of elements according to the above coset formulas of T.

The right-upper part of multiplication table of 0

Due to the following formulas are just the opposite to the secondset of formulas:

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Theory of Representations 61

The left-lower part of multiplication table of 0

SZ s4 s6

The right-lower part of multiplication table of 0

T,” R f R4

2 Rl R4

Ty”

The group 0 has five classes: C1 = {E}, C2 = {TZ, T t , T:}, C3 = {Tz,T2,T,,T,”,Tz,T,3), C4 = {Rj , R$l 5 j 5 4}, and C5 = {Sk , 1 5 k 5 S}. From l2 + l2 + 22 + 32 + 32 = 24, we know that the group 0 has five inequivalent irreducible representations, denoted by A, B, E, TI, and T2, with dimensions 1, 1, 2, 3, and 3, respectively. The quotient group O/T is isomorphic onto V2. From the quotient group, we obtain two one- dimensional representations: The identical representation D A = 1 and the antisymmetric representation D with the characters:

XB(Cl) = XB(C2) = XB(C4) = 1, XB(C3) = X B ( C 5 ) = -1.

For a one-dimensional representation, the character is the same as the rep- resentation matrix.

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62 Problems and Solutions an Group Theory

The group 0 contains another invariant subgroup D2 of order 6 , com- posed of four elements E, TZ, Ty" and TZ. Its cosets are R1D2, R:D2, S1D2, S3Dz and S5D2. As we know that two cosets RH and SH are different if R-lS does not belong to H . Now, it is easy to check that these five cosets all are different. The quotient group contains two three-order elements (R1D2 and R:D2) and three two-order elements (S1D2, S3D2 and S5D2) so that it is isomorphic onto the group D3. Let R1D2, R:D2 and SlD2 respectively correspond to D , F and A in D3. S5D2 has to correspond to B in D3, because RlSl = Ty" = S5T: E S5D3. Then, S3D2 corresponds to C in D3. In constructing the multiplication table of D3, the only independent multiplication formula is DA = B. The remaining formulas can be derived from it and the orders of elements. Therefore, the correspondence we just established is invariant to the product of elements.

The group D3 has a two-dimensional irreducible representation, which is a non-faithful representation of 0. The elements in 0 belonging to the invariant subgroup D3 correspond to the same representation matrix in this representation, so do those belonging to its coset. Note that R1D2 = {R i , R2, R3, R4}, R:D2 = {R:, Ri7 Rg7 Rj} , S1D2 = {S17 S2,Tz,TZ}, s3D2 = {s37s47Tz,T23)7 and s5D2 = {s5,s6,Ty,Ty3}*

f ( E ) = 2, XE(R1) = xE(R?) = -1,

XE(S1) = XE(S3) = XE(S5) = 0,

( -' "> D"(S5) = f (2 7) ,

DE(Sl) = (' ) D ~ ( R : ) = - 2 -a-1 ' 0 - 1 ' -1 -fi

We can establish one three-dimensional representation of 0 by the co- ordinate transformation matrices. In fact, being a rotation in the three- dimensional space, the representation matrices of some elements in 0 are easy to obtain:

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Theory of Representations 63

0 0 1 DT1(R1) = (I..;) ) DT'(S1) = 1 0 0 . (00: 01)

Thus, the characters of classes in 0 are

XT'(C1) = 3) XT1(C2) = -1) XT1(C3) = 1,

XT1(C4) = 0, XTl(C5) = -1.

Another three-dimensional irreducible representation DT2 (0) can be ob- tained by the direct product of the antisymmetric representation DB (0) and the three-dimensional representation DT1 (0) :

XT2(C1) = 3) XT2(C2) = -1, XT2(C3) = -1,

XT2(C4) = 0, XT2(C5) = 1,

13. The multiplication table for a group G of order 12 is as follows.

I E A B C D F I J K L M N E A B C D F I J K L M N

E A B C D F I J K L M N A B E I L K N D M J F C B E A N J M C L F D K I C K L D E B M I N F J A D N F E C L J M A B I K F D N K M I E B L C A J I M J L A E F N C K D B J I M B N D L K E A C F K L C M F N A E J I B D L C K A I J D F B E N M M J I F K C B A D N E L N F D J B A K C I M L E

a) Find out the inverse of each element in G; b) Point out which elements can commute with any element in the group; c) List the period and the order of each element; d) Find out the elements in each class of G; e) Find out all invariant subgroups in G. For each invariant subgroup, list its cosets and point out onto which group its quotient group is iso- morphic; f ) Establish the character table of G;

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64 Problems and Solutions in Group Theory

g) Decide whether G is isomorphic onto the symmetry group T of the tetrahedron, or the symmetry group D6 of the regular six-sided polygon.

Solution. a) E-' = E , A-' = B , C-' = D , F-l = I , J-l = K , L-' = L, M- l = M , and N-l = N .

b) Only the identity E can commute with any element in G. c) The order of the identity E is one. The orders of L, A4 and N are

two, and the orders of A , B , C , D, F , I , J and K all are three. d) The group G contains four classes, { E } , { L , M , N}, { A , C, F, J } ,

and { B , D , I , K } . The first two are self-reciprocal classes, but the last two are the mutual reciprocal classes.

e) G contains only one non-trivial invariant subgroup { E , L , M, N } with index 3. Its cosets are { A , C, F, J } and { B , D , I , K } , and its quotient group is isomorphic onto the cyclic group Cs.

f ) From the number of the classes in G, we know there are four inequiv- alent irreducible representations in G. Since l2 + l2 + l2 + 32 = 12, there are three one-dimensional representations and one three-dimensional irre- ducible representation in G. Three one-dimensional representations can be calculated from the quotient group of the invariant subgroup. The charac- ters in the three-dimensional representation of G can be determined by the orthogonality.

g) From the orders of elements and the classes in G, we know that G is isomorphic onto the group T, but not isomorphic onto the group D6.

14. Calculate the character table of the group G given in Problem 17 of Chapter 2.

Solution. The group G in Problem 17 of Chapter 2 has 12 elements, di- vided into 6 classes. Since l2 + l2 + l2 + l2 + 22 + 22 = 12, there are four one-dimensional and two two-dimensional inequivalent irreducible rep- resentations in G. There are three invariant subgroups in G. The invariant subgroup { E , K , N , A, M, L } has index 2. Its coset is { B , I , D, F, C, J } . Its quotient group is isomorphic onto the two-order inversion group V2.

From the irreducible representations of the quotient group we obtain

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Theory of Representations 65

x' x 2

x4

x5

x6

x3

two one-dimensional representations, including the identity representation D ( l ) ( R ) = 1. The invariant subgroup { E , N , M } has index 4. Its cosets are { A , K , L } , { B , C, D } and { F, I , J } . The quotient group is isomorphic onto the cyclic group C4, where { A , K , L } is a two-order element. From this quo- tient group we obtain four one-dimensional representations, including two which are known. The invariant subgroup { E , A } has index 6. Its cosets are { B , F } , { C , I } , { D , J } , { K , M } and { L , N } . The quotient group is isomorphic onto the group D3, where { K , M } and { L , N } are three-order elements. From the two-dimensional irreducible representation of D3 we obtain the irreducible representation D5 of G. The representation matrices for the generators K and 23, which correspond respectively to D and A in D3, are

E A B C D F I J K L M N 1 1 1 1 1 1 1 1 -1 -1 1 1 1 -1 i -i -1 1 1 -1 -i i -1 1 2 2 0 0 -1 -1 2 -2 0 0 1 -1

Another two-dimensional irreducible representation D6 of G is the direct product D3 x D5,

D 6 ( B ) = i D 5 ( B ) , D 6 ( K ) = -D5(K).

Thus, we obtain the character tables of G.

3.3 Subduced and Induced Representations

* Let the order of a finite group G be g , and H = {TI = E , T2, . . . , Th} be a subgroup of G with the order h and the index n = g / h . Denote its left- cosets by R,H, 2 < r < n. For unification, the subgroup is denoted by R1H with R1 = E. Although Rj are not determined uniquely, we make choice of them arbitrarily. Then, any element in the group G can be expressed as RTTt. Let Dj (G) be an mj-dimensional irreducible representation of G. The set of those matrices Dj(T,) corresponding to the elements in H

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66 Problems and Solutions in Group Theory

constitutes a representation of the subgroup H , denoted by D j ( H ) , which is called the subduced representation from an irreducible representation Dj(G) of the group G with respect to the subgroup H . Generally, the subduced representation is reducible, and can be reduced with respect to the irreducible representations Dk ( H ) of the subgroup H

(3.14)

where mk is the dimension of the representation D k ( H ) of H , and E ( P ) is the number of elements contained in the class cp of H .

* Denote by q!Jp the bases in the representation space of D k ( H )

We define an extended space of dimension nmk with the bases qTp = PR,+~, where = +/.L. The extended space is invariant to the group G. We can calculate an n%k-dimensional representation Ak(G) of G in the following way. For any given element S in G, we calculate SR, for each R,, which can be expressed in the form of RUTt, where u and t are completely determined by S and T . Because

we obtain

This representation Ak (G) is called the induced representation from the irreducible representation D k ( H ) of the subgroup H with respect to the group G. In general, the induced representation is reducible and can be reduced with respect, to the irreducible representation @(G) of G:

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Theory of Representations 67

where n(a) is the number of the class C, of G.

* Calculate the character xk(S) of the element S in the induced repre- sentation A k ( G ) . Let S belong to the class C, in G. In general, some elements in C, belong to the subgroup H , and constitute a few classes of the subgroup H , denoted by cp. It is possible that no element in C, belongs to the subgroup H . For this case we say that no p exists. F'rom Eq. (3.15), the diagonal element of A k ( S ) may appear only when T = u, i.e., SR, = R,Tt. Namely, xk(S) is non-vanishing only when the class C, contains a few elements belonging to the subgroup H . Denoting by lcp the number of the different R, satisfying SR, = R,Tt, where Tt belongs to the class cp with the character j$ in the irreducible representation D k ( H ) of H , we have xk = Ca q$.

horn Problem 14 in Chapter 2, the number of elements R in G satisfying SR = RTt is m(a) = g/n(a). Such elements R, in general, can be expressed in the form of R,T,, so SR, = R, (T,TtTgl), while xk(T,TtTL1) = Zk(Tt). On the other hand, the number of TP in the subgroup H which satisfy TYTtT;l = Tt is ~ ( p ) = h/%(P). If R,Tz satisfy S(R,T,) = (R,T,)Tt, then R,T,T, also satisfy this formula. However, the latter does not make new contribution to the characters x"(S), so ~p = m(a)/E(P),

(3.17)

By the way, we do not need to consider the case of SR' = R'Ttt with another Tt/ E ED, because its contribution to the characters xk(S) has been calculated. In fact, letting Ttt = T,TtT;l, we have S(R'T,) = (R'T,)Tt.

Note that the elements Tt in the class c p of H all belong to the class C, of G. Obviously, the different classes Ca correspond to the different classes c p . Therefore, the sum over the class a in Eq. (3.16) is equivalent to the sum over all classes ??p in H . Due to xj (S) = xj (R-l SR) , it is easy to show from Eq. (3.17) that two multiplicities a j k in Eq. (3.14) and b j k

in Eq. (3.16) are equal:

The formula (3.18) is called the F'robenius theorem.

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68 Problems and Solutions in Group Theory

15. Calculate all inequivalent irreducible representations of the group

Solution. The group D2n+l contains a (2n+ 1)-fold axis, called the princi- pal axis, and (2n+ 1) equivalent twofold axes, located in the plane orthogo- nal to the principal axis. Denote by C2n+l the generator of the (2n+ 1)-fold axis and by C2l the generator of one twofold axis. and C2l are two generators of D2n+l. Due to the orthogonality, Czn+1 C2‘ = C2’ CFi+l. The order of Dzn+l is g = 4 n + 2. The number of the classes in D2n+l is gc = n + 2. There are two one-dimensional and n two-dimensional inequiv- alent representations of D2n+l.

The group D2n+l contains an invariant subgroup C2n+l with index two. Its coset consists of all twofold rotations whose axes are orthogonal to the principal axis. The quotient group is isomorphic onto the two-order in- version group V2, which gives two one-dimensional inequivalent irreducible representations of D2n+l, respectively denoted by A and B:

D2n+l with the method of induced representation.

DA(C2n+l) = DB(Gn+1) = DA(Cz’) = 1, DB(C2’) = -1.

The invariant subgroup C2n+l is a cyclic group. It has ( 2 n + 1) one- dimensional inequivalent representations. Denote the basis in each one- dimensional representation space by @,

Extending the one-dimensional space by defining another basis q5j = C2l@,

we have

where the formula C2n+lC2/ = C2lC;:+, is used. Thus, we obtain n two- dimensional inequivalent irreducible representations of the group D2n+l, denoted by Ej, 1 5 j 5 n,

The remaining representations with the different j are either equivalent to one of the above representations or reducible.

16. Calculate all inequivalent irreducible representations of the group D2n with the method of induced representation.

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Theory of Representations 69

Solution. The group D272. contains a (2n)-fold axis, called the principal axis, and (2n) twofold axes, located in the plane orthogonal to the principal axis. The twofold axes are divided into two sets, each of which contains n equivalent twofold axes. They form two classes, respectively. Denote by C2,

the generator of the (2n)-fold axis and by C2' the generator of one twofold axis belonging to one class. Define C2" = C2nC2/ = C2/ CT,, belonging to another class. The angle between two twofold axes for and C2n is x/n. C2n and C2t are two generators of D2,. The order of D2n is g = 4n. The number of the classes in Dan is gc = n + 3. There are four one-dimensional and n - 1 two-dimensional inequivalent irreducible representations of D2,.

The group D2, contains four invariant subgroups. One subgroup is C2n

composed of all rotations around the principal axis. Its index is two, and its quotient group is isomorphic onto the two-order inversion group V2, which gives two one-dimensional inequivalent representations of D2,, denoted by A1 and A2:

The subgroup C, of Can composed of the even powers of C2, is also an invariant subgroup of D2n. Its index is four. One of its coset consists of the odd powers of C2,. The remaining cosets are two classes of the twofold rotations, respectively. The quotient group is isomorphic onto the four- order inversion group V4, which gives four one-dimensional inequivalent representations of Dan including two known. The new representations are denoted by B1 and B2:

The set composed of C, and one class of the twofold rotations also forms an invariant subgroup of D2,. These two subgroups are both the two-order groups. The representations of D2, provided by their quotient groups are nothing but Al, B1 and A1 , Bz, respectively.

I t has (2n) one- dimensional inequivalent representation. Denote the basis in each one- dimensional representation space by $j,

The invariant subgroup C2, is a cyclic group.

Extending the one-dimensional space by defining another basis qY' = C2/ $ j ,

we have

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70 Problems and Solutions i n Group Theory

Thus, we obtain ( n - 1) two-dimensional inequivalent irreducible represen- tations of the group D2n, denoted by Ej, 1 5 j 5 n - 1,

The remaining representations with the different j are either equivalent to one of the above representations or reducible.

17. Calculate all inequivalent irreducible representations of the symmetry

Solution. In Problem 12 we have calculated all inequivalent irreducible representations of the group 0 with the method of the quotient groups and the method of coordinate transformations. In the present problem we will calculate them with the method of induced representation. In Problem 12 we have given the multiplication table of the group 0. Now, we use the same notation as that used there.

The group 0 contains a subgroup D4 of order 8, composed of one four- fold axis along the z axis and four twofold axes in the zy plane. The generator of the fourfold axis is denoted by T', and four twofold rotations are denoted by T:, S1, T i , and S2. The angle of two neighboring twofold axes is 7r/4. The index of D4 in 0 is 3. The characters table of the subgroup D4 is as follows.

group 0 of a cube with the method of induced representation.

The characters table of 0 The characters table of D4

-1 -1 -1 -1 1 -1 -1 1 -1

-2

We calculate the character of the induced representation from B1 of D4

with respect to 0 by Eq. (3.17),

24 x 1 24 x 2 x (-1) X y E ) = ~ - - 3, XB1(TZ) = = -1,

X B 1 (S1) =

8 8 x 6 24 X B V 3 = 8x3 (1 x 1 + 2 x 1) = 3 ,

24 x 2 x (-1) = -1, xB'(R1) = 0,

8 x 6 1 x 32 + 6 x (-1)2 + 3 x (3)2 + 6 x (-1)2 + 0 = 48.

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Theory of Representations 71

Because the characters is orthogonal to that of the identical representation A , this representation is the direct sum of a one-dimensional representation B and a two-dimensional representation E of the group 0. Denote by $1 the basis of the representation B1 of Dq. Extend the space by defining new bases $2 = R1$1 and 43 = RT$1. By making use of the multiplication table of 0 given in Problem 12, we have TzR1 = S3 = RTT; and T,Rq = Ti = RIS1. Thus,

Tz41 = -41, Tz$2 = -43, Tt.43 = - 4 2 ,

-1 0 0 0 0 1

0 -1 0 0 1 0 D(Tt.)= ( 0 0 - I ) , D(R1) = (1 0 0 ) .

We obtain the reduced form of the direct sum of B and E:

D4 We calculate the characters for the induced representation from A2 of with respect to 0 by Eq. (3.17)

24 x 1 2 4 x 2 ~ 1 xA2(E> = - - - 3, X A 2 ( T Z ) = 8 6 = 1,

t 4 xA2(T;) = - (1 x I + 2 x (-1)) = -1,

8 x 3 24 x 2 x (-1)

= -1, XA2 (R1) = 0, 8 x 6 XA"S,> =

1 x 32 + 6 x l2 + 3 x (-1)2 + 6 x (-1)2 + 0 = 24.

This is the irreducible representation TI of 0. Denote by $1 the basis of the representation A2 of Dq. Extending the space by defining new bases $2 = R1 $1 , $3 = R:$I , we have

R1$1 = $2, R1$2 = $3, R1$3 = $1,

TZ$l = $ 1 7 T d 2 = $3 7 Tz+3 = -$2.

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72 Problems and Solutions in Group Theory

By a simple similarity transformation (the cyclic of 1,2,3), this form of representation coincides with the form given in Problem 12.

We calculate the characters for the induced representation from B2 of Dq with respect to 0 by Eq. (3.17)

24 x 2 x (-1) = -1,

24 x 1

24 X B 2 ( ( T Z ) = 8 6

(1 x 1 + 2 x (-1)) = -1,

= 1, XB"(R1) = 0 , 2 4 x 2 ~ 1

8 x 6 XB2(S , ) =

1 x 32 + 6 x (-1)2 + 3 x (-1)2 + 6 x l2 + 0 = 24.

This is the irreducible representation T2 of 0. The characters are the same as those in TI except for the signs of the characters of Tz and S1. For the group 0, 7'2 = TI x B. The calculation is similar.

18. The regular icosahedron is shown in Fig. 3.1. The opposite vertices are denoted by Aj and Bj, 0 5 j 5 5. Choose the coordinate frame such that the origin is at the center 0 of the regular icosahedron, and the x axis is in the direction from the vertex Bo to the vertex Ao. Aj

are located above the zy plane.

A3

B5

BO Fig. 3.1 The regular icosahedron.

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Theory of Representations 73

In the regular icosahedron, there are six fivefold axes around the di- rections from Bj to A, with the generators Tj, 0 5 j 5 5. Except for one fivefold axis ( j = 0) which is along the positive z axis, the polar angles of the remaining fivefold axes all are 01, and their azimuthal angles are 2 ( j - 1)7r/5, respectively. There are 10 threefold axes along the lines connecting the centers of two opposite triangles with the gen- erators Rj, 1 5 j 5 10. The polar angles of the threefold axes are 0 2

when 1 5 j 5 5, and 03 when 6 5 j 5 10. Their azimuthal angles respectively are ( 2 j - 1)7r/5. There are 15 twofold axes along the lines connecting the central points of two opposite edges with the genera- tors Sj, l 5 j 5 15. The polar angles of the twofold axes are 04 when 1 5 j 5 5, 05 when 6 5 j 5 10, and 7r/2 when 11 5 j 5 15. Their azimuthal angles are 2 ( j - 1)7r/5 when 1 5 j 5 5, ( 2 j - l )n/5 when 6 5 j 5 10, and (4 j - 3)7r/lO when 11 5 j 5 15, respectively.

tanel = 2,

tan04 = (& - 1)/2,

tano2 = 3 - &, tano3 = 3 + &, (3.19)

t a d 5 = (& + 1)/2.

All axes are the non-polar axes, and any two axes with the same fold are equivalent to each other.

The proper symmetry group of an icosahedron is denoted by I, which contains 60 elements and five classes. The classes, denoted by E , C5, Cz, C3 and C2, contain 1, 12, 12, 20, and 15 elements, respectively. Calculate the character table of the proper symmetric group I of an icosahedron with the method of induced representation.

Solution. Since l2 + 32 + 32 + 42 + 52 = 60, the dimensions of five inequiva- lent irreducible representations respectively are 1 ,3 , 3, 4 and 5, denoted by A , T I , T2, G and H . The representation A is the identical representation.

The character table of T The character table of D5

A2 -1

P 0

The group I does not contain any non-trivial invariant subgroup. There

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74 Problems and Solutions in Group Theory

are two bigger subgroups T and D5. The subgroup T consists of three twofold axes (SS, 5’12 and S1) and four threefold axes (Rc, RZ’, R4, and R,-d.), and the subgroup D5 contains one fivefold (To) and five twofold axes (S j , 11 5 j 5 15). Their character tables are given in the Table.

For the induced representation from the identical representation A of T with respect to I, we have from Eq. (3.17)

6 0 x 3 ~ 1 12 x 15

XA(E) = 5, XA(C5) = XA(CZ) = 0 , XA(C2) = = 1,

XA(C3) = - 6o ( 4 x l + 4 x 1 ) = 2 , 12 x 20

1 x 52 + 0 + 0 + 15 x l2 + 20 x 22 = 120.

Because there is no negative character, this representation must be equiva- lent to the direct sum of an identity representation A and a 4-dimensional representation G. Subtracting the characters of A , we obtain the characters of the irreducible representation G of I:

XA(E) = XA(C5) = XA(CZ) = XA(C2) = XA(C3) = 1,

xG(E) = 4, xG(C5) = xC(C5”) = -1, XG(C2) = 0, XG(C3) = 1.

For the induced representation from the representation E of T with respect to I, we have

= 1, 60 x 3 12 x 15 XE(E) = 5, XE(C5) = XE(CZ) = 0 , XE(C2) = -

60 12 x 20

XE(C3) = - (4 x bJ + 4 x b J 2 ) = -1,

1 x 52 + 0 + 0 + 15 x l2 + 20 x (-1)2 = 60.

It is the five-dimensional irreducible representation H of I:

XH(E) = 5, XH(C5) = XH(C5”) = 0 , XH(C2) = 1, XH(C3) = -1.

By the induced representations from the irreducible representation of T, it is impossible to distinguish two classes C5 and Cl.

For the induced representation from El of D5 with respect to I, we have

60 x 2 6 0 x 2 ~ ~ l o x 1 = P, XEl(E) = - - - 12, XE1(C5) =

x”l(C,”) =

p = (A - 1)/2,

10 x 12

7 X E 1 (C2) = X E 1 (C3) = 0, -1 = -p 60 x 2 x (-p-’) 10 x 12

p-1 = (fi + 1)/2, 1 x 122 + 12 x p2 + 12 x (-P-’)~ + 0 + 0 = 180.

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Theory of Representations 75

A2 A3 A4 A5

Ti T2 G

It is the direct sum of three irreducible representations. Because the char- acters are not orthogonal to those of the representations G and H , so this representation contains one representation G and one representation H . Subtracting the characters of G and H , we obtain the characters of the three-dimensional irreducible representation T I of I:

E c5 cg c: c5“

1 v2 714 77 r13 1 v3 rl r14 r12 1 714 713 V Z 71

3 p-’ -P -P P-’

A 1 1 1 1 1 1 1 rl r12 r13 r14

A l l 1 1 1

3 -p p-‘ p-’ -p 4 -1 -1 -1 -1

H 5 O 0 0 0

xT1(E) = 3, X T l ( C 5 ) =p- l , XTl(C5”) = -p,

X T 1 ( C 2 ) = -1, X T 1 ( C 3 ) = 0.

By the induced representation from E2 of D5, we can calculate the char- acters of T2 similarly. In compari- son with the characters of TI, only the characters of C5 and Cz are in-

table of the group I is listed in the table.

The characters table of the group 1

-p p-1 -1 terchanged. Finally, the character G -1 -1 0

0 1 -1

19. Calculate the reduction of the subduced representation from each ir- reducible representation of the group I with respect to the subgroups Cg, D5 and T.

Solution. The character tables of I, D5 and T have been given in Problem 18. C5 is a cyclic group, whose character table is easy to write. The important thing in calculation is to identify which class of I each element in the subgroup belongs to. Then, the character formula (3.5) can be used to calculate the reduction of the subduced representation from each irreducible representation of the group I with respect to the subgroups C5, D5 and T. Here we neglect the calculation process, but only list the results.

The reduction of the subduced representation from each

irreducible representation of I with respect to the subgroup C5

Reduction of the subduced representation

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76

A1

A2 El E2 A 1 Ti T2 G H 5

Problems and Solutions in Group Theory

1 1 1 1 1 1 1 -1 2 p -p - ’ 0 2 - p - l P 0

3 p-’ - p -1 A2 @ Ei 3 - p p - ’ -1 A2 @ E2 4 -1 -1 0 El @ E2

1 1 1 A1

0 0 1 A1 @ Ei @ E2

The reduction of the subduced representation from each

irreducible representation of I with respect to the subgroup D5

I E 2C5 2Cz 5C4 I Reduction of the subduced representation

1 1 1 1 1 1 w w 2 1 1 w 2 w 3 -1 0 0 1 1 1 1 3 -1 0 0 3 -1 0 0 4 0 1 1 5 1 -1 -1

A1 T T

A1 CBT E $ E ’ $ T

I E 3 c 2 4C4 4C4’ I Reduction of the subduced representation

T2

H

q = exp{ - i2~ /5} , w = exp{- i2~/3} ,

p = q + q 4 =(&-1) /2 , p-1 = - 7 + q 3 = ( & + 1 ) / 2 .

20. Calculate the reduction of the subduced representation from the regular representation of the improper symmetry group I h of an icosahedron with respect to the subgroup Csi, D5d and T h .

Solution. The I h group is an improper point group of I-type. It is the direct product of the group I and the two-order inversion group V2 = { E , o}. Any irreducible representation of I h can be expressed as the direct product of two irreducible representations respectively belonging to I and V2. Since the space inversion o is commutable with all elements in I h and its square is equal to the identity, the representation matrix of o in an irreducible representation of Ih is a constant matrix fl. Based on D ( o ) = 1 and -1, the irreducible representations of I h are divided into rg and ru) where r denotes the irreducible representations of I. g and u are the abbreviation of the Germanic gerade and ungerade, respectively.

The subgroups C5i, D5d and T h are also the improper point groups of

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Theory of Representations 77

I-type, their irreducible representations are the direct product of an irre- ducible representation of V2 and an irreducible representation of respec- tively the proper point groups C5, D5 or T. Similarly, based on D ( a ) = 1 and -1 the irreducible representations of the subgroups are also divided into yg and y,, where y denotes the irreducible representations of C5, D5 or T. Therefore, the reduction of the subduced representation from the regular representation of I h with respect to the subgroup C5i, D5d and T h is similar to the reduction of the subduced representation from the regular represen- tation of I with respect to the subgroup C5, D5 and T, where the only difference is that all the representations are duplicated by the subscripts g and u.

Remind that the character of each element in the regular representation is vanishing except for the character of the identity which is equal to the order of the group. In the reduction of the subduced representation from the regular representation of a group G of order g with respect to its subgroup H of order h, the multiplicity of an irreducible representation of H is equal to its dimension multiplied by n = g/h. The results of the reduction are as follows.

For group Csi ,

For group DSd,

For group T h ,

In all formulas, the series in the bracket is just the reduction series of the regular representation of the subgroup.

21. Calculate the representation matrices of R6 and s2 of 1 in its irre- ducible representations by R6 = SIT:SITcl and S12 = ToS1TiRG [see Problem 4 in Chapter 41, where the representation matrices of the generators TO and S1 are [see Problem 12 in Chapter 41:

DT1(To) = diag {q71,q-'},

DTZ(To) = diag{q2717q-2}

DG(To) = diag{q2,q7q-1,q-2},

DH(To) = diag{q2,q,l,q-1,q-2},

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78 Problems and Solutions an Group Theory

p-1 J2 -p D T ' ( S 1 ) = -

d3 p -Jz p-1

1 DG(S1) = -

lh

P-2 2p-1 fi 2P P2

1 5

P ( S 1 ) = -

-p 1 -1 -p-1 1 -P

-1 -p -p-l

-p-1 -1 1 -p-1 -p -1 1,

2p-' & 2p p2

-fi -1 fi - p-2 fi p2 -2p-' -2p l/G - 2p-1 p-2

p2 -45 -P-2 $3

Solution. Any element in I can be expressed by the product of two el- ements respectively belonging to the subgroup C5 and T: To" (RgS;Sf2), where TO and S1 are two generators of I and R6 and S2 can be expressed as the product of the generators. In this problem we are going to calcu- late the representation matrices of R6 and S2 explicitly. In the calculation we express R6 and S12 as a product of two elements and make use of the property that the representation matrix of To is diagonal. Remind that q = exp{ - i 2 n / 5 } satisfies the following useful formulas:

4 C q a = o , a+5n = q a , 1 + 2 ~ + 2 ~ ~ = - 1 - 2 ~

a=O

p = q + q 4 = (A - 1)/2, p-1 = - q 2 - q 3 = (& + 1 ) / 2 .

For the identical representation, oA(To) = oA(S1) = D A ( & ) = DA(S12) = 1. The results for other representations are as follows.

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Theory of Representations 79

0 0 1 0 0 .

' 0 0 0 0 1 0 0 0 - 1 0

0 - 1 0 0 0 , 1 0 0 0 0 1

3.4 The Clebsch-Gordan Coefficients

* If a quantum system consists of two subsystems, the wave function of the total system can be expressed as the combination of the products of the wave functions of two subsystems. Denote by G the common symmetry group of the system and two subsystems. We choose the wave function of each subsystem belonging to a given row of a given irreducible representa- tion of G

Then, the wave function for the total system is expressed as their product, transforming according to the direct product representation:

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80 Problems and Solutions in Group Theory

The character in the direct product representation is equal to the product of the characters in two representations.

The direct product of two irreducible representations is reducible in general, and can be reduced by a similarity transformation [see Eqs. (3.10- 12)] :

For a finite group or a compact Lie group, all representations in Eq. (3.22) are chosen to be unitary, then Cjk can be chosen to be a unitary matrix with determinant one. The series in the right-hand side of Eq. (3.22) is called the Clebsch-Gordan (C-G) series, the matrix entries in the similarity transformation matrix Cjk are called the C-G coefficients. Rewrite Eq. (3.22) in the form of characters, we may calculate the multiplicities a~ by the orthogonal relation (3.5) of characters:

(3.23)

Substituting Eq. (3.23) into Eq. (3.22), we are able to calculate the C-G coefficients CPv,JMr. j k When U J > 1, the parameter T is needed to distin- guish these UJ irreducible representation DJ(G) . The method has been demonstrated in section 2 in detail. The total wave function @hT(l, 2) with the given symmetry is combined from the wave functions !Pg(l, 2) by the C-G coefficients C;:,,,,:

P" (3.24)

In physics, it is convenient to express Eq. (3.24) in the Dirac notations. $i ( l )$ i (2) is denoted by I j , p) Ik, v), where the indices j and k are some- times neglected. @Lr(l, 2) is denoted by (IJ, ( T ) , M):

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Theory of Representations 81

The action of the symmetric operator PR on the wave function is

First, we choose R to be the generator A whose representation matrices are diagonal such that Eq. (3.26) becomes an eigenequation, and gives some relations of the C-G coefficients. Sometimes, the method of projection operators is useful in this step. Then, the C-G coefficients can be further determined in terms of the remaining generators. Before completion of the calculation, some undetermined parameters should be chosen arbitrarily.

* The inverse transformation of Eq. (3.24) is

y!$ (1)$,” (2) contains some functions @hr (1,2) belonging to different irre- ducible representation. In terms of the projection operator

(3.27)

one can pick up from $$(1)$:(2) the part belonging to the Mth row of the irreducible representation D J ( G ) . In fact,

The projection operator is helpful in the calculation of the Clebsch-Gordan coefficients.

22. Please use the projection operators to reduce the regular representation

Solution. In the notation for the group T given in Problem 16 of Chapter 2, R1 is a three-order element. Its eigenvalues are denoted by u p , where w = exp{-i2~/3}, = u p . The value of p can be taken as module 3, namely p + 3 is equal to p. Taking the representation matrix of R1 to be diagonal, we can designate the rows and columns of each irreducible representation by the power p of the eigenvalue u p for R1. By making use

of group T.

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82 Problems and Solutions in Group Theory

of the projection operator for the cyclic subgroup of R1, we can calculate the common eigenfunctions iak; for left- or right-multiplying of R1:

where ca is the normalization factor and can be chosen for convenience. Take R(’) = E and c1 = 9 ,

= 9PpEPu = (1 + w”-p + W P - ~ ) { E + R1w-p + R:wp) = 3Spu { E + Rid-’” + R : w ~ } .

Then, taking an element R(2) which does not appear in a,,, (1) , say R(2) = T:, we have

Now, all elements in T appear in the functions @rJ and a$. One will not obtain any other linearly independent function @$) by choosing new R(”).

Generally, QfJ does not belong to a given irreducible representation, and can be combined to new basis functions belonging to the irreducible representation by the class operator W = R1 + R2 + R3 + Rq, which take constant values 4Xj(Rl)/mj (see Problem 8) in an irreducible representa- tion. In the representation A, E , E‘ and T of the group T, the values of W are 4, 4w, 4w2 and 0 , respectively. In fact, for given subscripts p and v, we calculate the matrix form of W in the basis functions (at;, and then diagonalize it. Due to completeness of the function bases, we only need to list those terms of E and T; in the calculation

W@FJ = 3wp { E + T:} + - = wp { @?J + 3QfJ ) , Waf; = { E + T:} w” { 1 + w”-p + W P - ~ } + - . -

= SP”WP { @FJ + 3@;J . 1 When p = v , the matrix form of W in these basis functions is

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Theory of Representations 83

The eigenvalues are 4wp and 0, and the transpose of the corresponding eigenvectors respectively are (1 /3 ,1) and (1, -1). Combining the basis functions according to the eigenvectors, we obtain the irreducible basis functions which respectively belong to representations A, E, E‘ and T .

When p # u, there is only one basis function @fJ. representation T :

belongs to the

In the calculation, being the eigenfunctions, each basis function can con- tain an arbitrary constant factor. For a one-dimensional representation, this factor is the normalization factor, which is irrelevant to the representation matrix and can be omitted. But for the three-dimensional representation T , the choice of the factor is related to the representation matrix. In order to make the representation unitary, we choose the basis functions to be nor- malized, COO = 1/6 and c10 = c20 = 1/3. In calculating the representation matrix of another generator T’ with respect to the basis function QTo, we only need to list those terms of E , T:, and T:, and notice T:Ty” = T:.

QgTg = (3E + 3R1 + 3Rf - T: - Ty” - T: - Rq - R3 - R2 - R: - Rg - R i } / 6 ,

(7‘; + R2 + R: + w2 (T: + R4 + Rg) + u ( T i + R 3 + R i ) } / 3 ,

S;lb =

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84 Problems and Solutions in Group Theory

*To = {T: + R2 + Rz + w (Tz + R4 + Ri) + w2 (Ty” + R3 + R;)} /3.

Thus,

The representation matrices of the generators are

-T 1 0 0

0 0 w2

-T D (R1)= (0 w 0 ) , D

The factors cpY in other basis functions are determined by the representa- tion matrices and the following formula:

q v T ; = c D;p(T,2)qp. P

= {T: + R4 + Ri + w (T i + R3 + Ri) + w2 (T i + Ra + R i ) } /3,

*& = {T: + R4 + Ri + w (Ty” + R2 + R:) + d 2 (T’: + R3 + R:)} /3,

= {3E - TZ - Ty” - TZ + w (3R: - R; - R: - R:) + w2 (3R1 - R2 - R3 - Rd)} /6,

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Theory of Representations 85

Usually, the three-dimensional irreducible representation T of the group T is calculated by the coordinate transformation method in three- dimensional space. The results are as follows:

0 0 1 -1 0 0

0 1 0 0 0 1 DT(R1)= ( 1 0 o ) , DT(T,2>= ( 0 -1 .).

We can find a similarity transformation X to relate two equivalent repre- sentation. For the generators R1,

we have

X = (" a bw2 " ) . a bw cw2

Substituting it into DT(T,")X = X D T ( T 2 ) ,

- a + 2 b + 2 c 2 a - b + 2 c 2 a + 2 b - c -a + 2bw2 + 2cw 2a - bw2 + 2cw 2a + 2bw2 - cw

bw cw2 -a + 2bw + 2cw2 2a - bw + 2cw2 2a + 2bw - cw2

we obtain a = wb = w2c = m, x='(l 1 w2 w w 2 ) , w x - . - q ; 2 1 1 1 f ;). A 1 1 1

23. Please use the projection operators to reduce the regular representation of group 0.

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86 Problems and Solutions an Group Theory

Solution. We follow the notations and formulas given in Problem 16 of Chapter 2 and Problem 12 of this Chapter. Just like the method used in the preceding problem, we first calculate the common eigenfunctions @$ = c, Pp R(,)PV of left- or right-multiplying R1 in terms of the projection operators. Letting R(l) = E and R(2) = T:, we have

There still exist some elements in 0 which do not appear in the functions @$ and @pJ. Letting R(3) = Tz and R(4) = S2, we have

where 6,(-”) is equal to 1 when ( p + v) is a multiple of 3, otherwise it is zero. Thus, all elements in 0 appear in the four kinds of basis functions, so they are complete.

Now, we will combine the basis functions by making use of the class operator W = Tz + T: + Ty + TZ + T, + T: such that they belong to the irreducible representations. W takes a constant value 4xj(Rl)/rnj (see Problem 8 ) in an irreducible representation. Through calculation, the con- stants in the irreducible representations Al, A2, E , TI and T2 of 0 are

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Theory of Representations 87

6 , -6, 0, 2 and -2, respectively. In fact, for given subscripts p and v , we calculate the matrix form of W in the basis functions at?, and then diagonalize it. Due to completeness of the basis functions, we only need to list those terms of E , T:, T, and S2. In the calculation of left-multiplying W on the basis functions @E, the following formulas obtained from the multiplication table will be useful:

Hence,

When p = v = 0, there are four bases, a::’, @;I, and @g). The matrix form of W in the basis functions and its eigenvectors for the eigen- values 6, -6, 2 and -2 are

The corresponding eigenfunctions respectively belong to the representations

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88 Problems and Solutions in Group Theory

Al , A2, TI and T2:

Each eigenfunction may contain a constant factor. For one-dimensional representation, this constant is the normalized factor, which is irrelevant to the representation matrix and may be omitted. But for two-dimensional or three-dimensional representations, the choices of factors are related to the explicit forms of the representation matrices.

When p = v # 0, there are three bases, a,, , @,, and @rJ. The matrix form of W in the basis functions and its eigenvectors for the eigenvalues 0, 2 and -2 are

( 1 ) ( 2 )

The corresponding eigenfunctions respectively belong to the representation

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Theory of Representations

E , TI and T2:

89

q, = a,, { $ 9 3 + QfJ}

= a,, ( E + T i + Tl + T; + w-, (R1+ R2 + R3 + R4)

+w, (R;+R;+R;+R: ) } ,

1 QT1 ,, = b ,, { @FJ - +fJ - 2 W W t 3 , )

= b p , (3E - T: - Ty” - T: - 25’1 - 2S3 - 2S5

+ LJ-’ (3R1 - R2 - R3 - R4 - 2T: - 2T’ - 2Tz)

+up (3R: - Ri - Ri - Ri -2Tz -2Ty - 2 T z ) } ,

1 “2 = c,, {a!; - @pi + 2 W W f J

= cp, (3E - T: - T i - T: + 2S1 + 2S3 + 2S5

+ w-’ (3R1 - R2 - R3 - R4 + 2T: + 2Ty” + 2T:)

When p # v and p + v = 3, there are three bases, af;, @f; and ipp!.

The matrix form of W in the basis functions and its eigenvectors for the eigenvalues 0, 2 and -2 are

0 1 - 3

: ) , (i). (!J (3)- The corresponding eigenfunctions respectively belong to the representation E , TI and T2:

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Problems and Solutions in Group Theory

(2@$ - @f; + a:;} (2T: + 2R3 + 2Rg - Tz - T: - 5’1 + 35‘2

+ W-’ (2Ti + 2R4 + 2Ri - Ty - Ty” - 5’5 + 3S6)

+ W’ (2T: + 2R2 + 2Ri - Tx - T: - S3 + 3 5 4 ) ) .

When p = 0 and v # 0, there are two bases, @$ and Q O u . ( 3 ) The matrix form of W in the basis functions and its eigenvectors for the eigenvalues 2

and -2 are

The corresponding eigenfunctions respectively belong to the representation TI and T2:

*:; = bOv {@g) - L P @ g ) }

= bOv {T: + R4 + RZ - TX - Ty” - S1 + o-. (TZ + R2 + R: - T, - T: - S,)

+d (Tz + R3 + Rg - 3”’ - T: - S 3 ) } ,

1 !Pov T2 - - cov {@I;“? + W-,@g)

= coU {T: + R4 + RZ + Tx + Ty” + S1 + w-’ (T i + R2 + R: + T, + T: + 55) + w v (T: + R3 + Ri + Ty + 3”: + S3) } .

When u = 0 and p # 0, there are two bases, @$) and @$I. The matrix form of W in the basis functions and its eigenvectors for the eigenvalues 2 and -2 are

( - 2 L -2w’) 0 , (-;-’) (u’’) *

The corresponding eigenfunctions respectively belong to the representation TI and 7’2:

90

form

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Theory of Representations 91

+up ( T l + R3 + Ri + Tz + S5 + T’)} . Now, we calculate the representation matrix of the generator T,. The

representation matrix of the generator R1 is known. First, for the repre- sentation E , we take the normalized bases, all = a21 = ,/m, to make the representation unitary,

We only need to list the terms of E and T,:

Hence, the representation matrices are

According to these representation matrices we can calculate the remaining factors a12 = a22 = d m :

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92 Problems and Solutions in Group Theory

In comparison with the representation matrices calculated in Problem 12 with the method of the quotient group,

-E . the similarity transformation X satisfying X-’DE(R)X = D (R) is

Second, for the representation T I , we take the normalized bases, boo = d@ and blo = b20 = d m ,

*:A = d m ( 3 E + 3 R 1 +3Rf - Tz -T: - TZ - R2 - R: - R3 - Rg - R4 - RZ + Tx + T: + Ty + Ty” + Tz + T: + S1 + S3 + S5 - 3S2 - 3Sq - 3Sc} ,

QT; = d m {TZ + R2 + Ri - Ty - T: - S1 + w2 (Ti + R4 + Ri - Tz - Ty” - 5 ’3)

+ W (T2 + R3 + Rg - Tx - Tz - S5)} )

+ w (Ti + R4 + Rg - Tz - Ty” - 5 3 )

+ w2 ( T i + & + R: -Tx - T: - S 5 ) } .

= dWy{T; + R2 + Ri - Ty - T: - S1

We only need to list the terms of E , T: and T:. Notice that TzS2 = SIT, = T; .

Hence, the representation matrices are

1 0 0 1 -2w

0 0 w2 -2w2 1 -2w

-Ti D (R1)= ( 0 w 0 ) , oT’ (Tz)= ( - 2 w -2w2 - y 2 ) . They are symmetric. The factors in the remaining basis functions can be

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Theory of Representations 93

calculated by the following formulas:

1 1 18

SPFhT, = J- {-WE - w2T; - T: + - . - } = { 9;; - 2 ~ 9 ; ; - 2w29:;},

Thus,

= d m { T j + R4 + R; - Tx -Ty” - S1

+ w2 (TZ + R2 + Ri - T, - TZ - S5)

+ W (T i + R3 + Ri - Ty - T; - S3)} ,

Q r i = d m { T : + R4 + R; - Tx - T i - S1

+ w (T i + R2 + Ri - T, - T: - S5)

+ w2 (T: + R3 + Rz - Ty - T: - S3)} ,

9:; = d m {3E - T: - T; - T; - 2S1 - 2S3 - 2S5

+ w 2 (3R1 - R2 - R3 - R4 - 2T: - 2T; - 2T:)

+ W (3R: - R; - Ri - Ri - 2Tx - 2Ty - 2Tz ) } ,

12 - d m { 2T; + 2R3 + 2Rg + T, + T: + S1 - 3s2 QTl -

LJ2 (2Ty” + 2R4 + 2R; + Ty + Ty” + Ss - 3S6)

+ w (2T: + 2R2 + 2R,” + Tx + T: + S3 - 3S4)},

= I/m (2T: + 2R3 + 2Ri + T, + T: + S1 - 3s2 W (2Ty” 4- 2R4 + 2Ri + Ty + Ty” + S5 - 356)

+ w2 (2T: + 2R2 + ZR,” + Tx + T: + S3 - 3S4)},

9;; = I / m { 3 E - T: - T i - T: - 2S1 - 2S3 - 2S5

+ w ~ ( ~ R ~ - R ~ - R $ - R ~ - ~ T ~ -

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94 Problems and Solutions in Group Theory

\ r igh t left \

E A B C F K M N

In comparison with the representation matrices calculated in Problem 12 with the method of coordinate transformation in three-dimensional space,

E A B C F K M N

E A B C F K M N A E M K N C B F B N E M K F C A C K N E M A F B F M K N E B A C K C F A B E N M M F A B C N K E N B C F A M E K

0 0 1 -1 0

0 1 0 0 0 1 DT1(R1) = (1 0 0 ) , DT1(Tz) = (: 0 0 ) .

The similarity transformation Y , satisfying Y-lDT1 (R)Y = BT1 (R) is

1 w2 w 1 1 1 Y = - k ( l 1 w 1 w2 1). Y- l=%(: . w,’ ;)-

At last, for the representation T2, if we choose cpY = b p Y , the coefficients in the expansions of the basis functions @% are the same as that of @% except for the terms related to the elements in the coset of the subgroup T, which change signs. So, the representation matrix DT2((R) is just the direct product of DT1(R) and D A 2 ( R ) .

24. The multiplication table of the group G is given as follows.

a) Write the character table for the group G. b) If we know that the representation matrices of the generators A and B in a two-dimensional irreducible representation D are

and two sets of basis functions qp and @Y respectively transform ac- cording to this two-dimensional representation of G:

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Theory of Representations 95

$ $ x 3

combine the product function q!~~$, , such that the basis function belongs to the irreducible representation of G.

Solution. The group G consists of 8 elements, where the order of the identity E is one, the orders of A, B , C, F and K are two, and the orders of M and N are four. E and K are commutable with any element in G. G contains five classes, { E } , { K } , { A , C}, { B , F } and { M , N } . They are all self-reciprocal classes. Since l1 + l2 + l2 + l2 + 22 = 8, .the group G has four one-dimensional and one two-dimensional inequivalent irreducible representations. These representations all are self-conjugate.

There are four invariant subgroups in G. The cosets of the invariant subgroup { E , K } are { A , C}, { B , F } and { M , N } . Its quotient group is isomorphic onto the group Vq. The indices of the remaining three invariant subgroups, { E , A, K , C } , { E , B , K , F } and { E , M , K , N } all are two. We can obtain four one-dimensional representations of G from these quotient groups. The characters in the two-dimensional representation can be obtained by Eqs. (3.5-6). The character table of G is as follows. The characters for the self-direct product of the two-dimensional representation are also listed in the table.

K ( A n ( B , F ) (M,W 1 1 1 1 1 1 1 1 -1 -1 1 1 -1 1 -1 1 1 -1 -1 1 2 -2 0 0 0

x 4 4 0 0 0

Substituting the matrix forms of A and B into the above formula, one has

1 0 0 0 1 0 0 0

X-l( ; 0 0 1 0 - 1 0 ; 0 0 - 1 0 1 O 0 ) ) 0 0 0 - 1

X-l(O 0 0 0 1 0 1 o ) x = ( o - l o 1 0 0 0 0 )

0 1 0 0 0 0 1 0 - 1 0 0 0 0 0 0 - 1

From Eqs. (3.10-11), one obtains:

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96 Problems and Solutions in Group Theory

The eigenvectors of the product matrices for the eigenvalues f l are:

eigenvalue 1 -1 1 -1

Thus, the similarity transformation matrix X is

1 1 0 0 x=-( 1 0 0 1 1 )

Jz 0 0 1 - 1 - 1 - 1 0 0

The new basis functions belonging to the irreducible representations are calculated by X:

* ( l ) = ($141 + $ 2 4 2 ) /a, = ($142 + $ 2 4 1 ) /Jz,

!P(2) = ($141 - $ 2 4 2 ) /Jz, *1(4) = ($142 - $241) /4.

25. Calculate the unitary similarity transformation matrix X for reduc- ing the self-direct product of the three-dimensional irreducible unitary representation DT of the group T:

x-l { D T ( R ) x D T ( R ) } x = c UjDj (R> .

Solution. This problem will be solved by two methods. One is to calculate the similarity transformation matrix directly. Another is to calculate the expansions of states in the product space. These two methods are the same in principle. However, when the dimension of the representation is quite large, the first method may not be convenient. The aim of this problem is to show the relation between these two methods, and to demonstrate the merit of the second method.

To begin with, we solve this problem with the first method. Take the generators of the group T to be T: and R1. Their representation matrices in the three-dimensional representation T are

-1 0 0

0 0 1 0 1 0 DT(T,”) = ( 0 -1 0 ) , D T ( R 1 ) = (; : i) .

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Theory of Representations 97

Denote by D ( R ) the self-direct product representation of DT,

The multiplicities aj can be calculated with the character formula (3.11):

Namely,

X-'D(R)X = D A ( R ) @ D E ( R ) @ DE' (R) @ D T ( R ) @ DT(R) .

Let R = T; in the formula of the similarity transformation.

D(T:) = diag {1,1, -1,1,1, -1, -1, -1,l)

X-lD(T;)X = diag { l , l , l , -1,-1,1, -1,-1,l).

We conclude that for the ls t , 2nd, 3rd, 6th and 9th columns of X , the matrix entries in the 3rd, 6th, 7th and 8th rows are zero, while for the 4th, 5th, 7th and 8th columns, the matrix entries in the ls t , 2nd, 4th, 5th and 9th rows are zero. Then, let R = R1. The action of DT(R1) is to cycle the matrix entries in three rows as 1 + 2 -+ 3 --+ 1. In the direct product space, the rows (columns) can be designated by two indices ab, and the action of D(R1) is to change the 12-th component in the vector to the 23-th position, etc. If we change the designation of the rows (columns) by one index (u running from 1 to 9, corresponding to the order of the indices ab as 11, 12, 13, 21, 22, 23, 31, 32, 33, then the action of D(R1) is

11 -+ 22 ---+ 33 --+ 11, 12 -+ 23 -+ 31 -+ 12, 1 - 5 -+ 9 -+l, 2 -6 --i 7 -2 ,

13 -+ 21 -+ 32 -+ 13, 3-4 -+ 8 -3.

The matrix after the similarity transformation is

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98 Problems and Solutions in Group Theory

X - l D ( R 1 ) X =

where w = exp{-i2~/3}. W

‘ 1 0 0 0 0 0 0 0 0 o w 0 0 0 0 0 0 0 o o w 2 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0

, o o 0 0 0 0 0 1 0

3 have known that the thirG and the 6t il matrix entries in each of the first three columns of X are zero. The three column matrices are also the eigenvectors of D(R1) . From the above cyclic property, their matrix entries in the 2nd, 4th) 7th and 8th rows must also be zero, namely only the matrix entries in the lst, 5th and 9th rows can be nonzero. The values of those matrix entries can be determined from the eigenequation:

From the unitary property of X , for the 6th and 9th columns, only the matrix entries in the 2nd and 4th rows may be nonvanishing. Since the multiplicity of the representation T in the reduction of the direct product representation is two, we can simply choose the matrix entries in the 6th and 9th columns to be zero except for x26 = X49 = 1. Substituting them into the formula for the similarity transformation on D(R1) , one obtains

D(R1)X.g = X.7, D(R1)X.7 = x.8.

The solutions are

x64 = x75 = x87 = x38 = 1,

and the remaining matrix entries all are zero. Finally, the matrix X is

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Theory of Representations 99

X =

’ m m ~ 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1

0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0

w”J3 w/J3 0 0 0 0 0 0

,m u/& w2/& 0 0 0 0 0 0

It can be checked that X does satisfy the similarity transformation formulas. Now we turn to the second method. For convenience we use the Dirac

notation to express the basis state in the representation space before and after the similarity transformation. The representation space before the transformation is the direct product of two subspaces, and the basis state is denoted by [T,p)lT,v). Sometimes, two T in this basis state can be omitted. The actions of the generators T: and R1 on the basis state are:

The representation space after the transformation is the direct sum of five subspaces, where the basis state is respectively denoted by IIA), llE), IlE’), llT, ( l ) ,p ) and llT, (2 ) ,p ) , p is 1, 2 or 3, satisfying

where the index T = 1 or 2 is used to identify two representations T. The basis state which is invariant in the action of T: must be in the following form:

The basis state which changes its sign in the action of T; must be in the

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100 Problems and Solutions in Group Theory

following form:

Applying R1 to the basis state IIA), we have

By comparison, we obtain cf = c t = 0, and cf = cf = cf. After normal- ization, cf = 0, and the normalized basis state is

Applying R1 to the basis state IIE), we have

By comparison, we have cf = cf = 0, and cf = cfw = c f u 2 . Taking cf = 0, we obtain the normalized basis state

Similarly, we have

Applying R1 to the basis state llT, (r),3), we have

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Theory of Representations 101

Applying R1 to them, we have

These results are consistent with that obtained by the first method.

26. Calculate the C-G series and the C-G coefficients for the direct product representation of each pair of two irreducible representations of the group 0.

Solution. In Problem 17, we have calculated the representation matrices of the generators T, and R1 of the group 0 in five inequivalent irreducible representations A , B, E , 2'1 and Tz. Now take a similarity transformation to diagonalize the representation matrices of R1:

-2w2 2w -1

-1 -2w2 -2w DT1(Tz) = -DT2(Tz) = - 2w 1 - 2 u 2 ) ,

3

DT1(R1) =DT2(R1) = 0 1 0 , (; w : l )

where w = exp{ - i27r/3}. Since the representation matrix of R1 is diagonal, and the diagonal elements wm are different in each irreducible representa- tion, this power m can be used to designate the row (column) in each irreducible representation. Notice that the value of the power m is module 3, m + 3 = m. The representations A and B are one-dimensional, so that the index 0 for the row (column) can be omitted. The index for the row (column) in the representation E takes the values 1 and -1. The index in the representations 571 and T2 takes the values 1, 0 and -1.

By comparison, we have We may takeand

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102 Problems and Solutions in Group Theory

B E Tl T2

E x E E X Ti N_ E x T2

T ~ x T ~ E T ~ x T ~ Tl x T2

The direct product of the identical representation A and any repre- sentation is still equal to that representation. The direct product of the antisymmetric representation B and any representation is easy to calcu- late:

1 1 -1 1 -1 2 2 0 - 1 0 3 -1 1 0 -1 3 -1 -1 0 1 4 4 0 1 0 A @ B @ E 6 - 2 0 0 0 Tl @ T2 9 1 1 0 1 A @ E @ T l @ T z 9 1 -1 0 -1 B @ E @ T z @ T l

where 0 3 is the Pauli matrix. The Clebsch-Gordan series for the direct product representations is listed in the Table, and the Clebsch-Gordan co- efficients are calculated in the following.

T h e representation of 0 I E 3T: 6Tz 8R1 6S1 I T h e reduction I I

A 1 1 1 1 1 1 1

( 1 ) E x E z A e B e E . Since the representation matrices of R1 are all diagonal, the eigenvalue

of R1 for the product state is the product of the eigenvalues of two states, i.e., the sum of rn‘s. Thus, we have:

Applying T, to them, we have

So, a1 = u2, bl = -b2, and c1 = c2. After normalization we have

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Theory of Representations 103

(2) E x Ti N E CB T2. From the eigenvalues of R1, we have:

Applying Tz to them, we have

The solution is a1 = b2 = -c2 = a2 = c1 = -bl . After normalization we have

For the representation of T2, we obtain a1 = -b2 = c2 = -a2 = c1 = -bl by the similar calculation. After normalization we have

For the C-G coefficients in E x T2, one only needs to exchange 2'1 and T2 in the above results. Namely,

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104 Problems and Solutions an Group Theory

(3) Ti x Ti N T2 x T2 N A @ E @ Ti @ 2'2.

From the eigenvalues of R1, we have:

Applying T, to them, we have

Thus, a1 = -a2 = a3, bl = b2 = -b3 = -b4 = -b 5 - - -be. After normalization we have

The states belonging to the representation TI must be orthogonal to the above states,

Applying T, to them, we have

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Theory of Representations 105

From the orthogonality of states we have

Applying T, to them, we have

Thus, dl = -& = -d3. After normalization we have

Thus, After normalization we have

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106 Problems and Solutions in Group Theory

For the case of T2 x T2, the term lT1, p) 12’1, v) in the expansion is replaced with JT2, p)JT2, Y), and the coefficients are exactly the same.

(4) 571 x T2 -“B$E$T2$Tl. Multiplying the antisymmetric representation B on the formula in the

case (3), we obtain the formula for the present case. Note that the similarity transformation o3 appearing in the direct product of B and E. The results are as follows:

27. Calculate the C-G series and C-G coefficients in the reduction of direct product representation in terms of the character table of the group I given in Problem 18:

(1) D T ~ x DT1; ( 5 ) D*2 x DG; (9) DG x D H ;

(2) DT1 x DT2; (6) DT1 x D H ; (10) DH x D H .

(3) DT2 x DT2 ; (7) DTz x D H ;

(4) DT1 X DG; (8) DG X DG;

Solution. We will only list the results for this problem. Notice that the character for the direct product representation is equal to the product of characters of two representations.

In Problem 23 the representation matrices of the generators T and S1 in each irreducible representation of the group I were given, where the representation matrix of T is diagonal. Since the order of T is five, the diagonal element of D ( T ) takes the power qm where q = exp{-i2~/5). The power m is different in each irreducible representation of I, so this power rn can be used to designate the row (column) in each irreducible representation. The possible values of the power in each representation are as follows: 0 for the representation A , 1, 0 and -1 for the representation T I ,

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Theory of Representations 107

2,O and -2 for the representation T 2 , 2 , 1 , -1 and -2 for the representation G, and 2, 1, 0, -1 and -2 for the representation H . Note that the power m is module 5 , m + 5 = m.

The reductions of the direct product representations in 1

A Ti T2 G H

Ti x Ti Ti x T2

T2 x T2 Ti x G T2 x G Ti x H T2 x H G x G G x H H x H

1 1 1 1 1 3 p-1 -p -1 0 3 -p p-1 -1 0

9 P-2 P2 1 0

9 p2 p-2 1 0 12 -p-1 P 0 0 12 p -p-1 0 0

4 -1 -1 0 1 5 0 0 1 -1

9 -1 -1 1 0

15 0 0 -1 0 15 0 0 -1 0 16 1 1 0 1 20 0 0 0 -1 25 0 0 1 1

The reduction

(1) DT1 x DT1 Y D A @ DT1 @ D H .

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108 Problems and Solutions in Group Theory

(3) DT2 x DT2 fi D A @ DT2 @ D H .

(4) DT1 x DG N- DT2 @ DG @ D H .

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Theory of Representations 109

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110 Problems and Solutions in Group Theory

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Theory of Representations 111

(8) DG x DG 2: D A @ DT1 @ DT2$ D G $ D H .

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112 Problems and Solutions in Group Theory

(9) DG x DH 2 DT1 @ DT2 @ DG @ 2DH.

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Theory of Representations 113

(10) DH x DH E DA @ DT1 @ DT2 @ 2DG @ 2DH.

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114 Problems and Solutions in Group Theory

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Chapter 4

THREE-DIMENSIONAL ROTATION GROUP

4.1 SO(3) Group and Its Covering Group SU(2)

* In the real three-dimensional space, a spatial rotation which keeps both the position of the origin and the distance of any two points invariant:

R11 R12 R13 (i) = (R21 R22 "3) (zi) , - x' = R:, R31 R32 R33

(:) I T J: I - - gTg, RTR = 1, R* = R,

is described by a real orthogonal matrix R. R is a proper rotation if detR = 1, and an improper rotation if detR = -1. The set of all three- dimensional real orthogonal matrices with detR = 1, in the multiplica- tion rule of matrices, satisfies the four axioms and forms the simplest non- Abelian compact Lie group SO(3). It is a very important Lie group from the viewpoints of both physics and mathematics. The element R ( n , w ) in SO(3) is a rotation around the direction n through the angle w , satisfying

R(n, w + 27r) = R(n, w ) = R(-n, 27r - w ) , ( 4 4

R(n ,x ) = R(-n , r ) .

When n is along the coordinate axis, the rotation matrix can be written in the exponential form of matrix:

0 0

0 sinw cosw

115

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116 Problems and Solutions in Group Theory

cosw 0 sinw

-sinw 0 cosw R(e2,u) = ( 0 1 0 ) =exp{-iwT2},

cosw -sinw 0 R(e3,w) = (sinow co;w :) = exp{-iwT3},

where

0 -i 0

o i 0 -i 0 0 0 0 0 ( 4 4

( T a ) ) ~ d = -iCabd.

They are the generators in the self-representation of S0(3), satisfying the typical commutation relations for the angular momentum operators

* The rotation S(cp, 0)

transforms e3 to the direction n(0,cp). Since

The rotation R ( n , w ) around n through w can be expressed as

R(n, W) = S(cp, 0)R(e3, ~)S(cp, 0)-' = exp { - i d - T} . (4.5)

Thus, the rotations with the same rotational angle LJ are conjugate to each other. The class of SO(3) is described by the rotational angle w. The SO(3) group is a simple Lie group because it does not contain any nontrivial invariant Lie subgroup.

Define a vector 2, which is along the direction n with the length w. Let the polar angle and the azimuthal angle of n be 0 and cp, respectively. R(n,w) can be described by the spherical coordinates (u,O,cp) of i3, or its rectangular coordinates ( ~ 1 , w2, wQ), which are the group parameters of SO(3). The varied region of the group parameters is called the group space. The group space of SO(3) is a spheroid with radius T, where in the sphere,

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Three-dimensional Rotation Group 117

two end points of a diameter describe the same rotation. Namely, the group space of SO(3) is a doubly-connected closed region. It is the reason why SO (3) has the double-valued representations.

* The set of all two-dimensional unitary matrices u with det u = 1, in the multiplication rule of matrices, form the Lie group SU(2). Its element can be generally expressed by

u(n, w ) = 1 cos(w/2) - i(3 - n) sin(w/2)

) (4*6) -i sin(w/2) sin Be-+

cos(w/2) + i sin(w/2) cos e 7

cos(w/2) - i sin(w/2) cos e = ( -i sin(w/2) sin

satisfying

u(h, wl)u(fii, wz) = u(n, w1 + w2),

u(n,47r) = 1, u(ii,27r) = -1, (4.7)

.(n,w) = u(-ii,47r - w ) = -u(-ii,27r - w ) .

Choosing the group parameters of SU(2) to be the vector G along the di- rection n(6,cp) with the length w, we obtain the group space of SU(2) to be a spheroid with radius 27r, where all points on the sphere describe the same element -1. The group space for SU(2) is a simply-connected closed region, and the SU(2) group is a compact simple Lie group. The gener- ators in the self-representation of SU(2) are a,/2. There is a two-to-one correspondence between Au(n, w ) E SU(2) and R(n, w ) ~ s O ( 3 ) :

3

u(n, w)aau(n,w)- l = c abRba(n, w ) . (4.8) b= 1

So, the SU(2) group is the covering group of the SO(3) group. The subset of u ( n , w ) with the same w forms a class of SU(2).

1. Prove the preliminary formula by induction:

1 1 = P + g [a, P] + +% [a, [a, PI] + * - *

where a! and P are two matrices with the same dimension. Then, show Eq. (4.8) and prove that SU(2) is homomorphic onto SO(3).

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118 Problems and Solutions in Group Theory

Solution.

We are going to show by induction that the expression in the brackets is equal the following commutator with n-multiplicity

The above formula is true for n = 1 obviously. Now, if it is true for n - 1, we will show that it is also true for n.

n

n-1 n-I

(-1)"-"(n - l)! am-l = o x Pan-m

(m - l)!(n - m)! m=l

(-1)"-"(n - l)! (m)!(n - m)!

ampQn-m.

n-1

[m + (n - m)] ampan-" + (-I)npan = anp + c = C m!(n-m)!

m=l (-l)n-"n!

m=O

This completes the proof. Further, due to (Tc)ba = iEcab ,

3

2-1 [Z - ii, a,] = i C ncEcabOb = C 0 6 (ii - T ) ~ ~ , bc b=l

b= 1

Repeating the commutation relation by n times, we have n

b= 1

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Three-dimensional Rotation Group

By making use of the preliminary formula and

u(ii, w ) = exp{ -iwa' - n/2}, R(ii, w ) = exp ( - i w n - T) ,

we obtain

119

Now, Eq. (4.8) is proved. Any u ( n , w ) E SU(2) can uniquely determine a matrix R(n,w) E SO(3) by Eq. (4.8). If both u1 E SU(2) and u2 E SU(2) determine the same R E SO(3) by Eq. (4.8), u,'ul can commute with three Pauli matrices g a , so it has to be a constant matrix. Due to their determinants, we have u1 = fu2. Therefore, Eq. (4.8) gives a two-to-one correspondence between &u(ii,u) E SU(2) and R(n,w) E SO(3), which is obviously invariant in the product of elements. It means that SU(2) is homomorphic onto SO(3).

2. Expand R(n, w ) = exp ( - i w n . T) as a sum of matrices with the finite

Hint: (n - T)3 = n 0 T.

Solution. From

terms.

one has

R(n, w ) = exp ( - i w n - T)

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120 Problems and Solutions in Group Theory

This problem can also be solved with the method of similarity transfor- mation:

ii . T = ST3S-l, S = R(e3, cp)R(e2, O ) ,

where 8 and cp are the polar angle and azimuthal angle of the direction n(O,cp) , and S is the rotation transforming the X3 axis to the direction A. Since R(e3, w ) can be expanded in the following form

R(e3, w ) = exp (-iwT3) = 1 - Tl + T: cos w - iT3 sin w ,

we have

R(n, w ) = SR(e3, w)S-l

= 1 - ST?SP1 + STzS-l cos w - iST3S-l sin w

= 1 - (ii - T ) ~ + (ii. T ) ~ COW - i (ii. T) sinw.

3. Check the following formulas in the group 0 [see the notation given in Problem 12 of Chapter 31 in terms of the homomorphism of SU(2) onto SO(3):

TzR1 = S3, TzTx = R1, RlR2 = Ri.

Solution. Since the elements in S U ( 2 ) are two-dimensional matrices, the homomorphism of S U ( 2 ) onto SO(3) can be used to simplify the calculation for the multiplication of elements in SO(3). The elements TZ, Tx, R1, R2, RZ and S3 in the group 0 correspond to those elements in the group S U ( 2 ) ) where a sign in u(R) is irrelevant to us:

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Three-dimensional Rotation Group 121

Now, replacing the elements R of 0 in the multiplication formulas with the corresponding elements u(R) of SU(2), we are able to check whether the multiplication formulas are correct up to a sign.

u V Z ) u ( ~ 1 ) = m<1- i 0 3 ) (1/2) {I - i (01 + 0 2 + 0 3 ) )

= 4 7 5 ( - i 0 2 - i 0 3 ) = 4 ~ 3 1 ,

U ( T , ) U ( T , ) = m(1 - i 0 3 ) m { 1 - ia1)

= (1/2) (1 - i (01 + 0 2 + 0 3 ) ) = u(R1), U(Rl)U(R2) = (1/2) (1 - i (01 + 0 2 + 0 3 ) ) (1/2) (1 - i (01 - 0 2 - 0 3 ) )

= (1/2) (1 - i (a1 + 0 2 - 0 3 ) ) = -u(R$).

4. Prove the following formulas for the group I in terms of the homomor- phism of SU(2) onto SO(3)

R6 = s l T i s l T r l , s 1 2 = ToSiTiR6,

which were used in Problem 21 of Chapter 3. Solution. According to the notations given in Problem 18 of Chapter 3, we are able to calculate the matrices u(R) in SU(2) which corresponds to elements of I, where a sign in u(R) is irrelevant to us.

where the polar angle and azimuthal angle of n are 03 and n/5, and the polar angle and azimuthal angle of m are 0 4 and zero, tan 03 = 3 + &, and tan04 = (& - 1)/2. Thus,

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122 Problems and Solutions in Group Theory

Let A = S1 (T,SITcl) and B = (T:S1Tc1) S 1 2 . We are going to calculate the matrix entries of u(A) and u(B) to show that they are both equal to -u( &) . Because

s in(2~/5) = s in (3~ /5 ) = 6 - 1

C O S ( ~ T / ~ ) = -- 4 '

we have

45-1 (10 - 2 h ) -

- - - - 1 [(lo + 2 d q 7 & + l - 4 20

[(lo + 2&)(5 - &) - (10 - 2&)2&] 1 i - -

- 2 + 4 0 I / i c %

where d m m = 3& - 5 is used.

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Three-dimensional Rotation Group 123

4.2 Inequivalent and Irreducible Representat ions

* In the three-dimensional space, any rotation can be expressed as a product of three rotations around the coordinate axes

where ca = C O S ~ , s, = s ina , etc. Three angles a, /? and y are called the Euler angles. There are two methods for calculating the Euler angles. One is based on the matrix form of R. The third column of R is a unit vector, where its polar angle is p and its azimuthal angle is a. The third row of R is also a unit vector, where its polar angle is /? and its azimuthal angle is ( T - y ) . The other is based on the relative position of the coordinate frames before and after the rotation R. If R transforms the coordinate frame K to K‘, then in the K coordinate frame, the polar angle and the azimuthal angle of the 2‘ axis of K‘ are /? and a, respectively. In the K‘ coordinate frame, the polar angle and the azimuthal angle of the 2 axis of K are ,8 and (7r - 7). The domain of definition for the Euler angles in SO(3) are

- n S a Q n , o < p 5 7 r , - n < y < 7 r . (4.10)

For the SU(2) group, -27r < y 5 27r. When /? = 0 or n, only one parameter between a and y is independent.

* For the SU(2) group, its inequivalent irreducible representation can be designated by a half-integer j = 0, 1/2, 1, 3/2, . . .,

(-1)n { ( j + v)!(j - v)!(j + p ) ! ( j - p)!}1’2

( j + v - n)!( j - p - n)!n!(n - v + p) ! =

* {cos(w/2)}2j+Y-’-2n {~ in (w/2 )}~~-”+” ,

n : ,ax( O ) , e m - , rn in ( j+ ’ ) . j - p V - - P

(4.11)

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124 Problems and Solutions in Group Theory

The matrix d j (w ) satisfies the following symmetric relations:

(-l)P-”d&&) = d$J-bJ) = dj,,(w) = dt”-P(Ld))

d j ”P (7r - w ) = (-l)j-W:&J) = (-l)j+”dj ”-,(W).

d i p @ ) = ( - l )~-P&/( -p) ) dj,,(27r) = (-l)”ja,,, (4.12)

The character of the element u(n,w) in the representation Dj is

* D j is a unitary representation of SU(2) with dimension (2j + 1). When j is an integer, Dj is a single-valued representation of SO(3) and a non- faithful representation of SU(2). It is equivalent to a real representation by a similarity transformation. When j is a half-odd-integer, Dj is a double- valued representation of SO(3) and a faithful representation of SU(2). It is a self-conjugate representation, but cannot be transformed into a real representation by a similarity transformation. Do(.) = 1 is the identical representation, D1I2(u) = u is the self-representation of the SU(2) group. D1(u) is equivalent to the self-representation of the SO(3) group:

M-’R(a, p, Y ) M = D’(a, p, 7)) M = - ’ (1; ‘i) . (4.14) O J Z O

The generators in the representation D j are

Letting c = cos(p/2) and s = sin(P/2), we have

do(@) = 1) d”“(a) = ( -“) , s c

(4.15)

) i s2 Jzcs C2

c2 -Jzcs s2 d’ (@ = J z c s c2 - s2 -JZcs ,

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Three-dimensional Rotation Group 125

5 . Calculate the Euler angles for the following transformation matrices R, S and T , and write their representation matrices in Dj of SO(3):

a) ~ ( a , p , r ) = ; 1

-Jz J2 2 d 3

& + 2 6 3 a - 2 2& b) J z - 6 & + 2 d 3 '2Jz

- 2 J 2 -2& 4Jz

Solution. a) From the third column of R, we have

C O S ~ = -(J2/4) * 2 = -m, a = 3n/4.

s ina = ( f i /4 ) - 2 = m, From the third row of R, we have

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126

Hence,

Problems and Solutions in Group Theory

b) From the third column of S, we have

cosa = (@4) - Jz = &/2, a = 7r/6.

s ina = (Jz/4) - Jz = 1/2,

From the third row of S, we have

Hence,

c) From the third column or the third row of T , we have p = 7r, but a and p are undetermined. In fact, when ,f3 = 7r , only (a - y) is independent. For definiteness, we may choose a = 0 and calculate R(e2,7r)-lT

-A 1 0

= 1 ( --? i) =R(e3,-57r/6).

Hence,

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Three-dimensional Rotation Group 127

6. Calculate the Euler angles for the following rotations R, S and T , and write their representation matrices in Dj of SO(3). a) R is a rotation around the direction n = el sin8 + e3 cos8 through an acute angle 8; b) S is a rotation around the direction n = (e l + e2 + es) /& through 27r/3; c) T is a rotation around the direction n = (el + e2) /fi through 7r.

Solution. If a rotation R transforms the coordinate frame K to the frame K', the polar angle and the azimuthal angle of the 2' axis of K' in the coordinate frame K are p and a, respectively. In K', the polar angle and the azimuthal angle of the 2 axis of K are /3 and (i7 - y), respectively.

a) The X' axis and the 2' axis of K' are both in the X Z plane of K , and the angle between two Z axes is 28, so p = 28 and a = 7r - y = 0,

b) The X', Y' and 2' axes of K' are located in the Y , 2 and X axes of K , respectively, so p = 7r/2, a = 0 and 7r - y = 7r/2,

c) The 2' axes of K' is along the negative 2 axis of K , so that p = 7r and only (a - 7) is independent. Since the Y' axis of K' coincides with the X axis of K , we can choose y = 0 and obtain a = -7r/2.

7. Calculate the Euler angles for the rotations TO, T2, R1, R2, R6, S1, S2, $3, 5 ' 1 1 , and S12 in the icosahedron group I, where the notations for the elements are given in Problem 18 of Chapter 3.

Solution. The key for solving this problem is to determine the directions of the coordinate axes after rotation.

The Euler angles of TO are obviously /3 = 0 and a + y = 27r/5. One may choose a = 0 for definiteness.

After the rotation T2, the 2 axis is rotated to the direction O A l , so we have a = 0 and p = 81. The Y axis is rotated to the direction pointing from 0 to the central point D1 of the edge A2A3. The angle between two planes AlOAo and AlODl is 37r/10. So, 7r-y = n/2+37r/10, and y = 7r/5.

After the rotation R1, the 2 axis is rotated to the direction of O A l , so we have cy = 0 and ,8 = 81. The Y axis is rotated to the direction pointing

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128 Problems and Solutions an Group Theory

from the origin 0 to the central point 0 2 of the edge AoA4. The angle between two planes AlOAo and A 1 0 D 2 is 7r/10. SO, 7r - 7 = 7r/2 - 7r/10, and y = 37r/5.

After the rotation R2, the 2 axis is rotated to the direction of OA2, so a = 27r/5 and ,f3 = 81. The Y axis is rotated to the direction pointing from 0 to the central point 0 3 of the edge A3A4. The angle between two planes A2OAo and A 2 0 0 3 is 3n/10. So, n - y = n/2 + 3n/10, and y = n/5.

After the rotation Rg, the 2 axis is rotated to the direction OB3, so we have Q = -n/5 and p = 285. The Y axis is rotated to the direction pointing from 0 to the central point Q of the edge AoAl. The angle between two planes B30Ao and B3OQ is 7r/10. So, n - y = n/2 + n/10, and y = 2n/5.

From the result of the preceding problem, the Euler angles for S1 are p = 284, a = 0 and y = n.

After the rotation S2, the 2 axis is rotated to the direction OA2, so we have a = 2n/5 and p = 81. The Y axis is rotated to the direction pointing from 0 to the central point 0 4 of the edge AoA5. The angle between two planes A2OAo and A2004 is n/lO. So, n - y = 7r/2 - n/lO, and y = 3n/5.

After the rotation s6, the 2 axis is rotated to the direction OB4, so we have a = n/5 and p = 285. The Y axis is rotated to the direction pointing from 0 to the central point 0 5 of the edge A1A5. The angle between two planes B4OAo and B40D5 is 3n/10. So, n - y = n/2 - 3n/10, and

After the rotation ,511, the 2 axis is rotated to the direction of the negative 2 axis, so we have p = 7r and only a - y is independent. We may choose y = 0. The Y axis is rotated to the direction pointing from 0 to the central point D6 of the edge B3A5. The angle between OD6 and the Y axis in the XY plane is 47r/5. So, Q = -47r/5.

S12 is a rotation around the Y axis through n, so its Euler angles are ,8 = 7r and a = y = 0.

8. Express the representation matrix D j ( n , w > of a rotation around the

Solution. Letting S be a rotation with the Euler angles a = cp, /3 = 8 and y = 0, we have SR(e3,w)S-' = R ( ~ , L J ) ,

y = 47r/5.

direction n(8, cp) through LJ in terms of Dj(e3, a ) and d j (p ) .

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Three-dimensional Rotation Group 129

9. Calculate all the matrix entries d;,(w) by Eq. (4.11) where j = 1/2, 1,

Solution. Due to the symmetric relation (4.12) of d j (w) , we only need to calculate the matrix entries d;p(w) with 0 5 v 5 p. The remaining matrix entries can be calculated by the following formulas

3 / 2 , 2, 5/2 and 3.

It is useful to obtain the formulas for dLp(w) with p = j , j - 1, j - 2, and 0 from Eq. (4.16):

For convenience, we use the brief notations c = cos(w/2) and s = sin(w/2). dip(7r - w ) can be calculated from d;,(w) by interchanging c and s.

d:,(w) = c2,

dAl(w) = Jzcs , d3/2

d3/2

dAo(w) = c2 - s2, d3/2

( 1 / 2 ) ( 3 / 2 ) (dl = ( 3 / 2 ) ( 3 / 2 ) ( w ) = c 3 9

( 1 / 2 ) ( 1 / & 4 = c3 - 2cs2, dq2(w) = 2c3s,

dql (w) = c4 - 3c2s2,

&(w) = c4 - 4c2s2 + 5 4 ,

4 2 ( 4 = c4,

d&(w) = fic2s2,

d&(w) = J G C S (c2 - 2) , d5 /2

( 5 / 2 ) ( 5 / 2 ) ( W ) = c 5 7

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130

d5/2

d5 /2

d5/2

d5 /2

(3/2)(5/2) (O) =

(3/2)(3/2) (w ) =

(1/2)(3/2) (w ) =

(1/2)(1/2) (4 =

Problems and Solutions in Group Theory

d5/2 (w) = m c 3 s 2 , (1/2)(5/2)

J5c4 s , c5 - 4c3s2,

d2 (2c4s - 3c2s3) , c5 - 6c3s2 + 3cs4, d;,(w) = c6,

d l 3 ( w ) = &c5s, d i 3 ( w ) = 2&c3s3,

dY2(w) = (c5s - 2c3s3),

&(w) = 2 4 (c5s - 3c3s3 + cs5) , d&(w) = c6 - 9c4s2 + 9c2s4 - s6.

d : 3 ( ~ ) = &c4s2, d l 2 ( w ) = c6 - 5c4s2,

d i2 (w) = J30 (c4s2 - c2s4), d ; l (w ) = c6 - 8c4s2 + 6c2s4,

10. Reduce the subduced representation from the irreducible representa- tion D3 of SO(3) with respect to the subgroup D3 and find the simi- larity transformation matrix.

Solution. In terms of the formula x 3 ( w ) = sin(7w/2)/sin(w/2) for the representation D3 of S0(3), the characters of the elements with the rota- tional angles 0 (the identity E ) , 2 ~ / 3 and T are calculated respectively to be x 3 ( E ) = 7, x 3 ( 2 r / 3 ) = 1 and x3(r) = -1. The subduced representation from D3 with respect to the subgroup D3 is reducible, and the multiplicities uj of the irreducible representations A l , A 2 and E in it can be calculated by Eq. (3.11):

U A ~ = (1/6) (1 x 7 x 1 + 2 x 1 x 1 + 3 x (-1) x 1) = 1,

U A ~ = (1/6) (1 x 7 x 1 + 2 x 1 x 1 + 3 x (-1) x (-1)) = 2,

U E = (1/6) {I x 7 x 2 + 2 x 1 x (-1) + 3 x (-1) x 0} = 2.

Namely,

The dimensions of the representations on two sides of the equation are both 7.

In order to obtain the similarity transformation matrix for reduction, we need the matrix forms of the generators D and A of the subgroup D3

before and after the similarity transformation. The forms after the trans- formation have been given in Problem 11 of Chapter 3. The forms before

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Three-dimensional Rotation Group 131

the transformation depend upon the Euler angles of two generators:

D:,(A) = D : p ( - ~ / 2 , T , ~ / 2 ) = D:,(O, T , T ) = -bv ( -p ) ,

D 3 ( D ) = D3(&3, 2 ~ / 3 ) = diag(1, w2,w, 1, w2, w, l},

w = exp{-i2~/3} = (-1 - i&) /2, w2 = (-1 + i&) / 2 .

X-'D3(A)X == diag(1, -1, -1,l, -1,1, -l},

1 2

X-lD3(D)X = -

/ 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 - 1 - & 0 0 0 0 0 ~ - 1 0 0 0 0 0 0 0 - 1 4

\ o o o o 0 a-1

The row index of X is the same as that of D 3 , which runs over 3,2; - -, ( -3) . The column index of X is the same as that of the reduced representation, denoted by the representations and their rows after reduction, which runs over Al, A2(1), A2(2), [E(1)1], [E(1)2], [E(2)1] and [E(2)2] , where the digits in the bracket are used to distinguish the multiple representations. D3(D) and X-lD3(A)X are diagonal, and D3(A) is a block matrix con- taining three -ol and one digit -1. The first three column-matrices of X [designated by A l , A2(1) and A2(2)] are the eigenvectors of D 3 ( D ) for the eigenvalue 1 and the eigenvectors of D3(A) for the eigenvalues 1, -1 and -1, respectively, so that only the lst, 4th and 7th components (designated by 3, 0, and -3) are not vanishing:

The 4th and 6th column-matrices of X (designated by [E(1)1] and [E(2)1]) are the eigenvectors of D3(A) for the eigenvalue 1. Due to the orthogo- nality, their ls t , 4th and 7th components are vanishing and the remaining components can be chosen as

Substituting them into the similarity transformation formula on D3 ( D ) , we can calculate the remaining colomns in X. In the calculation we are only interested in the rows where the matrix entries are not vanishing. For the 4th column of X (designated by [E(1)1]), we are only interested in the 2nd

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132 Problems and Solutions in Group Theory

and 6th rows (designated by 2 and -2):

For the 6th column of X (designated by [E(2)1]), we are only interested in the 3rd and 5th rows (designated by 1 and -1):

Finally, we obtain the similarity transformation matrix X :

1 x=- Jz

' 1 1 0 0 0 0 0 0 0 0 l i 0 0 0 0 0 0 0 1 4

0 0 0 0 0 - 1 4 0 0 0 -1i 0 0

0 o J z o o o 0

( - 1 1 0 0 0 0 0 1- X . , ~ ( 1 ) 2 and X . , E ( 2 ) 2 are the eigenvectors of D 3 ( A ) for the eigenvalue -1.

11. Reduce the subduced representations from the irreducible representa- tions D20 and D1* of SO(3) with respect to the subgroup I (the proper symmetry group of the icosahedron) , respectively.

Solution. From the character formula for Dj(SO(3))

we calculate the characters for the elements with the rotational angles 0 (the identity E ) , 2n/5, 47r/5, 27r/3, and 7r, respectively

sin( 4i+) sin (n / 5)

sin (82n/5) sin (4 17r/3)

x20(E) = 41, x20(27r/5) = = 1)

= 1, x20(27r/3) = = -1 x20 ( 4 ~ / 5 ) = sin(27r/5) sin( 7r/3)

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Three-dimensional Rotation Group 133

sin (4 1 7r / 2) sin( 7r/2)

X20(7r) = = 1, X18(E) = 37,

sin(377r/5) - sin(27r/5) sin(z/5) sin( 7r/5)

sin( 74n/5) sin(47r/5) sin(27r/5) sin(27r/5)

sin(377r/3) - sin(.rr/3) sin (n / 3)

x18(27r/5) = - - = -2 cOS(T/~) = -p-l,

x'*(4n/5) = - - = 2 C O S ( ~ T / ~ ) = p,

= 1, x18(27r/3) = -

p ( " ) = - sin(3fi77in$j3) sin(.rr/2) sin(7r/2) sin(7r/2) - = 1.

The multiplicity u{ of each irreducible representation I' of I can be calcu- lated with the character formula (3.11) and the character table of I (see Problem 18 in Chapter 3). For the representation D20, we have

a;' = (1/60) (1 x 41 x 1 + 12 x 1 x 1 + 12 x 1 x 1

a$! = (1/60) (1 x 41 x 3 + 12 x 1 x p-' + 12 x 1 x ( - p )

+ 20 x (-1) x 1 + 15 x 1 x 1) = 1,

+ 20 x (-1) x 0 + 15 x 1 x (-l)} = 2, a20 Tz - - (1/60) (1 x 41 x 3 + 12 x 1 x ( - p ) + 12 x 1 x p-l

+ 20 x (-1) x 0 + 15 x 1 x (-1)) = 2,

U? = (1/60) (1 x 41 x 4 + 12 x 1 x (-1) + 12 x 1 x (-1)

+ 20 x (-1) x 1 + 15 x 1 x 0} = 2, u20 H - - (1/60){1 x 41 x 5 + 1 2 x 1 x 0 + 1 2 x 1 x 0

+ 20 x (-1) x (-1) + 15 x 1 x 1) = 4.

For the representation Il l8, we have

= (1/60) (1 x 37 x 1 + 12 x (-p)-' x 1 + 12 x p x 1

+ 2 0 x 1 x 1 + 1 5 ~ 1 x I } = 1,

a;: = (1/60) { 1 x 37 x 3 + 12 x (-p)-' x p-l + 12 x p x ( - p )

+ 20 x 1 x 0 + 15 x 1 x (-1)) = 1, u18 T2 - - (1/60) (1 x 37 x 3 + 12 x (-p)-' x ( - p ) + 12 x p x p-'

+ 20 x 1 x 0 + 15 x 1 x (-1)) = 2,

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134 Problems and Solutions in Group Theory

UE = (1/60) (1 x 37 x 4 + 12 x (-p)-' x (-1) + 12 x p x (-1)

3-20 x 1 x 1 + 1 5 x 1 x 0) = 3 ,

u f = ( 1 / 6 0 ) { 1 ~ 3 7 ~ 5 + 1 2 ~ ( - p ) - ~ ~ 0 + 1 2 ~ p ~ O

+ 2 0 ~ 1 ~ ( - 1 ) + 1 5 ~ 1 ~ 1 } = 3 .

Thus,

The dimensions of the representations on both sides of each equation are the same.

12. Reduce the subduced representations from the irreducible representa- tions D1, D2 and D3 of SO(3) with respect to the subgroup I, respec- tively, and calculate the similarity transformation matrices. From the results, calculate the polar angles of the axes in the icosahedron.

Solution. The representation matrices of the generators of I in each ir- reducible representation have been given in Problem 21 of Chapter 3, and the polar angles for all twofold axes, threefold axes and fivefold axes in I have been listed in Problem 18 of Chapter 3. In the present Problem we are going to show how to obtain these results. We need to calculate the irreducible bases in the group space of I by the projection operator method, just like we have done for the groups T and 0 in Problems 22 and 23 of Chapter 3. However, in the calculation we have to use the multiplication table of I, which is neglected in this book due to limited space. In the following we will only explain the calculation method by using the results for the multiplication of elements. The reader can find the multiplication table in the textbook Group Theory for Physicists, Chap. 3 [Ma]. However, it is suggested to pay more attention to the calculation method introduced here by accepting the multiplication results.

Since the representation matrix of TO we choose is diagonal and T i = E , each diagonal matrix entry of TO must be a power of q = exp{-i2~/5}. From the character of TO in each representation, we have

DA(To) = 1, DT1(T0) = diag{q,l,q-'}, DTZ(To) = d i a g { ~ ~ , l , q -

DG(To) = diag{q2,q,q-1,q-2}, DH(To) = diag{q2,q,l,q-1,q-2}.

This power p can be used to designate the row and column. Note that p is

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Three-dimensional Rotation Group 135

module 5, p + 5 = p.

jection operator PI: We first calculate the vectors @F in the group space of I by the pro-

where c is the normalization factor. @$ satisfies

Letting = E , we obtain @rd, which is vanishing if ,u # v. Then, letting R(2) be an element in I which does not appear in @rd ) say S11, we calculate @jL2?, which is vanishing if p + v # 0. In this way, we calculate @ f ! and

by R(3) = S5 and R(4) = SIO.

Now, we consider the case of p = v = 1, where there are three indepen- dent bases

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136 Problems and Solutions an Group Theory

Following the method used in Problems 22 and 23 of Chapter 3, we in- troduce W , which is the sum of elements in the class C5, to pick up the linear combinations of the vectors such that the new bases belong to the irreducible representations:

5

w = c (Tj +Tj") ) Dr(w) = J l ,

,A = 12) ,T' = 4p-1 ) Q*2 = -4p, ,G = -3, ,H = 0 ,

p = (A - 1)/2 = q + 77-1,

j = O

p-1 = (& + 1)/2 = -112 - q - 2 .

In order to calculate the matrix form of W in the bases a::), we only need to calculate the coefficients of three terms of E , S1 and SS in the action of W ,

w@;) = { p E - p-lS1 + . * .} /&, w@\;) = { -5p-lE 4- 2p-1s1 4- p-2s6 4- . . .} 15,

wait' = (p-2s1 -p2s6 + . . .} 15.

The matrix form of W in the bases is py. -P2

The eigenvectors corresponding to the eigenvalues 4p-', -3 and 0 respec- tively are

*

:(-;-I)> 2 -P A ( -1 : ) ) L ( 2 ) . m p-2

Thus, we obtain the new basis belonging to the representation 7'1) where we are only interested in the terms of E and S1:

- par:)} /2 = E/(2&) - p-lS1/10 + * . .

Sl\II:i = -p-'E/10 + . . . = -p-'9*' 11 /&+ . . . .

Solving it, we have D Z ( S 1 ) = --p-'/&. On the other hand, following the method used in the preceding Problem,

we reduce the subduced representations from the irreducible representations D1, D2 and D3 of SO(3) with respect to the subgroup I, respectively.

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Three-dimensional Rotation Group 137

In fact, the subduced representations from D 1 and D2 with respect to the subgroup I are the irreducible representations, and that from D3 is equivalent to the direct sum of DT2((I) and D G ( I ) . Due to the forms of Dr(To) we have chosen, we are able to define

From the explicit form of d' (P) given in Problem 9, we have

where c = cos 64, s = sin 64 and O4 is an acute angle. Thus, we obtain

c2 = p- l /& , s2 = p/&, cs = c2 - s2 = l / &

tan64 = s/c = p = (& - 1 ) 12, 64 = 31.72".

Then,

-c2 - a c s -s2 - 4 c s c2 - s2 J2cs

-s2 a c s -c2

-p-1 -Jz - p -a 1 Jz = D T 1 ( S l ) ,

D2(Si) c4 2 3 fic2s2 2cs3 s4

2c3s -c4 + 3c2s2 -fics(c2 - s 2 ) -3c2s2 + 5 4 -2cs3

&2s2 -fics(c2 - 2) c4 - 4 2 2 + s4 fics(c2 - 2) fic2s2

2cs3 -3c2s2 + s4 fiCs(.2 - .2) -c4 + 3c2s2 - 2 ~ ~ s s4 -2cs3 fic2s2 -2c3s c4

& -& -1 & &

p - 2 2p-1 fi 2 p p2 2p-1 p2 -JG -p-2 -2p

2 p -p-2 4 p2 -2p-1 p 2 -2p fi -2p-1 p-2

By the unitary similarity transformation X ,

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138

we obtain

Problems and Solutions in Group Theory

X =

( 0 0 c 0 0 0 -d* U O O - b * O O 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 O O d 0 0 0 c*

\ b O O a * O O 0

= ( X - ' ) + ,

where 1ul2 + lbI2 = lcI2 + ld12 = 1. The row index of X runs from 3 to -3, and the column index runs over (T2,1), (T2,2), (T2,3), ( G , l), (G ,2 ) , (G ,3 ) , and (G,4). We calculate the first column and the third column in the fifth row of the formula X- lD3(S1)X = DT2(S1) @ DG(S1):

D;2(S1)a + D;(-3)(Sl)b = 0, D;,(Sl)c+ D?(-a)(Sl)d = 0.

From Problems 7 and 9, we have DE,(S,) = d2,(284) exp{ - - i p r } , and

D 3 S 1 ) = Jloc3s(c2 - 2s2) = J 2 7 2 5 p ,

D;(-3)(s1) = - f i c 2 s 4 = - J m p ,

D;(-2)(s1) = -Jlocs3(2c2 - 2) = -d@%p-l.

~ : 3 ( ~ 1 ) = -J15c4s2 = - d W p - l ,

The solution is a / b = have a = d = &@, b = -c = m, and

and c/d = -m. Due to normalization, we

DT2(S1) = - Jz -1 Jz , Ib rpJzp7 p-1 Jz -p

1 -1 -p -p-1 l \ -p 1 -1 -p-1 DG(S1) = -

-p-l -1 1 - p J - lI 1 -p-1 -p -1

Now, we are going to calculate the directions for the symmetric axes in a regular icosahedron. The azimuthal angles can be determined directly from Fig. 3.1. Here, we only calculate their polar angles. Following the notations used in Problem 18 of Chapter 3, we denote by 81 the polar angle for the fivefold axes except for To, by 82 and 83 the polar angles of two sets of the threefold axes, and by 84 and 8 5 the polar angles of two

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Three-dimensional Rotation Group 139

sets of the twofold axes, respectively. Another set of the twofold axes lies in the X Y plane with the polar angle 7r/2. We have calculate 04 from DT1(S1). The remaining polar angles can be calculated by the common geometry knowledge. Let the length of the edge of the regular icosahedron be 1. Denote by R the radius of the circumcircle, and by r the radius of the incircle. The face of the regular icosahedral is a regular triangle. The distance from the center of the triangle to the vertex is d = m. From Fig. 3.1 we see that sine4 = (2R)-l, t an& = d / r , and R2 = d2 + r 2 . Thus,

= 3 - &i = 2p2, d 4

t an& = - = - r 3 + &

1 /2 (y) = 0.9511,

& = 0.7558, 4d3

e2 = 37.380,

2 tan O4 1 - tan2 e4

d 5 + 1 tane5 = cote4 =p-l = - = 2, 2 tanel =

e2 + e3 = 2e5 = 7r -e l , e3 = -e l - e2 = 79.190 tan O1 + tan 0 2

1 - tan81 tan02 tan03 = - = 3 4- & = 2p-2.

13. Prove that the elements u(n,w) with the same w form a class of the

Solution. It has been proven that the SU(2) group is homomorphic onto the SO (3) group through a two-to-one correspondence:

SU(2) group.

Denote by 8 and cp the polar angle and azimuthal angle of the direction n, respectively. We define

Thus, Sb3 = n b ,

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140 Problems and Solutions an Group Theory

Since u(n, w ) = 1 cos(w/2) - i (5 - n) sin(w/2), we have

v-'u(n, w)v = 1 cos(w/2) - ia3 sin(w/2) = u(e3, w>.

Namely, all elements u(n, w ) with a given w are conjugate to u(eg, u). They belong to the same class. Since Tr u(n,w) = 2cos(w/2), 0 5 w 5 27r, those u(n,w) with different w must not be conjugate to each other. They belong to the different classes. This completes the proof.

4.3 Lie Groups and Lie Theorems

* A group G is called a continuous group if its element R can be described by g independent and continuous parameters r p , 1 5 p 5 9, varying in a g-dimensional region. g is called the order of a continuous group G, and the region is called the group space of G. It is required that the set of parameters r , is one-to-one correspondence onto the group element R at least in the region where the measure is not vanishing. The parameters t , of the product T = RS are the functions t , = fp(rl . . . , r g ; s1 . . . , s g ) = j,(r; s) of the parameters of the factors R and S. fp(r; s) are called the composition functions. The domain of the definition of f p ( r ; s ) is square of the group space, and the domain of function is the group space. The composition functions have to satisfy some conditions such as

where F, are the parameters of the inverse R-l, and e, are the parameters of the identity E. e, are usually taken to be zero for convenience. The continuous group is called the Lie group if the composition functions are analytic, 'or at least piecewise continuously differentiable. A Lie group is said to be a mixed Lie group if its group space falls into several disjoint regions. A mixed Lie group contains an invariant Lie subgroup G whose group space is a connected region in which the identity element of the group lies. The set of elements related to the other connected region is the coset of the invariant Lie subgroup. If the group space of a Lie group G is multiply- connected, there exists another Lie group G' with simply-connected group space such that G' is homomorphic onto G. G' is called the covering group of G. The group space of SO(3) is doubly-connected, and its covering group is SU(2). The Lie group is said to be compact if its group space is compact. If the group space of G is a closed Euclidean region, G is compact.

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Three-dimensional Rotation Group 141

* The group elements are said to be adjacent if their parameters differ only slightly from one another. An element is said to be infinitesimal if it is adjacent with the identity E. The infinitesimal elements describe the local property of the Lie group. The representation matrix of an infinitesimal element A ( a ) is determined by g generators I ,

9

D(A) = 1 - ix a,Ip, I, = i - (4.17) ,=l

I, are linearly independent of one another if the representation D(G) is faithful . * The First Lie Theorem: The representation of a Lie group G with a

connected group space is completely determined by its generators. Namely, the representation matrix of any element R in G can be solved from the following differential equation and the boundary condition:

where S,, ( r ) is independent of the representation. Spy ( r ) depends upon the choice of the group parameters and the composition functions of the Lie group:

* The Second Lie Theorem: The generators of any representation of a Lie group satisfy the common commutation relations

I,Iv - I,I, = i x c,pr, (4.19) e

where C,: are called the structure constants, independent of the represen- tation:

Usually, they are calculated by the commutation relations of the generators in a known representation. The structure constants depend upon the choice of the group parameters.

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142 Problems and Solutions in Group Theory

The Third Lie Theorem: The necessary and sufficient condition for a set of parameters C,; to be the structure constants of a Lie group is the the parameters satisfy

P

A representation D(G) and its generators I , of a Lie group G satisfy

D(R)I,D(R)- ' = c IuD;",jR), (4.21) U

where Dadj(G) is called the adjoint representation of the Lie group. The generators in the adjoint representation is related to the structure constants:

The self-representation of SO(3) is the adjoint representation of both the SU(2) group and the SO(3) group:

(4.23)

where PR is the transformation operators for the scalar functions, and OR = QRPR is the transformation operators for the spinor functions of rank s:

La, Sa, and J a are the orbital, the spinor, and the total angular momentum operators, respectively.

If the spinor function S ~ ( X ) of rank s belongs to the pth row of the irreducible representation Dj(SO(3)),

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Three-dimensional Rotation Group 143

@i(z) is the simultaneous eigenfunctions of J 2 and J3 for the eigenvalues j ( j + 1) and p. Similar formulas hold for the orbital space and the spinor space.

14. Calculate the representation matrices of the generators in an irre-

Solution. This is a fundamental problem both in the matrix mechanics theory of quantum mechanics and in the representation theory of the Lie groups. The angular momentum operators are the generators in the trans- formation operators of SU(2). From the second Lie theorem, the generators in an irreducible representation of SU(2) satisfy the common commutation relations like that of the angular momentum operators:

ducible representation of SU(2) by the second Lie theorem.

J 2 = J i + J3 + J-J+ = J: - 5 3 + J+J-, J* = 51 A i J Z ,

[J3, Jj-] = k J * , [J+, J -] = 2 J 3 , [ J 2 , Ja] = O .

Since J 2 is commutable with all generators Ja, it takes a constant in an irreducible representation due to the Schur theorem. In terms of the Dirac notation, we denote by l j ) the eigenfunction of J3 with the highest eigen- value j in the representation space:

If the dimension of the representation is finite, there must exist a positive integer n such that

(J-)"" l j ) = 0, ( J - ) @ l j ) # 0, o 5 p 5 n.

Thus,

5 3 { J - Id} = ( J - J3 - J - ) l j ) = ( j - 1) { J - l j ) } , 5 3 (J-Y Id = ( j - n) (JY l j ) ,

J 2 ( J - ) n IA = ( j - n)( j - n - 1) (J - )" l j ) = j ( j + 1) (J-)" l j ) .

The positive root for n is 2 j . Therefore, for an irreducible representation with a finite dimension, j must be an integer or a half-odd-integer. Let

where, except for A- j = 0, A , is a nonzero constant to be determined. We are going to show by induction that this 2 j + 1-dimensional space spanned

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144 Problems and Solutions an Group Theory

by Ip) is closed for the angular momentum operators J,, namely,

where B,+1 is a constant to be determined. It is true when p = j because of Bj+l = 0. If it is true when v 5 p 5 j , we want to show that it is also true for p = v - 1 2 - j ,

This completes the proof. The above equation gives a recursive formula:

Choosing the suitable factor in the basis, we are able to make A, = B, = I?; to be a positive real number,

15. For any one-order Lie group with the composition function f ( r ; s ) , please try to find a new parameter r' such that the new composition function is the additive function, f ' ( r ' ; s ' ) = r' + s'. The Lorentz transformation A(v) for the boost along the 2 axis with the relative velocity v is taken in the following form. The set of them forms a one-order Lie group:

-1/2

v1 f v 2

1 0 0 0 y = (1 - v2/c2) , 0

1 + v1v2/c2' A ( v ) = ( O 0 0 0 0 iyv/c y -iyv/c ; ) , f ( W V 2 ) =

Find the new parameter with the additive composition function. Solution. For any given one-order Lie group with the composition function

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Three-dimensional Rotation Group 145

From the first Lie theorem,

where I is the generator of the representation D ( r ) , e is the parameter of the identity, taken to be zero. Solving the differential equation with the boundary condition, we obtain

Define the new parameter

w ( r ) = lr S(t)dt.

The new parameter w(0) of the identity is still zero. Thus,

D ( r ) = exp { - i Iw(r)} , exp { - i I [w(r) + w ( s ) ] } = D(r )D(s ) = D ( r s )

D ( s ) = exp { - i Iw(s)} ,

= exp { - i Iw(rs )} , w(rs) = w ( r ) + w(s) .

For the multiplication of elements, the new parameter is additive.

v along the 2 axis, the composition function for the parameter v is As an example, for the Lorentz transformation with the relative velocity

Defining a new parameter w

w = 1' S(t)dt = arctanh = arctanh ( 5 ) , we have

tanhw = v/c, coshw = (1 - v2/c2)-1'2 = y, sinhw = yv/c,

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146 Problems and Solutions an Group Theory

A(w)=(" 0 0 ) .

0 0 coshw -i sinhw 0 0 isinhw coshw

In the multiplication of two Lorentz transformations, the parameter w is additive.

4.4 Irreducible Tensor Operators

* The set of (2k+l) operators L%(z) , -k 5 p 5 k , is called the irreducible tensor operators of rank k if those operators transform according to the irreducible representation D k ( R ) in the rotation R of SO(3):

For the infinitesimal rotation, we have

(4.26)

(4.27)

Equation (4.26) is equivalent to Eq. (4.27). They are both regarded as the definition of the irreducible tensor operators of rank k . L;(z) are called the irreducible tensor operators of rank k with respect to the orbital space or the spinor space, if replacing OR with PR or QR, respectively.

The irreducible tensor operator of rank one is called the vector operator. Since the representation D1 is equivalent to the self-representation R of SO(3) [see Eq. (4.14)], there are two kinds of components for the vector operators:

ORL;(z)O~l =

O,Va(Z)OR1 =

A=-1 3 (4.28)

b= 1

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Three-dimensional Rotation Group 147

L: = - I / v ( V 1 + iV2),

L; = v 3 ,

L l l = m ( v 1 - iV2),

v1 = ( L i l - L:) , v2 = im (LLl + L;) , v 3 = LA.

L;(z) are called the spherical components of the vector operators, and V a ( 2 ) are called the rectangular components of the vector operators. The examples for the vector operators in common use are the electric dipole operators Z a or Y: (n), the momentum operators p a , the angular momentum operators J a , L a and S a etc.

* Being a vector operator, J a satisfies

where Ja is the generator for the transformation OR, and R b a is the rep- resentation matrix in the adjoint representation of SO(3) [see Eq. (4.23)]. Letting a = 3, we have

OR J 3 ( x ) O R 1 = J n(0, cp) , R = R(cp,0, 7 ) . (4.29)

y is often taken to be zero. If !P$(x) is the function belonging to the pth row of the irreducible representation Dj [see Eq. (4.25)], @$(x) is the simultaneous eigenfunctions of J 2 and 5 3 for the eigenvalues j ( j + 1) and p. Now, from Eq. (4.29), OR!P~(Z) is the eigenfunction for J - n ( O , c p ) with the eigenvalue p.

* For a spherical symmetrical system, the eigenfunction of the energy can be chosen to belong to a certain row of an irreducible unitary representation:

The Wigner-Eckart theorem says that if OR is a unitary operator for the inner product of functions, then

where ( @ f l l ! P j ) , called the reduced matrix entry, is independent of p. Namely, the functions belonging to two inequivalent irreducible represen- tations are orthogonal to each other, the functions belonging to different

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148 Problems and Solutions in Group Theory

rows of an irreducible unitary representation are orthogonal to each other, and the inner product of two functions belonging to the same row of an irreducible unitary representation is independent of the row. The reduced matrix entry depends upon j as well as the explicit forms of @ and Q. The theorem can be generalized to the matrix entries of the irreducible tensor operators in such basis functions. Because

OR {Lt(x)Q;(x)} = {oRLt(x)o,’} { o R Q ~ ( x ) }

= c {G(s)Q!(4) {D$(W@p(R)} ?

X U

L$ (z) !Pi (z) is transformed according to the direct product representation, which can be reduced by the unitary similarity transformation C j k :

j + k

( C j y ( D j ( R ) x o k ( R ) ) c j k = @ D J ( R ) . J = l j - k l

The matrix entries of C j k are called the Clebsch-Gordan (C-G) coefficients. According to the phase convention, we make the C-G coefficients real and satisfying

‘ ; :J( j+u) and ‘ L ; - k ) J ( p - k ) are positive,

CpuJM j k # 0 only when A4 = p + u and Ij - kl 5 J < j + k -

2 J + 1 112 = ( - 1 ) k - J - p (-) - - ( - l ) j - J + ” (-) 1/2 2J+1

The Clebsch-Gordan (C-G) coefficients can be used to combine L: (z) !PL (2) into the functions F& (z) belonging to the irreducible representation:

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Three-dimensional Rotation Group 149

From the Wigner-Eckart theorem, we have

where the constant c is independent of the subscripts p‘ and Ad. c depends on the indices j ‘ , k and j as well as the explicit forms of a, L and Xk. c is still called the reduced matrix entry, denoted by (d I ILk I I Xkj). Thus,

There are (2j’ + 1)(2k + 1)(2j + 1) matrix entries in the form of (*$ (x)lL:(x)I!P$(x)). Due to their property in rotation, they are related by the C-G coefficients. The Wigner-Eckart theorem simplifies the calcula- tion of (2j‘ + 1) (2k + 1) (2j + 1) matrix entries to the calculation of only one parameter (d IILkllXkj). If there is one matrix entry with given subscripts p‘, p and p which can be calculated, the remaining matrix entries all are calculable. In most cases, even one matrix entry is hard to be calculated. For example, the explicit forms of the wave functions a:(’) and @cl(x ) are unknown. However, Eq. (4.32) gives some relations among the matrix entries, which sometimes can be compared with experiments.

16. Directly calculate the C-G coefficients for the direct product repre- sentation of two irreducible representations of SU(2) in terms of the raising and lowering operators J*: a) D112 x D1/2, b) D1/2 x D1, c) D1 x D1, d) D1 x D3j2 .

Solution. In the representation space of Dj x D k , the basis state l j , p ) l k , v) is the eigenfunction of 53 for the eigenvalue p+v. The state with the highest eigenvalue of 53 is l j , j ) l k , k ) . It means that in the reduction of Dj x D k there is an irreducible representation Dj+lC. I j , j ) Ik, k) is also the state with the highest eigenvalue of 53 in the representation space of Dj+lC,

Applying the lowering operator 5- to it, we can calculate all remaining basis states in the representation Dj+lC. In the subspace which is orthogonal to the basis states belonging to Dj+k, we try to find the state with the highest eigenvalues of 53, which belongs to another irreducible representation D in the reduction of Dj x D k . Then, calculate the basis states in DJ by the

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150 Problems and Solutions an Group Theory

lowering operator. In the calculation the following formulas are useful:

J - I I J , M ) = [ ( J + M ) ( J - M + 1 ) p 2 llJ, M - 1),

a) In the direct product representation space, the state with the highest eigenvalue of J3 is ~1 /2 ,1 /2 )~1 /2 ,1 /2 ) , which belongs to the irreducible representation D1.

= p / 2 , -1/2)11/2, -1/2).

The remaining state which is orthogonal to I

110,O) = 2-1/2 { \1 /2 ,1 /2)p /2 , -1/2) -

1,O) is

It belongs to the irreducible representation Do. From the phase conven- tion (4.31) for the C-G coefficients, we choose the coefficient in the term 11/2,1/2)11/2, -1/2) to be positive. Finally, we obtain the Clebsch-Gordan series for the direct product representation

D112 x D112 N D1 @ D o .

b) In the direct product representation space, the state with the highest eigenvalue of J3 is ~ 1 / 2 , 1 / 2 ) ~ 1 , l), which belongs to the irreducible repre-

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Three-dimensional Rotation Group 151

sentation D312

1 1 3 ~ ~ 3/2) = IW, 1/w, I),

I13/2,1/2) = 3-1/25- I13/2,3/2)

= 3-lj2 { JzI1 /2 ,1 /2)p ,O) + 11/2) -1/2)11, l ) } ,

= (12)-ll2 {211/2,1/2)11, -1) + JZ11/2, -1/2)11,0)

= 3-ll2 { 11/2,1/2)11, -1) + v q 1 / 2 , -1/2)11,0)},

= 3-1 {11/2, -1/2)11, -1) + 211/2, -1/2)11, -1)) = p / 2 , -1/2)11, -1).

( (3 /2 , -1/2) = 2-1J-113/2,1/2)

+ Jz11/2, -1/2)11,0)} 7

I13/2, -3/2) = 3-112 J- I13/2, - 1/2)

The basis state which is orthogonal to 113/2,1/2) belongs to the irre- From the phase convention for the C-G coef- ducible representation

ficients, we take the coefficient in the term ~ 1 / 2 , 1 / 2 ) ~ 1 , 0 ) to be positive:

111/2,1/2) = 3-1/2 { 11/2,1/2)p,O) - @/2, -1/2)11

1 /2. -1 /2\ = .7_ll1/2.1/2\ - 1 - I - 1 - 1 - - 1 1 - 1 - I - I - 1

= 3-1/2 { Jz11/2) 1/2)11, -1) + p / 2 , -1/2)

= 3-ll2 {Jz(1 /2) 1/2)11, -1) - 11/2) -1/2)11)0)}. - 2l1/2, -1P)I1,0)},

Finallv. we obtain the Clebsch-Gordan series for the direct product repre-

u , x u - u u/u .

c) In the direct product representation space, the state with the highest - I -

eigenvalue of J3 is I1,l) 11, 1), which belongs to the irreducible representa- tion D2.

sentation

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152 Problems and Solutions in Group Theory

The remaining basis states can be calculated similarly. They can also be calculated by the symmetry (4.31) of the C-G coefficients:

The basis state which is orthogonal to 112,l) belongs to the irreducible representation D1. From the phase convention for the C-G coefficients, we take the coefficient in the term I1,l) I 1 , O ) to be positive:

In the direct product representation space, there are three basis states with the zero eigenvalue of J3. We have obtained two orthogonal basis states 112,O) and 111,O). The third state Il0,O) can be calculated from the orthog- onal condition. But here we prefer to use another method. This method is based on the condition for the state with the highest eigenvalue of 53 in an irreducible representation of SO(3) that it is annihilated by the raising operator J+. Let

110,O) = all, 1)11, -1) + b l l , O ) l l , 0) + cp , -1)11, 1).

J+ annihilates this state:

The solution is a = - b = c. After normalization we obtain

Il0,O) = 3-li2 { I1)1)11, -1) - p , O ) l l , 0) + 11, -1)11,1)}.

Finally, we obtain the Clebsch-Gordan series for the direct product repre- sentation

D1 x D ~ - D ~ c % D ~ $ D ~ .

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Three-dimensional Rotation Group 153

d) In the direct product representation space, the state with the highest eigenvalue of 53 is 11,1)13/2,3/2), which belongs to the irreducible repre- sentation D512.

The basis state which is orthogonal to 115/2,3/2) belongs to the irre- ducible representation D312. According to the phase convention of the C-G coefficients, we take the coefficient in the term I1,l) 13/2,1/2) to be positive.

In the direct product representation space, there are three basis states with eigenvalue 1 /2 of J3. We have obtained two orthogonal basis states 115/2,1/2) and 113/2,1/2). The third state belongs to the representation D112, and can be calculated by the orthogonal condition or by the annihi- lation condition of J+. Now, we use the latter method. Let

J+ annihilates this state:

The solution is u& = -b& = c a . After normalization we obtain

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154 Problems and Solutions in Group Theory

111/2,1/2) = 6-lI2 { 11,1)13/2, -1/2) - fi11,0)13/2,1/2)

+ 611, -1)p/2, 3/2)}.

The expansions for the remaining basis states can be obtained by the sym- metry (4.31) of the C-G coefficients:

Finally, we obtain the Clebsch-Gordan series for the direct product repre- sentation

17. Prove two sets of formulas for the C-G coefficients in terms of the raising and lowering operators J k : a) In the reduction of D1I2 x Dj ,

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Three-dimensional

b) In the reduction of D1 x Dj,

Rotation Group 155

( j + M ) ( j + A4 + 1) + { 2 j ( 2 j + 1)

Solution. We proof them by induction. a) We only prove the first equation. The second equation can be obtained by orthogonality of the basis states and the phase convention. In the direct product representation space, the state with the highest eigenvalue of 53 is I1/2,1/2) I j , j ) , which belongs to the irreducible representation Dj+1/2,

It satisfies the first equation with A4 = j + 1/2. Suppose that the first equation holds for M = m. We are going to show it also holds for M = m - 1.

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156 Problems and Solutions in Group Theory

The first equation is proved. b) In the direct product representation space, the state with the highest eigenvalue of J3 is 11) 1) l j, j ) , which belongs to the irreducible representation Dj+ 1

It satisfies the first equation with M = j+l. Suppose that the first equation holds for M = rn. We are going to show it also holds for M = m - 1.

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Three-dimensional Rotation Group 157

The first equation is proved. We substitute M = j in the second equation:

This equation is correct because J+Ilj, j) = 0. Suppose that the second equation holds for M = m. We are going to show it also holds for M = m - 1.

The second equation is proved. The third equation can be proved with the same method. However, it can also be proved by orthogonality of the basis states and the phase convention.

18. Calculate the eigenfunctions of the total spinor angular momentum in

Solution. We first calculate the eigenfunctions of the spinor angular mo- mentum in a two-electron system. The spin of each electron is 1/2. The spinor state for a two-electron system is the product of the spinor states of two electrons, lp)lu), where p and u are taken to be f 1 / 2 , denoted by f. The eigenfunction of the total spinor angular momentum in a two-electron system is denoted by IS12, M12), where 5'12 takes 0 or 1, IM12l 5 ,912. From

a three-electron system.

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158 Problems and Solutions in Group Theory

the result given in Problem 16 we have

One may regard the three-electron system as a compound system com- posed of a two-electron subsystem and a one-electron subsystem. Its state is the product of the states of two subsystems, IS12, M12)lp). Let IlS, S12, M ) be the state in the three-electron system with the spin S and A4, calculated in terms of the results given in Problem 16:

+ > I - > + 1 + > 1 - ) 1 + ) + -> + m11, -1>1+>

- > I - > + 1 - ) 1 + > 1 - > +

If the two-electron subsystem consists of the last two electrons, we cal- culate the eigenstate IS, S23, M ) in the three-electron system with the spin S and A4 in terms of the results given in Problem 16:

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Three-dimensional Rotation Group 159

Comparing two sets of results, we find that the expressions for the states with total spin 3/2 are the same, but the expressions for the states with total spin 1/2 are mixed together. Actually, two sets of solutions are related by the Racah coefficients.

19. The spherical harmonic function YA(n) belongs to the mth row of the representation De of SO(3) so that it is the eigenfunction of the or- bital angular momentum operator L3 for the eigenvalue m. Calculate the eigenfunction for the eigenvalue m of the orbital angular momen- tum operator L - ii along the direction ii = (el - ea) /fi in terms of combining Y$ (n) linearly.

The polar angle and the azimuthal angle of the direction Letting

Solution. Ci = (el - e2)/2 are 8 = n/2 and cp = -n/4, respectively. S = R( -n/4,7r/2,0), we have

The eigenfunction of L . a for the eigenvalue m is

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160 Problems and Solutions in Group Theory

because

20. Let the function + k ( x ) belong to the mth row of the irreducible repre- sentation D‘ of SO(3). Calculate the eigenfunction for the eigenvalue rn of the orbital angular momentum operator L . b along the direction b = (fie2 + e 3 ) /2 in terms of combining $k(x )* linearly.

Hint: Use the similarity transformation between the representation D j of SO(3) and its complex conjugate representation.

Solution. The representation Dj (R)* is equivalent to the representation D j ( @ ,

Combine the functions $&(x)* into the function + k ( x ) belonging to the irreducible representation De:

Thus, +L(x) is the eigenfunction of the orbital angular momentum L3 along the 2 axis for the eigenvalue m. The polar angle and azimuthal angle of b are n/3 and 7r/2, respectively. By the method used in Problem 19, we obtain the eigenfunction of the orbital angular momentum L - b along the direction b as follows, where S = R(7r/2, n/3,0):

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Three-dimensional Rotation Group 161

Another method of calculation is based on the property that L3t,bem(z)* = m$L(z)*. Thus, Pst,bLm(z)* is the eigenfunction of PsL3Pi1 for the eigenvalue m. Letting S = R ( x / 2 , ~ / 3 , 0 ) , we obtain the eigenfunction of the orbital angular momentum L - b along the direction b for the eigenvalue m:

Two results are the same up to the phase factor. index m' can be replaced with -m'.

21. QR is the rotational transformation operator

Note that the summing

in the spinor space. In the rotation QR, the spinor basis e ( S ) ( p ) belongs to the pth row of the irreducible representation D" so that it is the common eigenfunction of the spinor angular momentum S2 and S3 for the eigenvalues s(s + 1) and p, respectively. Based on this property, calculate the eigenfunction of the spinor angular momentum S - 2 along the radial direction, where i. is the unit vector in the radial direction.

Solution. The operator S . i is a scalar operator in the whole spatial rotation, but a vector operator in the rotation of the spinor space. Letting the polar angle and azimuthal angle of r be 8 and cp, respectively, we have

S * i. 1 Q R S ~ Q R ~ ,

Its eigenfunction for the eigenvalue p is

&Re(')((P) = C e('))(J)&(R)

(s - 6 ) QRe(')(p) = pQRe(S)(p). x

R = qcp, e, 0).

22. There are three sets of the mutual commutable angular momentum operators. One set consists of L 2 , L3, S2, and S3. The other set consists of J 2 , 5 3 , L2 and S2, and the third set consists of J 2 , J3,

S2 and S - i.. Calculate the common eigenfunctions of three sets of operators, respectively.

Solution. The common eigenfunction of L2, L3, S2, and S3 is the prod- uct of the spherical harmonic function Yh(n) and the spinor basis e ( ' ) ( p ) . Combining the products Y:(n)e(')(p) by the C-G coefficients, we obtain the common eigenfunction of J 2 , J3, L2 and S2, which is called the spherical

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162 Problems and Solutions in Group Theory

spinor function yjtS (n):

where the spinor basis e( ' ) (p ) is a spinor of rank s. The above formula is a spinor equation, namely, a (2s + 1) x 1 matrix equation. Since the sum runs over the azimuthal quantum numbers, the spherical spinor function Yies(n) is also the eigenfunction of L2 and S 2 :

J2Yies(n) = j ( j + l)y;ts(fi) , J 3YP jes(ii) = pYits(n) ,

JkYies (n) = I$pY$l (n) , L 2 Y i y n ) = e(e + 1 ) Y j y n ) , s " Y y ( n ) = s(s + 1 ) Y i y n ) .

By making use of the Clebsch-Gordan coefficients calculated in Problem 17, we obtain the explicit forms of spherical spinor function Yics(n), where s = 1/2 and e = j 1/2:

The common eigenfunction of J2, 5 3 and S2 is a spinor of rank s and belongs to the irreducible representation Dj of SO(3) in the whole spatial rotation,

Let the spherical coordinates of the position vector r be (r,e,cp). T = R(cp,B,y) is a rotation transforming the 2 axis to the radial direction r. Tro = r, where ro = (r,O,O) is the position vector r along the 2 axis.

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Three- dimensional Ro t at i o n Group 163

Replacing r with ro and R with 5"-' in the above formula, we have

Write it in the component equation,

Since the left-hand side of the equation is independent of y, its right-hand side has to be independent of y, too. Namely, !I!i(ro)g has to be zero except for the case u = (T,

Substituting it into !Pi (r)p, we have

where TO = R(p, 0,O). Remind that the quantity in the square bracket is nothing but the eigenfunction of S - f for the eigenvalue v [see Problem 211. Qi(r) is the eigenfunction of S - f if and only if the sum contains only one term with a fixed u. Namely, &(r) is nonvanishing only when u takes a given value. Therefore, the common eigenfunction of J 2 , J3, S2 and S - t respectively for the eigenvalues j ( j + 1)) p, s(s + 1) and v is

[ Q T o e ( s ) ( 4 ] qu(TO)*+j,b) = c e ( S ) ( P ) q u ( c p , 6, O P ; , ( c p , &O)*$j,(r), P

where &(r) plays the role of the radial function, or simply the role of the normalization factor.

23. Calculate de(0) I: 2de(S)- ' } , where de(e) is the representation

matrix of R(e2,O) in De of SO(3) and I: is the third generator in the representation.

{ 0 mm

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164 Problems and Solutions in Group Theory

Hint: Use the property of the adjoint representation.

Solution. The self-representation of SO(3) is just its adjoint representa- tion. From the definition (4.21) for the adjoint representation, we have

Since I! is diagonal, and I; = (1: + I!) /2 does not contain any nonvan- ishing diagonal matrix entry, so

sin2 t~ 4 - { ( t + m)(t - m + 1) + (t + m + l)(t - m)} + m2 C O S ~ e - -

sin2 e - - (e2 + t - m2} + m2 cos2 8.

2

24. Establish the differential equation satisfied by the matrix entries @,(a, P,y) of the representation Dj of SO(3).

Solution. D:,(a,P,r) is the matrix entry of the rotational operator OR(^,^,^) between the eigenstates of the angular momentum. Using the Dirac notation, we have:

Thus, OR(u,p,y) can be expressed by the angular momentum operators J,,

Define

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Three-dimensional Rotation Group 165

Ja is a vector operator, satisfying:

Letting s, = s ina , c, = COSQ etc., we have

Hence,

Noticing the order of J+ and J - , we have

Finally, we obtain

Calculating the matrix entry between two eigenstates of the angular mo- mentum, (j, v( and l j , p ) , we obtain the differential equation satisfied by

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166 Problems and Solutions in Group Theory

4.5 Unitary Representations with Infinite Dimensions

* SO(3) group is a compact Lie group. It has irreducible unitary represen- tations Dj of finite dimensions. However, SO(2,l) group is not a compact Lie group. Except for the identical representation, any irreducible represen- tation of SO(2,l) with a finite dimension is not unitary, and the dimension of any irreducible unitary representation of SO(2,l) is infinite. In this sec- tion we will study the irreducible unitary representations of SO(2,l) in some detail as an example of the Lie groups which are not compact.

The set of all three-dimensional real matrices R satisfying

RTJR = J , detR = 1, J = diag(1, 1, o}, (4.33)

in the multiplication rule of matrices, forms the SO(3) group if 0 = 1 and the SO(2,l) group if u = -1. We will denote these two groups by G uniformly. The following inner product is invariant for the group G:

(4.34)

It can be seen from Eq. (4.33) that R& = 1 - o (R23 + R;, ) . IR331 5 1 for S0(3) , but IR331 2 1 for SO(2,l) . Therefore, SO(3) group is a compact Lie group, while SO(2,l) group is not compact. The group space of SO(2,l) falls into two disjoint regions. The region with R33 2 1 corresponds to the invariant subgroup SO+(2,1). For convenience we will still denote this subgroup S0+(2 ,1 ) by the same symbol SO(2,l) if no confusion arises.

Discuss the infinitesimal elements and the generators of G:

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Three-dimensional Rotation Group 167

R = 1 - iaX,

J = R T J R = J - i a ( X T J + J X ) ,

1 = detR = 1 - iaTr X .

Thus,

3

Tr X = 0, X = - J X T J = C uaTU. a= 1

Denoted by La three generators, and by Ta their representation matrices in the self-representation:

0 0 0 0 Oia 0 -i 0 .=( -iO o o o ) , 0 T 3 = ( ; ; ; ) .

They satisfy the following commutation relations

Define the raising and lowering operators

L* = L1 f iL2, [L3 , L*] = fL*, [L+ , L-I = 2aL3. (4.35)

The Casimir operators are

L2 = 0 (L? + L;) + L; = aL+L- + L3 (L3 - 1)

= aL-L+ + L3 (L3 + 1))

a = 1, 2, 3.

(4.36)

[L2 , La] = 0,

* As we have known, the covering group of SO(3) is SU(2). Similarly, we will show that the covering group of SO(2,l) is SL(2,R)) the set of two- dimensional real matrices with determinant +1 where the multiplication rule of elements is the matrix product.

The space composed by two-dimensional traceless real matrices is a real space with respect to three real bases

7 1 = 0 3 , r2 = a l , 73 = i a2. (4.37)

Any two-dimensional traceless real matrix X maps onto the position vec- tor x = ( q , z 2 , q ) in the real three-dimensional space by a one-to-one

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168 Problems and Solutions in Group Theory

correspondence:

1 1 -1 (4.38) 2 2 2

21 = -Tr (XT~), 2 2 = -Tr (XT~), x3 = ---"I- (XT~).

After a similarity transformation uXu-' = X', u €SL(2,R), X' is still a traceless real matrix with the same determinant. Thus, the position vector x' corresponding to X' is related to the position vector x corresponding to X by a transformation R ~ S 0 ( 2 , 1 ) .

Similar to the proof for the covering group of S0(3), we can show the two-to- one correspondence between two matrices fu ~ s L ( 2 , R) and R ~ S 0 ( 2 , 1 ) . This correspondence is invariant in the product of elements. Therefore, SL(2,R) group is the covering group of SO(2,l). The representations for the SO(3) group and the SO(2,l) group including the double-valued rep- resentations are actually the representations of the SU(2) group and the SL(2, R) group, respectively.

25. Discuss all inequivalent and irreducible unitary representations of the

Solution. Choose the basis states IQ, m) in the irreducible representation space where the representation matrices of L2 and L3 are diagonal,

SO(3) group and the SO(2,l) group.

In a unitary representation, the representation matrices of L3 and L2 are both hermitian, so Q and m are both real numbers. Let

Q = j ( j + 1).

There are two solutions for j

(4.40)

Two solutions are related with each other by the following transformation

j++- j - l . (4.41)

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Three-dimensional Rotation Group 169

Recall that the irreducible representation is described by the real parameter Q . Two j 's related by the transformation (4.41) describe the same repre- sentation. j is real when Q 2 -1/4, and j is complex when Q < -1/4.

From Eqs. (4.35) and (4.36), we have

Thus, beginning with a given basis state I&, mo), by the raising and low- ering operators &, we can obtain a series of basis states IQ, m) where the neighboring m are different by f l :

(4.42)

until AQm or B Q ~ equals to zero. Equation (4.42) is the definition for these basis states. They are all common eigenstates of L2 and L3, satisfying Eq. (4.39). We need to show that the set of basis states is closed for Lh, namely

(4.43)

The proofs for two equations are similar. We will prove the first equation in Eq. (4.43) as example. When m > mo, we have

= {aL-L+ + L3 (L3 + 1)) I Q , m - 1)

- OAQm (L-IQ, m ) ) + (m2 - m) IQ, m - 1). -

(4.44)

From Eq. (4.42), AQm is nonvanishing if I & , m ) exists, so L-IQ, m) is pro- portional to I Q , m - 1). Thus, Eq. (4.43) is proved. So, these basis states form a complete set, spanning an irreducible representation space of G. Note that the definition (4.42) for the basis state I&, m ) allows a multiplied factor in front of it. Namely, the values of AQm and BQm are undetermined. On the other hand, the basis state IQ, m ) is also the eigenstate of the oper- ators L+L- and L-L+ for the eigenvalue AQmBQm and AQ(m+llBQ(m+l), respectively. In a unitary representation, the representation matrices for L+ and L- are conjugate to each other, so the representation matrices of L+L- and L-L+ are both positive semi-definite. From Eq. (4.44), when

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170 Problems and Solutions in Group Theory

m > mo, we have

Similarly, it is also true for m 5 mo. We may choose the factor in the basis state IQ,rn) such that

According to whether the state chain is truncated, i.e., whether AQ, (BQ,) is vanishing at some m, the irreducible unitary representations of the SO(3) group and the SO(2,l) group are classified into the unbounded spec- trum D(Q,mo), the spectrum D s ( j ) bounded below, the spectrum D - ( j ) bounded above, and the bounded spectrum D ( j ) [Adams et al. (1987)I.

m

j - m + l = O I j + m + l = O

Fig. 4.1 The varied region for the parameters j and m in the irreducible unitary representation of SO(3) and SO(2,l).

Recall that the values of o are different for two groups: 0 = 1 for the SO(3) group and o = -1 for the SO(2,l) group. From Eq. (4.45), the varied regions of the parameters j and m for two groups are different due to different 0. For S0(3), Q 2 m2 - m 2 -1/4, and j is real. The possible values of j and m are restricted in the intersection of two regions: the region above or below two crossed lines j + m = 0 and j - m + 1 = 0, and

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Three-dimensional Rotation Group 171

the region above or below two crossed lines j - m = 0 and j + m + 1 = 0. The intersection is divided into two infinite regions indicated by SO(3) in Fig. 4.1 [Adams et al. (1987)]. Two regions are equivalent because they are related by the transformation (4.41). In the above region of the intersection, - j 5 m 5 j , and j >_ 0, namely, the values of m are finite. Letting the upper boundary of m be M , from

AQ(M+1) = [ ( j + M + l)(j - jk!)]”2 = 0,

we obtain M = j >_ 0. Here and the remaining part in this Problem we take k to be the non-negative integer. Letting the lower boundary of m be j - k, from

we obtain j = k / 2 = 0, 1/2, 1, 3/2, .... Therefore, the bounded spectra D ( j ) , where j is a non-negative half integer and m = j, j - 1, - * ., - j , are the only unitary representations for the SO(3) group. Obviously, they are the commonly used finite dimensional representations Dj (SO(3)).

For the SO(2,l) group, j may be real or complex. When j is a complex number, j = -1/2 + ip, p > 0, one may have

Q = -1/4 - p2, m = mo f k, - 1/2 < mo 5 1/2. (4.47)

This is called the principal series of representations D,(P,mo) for the un- bounded spectrum.

When j is a real number, the possible values of j and m are restricted in the intersection of two regions: the region in the left or right side of two crossed lines j + m = 0 and j - m + 1 = 0 and the region in the left or right side of two crossed lines j - m = 0 and j + m + 1 = 0. The intersection is divided into two infinite areas and a finite area indicated by SO(2,l) in Fig. 4.1. Considering the equivalent transformation (4.41), we obtain a few kinds of representations of SO(2,l) . First, if AQ, (BQ,) is always nonvanishing for any m in the representation, there is no truncated in the series of m. In this case, -1/4 5 Q < 0, and it is the representation D s ( Q , mo) in the supplementary series for the unbounded spectrum :

-1/2 5 j < 0, m = mo f k, lmol < - j = 1/2 - (& + 1/4)’12. (4.48)

Second, if Q = j = m = 0, this is the identical representation. Third, if there is a lower boundary mo for m in the representation, the lower

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172 Problems and Solutions in Group Theory

boundary lies only on the line of A 0 (or the equivalent line CB). It is the representation D+(j ) for the spectrum bounded below where j < 0, m > 0, and mo = -j:

j < O , m = - j + k .

Finally, if there is an upper boundary mo for rn in the representation, the upper boundary lies only on the line DO (or the equivalent line FB). It is the representation D - ( j ) for the spectrum bounded above, where j < 0, m < 0, and mo = j:

j < O , m = j - k .

Therefore, the dimension of any irreducible unitary representation of SO(2,1), except for the identical representation, is infinite.

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Chapter 5

SYMMETRY OF CRYSTALS

5.1 Symmetric Operations and Space Groups

* The fundamental character of a crystal is the spatial periodic array of the atoms composing the crystal, called the crystal lattice. By the periodic boundary condition, the crystal is invariant in the following translation:

where [is called the vector of the crystal lattice. Three fundamental periods aj of the crystal lattice, which are not coplanar, are taken to be the basis vectors of the crystal lattice, or briefly called the lattice bases. The lattice bases are said to be primitive if any vector of the crystal lattice is an integral combination of the lattice bases. For simplicity, we only use the primitive lattice bases if without special indication.

* The multiplication of two translations is defined to be a translation where two translation vectors are added. The set of all translations T ( 4 which leave the crystal invariant forms the Abelian translation group T of the crystal. Usually, in addition to the translation symmetry, a crystal has some other symmetric operations which leave the crystal invariant. A general symmetric operation may be a combinative operation composed of the spatial inversion, the rotation and the translation. Denoted by g(R, 6) a general symmetric operation, where R is a proper or improper rotation, R E0(3), and d is a translation vector, not necessary an integral combina- tion of the lattice bases aj:

g(R,G)r = Rr + 6. ( 5 4

Moving out the vector of the crystal lattice l f rom G, we express the general

173

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174 Problems and Solutions in Group Theory

symmetric operation as

g(R,a') = T(i+)g(R, t) ,

t = C ajt j ,

a' = e'+ t 3

o 5 tj < 1. (5-3) j = l

When R = E , a'has to be a vector of the crystal lattice zand g(E, 4 = T ( 4 . From the definition (5.2), the multiplication of two symmetric opera-

tions satisfies

Thus, the inverse g(R,d)-l and the identity E are

The set of all symmetric operations g(R, a') for a crystal with the multiplica- tion rule (5.4) forms a group S, called the space group of the crystal. In the multiplication (5.4) of two symmetric operations, the multiplication RR' of two rotations does not matter with whether there are the translations a' and p or not. Therefore, the set of the rotational parts R in g(R, a') forms a group G, called the crystallographic point group. For a given crystal with the space group S, it is easy to show by reduction to absurdity that t in the symmetric operation g(R, t) depends upon R uniquely. Generally, R is not an element of S, and G is not a subgroup of S. The space group S is called the symmorphic space group if G is the subgroup of S. In a symmorphic space group, any element can be expressed as g(R,o = T(C)R.

From Eq. (5.4) we have

g(R, a')T(Gg(R, a ' ) -1 = T(Ri+). (5.6)

Namely, the translation group T is the invariant subgroup of the space group S. Due to Eq. (5.3), the coset of T is completely determined by the rotation R. Thus, the quotient group of T with respect to S is the crystallographic point group, G = T / S . From Eq. (5.6), the action of any element R of G on any vector of the crystal lattice {has to be a vector of the crystal lattice 8

Re'= 2. (5.7)

This is the fundamental constraint for the possible crystallographic point groups, the crystal systems, and the Bravais lattices.

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Symmetry of Crystals 175

* In the crystal theory, the lattice bases aj are usually chosen as the basis vectors in the real three dimensional space. The merit for this choice is that, due to Eq. (5.7), any matrix entry Rij of R in the bases is an integer:

i= 1

The shortcoming is that aj are generally not orthonormal. Define a set of basis vectors bj satisfying

bj is called the basis vector of the reciprocal lattice, or briefly called the reciprocal lattice basis. I t is convenient in the crystal theory to express a rotation R in a double-vector form:

1. In the rectangular coordinate frame, write the double-vector forms and the matrix forms of the proper and improper six-fold rotations around the z axis.

Solution. An improper six-fold rotation s6 around the z axis is equal to the corresponding proper one c6 multiplying by a spatial inversion. Therefore, their double-vector forms and the matrix forms are different only by a sign.

2. Express the directions of the proper and improper rotational axes in the groups T d and Oh by the basis vectors ej of the rectangular coordinate frame. Express the double-vector forms of the generators of the proper rotational axes in O h by ej.

Solution. The group T d is a subgroup of the group O h . T d contains four proper three-fold axes, three improper four-fold axes, and six improper two-fold axes. For Oh, all those axes are both the proper and improper ones.

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176 Problems and Solutions in Group Theory

The directions and the double-vector forms of the generators of the proper three-fold axes are

-#

Direction :

Direction :

Direction :

m{e1 + e2 + e3},

m{e1 - e2 - e3},

J17'5 {-el - e 2 + e3},

21 = e2e1 + e3e2 + ele3,

2 2 = -281 + e3e2 - eie3,

R3 = e2e1 - e3e2 - e1e3,

-#

-D 4

-# -#

Direction : J17'5 {-el + e 2 - e3}, Rq = -e2e1 - e3e2 + 6183.

The directions and the double-vector forms of the generators of the proper four-fold axes are

-+ -+

Direction : el, 2'1 = elel + e3e2 - e2e3,

Direction : e2,

Direction : e3, T3 = e2e1 - e1e2 + e3e3.

-#

T2 = -e3e1 + e2e2 + e1e3, +

The directions and the double-vector forms of the generators of the proper two-fold axes are

-#

Direction : m(e1 + e 2 } ,

Direction : m{e1 - e2},

31 = e2e1 + 8162 - e3e3,

3 2 = -e2e1 - e1e2 - e3e3, -+

-#

Direction : J172 { e 2 - e3},

Direction : m{e1 + e3},

Direction : @ {el - e3) ,

34 = -elel - e3e2 - ~ 8 3 ,

$5 = e3e1 - e 2 e 2 + e1e3,

96 = -e3e1 - e2e2 - e1e3.

-4

-#

3. Write the double-vector form of a rotation R(n, w ) around the direction

Solution. Let m be a unit vector orthogonal to n. Then, n x m is a unit vector orthogonal to both n and m . Applying R ( n , w ) to those three unit

n through the angle w by the unit vector i and the unit vector n.

Direction :

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Symmetry of Crystals

vectors, we have

177

-t

In terms of the project operators nn and 1'- nn onto n and the plane orthogonal to n, respectively, we have

+

where the following formulas are used:

x (n x m) = -m.

5.2 Symmetric Elements

* A point is called the symmetric center of a symmetric operation g(R, t) if the point is invariant in g(R, t) . A straight line is called the symmetric straight line of a symmetric operation g(R, t) if the straight line is invariant in g(R, t) . A plane is called the symmetric plane of a symmetric operation g(R, t) if the plane is invariant in g(R, t).

* A symmetric operation g(R, t) is called closed if its power may be equal to the identity, g(R, t)" = E , otherwise, it is called an open operation. There are only two kinds of open operations. One is g ( C N , t), N # 1, where CN is a proper N-fold rotation, and the nonvanishing component tll o f t along the direction n of the rotational axis is a multiple of the smallest vector of the crystal lattice in the direction n divided by N. This axis is called the screw axis, which is parallel to n through the point ro, where ro satisfies

( B - CN)ro = t - tll = tl.

tll # 0 is the gliding vector. A screw axis is a symmetric straight line of ~ ( C N , t) . The other is g(S2, t) , where S2 is an improper twofold rotation (reflection), and the component t l of t along the reflection plane is half of the smallest vector of the crystal lattice in the direction. This reflection

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178 Problems and Solutions an Group Theory

plane is called the gliding plane, which is orthogonal to the rotational axis n of the improper rotation S2 through the point ro, satisfying

( E - S2) ro = t - t l = tll.

t l is the gliding vector on the plane. A gliding plane is a symmetric plane of g(S2, t) . The closed transformation has a symmetric center ro, satisfying

( E - R)ro = t .

When R is a proper rotation CN ( N # l), t has to be orthogonal to the rotational axis. The straight line which is through the point ro and parallel to n is a symmetric line. Each point on the symmetric line is the symmetric center for the closed operation. When R is the reflection 5’2, t has to be parallel to the rotational axis, i.e. orthogonal to the reflection plane. The plane which is through the point ro and orthogonal to the direction n of the rotational axis is a symmetric plane. Each point on the symmetric plane is the symmetric center for the closed operation. When R is an improper rotation S N , N # 2, the closed transformation only has a symmetric center.

* A symbol for a space group of a crystal in the international notations contains all information on the symmetry of the crystal. In the international notations, the first character indicates the Bravais lattice, f denotes the two order inversion group Ci, and the digit N or N denotes the proper or improper axis. Except for the rhombohedra1 system, the axis denoted by a digit without prime indicates the principal axis, along which the lattice basis a3 directs, the axis denoted by the digit 2 with a prime indicates a twofold axis orthogonal to the principal axis, along which the lattice basis a1 directs, and the axis denoted by a digit N > 2 with a prime or denoted by the digit 2 with double primes indicates the rotational axis along other direction.

In analysis of the symmetry of a crystal from its international symbol, one has to know the rule for choosing the lattice bases aj in the given crystal system. Whether aj are primitive or not depends upon the Bravais lattice. Denote by aj the length of aj, and by a3 the angle between a1 and a2, and so on. Let al, a2 and a3 construct the right-handed system. The rule for choosing the lattice bases aj is as follows. (1). Triclinic system contains P1 and Pi.

on the lengths and the angles of the lattice bases. (2). Monoclinic system contains P2, PZ, P f 2, A2, A?, and A f 2.

The lattice bases aj are taken to be primitive. There is no restriction

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Symmetry of Crystals 179

Define the shortest vector of the crystal lattice along the twofold axis to be a3, and take two non-collinear smallest vectors of crystal lattice in the plane orthogonal to a3 to be a1 and a2, respectively. Two non-collinear vectors of the crystal lattice are called the smallest in a plane if any vector of the crystal lattice in the plane is their integral combination. The restriction for the lattice bases is a1 = a2 = n / 2 . ( 3 ) . Orthorhombic system contains P22', P2?, P f 22', C22 / , C22', A22', C f 22', I22', I22' , I f 22') F22', F25', F f 22'.

Define the shortest vectors of the crystal lattice along three twofold axes to be a1 , a2 and a3, respectively. For CaV (23')) a 3 is along the proper twofold axis. The restriction for the lattice bases is a1 = a 2 = a 3 = ~ / 2 . (4 ) . Tetragonal system contains P 4 , Pa, P f 4 , P42', P42', P32', P32', P f 42', 14, 14, I f 4 , I42', I42', I32', I32', and I f 42'.

Define the shortest vector of the crystal lattice along the fourfold axis to be a 3 , and take one shortest vectors of the crystal lattice in the plane orthogonal to a 3 to be al. Define a 2 = C4al. The restriction for the lattice bases is a1 = u2 and a1 = a 2 = a 3 = n / 2 . For the point group D2d, there are two different symbols 32' and 42", according to whether a1 directs along proper or improper twofold axis. (5). Cubic system contains P3'22', P3'22', P3'42', P3'32', P3'42', 13/22', I3'22', I3'42', I3'$2', I3'42', F3'22') F3'22', F3'42', F3'32', and F3'42'.

Define three shortest vectors of the crystal lattice along three orthogonal twofold axes [for T (3'22') and T h (3'22')l or fourfold axes [for 0 (3'42'), T d (3'32') and O h (3'42')] to be al, a2 and a3, respectively. The restriction for the lattice bases is a1 = a2 = a3 and a1 = a2 = a 3 = n / 2 . (6 ) . Hexagonal system contains P3, P3, P32', P32', P32', P32', P32', P32", F 6 , P6, P f 6 , P62', P62', P62', P62' and P f 62'.

Define the shortest vector of the crystal lattice along a threefold axis or a sixfold axis to be a3, and take one shortest vectors of the crystal lattice in the plane orthogonal to a 3 to be al. Define a2 = C3a1. The restriction for the lattice bases is a1 = u2, a1 = a2 = n / 2 and a 3 = 2n /3 . For D3, C3v

and Dsd, there are two kinds of symbols 32' or 32', 32' or 32') and 32' or 32', where 2' (or 2') means that a1 directs along a twofold axis, and 2' (or 5') means that a1 directs along the angular bisector of two twofold axes. For D6 (62 ' ) ) c 6 v (62') and D6h ( f 6 2 ' ) , a1 directs along a twofold axis. For D 3 h , there are two kinds of symbols 62' and 62", according to whether al directs along a proper or improper twofold axis. (7). Rhombohedra1 system contains R 3 , RS, R32', R32' and R32'.

-

Three lattice bases are distributed symmetrically around the threefold

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180 Problems and Solutions in Group Theory

axis with the same acute angle. The sum of three lattice bases is equal to the shortest vector of the crystal lattice along the threefold axis. The restriction for the lattice bases is a1 = a2 = a3 and a1 = a2 = a 3 . For D3

(32‘), CsV (32’) and D 3 d (32’), the map of any lattice basis onto the plane orthogonal to the threefold axis directs along the angular bisection of two twofold axes.

For different Bravais lattices, in addition to the integral combinations t?, the vectors of the crystal lattice are allowed to be some fractional com- binations f:

P type of Bravais lattice : f = 0,

R type of Bravais lattice : f = 0,

A type of Bravais lattice : f = (a2 + as) /2,

B type of Bravais lattice :

C type of Bravais lattice :

I type of Bravais lattice :

F type of Bravais lattice :

f = (a1 + as) /2,

f = (a1 + a2) /2,

f = (a1 + a2 + as) /2,

f = ( a 2 + as) /2,

and (a1 + a2) /2.

(a3 + al) /2,

(5.10)

4. Let R be a rotation around the direction n = (el +e2) / A through 2r/3. Find the symmetric straight line for g(R, t) where (a) t = e3, (b) t = el + e 3 , If the symmetric straight line is a screw line, find its gliding vector. Simplify g(R, t) by moving the origin to the symmetric straight line for checking your result.

Solution. Write the double-vector form of R = R(n,2~/3) with n = (el + e 2 ) / \ / 2 in the rectangular coordinate frame:

(a) t = t l = e3.

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Symmetry of Crystals 181

g(R, t ) has a symmetric straight line through the point ro and parallel to n, where ro satisfies

g(R, t)ro = Rro + t = ro.

Write the equation in the rectangular coordinate frame:

Taking 7-02 = 0, we have 7-01 = and ~ 0 3 = 1/2. Each point on the symmetric straight line is the symmetric center, Moving the origin to the point ro, we transform g(R, t) to g'

g' = T(-ro)g(R, t)T(ro) = g(R, -ro + t + Rro) = R.

(b) t = el + e3

g ( R , t) has a screw straight line which is through the point ro and parallel to n with the gliding vector tll, where ro satisfies

g(R, t l ) ro = Rro + t l = ro.

Write the equation in the rectangular coordinate frame:

Taking 732 = 0, we have r01 = 1/2+ the origin to the point ro, we transform g(R, t ) to g'

and 7-03 = (1- m ) / 2 . Moving

5 . Let R be a reflection with respect to the zy plane. Find the symmetric plane of g ( R , t ) where (a) t = e3, (b) t = el +e3. If the symmetric plane is a gliding plane, find its gliding vector. Simplify g(R, t) by moving the origin to the symmetric plane for checking your result.

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182 Problems and Solutions an Group Theory

Solution. R = aR(e3, n), where a is the spatial inversion. The double- vector form of R in the rectangular coordinate frame is

(a) t = tll = e3.

the z axis, where ro satisfies g(R, t) has a symmetric plane through the point ro and orthogonal to

g(R, t)ro = Rro + t = ro.

Letting ro be on the z axis, we have ro = e3/2. Each point on the sym- metric plane is the symmetric center. Moving the origin to the point ro, we transform g(R, t ) to g’

g’ = T(-ro)g(R, t)T(ro) = g(R, t + ( R - E)ro) = R.

(b) t = el + e3.

tll = e3 and t l = el. g(R, t) has a gliding plane which is through the point ro and orthogonal to the z axis with the gliding vector t l = el, where ro satisfies

Letting ro be on the z axis, we have ro = e3/2. Moving the origin to the point ro, we transform g(R, t) to g‘

6. Analyze the symmetry property of the crystal with the following space

and No. 199 [T5, 1 3 ’ 2 ( i O ; ) 2 ’ ( i i O ) ] . Point (a) The general form of the symmetric operation; (b) The relations of directions and lengths among three lattice bases; (c) The symmetric straight line, symmetric plane and the gliding vector of the generator in each cyclic subgroup of the space group, if they exist; (d) The equivalent point to an arbitrary point r = alxl + a222 + a323 in the crystal cell.

group: No. 52 [D”,, P f 2( f $0)2‘( f f f)], No. 161 [civ, R32 -’ ( 5 111 2)] ,

Solution. Let us analyze the property of the space groups, respectively. (1) The space group D;h = P f 21/2,1/2,02~~2, , /2 , , /2 belongs to P type of Bravais lattice of the tetragonal system. Its crystallographic point group is D2h, which contains three orthogonal twofold axes, both proper and

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Symmetry of Crystals 183

improper. Three primitive lattice bases are along three twofold axes, re- spectively. a1 = a2 = a3 = 7r/2 but the lengths of the lattice bases may not be equal to each other. The double-vector forms of the generators are

The arbitrary element in the space group is expressed as

where S1 denotes the spatial inversion, and n1, n2 and 7x3 are taken to be 0 or 1, respectively. The origin is the symmetric center of the spatial inversion.

g(C2,q) with q = q l = (a1 + a2)/2 has a symmetric straight line through the point ro and parallel to a3. Each point on the symmetric straight line is the symmetric center. Let the components of ro along the lattice bases be r01, 7 3 2 , and ~ 0 3 . Take ~ 0 3 = 0. Thus,

We have ro1 = ro2 = 1/4. From another viewpoint, since the spatial inversion S1 is a symmetric operation, there is also an improper twofold axis around the direction a3. Slg(C2, q) has a gliding plane, which is through the origin and orthogonal to a3 with the gliding vector q.

9Wi, P) with P = PI1 + P1,

has a screw axis which is through the point rb and parallel to a1 with the gliding vector PI[ . The component of rb along the direction a1 can be taken to be zero, and the components along a2 and a 3 are denoted by rb2 and rb3, satisfying:

Thus, r& = rb3 = 1/4. Similarly, there is also an improper twofold axis along the direction al. SIg(Ci, p) has the gliding plane which is through the point p11/2 = a1/4 and orthogonal to a1 with the gliding vector PI.

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184 Problems and Solutions an Group Theory

From an arbitrary point r = alxl + a 2 2 2 + a 3 2 3 in the crystal cell, we can obtain eight equivalent points by the symmetric operations. Due to the spatial inversion, the equivalent points are paired by different signs:

(2) The space group Cgw = R3?i/2,1/2,1/2 belongs to the R type of Bravais lattice in the Rhombohedra1 system. The crystallographic point group CsW contains a proper threefold axis, which is the principal axis, and three improper twofold axes, distributed symmetrically on the plane orthogonal to the principal axis. Recall that those improper twofold rotations are reflections with respect to the planes containing the principal axis. Three lattice bases have the same length and distributed around the principal axis symmetrically :

The double-vector forms for the generators are

The general element in the space group is

where m is taken to be 0, 1 or 2, n is 0 or 1. The proper threefold axis is the principal axis. The origin is taken on the principal axis. Each point on the principal axis is the symmetric center for the threefold rotation.

g(Sa, (a1 + a2 + a3)/2) with p = p l = (a1 + a2 + a3)/2 has a gliding plane which is through the origin and orthogonal to the direction a1 - a2 with the gliding vector PI.

From an arbitrary point r = alxl + a 2 2 2 + a 3 2 3 in the crystal cell, we

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Symmetry of Crystals 185

obtain six equivalent points by the symmetric operations:

(3) The space group T5 = 1 3 ' 2 1 ~ ~ , 0 , 1 ~ 2 2 ~ ~ 2 , 1 ~ 2 , 0 belongs to the I type of Bravais lattice in the cubic system. The crystallographic point group T contains three orthogonal proper twofold axes and four proper threefold axes distributed symmetrically. Three lattice bases with the same length are along three twofold axes, respectively. The lattice bases are not primitive. The following fractional combination is allowed to be a vector of the crystal lattice:

f = (a1 + a 2 + as) /2.

where 121,122 and 123 are taken to be 0 or 1, M is 0, 1 or 2. The origin is on the threefold axis. Ci is a pure rotation. g(C2, q) with q = 911 + q i

911 = a3/2, ql = a l p ,

has a screw axis which is through the point ro and parallel to the direction a3 with the gliding vector 911. The components of ro along the lattice bases are denoted by ~ 0 1 , 7-02 and ~ 0 3 . We may choose r03 = 0. 7-01 and r02

sat is fy :

The double-vector forms of generators are

The general element of the space group is

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186 Problems and Solutions in Group Theory

Thus, POI = 1/4 and r02 = TO3 = 0. g(Ci,p) with p = p11 + p i ,

has a screw axis which is through the point rb and parallel to the direction a1 with the gliding vector p11. The components of rb along the lattice bases are denoted by r&, 7-b2 and rb3. We may choose rbl = 0. 7-b2 and rb3 satisfy:

O = (Ci - E ) rb + p i = {-2rb2 + 1/2} a2 - 2rb3a3.

Thus, rb2 = 1/4 and rbl = rb3 = 0. From an arbitrary point r = a121 + a222 + a 3 2 3 in the crystal cell we

obtain 24 equivalent points by the symmetric operations, where half points can be calculated by a translation f . The remaining 12 points are:

5.3 International Notations for Space Groups

* Attaching as the subscripts the suitable translation vectors to the in- ternational symbol for the symmorphic space group, we obtain the interna- tional symbol for the general space group. The general element of a space

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Symmetry of Crystals

group can be expressed as

187

(5.11)

where n, n1 and n2 all are integers, L = t+ f is a vector of crystal lattice, f depends upon the Bravais lattice, and the translation vectors t , p and q are the fractional combinations of the lattice bases aj:

The translation vectors t , p and q should satisfy the group property, i.e., the multiplication of two elements of the space group has to be expressed in the same form (5.11). First, due to the property of a cyclic subgroup, the translation vector t in g(R, t) has to satisfy

t = O when R = C1 = E ,

{n (n . t)} = tll = mall/N when R = C N , N # 1, (5.12) \ I

{t - ii (n - t)} = t l = ma1/2

t no restriction when R = S N , N # 2.

when R = S2,

Second, there are constraints resulted from the multiplication of R, R1 and R2. For example, due to R1 = RRIR, we have

g(R1, p + L) = g(R, t)g(R1, p)g(R, t) = S(RRlR, t + RP + RRlt)

( R - E)p = L - ( E + RR1)t. (5.13)

At last, we have to avoid the equivalent symbols related to the different choice of the origin. Let the vector from the origin 0 to the new origin 0' be ro. If the symmetric operation is denoted by g(R, t) for the origin 0, it is changed to g' in the new origin 0'

g' = T(-ro)g(R, t)T(ro) = g(R, t + ( R - E)ro). (5.14)

The different choice of the origin does not change the space group, but changes its symbol. Those two symbols are called the equivalent symbols. We have to remove the equivalent symbols for the same space group.

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188 Problems and Solutions in Group Theory

* Usually, we choose the origin such that the translation vector t in Eq. (5.11) becomes as simple as possible. In other words, we choose the origin such that the part of t , which is not restricted in Eq. (5.12), vanishing. t can be more simplified if the remaining t is a component of L. Then, we may choose the origin again to simplify the translation vectors p and q under the condition that t keeps invariant. This is the general method to determine the symbol of a space group.

Since the choice of the origin first depends upon the simplification of t in g(R, t) , The space groups are classified as type A ( R = S N , N # a), type B ( R = C N ) and type C ( R = ,572). There are only two space groups in type C: PZ;,, and AZgoo.

7. Calculate the 12 space groups related to the crystallographic point group

Solution. The point group D2d contains an improper fourfold axis (the principal axis), two proper twofold axes and two improper twofold axes. Four twofold axes are distributed symmetrically on the plane orthogonal to the principal axis. The lattice basis a3 is along the principal axis. According to whether the lattice basis a1 is along a proper or an improper twofold axis, the international symbol of D2d is denoted by 32’ or 32”, namely, denoted by the product of two subgroups S4C4 or S4Cg. The space groups belong to the tetragonal system, where there are two Bravais lattices P and I . The restriction for the lattice bases are a1 = a 2 = a 3 , and a1 = u2. The double-vector form of the fourfold rotation is

D 2 d -

-#

g4 = -a2bl + alb2 - a3b3.

The space groups with the crystallographic point group D 2 d belong to type A. Choosing the origin on the principal axis, we express the general element in the space group as

where m is taken 0, 1, 2 or 3, n is 0 or 1. CL in the formula may be changed to C;, depending on the discussed space groups. We may translate the origin by ro, ro = alrO1 + a27-02 + a37-03, satisfying

Thus, ro may be taken the following values, in addition to L (a1 + a 2 ) / 2 , a3/2 Bravais lattice P or I (2al + as) /4, (2a2 + as) /4, Bravais lattice I . r o = {

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Symmetry of Crystals 189

In the following we study two kinds of space groups with the crystallo-

(1) For the crystallographic point group 32', a1 is along the direction of graphic point groups 32' and ;?2", respectively.

a proper twofold axis. The double-vector form of the twofold rotation is

Since S4C;S4 = Ci, we have

For the P type of Bravais lattice, we have solutions pl = p2 = 0 or 1/2, and p3 = 0 or 1/2. For the I type of Bravais lattice, because f = (a1 + a2 + a s ) /2 is a vector of crystal lattice, the solution p = f is ruled out, and two solutions p = (a1 + a 2 ) /2 and p = a3/2 become equiv- alent to each other. In addition, there are new solutions p = (2al + as) /4 and p = (2a2 + as) /4. In all those solutions, the parallel component p11 is equal to multiple of half vector of crystal lattice along the rotational axis.

Moving the origin by ro, p may be changed by:

For the P type of Bravais lattice, p is not simplified. But for the I type of Bravais lattice, the solution p = a3/2 is ruled out, and two solutions p = (2al + as) /4 and p = (2a2 + a 3 ) /4 become equivalent. Therefore, we have the following space groups:

2 2 2 2 4

which are listed in the table of space groups with Nos. 111, 112, 113, 114, 121 and 122.

(2) For the crystallographic point group 32", a1 is along the direction of an improper twofold axis. The double-vector form of the twofold rotation is

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190 Problems and Solutions in Group Theory

Since S4C;S4 = C;, we have

Although this equation is the same as the former case (l), the solution may be different from the former case because of the different lattice bases. For the P type of Bravais lattice, we have the same solutions as the former case: pl = p2 = 0 or 1/2, and p3 = 0 or 1/2. For the I type of Bravais lattice, we only have solutions p = a3/2 or 0. The solutions p = (2al + a s ) /4 and p = (2a2 + a s ) /4 are ruled out because their parallel component pi1 = (a1 - a2) /4 is less than half of the vector of crystal lattice along the rotational axis.

Moving the origin by ro, p may be changed by:

Both for the P type and the I type of Bravais lattice, p is not simplified. Therefore, we have the following space groups:

P42", P42bl,$ , p42; f r o , pq2'i L L , 142", 142i0$, 2 2 2

which are listed in the table of space groups with Nos. 115, 116, 117, 118, 119 and 120.

8. Please calculate the 9 space groups related to the crystallographic point

Solution. The international symbol of the crystallographic point group 0 2 is 22'. It contains three orthogonal proper twofold axes and can be expressed as the product of two subgroups: C2Ci. The space groups with the crystallographic point group 0 2 belong to the orthorhombic system with P , A, F and I types of Bravais lattices. Three lattice bases are orthogonal to each other, but have different lengths. The double-vector forms for twofold rotations are

group D2.

-. -+ -+

6 2 = -albl - a2b2 + a3b3, Ch = albl - a2ba - a3b3.

The space groups with the crystallographic point group D2 belong to type B. Choosing the origin on one of the twofold axes, we express the general element in the space group as

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Symmetry of Crystals 191

where both rn and n are taken to be 0 or 1. t = ta3 with t = 0 or 1/2. For the A type, F type and I type of Bravais lattices, L contains the component a3/2 so that t can be removed by a translation ro of the origin

Under the condition that the simplified t keeps invariant, the origin can be made a further translation ro

ro = { ~ 1 ' ~ ) a2/2, r03a3 Bravais lattices P, A, F and I a1 + a21 /4, Bravais lattice F,

where ~ 0 3 is arbitrary. Since C2C~C2 = Ci and C2Cit = -t, we have

Letting p = alp1 + a2p2 + a3p3, we have:

For P , A , F and I types of Bravais lattices, we obtain that pl and p2 are respectively taken to be 0 or 1/2, but p3 is arbitrary. There is an additional solution for the F type of Bravais lattice: p = (a1 + a2) /4. However, the latter is ruled out by the constraint that p11 has to be a multiple of half vector of crystal lattice along the rotational axis.

In translating the origin by ro, p is made the following transformation:

Thus, p3 is removed. Now, we study the space groups for different Bravais lattices.

For the P type of Bravais lattice, t , pl and pz are allowed to be 0 or 1/2 independently. However, some space groups may be equivalent to each other. Obviously, two space groups with t = pl = p2 = 0 and 1/2 are inequivalent. They are denoted by P22' and P200+2/,,,, listed in the table of space groups with Nos. 16 and 19. If one of t , pl and p2 is 1/2 and the remaining are vanishing, we obtain three equivalent space groups because three twofold axes are distributed symmetrically. In other words, three solutions t = 1/2 (pl = p2 = 0), pl = 1/2 ( t = p2 = 0) and p2 = 1/2 ( t = pl = 0) are equivalent because one can be obtained from the other by interchanging the lattice bases. Those three equivalent space groups are

2 2

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192 Problems and Solutions an Group Theory

denoted by P22b4,, listed in the table of space groups with No. 17. If two of t , pl and p2 are 1 /2 and the third one is vanishing, we also obtain three equivalent space groups. In other words, the solution with t = pa = 1/2 can be obtained from the solution t = pl = 1/2 by changing a1 to the direction of a2. For the solution with pl = p2 = 1/2 and t = 0, two twofold rotations are expressed as

After moving the origin by ro = a2/4, they become

Then, interchanging a1 and a3, the solution becomes that with t = p2 = 1/2 and pl = 0. Those three equivalent space groups are denoted by P22',,,, listed in the table of space groups with No. 18.

For the A type of Bravais lattice, t = 0. Since f = (a2 +as ) / 2 is a vector of crystal lattice, the solution with p2 = 1/2 can be removed. Thus, there are only two space groups for the A type of Bravais lattice A22' and A22iO0, listed in the table of space groups with Nos. 21 and 20.

For the F type of Bravais lattice, t = 0. Since f = (a1 +a3) /2 and (a2 + as) / 2 are the vectors of crystal lattice, the solutions with pl = 1/2 and (or) p2 = 1/2 can be removed. Thus, there is only one space group for the F type of Bravais lattice, F22', listed in the table of space groups with No. 22.

For the I type of Bravais lattice, t = 0. Since f = (a1 + a 2 + a s ) /2 is a vector of crystal lattice, the solution with pl = p2 = 1/2 can be removed. The solution with pl = 0 and p2 = 1/2 can be obtained from that with pl = 1/2 and p2 = 0 by changing a1 to the direction of a2.

Thus, there are only two space groups 122' and 122; for the 1 type of Bravais lattice, listed in the table of the space groups with Nos. 23 and 24. Usually, the space group with No. 24 in the table of the space groups are denoted by 1200t26t0, which is equivalent to 122iO0. One will convince himself by subtracting f from p, removing p3 through moving the origin, and interchanging a1 and a 3 .

In summary, there are nine inequivalent space groups related to the crystallographic point group D2, listed in the table of space groups with

2 2

200

NOS. 16-24.

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Chapter 6

PERMUTATION GROUPS

6.1 Multiplication of Permutations

by a 2 x n matrix

R =

In the matrix for R the order

* A rearrangement of n objects is called a permutation. Denote a permu- tation R which transforms the object in the position j into the position rj

( r t r : : : : : ) .

of columns does not matter, but the cor- responding relation between two digits in each column is essential. The multiplication rule for two permutations is defined as successive applica- tions of the two transformations.

There are n! permutations for the system of n objects. The set of those n! permutations forms a group, called the permutation group S n of n objects, where the identity and the inverse of R are

If a permutation S keeps (n - l ) objects invariant and changes the remaining l? objects in order, S is called a cycle with length l , described by a one-row matrix:

193

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194 Problems and Solutions in Group Theory

In the one-row matrix for a cycle, the order of digits is essential, while the transformation of digits in sequence is permitted. The order of a cycle S with length l is l , namely, l is the smallest power with S' = E. A cycle with length 2 is called a transposition.

* Any permutation can be decomposed into a product of cycles which contain no common object. The product of two cycles without any common object is commutable. Up to the order, the decomposition of a permutation into a product of cycles without any common object is unique. The set of lengths of those cycles is called the cycle structure of the permutation.

A cycle is equal to a product of two cycles by cutting it in any digit and repeating the digit:

( a - a * b c d f) = (a . . . b c) (c d . - . f).

Conversely, two cycles which contain one common object may be connected to be one cycle through Eq. (6.1). The product of two cycles which con- tain more common objects can be simplified as a product of several cycles without any common object by using Eq. (6.1) repeatedly. For example,

(a1 . . . a2 c a2+1 . . . aj d)(d bl . . . b, c b,+l . . . b,) = (a1 . . . ai c) (c ai+1 . . . aj d) (d bl . . . b, c)(c b,+l . . . b,)

= (a1 . . . ai c) (c b,+l . . . b,)(Ui+l . . . aj d)(d bl . . . b,) = (a1 . . . ai c)(ai+l . . . aj d)(d C ) ( C d)(d bl . . . b,)(c b,+l ... b s )

= (a1 . . a ai c b,+l ... bs)(ai+l . . . aj d bl . . . b,). (6.2)

A cycle with length l may be decomposed into a product of ( l - 1) transpo- sition by Eq. (6.1). Any permutation may be decomposed into the product of transpositions by different ways, where the number of transpositions in the product is not unique, but it is determined whether the number is even or odd. A permutation is said to be even (or odd) if its decomposition contains even (or odd) number of transpositions. The permutation parity S(R) of an even permutation is 1, and that of an odd permutation is -1. The permutation parity of a cycle with length l is ( l - 1).

* When a permutation S is moved from the left-side of a permutation R to its right-side, the digits (objects) in R are made the permutation S. Conversely, when a permutation S is moved from the right-side of R to its left-side, the digits in R are made the inverse permutation S-l. In other words, if the permutation R is a cycle (or the product of cycles) containing some digits rj, and the permutation S transforms rj into t j , then, T =

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SRS-l is the same cycle as R except that the containing digits rj are changed to t j , respectively. This property in the product of permutations is called the interchange rule. From the rule, two permutations with the same cycle structure are conjugate to each other. The class in a permutation group is described by the cycle structure of the permutations in the class.

* Denote by Pa = (a a + 1) the transposition of two neighbored objects a and (a + 1). It is easy to know from the interchange rule in the product of permutations that Pa satisfy

1. Simplify the following permutations into the product of cycles without any common object:

(1) : (1 2)(2 3)(1 2), (2) : (1 2 3)(1 3 4)(3 2 l),

(3) : (1 2 3 4)-1, (4) : (1 2 4 5)(4 3 2 6),

(5) : (1 2 3)(4 2 6)(3 4 5 6).

Solution: The first two problems can be solved by the interchange rule, and the last three problems can be solved by Eq. (6.1).

(1) : (1 31, (2) : (2 1 4 ) , (3) : (4 3 2 l),

(4) : (5 1 2 6)(4 3), (5) : (1 2 6)(4 5).

2. Decompose an arbitrary transposition ( b d) as a product of the trans-

Solution. Without loss of generality, we suppose b > d and obtain positions P a of the neighbored objects.

3. Prove that the order of a cycle R with length l is !, namely, Re = E.

Solution. Let R = (a1 a2 . . - a [ ) and b be an object different from any am, 1 5 m 5 l . Then, R transforms am into am+l, and into a l . Thus, Re--" transforms am into at , R transforms into a l , and Rm-l transforms a1 into am- Altogether, Re transforms an arbitrary am into itself, while keeping b invariant, namely, Re is the identity E.

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196 Problems and Solutions in Group Theory

4. Prove that PI = (1 2) and W = (1 2 3 . . . n) are two generators of the

Solution. From Problem 2, any permutation can be decomposed as a product of the transpositions P, of the neighbored objects. From the in- terchange rule, WP, = P,+1W. From Problem 3 we have W-' = Wn-l,

permutation group S,.

so P, = WPu_1W"-1 = Wa-lPlW,-a+l.

5 . There are 52 pieces of playing cards in a set of poker. The order of cards is changed in each shuffle, namely, the cards are made a permutation. If the shuffle is "strictly" done in the following rule: first separate the cards into two parts in equal number, then pick up one card from each part in order. After the strict shuffle the first and the last cards do not change their positions, while the remaining cards are rearranged. Try to find this permutation, to decompose it into a product of cycles without any common object, to write the cycle structure of the permutation, and to explain at least how many times of the strict shuffles will make the order of cards into its original one.

Solution. After the strict shuffle, the cards in the first part are arranged into the odd positions, while the cards in the second part are arranged into the even positions. In the mathematical language, after a shuffle, the nth card is changed to the (2n - 1)th position, while the (26 + n)th card is changed to the (2n)th position, where 1 5 n 5 26. By this rule, we want to decompose the permutation corresponding to the strict shuffle into the product of cycles without any common object. First, this permutation contains two cycles (1) and (52) with length 1. Second, according to the rule, the shuffle transforms 2 into 3, 3 into 5, 5 into 9, 9 into 17, 17 into 33, 33 into 14, 14 into 27, and 27 into 2, so we obtain a cycle with length 8

(2 3 5 9 17 33 14 27).

We arbitrarily choose a card in the remaining cards, say the fourth card. The shuffle transforms 4 into 7, 7 into 13, 13 into 25, 25 into 49, 49 into 46, 46 into 40, 40 into 28, and 28 into 4, so we obtain another cycle with length 8

(4 7 13 25 49 46 40 28).

Similarly, in the remaining cards, we choose the sixth card. The shuffle transforms 6 into 11, 11 into 21, 21 into 41, 41 into 30, 30 into 8, 8 into 15,

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Permutation Groups 197

15 into 29, and 29 into 6, so we obtain a cycle with length 8

(6 11 21 41 30 8 15 29).

In the remaining cards, we choose the tenth card. The shuffle transforms 10 into 19, 19 into 37, 37 into 22, 22 into 43, 43 into 34, 34 into 16, 16 into 31, and 31 into 10, so we obtain a cycle with length 8

(10 19 37 22 43 34 16 31).

In the remaining cards, we choose the twelfth card, the shuffle transforms 12 into 23, 23 into 45, 45 into 38, 38 into 24, 24 into 47, 47 into 42, 42 into 32, and 32 into 12, so we obtain a cycle with length 8

(12 23 45 38 24 47 42 32).

In the remaining cards, we choose the 18th card, the shuffle transforms 18 into 35, and 35 into 18. This time we obtain a cycle (18 35) with length 2. In the remaining cards, we choose the 20th card, the shuffle transforms 20 into 39, 39 into 26, 26 into 51, 51 into 50, 50 into 48, 48 into 44, 44 into 36, and 36 into 20, so we obtained a cycle with length 8

(20 39 26 51 50 48 44 36).

Now, 52 cards are exhausted, and the permutation corresponding to the shuffle is decomposed into the product of two cycles with length 1, one cycle with length 2 and six cycles with length 8. All those cycles do not contain any common digit. The cycle structure of this permutation is (12, 2, @). The least common multiple of the cycle lengths is 8, namely, after 8 times of the strict shuffles, the order of playing cards turns into its original one. In fact, the shuffle is not so strict. Since many accidental cases occur, there is no worry about.

6.2 Young Patterns, Young Tableaux and Young Operators

* A class in the permutation group S, is described by the cycle structure (e) = (el , &, . . .) of the permutations in the class. Since Cj l!j = n, ( l ) is a partition number of n. For a finite group, the number of the inequivalent irreducible representations is equal to the number of the classes. Therefore, the irreducible representation of S, can also be described by a partition

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198 Problems and Solutions in Group Theory

number of n, denoted by [A] = [ A 1 , A 2 , - . - ] , A1 2 A2 2 . - - , C.A. = n. Usually, when some k‘j or A j are duplicated, one may denote them by an exponent, for example, (23, 32) = (2 ,2 ,2 ,3 ,3) .

* From a partition number [A] of n, we define a Young pattern which consists of n boxes lined up on the top and on the left, where the j t h row contains A j boxes. The Young pattern is also denoted by [A]. Remind that in a Young pattern, the number of boxes in the upper row is not less than that in the lower row, and the number of boxes in the left column is not less than that in the right column.

A Young pattern [A] is said to be larger than a Young pattern [A’] if X i = A:, 1 5 i < j, and X i > A>. There is no analytic formula for the number of different Young patterns with n boxes. However, one can list all different Young patterns with n boxes from the largest to the smallest.

3 3

* Filling n digits 1, 2, . . ., n arbitrarily into a given Young pattern [A] with n boxes, we obtain a Young tableau. A Young tableau is said to be standard if the digit in each column of the tableau increases downwards and the digit in each row increases from left to right.

For two standard Young tableaux with the same Young pattern, if the digits filled in the first (i - 1) rows and in the first ( j - 1) columns of the i th row are the same as each other, respectively, but the digits filled in the box at the j t h column of the i th row are different, the Young tableau with the larger digit in this box is said to be larger than the other. This increasing order for the standard Young tableaux is the so-called dictionary order. We can enumerate the standard Young tableaux from the smallest to the largest by an integer p. The number of the standard Young tableaux with a given Young pattern [A] may be calculated by the hook rule,

where hij is the hook number of the box in the j t h column of the i th row of the given Young pattern, which is equal to the number of boxes at its right in the i th row, plus the number of boxes below it in the j t h column, and plus one. Two Young patterns related by a transpose are called the associated Young patterns. The corresponding standard Young tableaux for two associated Young patterns are related by a transpose such that the larger Young tableau for one Young pattern becomes the smaller for the associated Young pattern, but the numbers of the standard Young tableaux for two associated Young patterns are the same.

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* A permutation of the digits in the same row of a given Young tableau is called the horizontal permutation P of the Young tableau, and a permu- tation of the digits in the same column is called its vertical permutation Q. Generally, a horizontal permutation P may be a product of the horizontal permutations Pj, where Pj is a permutation of digits in the j t h row. Sim- ilarly, Q may be a product of Q k , where Q k is a permutation of digits in the Icth column. The sum of all horizontal permutations of a given Young tableau is called its horizontal operator P = C P = nj C P’, and the sum of all vertical permutations multiplied by their permutation parities is called its vertical operator Q = CS(Q)Q = nk C d ( Q k ) Q k . The Young operator Y corresponding to the Young tableau is equal to the product of the horizontal operator and the vertical operator, Y = PQ. A Young operator is said to be standard if its Young tableau is standard. Since for a given Young operator Y , its Young tableau and its Young pattern have been determined, we often say the Young pattern Y and the Young tableau Y for convenience. P and Q are said to be the horizontal permutation and the vertical permutation of the Young operator Y (or the Young tableau Y ) , respectively, and PQ is said to be a permutation belonging to the Young operator Y (or the Young tableau Y ) . Except for the identity, no permuta- tion can be both the horizontal permutation and the vertical permutation belonging to the same Young tableau.

* The Young operator Y can be expressed as a combination of its permu- tations, Y = CG(Q)PQ, and satisfies

[,, c x (a , h,)] Y = 0, Y [ E - (c, 4 = 07 (6.5)

,=1 p=l

where the last two formulas are called the Fock conditions. The sum in the first Fock condition runs over all X objects a, in a given row, while b, is any object in another row with less boxes, A’ 5 A. Similarly, the sum in the second Fock condition runs over all r objects c, in a given column, while d, is any object in another column with less boxes, r‘ 5 r.

If there exist two digits filled in the same row of a Young tableau y and in the same column of another Young tableau Y’, then Y’Y = 0. Two Young operators Y and Y‘ with different Young patterns are orthogonal to each other, YY’ = Y‘Y = 0. For a given Young pattern, if a standard Young tableau Y, is larger than another standard Young tableau y,, then

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200 Problems and Solutions in Group Theory

YpY, = 0.

* If a permutation R ES, transforms one Young tableau y into another Young tableau y’ with the same Young pattern [A] for the permutation group S,, the corresponding Young operators satisfy y‘ = RyR-l. R can be calculated in the following way: Put the digits filled in the Young tableau y to the first row of the two-row matrix for R, and the digits in the Young tableau y’ to the second row in the same order, namely, in each column of the two-row matrix for R, the upper digit and the lower digit are filled in the same box of two Young tableaux, respectively. The necessary and sufficient condition for y’y # 0 is that R belongs to the Young tableau y , namely, R is a product of the horizontal permutation P and the vertical permutation Q of the Young tableau Y . Note that P‘ = RPR-l and Q’ = RQR-’ are respectively the horizontal one and the vertical one of the Young tableau y‘, and R = PQ = P‘Q’ = Q‘P. The square of a Young operator y corresponding to the Young pattern [A] for S, is

where d [ x ~ (S,) is the number of the standard Young tableaux for the Young pattern, given in Eq. (6.4).

6. Write all Young patterns of the permutation groups s 5 , s6 and s 7 from

Solution. For the permutation group S g , its Young patterns are [5], [4,1],

For the permutation group s6, its Young patterns are [6], [5,1], [4,2],

For the permutation group S7, its Young patterns are [7], [6,1], [5,2],

the largest to the smallest, respectively.

[3,21, [3, i23, [z2, 11, P, i31, p51.

[4,121, ~ 2 1 , [3,2,11, [3,131, 1231, ~ 2 , i21, 12,141, ~161.

[5, i21, [4,31, [4,2,11, [4, i31, [32, 11, 13,221, [3,2,121, [3,141, [23,11, [22,131, P, 151, ~ 7 3 .

7. Calculate the number n(C) of elements in each class (C) of the permuta- tion groups S4, S5, s6 and S7, respectively.

Solution. The number n(C) of elements in the class (C) for S4: n(4) = 6, n(3,l) = 8, n(22) = 3, n(2, 12) = 6, n(14) = 1.

The number n(C) of elements in the class (C) for S5: n(5) = 24, n ( 4 , l ) =

The number n(C) of elements in the class (C) for SS: n(6) = 120, n ( 5 , l ) = 144, n(4,2) = 90, n(4, 12) = 90, 71(3~) = 40, n(3,2,1) = 120, n(3, 13) = 40,

30,n(3,2) = 2 0 , ~ ( 3 , 1 2 ) = 2 0 , 4 2 2 , q = 1 5 , ~ ( 2 , 1 3 ) = i 0 , ~ ( 1 5 ) = 1.

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Permutation Groups 20 1

4 2 3 ) = 15,~(22 ,12) = 45,42 ,14) = 15, n(i6) = 1. The number n(l) of elements in the class ( l ) for S7: n(7) = 720,

n(6, l ) = 840, n(5,2) = 504, n(5, 12) = 504, n(4,3) = 420, n(4,2,1) = 630, n(4, 13) = 210, n(32, 1) = 280, ~ i ( 3 , 2 ~ ) = 210, n(3,2, 12) = 420, n(3, 14) = 7 0 , 4 2 3 , i ) = 105,n(22,13) = 1 0 5 , ~ ( 2 , 1 5 ) = 21,417) = 1.

8. Calculate the number dlA](S,) of the standard Young tableaux for each

Solution. We calculate the number d[A] (S,) of the standard Young tableaux for the Young pattern [A] by the hook rule (6.4). Remind that the associate Young patterns [A] and [i] have the same number of the standard Young tableaux.

Young pattern [A] of the permutation groups S,, 5 5 n 5 9.

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202 Problems and Solutions in Group Theory

9. Write the Young operators corresponding to the following Young tableaux:

(1) : (2) : Fl (3) : p-l Solution.

(1) : [E + (1 2) + (1 3) + (2 3) + (1 2 3) + (3 2 l ) ] [E - (1 4)] = E + (1 2) + (1 3) + (2 3) + (1 2 3) + (3 2 1) -(1 4) - (2 1 4 ) - (3 1 4 ) - (2 3)(1 4) - (2 3 1 4 ) - (3 2 1 4 ) .

(2) [E + (1 211 [E + (3 4)1 [E - (1 3)1 [E - (2 4)i = E + (1 2) + (3 4) + (1 2)(3 4) - (1 3) - (2 1 3) - (4 3 1) - (2 1 4 3) - (2 4) - (1 2 4) - (3 4 2) - (1 2 3 4) + (1 3)(2 4) + (1 3 2 4) + (3 1 4 2) + (3 2)(1 4).

(3) : E + (1 2) + (1 3) + (1 4) + (2 3) + (2 4) + (3 4) + (1 2)(3 4) +(1 3)(2 4) + (1 4)(2 3) + (1 2 3) + (1 3 2) + (1 2 4) + (1 4 2) +(1 3 4) + (1 4 3) + (2 3 4) + (2 4 3) + (1 2 3 4) + (1 2 4 3) +(1 3 4 2) + (1 3 2 4) + (1 4 2 3) + (1 4 3 2) - (1 5) - (2 1 5)

-(3 2 1 5) - (2 4 1 5) - (4 2 1 5) - (3 4 1 5) - (4 3 1 5)

-(3 4 2 1 5) - (3 2 4 1 5) - (4 2 3 1 5) - (4 3 2 1 5).

-(3 1 5) - (4 1 5) - (2 3)(1 5) - (2 4)(1 5) - (3 4)(1 5) -(3 4)(2 1 5) - (2 4)(3 1 5) - (2 3)(4 1 5) - (2 3 1 5)

-(2 3 4)(1 5) - (2 4 3)(1 5) - (2 3 4 1 5) - (2 4 3 1 5)

10. Write five standard Young tableaux Y p corresponding to the Young pattern [3,2] of S5 from the smallest to the largest, and calculate the permutations R,, transforming the Young tableau y , into Young tableau &.

Solution. The standard Young tableaux Y, corresponding to the Young pattern [3,2] of S5 are as follows:

Yl Y2 Y3 Y4 Y5

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11. Calculate the permutation R12 transforming the following Young tableau & to Y1, and check the formulas 7'1R12 = R127'2, QlR12 = R12Q2 and YlR12 = R12Y2, where

1 2 4 3 ( 1 2 3 4 ) = ( 3 4 ) ' Solution. R12 =

7'1R12 = [E + (1 2) + (1 3) + (2 3) + (1 2 3) + (3 2 l)] (3 4)

= (3 4) + (1 2)(3 4) + (1 3 4) + (2 3 4) + (1 2 3 4) + (2 1 3 4),

R127'2 = (3 4) [E + (1 2) + (1 4) + (2 4) + (1 2 4) + (4 2 l ) ] = (3 4) + (1 2)(3 4) + (3 4 1) + (3 4 2) + (3 4 1 2) + (3 4 2 l),

QlRl2 = [E - (1 4)] (3 4) = (3 4) - (1 4 3),

R12Q2 = (3 4) [E - (1 3)] = (3 4) - (4 3 1).

Thus,

12. It can be seen that there is no pair of digits filled in the same row of the Young tableau y and in the same column of the Young tableau y'. Calculate the permutation R transforming the Young tableau y into the Young tableau Y', and express R as PQ belonging to the Young tableau y , and as P'Q' belonging to the Young tableau Y':

The Young tableau Y The Young tableau y'

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204 Problems and Solutions in Group Theory

Solution.

R = ( 1 2 4 7 3 5 9 6 8 ) 1 2 3 4 5 6 7 8 9 = ( 4 3 5 6 8 9 7 )

= (7 4)(4 3 5 6 8 9) = (7 4)(3 5)(5 6 8 9 4) = (7 4)(3 5)(5 6 8 9)(9 4) = (7 4)(3 5)(9 5)(5 6 8)(9 4)

= (7 4)(3 5 9)(6 8)(8 5)(9 4) = PQ, P = (7 4)(3 5 9)(6 8), Q = (8 5)(9 4),

R = (4 3 5 6 8 9 7) = (4 3)(5 6 8 9 7 3) = (4 3)(5 6)(8 9 7 3 6) = (4 3)(5 6)(8 9)(6 9 7 3) = (4 3)(8 9)(5 6)(7 6 9)(7 3) = (4 3)(8 9)(5 6 7)(6 9)(7 3) = P'Q',

P' = (4 3)(5 6 7)(8 9) = RPR-l, Q' = (9 6)(7 3) = RQR-l.

13. Try to expand all non-standard Young tableaux y of the Young pattern [2,1] in terms of the standard Young tableaux Y,, Y = C, t,y,, where t , is the vector in the group space of S3.

Solution. There are six Young tableaux corresponding to the Young pat- tern [2,13, where two Young tableaux & and & are standard, and the remaining are non-standard.

The symmetry and the Fock conditions of a Young operator lead to

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Permutation Groups 205

6.3 Primitive Idempotents in the Group Algebra

* A vector e in the group algebra L of a finite group G is called an idempotent if e2 = e E C . n vectors e, in C are called a set of the mutually orthogonal idempotents if e,ev = S,,e,. Le, = L, and e,L = R, are respectively called the left ideal and the right ideal of ,C generated by the idempotent e,. Two left (or right) ideals generated by two orthogonal idempotents contain no common vector.

For a complete set of the arbitrarily chosen bases IC, in a left ideal L, = Le,, the matrices D ( S ) of the group elements S in the bases, Sz, = C, I cTDT~(S) , form a representation of the group G, which is called the representation corresponding to the left ideal L,. Two left ideals, or two idempotents generating those two left ideals, are said to be equivalent if the representations corresponding to two left ideals are equivalent. A left ideal C, is said to be minimum and its idempotent is said to be primitive if the representation corresponding to the left ideal is irreducible. The necessary and sufficient condition for an idempotent e, to be primitive is that e,te, = Ate, holds for any vectors t in the group algebra L, where At is a constant depending on t , which is allowed to be zero. The necessary and sufficient condition for two primitive idempotents e, and e, to be equivalent is that there exists at least one group element R such that e,Re, # 0. These conclusions are also suitable for the right ideals, except for the calculation of the representation matrix of S in a right ideal R, by z,S = c, D,,(S)z,.

* The Young operator is proportional to the primitive idempotent of the permutation group. Two Young operators corresponding to different Young patterns are orthogonal to each other. The representations generated by them are inequivalent and irreducible. Thus, an irreducible representation for a permutation group is described by a Young pattern.

When n 5 4, those standard Young operators corresponding to the same Young pattern for the permutation group S, are orthogonal to one another. This conclusion is not generally true. When n 2 5 , for a given Young pattern [A], we define 4 x 1 (S,) mutually orthogonal primitive idem- potents in the following way. For simplicity, the index [A] and S, are omit- ted. Denote by R,, the permutation transforming the standard Young tableau y, into the standard Young tableau y,. If y,yv # 0, then R,, = P;’)Q?) = PL”)Qf), where P;’) and Q?) are respectively the horizontal and the vertical permutations belonging to the Young tableau y,, while Pi”’ and Qt) are respectively the horizontal and the vertical

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206 Problems and Solutions in Group Theory

permutations belonging to the Young tableau yp. Let

d primitive idempotents ep are mutually orthogonal. In another way, let

p= 1

d primitive idempotents e; are mutually orthogonal. The identity E of the permutation group may respectively expand in terms of two sets of the primitive idempotents

Note that n ! / d l x ] is nothing but the product of the hook numbers hij for all boxes in the Young pattern [A].

14. Expand explicitly the identity of the Sq group in terms of the Young operators.

A B C Solution. E = - + - + - where

24 12 8 ’

= 2 { E + (1 2)(3 4) + (1 3)(2 4) + (1 4)(2 3) + (1 2 3) + (1 3 2) + (1 2 4) + (1 4 2) + (1 3 4) + (I 4 3) + (2 3 4) + (2 4 3)) ,

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Permutation Groups 207

= E + (1 2) + (3 4) + (1 2)(3 4) - (1 3) - (2 1 3)

- (4 3 1) - (2 14 3) - (2 4) - (1 2 4) - (3 4 2) - (1 2 3 4)

+ (1 3)(2 4) + (1 3 2 4) + (3 1 4 2) + (1 4)(3 2) + E + (1 3) + (2 4) + (1 3)(2 4) - (1 2) - (3 1 2) - (4 2 1) - (3 1 4 2) - (3 4) - (1 3 4) - (2 4 3) - (1 3 2 4)

+ (1 2)(3 4) + (1 2 3 4) + (2 14 3) + (1 4)(2 3) = 2 {E + (1 2)(3 4) + (1 3)(2 4) + (1 4)(3 2)) - (1 2 3) - (1 3 2)

- (1 2 4) - (1 4 2) - (1 3 4) - (1 4 3) - (2 3 4) - (2 4 3),

+ y ~ + y F 4

= 2 {E + (1 2 3) + (1 3 2) - (2 3)(1 4)) - (2 1 4) - (4 1 2) - (3 1 4) - (4 1 3) - (2 3 1 4) + (4 1 2 3) - (3 2 1 4) + (4 1 3 2)

- (4 1 3) - (3 1 4) - (2 4 1 3) + (3 1 2 4) - (4 2 1 3) + (3 1 4 2)

- (3 1 2) - (2 1 3) - (4 3 1 2) + (2 1 4 3) - (3 4 1 2) + (2 1 3 4)

+ 2 { E + (1 2 4) + (1 4 2) - (2 4)(1 3)) - (2 1 3) - (3 1 2)

+ 2 {E + (1 4 3) + (1 3 4) - (4 3)(1 2)) - (4 1 2) - (2 1 4)

= 6E - 2(2 3)(1 4) - 2(2 4)(1 3) - 2(4 3)(1 2).

15. Calculate the orthogonal primitive idempotents for the Young pattern

Solution. There are five standard Young tableaux yp for the Young pat- tern [2,2,1] of S5:

[2,2, I] of the permutation group S5.

Yl Y2 Y3 Y4 Y5 pq Fl Fl After checking, the product of each pair of the standard Young operators is vanishing except for &y5 # 0. The permutation transforming the Young tableau ys into the Young tableau Yl is R15 = (4 2 3 5) = (2 5) (5 4)(2 3). horn Eq. (6.7) we obtain y1 = E - (2 5), and the remaining ycL are E.

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208 Problems and Solutions in Group Theory

Thus, the orthogonal primitive idempotents are

From Eq. (6.8) we may decompose R15 as R15 = (3 4) (4 2)(3 5), and obtain another set of the orthogonal primitive idempotents

16. Calculate the orthogonal primitive idempotents for the Young pattern

Solution. There are nine standard Young tableaux yp for the Young pattern [4,2] of s6, listed from the smallest to the largest:

[4,2] of the permutation group s6.

1 2 3 4 1 2 3 5 1 2 3 6 1 2 4 5 1 2 4 6 5 6 4 6 4 5 3 6 3 5

1 2 5 6 1 3 4 5 1 3 4 6 1 3 5 6 3 4 2 6 2 5 2 4

After checking, only the products of three pairs of the Young operators are non-vanishing:

From Eq. (6.7) we obtain the orthogonal primitive idempotents as

1 1 80 80 1 1

80

el = -Y1 [E - (2 5)(3 6 4)]

e3 = -Y3 [E - (2 4)(3 5)] ,

e2 = - 9 2 [E - (2 4)(3 6 511

e p = 4 9 5 9 .

From Eq. (6.8) we may decompose the above R,, as R18 = (4 3 2)(5 6) (6 2), R29 = (5 3 2)(4 6) (6 2), R39 = (3 2)(4 5) (5 2), and obtain another set of orthogonal primitive idempotents

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17. Calculate the orthogonal primitive idempotents for the Young pattern

Solution. There are 16 standard Young tableaux Y, for the Young pattern [3,2,1] of s6, listed from the smallest to the largest:

[3,2,1] of the permutation group SS.

1 2 3 1 2 3 1 2 4 1 2 4 1 2 5 1 2 5 1 2 6 1 2 6 4 5 4 6 3 5 3 6 3 4 3 6 3 4 3 5 6 5 6 5 6 4 5 4

1 3 4 1 3 4 1 3 5 1 3 5 1 3 6 1 3 6 1 4 5 1 4 6 2 5 2 6 2 4 2 6 2 4 2 5 2 6 2 5 6 5 6 4 5 4 3 3

After checking, there are 8 pairs of the Young operators whose products are non-vanishing:

R1,ll = (2 4 5 3) = (2 4)(5 3) (3 4) = (4 5)(3 2) (2 5), R1,12 = (2 4 6 5 3) = (5 3)(6 2) (2 4)(6 3) = (3 2)(5 4) (4 6)(5 2), R2,13 = (2 4 6 3) = (2 4)(6 3) (3 4) = (4 6)(3 2) (2 6), R2,14 = (2 4 5 6 3) = (6 3)(5 2) (2 4)(5 3) = (3 2)(6 4) (4 5)(6 2),

= (2 3 6 5 4) = (5 4)(6 2) (2 3)(6 4) = (4 2)(5 3) (3 6)(5 2), R 4 , 1 6 = (2 3 5 6 4) = (6 4)(5 2) (2 3)(5 4) = (4 2)(6 3) (3 5)(6 2), R5,15 = (2 3 6 4) = (6 2) (2 3)(6 4) = (4 3) (3 6)(4 2), R7,16 = (2 3 5 4) = (5 2) (2 3)(5 4) = (4 3) (3 5)(4 2).

Thus, we obtain the orthogonal primitive idempotents from Eq. (6.7):

1 45

el = -71 [E - (2 4)(3 5) - (2 6)(3 5)] ,

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210 Problems and Solutions in Group Theory

Another set of the orthogonal primitive idempotents is obtain from Eq. (6.8)

18. Calculate the orthogonal primitive idempotents for the Young patterns

Solution. There are five standard Young tableaux Y, for the Young pat- tern [3,3] of s 6 :

[3,3] and [4,1,1] of the permutation group s6, respectively.

Yl Y2 Y3 Y4 y5 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 4 5 6 3 5 6 3 4 6 2 5 6 2 4 6

The digit 6 can only be filled in the last box of the second row, so the standard Young tableaux for the Young pattern [3,3] of s6 are the same as those for the Young pattern [3,2] of S 5 , respectively, except for adding the last box filled with 6. The calculation for the orthogonal primitive idempotents is also the same.

After checking, the product of each pair of the standard Young operators is vanishing except for y1y5 # 0. The permutation transforming the Young tableau y5 into the Young tableau Y1 is R15 = (2 4 5 3) = (2 4)(5 3) (3 4) = (4 5)(3 2) (2 5). From Eqs. (6.7) and (6.8) we obtain two sets of the orthogonal primitive idempotents as

There are ten standard Young tableaux yp for the Young pattern [4,1,1] of s 6 , listed from the smallest to the largest:

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Permutation Groups 211

1 2 3 4 1 2 3 5 1 2 3 6 1 2 4 5 1 2 4 6 5 4 4 3 3 6 6 5 6 5

1 2 5 6 1 3 4 5 1 3 4 6 1 3 5 6 1 4 5 6 3 2 2 2 2 4 6 5 4 3

After checking, each pair of the Young operators are orthogonal. Y, /72 are ten orthogonal primitive idempotents for the Young pattern [4,1,1] of S6

19. Give an example to demonstrate that the primitive idempotent gener-

Solution. Take the S3 group as an example. Let Y1 be the standard Young operator corresponding to the Young tableau w! & generates the minimal left ideal L1 = L y l . Let & be the non-standard Young operator corresponding to Young tableau F l . It is easy to see that J$ = (1 3 )&( l 3) = -(1 3)&. Being a group algebra, L = -L(l 3). Therefore, the left ideals generated by & and Y2 are the same, L1 = LCy, = LCy2. & / 3 and &/3 are both the primitive idempotents generating the same minimal left ideal L1.

ating a minimum left ideal is not unique.

6.4 Irreducible Representations and Characters

In this section we will discuss the irreducible representation of S, denoted by a given Young pattern [A]. Sometimes, we neglect the index [A] for simplicity.

* Let the permutation R,, transform the standard Young tableau y, into the standard Young tableau y,, and d be the number of the standard Young tableaux corresponding to the Young pattern [A]. Define the standard basis as

For a given v , the d bases b,, span the left ideal L,, while for a given p, the d bases b,, span the right ideal R,. The representation matrix of the group element S in the standard bases take the same form both for the left

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212 Problems and Solutions in Group Theory

ideal and for the right ideal:

Of course, another set of the standard bases can also be used:

For definiteness, we will only discuss the former set of bases.

* The representation matrix of the group element S in the standard bases can be calculated by the tabular method. Denote by Y,(S) the Young tableau transformed from the standard Young tableau y, by the permuta- tion S. Let y p = C S k T k , and denote by Ypk the Young tableau transformed from the standard Young tableau yp by the permutation T;'. Define the coefficient ALk(S) by comparing two Young tableaux Y,, and Yv(S). If there is a pair of digits filled in the same row of the Young tableau Y,(S) and in the same column of the Young tableau Y,k, then ALk(S) = 0. Otherwise, ALk(S) is defined to be the permutation parity of the vertical permutation R of the Young tableau Y p k which transforms the Young tableau yPk into the Young tableau y' such that the digits in each row of the Young tableau y' are the same as those in the same row of the Young tableau Y,(S). The representation matrix entry D p v ( S ) of the permutation S in the standard bases b,, is equal to C k 6kALk(S), which can be calculated by the tabular method. In fact, one may draw a table, where each row is designated by the algebraic sum of the Young tableaux, Ed, {the Young tableau Y p k } , and each column is designated by the Young tableau y,(S). A$) can be calculated by comparing the Young tableau Ypk in the row and the Young tableau yv(S) in the column. The irreducible representation D ( S ) in the standard bases is generally not unitary.

* In the orthogonal basis ap, the representation matrix D(P,) of a trans- position P, for two neighbored objects, P, = ( a a + l), is the direct sum of several one- or two-dimensional submatrices, which can be calculated as follows. If a and a + 1 are filled in the same row or the same column of the standard Young tableau Yp, we have a one-dimensional submatrix D,,(P,) = 1 or -1, respectively. Otherwise, we have a two-dimensional submatrix. Without loss of generality, we assume that a is filled in the upper row of the standard Young tableau yp than (a + l), and the stan- dard Young tableau ypa is obtained from the standard Young tableau y,

-

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Permutation Groups 213

by interchanging a and a + 1. Since a and a + 1 are filled in the different rows and in the different columns of the standard Young tableau y,, the Young tableau ypa must be standard and larger than the standard Young tableau y,. Now, the two-dimensional submatrix is

(6.12)

where the digit a is filled in the c,(a)th column of the r,(a)th row of the standard Young tableau y,, and similar for the digit a + 1. m,(a) is called the content of a in the Young tableau y,. If one goes from the box filled with a in the Young tableau y, downwards or leftwards and at last reaches the box filled with a + 1, the number of steps is equal to m.

* The merit of the representation D ( S ) of Sn in the standard bases is that the bases are explicitly obtained, but the representation is not unitary. The merit of the representation B(Pa) of Sn in the orthogonal bases is that the representation is real orthogonal, but its bases have to be calculated from the standard bases. The similarity transformation matrix X from the standard bases to the orthogonal bases is an upper triangle matrix:

D(Pa)X = XD(Pa) , X,, = 0, when p > v. (6.13)

If only one left ideal L, is considered, the orthogonal basis 4pv is

(6.14) P P

If both left- and right-ideals are considered, we have to calculate the or- thogonal bases @,”

P P

(6.15)

* There is a graphic method for calculating the character of a class (l) in the irreducible representation [A]. Rearrange the partition number l j in an arbitrary order, but it is convenient to arrange l j from the smallest to

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214 Problems and Solutions an Group Theory

the largest. According to the following rule, we fill n digits into the Young pattern [A] in order: first el digits 1, then & digits 2 and so on:

a) After each set of e j digits j is filled, the boxes which have been filled must constitute a sub-Young pattern, namely the boxes are lined up on the top and on the left, where the number of boxes on the upper row is not less than that on the lower row, and the number of boxes at the left column is not less than that at the right column. In each row, there is no unfilled box embedding between two filled boxes. b) The boxes filled with the same digit are connected such that from the lowest and the leftist box one can go rightwards or up- wards, without going leftwards or downwards, through all boxes filled with the same digit.

If all digits are filled into the Young pattern according to the rule, we say there is one regular application. The filling parity for the digit j is defined to be 1 if the number of rows of the boxes filled with j is odd, and to be -1 if that is even. The filling parity of a regular application is defined to be the product of the filling parities for all digits. The character $'I {(C)} of the class (1) in the representation [A] is equal to the sum of the filling parities of all regular applications. If the digits cannot be all filled in the Young pattern satisfying the above rule, we say there is no regular application and obtain x['] { ( 1 ) } = 0. For the class (1") composed of only the identity, each standard Young tableau corresponds to one regular application, so its character is just the dimension of the representation. Usually, the character of the identity is calculated by the hook rule instead of the graphic method.

* Two Young patterns related by a transpose are called the associated Young patterns. Each regular application of the class (C) in one Young pat- tern is a transpose of a regular application in its associated Young pattern. Since the sum of the number of rows and the number of columns of the boxes filled with j in a regular application is equal to Cj + 1. Note that (-l)'j+' is just the permutation parity of the cycle with length e j . Therefore, the product of two filling parities of the digit j in the corresponding regular applications for two associated Young patterns is equal to the permutation parity of the cycle, and the characters of the class (C) in two representations denoted by two associated Young patterns are the same if the permutation parity of the elements in the class is even, and different by a sign if that is odd. The Young pattern of the identical representation is [n] and its

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Permutation Groups 215

associated Young pattern [ l”] denotes the antisymmetric representation, where the representation matrix of S is equal to the permutation parity 6(S). A representation denoted by [A] is equivalent to the direct product of the antisymmetric representation [I”] and the representation denoted by the associated Young pattern [XI. 20. Calculate the Clebsch-Gordan coefficients for the reduction of the self-

product of the irreducible representation [2,1] of the S3 group in the standard bases and in the orthogonal bases, respectively.

Solution. There are three inequivalent irreducible representations for S3.

[3] is the symmetric (identical) representation. [13] is the antisymmetric representation. Take the transpositions PI and P2 of the neighboring ob- jects as the generators of S3. Their representation matrices are known:

[2,1] is the mixed representation in which the representation matrices of the generators can be calculated by the tabular method. There are two orthogonal standard Young operators for the Young tableau [2,1]. The rows in the table are designated by the standard Young tableaux YcL, listed in the first column of the table. Apply Pa, a = 1 or 2, to the standard Young tableau y,, one obtains the Young tableau yv(Pu). The columns in the table are designated by the Young tableau Y,(P,), listed in the first row of the table. The coefficients AL(Pu) is determined by comparing the Young tableaux indicating the pth row and the vth column. AL(Pa) is nothing but the representation matrix entry of Pa, a = 1 or 2.

The representation matrices for the generators in the mixed representation [Z, 11 are

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216 Problems and Solutions in Group Theory

The reduction for the self-product of the representation [2,1] is

Denote by X the similarity transformation matrix for reduction

(o - l 1-1 -1 0 1 l ) x ~ x ( o - l o 1 0 0 0 0 )

(:; ;)x=x(:-d: 0 0 1 0 ;).

0 0 -11 0 0 1-1 ' 0 0 0 1 0 0 0 - 1 0 0 0 1 1 0 0 0

Due to Pz = E , the eigenvalues for two matrices are both f l . The eigen- vectors for the eigenvalue 1 of two representation matrices before transfor- mation respectively are

The transpose of the common eigenvector is X r = ( 2 1 1 2 ) . The eigen- vectors for the eigenvalue -1 respectively are

The transpose of the common eigenvector is XT = (0 1 - 1 0). XI and X, are respectively the first and the second columns of X. Denote by X3 and X4 the third and the fourth columns of X. F'rom the first formula of the similarity transformation, X3 is an eigenvector with the eigenvalue 1 of the representation matrix before transformation, XT = ( a b b 2b ) . Substituting X3 into the second formula, we calculate the third column of the matrix, and obtain XT = (2b b b u) . Substituting them into the first formula, we

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Permutation Groups 217

calculate the fourth column of matrix, and obtain

-a - 2b

-a - 2b

Thus, a = -b. Taking -a = b = 1, we have

2 0 - 1 2

x = ( 1 1 1-1 1 1 ) 1 - 2 0 2 - 1

The reader is suggested to be familiar with this method. If not, he can calculate the matrix X by a direct way

Substituting it into the formulas of the similarity transformation, he will obtain the same result. I t is worthy to mention that the column-matrices in X are not orthogonal to one another because the representation before transformation is not unitary.

Denoting by Ip) Iv) the bases before transformation, and by I I [3]), I I [ 13]) and I I [2,13, p) the bases after transformation, where p, u and p are respec- tively taken 1 and 2, we have

For the representations [3] and [13], the representation matrices in the orthogonal bases are the same as those in the standard bases. For the representation [2,1], the representation matrices of the generators in the orthogonal bases are

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218 Problems and Solutions in Group Theory

Denote by 2 the real orthogonal similarity transformation matrix in the reduction of the self-product of the representation [2,1]:

1 0 0 0 1 0 0 0 (1 0 0 - 1 0 1 :).=.(. 0 0 1 0 ) -l O 0 ) 1(" - 4 4 3 -1 3 d) 3 ; (; x 2 : :) 0 0 0 - 1

-1 6 - z = - z 4 -a 3 -1 fi

f i f i 1 0 0 & 1

Due to P: = E , the eigenvalues of two matrices are both f l . The eigen- vectors for the eigenvalue 1 of two representation matrices before transfor- mation respectively are

The normalized common eigenvector is ZT = (1 0 0 1)/& The eigenvec- tors for the eigenvalue -1 respectively are

0 0 (i). (si;.: ($ (-The normalized common eigenvector is 2; = (0 1 - 1 O)/ f i . 21 and 2 2

are the first and the second columns of 2, respectively. From the condition of real orthogonal matrix, one may write the matrix 2 as .=.i" ! 0" 'i

fi 0 - 1 0 1 ' 1 0 - a 0

Substituting it into the formulas for the similarity transformation related to Pz, we obtain a = -1 from the first row of the fourth column. Denoting again by 1p)Iv) the bases before transformation and by II[3]), 11[13]) and

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Permutation Groups 219

11[2,13, p ) the bases after transformation, where p, v and p are respectively taken 1 and 2, we have

21. Calculate the standard bases for the irreducible representation [3,1] in the group space of Sq, and calculate the representation matrix of the transposition Pa for the neighbored objects in the standard bases by the tabular method.

Solution. There are three standard Young tableaux Y1, &, and Y3 for the Young pattern [3,1] of the S3 group

The corresponding Young operators yF are orthogonal to one another. The permutations between three standard Young tableaux respectively are

Removing the factor 1/8, we obtain the standard bases

where each column belongs to one left ideal, each row belongs to one right ideal, and

Y1 = E + (1 2) + (1 3) + (2 3) + (I 2 3) + (3 2 1) - (1 4) - (2 1 4) - (3 1 4) - (2 3)(1 4) - (2 3 1 4) - (3 2 1 4))

Y2 = (3 4)Y1(3 4)) Y3 = (2 4)%(2 4).

The representation matrices for P a are calculated by the tabular method:

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220

Pl 2 1 3 2 1 4 2 3 4

RW. [ 3 ~ 1

1 0 -1

0 1 -1

0 0 -1

1 2 3 4 1 2 4 3 1 3 4 2

Problems and Solutions in Group Theory

p2 p3 1 3 2 1 3 4 1 2 4 1 2 4 1 2 3 1 4 3 4 2 3 3 4 2

1 0 0 0 1 0

0 0 1 1 0 0

0 1 0 0 0 1

Thus, the representation matrices of the transpositions Pa for the neigh- bored objects in the representation [3,1] are:

22. Calculate the real orthogonal representation matrix of the transposi- tions Pa for the neighbored objects in the irreducible representation [3,1] of the S4 group in the orthogonal bases by the formula (6.12). Calculate the similarity transformation matrix between two represen- tations in the standard bases (Problem 21) and in the orthogonal bases and write explicitly the orthogonal bases <p,, in the group space of Sq.

Solution. The representation matrices of Pa in the orthogonal bases of the irreducible representation [3,1] of S4 can be calculated by Eq. (6.12))

- D(P1) = (:; : ) ) a(&) = (: -:iz &2) )

0 0 - 1 0 J3/2 1/2

Denote by X the similarity transformation matrix transforming the rep- resentation D(Pa) in the standard bases into the representation a ( P a ) in the orthogonal bases. X is an upper triangular matrix with three column matrices X I , X2 and X3:

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Permutation Groups 221

where the coefficient J8 is added for simplicity. In calculation we neglect the rows where all entries are vanishing. From

we obtain a1 = 1 and a2 = 3. From

D(P2)X2=--X2+-X3, 2 1 2 (;) 0 = - - (1) 3 +- qwe obtain bl = b2 = & and b3 = 2&. Finally, the similarity transforma- tion matrix X and its inverse matrix are

J g 1 fi

According to Eqs. (6.14) and (6.15), we first calculate the bases +clv in a left ideal by X, then calculate the nine orthogonal bases @,,([3,1]) by X-l. In the results the normalization factors di@ and 2& are used.

411 = JW'bl1 = d m { E + (1 2) + (1 3) + (2 3) + (1 2 3) + (3 2 1) - (1 4)

- (2 1 4 ) - (3 1 4 ) - (2 3)(1 4) - (2 3 1 4 ) - (3 2 1 4 ) ) ,

4 2 1 = ~/m{hi -t 3hi ) = d m { E + 3(3 4)) hi = d m { E + (1 2) - 2(1 3) + (2 3) + (1 2 3) - 2(3 2 1)

- (1 4) - (2 1 4) + 2(3 1 4 ) - (2 3)(1 4) - (2 3 1 4) + 2(3 2 1 4) + 3(3 4) + 3(3 4)(1 2) + 3(4 3 2) + 3(4 3 1 2) - 3(3 4 1) - 3(3 4 2 1) - 3(3 2 4 1) - 3(3 1)(4 2)) ,

431 = d m { b i i + b2i 4- 2b31) = d m { E 4- (3 4) 2(4 2)) hi = d W { E - (1 2) + (2 3) - (1 2 3) - (1 4) + (2 1 4 )

- (2 3)(1 4) + (2 3 1 4) + (3 4) - (3 4)(1 2) + (4 3 2) - (4 3 1 2) - (3 4 1) + (3 4 2 1) - (3 2 4 1) + (3 1)(4 2) + 2(4 2) + 2(4 2 3) - 2(2 4 1) - 2(2 3 4 l )} ,

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222 Problems and Solutions in Group Theory

4 1 2 = dmbi2 = d m ( 3 4)b22 = d n ( ( 3 4) + (3 4)(1 2) + (3 4 1) + (3 4 2) + (3 4 1 2)

+ (3 4 2 1) - (4 3 1) - (4 3 2 1) - (4 1) - (4 2 3 1)

- (4 1x3 2) - (4 2 1)) 7

4 2 2 = dW(bi2 + 3322) = J 9 6 ( ( 3 4) + 3E} b22

= J m ( ( 3 4) + (3 4)(1 2) - 2(3 4 1) + (3 4 2) + (3 4 1 2) - 2(3 4 2 1) - (4 3 1) - (4 3 2 1) + 2(4 1) - (4 2 3 1) - (4 1)(3 2) + 2(4 2 1) + 3E + 3(1 2) + 3(2 4) + 3(1 2 4) - 3(1 3) - 3(2 1 3) - 3(2 4)(1 3) - 3(2 4 1 3)},

432 = d m (b i2 + b22 + 2b32) = Jm { (3 4) + E + 2(2 3)) b22

= d D ( ( 3 4) - (3 4)(1 2) + (3 4 2) - (3 4 1 2 ) - (4 3 1) + (4 3 2 1) - (4 2 3 1) + (4 1)(3 2) + E - (1 2) + (2 4) - (1 2 4) - (1 3) + (2 1 3) - (2 4)(1 3) + (2 4 1 3) + 2(2 3) + 2(3 2 4) - 2(2 3 1) - 2(2 4 3 1)) ,

413 = dnbl3 = d w ( 2 4)b33 = d m ( ( 2 4) + (2 4)(1 3) + (2 4 1) + (2 4 3) + (2 4 1 3 )

+ (2 4 3 1) - (4 2 1) - (4 2 3 1) - (4 1) - (4 3 2 1)

- (4 1" 3) - (4 3 1)) 7

$23 = J1 /96{b13 + 3b23) = J m ( ( 2 4) + 3(2 3)) b33

= d W { ( 2 4) - 2(2 4)(1 3) + (2 4 1) + (2 4 3) - 2(2 4 1 3 ) + (2 4 3 1) - (4 2 1) + 2(4 2 3 1) - (4 1) - (4 3 2 1)

+ 2(4 1)(2 3) - (4 3 1) + 3(2 3) + 3(2 3 1) + 3(2 3 4) + 3(2 3 4 1) - 3(3 2 1) - 3(3 1) - 3(3 4 2 1) - 3(3 4 I ) ) ,

433 = dm{b13 + b23 + 2333) = d m { ( 2 4) + (2 3) + 2 E ) b33

= d i 7 5 { ( 2 4) - (2 4 1) + (2 4 3) - (2 4 3 1) - (4 2 1)

+ (4 1) - (4 3 2 1) + (4 3 1) + (2 3) - (2 3 1) + (2 3 4) - (2 3 4 1) - (3 2 1) + (3 1) - (3 4 2 1) + (3 4 1)

+ 2 E + 2(3 4) - 2(1 2) - 2(3 4)(1 2)},

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Permutation Groups 223

a11 ([3,1]) = &@ { 3 h - $12 - $13)

= Jm {3bll - (3 4)b22 - (2 4)b33) = Ji752(3E + 3(1 2) + 3(1 3) + 3(2 3) + 3(1 2 3) + 3(3 2 1) - (1 4) - (2 1 4 ) - (3 1 4 ) - (2 3)(1 4) - (2 3 1 4 ) - (3 2 1 4 ) - (3 4) - (3 4)(1 2) - (3 4 1) - (3 4 2) - (3 4 1 2) - (3 4 2 1) - (2 4) - (2 4)(1 3) - (2 4 1) - (2 4 3) - (2 4 1 3) - (2 4 3 1)) ,

a21 ([3,1]) = (3421 - $22 - $23)

= (1/24) (3 [E + 3(3 4)] b i i - [(3 4) + 3E] b22 - [(2 4) + 3(2 3)] b33)

= (1/6) {-(1 4) - (2 1 4) - (2 3)(1 4) - (2 3 1 4) - (3 4 1) - (3 4 2 1) - (3 2 4 1) - (3 1)(4 2) - (2 4) - (3 4 1 2) - (1 2 4) - (2 3 4) + 2(3 1 4) + 2(3 2 1 4) + 2(3 4)

+ 2(3 4)(1 2) + 2(4 3 2) + 2(4 3 1 2)} , %i([3,1]) = &@{3$3i - $32 - $33)

= dW-{3 [E + (3 4) + 2(4 2)] b i i - [(3 4) + E + 2(2 3)] b22

- “2 4) + (2 3) + 2EI b33)

= dm{-(l 4) + (2 1 4 ) - (2 3)(1 4) + (2 3 1 4 ) - (3 4 1) + (3 4 2 1) - (3 2 4 1) + (3 1)(4 2) + (4 2) + (4 2 3) - (2 4 1) - (2 3 4 1)) ,

@12([3,1]) = (2412 - 613) = (1/6) (2(3 4)b22 - (2 4)b33} = (1/6) {2(3 4) + 2(3 4)(1 2) + 2(3 4 1) + 2(3 4 2)

+ 2(3 4 1 2) + 2(3 4 2 1) - (4 3 1) - (4 3 2 1) - (4 1) - (4 2 3 1) - (4 1)(3 2) - (4 2 1) - (2 4) - (2 4)(1 3) - (2 4 1) - (2 4 3) - (2 4 1 3) - (2 4 3 I ) ) ,

a22 ([3, I]) = (2422 - $23}

= J m ( 2 [(3 4) + 3E] b22 - [(2 4) + 3(2 3)] b33)

= J m { 2 ( 3 4) + 2(3 4)(1 2) - (3 4 1) - (3 4 2) - (3 4 1 2) - (3 4 2 1) - (4 3 1) - (4 3 2 1) + 5(4 1) - 4(4 2 3 1) - 4(4 1)(3 2) + 5(4 2 1) + 6 E + 6(1 2) + 5(2 4) + 5(1 2 4) - 3(1 3) - 3(2 1 3) - 4(2 4)(1 3)

- 4(2 4 1 3) - (2 4 3) - (2 4 3 1) - 3(2 3) - 3(2 3 1)) ,

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224 Problems and Solutions in Group Theory

@32([3,1]) = (2432 - 433)

= d m { 2 [(3 4) + E + 2(2 3)] b22

- [(2 4) + (2 3) 4- 2E] b33)

= J i p { ( 3 4 2) - (3 4 1 2) + (2 4) - (1 2 4) + (4 2 1) - (4 1) + (3 4 2 1) - (3 4 1) - 2(4 2 3 1) + 2(4 1)(3 2) - 2(2 4)(1 3) + 2(2 4 1 3) - 3(4 3 1) - 3(1 3) + 3(2 1 3)

+ 3(4 3 2 1) - 3(2 3 1) - 3(2 4 3 1) + 3(2 3) + 3(3 2 4)},

@i3([3,1]) = $13 = d m ( 2 4)b33 = d@{(2 4) + (2 4)(1 3) + (2 4 1) + (2 4 3) + (2 4 1 3 )

+ (2 4 3 1) - (4 2 1) - (4 2 3 1) - (4 1) - (4 3 2 1)

- (4 1)(2 3) - (4 3 1)) I

@23([3,1]) = $23 = d W ( ( 2 4) + 3(2 3)) b33

= d P ( ( 2 4) + (2 4 1) + (2 4 3) + (2 4 3 1) - (4 2 1) - (4 1) - (4 3 2 1) - 2(2 4)(1 3) - 2(2 4 1 3) + 2(4 2 3 1)

+ 2(4 1)(2 3) - (4 3 1) + 3(2 3) + 3(2 3 1) + 3(2 3 4) + 3(2 3 4 1) - 3(3 2 1) - 3(3 1) - 3(3 4 2 1) - 3(3 4 I)},

@33([3,13) = $33 = d m { (2 4) + (2 3) + 2 E ) b33

= J i 7 E { ( 2 4) - (2 4 1) + (2 4 3) - (2 4 3 1) - (4 2 1)

+ (4 1) - (4 3 2 1) + (4 3 1) + (2 3) - (2 3 1) + (2 3 4) - (2 3 4 1) - (3 2 1) + (3 1) - (3 4 2 1) + (3 4 1) + 2 E + 2(3 4) - 2(1 2) - 2(3 4)(1 2)) .

23. Calculate the representation matrices of the transpositions Pa for the neighbored objects both in the standard bases and in the orthogonal bases of the irreducible representation [2,1, I] of the S4 group. Calcu- late explicitly the orthogonal bases aPv of the irreducible representa- tion [2,1,1] in the group space of S4.

Solution. There are three standard Young tableaux yl, y2 and ys corre- sponding to the Young pattern [2,1, I] in S3:

Yl : Y2 : Y3 :

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Permutation Groups 225

Rep. [2,1,1]

1 2 3 4 1 3 2 4 1 4 2 3

The corresponding Young operators yCl are orthogonal to one another. The permutations between three standard Young tableaux respectively are R 1 2 = (2 3), R 1 3 = (2 3 4) and R 2 3 = (3 4), so the standard bases are

Pl p2 p3

2 1 2 3 2 4 1 3 1 2 1 4 1 2 1 4 1 3 3 1 1 2 3 3 4 2 2 4 4 3 4 4 2 3 3 4

1 - 1 1 0 1 0 -1 0 0

0 - 1 0 1 0 0 0 0 1

0 0 - 1 0 0 - 1 0 1 0

The representation matrices of the transpositions Pa of the neighbored ob- jects in the representation [2,1,1] can be calculated by the tabular method:

Namely,

1-1 1 0 1 0 - 1 0 0

0 0 -1 0 0 - 1 0 1 0 D(P1) = ( 0 - 1 0 ) , D ( P 2 ) = (1 0 0 ) , D ( P 3 ) = ( 0 0 1 ) .

The representation matrices of Pa in the orthogonal bases of the irreducible representation [2,1,1] is calculated by the formula (6.12):

0 0 - D ( P 3 ) = (-d - 1/3 a / 3 ) .

0 4 / 3 1/3

From D(P,)X = XD(P,), the similarity transformation matrix X and its inverse matrix can be calculated to be

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226 Problems and Solutions in Group Theory

The orthogonal bases with the normalization factor are calculated by Eqs. (6.13-14):

([2, l]) = d m (2bll - b12 + b13)

= d W ( 2 b i i - (2 3)b22 - (4 2)b33) = drn{-(l 3) - (1 4) + (1 3 4) + (4 3 1) - (2 13) - (2 1 4 )

+ (2 1 3 4) + (2 1 4 3) - (2 3) + (3 2 4) - (3 1)

+ (3 1 2 4) - (4 2) + (4 2 3) - (2 4 1) + (4 1 2 3) + 2 E - 2(3 4) + 2(1 2) - 2(1 2)(3 4)},

a21 ([2,1,1]) = d m {&l + 4b21 - b12 - 2b22 + b13 + 2b23) = d m ( 2 [E + 2(2 3)] b l l - [(2 3) + 2E] b22

+ [-(4 2) + 2(3 411 b33) = d57G{-(3 2 4) - (3 2 1 4 ) + (3 1 2 4) + (3 1 4 ) + (2 4)

- (1 2 4) - (1 4) + (4 2 1) - 2(2 3)(1 4) + 2(2 3 1 4)

- 2(3 2 4 1) + 2(3 1)(2 4) - 3(1 3) + 3(1 3 4) + 3(2 1 3) - 3(2 1 3 4) + 3(2 3) - 3(2 3 1) - 3(2 3 4) + 3(2 3 4 1)) ,

a31 ([2,1,1]) = d m { - 2 h i + 2b21 + 6b31 + b12 - b22

- 3b32 - b13 + b23 + 3b33)

= J1/192{2 [-E + (2 3) - 3(2 4)] b l l - [-(2 3) + E + 3(3 4)] b22

+ [(4 2) + (3 4) + 3E] b33)

= dm{(l 4) - (4 3 1) - (2 1 4 ) + (2 1 4 3) - (2 3)(1 4)

+ (2 3 1 4) - (2 4) + (2 4)(1 3) + (2 4 1) + (2 4 3) - (2 4 1 3) - (2 4 3 l)},

@12([2,1,1]) = d n ( 3 b i 2 - bi3) = d m { 3 ( 2 3)b22 + (4 2)b33} = J57G{(4 2) - (4 2 1) - (4 2 3) + (4 2 1 3 ) + (2 4 1)

- (4 1) - (4 1 2 3) + (4 13) - 2(2 3)(1 4) + 2(3 2 4 1)

- 2(2 3 1 4) + 2(3 1)(2 4) + 3(2 3 1) - 3(3 1) - 3(3 1 2 4) + 3(3 1 4) + 3(2 3) - 3(3 2 1) - 3(3 2 4) + 3(3 2 1 4)},

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Permutation Groups 227

@22([2, 1,1]) = dm{3b12 + 6322 - b13 - 2b23) = Jm(3 [(2 3) + 2E] b22 - [-(4 2) + 2(3 4)] b33)

= d i p z { - ( 3 2 4) + (3 2 1 4 ) - (4 2 3) + (4 2 13) + (4 1 2 3) - (4 1 3) + (3 1 2 4) - (3 1 4) - 2(3 4) + 2(3 4)(1 2) + 3(2 3) - 3(3 2 1) - 3(2 3 1) + 3(3 1) - 4(2 3)(1 4) + 4(3 2 4 1) + 4(2 3 1 4) - 4(3 1)(2 4) - 5(1 4) - 5(2 4) + 5(1 2 4) + 5(4 2 1) + 6 E - 6(1 2)) ,

$)32([2,1,1]) = (1/24) {-3bi2 + 3b22 + 9332 + b13 - b23 - 3333) = (1/24) (3 [-(2 3) + E + 3(3 4)] b22 - [(4 2) + (3 4) + 3E] b33)

= (1/6) ((2 3)(1 4) + (3 2 4) - (3 2 4 1) - (3 2 1 4) - (2 3 1 4) - (3 1 2 4) + (3 1)(2 4) + (3 1 4) - (1 4) - (2 4) + (1 2 4) + (4 2 1) + 2(3 4) - 2(3 4)(1 2) - 2(3 4 1)

- 2(3 4 2) + 2(3 4 1 2) + 2(3 4 2 l)} , @13([2, 1713) = dmh3 = - d m ( 4 2)b33

= d i p 3 { - ( 4 2) + (4 2 1) + (4 2)(1 3) + (4 2 3) - (4 2 3 1) - (4 2 1 3) - (2 4 1) + (4 1) + (2 4 1 3) + (4 1 2 3)

- (4 1)(2 3) - (4 1 3)) 9

@23([2,1,1]) = (1/6) (b13 + 2b23} = (1/6) {-(4 2) + 2(3 4)) b33

= (1/6) {-(4 2) + (4 2 1) + (4 2)(1 3) + (4 2 3) - (4 2 3 1) - (4 2 1 3) + (2 4 1) - (4 1) - (2 4 1 3) - (4 1 2 3) + (4 1)(2 3) + (4 1 3) + 2(3 4) - 2(3 4)(1 2) - 2(4 3 1) - 2(4 3 2) + 2(4 3 1 2) + 2(4 3 2 l )} ,

@33([2,1, 13) = d m { - h 3 + b23 + 3b33)

= d W ( ( 4 2) + (3 4) + 3E) b33

= d r n { ( 4 2) - (4 2 1) - (4 2)(1 3) - (4 2 3) + (4 2 3 1) + (4 2 13) - (2 4 1) + (4 1) + (2 4 1 3 ) + (4 1 2 3)

- ( 4 3 2 ) + ( 4 3 1 2 ) + ( 4 3 2 1 ) - (4 1)(2 3) - (4 1 3) + (3 4) - (3 4)(1 2) - (4 3 1)

+ 3E - 3(1 2) - 3(1 3) - 3(2 3) + 3(1 2 3) + 3(3 2 1 ) ) .

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228 Problems and Solutions in Group Theory

Rep- [2,21

24. Calculate the standard bases and the orthogonal bases in the group space of S4 for the irreducible representation [2,2], and calculate the representation matrices of the transpositions Pa for the neighbored objects in these two sets of bases.

Sohtion. There are two standard Young tableaux Y1 and Y2 correspond- ing to the Young pattern [2,2] of S4

2 1 2 3 1 3 1 2 1 2 1 4 3 4 1 4 2 4 3 4 4 3 2 3

Two Young operators Y1 and Y2 are orthogonal to each other. The per- mutation relating two standard Young tableaux are R12 = R21 = (2 3). Therefore, removing the normalization factor 1/12, we obtain the standard bases

1 - 1 0

0 - 1 1

1 2 3 4 1 3 2 4

bll = Y1 = E + (1 2) + (3 4) + (1 2)(3 4) - (1 3) - (2 1 3) - (4 3 1) - (2 1 4 3) - (2 4) - (1 2 4) - (3 4 2) - (1 2 3 4) + (1 3)(2 4) + (1 3 2 4) + (3 1 4 2) + (3 2)(1 4))

b2i = (2 3)yi b12 = (2 3)&, b22 = y2 = (2 3)&(2 3).

1 1 - 1

0 0 - 1

The representation matrices of the transposition Pa of the neighbored ob- jects in the standard bases for the representation [2,2] are calculated by the tabular met hod:

On the other hand, the representation matrices of Pa in the orthogonal bases of the irreducible representation [2,2] are calculated by Eq. (6.12)

We calculate the similarity transformation matrix X and its inverse matrix from D(P,)X = x D ( P ~ ) :

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Permutation Groups 229

Thus, from Eqs. (6.13-14), we obtain the orthogonal bases with the nor- malization factor d w :

@ii([2,2]) = d m ( 2 b i i -312) = d ~ ( 2 h i - (2 3)b22) = J m ( 2 E + 2(1 2) + 2(3 4) + 2(1 2)(3 4) + 2(1 3)(2 4)

+ 2(1 3 2 4) + 2(3 1 4 2) + 2(3 2)(1 4) - (1 3) - (2 1 3) - (4 3 1) - (2 1 4 3) - (2 4) - (1 2 4) - (3 4 2) - (1 2 3 4) - (2 3) - (2 3 1) - (3 2 4) - (3 1 2 4) - (2 1 3 4) - (3 4 1) - (2 1 4 ) - (1 4 ) ) ,

@21 ([2,2]) = d m (2311 4- 4b2i - 312 - 2322) = d m (2 [E + 2(2 3)] 311 - [(2 3) + 2E] b22)

= (1/4) (-(1 3) + (2 1 3) + (4 3 1) - (2 1 4 3) - (2 4) + (1 2 4) + (3 4 2) - (1 2 3 4) + (2 3) + (2 1 3 4) - (2 3 1) - (2 1 4) - (3 2 4) - (3 4 1) + (3 1 2 4) + (1 4)},

@12([2,2]) = ( 1 / 4 ) b = (1/4)(2 3 ) h = (1/4) ((2 3) + (2 3 1) + (3 2 4) + (3 1 2 4) - (3 2 1) - (3 1) - (3 2 1 4) - (3 1 4 ) - (2 3 4) - (2 3 4 1) - (2 4)

- (2 4 1) + (2 1 3 4) + (3 4 1) + (2 1 4) + (1 4)) , @22([2,2]) = d77@(312 + 2322) = J r n ( ( 2 3) + 2E) b22

= J i p i ( ( 2 3) - (2 3 1) - (3 2 4) + (3 1 2 4) - (3 2 1) + (3 1) + (3 2 1 4) - (3 1 4) - (2 3 4) + (2 3 4 1) + (2 4) - (2 4 1) + (2 1 3 4) - (3 4 1) - (2 1 4) + (1 4) + 2 E + 2(1 3)(2 4) - 2(1 2) - 2(3 1 4 2) - 2(3 4) - 2(1 3 2 4) + 2(1 2)(3 4) + 2(2 3)(1 4 ) ) .

25. Calculate the symmetric bases of the oscillatory wave function of a molecule with the symmetry of a regular tetrahedron, for example, the methane CH4.

Solution. The molecule CH4 has the symmetry of a regular tetrahedron. The carbon atom is located at the center of the regular tetrahedron denoted by 0, while four hydrogen atoms lie in four vertices of the tetrahedron denoted by Aj, 1 5 j 5 4, respectively. The stretching oscillators of the C-H bonds OA, are described by the state lala2a3a4) with four oscillatory

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230 Problems and Solutions in Group Theory

quantum numbers. Due to the symmetry, the frequency CJ of four oscillators are the same, and the energy for the state is E = Cj=l (aj + 2)tiw.

The symmetry group for a regular tetrahedron is the T d group, which is isomorphic onto the permutation group S4. The correspondence between the elements of two groups can be established as follows. If an element R of T d transform four vertices Aj of the tetrahedron respectively into the positions of Arj, R corresponds to the following permutation:

R - ( l f l f 2 7-3 7-4 4 ) .

Especially, from Fig. 6.1 we see that three improper twofold rotations around the directions (el + e2) /a, (el - e3) /a, and (el - e 2 ) /fi cor- respond to the transpositions PI, P2, and P3 of the neighbored objects, respectively. Since two groups are isomorphic, we will not distinguish the elements in the Td group and in the permutation group Sq.

A3

Fig. 6.1 Diagram of a molecule with the Td symmetry.

We have obtained the orthogonal bases Q P v ( [ X ] ) for the inequivalent irreducible representations of S4 in Problems 21-24, in addition to the bases for two one-dimensional representations

@([4]) = d m { E + (1 2) + (1 3) + (1 4) + (2 3) + (2 4) + (3 4)

t (1 2)(3 4) t (1 3)(2 4) + (1 4)(2 3) + (1 2 3) + (1 3 2) + (1 2 4) + (1 4 2) + (1 3 4) + (1 4 3) + (2 3 4) + (2 4 3) + (1 2 3 4)

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Permutation Groups 231

+ (1 2 4 3) + (1 3 4 2) + (1 3 2 4) + (1 4 2 3) + (1 4 3 2)},

<p([i4]) = ,/W{E - (1 2) - (1 3) - (1 4) - (2 3) - (2 4) - (3 4) + (1 2)(3 4) + (1 3)(2 4) + (1 4)(2 3) + (1 2 3) + (1 3 2) + (1 2 4) + (1 4 2) + (1 3 4) + (1 4 3) + (2 3 4) + (2 4 3) - (1 2 3 4) - (1 2 4 3) - (1 3 4 2) - (1 3 2 4) - (1 4 2 3) - (1 4 3 2)}.

Applying those operators to the oscillatory state la1 u ~ u ~ Q ) , we obtain the symmetric bases of the oscillatory wave function of the molecule with T d

symmetry :

R@,v([AI)la1.2a3a4) = c ~X,'(R)a,,([x])lalaza3a4). P

The application of an element R of S4 to the defined as follows. For the permutation R,

R = ( 1 2 3 4 ) = (" T1 7-2 7-3 T4 1

we have

oscillation state ~ a ~ a 2 a 3 a 4 ) is

Recall that the application of a permutation R to the state only changes the order of four oscillatory quantum numbers a j , but does not change their values, In fact, there exist 24 linearly independent symmetric bases belonging to the inequivalent irreducible representations if four oscillatory quantum numbers aj all are different. The number of the linearly indepen- dent bases will decrease if some oscillatory quantum numbers aj are equal to each other. For example, when two and only two quantum numbers co- incide with each other, there exist 12 linearly independent symmetric bases belonging to five irreducible representations ([4] @ [2,2] @ 2 [3,1] @ [2,1,1]).

26. Calculate the representation matrices of the generators (1 2) and (1 2 3 4 5 ) of the S5 group in the irreducible representation [2,2, I] by the tabular method.

Solution. From Problem 15, the orthogonal standard Young operators for the Young pattern [2,2,1] of S5 are J+ [E - (2 5)] and yp , 2 5 ,!A 5 5. Now we will calculate the representation matrices of the elements (1 2) and (1 2 3 4 5) in this representation [2,2,1] by the tabular method.

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232 Problems and Solutions in Group Theory

From the tables, the representation matrices for the generators (1 2) and (1 2 3 4 5) in the representation [2,2,1] respectively are

1 0 - 1 0 1 1 - 1 - 1 0 0 [ g fl ; , [; p1; ; .

0 0 0 0 - 1 0 0 1 0 0

27. Respectively calculate the real orthogonal representation matrices of the transpositions Pa of the neighbored objects in two equivalent irre- ducible representations [23] 21 [16] x [3,3] of SS, and find the similarity transformation matrix X between them.

Solution. The standard Young tableaux for the Young pattern [23] and the Young pattern [3,3] are listed from the smallest to the largest, respectively

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Permutation Groups 233

1 2 1 2 1 3 1 3 1 4 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 3 4 3 5 2 4 2 5 2 5 4 5 6 3 5 6 3 4 6 2 5 6 2 4 6 5 6 4 6 5 6 4 6 3 6

Obviously, each standard Young tableau in [23] is the transpose of the cor- responding one in 3,3], but the larger in [23] becomes the smaller in [3,3]. According to Eq. t 6.12), the representation matrices of the transpositions P, for the neighbored objects in the representations p3] and [16] x [3,3] respectively are

represent at ion [ 23] representation [16] x [3,3]

PI :

1 0 0 0 0 [; H ;Al i ) , 0 0 0 0 - 1

-1 o & 0 0 - 2 0 0 0

O A -a 0 -1 0 0 0 0 - 2 0 0 -a 0 - 1

P2+ - 1 0 ; o a o ; ;) , +[; 0 1 0-a 1 0 -0

3 0 0 0 0 1 - d o 0 0 0 - 3 0 0 0 -d -1 0 0 0

0 0 0 4 1 0 0 3

P 3 : g [o 0 0 0 -3 0 - 1 4 ) > 0 0 i( i i i3:3:)

P5 : [;;;!l 0 0 1 0 ;) 0 ,

0 0 0 0 - 1

[ -'P,H i i] 0 0 0 - 1 0 0 0 0 0 1

Each matrix in [16] x [3,3] can be obtained from the corresponding one in [23] by a similarity transformation X = Y Z . Y is a diagonal matrix diag{ - 1, 1, 1, - 1 , l ) . Since the non-diagonal matrix entries of P, in the representation [23] appear only in 12, 13, 24, 34 and 45 row-columns, Y changes the signs of all non-diagonal matrix entries of P, in [23]. The

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234 Problems and Solutions in Group Theory

class ( I5> P 3 J > ( 1 J 2 > (12 ,3> (2,3) ( L 4 ) ( 5 )

1 4 1 3 1 2 1 1 1 2

3 2 3 2 2 2 3 3 3 2 2 2 2 no Regular

application neglected 2 4

Application -1 1 -1 -1 1 , parity

[(e>l 5 -1 1 -1 -1 1 0

matrix entries in 2 are all vanishing except for those on the minor diagonal line which are 1. 2 transpose the matrix in [23] with respect to the minor diagonal line. Namely,

0 0 0 0 - 1 0 0 0 1 0

X-1[23]X = [16] x [3,3], X =

This method is suitable for other pair of the representations with the asso- ciated Young patterns.

28. Calculate the character of each class of the permutation group S5 in

Solution. The characters of each class of S5 in the representation [2,2,1] are calculated in the following table.

the representation [2,2,1] by the graphic method.

29. Calculate the characters of each class of the permutation group $3

in the representations [3,2,1], [3,3], and [23] by the graphic method, respectively.

Solution. We calculate the characters in three representations of SS by the graphic method in the following table, where we also list the permitted reg- ular application. The regular application of the identity (1") is neglected. The Young pattern [23] is the one associated with the Young pattern [3,3]. The regular applications for [Z3] are neglected because they can be ob- tained from those for [3,3] by transpose. The Young pattern [3,2,1] is a self-associated Young pattern, so the characters of the classes composed of the odd permutations are vanishing.

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235 Permutation Groups

Characters in the ii

I (3113) I -2 -l -l

(3,2,1) 0 1 -1

(4,2) 1 0 -1 -1

reducible representations of s6 Regular application

neglected 1 2 3 1 2 4 1 3 4 1 2 5 1 3 5 4 5 5 3 5 5 2 5 5 3 4 5 2 4 5

1 2 4 1 3 3 1 3 4 3 3 4 2 4 4 2 3 4 1 1 3 1 2 2 1 2 3 2 2 3 1 3 3 1 2 3

1 2 3 1 4 4 1 2 3 1 2 4 1 3 4 4 4 2 4 4 4 4 3 4 4 2 4 4 4 3 1 2 2 1 3 3 1 2 2 3 3 2 3 3 3 3 3 2 1 1 1 1 2 2 1 1 1 1 1 2 2 2 1 2 2 2 2 1 2 2 2 1

1 2 3 3 3 3 1 1 2 2 2 2

1 2 2 2 2 2

30. Calculate and fill in the character tables of S3, Sq, Sg, SS and S7 by

Solution. Here we will not give explicitly the regular applications, but only list the calculation results in the table. The character of each class in the antisymmetry representation [In] shows whether the permutation in the class is even or odd. n(l) denotes the number of elements in the class ( l ) . Instead of showing the pair of associated representations, only one of them is shown for simplicity.

the graphic method.

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236

\ I

( 3 , 2 j ( 4 , l )

( 5 )

Problems and Solutions in Group Theory

20 -1 -1 1 0 30 -1 0 -1 0 24 1 -1 0 1

1 -2 (3.12) 20 I 1 1 -1 0

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Permutation Groups 237

6.5 The Inner and Outer Products of Representations

* The direct product of two irreducible representations of a permutation group S, is called the inner product of two representations for S,, which is reducible in general. There is no special graphic rule for the reduction of the inner product of representations for the permutation group. The C-G series and the C-G coefficients for S, can be calculated by the character method just like the usual finite group:

Since the irreducible representations of S, are real, the multiplicity .([A], [TI, [cJ]) is totally symmetric with respect to three representations. This symmetry leads to some common property for the C-G series of S,. Letting [A] and [?;I be two associated Young patterns, we have

(6.17)

In the reduction of the inner product of two irreducible representations of S,, there is one identical representation [n] if and only if two representations are equivalent, and there is one antisymmetric representation [ln] if and only if the Young patterns of two irreducible representations are associated.

* We are going to apply the concepts of the subduced representation and the induced representation, discussed in $3.3, to the permutation group. The subgroup H = S, x S, of the permutation group G = Sn+, is the direct product of two subgroups S, and S,, where S, consists of the permu- tations among the first n objects in the (n + m) objects and S, consists of the permutations among the last r n objects. The order of G is g = (n+rn)!. The order of the subgroup H is h = n!m! and its index is n = g / h . Denote by R j H , 2 5 j 5 n, the left cosets of H in G where the element Rj is a permutation transforming the first n objects to n different positions in all (n + rn) objects. Rj is not unique, but we assume that Rj has been chosen. For convenience, we denote the subgroup H by R1H with R1 = E.

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238 Problems and Solutions in Group Theory

Let L be the group algebra of G. Denote by e[W] the primitive idem- potent in L, and by Le[W] the minimal left ideal generated by &", where the Young pattern [w] with (n + m) boxes describes the irreducible rep- resentation of G. Let P" c LC be the group algebra of the subgroup H . Denote by e[A]e[p] the primitive idempotent in Lnm, and by Ln"e[A]e[p] the minimal left ideal generated by e[A]e[p] , corresponding to the irreducible representation of H described by the direct product of two Young patterns [A] x [p] , where the numbers of the boxes in two Young patterns [A] and [p] are n and m, respectively.

* A subduced representation from an irreducible representation [w] of G with respect to the subgroup H is generally reducible and can be reduced with respect to the irreducible representations [A] x [p] of H ,

(6.18)

From the viewpoint of the group algebra, Eq. (6.18) shows that there are ayp linear independent vectors e[A]e[pltae[w] , which map the left ideal Lce[W] onto the minimal left ideals for H , respectively,

Equation (6.19) gives the linear expansion of the new basis in the minimal left ideal in terms of the bases in Le[W] . The coefficients in the expansion constitute the column-matrices in the similarity transformation matrix X.

* The primitive idempotent e[']e[p] in Lnm is also an idempotent in L, but generally not primitive. The left ideal Ce[A]e[p] generated by e[A]e[p] in C corresponds to a reducible representation of G, which is the induced representation from the representation [A] x [p] of H with respect to G. For the permutation group, the induced representation is called the outer product of two representations, denoted by [A] 8 [p] . This representation can be reduced with respect to the irreducible representations [u] of G,

From the viewpoint of the group algebra, Eq. (6.20) shows that there are b'fD linear independent vectors e [ w ] t ~ e [ x ] e [ ~ ] , which map the left ideal

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Permutation Groups 239

Le[x]e[p] onto the minimal left ideals for G, respectively,

(6.21)

Equation (6.21) gives the linear expansion of the new basis in the minimal left ideal in terms of the bases in Le[’]e[p]. The coefficients in the expansion constitute the column-matrices in the similarity transformation matrix Y . Due to the symmetry between left- and right-ideals, the numbers of the linear independent vectors e[Xle[plt,e[W] and e[w]t~e[x]e[pl are equal, a& = b!p. This is an example of the Frobenius theorem (see Chapter 3).

* There is a graphic rule, called the Littlewood-Richardson rule, for cal- culating the multiplicity b z p in the reduction of the outer product of two irreducible representations for the permutation group, although it can be calculated by the character method as usual. The calculation method is as follows.

Arbitrarily choose one of the Young patterns in an outer product [A] @

[p] , say [p] . Usually, choose the Young pattern with the less boxes. Assign each box in the Young pattern [p] with its row number, namely assign the box in the j t h row with j . Now, attach the boxes assigned with 1 to another Young pattern [A] from its right or from its bottom, and then attach the boxes assigned with 2, and so on, in all possible way subject to the following rule. After attaching the boxes assigned with each digit, say j , the resultant diagram satisfies the following conditions:

1) The diagram constitutes a Young pattern, namely, the diagram is lined up on the top and on the left without unconnected boxes in each row such that the number of boxes in the upper row of the diagram is not less than that in the lower row, and the number of boxes in the left column is not less than that in the right column. 2) No two j appear in the same column of the diagram. 3) We read the attached boxes from right to left in the first row, then in the second row, etc., such that in each step of the reading process, the number of boxes assigned with the smaller digit must not be less than that assigned with the larger digit.

In each way subject to the above rule, the resultant diagram [w] after at- taching all boxes assigned with digits denotes an irreducible representation [w] of Sn+m which appears in the reduction of the outer product [A] @ [p] .

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240 Problems and Solutions an Group Theory

If the same Young pattern [w] is obtained in different ways, the number of the ways is the multiplicity of the irreducible representation [w] in the reduction.

* Due to the Fkobenius theorem, the Littlewood-Richardson rule can also be used for calculating the reduction of the subduced representation from the irreducible representation [w] of Sn+, with respect to the subgroup S, x S,. In the calculation, all the Young patterns [A] with n boxes and all the Young patterns [p] with rn boxes are taken to constitute the irre- ducible representation [w] of Sn+m with the multiplicity b;p, if [w] appears in the reduction of the outer product [A] 8 [p] according to the Littlewood- Richardson rule with the same multiplicity.

31. Calculate the Clebsch-Gordan series of the inner product of each pair of the irreducible representations for the permutation groups S3, S4,

S5, s6 and S7 by the character method. Solution. We can calculate the C-G series for the reduction of the inner product of two representations by Eq. (6.16). We only list some results for the C-G series, and neglect the remaining C-G series which can be obtained from them by Eq. (6.17).

For the S3 group:

For the S4 group:

[2, I] x [’&I] N [3] @ [I3] @ [2, I].

[3,1] x [3,1] = [4] @ [3,1] @ [221 @ [2, 121,

[3,1] x [Z2] = [3,11 @ [2, 121, p”3 p 2 1 [41 CB [221 @ [141.

For the S5 group:

[4,1] x [4,13 = [5] @ [4,13 @ [3,21 @ [3, 121, [4, 11 x [3,2] = [4,1] @ [3,2] @ [3, 121 @ P2, 11, [4, I] [ 3 , 1 2 1 = [4, 13 @ [3, 21 @ ~ 3 , 121 CB [22,11@ [2, 131,

[3, 21 [3, 21 = [51 p, 11 @ [3, 21 (3,121 @ [22,11 @ [2,131,

p, 21 [3, 121 = p, 13 @ p, 21 2[3,121@ p, 13 @ p, 131,

[ 3 , q [3, 121 [5] @ [4, 11 @ 2[3,21@ [ 3 , q @ 2[22,11 @ [2,131@ ~ 5 1 .

For the s6 group:

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Pe rrnu t at ion Group

For the S7 group:

241

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242 Problems and Solutions in Group Theory

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Permutation Groups 243

32. Calculate the Clebsch-Gordan coefficients for the reduction of the self- product of the irreducible representation [3,2] of S g .

Solution. This is a typical example for calculating the reduction of a re- ducible representation with a higher dimension for a finite group. Usually in the reduction of a reducible representation, one may try to find one or a few commutable elements in the group such that their eigenvalues can designate the bases for each irreducible representation without degenera- tion. This method has been used in the calculation for the Clebsch-Gordan coefficients of the groups T, 0 and I in Problems 25-27 of Chapter 3. For the permutation group S,, when n > 5, the dimension of the irreducible representation is quite high. It is hard to find the eigenvalues of some commutable elements to designate the bases of each irreducible represen- tation without degeneration. For example, one can designate the bases by the eigenvalues of two commutable elements (1 2) and (3 4 5), where the number of the different pairs of the eigenvalues is six. However, there is a 16-dimensional representation [3,2,1] in Sg. In the present Problem we are going to introduce another method for calculating the C-G coefficients,

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244 Problems and Solutions in Group Theory

which will be suitable for the representations with higher dimension. From Problem 31 we have

We take the orthogonal bases where the representation matrix of each trans- position Pa of two neighbored objects is known. The basis state before reduction is denoted by Ip)lv), where p and Y both run from 1 to 5, respec- tively. Denote by I ][A], p) the basis state after reduction, where p runs from 1 to the dimension of [A]. There is a characteristic in this example that two representations in the direct product are the same. The expansion of the new basis I ][A], p) can be chosen to be symmetric or antisymmetric in the permutation of two states in the product. For simplicity, we use the short notation for the basis state with the given symmetry in the permutation:

symmetric state exists, and we just use the basis state 1p)lp) without the subscript S. The method is outlined as follows. For a given representation, the representation matrix of Pi is diagonal. The eigenvalue of PI is used to classify the basis states. The basis state with the eigenvalue 1 of Pi is called the A-type basis, which has the following expansion

IP ,v )S = IP)IY) -k IY)IP) and I P , V ) A = IP)Iv) - Iv ) IP)- When P = v, only

The action of the transposition Pa on the basis state is

where the representation matrix of Pa in [A] is known. The representation matrices of Pa in the representation [3,2] are

The basis state with the eigenvalue —1 of P1 is ccalled the B-type basis,which has the following expansion

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Permutation Groups 245

1 0 0 0 0 2

P I : 1 1 0 0 0 0 0 0 0 - 1 1 0 0 0 o ) , p2:;[: 0

P 3 2 [ Js 0 0 1 0 0 3 0 0 0 1 , P 4 2

0 0 0 0 - 1 0 - 1 J s o o 0

2 0 0 0 3 0 0 0 0 0 - 3

- 1 o f i o 0 - 1 0 4

4 0 1 0

0 - 1 4 0 0 0 & 1 0 0 0 0 0 -la

t o o 0 4 1

' 2 0 0 4 0 o o o o l 0 0 1 0

For each irreducible representation [A], we first choose a basis state which is the common eigenstate of as many transpositions P a as possible, and write its expansion according to its eigenvalue of PI. Then, in terms of the representation matrices of Pa and the orthonormal condition we can determine the coefficients in the expansion. From the Schur theorem, two basis states belonging to two inequivalent irreducible representations or belonging to the different rows of a unitary irreducible representation are orthogonal. If the multiplicity of [A] in the reduction is one, the expansion must have a definite symmetry in the permutation. If it is larger than one, we can choose the expansions having definite symmetry. From the first basis state we are able to calculate the remaining basis states by Eq. ( c ) , which have the same symmetry in the permutation as that the first basis state has. The orthonormal condition and the symmetry in permutation can be used for check.

First, we calculate the basis state in the representation [5], which is the identical representation with dimension one. The basis state I I [5]) is given in Eq. (a). All representation matrices of Pa in [5] are 1. According to Eqs. (c) and (d), we have from the action of P2

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246 Problems and Solutions in Group Theory

Comparing the coefficient of each term, especially the coefficients of the terms which do not appear in 11[5]), we obtain a2 = a3 = a5 = a6 = 0, a4 = ale, a7 = all, a8 = a12 and a9 = ~ 1 3 , namely

Applying P 3 to 11[5]), we have

Applying P 4 to 11[5]), we have

Thus, a1 = ag. After normalization we obtain

Second, for the four-dimensional representation [4,1], we obtain the representation matrices of Pa from Eq. (12)

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Permutation Groups 247

P1 : O 1 0 0 O O ) ) 0 p2:1(o2 2 0 0 O :) 0 0 1 0 2 0 0 - 1 & 0 0 0 - 1 o o f i 1

3 0 0 0 -1 a 0 0 p 3 ' 3 1 ( O - l f i O 0 Jg 1 0 ) 7 .:(aj ; ;:).

0 0 0 3 0 0 1

The solutionm is

The following calculation is similar. We only list the method and theresults. Applying P3 to ||[4,1], 2} gives

Thus,

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248 Problems and Solutions in Group Theory

Applying P2 to 1J[4,1],3), we have

Thus,

Note that 11[4,1], 4) is a B-type basis state. Third, we calculate the basis states in the representation [3,2] where

the representation matrices of Pa were given in Eq. (d). 11[3,2], 1) is an A-type basis state. Applying PI and P2 to 11[3,2], 1) one by one leads to a formula similar to Eq. (e):

Applying P 4 to 11[3,2], 1)) we have

a1 + 2a4 + 2a9 = 0, 2al + 4a4 - 6 ~ 9 = 0.

Thus, a1 = -2a4 and a9 = 0. The normalization condition leads to

Applying P3 to 11[3,2], 1), we have

Thus,

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PI :

6 1 0 0 0 0

0 0 0 0 - 1 6

0 0 - 3 0 0 0 0 0 0 3 0 0

1 3

P 3 : - 1

' p 4 : - 4

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 - 1 0 0 0 0 0 0 -1 0 0 0 0 0 0 -1

1 P 2 : -

2 o f i o 1 0 0 '

0 - 1 0 &i 0 0

o o f i o 1 0

0 - 1 6 0 0 0

0 0 0 - 1 6 0 *

0 0 o a 1 0

1 0 0 O a l 0 0 O 0 4 O 0 1

2 0 0 0 0 0

0 0 -1 0 0

0 0 0 0 0 - 1 - 4 0 0 0 0 0

11[3,12], 1) is an A-type basis. Applying PI and P 2 to 11[3, 12], 1) one by one leads to a formula similar to Eq. (e):

Permutation Groups 249

Thus,

Fourth, we calculated the basis strates in the six-dimensional representa-tion [3,1], where the representation matrices of Pa are

Thus,

Thus,

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250 Problems and Solutions an Group Theory

Applying P 4 to 11[3, 1’3, 1) changes its sign

Thus, a1 = 0 and a4 = -a9 = &(a7 + a8)/2. The orthogonal condition between 11[3,12],1) and 11[22,1],1) leads to a4 = a9 = 0. Then, a7 = -a8. The basis state 11[3, 12], 1) is antisymmetric in the permutation. After normalization we have

Applying P3 to 11[3, 12], 1)) we have

Thus,

Applying P 4 to “[3, 1’3, 2), we have

Thus,

Applying P2 to 11[3, 12], 2)) we have

Thus,

11[3, 1’1, 4) is a B-type basis state. Applying P4 to 11[3, 1’1, 4), we have

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Permutation Groups 25 1

Thus,

Applying P3 to 11[3, 12],5), we have

Thus,

Fifth, we calculate the basis states in the five-dimensional representation [22, 11, where the representation matrices of Pa are

- 1 o & i o 0

0 0 0 -1 0 o & i o 1 0

3 0 0 0 0 - 1 A o 0 0

0 0 0 - 1 6

0 0 0 0 -1 0 0 0 0 - 2

0 - 3 0 0 0

The basis state 11[22, 1],5) is the common eigenstate of PI, P 2 and P 4 . We begin the calculation with this state. 11[22, 13,5) is a B-type basis with the form giver

p 2

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252 Problems and Solutions an Group Theory

The solution is bl = b2 = b3 = b4 = 0, b5 = -b6, b7 = -blo, b8 = - b g and b l l = -b12, namely

Applying P 4 to 11[2” 1],5), we have

The solution is b5 = -bl l = f i ( b 7 - b8)/2. The orthogonal condition between 11[22, 1 ] ,5 ) and 11[3,12], 6) leads to b5 = b l l = 0 and b7 = b g . After normalization we obtain the symmetric expansion

The solution is

Thus,

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Permutation Groups 253

Thus,

IP2, 11,1) = (2&)-' { IW4 - 13)13) - 1414) + 15)15) + & l V ) s } - Finally, we calculate the basis states in the four-dimensional represen-

tation [2, 13], where the representation matrices of Pa are

1 0 0 0 0 - 1 0 0

0 0 0 - 1 -3 0 0

p1: ( 0 0 - 1 0

P 3 4 ; 2:

-1ao 0 , p 2 : - ( 1 d l 0 0 ) ' ) , 0 p4:;( 0 0 -4 0 - 1 4 5 0 ' ) .

2 0 0 - 2 0 ' 0 0 0 - 2

-4 0 0

\ O 0 0 - 3 0 o a 1

The basis state )/[2, 13], 4) is the common eigenstate of P I , P2 and P3. We start the calculation with this state. 11[2, 13], 4) is a B-type basis. Applying P 2 to 11[2, 13],4), we obtain a similar formula to Eq. (g):

Applying P3 to 11[2, 13],4), we have

Thus, b5 = b7 = b8 = 0. After normalization we obtain the antisymmetric expansion

Applying P4 to I1[2, 13], 4), we have

Thus,

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254 Problems and Solutions in Group Theory

Thus,

33. Calculate the reduction of the following outer products of the repre- sentations in the permutation groups by the Littlewood-Richardson rule: (1) [3 ,2 ,118 ~31, (2) [3,218 ~ 2 , 11, (3) P, 11 8 [4, z31.

Solution. (1)

The dimensions of the representations on both sides are

9! - x 16 x 1 = 1344 = 105 + 162 + 120 + 189 + 168 + 216 + 216 + 168. 6! 3!

The solution is

Thus,

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Permutation Groups 255

The dimensions of the representations on both sides are

8! 5! 3! ~ 5 ~ 2 = 5 6 0 = 2 8 + 6 4 + 1 4 + 2 ~ 7 0 + 5 6 + 9 0 + 4 2 + 5 6 + 7 0 .

(3)

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256 Problems and Solutions an Group Theory

The dimensions of the representations on both sides are

13! x 2 x 300 = 171600 = 12012 + 9009 + 12870 + 11583 3- 2 x 21450 3! l o !

+ 5005 + 10296 + 8580 + 12870 + 15015 + 8580 + 17160 + 5720.

34. Calculate the reduction of the subduced representations from the fol- lowing irreducible representations of S6 with respect to the subgroup s3 6x33 :

Solution. (1)

[4,2] 2 [3] x [3] @ [3] x [2,1] @ [2,1] x [3] @ [2,1] x [2,1].

The dimensions of the representations on both sides are

9 = 1 x 1 + 1 x 2 + 2 x 1 + 2 x 2 .

(2)

[22,121 fi [2, 13 x [2, 11 CD [2, 11 x [ i 3 1 CB [PI x [2, 11 @ [ i 3 1 x p31. The dimensions of the representations on both sides are

9 = 2 x 2 + 2 x 1+1 x 2 + 1 x 1.

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Permutation Groups 257

(3)

The dimensions of the representations on both sides are

5 = 1 x 1 + 2 x 2.

35. Calculate the similarity transformation matrix in the reduction of the subduced representation from the irreducible representation [3,3] of SS with respect to the subgroup S3BS3.

Solution. The reduction of the subduced representation from [3,3] of SS with respect to the subgroup S36& was calculated in Problem 34:

The standard Young tableaux for the Young pattern [3,3] were given in Problem 18. Denote by Y1 the Young operator for the standard Young tableau

The standard bases in the representation space C of [3,3] are

From the Fock condition (6.5) we have

{ E + (1 4) + (2 4) + (3 4)) Yl = 0, { E + (1 5) + (2 5) + (3 5)) Yi = 0,

{ E + (1 6) + (2 6 ) + (3 6)) Yi = 0,

{E + (3 4) + (3 5) + (3 6)) Yl = 0.

The idempotent of the one-dimensional representation [3] x [3] of the subgroup S38S3 is

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258 Problems and Solutions an Group Theory

From Eq. (6.19), the basis state for the representation [3] x [3] is

The idempotent of the four-dimensional representation [2,1] x [2,1] of the subgroup S38S3 is

Since y3y1 = 0, from Eq. (6.19), we choose the basis state for the repre- sentation [2,1] x [2,1] as

'$1 = Y3(3 4)Yl = (3 4) {E + (1 2) - (1 4) - (2 1 4)) {E + (3 5) - (3 6) - (5 3 6)) YI = (3 4) {2E - (1 4) - (4 2)) { E + (3 5) - 2(3 6)} Y1

= {2(3 4) + 2(4 3 5) - 4(4 3 6) - (3 4 1) - (4 1 3 5) + 2(4 1 3 6) - (3 4 2) - (4 2 3 5) + 2(4 2 3 6)} Y1

= 2b2 + 2b3 - 4(3 6)bl - (1 4)bl - (1 5)bl + 2(6 l ) b l - b4 - b5 + 2(6 2)bi = 2b2 + 2b3 - b4 - b5 + [bl + b2 + b4] + [bl + b3 + b5] - 2bl

+ 6 [bi + b2 + b3]

= 6b1 + 9b2 + 9b3.

Then, we have

Thus, we obtain the similarity transformation matrix X:

36 6 - 3 6 - 3 0 9 0 0 0

0 0 -9

The generators of S38S3 are PI , P 2 , P 4 , and P5. Their representation

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Permutation Groups 259

matrices in [3,3] can be calculated by the tabular method:

1 0 0 -1 -1 1 0 0 0 0 0 1 0 - 1 0 0 0 0 1 0

0 0 0 0 1 0 0 0 1 - 1 0 0 0 1 0 0 0 0 0 - 1

Their representation matrices in [3] x [3] @ [2,1] x [2,1] are

1 0 0 0 0 0 1 0 - 1 0

- ~ ( ~ l ) = l @ ( ~ - ~ ) ~ ( ~ ~ ) = ~ o o 0 -1 0 0 0 - 1 1 0 j, 0

1 - D ( P 4 ) = 1 @ ~ ~ ; ) x ( ~ - ~ ) ( 0 0 0 -1 0 0 -1 0 0 1 - 1 o ) ,

1

0 0 0 0 -1

0 0 0 0 1 ,

1 0 0 0 0 0 0 0 1 0

0 1 0 0 0 0 0 1 0 0

1 0 0 0 0 0 1 - 1 0 0

0 0 0 0 - 1

0 1 0 0 0 .

1 0 0 0 0 0 0 1 0 0

0 0 0 0 1 0 0 0 1 0

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260 Problems and Solutions an Group Theory

It is easy to check that X satisfies the similarity transformation

36. Calculate the representation matrices of the generators of S4 in the in- duced representation from two-dimensional irreducible representation [2,1] of S3 with respect to Sq.

Solution. For simplicity, let A and B respectively be the representation matrices of the generators of S3 in the two-dimensional representation [2, I]:

Denoting by &, p = 1 and 2 , the state bases in the representation space, we have

2 2

u=l u=l

This two-dimensional space is invariant for the S 3 group, but not for Sq.

The S 3 group is a subgroup of S4 with index 4 . Extend the two-dimensional space:

The eight-dimensional space is invariant for S4. In fact,

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Permutation Groups 26 1

Therefore, the representation matrices of the generators (1 2) and (1 2 3 4) of Sq in the induced representation are:

A 0 0 0 0 0 O B

O A O O O A B O O *

O O O A 0 O A O

D ( 1 2 ) = ( A O ) , D ( 1 2 3 4 ) = ( B O 0 0 ) It is a reducible representation of Sq, which can be reduced into the direct sum of the representations [3,1], [2,2] and 12, 12].

37. Calculate the similarity transformation matrix in the reduction of the outer product representation [2, I] 8 (21 of the permutation group.

Solution. The outer product representation [2, I] @ [2] is the induced rep- resentation from [2,1] x [2] of the subgroup Ss xS2 with respect to the group S5. Its reduction can be calculated by the Littlewood-Richardson rule:

[2,1] 63 [2] = [4,13 @ [3,2] 43 [3, 12] @ p2, 11 ~ 2 ~ 1 = 2 0 = 4 + 5 + 6 + 5 . 120

6 x 2 -

The idempotent of the two-dimensional representation [2,1] x [2] of the subgroup S3@S2 is

The standard bases in the representation space of [2,13 x [2] are

From the Fock condition (6.5) we have

and the useful formula for the later calculation,

[E + (1 2)] [E - (1 3)] bl = [2E - (1 3) - (1 2)(1 3)] bl = 3h-

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262 Problems and Solutions in Group Theory

Extending the two-dimensional space, we obtain the bases

b3 = ( 1 4)bl , b5 = (2 4 ) h ,

b4 = (1 4)bz = ( 1 4) (2 3)bl, be = ( 2 4)bz = (4 2 3)bl,

After the similarity transformation X, the out product representation [2, I] 8 [2] becomes a direct sum of four irreducible representations:

The representation matrices of the generators P1 = (1 2) and W = (12345) of S5 in the out product representation [2,1] []2 for the bases

are

Through a direct calculation we have

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Permutation Groups 263

where R = Pi and W . By the tabular method we obtain

1 0 0 - 1 - 1 1 0 0

0 0 1 - 1 0 0 0 - 1 - 1 0 0 0

0[4qp1) = ( -’) , 0[4111(W) = (-’ - 1 0 0 1 O ) ’

0 -10 0 0 ,

-1-11 1 0 - 1 0 0 0 1

- 1 0 0 1 0 0 - 1 0 1 0 i

1-11 0 0 0 1 0 0 - 1 1 0 0 1 0 - 1 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0

[ o l 1 0 - 1 0 0 - 1 1 ) 1 [;m;mcl) 1-1-1 0 0 D[22Jl(q) = 0 0 -1 0 0 D[22Jl(w) = 1-1 -1 1 0 .

0 0 0 - 1 0 0 0 0 0 - 1 0 0 1 0 0

Now, we calculate the similarity transformation matrix X by Eq. (6.21). The idempotent of the four-dimensional representation [4,1] of S5 is

From Eq. (6.21) we obtain the expansions of the basis states T+@”’ of [4,1] in the extending space:

20

$ ~ ” ] = C bvXv3 = Y2Y1

= [ETtl 4) + (2 4) + (1 5) + (2 5) + (1 4)(2 5)]

x [E + (1 211 [E + (4 511 [E - (1 311 bi = 6 [E + (1 4) + (2 4) + (1 5) + (2 5) + (1 4)(2 5)] bi = 6 [bi + b3 + b5 + b g + b i i + b15] ,

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264 Problems and Solutions in Group Theory

The idempotent of the five-dimensional representation [3,2] of S5 is

From Eq. (6.21) we obtain the expansions of the basis states $F’21 of [3,2] in the extending space:

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Permutation Groups 265

20

$!I2’ = C buXu7 = (4 5)$F’21 u=l

= 3 [bl + b9 + b l l + 2b4 - 2b5 - 2b6 + b7 + 2b16 - b17

20

+!”I = C b,X,g = (2 3)(4 5)$~!’~]

The idempotent of the six-dimensional representation [3, 12] of S5 is

From Eq. (6.21) we obtain the expansions of the basis states $ J ~ ’ ~ ’ ’ of [3, 12] in the extending space:

20

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266 Problems and Solutions in Group Theory

20 $,F912] = bvXu13 = ( 2 3)q!Jf3”21

The idempotent of the five-dimensional representation [22, 11 of S5 is

Y5 =

From Eq. (6.21) we obtain the expansions of the basis states +F2’l1 of [22, 11 in the extending space:

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Permutation Groups 26 7

20 $ f2’11 = bvXv19 = ( 2 3)(4 5)llfz2’11

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Chapter 7

LIE GROUPS AND LIE ALGEBRAS

7.1 Classification of Semisimple Lie Algebras

* An element R in a Lie group G can be described by a set of independent real parameters ( T I , r2, . . .). Those parameters vary continuously in a given Euclidean region, called the group space. The number g of the parameters, which is equal to the dimension of the group space, is called the order of the group G. In this chapter we only discuss the Lie group whose group space is simply-connected.

A Lie group is said to be compact if its group space is a closed region in a Euclidean space. A Lie group is said to be simple if it does not contain any non-trivial invariant Lie subgroup. A Lie group is said to be semisimple if it does not contain any Abelian invariant Lie subgroup including the whole group. Any Lie group of order one is a simple Lie group, but not a semisimple Lie group. A simple Lie group whose order is more than one must be a semisimple Lie group.

* The generators I A in a representation D ( R ) of a Lie group G are defined as

A

The generators I A in any representation D ( R ) of commutation relation:

[ I A , 1131 =iC cA,”ID, D

G satisfy the common

(7.2)

where the real parameters CA,”, independent of the representation, are called the structure constants of the Lie group.

269

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270 Problems and Solutions in Group Theory

* The real linear space spanned by the bases ( 4 1 ~ ) is called the real Lie algebra if the product of two vectors X and Y in the space is defined by the Lie product:

If the linear space is complex, the algebra is called the complex Lie algebra, or briefly, the Lie algebra. A real Lie algebra of a compact Lie group is compact. A Lie algebra of a simple or a semisimple Lie group is simple or semisimple, respectively. In fact, a simple Lie algebra does not contain any non-trivial ideal, and a semisimple Lie algebra does not contain any Abelian ideal, including the whole algebra. Therefore, a one-dimensional Lie algebra is simple, but not semisimple. In the rest of this chapter, if without special indication, “a simple Lie algebra with the dimension larger than one” will be simply called as “a simple Lie algebra” for simplicity. It is proved that a semisimple Lie algebra can be decomposed into the direct sum of some simple Lie algebras.

* The Killing form QAB of a Lie group or its Lie algebra is defined from its structure constants cA::

QAB is a real symmetric matrix of dimension 9. When the parameters of a given Lie group are changed, the generators transform like a vector, and the structure constants as well as the Killing form transform like a tensor:

* The Cartan Criteria says that a Lie algebra is semisimple if and only if its Killing form is non-singular, and a real semisimple Lie algebra is compact if and only if its Killing form is negative definite.

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Lie Groups and Lie Algebras 271

* If { H i , 1 5 j 5 !} is the largest set of the mutually commutable gener- ators in a simple Lie algebra L, the set constitutes the Cartan subalgebra and e is called the rank of the Lie algebra L. It is proved that the group parameters in a simple Lie algebra L can be chosen such that the generators Hj and the remaining generators E, satisfy

[ H j , Ea] = @a,

( N,,pE,+p when a' + p is a root, (7.6)

l o the remaining cases,

and the Killing form becomes

9jk = - s j k , Sap = -S(-CY)p, gj, = gaj = 0. (7.7)

The real vector d in the !-dimensional space is called the root in L, and the space is called the root space. The space spanned by the roots in a simple Lie algebra of rank l is &dimensional. The roots kG appear in pair. Except for the pair of roots, any two roots are not collinear. There is a one-to-one correspondence between the generator E, and the root a'. Hj and E, are the Cartan-Weyl bases for the generators in the simple Lie algebra L.

* In a semisimple Lie algebra, we can constitute a root chain from a root a' by adding or subtracting another root p several times. Without loss of generality, if d+np, -qag 5 n 5 pap, are the roots, and a'+ (pap + 1)p and a' - (qap + 1)p are not the roots, where pap and qap are both non-negative integer, it is proved that

When d = p, the root chain consists of a', 0 and -6.

* A root is called positive if its first nonvanishing component is positive, otherwise it is negative. Sometimes, Hj is called the generator correspond- ing to the zero root, which is l degeneracy. A positive root is called a simple root if it cannot be expressed as the non-negative integral combination of other positive roots. Therefore, any positive root is equal to a non-negative integral combination of the simple roots, and any negative root is equal to

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272 Problems and Solutions an Group Theory

a non-positive integral combination of the simple roots. The difference of two simple roots is not a root. From Eq. (7.8), the simple roots are linearly independent. Thus, the number of the simple roots in a t-rank simple Lie algebra is equal to t . The angle between two simple roots has to be equal to 5.n/6, 3n/4, 2n/3 or n/2, and correspondingly, the ratio of the length squares of the two simple roots is 3, 2, 1, or no restriction, respectively.

* There are four series of the classical simple Lie algebras (At, Be, Ce and De) and five exceptional simple Lie algebras (G2, Fq, Eg, E7 and Eg),

where the Lie group of At is SU(t + l), the Lie group of Be is S 0 ( 2 t + 1), the Lie group of Ce is Sp(2t), and the Lie group of De is SO(2l).

The Dynkin diagram for a simple Lie algebra is drawn by the following rule. There are one or two different lengths among the simple roots in any simple Lie algebra. Denote each longer simple root by a white circle, and denote each shorter simple root, if it exists, by a black circle. Two circles denoting two simple roots are connected by a single link, a double link, or a triple link depending upon their angle to be 2n/3, 3n/4 or 57r/6, respec- tively. Two circles are not connected if two simple roots are orthogonal and the ratio of their lengths is not restricted. The Dynkin diagrams for the simple Lie algebras are listed in Fig. 7.1.

G2

F4 Ap, t 2 1 ----- 1 2 3 c - 2 e - 1 e

E6 Bp, t 2 3 O-G-O-----

1 2 3 e - 2 e - 1 c

E7 ce, e 2 2 t--t--+---+-~==~ 1 2 3 e - 2 e - 1 e

Dp, .f 2 4 --o---<'-~ 1 2 3 e - 3 c - 2 e &3

o==e 1 2 - L-2.L 1 2 3 4 5

& 1 2 3 4 5 6

1 2 3 4 5 6 7

Fig. 7.1 The Dynkin diagrams of the simple Lie algebras.

* The simple Lie algebra can also be described by the Cartan matrix. The Cartan matrix for a simple Lie algebra with t simple roots r, is an &dimensional matrix A:

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Lie Groups and Lie Algebras 273

The diagonal element of A is always equal to 2, and its non-diagonal element can be equal to 0, -1, -2 or -3. In comparison with the Dynkin diagram, if two simple roots r, and r, are disconnected, A,, = A,, = 0, and if they are connected by a single, double or triple link, and the length of r, is not less than the length of r,, A,, = -1 and A,, = -1, -2 or -3, respectively.

The Dynkin diagram or the Cartan matrix gives all the information of the simple Lie algebra, such as the angles and the lengths of the simple roots as well as all roots. The roots can be calculated recursively as follows. Any positive root can be expressed as a sum of the simple roots:

e a'= C C,r,, C, 2 0, (7.10)

C, Cv is called the level of the root d. Any simple root is a root of level one. If all positive roots with the level less than n have been found, we want to judge whether a' + r, is a root where 6 is a root of level (n - 1). C alculat e

e e q - p = r (a'/.,) = A&,, if a'= Cur,. (7.11)

u=l u=l

Since q is known, one can calculate p from Eq. (7.11) to determine whether a' + rP is a root or not.

1. Prove the number pap + qap + 1 of roots in a root chain a'+ nB, -qap 5

Solution. Prove this problem by reduction to absurdity. If the length of the root chain is larger than four, one can redefine the root a' such that the following five vectors are all non-zero roots:

n 5 pap, in a simple Lie algebra is less than five.

a' - 2& a'- 6, a', a'+ /?, a'+ 28.

However, the following four vectors are not roots:

(a' f 2B) + a' = 2(a' f B), (a' f 28) - a' = *2B.

Thus,

o = r[(a'f 2 f i ) / a ' ]= di1(a'* 2B) -a' = 2 f 2d;l p . a' . r ) Adding two equations leads to contradiction. This completes the proof.

Sin:e qap +pap + 1 5 4, both qap and pap are less than four. Thus, I r ( w > I L 3.

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274 Problems and Solutions in Group Theory

2. Prove that the difference of two simple roots is not a root, and there are

Solution. We first prove the difference of two simple roots rcl and r, not to be a root by reduction to absurdity. Suppose rp - r, = a'. If a' is a positive root, rp = r, + a' is the sum of two positive roos. If a' is a negative root, r, = rP + (-a') is the sum of two positive roos. It contradicts with the definition of a simple root. Thus,

l simple roots in a simple Lie algebra of rank l .

The angle of two simple roots is not an acute angle. It is obvious that the number of the simple roots is not less than l?,

because any root can be expressed as a linear combination of the simple roots. If the simple roots are linearly independent to each other, the number of the simple roots must be l . Now, we prove it by reduction to absurdity. If there is a linear relation among simple roots, we separate the terms with the negative coefficients from those with the positive ones:

P v

where cp and d, are all positive, and the values of p and u are different. Since a # 0,

0 < a2 = Cc,d,r, ' r , 5 0.

This completes the proof.

3. Prove that the angle between two simple roots has to be equal to 57r/6, 37r/4, 2 ~ / 3 or 7r/2, and correspondingly, the ratio of the length squares of the two simple roots is 3, 2, 1, or no restriction, respectively.

Solution. Calculate the square of cosine of the angle between two simple roots:

Since rP and r, is not collinear, 1 cos2 81 < 1, and the integer can only be equal to 3, 2, 1 or 0. Namely, the angle 9 can be 57r/6, 37r/4, 27r/3 or 7r/2. Without loss of generality, let the length of rp be not less than that of r,,

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Lie Groups and Lie Algebras 275

Due to the restriction for their product, r'(r,/rp) has to be equal to -1 or 0, and I'(r,/r,) is equal to -3, -2, -1 or 0. The ratio of those two number is

depending on the angle of two simple roots to be 5 ~ / 6 , 3 n / 4 , 2 ~ / 3 or ~ / 2 .

4. Calculate the Cartan matrix of the Lie algebra E6.

Solution. The Cartan matrix of E6 is a six-dimensional matrix with the diagonal elements to be 2. According to the enumeration for the simple roots in its Dynkin diagram, the row-column numbers of the positions of the non-diagonal elements, which are -1, are 12, 23, 34, 36 and 45, or vice versa,

A = I j l - l o - 1 0 0 0 0 0 0 )

0 -1 2 -1 0 -1 0 0 - 1 2 - 1 0 '

0 0 0 - 1 2 0 0 0 - 1 0 0 2

5 . Draw the Dynkin diagram of a simple Lie algebra where its Cartan matrix is as follows, and indicate the enumeration for the simple roots.

Solution. The Dynkin diagram consists of four circles. Since A32 = -2, the circle designated by 2 is white corresponding to a longer simple root, and the circle 3 is black corresponding to a shorter simple root. The circles 2 and 3 are connected by a double link. Since A12 = A21 = A34 = A43 = -1, the circles 1 and 2 are connected by a single link, so be the circles 3 and 4. Thus, the circle 1 is white and the circle 4 is black.

r\ 1 2 3 4

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276 Problems and Solutions an Group Theory

6. Calculate all positive roots in the C2 Lie algebra.

Solution. There are two simple roots in C2

where r2 is the longer simple root, and r1 is the shorter. Their angle is 3n/4. Usually, the length square of the longer root is taken to be 2. The Cartan matrix of C2 is

2 -2 A = ( - 1 2 ) .

First, we calculate pl in the root chain rl + plr2. From Eq. (7.11)) we have

p1 = -I'(rl/r2) = -A21 = 1.

(el + e2) is a root, but rl + 2r2 is not a root. Then, Thus, rl + r2 = we calculate p2 in the root chain r2 + p2r1. Since

in addition to r1 + r2, we know 21-1 + r2 = fie1 is a root. 3rl + 1'2 is not a root. rl + r2 is the only root of level two, and 21-1 + r2 is the only root of level three. Since (2rl + 21-2) is twice of rl + r2, it is obviously not a root. Therefore, the C2 Lie algebra contains four positive roots and four negative roots. The rank of C2 is 2, and its order is ten. This result can be generalized to the Ce Lie algebra. Ce contains 2C2 roots: f f i e j , & m ( e j + ek) and m ( e j - ek). The rank of Ce is C and its order is C(2C+ 1).

7. Calculate all positive roots in the B3 Lie algebra.

Solution. There are three simple roots in B3:

rl = el - e2, r2 = e2 - e3, r3 = e3,

where rl and r2 are the longer roots with the length square 2, and r3 is the shorter root. The angle between rl and r2 is 2 ~ / 3 . The angle between r2 and r3 is 3n/4. rl is orthogonal to r3. The Cartan matrix of B3 is

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Lie Groups and Lie Algebras 277

We first calculate pl, p2 and p3 in the root chains rl + plr2, r1 + p2r3,

r 2 + p3r3. From Eq. (7.11)) we have

Thus, there are two roots of level two: rl + r 2 and r 2 +r3. We also know that r 2 + 2r3 is a root of level three, and rl + 2 1 2 and r2 + 331'3 are not the roots. Then, we calculate p4, p5, p6 and p7 in the root chains (rl + r 2 ) + p4r1,

(ri + r 2 ) + p5r3, (r2 + 1-3) + p6rl and (1-2 + r3) + p7r2. Since

in addition to r 2 + 2r3, there is another root of level three: rl + r 2 + 1-3.

Besides, rl + r2 + 2r3 is a root of level four. Adding r 2 or r3 to the root r 2 + 21'3 is not a root. w e are going to calculate p8 in the root chain (rl + r 2 + 1-3) + ~ 8 1 - 2 . Since rl + r3 is not a root, we have

Thus, rl + r 2 +2r3 is the only root of level four. Adding r3 to it is not a root. We want to calculate pg and plo in the root chains (rl + r 2 + 2r3) + pgrl

and (rl + r 2 + 2r3) + ~101-2 . Since

we obtain the only root of level five: rl + 2r2 + 213. Besides, rl + 31-2 + 213 is not a root. 2rl + 2r2 + 2r3 is twice of a known root, so it is not a root. Now, we are going to calculate pll in the root chain (rl + 2 1 2 + 213) +pllr3.

Since

there is no root of level six. In summary, the B3 Lie algebra contains three roots of level one (three

simple roots: rl = el - e 2 , r 2 = e2 - e3 and r3 = e3), two roots of level two (rl + r 2 = el -e3 and r 2 +r3 = e 2 ) , two roots of level three (rl + r 2 + r 3 = el and r 2 + 2r3 = e2 + e3), one root of level four (rl + r 2 + 213 = el + e3),

and one root of level five (rl + 2r2 + 21'3 = el + e 2 ) . The rank of B3 is 3, and its order is 21. This result can be generalized to the Bc Lie algebra. Bc contains 212 roots: fej, f (e j + e k ) and (ej - e k ) . The rank of Bc is l , and its order is l ( 2 l + 1).

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278 Problems and Solutions in Group Theory

8. Calculate all positive roots in the D4 Lie algebra.

Solution. There are four simple roots in D4:

rl = el - e2, r 2 = e 2 - e3, r3 = e3 - e 4 , r4 = e3 + e 4 ,

where the length squares of four simple roots are all 2. The angle between r 2 and any other simple root ra, a # 2, is 27r/3, and the other pair of simple roots is orthogonal to each other. The Cartan matrix of D4 is

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Lie Groups and Lie Algebras 279

r3 + 1-4) + p20r1 and ( r 2 + r3 + r 4 ) + p21r2. Since

there is only one root of level four rl + r 2 + r3 + r 4 = el + e3. Adding rl, r3 or 1-4 to the root of level four is not a root. We calculate p22 in the root chain (rl + 1-2 + r3 + 1-4) + ~221-2. Since

we obtain the only root of level five: rl + 2r2 + r3 + r 4 = el + e2. rl + 3r2 + 1-3 + 1-4 is not a root. It is easy to know those vectors by adding rl, r3 or r 4 to the root of level five is not a root, because

Therefore, the D4 Lie algebra contains four roots of level one, three roots of level two, three roots of level three, one root of level four, and one root of level five. The rank of D4 is 4, and its order is 28. This result can be generalized to the Dc Lie algebra. Dc contains 2 l ( l - 1) roots: f ( e j + e k )

and (ej - e k ) . The rank of Dc is l , and its order is l ( 2 l - 1).

7.2 Irreducible Representations and the Chevalley Bases

* Discuss an irreducible representation of a simple Lie algebra where the representation matrices of Hj are diagonal:

where Im) is the basis state in the representation space, and the vector m in an l-dimensional space is called the weight of Im). The &dimensional space is called the weight space. A weight m is said to be multiple if the number of the basis states Im) satisfying Eq. (7.12) in the representation space is larger than one. Otherwise, it is a single weight. From Eq. (7.6) we have

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280 Problems and Solutions in Group Theory

Therefore, E, is called the raising operator and E-, the lowering operator, if a' is a positive root.

* In a given irreducible representation, one can construct a weight chain from a given weight m and a root d:

m + nd, - q 5 n 5 p , are weights, p 2 0, q 2 0,

m + ( p + 1)G and m - (q + 1)G (7.13)

are not the weights.

It can be proved that

I? (m/Z) = q - p = integer. (7.14)

The weight m' = m - GI' (m/d) is related to the weight m by a reflection with respect to the plane through the origin and orthogonal to the root G. m and m' have the same multiplicity and are called the equivalent weights. This reflection is called the Weyl reflection. All the mutually equivalent weights constitute the Weyl orbit, and their number is called the Weyl orbital size.

* The Chevalley bases H p , Ep and Fp, 1 5 p 5 l , are another set of bases for the generators in a simple Lie algebra, which can be expressed as the linear combinations of the Cartan-Weyl bases Hj and E,:

They satisfy the following commutation relations:

(7.16)

(7.17)

The generator E, corresponding to a non-simple root G can be calculated by the commutator (7.6). The main merit of the Chevalley bases is that three Chevalley bases H p , E, and Fp with the same subscript p span a subalgebra, denoted by dp, which is isomorphic onto the Lie algebra A1

and the Serre relations:

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Lie Groups and Lie Algebras 281

(the SU(2) group). In fact, they satisfy the same commutation relations as those satisfied by the generators 22'3, T+ and 2'- in A1:

* In the weight space, define a set of new bases w,, called the fundamental dominant weight, satisfying

r (w,/r,) = d;' (wp - r,) = (2w, - r,) / (r, - r,) = b,,. (7.18)

In comparison with Eq. (7.9), we can relate the fundamental dominant weight wp with the simple root r, by the Cartan matrix

L e

p=l u=l

In the bases w,, all components of the roots and the weights are integers:

(7.20)

The weight m' equivalent to m with respect to a simple root r, can be expressed as

m' = m - r,r (m/rp) = m - mpr,. (7.21)

* Rewrite the weight chain in Eq. (7.13) with respect to a simple root rp where m+prp is redefined as m:

m, m - r p , . . . , m - qr,, q = r(m/rp) = m,. (7.22)

The states corresponding to the weight chain constitute a (q+l)-multiplet of the subalgebra d,. The basis states in the irreducible representation space constitute some multiplets of the different subalgebras dp. If all weights in the weight chain (7.22) are single, the representation matrix entry of Fp for the multiplet is

The representation matrix of E, is the transpose of that of F,. If the multiple weight exists, the state in the multiplet of A, is a suitable com-

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282 Problems and Solutions in Group Theory

bination of the basis states, calculated by the commutators (7.16) and the ort honor ma1 condition.

* For an irreducible representation with a finite dimension, there is a state JM) with the highest weight M:

The highest weight M in an irreducible representation must be a single weight. Usually, an irreducible representation can be denoted by its highest weight M. A representation is called fundamental if its highest weight is the fundamental dominant weight. Any basis state Im) with the weight m in the irreducible representation space can be obtained from the highest weight state by the actions of the lowering operators. Thus,

Cm is called the height of the basis state Im). The height of the highest weight state is zero.

* Two irreducible representations are conjugate with each other if the lowest weight of one representation multiplied by -1 is equal to the highest weight of another representation. The irreducible representations is self- conjugate if its lowest weight multiplied by -1 is equal to its highest weight.

* A weight is called a dominant weight if its components are non-negative integers. Due to Eq. (7.14), the highest weight must be a dominant weight, and any dominant weight can be the highest weight in one irreducible repre- sentation. However, for a given irreducible representation, there may exist a few dominant weights with multiplicities, among them only one domi- nant weight is the highest weight of this representation. The dimension of an irreducible representation is equal to the sum of the multiplicities of the dominant weights multiplied by their Weyl orbital sizes. Therefore, the dominant weights play a very important role in the calculation for the basis states and the representation matrices of the generators.

* The method of the block weight diagram is effective for calculating the representation matrices of the generators in an irreducible representation. In a block weight diagram for an irreducible representation, each basis state corresponds to a block filled with its weight. For convenience, the negative component of the weight is denoted by a digit with a bar over it: Fi = -m. The blocks corresponding to the multiple weight are enumerated. The block

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Lie Groups and Lie Algebras 283

for the highest weight is located on the first row. The remaining blocks are arranged downwards as the height increases. The blocks for the weights with the same height are put in the same row.

Beginning with the highest weight, from each positive component my > 0, we draw m, blocks downwards. They constitute a weight chain of length (m, + 1) with respect to the simple root r,. The states corresponding to the weight chain constitute a multiplet of d,. Two neighbored blocks are related by a link denoting the application of the lowering operator F,. Usually, in the block weight diagram we connect two blocks by different links if their basis states are related by the different lowering operators. I t is better to indicate the corresponding representation matrix entry of the lowering operator behind the link. Usually, the matrix entry is neglected when it is equal to one.

If all weights in a weight chain are single, the representation matrix entries of the lowering operator F, between two basis states corresponding to the neighbored weights can be calculated by Eq. (7.23). If a multiple weight appears in the weight chain, the state in the multiplet may be a com- bination of the basis states. The combination as well as the representation matrix entries of F’ can be determined by Eq. (7.16) and the orthonormal condition.

When a dominant weight appears, we determine its multiplicity which is equal to the number of paths along which the dominant weight is obtained from the highest weight. If the representation matrix entry in a path is calculated to be zero, the multiplicity decreases. All weights equivalent to the dominant weight has the same multiplicity as the dominant weight has. Continue this method until all components of the weight are not positive. In the completed block weight diagram, the number of blocks with the same height first increases and then decreases as the height increases, symmetric up and down like a spindle.

* For a two-rank Lie algebra the weights can be drawn in a planar di- agram, called the planar weight diagram. The planar weight diagram of an irreducible representation is the inversion of that of its conjugate rep- resentation with respect to the origin. The planar weight diagram of a self-conjugate irreducible representations is symmetric with respect to the inversion. Note that the simple roots as well as the fundamental dominant weights are not required to be along the coordinate axes.

* The representation in whose space the basis is the generator of the Lie algebra is the adjoint representation. Therefore, the root coincides with

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284 Problems and Solutions in Group Theory

the weight in the adjoint representation, and the root space coincides with the weight space. To draw the block weight diagram or the planar weight diagram for the adjoint representation is another method for calculating all roots in a simple Lie algebra.

9. Draw the block weight diagrams and the planar weight diagrams of the representations (1,0), (0, l), (1, l), (3,0), and (0,3) of the A2 Lie algebra (the SU(3) group).

Solution. The Cartan matrix A and its inverse A-' of A2 are

2 -1 A - 1 = - ( 1 2 1 ) 3 1 2 *

The relations between the simple roots and the fundamental dominant weights are

rl = 2wl - w2, r2 = -w1+ 2w2,

w1 = (1/3) (2rl + ra) , w2 = (1/3) (rl + h).

The Weyl orbits of the dominant weights ( l , O ) , ( O , l ) , ( l , l ) , (3,O) and (0,3) respectively are:

(1,o) rl, ( i , i ) rZ, (o , i ) ,

(0,1) r2, ( 1 , i ) -% ( i ,o ) ,

(3,O) rl, (3,3) -% (0,3),

(0,3) 3 (3 , s ) -% (5,O).

The block weight diagrams of the representations ( l , O ) , (0, I), (1, l), (3 ,0) , and (0,3) are given in Fig. 7.2. In the diagrams, the single link means the action of Fl and the double link means that of F2.

First, we discuss the fundamental representation (1,O). From the high- est weight (1,O) we obtain a doublet of A1 with (i , l) . Then, from the weight (i, 1) we obtain a doublet of .Ap with ( 0 , i ) . The weight (0,T) con- tains no positive component. It is the lowest weight in the representation (1,O). The dimension of the representation ( 1 , O ) is three, where three weights are equivalent to each other. The representation matrix entries of

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Lie Groups and Lie Algebras

the lowering operators are

285

Second, we discuss the fundamental representation (0,l). From the highest weight (0 , l ) we obtain a doublet of d 2 with (1,i). Then, from the weight (1,i) we obtain a doublet of ,A1 with (T,O). The weight (T,O) contains no positive component. It is the lowest weight in the representation (0 , l ) . The dimension of the representation (0, l ) is three, where three weights are equivalent to each other. The representation matrix entries of the lowering operators are

It is worthy to notice that the block weight diagram of (0 , l ) can be obtained from that of ( 1 , O ) by turning upside down and changing the signs of all weights. In fact, the lowest weight (0,i) in the representation ( l , O ) , times by -1, is the highest weight in the representation (0,l) . Two repre- sentations are conjugate to each other.

0, i

Fig. 7.2 The block weight diagrams of some representations of A2

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286 Problems and Solutions in Group Theory

Third, we discuss the adjoint representation (1,l) of A2. From the highest weight (1,l) we obtain a doublet of A1 with (i,2) and a doublet of A2 with ( 2 , i ) . Then, from the weight ( 2 , i ) we obtain a triplet of A1 with (0,O) and (2, l), and from the weight (i, 2) we obtain a triplet of A2 with (0,O) and (1,2). There are two paths to the dominant weight (0, 0), which may be a double weight. We define two orthonormal basis states (0,O)l and (0, O),, where (0,O)l belongs to the triplet of A1 with (2, i) and (2, l), and (0, O), is a singlet of -41. The state (0,O) in the triplet of A2 with (i, 2) and (1,2) has to be a combination of (0, O), and (0,0)2:

Applying El F2 = F2E1 to I (1, 1), (T,2)), we have

Thus, a = m. Choosing the phase of 1(1,1),(0,0)2) such that b is a positive real number, we have b = d m = m. Let

Applying E2F2 = F2E2 + H 2 to l(1, l), (0,0)1), we have

Choosing the phase of l(1, l), (1,g)) such that c is a positive real number, we have c = @, and then, d = @. From the weight (1,T) we obtain a doublet of A1 with (i,i), and from the weight (2,l) we obtain a doublet of A2 with (1, i). Since the weight (i,i) is equivalent to (1, I), it is a single weight and is the lowest weight in the representation (1,l). The representation (1,l) is self-conjugate. The dimension of the representation (1,l) is eight containing six equivalent single weights and one double weight (0,O). The representation matrix entries of the lowering operators are

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We see from the block weight diagram of the adjoint representation (1,l) that the A2 Lie algebra contains three positive roots:

2 w l - w2 = r l , - w1+2w2 = 1-2, w1+ w2 = r1 +r2.

Fourth, we discuss the representation (3,O). From the highest weight (3,O) we obtain a quartet of A1 with (1, l), ( i , 2 ) and (3,3), where the representation matrix entries of F1 are a, 2, and a, respectively. The weight (1 , l ) is a dominant weight, which is a single weight because there is only one path to (1 , l ) lowered from the highest weight. All weights equivalent to (3,O) or (1 , l ) are single, too. Then, from the weight (1,l) we obtain a doublet of A2 with (2, i). From the weight (T, 2) we obtain a triplet of d 2 with (0,O) and ( l , z ) , and from the weight (2 , i ) we obtain a triplet of d1 with (0,O) and (2,l). There are two paths to the dominant weight ( O , O ) , which may be a double weight. We define two orthonormal basis states (0, O), and (0,0)2, where (0,O)l belongs to the triplet of d1 with ( 2 , i ) and (2, l), and (0, O), is a singlet of d1. The state (0,O) belonging the triplet of A2 with ( i , 2 ) and (1,2) has to be a combination of (0, O), and (0,0)2:

F21(3,0),(T,2)) =a1(3,0),(0,0)1) +b1(3,0),(0,0)2), a2 + b 2 = 2 .

Applying E1F2 = F2E1 to 1(3,0), (i, 2)), we have

~ i 1 ( 3 , 0 ) , (T,2)) = d N 3 , o), (2 , i ) )

= F2Ei1(3,0), ( i , 2 ) ) = 2F21(3,0), (1,1)) = 21(3,0), (2,T))-

Thus, a = fi and b = 0. It means that the weight (0,O) is single. The basis state ( (3 ,0) , (0,0)2) does not exist, and ((3,0), (0, O), ) can be denoted

From the weight (1,2) we obtain a doublet of A1 with (i,i). From the weight (3,3) we obtain a quartet of d2 with (2, 1), (7, T), and (0, Z), where the representation matrix entries of F2 are a, 2, and a, respectively. The weights (2 , l ) and (T, 1) are single because they are equivalent to (1 , l ) . Thus, the state (2 , l ) coincides with the state in the triplet of A1 with (2,T) and (0, 0), and the state (T, T) coincides with the state in the doublet

by i(3, o), (0,o)).

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288 Problems and Solutions in Group Theory

of d1 with (1,2). The block weight diagram of the representation (3,O) is complete. The dimension of the representation (3,O) is ten, containing three single dominant weights (3,0), ( 1 , l ) and (0, 0), with the Weyl orbital sizes 3, 6 and 0, respectively. The representation matrix entries of the lowering operators are

The lowest weight (0,z) in the representation (3,0), multiplied by -1, is the highest weight of the representation (0,3). Therefore, the representa- tion (0,3) is conjugate with the representation (3,0), and its block weight diagram is upside down of that of (3,0), where all weights change their signs.

t"'

(i,i)

Fig. 7.3 The planar weight diagrams of some representations of A2.

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Lie Groups and Lie Algebras 289

In the rectangular coordinate frame, the simple roots and the funda- mental dominant weights of A2 are

Physically, people prefer to put el along the ordinate axis, and e2 along the abscissa axis for A2. The planar weight diagrams for the representations ( l , O ) , (0,1), (1,1), (3,0), and (0,3) are given in Fig. 7.3. Generally, the planar weight diagram for the representation ( 0 , M ) of A2 is a straight regular triangle where all weights are single, and that of ( M , 0) is a regular triangle upside down. The planar weight diagram of ( M , M ) of A2 is a regular hexagon, where the weights on the boundary are single, and the multiplicity increases as the position of the weight goes to the origin.

10. Draw the block weight diagrams and the planar weight diagrams of two fundamental representations and the adjoint representation (2,O) of the C2 Lie algebra.

Solution. The Cartan matrix A and its inverse A-' in C2 is

2 -2 A - l = - ( 1 2 2 ) 2 1 2 -

The relations between the simple roots and the fundamental dominant weights are

rl = 2 w l - w2, r2 = - 2 w l + 2w2,

The Weyl orbits of the dominant weights (1, 0)) (0, 1)) and (2,O) respectively are:

(1,o) rl, (i, 1) -3 (1,i) -% ( i ,o) , (0, 1) r2, ( 2 , i ) rl, (2,i) -3 (o , i ) , (2,O) rl, (2,2) -3 ( 2 3 I'l, (2,O).

Their Weyl orbital sizes are all 4. The block weight diagrams of the repre- sentations (1,0), (0, l), and (2,O) are given in Fig. 7.4. In the diagrams, the single link means the action of F1 and the double link means that of F2.

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290 Problems and Solutions in Group Theory

Fig. 7.4 The block weight diagrams of some representations of C2.

First, we discuss the fundamental representation ( 1 , O ) . From the high- est weight ( 1 , O ) we obtain a doublet of d1 with (i,1). Then, from the weight (7,l) we obtain a doublet of A2 with (1,i). From the weight (1,i) we obtain a doublet of dl with ( i , O ) , which contains no positive component. The dimension of the representation ( 1 , O ) is four, where four weights are equivalent to each other. The representation matrix entries of the lowering operators are

Second, we discuss the fundamental representation (0,l) . From the highest weight (0 , l ) we obtain a doublet of da with (2 , i ) . Then, from the weight (2,T) we obtain a triplet of d1 with (0,O) and (2,l). The representation matrix entries of F1 in the triplet are both a. The weight (0,O) is a dominant weight. There is only one path to (0,O) lowered from the highest weight, so (0,O) is single. From the weight (2,l) we obtain a doublet of A2 with ( O , i ) , which contains no positive component. The five-dimensional representation (0 , l ) contains two single dominant weights (0 , l ) and (0,O) with the Weyl orbital sizes four and one, respectively. The representation matrix entries of the lowering operators are

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At last, we discuss the adjoint representation (2,O). From the highest weight (2,O) we obtain a triplet of A1 with (0,1> and (2,2), where the representation matrix entries of F1 are both a. The dominant weight (0 , l ) is single because there is only one path to (0,l) lowered from the highest weight. From the weight (0 , l ) we obtain a doublet of A2 with (2, i ) . From the weight (2,2) we obtain a triplet of A2 with (0,O) and (2,2), and from the weight (2, i) we obtain a triplet of A1 with (0,O) and (2,l). There are two paths to the dominant weight (O,O), which may be a double weight. We define two orthonormal basis states (0, O), and (0, O),, where (0,O)l belongs to the triplet of A1 with (2, i ) and (2, l), and (0,0)2 is a singlet of -41. The state (0,O) in the triplet of A2 with (2,2) and (2,T) has to be a combination of (0,O)l and (0,0)2:

Thus, a = 1. Choosing the phase of ((2,0), (0,0)2) such that b is a positive real number, we have b = -\/2--i-" = 1. Applying E2F2 = F2E2 + H2 to P, 01, (O,O)l), we have

Choosing the phase of 1(2,0), (2,2)) such that c is a positive real number, we have c = 1, and then, d = 1. All weights in the representation (2,O) are single except for (0,O). From the weight (2,2) we obtain a triplet of d1 with (0,T) and (z,O), where the representation matrix entries of F1 are both a. From the weight (2,l) we obtain a doublet of A2 with (0,T). Since the weight (0,T) is single, it also has to belong to the triplet of -41 with (2,2) and (3,O). Thus, the block weight diagram of the representation (2,O) is complete. The representation (2,O) contains two single dominant weights (2,O) and (0,1), and one double dominant weight (0,O). The dimension of the representation (2,O) is 1 x 4 + 1 x 4 + 2 x 1 = 10. The representation matrix entries of the lowering operators are

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292 Problems and Solutions in Group Theory

We see from the block weight diagram of the adjoint representation (2,O) that the C2 Lie algebra contains four positive roots:

2wl - w2 = r1,

w2 = rl + 1-2,

-2wl+ 2w2 = 1'2,

2wl = 2rl + r 2 .

In three representations (1,0), (0, 1), and (2,O) of C2, the lowest weights are equal to the highest weight times by -1, respectively. It means that three representations are all self-conjugate.

In the rectangular coordinate frame, the simple roots (see Problem 6) and the fundamental dominant weights of C2 are

Thus, we obtain the planar weight diagrams for the representations ( l , O ) , (0 , l ) and (2,O) as follows.

11. Draw the block weight diagrams and the planar weight diagrams of three representations ( O , l ) , ( l , O ) , and (0,2) of the exceptional Lie algebra G2.

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Lie Groups and Lie Algebras 293

Solution. The Cartan matrix A and its inverse A-l in G2 are

A = ( -3 -'), 2 i ) . The relations between the simple roots and the fundamental dominant weights are

rl = 2wl - 3w2, 1'2 = -w1+ 2w2,

The Weyl orbits of the dominant weights (0, l ) , (1, 0), and (0,2) respectively are:

( o , i ) -3 ( i , i ) -% ( i , 2 ) r2, ( i ,2 ) rl, ( i , i ) -3 ( o , i ) , (1,o) -3 ( i ,3 ) -% (2,s) -3- (2,3) r2, (1,s) rl, ( i ,o) ,

(0,2) rz, ( 2 3 1p1, (2,4) r2, (2,;I) rl, (2,2) 3 (0)Z).

Their Weyl orbital sizes are all 6. First, we discuss the fundamental representation (0 , l ) . From the high-

est weight ( 0 , l ) we obtain a doublet of d 2 with (1 , i ) . Then, from the weight (1, i ) we obtain a doublet of d1 with ( i ,2 ) . From the weight (i, 2) we obtain a triplet of d 2 with (0,O) and (1)z). The representation matrix entries of F 2 in the triplet are both a. The weight (0,O) is a dominant weight. There is only one path to (0,O) lowered from the highest weight, so (0,O) is single. From the weight (1,Z) we obtain a doublet of A1 with (1, l ) . From the weight (1,l) we obtain a doublet of d 2 with (O,i), which contains no positive component. The dimension of the representation (0 , l ) is seven, where six weights are equivalent to each other and (0,O) is equivalent only to itself. The representation matrix entries of the lowering operators are

F21(0, (071)) = KO, 11, (l,i>>,

F21(0,1>, (i, 2)) = JZI(0, I) , ( O , O ) ) , FZl(0,1), (0,O)) = J Z l ( O , l > , (1,2)>,

F11(0,1>, (1,W = KO, 11, ( i ,1 )> ,

FlI(0, 11, ( l , i ) > = KO, 11, m),

F2I(O, 11, (i, 1)) = I(O,l),(O,i)).

Second, we discuss the fundamental representation ( 1 , O ) . From the highest weight (1,O) we obtain a doublet of d1 with ( i ,3 ) . Then, from the weight ( i , 3 ) we obtain a quartet of d 2 with (0, l), (1 , i ) and (2,s). The representation matrix entries of F1 in the quartet are a, 2, and arespectively. The weight (0 , l ) is a dominant weight. There is only one path to ( 0 , l ) lowered from the highest weight, so ( 0 , l ) is single. All weights

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294 Problems and Solutions in Group Theory

equivalent to (1,O) or (0,l) are single, too. From the weight (1, T) we obtain a doublet of A1 with (i, 2). From the weight (2,z) we obtain a triplet of A1 with (0,O) and (2,3), and from the weight (i, 2) we obtain a triplet of A2 with (0,O) and (1,2). There are two paths to the dominant weight (O,O) , which may be a double weight. We define two orthonormal basis states (0,O)l and (0,0)2, where (0,O)l belongs to the triplet of A1 with (2,s) and (2,3), and (0,0)2 is a singlet of d1. The state (0,O) in the triplet of A2 with (i,2) and (1,g) has to be a combination of (0,O)l and (0,0)2:

Thus, a = m. Choosing the phase of I(l,O), (0,0)2) such that b is a positive real number, we have b = d m = m. Applying E2F2 = F2E2 + H 2 to 1(1,0),(0,0)1), we have

Choosing the phase of 1(1,0), (1,2)) such that c is a positive real number, we have c = m, and then, d = m. The remaining part of the block weight diagram is easy to calculate. We only list the results in the block diagram and the representation matrix entries of the lowering operators in the following. The representation (1,O) contains two single dominant weights (1,O) and (0,l) with the Weyl orbital size 6 and one double domi- nant weight (0,O). The dimension of (1,O) is 1 x 6 + 1 x 6 + 2 x 1 = 14.

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Lie Groups and Lie Algebras 295

The representation ( 1 , O ) is the adjoint representation of G2. F'rom the block weight diagram we obtain all positive roots of G2 as follows:

At last, we discuss the representation (0,2). From the highest weight (0,2) we obtain a triplet of d 2 with ( 1 , O ) and (2,2), where the represen- tation matrix entries of F 2 are both a. The weight ( 1 , O ) is a dominant weight. There is only one path to ( 1 , O ) lowered from the highest weight, so ( 1 , O ) is single. From the weight ( 1 , O ) we obtain a doublet of d1 with (i, 3). F'rom the weight (2,2) we obtain a triplet of A1 with (0 , l ) and (2,4), and from the weight (i,3) we obtain a quartet of d 2 with (0, l), (1, i), and (2,s). There are two paths to the dominant weight (0,1), which may be a double weight. We define two orthonormal basis states (0,1)1 and (0,1)2, where (0, l ) l belongs to the triplet of d1 with (2,2) and (2,4), and (0,1)2 is a singlet of d1. The state (0 , l ) in the quartet of d2, composed of (T, 3), ( O , l ) , (l,i'), and (2,3), has to be a combination of (0,O)l and (0,0)2:

Applying E1F2 = F2E1 to I(0,2), (T,3)), we have

Thus, a1 = 1. Choosing the phase of [ ( O , 2)) (0,1)2) such that a2 is a pos- itive real number, we have a2 = d m = a. Since the weight (1,i) is equivalent to the dominant weight (0,1), it is a double weight. De- fine two orthonormal basis states I(0,2), (1,i)l) and I(0,2), ( l , i ) 2 ) , where I(0,2), (1,i)l) is proportional to F21(0,2), (0,1)1) with a positive real coef- ficient b l :

Applying E2F2 = F2E2 + H2 to [(O, 2), (0,1)1) and I(0,2), (0,1)2), we have

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296 Problems and Solutions in Group Theory

F21(0,2), ( 1 ) i ) i ) = b41(0,2), (2,3)), F21(0,2), ( l , i ) 2 ) = b51(0,2), (2,Z)).

Applying E2F2 = F2E2 + H2 to 1(0,2), (1 , i j l ) , we have

~%F21(0,2), ( 1 , i ) i ) = bz1(0,2), (1 , i ) i ) hb51(0,2), (1,i)z) = (F2E2 + H2) I(0,2), ( 1 , i ) l )

= (2 + 1 - l > l ( O , 2), (1,i)l) + JZI(O,2), (hQ2).

Choosing the phase of I(0,2), (2,s)) such that b4 is a positive real number, we have b4 = fi and b5 = 1.

F'rom the weight (2,4) we obtain a quintet of d 2 with ( i ,2 ) , (O,O) , ( l ,T) , and (2,4), where the representation matrix entries of F 2 are 2, &, fi, and 2, respectively. Since the weights (T, 2) and (1,T) are equivalent to the dominant weight (0, l ) , they are the double weights. We define the orthonormal basis states (T, 2)1, (i, 2)2, (1,2)1, and (1,2)2, where (i, 2)l and (1,2)1 belong to the quintet of A2, and ( i , 2)2 and (1,2), belong to a triplet of A2 with (0,O). Let

FlI(0, 21, (1J)l) = c11(0,2), ( i , 2)l) + c21(0,2), (i, 2)2), c; + c; = 1,

F11(o,2)7(17i)2) = C3I(o72),(i,2)i) +c41(0,2),(i,2)2), Ci +Ci = 1-

Applying E2F1 = F1E2 to /(0,2), (1 , i ) l ) and 1(0,2), (1 , i )2) , we have

E 2 J N m (1,i)l) = 2c11(0,2),(2,4))

Thus, c1 = 1 and c2 = c3 = 0. Choosing the phase of I(0,2), (i, 2)2) such that c4 is a positive real number, we have c4 = 1.

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Lie Groups and Lie Algebras 297

In addition to a quintet and a triplet of d 2 , which both contain the weight ( O , O ) , there is another triplet of d1 containing (0,O). The triplet of A1 consists of (2,3), (O,O) , (2,3). There are three paths to the dominant weight (0,O) lowered from the highest weight, so (0,O) may be a triple weight. We define three orthonormal basis states (0, O),, (0, O),, and (0, O),, where (0,O)l belongs to the quintet of A2 composed of (2,4), (T,2)1, (0,0)1, (1,2)1 and (2,3), (0,0)2 belongs to the triplet of A2 composed of ( i , 2 ) 2 ,

(0,0)2 and (l,a),, and (O,O)3 is a singlet of A2. Thus, we have

Let

Fl

Fl

Fl

Applying E2F1 = FIE2 to 1(0,2), (2,3)), we have

Thus, dl = and d2 = m. Choosing the phase of I(0,2), (0,0)3) such that d3 is a positive real number, we have d3 = m. Applying EIFI = FlEl + HI to I(0,2), (0, 0)3), we have

Choosing the phase of I(0,2), (2,3)) such that d6 is a positive real number, we have d6 = m, and then d4 = and d5 = m. It is easy to calculate the remaining part of the block weight diagram.

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298 Problems and Solutions an Group Theory

q 0, i

Ji Ji

lT,oI

Fig. 7.5. The block weight diagrams of some representations of G2 .

0

0

The results for the representation (0,2) as well as ( 1 , O ) and (0 , l ) are given in the block weight diagrams (Fig. 7.5). In the diagrams, the single link means the action of Fl and the double link means that of F2. The planar weight diagrams are also given, where the simple roots and the fundamental dominant weights of GZ in the rectangular coordinate frame

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Lie Groups and Lie Algebras 299

are

7.3 Reduction of the Direct Product of Representations

* The Clebsch-Gordan (CG) series for the reduction of the direct prod- uct M1 x M2 of two irreducible representations of a simple Lie algebra can be calculated by the dominant weight diagrams. The main point is to list all dominant weights N contained in the product space, to count their multiplicities n(M1 x M2,N) in the space, to calculate their Weyl orbital sizes OS(N), and to calculate the multiplicities n(M,N) of those dominant weights N in the representations M contained in the CG series, respectively. First, we list OS(N) and n(M1 x M2,N) of all dominant weights N contained in the product space at the first column of the di- agram. Let M1 + M2 - N = Cj cjrj, Cj cj is called the height of the dominant weight N. Arrange the dominant weights N downwards as the height increases. The height of M1 +M2 is zero and is listed in the first row. The representation MI+ M2 is contained in the CG series with multiplicity one. Then, the second column of the diagram lists the dominant weights N in the representation M1 + M2 and their multiplicities n(M1 + M2, N) (as the superscript). The difference n(M1 x M2,M’) - n(M1 + M2,M‘) for the dominant weight M’ with the next smallest height is equal to the multiplicity of the representation M‘ contained in the CG series. If it is not zero, the third column of the diagram lists the dominant weights N in the representation M‘ and their multiplicities n(M’, N) (as the superscript). Continue the calculation to determine the multiplicities of the representa- tions N in the CG series by comparing n(M1 + Ma, N) with the sum of the multiplicities of N in the representations, which we have known to be contained in the CG series, until all dominant weights N are calculated. At last, calculate the dimensions d(M‘) of the representations M’ contained in the CG series by

d(M’) = OS(N)n(M’,N), (7.26) N

to see whether their sum is equal to the dimension of the direct product

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300 Problems and Solutions an Group Theory

space:

d(M1 x M2) = C OS(N)n(M1 x M2,N) = C d(M‘). (7.27) N M‘

* Denote by JM1, ml)JM2, ma) the basis state before reduction, and Usually, the basis state by llM,m) the basis state after reduction.

IM1, ml)IM2, ma) is briefly written as Iml)lm2) for simplicity.

where CmM,lmM,2,Mm are the Clebsch-Gordan (CG) coefficients. Applying the lowering operator Fp to the basis state, we have:

In terms of Eq. (7.29) we are able to calculate the expansions of all basis states in the representation MI+ M2. Then, by the orthonormal condition or by Eq. (7.24) we calculate the highest weight state in the representation M’ contained in the CG series. At last, we calculate the expansions of all basis states in the representation M’ by Eq. (7.29).

When M1 = M2, the expansion of each basis state can be symmetric or antisymmetric with respect to the permutation between two basis states Iml)lm2) and Im2)lml). For simplicity, we denote by (sym. terms) or (antisym. terms) the remaining terms obtained from the preceding terms by the permutation, multiplied by 1 or -1, respectively.

12. Calculate the Clebsch-Gordan series and the Clebsch-Gordan coeffi- cients for the direct product representation ( 1 , O ) x ( 1 , O ) of the C2 Lie algebra.

Solution. In Problem 10 we have calculated the block weight diagrams of the representations ( l , O ) , (0 , l ) and (2,O) of C2. The representation ( 1 , O ) contains only one single dominant weight ( 1 , O ) . The representation (0 , l )

The main steps of the calculation are as follows. First, the highestweight state in the representation M1 + M2 is

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Lie Groups and Lie Algebras 30 1

contains two single dominant weights (0 , l ) and (0,O). The representation ( 2 , O ) contains two single dominant weights (2,O) and (0, 1), and one double dominant weight (0,O). The Weyl orbital sizes of the weights (2,0), (0, l), and (0,O) are 4,4, and 1, respectively. The linearly independent basis states for the dominant weights in the product space are

Dominant Weight Basis State No. of States

Thus, we draw the diagram of the dominant weights as follows.

0s x n

4 x 2

1 x 4

16 = 1 0 + 5 + 1

Now, we calculate the expansions of the basis states. First, for the representation (2,0), from Eq. (7.30) we have

From the block weight diagram of (2,O) of C2 (see Fig. 7.4) and Eq. (7.29), we have

According to the orthonormal condition, we obtain the highest weight state in the representation (0, l )

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302 Problems and Solutions in Group Theory

Applying the lowering operators to the basis states continuously, we obtain

For the basis states in the representation (0 , l ) we have

The expansion of the basis state in the representation (0,O) can be calculated by the orthonormal condition. But now, we prefer to calculate it by the condition (7.24) for the highest weight state. Let

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Lie Groups and Lie Algebras 303

Applying El and E2 to it, we have

Thus, a = -b = c = -d. After normalization we obtain

It can be checked that four basis states with the weight (0,O) are orthogonal to each other. In the permutation between lml) and Im2), the representa- tion (2,O) is symmetric, but (0 , l ) and (0,O) are antisymmetric.

13. Calculate the Clebsch-Gordan series and the Clebsch-Gordan coeffi- cients for the direct product representation (0, l ) x (0 , l ) of the C2 Lie algebra.

Solution. One may solve this problem following the way used in the pre- ceding problem. But now, we prefer to calculate the block weight diagram and the CG coefficients with another method, where the CG coefficients for (0 , l ) x (0,l) and the block weight diagram for (1,l) are calculated at the same time.

From the block weight diagram of (0 , l ) given in Fig 7.4, we know that the linearly independent basis states for the dominant weights in the prod- uct space are as follows, where we neglect the basis state which can be obtained by the permutation of two states Iml) and Im2).

We have calculated the block weight diagrams of the representations (0 , l ) and (2,O). However, we did not calculate the block weight diagram of the representation (0,2). The representation (0, l ) contains two single dominant weights (0 , l ) and (0,O). The representation (2,O) contains two single dominant weights (2,O) and (0, l ) , and one double dominant weight (0,O). The Weyl orbital sizes of the weights (2,0), (0, l), and (0,O) are 4,

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304 Problems and Solutions an Group Theory

4, and 1, respectively. The Weyl orbital of the weights (0,2) contains four weights: (0,2), (4,?), (3,2), and (O,?).

F'rom Eq. (7.30) we have the expansion of the highest weight state in the representation (0,2)

From the highest weight (0,2) we have a triplet of d2:

The dominant weight (2,O) is single in the representation (0,2) because there is only one path to it lowered from the highest weight. However, the multiplicity of the weight (2,O) in the product space is 2. It means that one representation (2,O) is contained in the CG series for (0 , l ) x (0,l) .

From the weight (2,O) we have a triplet of A1,

The dominant weight (0 , l ) is single in the representation (0,2) because there is only one path to it lowered from the highest weight. The repre- sentation (2,O) contains a single dominant weight (0, l), and the product space contains two basis states with (0,l) . It means that the representation (0 , l ) is not contained in the CG series for (0 , l ) x (0 , l ) .

F'rom the weight (4,2) we have a quintet of d1 with (2, i ) , (O,O) , (2, l), and (4,2), and from the weight (?,2) we have a triplet of d 2 with (0,O) and (2,?). There are two paths to the dominant weight (0,O) lowered from the highest weight, so (0,O) may be a double dominant weight in the representation (0,2). We define two orthonormal basis states for the weight (0,O). "(0,2), (0,O)l) belongs to the quintet of d1, and 11(0,2), (0,0)2) has to belong to a singlet of St1 because the weight (2,T) is a single weight.

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Lie Groups and Lie Algebras 305

We obtain the inner product between this basis state and Il(0,2), (0,O)l) to be m. Thus,

After normalization we have a = m. In fact, a can also be calculated by the property of a triplet: u2 + 1/3 = 2. Hence,

It is easy to calculate the remaining expansions of the basis states in the representation (0,2). We list the results as follows.

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306 Problems and Solutions an Group Theory

In addition, we have

From the above calculation, we are able to draw the dominant weight diagram for (0 , l ) x (0, l ) of Cz and the block weight diagram and the planar weight diagram of the representation (0,2) as follows.

0s x n

4 x 1

4 x 2

4 x 2

1 x 5

25 = 1 4 + 1 0 + 1

Now, we calculate the expansions of the basis states in the representation (2,O). Since 11(2,0), (2,O)) is orthogonal to Il(0,2), (2,0)), we have

The remaining basis states are

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Lie Groups and Lie Algebras 307

We may calculate the basis state I l(O,O), (0,O)) by the orthonormal con- dition to the preceding four basis states with the weight (0,O). Here we calculate it by Eq. (7.24). Let

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308 Problems and Solutions an Group Theory

From Eq. (7.29) we have

Thus, a = -b = c = -d = e. After normalization, we have

It can be checked that five basis states with the weight (0,O) are orthogonal to each other. In the reduction of the direct product representation (0 , l ) x (0 , l ) of C2, the representations (0,2) and (0,O) are symmetric with respect to the permutation between Iml) and Imz), and the representation (2,O) is antisymmetric.

14. Calculate the Clebsch-Gordan series for the direct product represen- tation (1,O) x (0 , l ) of the Ca Lie algebra and the expansion for the highest weight state of each irreducible representation in the CG series.

Solution. From the block weight diagrams for the representations ( 1 , O ) and (0 , l ) given in Fig 7.4, we know that the linearly independent basis states for the dominant weights in the product space are as follows.

There are a few methods to determine the dominant weights and their multiplicities contained in an irreducible representation. One method is to draw the upper part of the block weight diagram until all dominant weights in the representation appear. Another method is to look over the table book, for example, the table book [Bremner et al. (1985)] contains the information for the simple Lie algebras with rank less than 13. Anyway, we know that the representation (1, l ) contains one single dominant weight (1 , l ) and one double dominant weight (1,O). The representation (1,O) contains only one single dominant weight (1,O). The Weyl orbital size of the weight (1,O) is 4. The Weyl orbital of the weights (1 , l ) contains eight weights: (1, l ) , (1, 2)) (3 , i ) ) (3,2), (3,2), (3, l ) , (1,2) and (i,i). NOW, we

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Lie Groups and Lie Algebms 309

are able to draw the diagram of the dominant weights for the direct product representation (1,O) x (0,I) of Ca.

0s x n 8 x 1

4 x 3

20 = 1 6 + 4

Fig. 7.6. The diagram of the dominant weights for (1,O) x ( 0 , l ) of C2.

The highest weight states of the representations (1, l) and (1,O) can be calculated by Eqs. (7.30) and (7.24). The results are

15. Calculate the CG series for the following direct product representations of the Ga Lie algebra and the expansion for the highest weight state of each irreducible representation in the CG series:

where the dimensions d(M) of some representations M, the Weyl or- bital sizes OS(M) of M, and the multiplicities of the dominant weights in the representation M are listed in the following table.

6

6 6

12 6 6

1 1

2 1 1 3 2 1 1 4 4 2 2 1 5 4 3 2 1 1 5 3 3 2 1 1 1

Solution. In Problem 11 we have studied the block weight diagrams of the

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310 Problems and Solutions an Group Theory

representations (0 , l ) and ( 1 , O ) of G2. The representation (0 , l ) contains seven single weights

The representation ( 1 , O ) contains 12 single weights and one double weight:

In the following, we first list the linearly independent basis states Iml) Im2) for each dominant weight in the product representation space, where we neglect the basis state which can be obtained by the permutation of two states Iml) and Im2). Then, comparing the number of the basis states of the dominant weight in the product representation space with the multiplicities of the dominant weight in the relevant representations, we are able to calculate the Clebsch-Gordan series. The expansion for the high- est weight state of each irreducible representation in the CG series can be calculated by Eq. (7.24). The results are listed in the following.

a) The product representation (0 , l ) x (0 , l ) .

Dominant Weight Basis State No. of States

The expansions for the highest weight states of the irreducible represen-tations in the CG series are

In the permutation between |m1) and |m2), the representation (0,2)and(0,0) are symmetric, but (1,0) and (0,1 are antisymmetric.

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Lie Groups and Lie Algebras 311

O S x n

49 = 2 7 + 14 + 7 + 1

(0,1) x (0, I)= (0,2) + (170) + (0,1) + (0,O)

Fig. 7.7. The dominant weight diagram for (0 , l ) x ( 0 , l ) of G2.

b) The product representation (0, l ) x (1,O).

Dominant Weight Basis State No. of States

The expansions for thje highest weight states of the irreducible represen-tations in the CG series are

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312 Problems and Solutions in Group Theory

Fig. 7.8. The dominant weight diagram for ( 0 , l ) x ( 1 , O ) of G2.

c) The product representation (1,O) x (1,O).

The expansions for the highest weight states of the irreducible represen- tations in the CG series are

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Lie Groups and Lie Algebras 313

In the permutation between Iml) and Im2), the representation (2,0), (0,2) and (0,O) are symmetric, but (0,3) and ( 1 , O ) are antisymmetric.

Fig. 7.9. The dominant weight diagram for ( 1 , O ) x ( 1 , O ) of G2.

16. Calculate the Clebsch-Gordan series for the direct product represen- tation ( O , O , 0 , l ) x ( O , O , 0 , l ) of the F4 Lie algebra and the expansion for the highest weight state of each irreducible representation in the Clebsch-Gordan series, where the dimensions d( M) of some represen- tations M, the Weyl orbital sizes OS(M) of M, and the multiplicities of the dominant weights in the representation M are listed in the fol- lowing table.

0s 1

24

24

96

24

Solution. F’rom the Cartan matrix of F4 we obtain the relations between the simple roots rj and the fundamental dominant weights wj

The representation (0, 0, 0 , l ) of F4 contains 24 equivalent single weights

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314 Problems and Solutions in Group Theory

and one doublet weight ( O , O , 0,O):

First, we list the linearly independent basis states Im1)lmz) for each dom- inant weight in the product representation space (0, 0, 0 , l ) x (0, 0, 0, l ) , where we neglect the basis state which can be obtained by the permutation of two states Iml) and Im2).

Dominant Weight Basis States

No. of States

Then, comparing the number of the basis states of the dominant weight in the product representation space with the multiplicities of the dominant weight in the relevant representations, we are able to calculate the Clebsch- Gordan series, as given in the dominant weight diagram.

The expansion for the highest weight state of each irreducible represen- tation in the CG series can be calculated by Eq. (7.24). The results are listed in the following.

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Lie Groups and Lie Algebras 315

antisymmetric.

0s x n 24 x 1

96 x 2

24 x 6

In the permutation between |m1) and |m2), the representation (0,0,0,2),(0,0,0,1) and (0,0,0,0) are symmettric, but (0,0,1,0) and (1,0,0,0) are

Fig.7.10…The dominant weight diagram for (0,0,0,1)´(0,0,0,1) of F4.

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Chapter 8

UNITARY GROUPS

8.1 The SU(N) Group and Its Lie Algebra

* The set of all N-dimensional unimodular unitary matrices, in the mul- tiplication rule of matrices, constitutes the SU(N) group. It is a simply- connected compact Lie group with order ( N 2 - 1) and rank ( N - 1). The generators in its self-representation are traceless hermitian

6cd [2a(a - 1)]-1’2 when c < a,

matrices:

(8.1)

-&d [(a - 1 ) / ( 2 ~ ) ] ” ~ when c = a, 0 when c > a,

2 L a L N ,

where the subscripts a and b are the ordinal indices of the generator, while c and d are the row and column indices of the matrix. The matrix Ti:) is symmetric with respect to both ab and cd, but Tit) is antisymmetric. TA3) is diagonal. Usually, three types of the generators can be enumerated uniformly in the following order 2’1 = Ti:), T2 = Ti;), T3 = Ti3) , T4 = Ti:),

Satisfy the normalization condition: TAT') = 6,4B/2. Any element u in SU(N) can be expressed in the exponential matrix form:

T5 = Ti:), Ts = TJi ) , 2-7 = T23 (2) , Tg = Ti3) , and so on. The generators TA

N2-1 N

317

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318 Problems and Solutions an Group Theory

* Let N E l + 1, and define ( 1 + 1) vectors V a in the -!-dimensional space:

v1=

vz =

v 3 =

-/=, 0, . . . , ‘ ‘ 8 ) 0 , 1 1 1 .a={ d- 2 l ( l + 1 ) ’ @-(cq’ ”” 2a

) O , . . . . * . . . . . . . . . .

0, . . . . . . . . . . . . . . .

1

v N = { - d X 2 ( l + 1)’ 0,

The Cartan-Weyl bases for the generators in the self-representation of SU(N) can be expressed as

03-41 Hj = &T,(!\+,, Esab = Tab (1) + iTit),

when l! - j + 2 = a,

Ezob when a < l - j + 2 < b,

1-j+2

1

(e - j + 2)(-!- j + 1)

[ H j , Ea,,l =

when l - j + 2 = b,

when l - j + 2 < a , or l - j + 2 > b ,

where a < b. Namely,

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Unatary Groups 319

. I

When a < b, Ezab corresponds to the positive root Gab = &{V, - vb} = rP. Therefore, the Lie algebra of the S U ( l + 1) group is Ae. The

largest root in At is c3 = c/.L=l rP = w1 + we, which is the highest weight of the adjoint representation of At .

1

1. Calculate the anti-commutation relations for the generators TA in the

Solution. The anti-commutator for two generators TA and TB is hermitian, but not traceless. Let

self-representation of SU ( N ) .

P=l

Taking the trace, we have cN = ~ T ~ ( T A T B ) = B A B . Thus,

N2-1 'I

2. Calculate the structure constants and dABD in Eq. (8.7) for the su(3)

Solution. Write the commutation relations for the generators TA in the self-representation of SU(N):

group.

where CAj$' are the structure constants. Multiplying Eqs. (8.7) and (8.8) by To and taking the trace, we have

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320 Problems and Solutions in Group Theory

ABD

CA,” are totally antisymmetric with respect to three indices, but ~ A B D

are totally symmetric. Substituting the generators TA into Eq. (8.9), we obtain the nonvanishing structure constants CA,” and the coefficients ~ A B D

of SU(3) as listed in Table 8.1.

123 147 156 246 257 345 367 458 678

Table 8.1 The nonvanishing structure constants CAf and the coefficients ~ A B D of SU(3).

ABD 118 146 157 228 247 256 338 344

I ABD I 355 366 377 448 558 668 778 888 I

3. Calculate the Killing form for SU(N) group.

Solution. The generator Iidj in the adjoint representation is related to the structure constant CA,D by

Thus, the Killing form gAB is

(8.10)

(8.11)

Being an ( N 2 - 1)-dimensional matrix, g is commutable with any generator Iidj in the adjoint representation of SU(N):

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Due to the Schur theorem, g A B is a constant matrix, the constant c for the SU(3) group by taking A = 3:

= C ~ A B . Calculate

Note that T4 = Ti:) and TG = Ti:). When generalizing the result to the SU(N) group, the subscript 3 in Ti:) and Ti;) becomes a, 3 5 a 5 N , so that the number in the curve brackets becomes 1 + 2 ( N - 2)/4. Therefore,

g A B = - N ~ A B , for SU(N). (8.12)

8.2 Irreducible Tensor Representations of SU( N )

* The Chevalley bases for the generators in the self-representation of SU(N) are

(EP)cd = (Erp)cd = 'W6(p+l)d, (Fp)Cd = (E-'p)cd = d(p+I)cdpd,

1 5 p 5 e = N - 1. (8.13)

Note that for a given p, three generators H,, Ep and Fp constitute the subalgebra Ap, which is isomorphic onto the A 1 algebra.

* A covariant tensor T a l . , . a n of rank n with respect to SU(N) transforms as

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322 Problems and Solutions an Group Theory

Since the matrix entries ?&idi in Eqs. (8.14) and (8.15) are commutable, the symmetry among the indices of a tensor keeps invariant in the SU(N) transformation. Therefore, the tensor space is reducible, and can be reduced into the direct sum of some irreducible tensor subspaces with the definite permutation symmetry among the indices.

* In order to describe the symmetry of the indices of a tensor, we define the permutation operators R for the tensor. R is a linear operator by which a tensor T is transformed into a new tensor T'. The product of two permutation operators should satisfy the multiplication rule of elements in the permutation group. Let us begin with a simple example. If

R = (ii;) = (iii) = ( 1 2 3 ) = ( 1 2 ) ( 2 3 ) ,

then a tensor transforms as

On the other hand, a tensor basis transforms as

They transform in the different but compatible ways. In fact, expanding a tensor T in the tensor basis @&, we have

In general, if

we have

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It is easy to show that the product order of an SU(N) transformation 0, and a permutation R among indices of a tensor are commutable

ROUT = 0,RT. (8.17)

This property is called the Weyl reciprocity. It is the theoretical foundation for decomposing a tensor space by the Young operators.

* In general, the tensor space ‘T is reducible. The minimal tensor subspace ex] corresponding to an irreducible representation [A] of SU(N) can be obtained by the projection of a Young operator, = y117. Two tensor subspaces denoted by different Young patterns are inequivalent. Two tensor subspaces denoted by different standard Young tableaux of the same Young pattern are equivalent, but are linearly independent.

* An arbitrary tensor basis [email protected] in a tensor subspace can be denoted by a tensor Young tableau with the Young pattern [A], where the box filled with j in the standard Young tableau yF1 is now filled with the subscript aj . For example, for the Young operator $”’ corresponding to the standard Young tableau wl, the tensor Young tableau of the tensor

basis $’11@ala2a3 is F l . For convenience we write

(8.18)

Note that the same tensor Young tableaux in different tensor subspaces denote the different tensor bases. For example, if the Young operator $”I corresponds to the standard Young tableau El, the tensor Young

tableau Fl denotes the different tensors in two tensor subspaces:

Due to the symmetry and the Fock condition (6.5) satisfied by the Young operator, the tensor Young tableaux in a given tensor subspace satisfy some linear relations. For example, for the Young operator y?’l1 we have

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E The relations similar to Eq. (8.19) hold for the general cases. Those re- lations obviously depend upon the Young pattern, but are independent of the Young operators with the same Young pattern as well as the tensor subspaces. The first equality says that the tensor Young tableau is anti- symmetric with respect to any two digits filled in the same column. The tensor Young tableau is equal to zero if there are two equal digits filled in its same column. We may rearrange the digits in each column of the tensor Young tableau by the antisymmetric relation such that the digit increases downwards. The second equality is also called the Fock condition.

* Any tensor in a tensor subspace $’I is a combination of the tensor bases yF1Oa1...,,. yp [’I can be expressed as

Since yFIRFY are the standard bases in the minimal right ideal of the permutation group produced by the Young operator yI1, the following tensor bases

YFIR,uu@bl ... b , > bl 5 b2 5 - - - 5 bn, (8.20)

constitute a complete set of the tensor bases in the tensor subspace $’I, where R,, is the permutation transforming the standard Young tableau y;’] to the standard Young tableau ypl. We are going to discuss the characteristic of the basis (8.20). Since

the tensor Young tableau for the tensor basis Y ~ l R p , @ ~ l . . . , , in the tensor subspace $’I is the same as that for the tensor basis y~’]@b,...,,, in the ten- sor subspace d’]. In fact, if R permutates ~j to j, R ~ ~ o ~ ~ . . . , , = ~ b , . . .bp, .

Therefore, the box filled with j in the standard Young tableau yrl is filled with T j in the standard Young tableau y;’], and is filled with bj both in the tensor Young tableau y~lRp,@bl . . .b , and in the tensor Young tableau

[’I yu @b1 ... b , .

* A tensor Young tableau is called standard if the digit in each column of the tableau increases downwards and in each row does not decrease right-

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wards. Since YLxl is a standard Young tableau and bl 5 b2 5 . . . 5 b,, the tensor Young tableau yLxl@,,,..,n as well as the tensor Young tableau

tute a complete set of the tensor bases in the tensor subspace The standard tensor Young tableau is the eigenstate of HP where, due to Eq. (8.13), the eigenvalue is equal to the number of the digit p filled in the tableau, subtracting the number of the digit ( p + 1). Therefore, the repre- sentation matrix of the Chevalley basis HP is diagonal in the tensor bases. The action of F, on the standard tensor Young tableau is equal to the sum of all possible tensor Young tableaux, each of which is obtained from the original one by replacing one filled digit p with the digit ( p + 1).

If without special indication, we use the notation for the inner product of tensors that the bases @al...a, are orthonormal. Two standard tensor Young tableaux with the different sets of the filled digits are orthogonal, but may not be normalized to the same number. Two standard tensor Young tableaux with the equal set of the filled digits but different filling order may not be orthogonal. To find a set of the orthonormal bases, one need to make suitable combination and normalization from the standard tensor Young tableaux.

* Since RpvYLx1 are the standard bases in the minimal left ideal of the permutation group corresponding to the irreducible representation [A], Eq. (8.21) also shows that the same tensor Young tableaux in all tensor subspace 7;Ix1 with the same Young pattern [A] constitutes a complete set of the

yP [XI f&[email protected], is standard. The standard tensor Young tableaux consti-

standard bases in the irreducible group, denoted by [A]. Therefore,

representation space of the permutation in the decomposition of the tensor space

the tensor subspace corresponds to the irreducible representation [A] of SU(N), and for a given Young pattern [A], the tensor bases with the same tensor Young tableau in the d[xl(S,) subspaces $'I span the irreducible representation space of the permutation group S, denoted by [A].

* The dimension d[xl(SU(N)) of the representation [A] of SU(N) can be calculated by the hook rule. In this rule the dimension is expressed by a quotient, where the numerator is the product of ( N + mij) , and the

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326 Problems and Solutions in Group Theory

denominator is the product of hij. mij = j - i is called the content of the box located in the j t h column of the ith row of the Young pattern [A]. hij

is called the hook number of that box in the Young pattern, which is equal to the number of the boxes on its right in the ith row of the Young pattern, plus the number of the boxes below it in the j t h column, and plus one.

(8.23)

where the symbol Yyl is a tableau obtained from the Young pattern [A] by filling ( N + mij) into the box located in its ith row and j t h column, and the symbol YLX1 is a tableau obtained from the Young pattern [A] by filling hij into that box. The symbol Y y l means the product of the filled digits in it, so does Yh [XI .

* The one-row Young pattern [n] denotes the totally symmetric tensor rep- resentation. The one-column Young pattern [In] denotes the totally anti- symmetric tensor representation. The totally antisymmetric tensor space of rank N , denoted by [IN], contains only one standard tensor Young tableau, corresponds to the tensor basis

(8.24)

The tensor E is an invariant tensor in the SU(N) transformation, so [l”] denotes the identical representation of SU( N ) . The tensor subspace de- scribed by a Young pattern with more than N rows is a null space. The reduction of the direct product of two irreducible representations described by two Young patterns [A] and [p] can be calculated by the Littlewood- Richardson rule. However, the Young patterns with more than N rows contained in the reduction should be removed. The representation denoted by the Young pattern [A] with N rows is equivalent to the representation denoted by the Young pattern [A’] obtained from [A] by removing its first column.

* A contravariant tensor Tal...an of rank n with respect to SU(N) trans- forms as

(8.25) d l ... d,

= C Tdl...dn ( u - ~ ) ~ ~ ~ ~ . . . ( u - ’ ) ~ , ~ , , u E SU(N).

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d l ... d , (8.26)

d l ... d ,

The contravariant tensor space T can also be decomposed by the pro- jection of the Young operator, T = @$T = @$"*. The tensor subspace $"I' corresponds to the representation [A]* of SU(N), which is the conjugate one of the representation [A]. The basis in the tensor sub- space $"* is denoted by the standard tensor Young tableau with star, for example

(8.27)

* A mixed tensor T$;;t; of rank (n, m) with respect to SU(N) transforms as

d l dn ( o ~ T ) ~ ~ ~ * . ~ ~ ~ = u a l c l ' ' * U a n d n T c l : : : c n ( U - l ) d l b l ' ' ' ( u - ' ) d n b n '

C l . . . C n d l ... dn (8.28)

Obviously, the mixed tensor (D): = 6; is an invariant tensor of rank (1,l) in the SU(N) transformation.

A mixed tensor is said to be made a contraction (or the trace) of its one pair of indices, one covariant and one contravariant, if to take the pair of indices to be equal and then to sum over the indices. The trace tensor of a mixed tensor of rank (n,m) is a tensor of rank (n - 1,m - 1). A mixed tensor can be decomposed into the sum of a series of traceless tensors with different ranks. For example,

} + d t [; + D l ~ d : 1

c d

c d

T i ] 7

T:] 7

(8.29)

where the tensor in the curve brackets is a traceless mixed tensor of rank (1, l), and the tensor in the square brackets is a scalar [the tensor of rank (0, O)]. Therefore, a mixed tensor space can be decomposed into the direct

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328 Problems and Solutions in Group Theory

sum of a series of traceless tensor subspaces with different ranks, each of which is invariant in the SU(N) transformation.

The traceless tensor space can be decomposed into the direct sum of some minimal tensor subspace by the projection of the Young operator YF1 on the covariant indices and the projection of the Young operator yF1 on the contravariant indices, where the Young patterns [A] and [TI contain n and m boxes, respectively. The minimal tensor subspace corresponds to an irreducible representation of SU(N), denoted by [A]\[T]*. It can be shown that the tensor subspace corresponding to [A]\[T]* is null space if the sum of the numbers of rows of [A] and [TI is larger than N , and the representation [A]\[T]* of SU(N) is equivalent to the representation [A']\[T']* if [TI contains t rows, [T'] is obtained from [TI by removing its first column, and [A'] is obtained from [A] by attaching a new column with ( N - t ) boxes from its left. Thus, the representation [TI* of SU(N) is equivalent to the representation [A] if by rotating the Young pattern [TI upside down and then by attaching it to the Young pattern [A] from below, it just forms a rectangle with N row.

4. Calculate the dimensions of the irreducible representations denoted by the following Young patterns for the SU(3) group and for the SU(6) group, respectively:

Solution.

3 4 6 7

1 3 4 5

1 6 7 8

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5 . Calculate the Clebsch-Gordan series for the following direct product representations, and compare their dimensions by Eq. (8.23) for the SU(3) group and for the SU(6) group, respectively:

o o l l l $ o o l l $ o 0 0 1 1 0 0 1 $ 0 1 ,

1 1 0 0 1 Solution. a): O O 8 1 1 1 z 0

SU(3) : 8 x 10 = 80 = 35 + 27 + 10 + 8, SU(6) : 70 x 56 = 3920 = 1050 + 1134 + 840 + 896.

0 0 0 1 1 $ 0 0 0 1 $ 0 0 0 1 1 1 1 1 1’ b): 0 0 0 ~ 1 1 1 ~ 0 0 0 1 1 1 $

SU(3) : 10 x 10 = 100 = 28 + 35 + 27 + 10, SU(6) : 56 x 56 = 3136 = 462 + 1050 + 1134 + 490.

0 0 0 1 1 0 0 0 1 0 0 0 $ 0 0 0 $ 0 0 0 ,

1 1 1 1 1 1

0 0 0 N o o o l l l $ o o o 4: l l ~ @ o o o - o o o

SU(3) : 10 x 10 = 100 = 64 + 27+ 8 + 1, SU(6) : 56 x 490 = 27440 = 9240 + 11340 + 5880 + 980.

0 0 0 0 @ 1 1 N o o o o l l $ o o o o o o l l 0 0 0 0 1

0 0 2 - 0 0 2 r ) @ 0 0 1 2 d):

L

0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 $ 0 0 1 $ 0 0 2 $ 0 0

2 1 1 2

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330 Problems and Solutions in Group Theory

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 o o l 0 0 0 0 1

@ 1 1 ’ O 0 @ 0 0 1 1 @ 0 0 1 @ I 2 2

2 1 2 2

SU(3) : 27 x 8 = 216

SU(6) : 1134 x 70 = 79380 = 64+ 35 + 35 + 2 x 27+ 10 + O + 1 0 + 8 + 0 + 0,

= 9240 + 11550 + 5880 + 2 x 11340 + 6000 + 5670 + 4704 + 5880 + 4536 + 3240.

6 . Write all tensor Young tableaux in the tensor subspace 7;(2’11 = $’ll‘T of SU(3), and prove that the standard tensor Young tableaux constitute the complete bases in

Solution. The standard

this subspace.

Young tableau $’11 is F l . Thus,

If there is a pair of the filled digits to be equal in the tensor Young tableau, from Eq. (8.19) we have

where the left tensor Young tableau in each equality is standard. If all three filled digits in the tensor Young tableau are different, from Eq. (8.19) we

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have

331

where the left tensor Young tableaux in the first two equalities are standard. The dimension of the subspace 7;[2’11 is eight. There are eight standard tensor Young tableaux. The first six standard tensor Young tableaux are normalized to 6 , but the last two to 4. In addition, the last two standard tensor Young tableaux are not orthogonal to each other.

7. Try to express each nonzero tensor Young tableau for the irreducible representation [3,1] of SU(3) as the linear combination of the standard tensor Young tableaux.

Solution. For the SU(3) group, the digits filled in the tensor Young tableaux are 1, 2 and 3. The equal digits cannot be filled in the same column. There are three types of nonzero tensor Young tableaux. A tensor Young tableau in the first type contains three equal digits. From Eq. (8.19) we have

When a < b the left tensor Young tableaux is standard, while when a > b the right one is standard. There are 3 x 2 = 6 standard tensor Young tableaux in the first type. A tensor Young tableau in the second type contains two pairs of equal digits. From Eq. (8.19) we have

When a < b, the first tensor Young tableau is standard, while when a > b the fourth one is standard. There are 3 standard tensor Young tableaux in the second type. A tensor Young tableau in the third type contains one pair of equal digits. There are six standard tensor Young tableaux in the third

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332 Problems and Solutions an Group Theory

type. The other tensor Young tableaux can be expanded in the standard tensor Young tableaux by Eq. (8.19),

U

The dimension of the irreducible representation [3,1] of SU(3) is 15. There are 15 standard tensor Young tableaux.

8. Write the explicit expansion of each standard tensor Young tableau in the tensor subspace JfP1]7, where 5'- is the tensor space of rank four for

the standard Young tableau of the Young operator - Solution. First, we write the explicit formula for the application of the

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Similar to Problem 7, there are three types of standard tensor Young tableaux. In the first type there are six standard tensor Young tableaux, each of which contains three equal digits.

PI= -PI [3 11 = Y2 ’ Oaaba = 6@aaba - 2 0 b a a a - 2 0 a b a a - 2 0 a a a b .

When a < b the left tensor Young tableau is standard, and when a > b the right one is standard. The module square of the standard tensor Young tableaux in the first type is 48. In the second type there are three standard tensor Young tableaux, each of which contains two pairs of equal digits:

The module square of the standard tensor Young tableau in the second type is 24. In the third type there are six standard tensor Young tableaux, each of which contains one pair of equal digits:

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334 Problems and Solutions in Group Theory

where a and b are respectively taken 2 and 3. The module square of the first form, which includes two standard tensor Young tableaux, is 24, and that of the remaining standard tensor Young tableaux is 18. Two standard tensor Young tableaux with the equal set of the filled digits are not orthogonal.

9. Transform the following traceless mixed tensor representations of the SU(6) group into the covariant tensor representations, respectively, and calculate their dimensions:

Solution. (1) [3,2, I]* N [i3]\[2, I]* N [23, 1]\[1]* 21 [33, 2,1],

6 7 8 5 6

6 7 8 5 6 7 4 5 6 3 4

6 7 8 9 10 11 5 6 7 8 9 4 5 6 7 3 4 5

= 147840. 9 8 7 5 3 1 7 6 5 3 1 5 4 3 1 3 2 1

d[6 , 5,4,3] (su(6) =

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6 7 8 9 10 11 12 5 6 7 8 9 10 4 5 6 7 3 4 5 2

d [ 7 , 6 , 4 , 3 , 1 ] ( S U ( 6 ) ) = 11 = 612500.

9 7 6 4 2 1 6 4 3 1 4 2 1 1

10. Prove the identity:

where (TA),, are the generators in the self-representation of the SU(N) group.

Solution. Define a mixed tensor T of rank (2,2) with respect to SU(N)

It is easy to show that T is invariant in the SU(N) transformation,

Therefore, T can be expressed as the product of the invariant mixed tensors d: of SU(N),

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336 Problems and Solutions in Group Theory

Since T’ is a traceless matrix, 0 = C, TE,d = ( N p + q) 6:, we have q = - N p . AS

when A # N 2 - 1

when A = N 2 - 1,

we have N - 1 -- - Tgg = p + q = - ( N - 1)p.

2N

The solution is p = - (2N)- ’ and q = 1 / 2 . This completes the proof.

8.3 Orthonormal Bases for Irreducible Representations

* Recall that the standard tensor Young tableau is the eigenstate of the Chevalley basis H p where the eigenvalue is equal to the number of the digit p filled in the tableau, subtracting the number of the digit ( p + 1 ) . The action of the lowering operator Fp on the standard tensor Young tableau is equal to the sum of all possible tensor Young tableaux, each of which is obtained from the original one by replacing one filled digit p with the digit ( p + 1 ) . Two standard tensor Young tableaux with different sets of the filled digits are orthogonal, but may not be normalized to the same number. Two standard tensor Young tableaux with the equal set of the filled digits but different filling order may not be orthogonal.

The merit of the method of Young operators is that the complete tensor bases in an irreducible representation space of SU(N) have been explicitly known. The shortcoming is that the tensor bases are not orthonormal, and the obtained representation is not unitary. On the other hand, the method of the block weight diagram gives the representaion matrices of the Chevalley bases of generators in the unitary irreducible representation, but the orthonormal bases are given only symbolically. To combine two methods will offset one’s shortcoming by the other’s strong point.

* In the standard tensor Young tableau corresponding to the dominant weight, the number of each digit p filled in the tableau must not be less than the number of the digit (p+ 1) filled there. It is the criterion for determining which standard tensor Young tableau corresponds to the dominant weight in the representation. The states for the multiple weight correspond to the

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lwab) =

standard tensor Young tableaux with the equal set of the filled digits but in different filling order. In the standard tensor Young tableau corresponding to the highest weight, each box located in the j t h row of the tableau is filled with the digit j , because each raising operator E, eliminates it. Therefore, the component MP of the highest weight M in the representation [A] is

W2N . . . w ( N - 1 ) N W ( N - 1) ( N - 1) . . . w1 N

W1(N-l) . . . . . . . . . w12 w22

(J11

M, = A, - &+I. (8.30)

* Gelfand presented the symbols for the orthonormal basis states of an irreducible representation [A] of SU(N), usually called the Gelfand bases, and calculated the analytic formulas for the representation matrix entries of the generators (the Chevalley bases) of SU(N) in the Gelfand bases. For a multiple weight, he chose the orthonormal basis states first according to the multiplets of the subalgebra d1, then, according to the multiplets of the subalgebra -A2 if there still exist the multiple states, and so on. Gelfand described the basis state with the weight m in the representation [A] of SU(N) by N ( N + 1)/2 parameters &b, 1 5 a 5 b 5 N , arranged as a regular triangle upside down:

where W ~ N = A,, WNN = AN = 0, and

(8.31) i

The representation entries of the generators are

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338 Problems and Solutions an Group Theory

A,, (Wab) r2 } I 2 -

P-1 P+l - (&(@-I) - Wvp - t + v - 1) n ( W p ( p + l ) - WV, - p + .)

n = { t = l p = l

, (Wdp - Wv, - d + v) (Wd, - Wv, - d + v - 1)

d#v,d=l .( (8.32) Namely, the eigenvalue of H, is equal to twice the sum of the parameters in the pth row from the bottom, minus the sum of the parameters in the ( p - 1)th row and the (p+l ) th row from the bottom. The application of E, (F,) to a Gelfand basis leads to a linear combination of the possible basis states, each of which is obtained by increasing (decreasing) one parameter wap in the pth row from the bottom under the condition (8.31). The parameters of the highest weight state in the Gelfand bases satisfy

11. Expand the Gelfand bases in the irreducible representation [3,0] of the SU(3) group with respect to the standard tensor Young tableaux by making use of its block weight diagram given in Fig. 7.2.

Solution. The tensor subspace described by the representation [A] = [3,0] consists of the completely symmetric tensors of rank three. There is no multiple weight in the representation. Its highest weight is M = (3,0), which corresponds to the standard tensor Young tableau and the Gelfand basis are

There are three types of the standard tensor Young tableaux which are normalized to 6, 12 and 36, respectively:

where a, b and c are all different. In the application of the lowering operator F, , all possible terms, which may be non-standard tensor Young tableaux,

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need to be considered. For example

Now, the orthogonal bases can be calculated as follows.

2 0 0 0 ) ) 2

2 0 0 0 ) ) 0

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340 Problems and Solutions in Group Theory

The expansions of the orthonormal basis states with respect to the stan- dard tensor Young tableaux give the wave functions of the hadrons. Substi- tuting the expansions into the planar weight diagram of (3,O) given in Fig. 7.3, we obtained the wave functions for the baryon decuplet. In the planar weight diagram, the ordinate axis and the abscissa axis are the eigenvalues of T3 and 2'8, respectively (see the end of Problem 14).

12. Expand the Gelfand bases in the irreducible representation [3,3] of the SU(3) group with respect to the standard tensor Young tableaux by making use of its block weight diagram given in Fig. 7.2.

Solution. The representation [A] = [3,3] of SU(3) is the conjugate repre- sentation of [3,0], [3,3] N [3,0]*. This problem can be solved in the same way as that used in Problem 11. However, in the present problem we prefer to calculate the standard tensor Young tableaux of the contravariant tensor instead of those of the covariant tensor. Both results for two kinds of the standard tensor Young tableaux are given here. Note that the action of the generator on the basis state of the conjugate representation is differ- ent from that on the basis state of the original representation. Denote by D(IA) and D'(IA) the representation matrices of the generator I A in the representations D ( R ) and those in its conjugate one D(R)* , respectively:

D(R) = 1 - i C D ( I A ) w A ,

D(R)* = 1 +ix D(IA)*wA = 1 - ix D'(IA)wA. A

A A

Thus, D'(IA) = -D(IA)* . For the Chevalley bases we have

Namely, the standard tensor Young tableau with star is the eigenstate of the Chevalley basis H p where the eigenvalue is equal to the number of the digit ( p + 1) filled in the tableau, subtracting the number of the digit p. The action of the lowering operator Fp on the standard tensor Young tableau with star is equal to the sum multiplied by -1, of all possible tensor Young tableaux with star, each of which is obtained from the original one by replacing one filled digit ( p + 1) with digit p. The order of the filled

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digits in the standard tensor Young tableau changed to 3, 2, 1 instead of 1,

The highest weight of the representation [3,3] is M = (0,3), which corresponds to the standard tensor Young tableau and the Gelfand basis are

2) 3.

1 1 1 2 2 2 1(0,3),(0,3)) =m]* =wl= l 3 3 : 3 O ) .

The normalization factor for the standard tensor Young tableau with star is similar to that for the standard tensor Young tableau. Now, the orthogonal bases can be calculated as follows.

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342 Problems and Solutions in Group Theory

13. Expand the Gelfand bases in the irreducible representation [2,1] of the SU(3) group with respect to the standard tensor Young tableaux by making use of its block weight diagram given in Fig. 7.2.

Solution. The representation [A] = [2,1] is the adjoint representation of SU(3). For the tensor subspace 7;12’11 = yi2’117, the standard Young operator is $’11 = E + (1 2) - (1 3) - (2 1 3), corresponding to the Young

tableau Fl . There are six single weight and one double weight (0,O)

in the representation [2,1]. The typical standard tensor Young tableaux and their normalization factors are

Due to the Fock condition (6.5), we have

In particle physics, a hadron is composed of three quarks (baryon) or a pair of one quark and one antiquark (meson). The expansions of the orthonormal basis states in [2,1] with respect to the standard tensor Young tableaux give the flavor wave functions for the baryon octet or for the meson octet. For the meson octet, the standard tensor Young tableau should be

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transformed into the traceless mixed tensor by the totally antisymmetric tensor &bc:

(8.35)

The highest weight in [2,1] is M = (1, l), which corresponds to the standard tensor Young tableau and the Gelfand basis are

The remaining orthonormal bases can be calculated from the block weight diagram given in Fig. 7.2:

U I

U I I

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344 Problems and Solutions in Group Theory

+FV

- d m * 2 = I 2

PI 0 l o 0 ) ,

where we use the formula calculated from Eq. (8.32):

F 2 I 2 2 : l 0 ) = & I 3 2 l l l o ) 1 +Ll 1 2 , l O O ) . 1

Each basis state in the representation [2,1] is normalized to 6 . Especially,

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the basis state

345

14. Express each Gelfand basis in the irreducible representation [4,0] of the SU(3) group by the standard tensor Young tableau and calculate the nonvanishing matrix entries for the lowering operators Fp. Draw the block weight diagram and the planar weight diagram for the rep- resentation [4, O] of SU(3).

Solution. The tensor subspace described by the representation [A] = [4,0] consists of the completely symmetric tensors of rank four. There are four single dominant weights in the representation:

The highest weight of the representation [4,0] is M = (4,0), which corre- sponds to the standard tensor Young tableau and the Gelfand basis:

)(4,0),(4,0)) = M I = l 4 4 0 "). There are four types of the standard tensor Young tableaux which nor-

malize to 48, 96, 144, and 576, respectively. The normalization them are m, &, 2, and I, respectively.

factors for

Obcac

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346 Problems and Solutions in Group Theory

where a, b, and c all are different.

[A] = [4,0] we have a quintet of d1:

Now, we calculate the remaining basis states. From the highest weight

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= dZ[11212131= 4 3 : 0 0 ) ,

1/2)F2 { dZ p-pp-p-1)

= 2ml=/4 O ) ,

= ml=14 o 0 O ) .

It is easy to check that

The block weight diagram and the planar weight diagram are now drawn as follows. From Eqs. (8.3) and (8.6), the simple roots and the fundamental dominant weights of the SU(3) group in the rectangular coordinate frame

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348

are

Problems and Solutions in Group Theory

From Eq. (8.4) we know that the first and the second components of the weight m in the rectangular coordinate frame are related to the eigenvalues of T8 = Ti3) and T3 = Ti3) in SU(3), respectively. Namely, the ordinate axis and the abscissa axis are the eigenvalues of T3 and T8, respectively.

15. Express each Gelfand basis in the irreducible representation [3,1] of the SU(3) group by the standard tensor Young tableau and calculate the nonvanishing matrix entries for the lowering operators F’. Draw the block weight diagram and the planar weight diagram for the rep- resentation [3,1] of SU(3).

Solution. There are two single dominant weights ( 2 , l ) and (0 ,2) , and one double dominant weight (1,O) in the representation [A] = [3,1] of SU(3):

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,

The highest weight of the representation [3,1] is M = (2,1), which corre- sponds to the standard tensor Young tableau and the Gelfand basis:

In the tensor subspace 7;[3”1 = 3/1[3’117, the standard Young tableau is on the tensor basis P I , and the action of the Young operator

Note that

where a may be filled in the second column or the third column. There are four types of standard tensor Young tableaux which normalize

to 18, 24, 24, and 48, respectively. The normalization factors for them are m, a, a, and 1, respectively.

J“(J-- 1/12 ( 3 G a a a d - O d a a a - O a d a a - e a a d a ) } -

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350 Problems and Solutions in Group Theory

From the highest weight state (2 , l ) we have a triplet of A1 with (0,2) and (2,3), and a doublet of A2 with (3 , i ) .

- I

2 1 1 ° ) ,

From the weight (3,i) we have a quartet of d1 with ( l , O ) , (i, l), and (3,2), where the dominant weight ( 1 , O ) and its equivalent weight (i, 1) are the double weights. We define one basis state for ( 1 , O ) or for (i,l) belongs to the quartet of d1, and the other belongs to the doublet.

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From the weight (0,2) we have a triplet of A2 with (1,O) and (2,2), where the basis state with the weight (1,O) is a combination of 1(1,0)1) and 1(1,0)2). Letting

we have

Thus, a1 = m, and a2 = &-!@ = m. Applying E2F2 = F2E2 + H2 to the basis state ((2, 1), ( l ,O)l) , we have

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352 Problems and Solutions in Group Theory

f iom the weight (2,2) we have a triplet of A1 with (0 , i ) and (2,0), where the weight (0, i) is a double weight. We define one state basis I(0, i),) belongs to the triplet of A1, and the other 1 ( O 7 i ) 2 ) belongs to the singlet.

From the weight (2,3) we have a quartet of A2 with (T, 1)) (0,T) and (l,s), where the states with the weights (i, 1) and (0, i ) are the combina- tions of the. basis states. Letting

we have

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Thus, b l = m, b2 = m, ci = m, and c3 = m. Applying E2F2 = F2E2 + H2 to the basis state l (2 ,1 ) , ( i , l )2 ) , we have

Choosing the phase of the basis state )(2,1), (0, i ) 2 ) such that c4 is real positive, we obtain c4 = fi, and then c2 = 0. Applying E2F2 = F2E2 + H2

to the basis state )(2,1), ( O , i ) 2 ) , we have

E2F2 P, 11, ( O , i ) 2 ) = dld2 1(2,1), (0 , i ) l ) + di; l(2,1), ( O , i ) 2 ) = (F2E2 + H2) ((2,1), ( O , i ) 2 )

= Jz I @ , 11, (0 , i ) l ) + (3 - 1) 1(2,1), (0,1)2).

Choosing the phase of the basis state l(2, l), (1,s)) such that d2 is real positive, we obtain d2 = a, and then dl = 1. From the above calculation, we have

l(27 I>, ( O A 2 ) = r n ( F 2 1 ( 2 , 1 > , ( i , 1 ) 2 ) - r n I ( 2 , 1 > , ( O J ) l ) }

=mF2 { m p ! - J . j ( 2 P l - PI}

= + p = 1 = 1 3 1 3 3 ; 0 ) ,

l(2, 1)) ( L a ) = l / W 2 1 ( 2 , 1 > , ( O , i ) 2 ) = F21(2,1), (0 , i ) l )

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354 Problems and Solutions in Group Theory

Note that those representation matrix entries of F 2 related to the basis states with the multiple weights can also be calculated with the formula of Gelfand.

From Eq. (8.32) we have

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The block in the figure.

355

weight diagram and the planar weight diagram are given The planar weight diagram of the representation [A,, A,] of

SU(3) is a hexagon with the lengths of two neighbored sides to be (A1 - A,) (i.e. M I ) and A2 (i.e. Mz) . The weights on the boundary is single, and the multiplicity increases as the position of the weight in the diagram goes inside until the hexagon reduces to a triangle.

16. Calculate the Clebsch-Gordan series for the direct product representa- tion [2,1] x [2,1] of the SU(3) group, and expand the highest weight state of each irreducible representation in the CG series with respect to the standard tensor Young tableaux.

Solution. The representation [2,13 is the adjoint representation of SU(3) and was studied in Problem 13. The highest weight of [2,1] of SU(3) is ( 1 , l ) .

Unitary Groups

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356 Problems and Solutions in Group Theory

Usually, the Clebsch-Gordan series for the direct product representation of the SU(N) group is best calculated with the Littlewood-Richardson rule, as used in Problem 5. For [2,1] x [2,1] of SU(3) we have

However, the method of the dominant weight diagram discussed in Chapter 7 is also a good method for it. In this method we need to know the dominant weights and their multiplicities in the relevant representations, which can be obtained, say, by the method of the block weight diagram, or by consulting the table book. They can also be obtained directly from the standard tensor Young tableaux. This is the aim of this problem.

Dominant weight

Standard tensor Young tableaux

X

X

No. of states

1

2

2

6

10

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We have known that the weight for a standard tensor Young tableau is dominant if and only if the number of each digit p filled in the tableau is not less than the number of the digit ( p + 1) filled there. We can easily count by this criterion which and how many standard tensor Young tableaux correspond to the dominant weights in the direct product representation space and in the relevant representation spaces. First, we list the standard tensor Young tableaux with the dominant weights in the direct product representation space [2,1] x [2,13, where only one standard tensor Young tableau in the pair related by a permutation of two product factors is listed for simplicity. Then, we calculate the standard tensor Young tableaux with the dominant weights in the representation (2,2) which is contained once in the CG series because the weight (2,2) is the highest weight in the product space.

Dominant weight Standard tensor Young tableaux 1 1 1 1 1 1 1 1 1

1111 1 2 1 2 1

No. of states

By comparing respectively the multiplicities of the dominant weights (3 ,O) and (0,3) in the product space with those in the representation (2,2), 2- 1 = 1, we conclude that there is one representation (3,O) and one representation (0,3) contained in the CG series. Third, we calculate the standard tensor Young tableaux with the dominant weights in the representations (3,O) and (0,3). Comparing the numbers of the standard tensor Young tableaux with the dominant weight (1,l) in the product space and in the three representations (2,2), (3 ,O) and (0,3), 6 - 2 - 1 - 1 = 2, we conclude that the representation (1,l) is contained in the CG series twice.

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358 Problems and Solutions in Group Theory

Dominant weight

~~

Standard tensor Young tableaux in (3,O)

TTTT-Tq in ?0,3)

No. of states in (3,O) in (0,3)

Fourth, we calculate the tensor Young tableaux with the dominant weights in the representation (1,l). Comparing the numbers of the stan- dard tensor Young tableaux with the dominant weight (0,O) in the prod- uct space and in the representations (2,2), (3,0), (0,3), and two ( l , l ) , 10 - 3 - 1 - 1 - 2 x 2 = 1, we conclude that the representation (0,O) is contained once in the CG series. Therefore, we complete the diagram of the dominant weights for the direct product representation [2,1] x [2, I], where the subscript S or A denotes the symmetric or antisymmetric combination in the permutation of two factors in the product, respectively.

Standard tensor Young tableaux No. of states -1 0s x n

6 x 1

3 x 2

3 x 2

6 x 6

1 x 10

R 2

0,o

At last, we calculate the expansion for the highest weight state of each

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Unitary Groups 359

representation contained in the CG series by Eq. (7.24), namely, each raising operator E, annihilates the highest weight state. The results are

- Fl x Fl+ (antisym. terms),

X X

U U U

In terms of the orthonormal basis states we have

terms)

U

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360 Problems and Solutions in Group Theory

17. A neutron is composed of one u quark and two d quarks. Construct the wave function of a neutron with spin S3 = -1/2, satisfying the correct permutation symmetry among the identical particles.

Solution. In particle physics, a hadron is composed of three quarks (baryon) or a pair of one quark and one antiquark (meson). The strong in- teraction among quarks is the gauge interaction of the colored SU(3) group. In addition to the color, a quark brings the flavor quantum number. The u and d quarks span the self-representation space of the flavor SU(2) group, the isospin symmetry group. The isospin is conserved in the strong in- teraction. The next quark is s quark, which is heavier than the u and d quarks. u, d and s quarks span the self-representation space of the flavor SU(3) group, which describes an approximate symmetry, but played an im- portant role in the history of particle physics. Even now, it is still useful in the rough estimation. The expansions of the orthonormal basis states with respect to the standard tensor Young tableaux give the flavor wave functions of the hadrons.

The flavor SU(3) group is a compact simple Lie group with order 8 and rank 2. The two commutable generators have important physical meanings. In the self-representation, T3 describes the third component of the isospin of a quark, and T8 is proportional to the supercharge Y of a quark:

The basis states in an irreducible representation space of the flavor SU(3) group can be described in the planar weight diagram with the coordinate axes T3 and Y , as shown in Fig. 7.3 and Problems 14 and 15 in this Chapter. Since the planar weight diagram is intuitive, it has widely applied in particle physics.

A neutron consists of three quarks. It is a system of three identical fermions. According to the Fermi statistics, the wave function has to be totally antisymmetric in the permutation of quarks. The internal orbital angular momentum of quarks is assumed to be zero. The total wave func- tion is composed of three parts: the color part, the flavor part, and the spinor part. Since the neutron has to be the color singlet, its wave function

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of the color part is totally antisymmetric in the permutation. Thus, the product of the wave functions of the flavor part and the spinor part should be combined into the totally symmetric state in the permutation. The neu- tron belongs to a flavor octet with T = -T3 = 1/2 and Y = 1, and to a spinor doublet. In this problem we discuss the state with S3 = -1/2. The Young patterns corresponding to two parts both are [2,1]. Choosing the standard Young tableau Fl, we obtain two basis states of the flavor

wave function by the standard tensor Young tableaux:

Fl = udd + dud - 2ddu, (23) Fl = udd + ddu - 2dud .

There are also two basis states of the spinor wave function:

Fl = (+ - -) + (- + -) - 2(- - +) , - (23) El = (+ - -) + (- - +) - 2(- f -) .

In this set of the standard bases, the representation matrices of the genera- tors of the permutation group S3 can be calculated by the tabular method (see Problem 20 in Chap. 6),

In the direct product representation,

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362 Problems and Solutions in Group Theory

Their eigenvectors for the eigenvalue 1 respectively

The totally symmetric state corresponds to the common eigenvector with eigenvalue 1, which is ( 2 1 1 2)T . Thus, the total wave function of the flavor part and the spinor part for the neutron with S3 = -1 /2 is

r

= {udd + dud - 2 d d ~ } * 3 {(+ - -) - (- - +)} + {udd + ddu - 2 d ~ d ) * 3{(+ - -) - (- + -))

= 3 { 2 ~ + d - d - - d + d - ~ - - d + ~ - d - - ~ - d - d + + 2 d - d - ~ +

- d-u-d+ - u-d+d- - d-d+u- + 2 d - ~ + d - } .

For normalization, one may replace the factor 3 with d m .

8.4 Subduced Representations

* Now, we discuss the reduction problem of the subduced representation of an irreducible representation of the SU(N) group with respect to its sub- group. There are three kinds of important subgroups for the SU(N) group. One is SU(N- 1) cSU(N), the second is SU(M)xSU(N/M) cSU(N), and the third is SU(M)xSU(N - M) cSU(N).

* The reduction of the subduced representation from an irreducible rep- resentation [w] of SU(N) with respect to the subgroup SU(N - 1 ) can be solved directly by the standard tensor Young tableau. The tensor indices taken the value N keep invariant in the subgroup SU(N - 1). In a stan- dard tensor Young tableau, the box filled with N can only be located in the lowest position of each column, where the neighbored right position has to be empty or a box filled with N . Removing the boxes filled with N in a standard tensor Young tableau of SU(N), we obtain a basis state of an

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irreducible representation of the subgroup SU(N - 1). Thus, by remov- ing several (including zero) boxes located in the lowest positions in some columns from the Young pattern [A], we obtain the Young pattern [p], de- noting an irreducible representation of SU(N - 1) in the Clebsch-Gordan series for the reduction of the subduced representation:

(8.37)

This method may be applied continuously to the reduction for the sub- groups in the group chain

SU(N) 3 SU(N - 1) 3 . ‘ . 3 SU(3) 3 SU(2) .

Thus, the subduced representation of an irreducible representation [A] of SU(N) with respect to its subgroup SU(N’) with N’ < N can be reduced.

* For the SU(NM) group, containing the subgroup SU(N) x SU(M), each tensor index a consists of two indices (a;) , where a is the tensor index of the subgroup SU(N), and i is the tensor index of the subgroup SU(M). In the reduction of a subduced representation from an irreducible representation [w] of SU(NM) with respect to the irreducible representations [A] 8 [p] of the subgroup SU(N)xSU(M), the numbers of the boxes in three Young patterns are the same n. The three Young patterns respectively describe the permutation symmetry among n indices a, n indices a, and n indices a. When n indices i are fixed, the n indices a have the permutation sym- metry denoted by [A]. When n indices a are fixed, the n indices i have the permutation symmetry denoted by [p] . When both sets of indices are transformed, n indices a! have the permutation symmetry denoted by [w]. Therefore, the representation [w] of the permutation group S, should be contained in the reduction of the inner product of two representations [A] and [p]:

Now, being an irreducible representation of SU(NM), the number of rows of the Young pattern [w] cannot be larger than N M . Correspondingly, the number of rows of the Young pattern [A] for SU(N) cannot be larger than N, and that of the Young pattern [p] for SU(M) cannot be larger than M .

* For the SU(N + M ) group, containing the subgroup SU(N)x SU(M), each tensor index a! takes two possible kinds of values, either the tensor in-

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364 Problems and Solutions in Group Theory

dex a of the subgroup SU(N), or the tensor index i of the subgroup SU(M). In the reduction of the subduced representation from the irreducible rep- resentation [w] of the SU(N + IM) group with respect to the irreducible representation [A] 8 [p] of the subgroup SU(N) xSU(M), the number n1 of boxes in the Young pattern [A] and the number 722 of boxes in the Young pattern [p] are not fixed, but the sum nl + n2 of their numbers of boxes is fixed and equal to the number n of boxes in the Young pattern [w]. In the permutation among the tensor indices, all indices are divided into two sets. The n1 indices a in the first set have the permutation symmetry denoted by [A], which contains n1 boxes, and the 722 indices i in the sec- ond set have the permutation symmetry denoted by [p] , which contains 122

boxes. Altogether, they have the permutation symmetry denoted by [w], which contains n = n1 + n2 boxes. Therefore, the representation [w] of the permutation group S, should be contained in the reduction of the outer product of two representations [A] and [p] of the permutation group:

[A] 8 [p] = a:, [w] (3 - * - -

The reduction of the subduced representation can be calculated by the Littlewood-Richardson rule.

Being an irreducible representation of SU(N + M ) , the number of rows of the Young pattern [w] cannot be larger than N + M . Correspondingly, the number of rows of the Young pattern [A] for SU(N) cannot be larger than N , and that of the Young pattern [p] for SU(M) cannot be larger than M .

18. Reduce the subduced representation from [3,1] of the SU(4) group with respect to the subgroup SU(3) and list the standard tensor Young tableaux as the basis states in each irreducible representation space of SU(3).

Solution. The dimension of the representation [3,1] of SU(4) is

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Its subduced representation with respect to SU(3) is reduced into the direct sum of six irreducible representations of SU(3). They and their basis states are listed as follows. 1). The representation [3,1] of SU(3) is 15-dimensional with the following standard tensor Young tableaux as the basis states

1 1 1 111 1 1 2 1 1 2 1 1 3 2 ' 3 ' 2 ' 3 l 2 ' 1 1 3 1 2 2 1 2 2 1 2 3 1 2 3 3 ' 2 ' 3 l 2 l 3 ' 1 3 3 1 3 3 2 2 2 2 2 3 2 3 3 2 l 3 ' 3 j 3 ' 3 *

2). The representation [3,0] of SU(3) is 10-dimensional with the following standard tensor Young tableaux as the basis states

1 1 1 1 1 2 1 1 3 1 2 2 1 2 3 4 ' 4 ' 4 ' 4 ' 4 ' 1 3 3 2 2 2 2 2 3 2 3 3 3 3 3 4 ' 4 l 4 l 4 ' 4 -

3). The representation [2,1] of SU(3) is 8-dimensional with the following standard tensor Young tableaux as the basis states

1 1 4 1 1 4 1 2 4 1 2 4 2 ' 3 l 2 ' 3 1 3 4 1 3 4 2 2 4 2 3 4 2 l 3 l 3 l 3 *

4). The representation [2,0] of SU(3) is 6-dimensional with the following standard tensor Young tableaux as the basis states

1 1 4 1 2 4 1 3 4 2 2 4 2 3 4 3 3 4 4 ' 4 ' 4 ' 4 ' 4 ' 4 -

5 ) . The representation [1,1] of SU(3) is 3-dimensional with the following standard tensor Young tableaux as the basis states

1 4 4 1 4 4 2 4 4 2 ' 3 ' 3 *

6). The representation [l, 01 of SU(3) is 3-dimensional with the following

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366 Problems and Solutions in Group Theory

standard tensor Young tableaux as the basis states

1 4 4 2 4 4 3 4 4 4 ) 4 ' 4 -

19. Reduce the subduced representations from the following irreducible representations of the SU(6) group with respect to the subgroup SU(3)@SU(2), respectively:

Solution. In particle physics, a representation of the flavor SU(3) group describes a flavor multiplet, and a representation of the spinor SU(2) group describes a spinor multiplet. All those representations are denoted by the state numbers contained in the multiplets.

6 E (3, 2)

15 N (3*, 3) @ (6, 1)

21 21 (3*, 1) @ (6, 3)

56 N (10, 4) @ (8, 2)

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EP 70 "-

20 21

35 N

n

Notice the reductions of the subduced representations from [3] and [2, 14] 2

[1]\[1]*. The former gives the observed multiplets of the low-energy baryons in the sixties of the 20th century, namely, a decuplet with spin 3/2 and an octet with spin 1/2. Since they belong to the same representation of SU(6), their parities have to be the same (to be positive relative to the parity of a proton). The latter gives the observed multiplets of the low-energy mesons in that years, namely, a singlet with spin 1, an octet with spin 0, and an octet with spin 1, whose parities are the same (to be negative from experiments). It is the success of the SU(6) symmetry theory for hadrons. In the SU(6) symmetry theory, the spin is described by the SU(2) group, so that this theory is only a nonrelativitic theory. This theory cannot explain many phenomena in high energy particle physics.

20. Reduce the subduced representations from the following irreducible representations of the SU(5) group with respect to the subgroup SU( 3) @SU( 2), respectively:

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Solution. In the grand unified model proposed in 1974, one needs to study the reduction of the subduced representations from some irre- ducible representations of the SU(5) group with respect to the subgroup SU(3)@SU(2). There is a U(1) subgroup in the SU(5) group with the gen- erator Y =diag{-1/3, - 1/3, - 1/3, 1/2, 1/2}, which is commutable with all elements in the subgroup SU(3)@SU(2). However, since there are five common elements in two subgroups in addition to the identity, the SU(5) group is not the direct product of two subgroups SU(3)@SU(2) and U(1). The color subgroup SU(3) describes the strong interaction of quarks, and the subgroup SU(2)@U( 1) describes the weak-electric interaction. Each representation in the reduction is denoted by two multiplets of SU(3) and SU(2) and by the eigenvalue of Y as the subscript. In the SU(5) grand unified model, the electric charge Q is equal to T 3 + Y , where T 3 is the third generator of the subgroup SU(2):

Q = 7'3 + Y = diag (0, 0, 0, 1/2, - 1/2}

+ diag{-1/3, - 1/3, - 1/3, 1/2, 1/2}

= diag{-1/3, - 1/3, - 1/3, 1, 0} .

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In the SU(5) grand unified model, the state of a particle is divided into the left-hand state and the right-hand state. They are filled into the 5-dimensional and 10-dimensional representations, respectively. The first generation of particles are filled as follows:

where the superscript c denotes the charge conjugate state, and the sub- script R and L denote the right-hand state and left-hand state, respectively. The 10-dimensional represent ation is the antisymmetric tensor represent a- tion, so only the upper half matrix is needed to be filled with the particle states. The 24-dimensional representation is the adjoint representation, describing the gauge particles, which is not listed here.

8.5 Casimir Invariants of SU(N)

* The generators TA in the self-representation of SU(N) satisfy the com- mutation relations and the anticornmutation relationship

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Due to the normalization condition, ” ~ ( T A T B ) = 6,4B/2, the coefficients cAi and d A B D can be expressed by T’ as given in Eq. (8.9)

Recall that the adjoint representation Dadj(u) of SU(N) satisfy

N2-1 D [ ’ ~ ( u ) - ~ I ~ ~ D [ ’ ~ ( u ) = C Di21(u ) I$1 , u E SU(N), (8.38)

A‘=l

where D[’] (u) and It1 are the representation matrices of the element u and the generator I A in the representation [A] of SU(N), respectively. When [A] is the self-representation [l] of SU(N), D[’](u) = u and I?] = TA. Let CA,” and ~ A B D be the totally symmetric and antisymmetric tensors of rank three with respect to the adjoint representation. We are going to show that they are the invariant tensors. Prove it with CA; as example.

CA,D + C Di$iD$$iD$DiCAiD,: A’B’D‘ = - 2 i n { U-~TAUU-~TBUU-~TDU - u - ~ T ~ u u - ~ T ~ u u -= CA! .

(8.39) According to the Littlewood-Richardson rule, except for N = 2, the C-G se- ries for the direct product of two adjoint representations of SU(N) contains, in addition to the other representations, only one identical representation and two adjoint representations. Therefore, the identical representation only appears twice in the reduction of the direct product of three adjoint representations of SU(N). Equation (8.39) says that CA: and ~ A B D are the only invariant tensors of rank three, up to a constant factor, with re- spect to the adjoint representation of SU(N). For the SU(2) group, since there is only one adjoint representation contained in the CG series for the direct product of two adjoint representations, and then, there is only one identical representation contained in the CG series for the direct product of

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three adjoint representations. On the other hand, d ~ g ~ = 0 for SU(2), and the invariant tensor of rank three with respect to the adjoint representation of SU(2) has to be proportional to CA,D.

Now, we replace TA on the right hand side of Eq. (8.9) with It1. They are still the symmetric or antisymmetric invariant tensors of rank three with respect to the adjoint representation. Thus, they are proportional to cA[ or dAgg, respectively

From the first equality we obtain

Tz([X]) and A([X]) are respectively related to the Casimir operators of rank two and rank three of SU(N), and sometimes are simply called the Casimir invariants of rank two and rank three in the literature. Both Tz([X]) and A( [A]) are independent of the similarity transformation of the representa- tion. It is easy to show from Eq. (8.40) that

(8.42)

In the calculation of T!([X]) with Eq. (8.41), we take A = D = 3 such that both generators contained in Eq. (8.41) belong to the subgroup SU(2). Denote T2([A]) in SU(2) by T2(')([X]), which is calculated to be

1 T 2 ( O ) ( [ X ] ) = Tr { = 12 X(X + 1)(X + 2) . (8.43)

Calculate T2( [A]) in the subduced representation from the irreducible rep- resentation [A] of SU(N) with respect to the subgroup SU(N - 2)@SU(2), which is equivalent to a direct sum of some irreducible representation [v] @I [p] of the subgroup,

(8.44)

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In the calculation of A([X]) with Eq. (8.40), we take A = B = 3 and D = 8 such that three generators are contained in Eq. (8.40) all belong to the subgroup SU(3), where d338 = m. Denote A([X]) in SU(3) by do) "XI),

In the planar weight diagram, the states located in the same horizontal

level have the same eigenvalues of Tix1, and the sum of (Tix1)2 can be calculated by Eq. (8.43). Then, we sum up the contribution for the states with different TiX1. For a one-row Young pattern, we have

x A(')([A,O]) = 2 C (A - 3m)Tz("'([A - m])

m=O x (8.45)

= (1/6) C (A - 3m)(X - m)(X - m + 1)(X - m + 2) m=O

= (1/120)X(A + 1)(X + 2)(X + 3)(2X + 3).

Generally,

A(O)([A + A', A']) = A(O)([A] CZI [A']*) - A(O)([A - 11 CZI [A' - 1]*)

= dpll ( s W ) ) A ( O ) ([XI) - 4 x 1 (sw~))A(O)([X'I) - ~ ~ ~ , - - ~ I ( S U ( ~ ) ) A ( ' ) ( [ X - 13) + ~ ~ ~ - ~ I ( S U ( ~ ) ) A ( O ) ( [ X ' - 11) 1

120 = - (A + 1)(A' + 1)(A - A')(A + A' + 2)(A + 2X' + 3)(2A + A' + 3).

(8.46) Calculate A( [A]) in the subduced representation from the irreducible repre- sentation [A] of SU(N) with respect to the subgroup SU(N - 3)@SU(3), which is equivalent to a direct sum of some irreducible representation [v] @ [p] of the subgroup,

21. Calculate the Casimir invariants T2([1']) and A([1']) in the represen- tation denoted by a one-column Young pattern ([l']) of the SU(N) group.

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Unitary Groups 373

Solution. From Eqs. (8.44) and (8.47), we obtain

1 N - 2 Tz([l‘]) = d[lr-~l(SU(N - 2)>T4O’([1]) = - ( ) 2 r - 1 ’

A([lr]) = {d[,r-1](SU(N - 3)) - d[lp-21(SU(N - 3))) A‘O’([1])

(N - 3)! - (N - 3)! - - (r - l)!(N - r - 2)! (N - 3)! {(N - r - 1) - (T - 1))

(r - l)!(N - r - l)!

(r - 2)!(N - r - l)!

- -

22. Calculate the Casimir invariants T2([3]) and A([3]) in the irreducible

Solution. From Eq. (8.43) we have

representation [3] of the SU(6) group.

T4O’([3]) = 5, Ti0’([2]) = 2, TiO’([l]) = 1/2 TiO’([O]) = 0.

From the reduction of the subduced representation from [3] of SU(6) with respect to the subgroup SU(4)@SU(2)

we obtain

From Eq. (8.45) we have

A(’)([3]) = 27, A(’)([2]) = 7, A(’)([l]) = 1 A(’)([O]) = 0.

From the reduction of the subduced representation from [3] of SU(6) with respect to the subgroup SU(3)@SU(3)

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374 Problems and Solutions in Group Theory

we obtain

23. Calculate the Casimir invariants T2([X]) and A([X]) in the representa-

Solution. The calculation method is the same as Problem 22. Here we only list the result.

tion ([A] = [A, O,O, 01) of the SU(5) group.

T 2 ( O ) ( [ X ] ) = X(X + 1)(X + 2)/12,

d[A](SU(3)) = (A + 2)(X + 1)/2, d[A](SU(2)) = + 1.

A(O)([X]) = X(X + 1)(X + 2)(X + 3)(2X + 3)/120,

Hence

Generalizing this result to the SU(N) group, we obtain

N - X ( X + l ) . - ( X + N )

2 ( N + l ) ! )

- n=O

N X(X + 1) - - - (A + N ) ( U + N ) ( N + 2)!

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Chapter 9

REAL ORTHOGONAL GROUPS

9.1 Tensor Representations of S O ( N )

* The set of all N-dimensional unimodular real orthogonal matrices, in the multiplication rule of matrices, constitutes SO(N) group. In this chapter we discuss only the case of N 2 3, because S O ( 2 ) group is an Abelian group which is isomorphic onto the U(1) group. The S O ( N ) group is a doubly-connected compact Lie group with order N(N - 1 ) / 2 . SO(N) is a subgroup of SU(N). The generators in the self-representation of S O ( N ) is usually taken as the second type of generators in the self-representation of SU(N) multiplied by two

(Tab) , , = 2 (TLt)) = - idarSbs + i basdbr , a < b, r s

Tr ( T a b T c d ) = 2SacSbd - 26adSbc, ( 9 4

where the indices a and b are the ordinal indices of the generator, while T

and s are the row and column indices of the matrix. N(N - 1 ) / 2 matrices Tab all are antisymmetric, and they are also antisymmetric with respect to the ordinal indices. Any matrix R in SO(N) can be expressed in the form of exponential matrix function:

R = e x p -iC WabTab . (9.2) { a<b 1 j , The S O ( N ) group with N = 2 l or N = 2 l + 1 contains l mutually

commutable generators,

375

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376 Problems and Solutions in Group Theory

which span the Cartan subalgebra of the Lie algebra of SO(N) . Thus, the ranks of both SO(2t) and S0(2C+ 1) are C. We want to find the common eigenvectors of H j , [ H j , Ea] = ajEa. When N = 2 4 the eigenvectors are

1 E,, ( 2 ) - - 5 [T(2a)(2b- l ) + iT(2a-1) (2b- l ) + i T ( 2 a ) ( 2 b ) - T ( 2 a - 1 ) ( 2 b ) ] )

(9.4) (3)

Eab = 5 [ T ( 2 a ) ( 2 b - 1 ) - i T ( 2 a - 1 ) ( 2 b - l ) + iT(2a)(2b) + T ( 2 a - l ) ( 2 b ) ] 7

with the eigenvalues { e, - eb} j , { -e, + eb}j, { e , + eb} j , and { -e, - eb} j , respectively, where a < b, and { e , } j = 6,j. When N = 2 t + 1, in addition to Eq. (9.4), there are another two types of eigenvectors:

= 8 [T(2,)(2l+l) + i ~ ( 2 a - l ) ( 2 C + l ~ l 3

(9.5)

with the eigenvalues { e , } j and - { e , } j , respectively. The simple roots for the S 0 ( 2 l + 1) group are

The remaining positive roots are

where a < b. The largest root, which is the highest weight of the adjoint representation of SO(2C + l), is

v=2

Therefore, the Lie algebra of S0(2C+ 1) is Bl, and L3 = w2. The simple roots for the SO(2t) group are

The remaining positive root are

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Real Orthogonal Groups 377

where a < b . The largest root, which is the highest weight of the adjoint representation of S0(2 l ) , is

Therefore, the Lie algebra of SO(2 l ) is De, and c3 = w 2 .

* The Chevalley bases in the self-representation of the S 0 ( 2 l + 1) group (the Be Lie algebra) are

H p = Ep =

Fp = He =

Ee = Fe =

where 1

q 2 e ) ( 2 e + l ) + i q 2 e - l ) ( 2 e + l ) , (9.10)

5 p < l. The Chevalley bases in the self-representation of the SO(2l) group (the DC Lie algebra) are the same as those of the S 0 ( 2 l + 1) group except for those when p = l , which are

He = T(2[-3)(2&-2) + 7'(2e-i)(2e),

Fe = $ { ~ ( 2 e - 2 ) ( 2 e - 1 ) + i ~ ( 2 e - 3 ) p e - i ) - i ~ ( 2 [ - 2 ) ( 2 e ) + T(ze -3 )pe ) ) -

* The definition for a tensor with respect to the S O ( N ) group is the same as that to the SU(N) group, except that the transformation matrix u ESU(N) is replaced with R &O(N) . The Weyl reciprocity still holds for the tensors of SO(N) . However, this replacement makes some new characteristics for the tensors of SO(N) . Since the transformation matrix R E S O ( N ) is real, there is no difference between the covariant tensor and the contravariant tensor for the SO(N) group. As a result, first, the tensor space of S O ( N ) group can be decomposed into a direct sum of a series of the traceless tensor subspaces 7, which are invariant in SO(N) . Each 7- can be projected by the Young operators into the subspace = $?IT.

= $ { T(21-2)(2e-l) - iT(21-3)(2l-l) + iT(21-2)(21) + T(2e-3)(21!)} 7

(9.11)

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378 Problems and Solutions in Group Theory

Second, since the tensor is traceless, the tensor subspace is a null space if the sum of the numbers of boxes in the first and the second columns of the Young pattern [A] is larger than N. Third, since there is no difference be- tween the covariant tensor and the contravariant tensor, the representation described by a Young pattern [A] with the row number m larger than N / 2 is equivalent to that by [A'], where [A'] is obtained from [A] by replacing its first column with a column containing ( N - m) boxes. Fourth, when the row number of [A] is equal to N/2 , the representation space '7$ ]̂ is reduced into two minimal invariant subspaces with the same dimension. They are the self-dual and the anti-self-dual tensor subspaces, whose representations are denoted by [+A] and [-A], respectively.

* A tensor of rank two for SO(N) is easy to be decomposed into the sum of the traceless tensors:

where the tensor in the curve bracket is a traceless tensor of rank two, and the tensor in the square bracket is a tensor of rank zero. However, generally speaking, the decomposition of a tensor of rank n into a sum of some traceless tensors of different ranks is straightforward, but tedious. As discussed in Chap. 8, the irreducible tensor bases for SU(N) are obtained explicitly. In fact, they are the standard tensor Young tableaux. The bases can be orthonormalized by making use of the block weight diagram. How- ever, due to the traceless condition, some standard tensor Young tableaux for SO(N) become linearly dependent. It becomes a new problem how to find the complete set of the independent irreducible tensor bases for S O ( N ) explicitly. Another problem is that the tensor basis Oal...,, is not the com- mon eigenstate of the Chevalley bases H,. Since a tensor basis is a direct product of n vector bases O,, -..O,,, we only need to find the new vector bases such that H p are diagonal. Namely, we want to diagonal- ize H p in the self-representation of S O ( N ) by a similarity transformation. Both problems are well solved for the SO(3) group. Now, we generalize the solution to the S O ( N ) group.

* In the self-representation space of SO(2C + 1) we defined a set of new orthonormal bases, called the spherical harmonic bases,

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(-l)e-a+l + i 0 2 , ) when 1 5 a 5 l , @a = @2e+1 when c t = l + l ,

(@4~-2 ,+3 - i@4C-2a+4) when ! + 2 5 a 5 2 l + 1. (9.12)

The tensor basis @,l . . . (y , of rank n for S 0 ( 2 l + 1) is the direct product of n vector bases a,, - - - @ a n . In the spherical harmonic bases a,, the nonvanishing matrix entries of the Chevalley bases are

i

(9.13)

where 1 5 p 5 l - 1. In the spherical harmonic bases, the standard tensor Young tableau $l+,,...(y,, in the representation [A] of S0(2!+ 1) corresponds to the highest weight if each box in its j t h row is filled with the digit j , because each raising operator Ep annihilates it. Therefore, the relation between the highest weight M and the Young pattern [A] for S 0 ( 2 l + 1) is

Although the standard tensor Young tableaux is generally not traceless, the standard tensor Young tableau with the highest weight in [A] of S 0 ( 2 l + 1) must be traceless, because it only contains with a < l + 1. For example, the tensor basis 0 1 0 1 is not traceless, but is traceless. The remaining traceless tensor bases in an irreducible representation space [A] of S0(2 l+ 1) can be calculated from the basis with the highest weight by the lowering operators Fp. In the calculation the block weight diagram is helpful.

* In the self-representation space of SO(21) we defined a set of new or- thonormal bases, called the spherical harmonic bases,

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380 Problems and Solutions an Group Theory

The tensor basis @)(Y1...CYn of rank n for SO(2t) is the direct product of n vector bases aal - - - a,, . In the spherical harmonic bases a,, the nonvan- ishing matrix entries of the Chevalley bases are

(9.16)

where 1 5 p 5 C - 1. In the spherical harmonic bases, the standard tensor

to the highest weight if each box in its j t h row, j < t, is filled with the digit j and if all boxes located in its t th row, if [A] contains k' row, are filled with the same digit l or k ' + 1, depending on whether the representation is self-dual [+A] or anti-self-dual [-A]. Therefore, the relation between the highest weight M and the Young pattern [A] for SO(2t) is

Young tableau yp [XI @al...Qn in the representation [A] of SO(2t) corresponds

Since the standard tensor Young tableau with the highest weight contains either @e or @!+I, in addition to with a < C, it is a traceless tensor. The remaining traceless tensor bases in an irreducible representation space [A] of SO(2l) can be calculated from the basis with the highest weight by the lowering operators Fp.

* The dimension d[xl(SO(N)) of the representation [A] of S O ( N ) can be calculated by the hook rule. In this rule the dimension is expressed as a quotient, where the numerator and the denominator are denoted by the

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Real Orthogonal Groups 381

symbols Ygl and YL”], respectively:

Let us explain the meaning of two symbols YL”] and YP]. The hook path (i, j) in the Young pattern [A] is defined to be a path which enters the Young pattern at the rightist of the i th row, goes leftwards in the i row, turns downward at the j column, goes downwards in the j column, and leaves from the Young pattern at the bottom of the j column. The inverse path (i, j) is the same path as the hook path (i, j) but with the opposite direction. The number of boxes contained in the path (i, j ) , as well as in its inverse, is the hook number hij. YL”’ is a tableau of the Young pattern [A] where the box in the j t h column of the i th row is filled with the hook number hij. Define a series of the tableaux Ygl recursively by the rule given below. Ypl is a tableau of the Young pattern [A] where each box is filled with the sum of the digits which are respectively filled in the same box of each tableau Ygl in the series. The symbol Ygl means the product of the filled digits in it, so does the symbol YL”].

The tableaux Ygl are defined by the following rule:

(a) Ygl is a tableau of the Young pattern [A] where the box in the j t h column of the i th row is filled with the digit

(b) Let [A(1)] = [A]. Beginning with [A(1)], we define recur- sively the Young pattern [A(”)] by removing the first row and the first column of the Young pattern until [A(”)] contains less than two columns.

(c) If [A(”)] contains more than one column, define Ygl to be a tableau of the Young pattern [A] where the boxes in the first ( a - 1) row and in the first (a - 1) column are filled with 0, and the remaining part of the Young pattern is nothing but [A(”)]. Let [A(“)] have r rows. Fill the first r boxes along the hook path (1, 1) of the Young pattern [A(”)], beginning with the box on the rightmost, with the

( N + j - i).

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382 Problems and Solutions in Group Theory

digits (Xi"' - l) , (A?) - l), "-, (A?) - l), box by box, and fill the first (Xi"' - 1) boxes in each inverse path (i, 1) of the Young pattern [A(")], 1 5 i 5 T , with -1. The remaining boxes are filled with 0. If a few -1 are filled in the same box, the digits are summed. The sum of all filled digits in the pattern Ygl is zero.

1. Calculate the dimensions of the irreducible representations of the SO(8) group denoted by the following Young patterns: (a ) [4,2], (b) [3,2], (c) ~ 4 ~ 4 1 , (d) [+3,2,1,11, (el [+3,3,1,11.

8 9 1 0 1 1 -1 - 1 1 3 7 8 1 1 1 4 5 8

= 4312. - - 7 8 + -2 0 Solution. (a). d[4,2~(S0(8)) =

5 4 2 1 5 4 2 1 2 1 2 1

8 9 1 0 . - 1 1 2 7 1 0 1 2

2 1 2 1

8 9 1 0 1 1 -1-1 3 3 0 0 0 0 7 8 1 3 14 7 8 9 1 0 + - 2 - 1 - 1 0 ' 0 - 1 - 1 2 - 5 6 7 12

- 5 4 3 2 5 4 3 2 (c). d[4,4](S0(8)) =

4 3 2 1 = 8918.

4 3 2 1

8 9 10 0 1 2 8 1 0 1 2 7 8 0 0 7 8 6 ' - 1 5

4 1 4 1

1 1 2 x 2 2 x 2

8 9 10 0 2 2 0 0 0 8 1 1 1 2 7 8 9 0 0 0 0 - 1 1 7 7 1 0 6 + - 2 + o 4

3 = 4312. -- -

5 -2 0 (el. d[+3,3,1,1](S0(6)) = 6 3 2 6 3 2

5 2 1 2 x 2

5 2 1 2 2 x

1 1

2. Calculate the Clebsch-Gordan series for the subduced representations of the following irreducible representations of the SU(N) group with respect to the subgroup S O ( N ) , and then check the results by their dimensions for N = 7:

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Real Orthogonal Groups 383

Solution. The following results for the reductions of the subduced repre- sentations of SU(N) with respect to the subgroup SO(N) are independent of N , except for the Young patterns in the CG series with the row number not less than N/2, or with the digit number in the first two column larger than N.

210 = 189 + 21.

+ +EP 77 + 105 + 7. @

1008 = 819

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384 Problems and Solutions an Group Theory

882 = 693 + 77 + 105 + 7.

'3 490

t- 224 = 189 +

756 = 616 + 105 + 35.

= 378 + 105 + 7.

35.

924 = 714+ 182 + 27+ 1.

2646 = 1911 + 182 + 168 + 330 + 2 x 27 + 1. p+p@r2100 = 1560 + 189 + 330 + 21.

1176 = 825 + 330 + 21.

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Real Orthogonal Groups 385

2352 = 1617

V@F@P+ 189 + 330 + 168 + 27 + 21.

140 = 105 t

840 = 616 + 189 + 35.

490 = 294+ 168+ 27+ 1.

= 378 + 189 + 21.

.35.

3. Calculate the Clebsch-Gordan series in the reductions of the following direct products of the irreducible tensor representations, and check the results by their dimensions for the SO(7) group:

(1) PI €3 PI, (2) PI €3 [I, 11, (3) [31 €3 [2,11*

Solution. (1):

27 x 27 = 729 = 182 + 330 + 168 + 27 + 21 + 1. m

27 x 21 = 567 = 330 + 189 + 27 + 21.

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386 Problems and Solutions in Group Theory

EF 77 x 105 = 8085 = 1750 + 1911 + 1560 + 1617

+ 182 + 330 + 168 + 330 + 189 + 27 + 21.

4. Calculate the orthonormal bases in the irreducible representation space [2,2] of SO(5) by the method of the block weight diagram, and then express the orthonormal bases by the standard tensor Young tableaux in the traceless tensor space of rank four for SO(5).

Solution. The Lie algebra of SO(5) is B2. The relations between the simple roots rp and the fundamental dominant weights wp are

rl = 2wl - 2w2, r 2 = -wl + 2w2,

The highest weight of the representation [2,2] of SO(5) is (0,4), and its dimension is

5 6 1 1 6 7

2 1 2 1

From the highest weight state (0,4) we have a quartet of A2 with (1,2), (2,0), (3,g) and (4, z), where the matrix entries of F 2 are 2, &, &, and 2, respectively. The weights (1,2) and (2,O) are the single dominant weights. From (1,2) we have a doublet of A1 with (T, 4). From the weight (2,O) we have a triplet of d1 with (0,2) and (z,4), and from the weight (T,4) we have a quintet of A2 with (0,2), (l,O), (2,2) and (3,a). The weight (0,2) may

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be a double dominant weight. Define the first basis state I(0,2)1) belonging to the triplet of A1 and the other I(0,2)2) belonging to a singlet of CAI. Let

Applying El F 2 = F2El to the basis state l(i, 4))) we have

Thus, a1 = &, and then, a2 = d m = 1. From the weight (3,2) we have a quartet of d1 with (1,0), (i, 2)) and (3,4). On the other hand, applying F2 to two states )(0,2)1) and 1(0,2)2), we obtain another two states with the weight (1,O). Thus, the dominant weight (1,O) may be a triple weight. We define the first basis state (1, O), belonging to the quartet of dl and the other two belonging to two doublets of d1. The action of F 2 on 1(0,2)1) is the combination of I ( 1 , O ) l ) and I(l,0)2). The weight (2,2) is the double weight because it is equivalent to (0,2). The weight ( i , 2 ) is equivalent to ( 1 , O ) . We have

Let

Applying E1F2 = F2El to the basis state )(0,2)1) and 1(0,2)2), we have

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388 Problems and Solutions an Group Theory

Choosing the phases of the basis states 1(1,0)2) and 1(1,O)3) such that a4 and a7 are real positive, we have a4 = 1, a6 = &, and a7 = 0. It means that the dominant weight ( 1 , O ) as well as its equivalent weight (T,2) in [2,2] are the double weights, not triple. The states 1(1,O)3) and I(T, 2)3) do not exist in [2,2]. From the weight (4, q) we have a quintet of -41 with the weights (2,2)1, (0,0)1, (2,2)1 and (3,4), where the matrix entries of F1 are 2, &, & and 2, respectively. Another basis state 1(2,2)2) belongs to a triplet of -41 with (0, O), and (2,2)2, where both matrix entries of F1 are Jz. Let

Applying E1F2 = F2El to the basis state I ( l , O ) 1 ) and 1(1,0)2), we have

Thus, a8 = 4 and a10 = 0. Applying E2F2 = F2E2 + H2 to the basis state I ( l , O ) 1 ) , we have

Choosing the phase of the basis state )(2,2)2) such that a9 is real positive, we have a9 = 1 and all = 2. Applying E2F2 = F2E2 + H2 to the basis state 1(2,2)1), we have

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Real Orthogonal Groups 389

Choosing the phase of the basis state 1(3,;?>) such that a12 is real positive, we have a12 = 1 and a13 = 4.

Applying F2 to the states with the double weight (T,2), we have two basis states with the weight (O,O), in addition to the state bases I(0,O)l) in the quintet and I(O,O)2) in the triplet of -41. Thus, the dominant weight (0,O) may be a quadruple weight. Both I(O,O)3) and I(0, O),) are singlet of A1. Let

Applying ElF2 = F2E1 to the basis states 1(2, 4))) I(i, 2)1) and [(i) 2)2) , we have

Thus, we have bl = b2 = b3 = b4 = b7 = fi and b6 = 0. Applying E2F2 = F2E2 + H2 to the state basis [(i, 2)1), we have

Thus, b5 = 0, and the basis state 1(0,0)4) does not exist. So, b g = 0. Applying E2F2 = F2E2 + H2 to the basis state l(i, 2)2) , we have

Choosing the phase of the basis state 1(0,0)3) such that b8 is real positive, we have bs = fi.

The remaining basis states and the matrix entries of the lowering oper- ators Fp can be calculated similarly. The results are given in Fig. 9.1.

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390 Problems and Solutions in Group Theory

Fig. 9.1 The block weight diagram for [2,2] of SO(5).

Now, we expand the orthonormal basis states in the standard tensor Young tableaux. The Young tableau yi2’21 is Fl . F’rom the Fock

condition (8.19) we have

The normalization factors for the typical tensor Young tableaux are

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Real Orthogonal Groups 391

From Problem 2 , we know that the number of linearly independent trace tensors in the tensor space of rank four projected by yPy2] is 50 - 35 = 15. In fact, they belong to the representation [2,0] and [O,O] of SO(5). We can calculate the trace tensors from the highest weight condition (7.24) and in terms of the lowering operators. In the following we only list the trace tensors with the dominant weights.

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392 Problems and Solutions in Group Theory

+ 2 p + q 2 5 - 2

Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (9.13) and the block weight diagram. Due to the Wigner-Eckart theorem, the state basis belonging to the representation [2,2] of SO(5) is orthogonal to the basis states belonging to [2, O] and [O,O], so it is a traceless tensor. For simplicity we neglect the index (0,4) for the highest weight in the basis state I(0,4), (ml , ma)).

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394 Problems and Solutions in Group Theory

5 . Calculate the orthonormal bases in the irreducible representation space [2,0,0,0] of SO(8) by the method of the block weight diagram, and then, express the orthonormal bases by the standard tensor Young tableaux in the traceless symmetric tensor space of rank two for SO(8).

Solution. The Lie algebra of SO(8) is D4. The relations between the simple roots rp and the fundamental dominant weights wp are

rl = 2wl- w2, 1-2 = -w1+ 2w2 - WQ - w4,

The highest weight of the representation [2,0,0,0] of SO(8) is (2,0,0,0), and its dimension is

From the highest weight state (2,0,0,0) we have a triplet of A1 with

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Real Orthogonal Groups 395

(0, l , O , 0), and (2,2,0,0), where both matrix entries of F1 are a. The weight (0, l , O , 0) is a single dominant weight. The weights equivalent to two dominant weights are

Therefore, the representation [2,0,0,0] contains two simple dominant weights (2,0,0,0) and (0, 1, 0,O) and one triple dominant weight (O,O, 0,O). The block weight diagram for the representation [2,0,0,0] is very easy except for the part related to the triple weight ( O , O , O , O ) , which will be explained below. In Fig. 9.2 we give the block weight diagram for the rep- resentation [2,0,0,0] where only those matrix entries which are not equal to 1 are indicated.

Let 1(2,%,0,0)), 1(0,0,0,0)1) and I(?, l , O , O ) ) constitute a triplet of A1, and the other two basis states I(O,O, 0,0)2) and I(O,O, O,O)3) are the singlets of C A I . We assume

where we neglect the index (2,0,0,0) for the highest weight in the basis state I(2,0,0,0), ( m l , m2, m3, m4)) for simplicity. Applying El F2 = F2E1 to the basis state I (i, 2, i, i)), El F3 = F3 El to the basis state I(0, i, 2, 0)), and ElF4 = F4E1 to the basis state I ( O , i , O , Z ) ) , we have

E1F2 ( ( i , Z , i , 1)) = J Z C l ((2, i, 0,O)) = F ~ E ~ ((i, 2 , i , i ) ) = F~ ((I , i,i,i)) = ](2,i,o,o)),

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396 Problems and Solutions an Group Theory

Thus, c1 = d1/2 and a1 = bl = 0. Choosing the phase of the basis state I ( O , O , O , O ) 2 ) such that a2 is real positive, we have a2 = a. Applying E3F2 = F2E3 to the basis state I(i, 2 , i , i ) ) , and E3F4 = F4E3 to the basis state I(0, T, 0,2)), we have

Thus, c2 = and b2 = 4. Then, b3 = 0. Choosing the phase of the basis state 1(0,0,0,0)3) such that c3 is real positive, we have c3 = 1.

Now, we expand the orthonormal basis states in the standard tensor Young tableaux. The Young tableau yi2’o’0’01 is ml . The normaliza- tion factors for two typical tensor Young tableaux are

There is only one trace tensor in the tensor space of rank two by y/2~0~0~01 . It belongs to the representation [O,O, O,O] of SO(8):

projected

Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (9.16) and the block weight diagram. Due to the Wigner-Eckart theorem, the basis state belonging to the representation [2,0,0,0] of SO(8) is orthogonal to the basis state belonging to [O,O, O,O], so it is a traceless tensor.

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Real Orthogonal Groups 397

I 2, ?,o( 01 I ii; , i, i I A = 18 + 27

B = 36 + 45

( ? , O , O , O l C=-18 + 27 + 36 - 45

Fig. 9.2 The block weight diagram for [2,0,0,0] of SO(8).

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398 Problems and Solutions in Group Theory

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Real Orthogonal Groups 399

6. Calculate the spherical harmonic functions Ygl in an N-dimensional

Solution. The relations between the rectangular coordinates xa and the spherical coordinates T and e b in an N-dimensional space are

space.

(9.19)

a=l

The unit vector along x is usually denoted by j;: = x / r . The volume element of the configuration space is

N N-1

The orbital angular momentum operators Lab are the generators of the transformation operators PR for the scalar function, R ESO(N),

F'rom the second Lie theorem, Lab satisfy the same commutation relations as those of the generators Tab in the self representation of SO(N). The Chevalley bases H,(L), E,(L) and F,(L) can also be expressed by Eqs. (9.10) and (9.11), where Tab are replaced with Lab. Because

N N

c=l c=l

where J a b are the generators of OR. The common eigenfunctions X, of H,(L) can be obtained from @a given in Eq. (9.12) for S 0 ( 2 l + 1) and in Eq. (9.15) for SO(2l) by replacing the vector basis 0, with the rectangular coordinate z a / T , where the factor T - ~ stands for removing the dimension of length.

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400 Problems and Solutions in Group Theory

The spherical harmonic function Ykl(ii) is the eigenfunction of the or- bital angular momentum for a single particle. Since there is only one co- ordinate vector x, the representation [A] for the spherical harmonic func- tion Ykl(P) has to be the totally symmetric representation denoted by the one-row Young pattern [A] = [A, 0, . . . ,O]. X,, multiplied by a normaliza- tion factor, is nothing but the spherical harmonic function Ykl(ii) in the self-representation [A] = [l]. The component mp of the weight m is the eigenvalue of H p ( L ) in the eigenfunction X,. Generally, for the highest weight state M = (A, 0,. . . ,0) we have

Y i l ( k ) = cjvxx: = CN,* { (-V ( 5 1 + is , )}* 7 (9.22) T d 5

where t is equal to l when N = 2 l + 1 and to ( l - 1) when N = 21. The remaining spherical harmonic function Y k l ( i ) with the weight m can be calculated by the lowering operators F p ( L ) , just like we have done in Problems 4 and 5. Here we have to calculate the normalization factor C N ,and the eigenvalue of the angular momentum square L2. Both problems can be solved from the highest weight state Y&](P).

Since z1 + ix2 = reiel sin 02 . . . sin ODM1 and

(n - 1) 1 . 3 . 5 . . . I,.

('* 1 . 3 . 5 . : . (4 An in de (sir

' when n is odd,

we have

For SO(2l + 1) group, we have

2 * 4 - 6 * * - (2A + 2 1 - 2) - - 2 ' 4 4 7 1 ) (z) . . . ( A + l - 1 7r ) 2 - 1 * 3 . 5 . . * ( 2 A + 2 l - l ) A + 1 x + 2

2 * + 2 W A ! ( A + l - l ) ! - - (2X + 2l - l)! ' *

(9.23)

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Real Orthogonal Groups

For SO(2l) group, we have

401

T'X! - - 21-x7r (1) ('> .. . ( 7r ) = 2 X - l ( X + & I)!. x + l x + 2 X + l - l (9.24)

Assuming both b and d are not equal to 1 and 2, we have

Thus, the eigenvalue of L2 in the spherical harmonic function Ykl(Ec) is k'(k'+ N - 2).

7. Calculate the complete set of the independent bases for the eigenfunc- tions of angular momentum in a three-body system which is spherically symmetric in an N-dimensional space.

Solution. After separating the motion of the center of mass, there are two Jacobi coordinate vectors in a three-body system, denoted by R1 = x and R 2 = y. Thus, the eigenfunction of angular momentum in the system belongs to an irreducible representation, denoted by a Young pattern with only one or two rows, [XI , X 2 , 0 , . . . , O ] = [XI, A,]. In a three-dimensional space, the eigenfunction of angular momentum is described by the represen- tation D' and the magnetic quantum number m. But, in the N-dimensional space, N > 3, the eigenfunction of angular momentum is described by the representation [A,, A,] and the weight m. Due to the spherical symmetry, we only need to calculate the eigenfunction of angular momentum with the highest weight, namely, m = d in the three-dimensional space and m = M = (A1 - X2, X 2 , 0 , . . . ,0) in the N-dimensional space. The remain- ing eigenfunctions can be calculated by the lowering operators Fp ( L ) , just like we have done in Problems 4 and 5. Therefore, the eigenfunction of angular momentum with the highest weight in the N-dimensional space is described by two parameters A1 and X2.

Denote by Q;lX2 (x, y) the independent bases of the eigenfunctions of angular momentum with the highest weight, where q is understood as an ordinal parameter. What the complete set means is that any eigenfunction $Jk (x, y) with the angular momentum [XI , A,] and the highest weight M can be expanded with respect to Q;IX2(x, y),

(9.25)

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402 Problems and Solutions in Group Theory

where the coefficients +q(c) only depend upon the internal invariants t. In the three-body system there are three internal variables: = x-x, 62 = y - y and 6 3 = x - y. What the independent bases means is that any basis cannot be expanded with respect to the other bases like Eq. (9 .25) . Obviously, a basis is not a new independent basis if it is a product of another basis and a function of the internal variables.

Consider the product of two spherical harmonic functions Ykl(ii) and Yk) (y). It belongs to the direct product [q] x [p3 of two irreducible repre- sentations, which is usually a reducible one. Its reduction can be calculated by the Littlewood-Richardson rule and the contraction of the components of x and y. In the contraction the factor of the internal variables appears. Therefore, the independent bases of angular momentum come from the calculation result by the Littlewood-Richardson rule.

The representations with t > 0 correspond to the bases which are not independent. Conversely, any independent basis of angular momentum [XI, A,] can be calculated by the product Ykl (k)Y$](y) where X2 5 q 5 X1 and p = X1 + X2 - q. Note that the highest weight state corresponds to the standard tensor Young tableau where each box in the first row is filled with 1 and each box in the second row is filled with 2 , and the tensor indices in the same column of the tensor Young tableau are antisymmetric. Now, the digit 1 means X1 = c(z1 + i 2 2 ) or Y1 = c(y1 + iy2) and the digit 2 means X2 = -C(Q + izd) or Y2 = - c ( y ~ + iy4), where c is a constant factor. Up to the normalization factor, the independent basis state Q?lX2(x, y) is expressed as a product of three parts:

For SO(4) group the irreducible representation denoted by a two-row Young pattern can be self-dual or anti-self-dual. The standard tensor Young tableau for the highest weight state of the self-dual representation is still in the form ( 9 . 2 6 ) , but that of the anti-self-dual representation is different. In fact, each box in the second row of that standard tensor Young tableau for the highest weight state of the anti-self-dual representation is filled with 3,

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Real Orthogonal Groups 403

not 2. In this case, the independent basis state Q;IX2 (x, y) is expressed as

For SO(3) group, due to the traceless condition, the sum of the numbers of boxes in the first two columns of the Young pattern describing the irre- ducible representation is not larger than 3, namely, A2 = 0 or 1. On the other hand, the representation [ A , l ] for SO(3) is equivalent to the repre- sentation [A, 01. But, the bases of angular momentum for [A, 01 and [A, 11 have different parities.

9.2 Spinor Representations of S O ( N )

* Let N matrices yu satisfy the anti-commutation relations,

The set of all their products constitutes a finite group, denoted by r N . I’N

is a matrix group. Usually, the matrix “(a is defined by its representation matrix in a faithful unitary irreducible representation of r N . The matrices yu, called the irreducible yu matrices, are unitary and hermitian. When N = 2l is even, define

yf = (-i)”2y1yz.. .yp/* (9.29)

yf is also unitary and hermitian, and yf is anticommutable with any matrix ya. Thus, the matrix yj and N matrices yu satisfy the anti-commutation relations (9.28). They are regarded as the definition of yu in rN+l. Note that

The left-hand side of Eq. (9.30) changes its sign if one transposes arbitrary two matrices ya and yb, or to change the sign of arbitrary one matrix yu. When N = 4rn + I, the right-hand side of Eq. (9.30) is not a new element, the group r4m+1 is isomorphic onto the group r4m. When N = 4m - 1, the right-hand side of Eq. (9.30) is a new element, the set r4m-2 and its

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404 Problems and Solutions an Group Theory

product with i is isomorphic onto the group l?4m-1.

r4m+l M r4m, r4m-1 = (r4m-2, i r4m-2) . (9.31)

When N = 21, there are 22'+1 elements in For any element of rN, except for the constant matrix, there exists at least one element which is anti-commutable with the given element. Thus, the trace of the element is zero. The dimension d2') of the irreducible ya matrices which satisfy Eq. (9.28) can be calculated to be 2' by the character formula. The matrices in r2', neglecting their signs, are linear independent, and form a complete set of bases for the 2'-dimensional matrices.

The equivalent theorem says that any two sets of irreducible Ta matrices which respectively satisfy Eq. (9.28) are equivalent to each other. Namely, they can be related one-to-one by a similarity transformation X :

(9.32)

When N = 21 + 1, the dimension of matrix ^la is still 2[. The equivalent condition for two sets of the irreducible ya matrices should include, in addition to Eq. (9.28), that two products y1 . . . y~ are equal to each other.

* The equivalent theorem of ya plays an important role in the spinor the- ory of SO(N) . The charge conjugate matrix C and the space-time inversion matrix B in particle physics can be defined based on it. When N = 21,

- = X - l y a X , 1 5 a 5 21.

c - l y a c = - ( y a y , CtC = 1, CT = (_l)W+1)/2C,

B ~ B = 1,

detC = 1,

B-lyaB = (%IT , detB = 1, BT = (-1)-!?('-1)/2B.

Due to the condition (9.30), only matrix C can be defined when N = 4m-1, and only matrix B can be defined when N = 4m + 1.

The fundamental spinor representation D["] (R) of SO( N ) , which is sim- ply called the spinor representation in literature, is also defined based on the equivalent theorem:

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Real Orthogonal Groups 405

Its generators Sab = -iYaYb/2, a < b, are called the spinor operators. The fundamental spinor representation, being a group, is homomorphic onto SO(N) by 2 1 correspondence. When N is odd, the fundamental spinor representation D['I(R), or denoted by [s], is irreducible. [s] is real when N = 8k f 1, but self-conjugate when N = 8k f 3. Its dimension is di,] = 2(N-1)/2. The fundamental spinor representation [s] is reducible when N is even. I t can be reduced into two inequivalent representations, denoted by D[*"](R) or [fs], [s] N [+s] @ [ -s ] , with the same dimension d [ k S ] = 2(N/2) -1 . [fs] are conjugate to each other when N = 4k + 2. [ fs] are real when N = 8k and self-conjugate when N = 8k + 4.

* In S O ( N ) transformation, Jr is called the fundamental spinor of S O ( N ) if it transforms by the fundamental spinor representation D['] (R) :

where Q is a column matrix with d[,] components. When N is even, @ can be decomposed into two parts by the project operators Pk = (1/2)(1 f yf), Q* = P*Q.

* The spin-tensor is a spinor with the tensor subscripts,

The spin-tensor space is reducible. In its minimal invariant subspace, the spin-tensor has to satisfy the usual traceless condition and the second trace- less condition:

and is projected by the Young operator. The Young pattern for the Young operator has the row number not larger than N/2, otherwise the subspace corresponds to the null space. The representation corresponding to the minimal invariant subspace is called the irreducible spinor-tensor repre- sentation or simply the spinor representation, which are described by a Young diagram with a symbol s: [s,X] = [ s , X ~ , X ~ , - - - ] for an odd N or [fs, A] = [ f s , XI, X2, - .] for an even N . The spinor representation is double-valued one of SO ( N ) . * The dimension of a spinor representation [s, A] of S 0 ( 2 l + 1) or [fs, A]

of SO(2l) can be calculated by the hook rule. In this rule the dimension

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406 Problems and Solutions in Group Theory

is expressed as a quotient, where the numerator and the denominator are denoted by the symbols Ygl and YiA1, respectively:

(9.35)

We still use the concept of the hook path (i, j ) in the Young pattern [A], which enters the Young pattern at the rightmost of the i th row, goes leftwards in the i row, turns downwards at the j column, goes downwards in the j column, and leaves from the Young pattern at the bottom of the j column. The inverse path (i, j ) is the same as the hook path (i, j ) except for the opposite direction. The number of boxes contained in the hook path (i, j ) is the hook number hij of the box in the j t h column of the i th row. YiA1 is a tableau of the Young pattern [A] where the box in the j t h column of the i th row is filled with the hook number hij. Define a series of the tableaux Yitl recursively by the rule given below. Ykl is a tableau of the Young pattern [A] where each box is filled with the sum of the digits which are respectively filled in the same box of each tableau Yg' in the series. The symbol Ypl means the product of the filled digits in it, so does the symbol YiA1.

The tableaux YE1 are defined by the following rule:

(a) Y t ' is a tableau of the Young pattern [A] where the box in the j t h column of the i th row is filled with the digit

(b) Let [A(1)] = [A]. Beginning with [A(1)], we define recur- sively the Young pattern [A(")] by removing the first row and the first column of the Young pattern [A("-')] until [A(")] contains less than two rows.

(c) If [A(")] contains more than one row, define YE1 to be a tableau of the Young pattern [A] where the boxes in the first (a - 1) row and in the first (a - 1) column are filled with 0, and the remaining part of the Young pattern is nothing but [A(")]. Let [A(")] be T row. Fill the first (T - 1) boxes along the hook path (1, 1) of the Young pattern [A(")], beginning with the box on the rightmost, with the digits A?', A t ) ,

( N - 1 + j - i).

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Real Orthogonal Groups 407

- - - , A?), box by box, and fill the first A!') boxes in each inverse path (i, 1) of the Young pattern [A(")], 2 5 i 5 r , with -1. The remaining boxes are filled with 0. If a few -1 are filled in the same box, the digits are summed. The sum of all filled digits in the pattern Y x l is zero.

SO(7) group denoted by the following Young patterns: 8. Calculate the dimension of the irreducible spinor representation of the

Solution.

5 4 3 1 3 2 1

6 7 8 0 1 1 5 $ 0 4 -2

5 2 1 2 1

= 8 x

6 7 8 0 2 2 0 0 0 5 6 + - I 0 $ 0 1 4 5 - 2 - 1 0 - 1

5 4 1 3 2 2 1

6 8 9 5 2 5 2 1 2 1

= 8~

= 1728.

6 9 10 4 7

L} 5 4 1 = 3024.

3 2 2 1

9. Calculate the highest weight of the fundamental spinor representations

Solution. Fkom the second Lie theorem, the spinor operators Sub satisfy the same commutation relations as those of the generators Tab in the self representation of SO(N) . The Chevalley bases H,(S), E,(S) and F,(S) in

of the S O ( N )

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408 Problems and Solutions an Group Theory

the fundamental spinor representation can also be expressed by Eqs. (9.10) and (9.11)) where Tab are replaced with S a b . It is convenient to choose an explicit forms of the ya matrices for calculating the highest weight of the fundamental spinor representations, although the calculated result is independent of the chosen forms. Since the dimension of ya for the SO(2l) group is 2e , we may define the ya matrices of l?2e to be the direct product of l Pauli matrices such that Eq. (9.28) is satisfied:

y 2 P - 1 = 1 x . . . x 1 x01 x 0 3 x . . . x 0 3 - - P - 1 1 -P

P--1 e - P

,

(9.36) y 2 P = t x .;. x ; x 0 2 x 0 3 x . . . x 0 3 , - yf = 0 3 x . . . x 0 3 . -

e

In fact, they are also ya matrices of r 2 0 + 1 with 7 2 e + 1 = yf. Thus, the Chevalley bases in the fundamental spinor representation of SO(2l + 1)) whose Lie algebra is Be, are

HP = 1 x . . x 1 xdiag(0, 1, - 1, 0) x 1 x . . x 1 - -)

," 1P - 1 e - P - i - - P - 1 e - P - i

H e = IX ~ 1 x 0 3 , - e- 1

03 x . * * x 0 3 XQ+ ) - - e - P - i P--1

E P = IX X ~ X ( O + X O - } X ~ X . . x l

F P = IX x ~ x { ~ - x o + } x ~ x x l ,

Ee = Fe = 0 3 x * . . x 0 3 XCT- )

e- 1 e- 1

(9.37)

where 1 5 p < l . The Chevalley bases in the fundamental spinor represen- tation of the SO(2l) group, whose Lie algebra is De, are the same as those of the S 0 ( 2 l + 1) group except for those when p = !, which are

He = 1 x x l xd iag (1 , 0, 0, -1) ) - e-2 - Ee= - l x . . . x l x { o + x a + } ,

e - 2

e - 2

Fe= - 1~ . . . x I x { a _ x a - } . - (9.38)

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Let a and p be two eigenfunctions of 03, respectively,

a = ( ; ) , P = ( ; ) ,

ff3a = a, ff+a = 0, f f -a = p, asp = -p, a+p = a, a-P = 0.

Denote by x(m) the basis states with the weight m in the fundamental spinor representation [s] of S 0 ( 2 & + 1) and by x*(m) the basis states with the weight m in the fundamental spinor representation [ks] of SO(2l). From the condition that each raising operator annihilates the highest weight state, we obtain the basis state x(M) for the highest weight for [s] of S0(2k‘+ 1) and the basis state x,t(M) for the highest weight for [ks] of SO(2C) as follows:

x(M) = C X X . - - X Q I 1’

e M = (0, 0, ..., O , l ) ,

-’ e

for [s] of SO(2&+ I),

x+(M) = C Y X . - . X ~

M = (0, 0, . . . , O , l ) , for [+s] of S0(2&),

x-(M) = C Y X . - . X C X X ~ , - e- 1

M = (0, 0, . . . , l ,O), for [-s] of SO(2l).

(9.39)

10. Calculate the highest weight of the spinor representation [s,A] of S0(2&+1), and the highest weights of the spinor representations [fs, A]

Solution. The spin-tensor of SO(N) corresponds to the irreducible repre- sentation [s, A] when N = 2 & + l or to [ks, A] when N = 2& if the spin-tensor satisfy the usual traceless condition and the second traceless condition, given in Eq. (9.34). The generators in the irreducible spinor representation are the total angular momentum operators Jab = L a b + Sub, a < b. The or- bital angular momentum operators Lab and the spinor angular momentum operators S,b are respectively applied to the tensor part and the spinor part of the spin-tensor. The total angular momentum operators Jab satisfy the same commutation relations as those of Lab and Sab. The Chevalley bases H,(J) , E,(J) and F,(J) in the irreducible representation [s, A] can also be expressed by Eqs. (9.10) and (9.11), where Tub are replaced with

of SO(2t)

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410 Problems and Solutions in Group Theory

Jab. Therefore, the weight of a spin-tensor basis is the sum of the weights of the tensor part and the spinor part. Let [A] = [A,, Aa, . . . , A,]. The highest weight M of the spinor representation [s,A] of S0(2e+ 1) is

The highest weight M of the spinor representation [+s,A] of SO(2e) is

The highest weight M of the spinor representation [-s,A] of SO(2e) is

11. Calculate the Clebsch-Gordan series for the direct product of the ten- sor representation [A] and the fundamental spinor representation [s] of S0(2C+ 1) or [fs] of SO(24, where [A] is a single row Young pattern or a single column Young pattern.

Solution. If the tensor part of a spin-tensor qal...,,, belongs to the tensor representation [A] but it does not satisfy the second traceless condition, namely, its tensor part satisfies the traceless condition for each pair of the subscripts ai and aj and is projected by a Young operator, then, the spin- tensor space corresponds to the direct product of the tensor representation (A] and the fundamental spinor representation. The method for reducing the direct product representation is to decompose the spin-tensor space into a series of subspaces satisfying the second traceless condition (9.34). In fact, xb TbQal...b...a,, is a new spin-tensor where the tensor rank decreases by one. If [A] is a single row Young pattern, the tensor part of the spin-tensor is a totally symmetric traceless tensor such that twice application of the second traceless condition is trivial, zbc Tb~c*al...b...c...a, = 0 due to Eq. (9.28). If [A] is a single column Young pattern, the tensor part of the spin-tensor is a totally antisymmetric tensor. Note that the antisymmetric tensor space of rank n of S O ( N ) is equivalent to that of rank N - n. Thus, the reductions of the direct product representations for SO(2.t + 1) are

[A,O,. . . , O ] x [s] 2 [s, A,O,. . * , O ] c3 [s, (A - l ) ,O , * * * ,O],

[PI x [s] 21 [s, I”] CEI [s, ln - - l ] CB .. . (9.40)

[s], n 5 .t , and those for SO(2l) are

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Real Orthogonal Groups 411

[A)O,. . . ,O] x [h] N [As, x ,o , . . . ,O] a3 [ys, (A - l ) , O , . . . ,O],

The sign in fs changes because P k y , = where P* = (1 f y f ) / 2 . Those formulas can be shown with the method of the dominant weight diagram. The reader is encouraged to check them for the groups SO(7) and SO(8). Equations (9.40) and (9.41) can be partly checked by their dimensions. From Eqs. (9.18) and (9.35), the dimensions of [A] and [s,X] of S O ( 2 t + l), for example, are

( 2 t ) ( 2 t + 1). . . ( 2 t + x - 2 ) ( 2 t + 2 x - 1) A! q x , o , ..., 01 =

( 2 t + x - 2 ) ! ( 2 t + 2 x - 1) (21 - l)!A!

- - >

( 2 l f l)! ' when n 5 t ,

d ~ l n ~ = ( 2 t - n + 1) !n ! 7

+,x,o ,..., 01 = 2' (2!)(2t + 1). . . ( 2 t + x - 1) ' (2!+ x - l)!

= 2 A!

Then, for Eq. (9.40) we have

( 2 t + x - 2 ) ! ( 2 t + 2 x - 1) ( 2 t - l ) !X!

2e ' ( 2 t + A - l)! + 2t ( 2 t + A - 2) ! = 2 ( 2 t - l)!X! ( 2 t - 1)!(A - l ) ! )

( 2 t + 1 ) ! (2 t - 2a + 2 ) ( 2 t - a + 2)!a!

= 2 e x when n 5 t . ( 2 t + l)! (21 + 1 - n)!n! )

a=O

2e

12. Calculate the basis states in the fundamental spinor representation [s]

Solution. The Lie algebra of the SO(7) group is B3. The highest weight of the fundamental spinor representation [s] of SO(7) is (0, 0 , l ) . Its dimension is 8. Its block weight diagram is as follows. All nonvanishing matrix entries of the lowering operators FD are 1.

of SO(7).

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412

i,i, 1

II x( i7 i7 1) = x p x a,

Problems and Solutions an Group Theory

JI - 1, i , i

x ( i , o , i ) = - x p x p,

x(o, i , i ) = - p x p x a,

~ ( o , o , i ) = - p x p x p.

Fig. 9.3 T h e block weight diagram of [s] of B3.

13. Calculate the basis states in the fundamental spinor representations [AS] of SO(8).

Solution. The Lie algebra of SO(8) group is Dq. The highest weights of the fundamental spinor representations [+s] and [-s] of SO(8) are (O,O, 0 , l ) and (0, 0,1,0), respectively. The dimensions of both representations are 8. Their block weight diagrams are as follows. All nonvanishing matrix entries of the lowering operators Fp are 1.

By the notation used in Problem 9, we have

x + ( O , O , O , 1) = a x a x a x a, x - ( O , O , l , O ) = a x x a x p, x+(o , i , o , i ) = - x x p x p, x- (o , i , i , o ) = a x a x p x 0,

x + ( i , i , i , o ) = - x p x x p, x-( i , i ,o , i ) = a x p x a x a,

x+( i ,o , i ,o ) = - p x a x x p, x-(i,o,o,i) = p x x x a,

x+( i ,o , i , o ) = - a x p x p x a, x-(i,o,o,i) = - a x p x p x P , x + ( i , i , i , o ) = - p ~ ~ ~ p ~ ~ , x - ( i , i , o , i ) = - p x c ~ x p x p ,

x+(o, i ,o , i ) = - p x p x a x a, x-(o , i , i ,o ) = - p x p x x p, X+(o,o,o, i ) = p x p x p x p, x-(o,o, i ,o) = - p x p x p x a.

By the notation used in Problem 9,we have

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Real Orthogonal Groups 413

a) [+sl b) [-Sl

Fig. 9.4 The block weight diagrams of [fs] of Dq.

14. Expand the eigenfunction of the total angular momentum with the highest weight in terms of the product of the spherical harmonic func- tion Y:’(EE) and the spinor basis x(m).

Solution. In Problem 11 the Clebsch-Gordan series for the direct product representation [X,O, . . . , O ] x [s] is given. For the SO(2t + 1) group, we express the representations in Eq. (9.40) by their highest weights,

(A,O.. . , O ) x (0 . . . ) O , 1) 21 (A,O.. . ) O ) 1) @ ( A - 1 , o . . . , O ) 1).

Let (J) = (A,O.,.)0,1), (L) = ( A , O , . , , O ) , @ + I ) = (A+1 ,0 ..., O), and (S) = (0 . . . , O , 1). There are two ways to obtain the total angular momentum state ( J ) : ( L ) x (S) and ( L + 1) x (S). The highest weight state can be calculated by the condition that it should be annihilated by each raising operator. The highest weight state for the first case is easy to calculate:

where C(z~+l),x is given in Eq. (9.23). The highest weight state for the second case is

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4 14 Problems and Solutions in Group Theory

For the SO(2C) group, we express the representations in Eq. (9.41) by their highest weights,

(A, 0 . . . , O ) x (0 . . . ,o , 1) 2 (A, 0 . . . ) 0 , l ) @ (A - 1) 0 . . . ,o , 1) O),

(A,O * * - , O ) x (0. - * ) O , 1 , O ) N (A,O. . . , O ) 1,O) 63 (A - 1 ) O . . . ,o , 1).

Let ( J+) = (A$. . . , O , l), ( J - ) = (A$. . . , O , 1,0), ( L ) = (A$. . . , O ) , ( L + 1) = (A + 1 , O . . . , O ) , (+S) = (0 . . . ,0 , l), and (-S) = (0 . . . ,O,l,O). There are two ways to obtain the total angular momentum state (J*): ( L ) x (fS) and ( L + 1) x ( r S ) . It is easy to calculate the highest weight state for the first case:

where C ( 2 1 ) , ~ is given in Eq. (9.24). The highest weight state for the second case is

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Real Orthogonal Groups 415

+ (z3 + i ~ & - [ ( i , i , 0 , . . . ,o , i)] + (21 + iz2)x-[(i, 0, * * * , 0, l)]} -

9.3 SO(4) Group and the Lorentz Group

* The inequivalent irreducible representations of the Lorentz group can be obtained from those of SO(4) group. This is the general method for study- ing the inequivalent irreducible representations of a non-compact group.

* Find new bases of generators of SO(4) from six generators Tab,

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416 Problems and Solutions an Group Theory

The new bases are separated into two sets. Two generators in different sets are commutable. Express them in the form of the direct product of the Pauli matrices:

By the similarity transformation N , we have

-10 0 1 N = - ( 1 -i 0 0 -i )

Jz 0 1 1 0 - 0 i - i 0

Therefore, any element R of SO(4) can be expressed in the direct product of two matrices

R = exp -i

= e x p l - i

( I

4 \

W u b T a b ) a<b=2 1

3

where

3

Thus, matrix R is explicitly expressed in the product u1 x 212 of two- dimensional unimodular unitary matrices. If two matrices u1 and u 2 change their signs at the same time, R keeps invariant. Therefore, Eq. (9.43) gives

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Real Orthogonal Groups 417

a one-to-two correspondence between an element R in SO(4) and two ele- ments u1 and u2 in SU(2). Furthermore, this correspondence keeps invari- ant in the multiplication of the group elements. Thus,

SO(4) - SU(2) x SU(2)'. (9.44)

We restrict the varied area of the parameters of the SO(4) group such that there is a one-to-one correspondence between a set of parameters and a group element, at least in the region with nonzero measure:

(9.45)

where O ( * ) and v(*) are the polar angle and the azimuthal angle of the direction n(*). Since the group space of the second SU(2)' group is reduced to that similar to the group space of the SO(3) group, namely, the two ends of a diameter in the group space correspond to the same element, the group space of SO(4) is doubly-connected.

Any irreducible representation of SO(4) can be expressed as the direct product of two irreducible representations of two SU(2) groups, denoted by Djk :

Djk (fi(+),u(+);~(-),u(-)) = ~ . i ( f i (+) ,J+)) ~k ( & - ) , J - ) ) .

(9.46) The row (column) index of Djk is denoted by two indices (pv ) . The di- mension of Djk is ( 2 j + 1)(2k + 1). Its generator I:"*) is expressed by the generator I: of Su(2):

(9.47)

Djk with different superscripts jIc are inequivalent to each other. When ( j + I c ) is an integer, Djk is a single-valued representation of S0(4), i.e., the irreducible tensor representation. When ( j + k - 1/2) is an integer, Djk is a double-valued representation of S0(4), i.e., the spinor representation. The correspondence between Djk and the irreducible representation denoted by

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418 Problems and Solutions in Group Theory

a Young pattern [X,, A,] is as follows. When j + k is an integer, we have

Djj [ 2 j , 01, when j = k ,

Djk N [+( j + k ) , ( j - k ) ] when j > k , Djk 21 [ - ( j + k), (k - j ) ] when j < k,

and when j + k - 1/2 is an integer, we have

Djk 21 [+s, ( j + k - 1/2), ( j - k - 1/2)], Djk 21 [-s, ( j + k - 1/2), (k - j - 1/2)],

when j > k, when j < k.

Note that [+X,,Xz] is the self-dual representation, and [-X,,Xz] is the anti-self-dual one. DOo is the identical representation, Dqo and Do* re- spectively are two fundamental spinor representations [+s] and [-s]) and D f 3 is equivalent to the self-representation.

* The Lorentz transformation matrix A between two inertial frames is a 4 x 4 orthogonal matrix:

A*A = A A ~ = I. (9.48)

The matrix entries of A satisfy the following condition:

a and b = 1, 2, 3. Aab and A44 are real

Aa4 and A4a are imaginary

This condition keeps invariant in the product of two matrices A. The set of all such orthogonal matrices A, in the multiplication rule of matrices, constitutes the homogeneous Lorentz group, denoted by O(3,l) or Lh. The orthogonal condition (9.48) gives

3

detA = f l , Ai4 = 1 + IA,4I2 2 1. a=l

These two discontinuous constraints separate the group space of O(3, l ) into four unconnected sections. The element in the section where the identity belongs to satisfies

detA = 1, A44 2 1. (9.49)

Those elements satisfying Eq. (9.49) constitute a simple Lie subgroup, called the proper Lorentz group, denoted by L,. Since there is no upper limit of A44, the group space of L, is an open area in the Euclidean space. Thus, L, is a non-compact Lie group. The representative elements in three

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Real Orthogonal Grozlps 419

cosets of L, are usually chosen to be the space inversion CT, the time inver- sion r and the whole inversion p:

o = diag(-1, - 1, - 1, 1) , r = diag(1, 1, 1, - 1) , p = diag(-1, - 1, - 1, - 1).

rt Let A be an infinitesemal element of L,:

A = I - i a X , 1 = ATA = 1 - ia ( X + X T ) , 1 = detA = 1 - i a T r X ,

AT = 1 - i a X T , X T = - X , Trx = 0.

Thus, X is a traceless antisymmetric matrix. Expand X with respect to the generators in the self-representation of SO(4):

3 3

a<b=2 a=l

a=l

where d a b is real, w a 4 is imaginary, and

3

b<c=2

Except for the imaginary parameters, the generators

(9.50)

(9.51)

in the self- representations of SO(4) and L, are completely the same, so are the gener- ators in the corresponding irreducible representations of two groups. The finite-dimensional inequivalent irreducible representations of L, are also denoted by Djk(L,), whose generators are given in Eq. (9.47). Since the parameters for SO(4) and L, are different, the global properties of the two groups are very different. Any finite-dimensional irreducible representation Djk of L,, except for the identical representation, is not unitary. There exist infinite-dimensional unitary representations for L,.

* The transformation generated by T a b is obviously a pure rotation, be- longing to the subgroup SO(3). Let us study the transformation generated

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420 Problems and Solutions in Group Theory

by T34 with the parameter w34 = iw:

0 0

A(G3,iw) rexp{-i(iw)T34} = (" 0 0 coshw -isinhw ) sinh w = (w/c) (1 - 'u2/c2) -1'2 .

0 0 isinhw coshw (9.52)

w = ctanhw, coshw = (1 - w2/c2)-1'2,

It describes the Lorentz boost along the direction of x axis with the relative velocity w = c tanh w.

Any Lorentz transformation A satisfying Eqs. (9.48) and (9.49) can be decomposed into the product of some rotational transformation and the Lorentz boost along the x direction. Let A44 = coshw, from which 13 is determined. Extracting a factor -i sinh w from Aa4, we obtain a unit vector n(8, cp) in the three-dimensional space, whose rectangular coordinates are (i/ sinhw)(A14rA24,A34):

A44 = COShw, iA24/ sinh w = sin 8 sin cp,

iA14/sinhw = sinOcoscp, iA34/ sinh w = cos 8.

(9.53)

Extending the rotational matrix into a 4 x 4 matrix, we have

coscpcos8 - sincp coscpsin8 0 sincpcoso coscp sincpsin8 0

0 0 0 1 0 R(g3 , p)R(s2) 0) =

Left-multiplying to A the inverse matrix of R(G3, cp)R(&, 8)A(G3) i w ) ) we obtain a rotational matrix R(a, p, T),

from which the Euler angles a, P and y can be calculated. Thus,

The geometrical meaning of the decomposition is evident. Two rotations in two-sides of A(&, iw) rotate two z axes in two frames before and after the Lorentz transformation A to the direction of the relative motion, and the remaining axes to be parallel to each other, respectively. After two rotations, the Lorentz transformation A is simplified into A(&, '1'13).

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Real Orthogonal Groups 42 1

* Since all inequivalent irreducible representations of L, have been known, the inequivalent irreducible representations of the homogeneous Lorentz group O(3,l) can be obtained by determining the representation matrices of r and p. The representation matrix of o can be calculated by the formula o = ~ p . Since p is commutable with any element in O(3,l) and the p2 is equal to the identity, the representation matrix of p in an irreducible representation of O(3,l) is a constant matrix, where the constant is fl in a single-valued representation. The property of the representation matrix of r can be determined by the relations in the self-representation

Let V ( x ) , X = 1, 2, 3 and 4, be four inequivalent irreducible represen- tations of the four-order inversion group V4. When j = k , The irreducible representation DJJ of L, induces four inequivalent irreducible representa- tion Ajjx of O(3,l):

. .

AjjX(A) = D j j ( A ) , A E L,, (7) V(’) (T)S,,~ SvcLj, (9.55)

Ajjx(p) = V(’)(p) 1, 1 5 X 5 4.

When j # k , while j + k is an integer, the representation Djk @ Dkj of L, induces two inequivalent irreducible representation Ajk* of O(3,l):

where a and P are used to identify two representation spaces of Djk and Dkj . Since Ajk*(r) has no diagonal entries, changing its sign leads to an equivalent representation.

When j + k is a half-odd number, we only discuss the Dirac spinor representation D(O(3,l)) here. For definiteness, the index p runs from 1 to 4, while the index a runs from 1 to 3. Let T~ be four anti-commutable matrices and C be the charge-conjugate transformation matrix:

c-ly,c = -y;, C+C = 1, CT = -c, de tC = 1.

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422 Problems and Solutions in Group Theory

Thus, D(A) satisfies

4

D(A)-'ypD(A) = Apuyu, detD(A) = 1,

C-lD(A)C = {D(A-l)}*. IA44 I

u = l (9.57)

The additional factor A44/IA441 comes from the physical reason. Its gener- ators are

The representation matrices of the representative elements in the cosets of L, are

D ( a ) = fiy4, D(7) = fy4y5, D ( p ) = fiy5. (9.59)

15. Discuss the classes in the SO(4) group and calculate their characters

Solution. Any element R of SO(4) can be transformed into the standard form by a real orthogonal similarity transformation X:

in the irreducible representation D j k .

where Eq. (9.43) is used. Therefore, the class of SO(4) is described by two parameters cpl and y2. Its character in the representation D j k ( R ) is

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Real Orthogonal Groups 423

16. Calculate six parameters of the following proper Lorentz transforma- tion A, and write its representation matrix in the irreducible represen- tation Djk (A) of the proper Lorentz group L,:

/1 0 0 0 \ 0 &/2 (cosh w ) / 2 -i(sinh w ) / 2

A(cp7 " w 7 " = 0 -1 /2 &(cosh w ) / 2 -i&(sinh w ) / 2

\ O 0 isinhw coshw 1

Solution. From the fourth column of A(cp, 8, w , a, P, r), the boost parame- ter is w , and 6' = 7r/6 and cp = 7r/2 because sin 6' cos cp = 0, sin0 sin cp = 1 / 2 and cos 0 = &/2. The matrix form of the rotation R(7r/2,7r/6,0) is

0 - 1 0 0 f i / 2 0 1 / 2 0 1 0 0 0

0 0 0 1 0 0 0 R(7r'2,n/6,0)= (0 0 l o ) ( -;/2 : &2 ;)

/ 0 &/2 - 1 / 2 o \ R(n/2, 7r/6,0)-' =

Thus,

From this we obtain and Thus, we obtain therepresentation matrix of A in

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424 Problems and Solutions in Group Theory

17. Prove that the Dirac spinor representation satisfy

Solution. The Dirac spinor representation is not unitary. The aim of this problem is to study the property of its conjugate representation. From the definition (9.57) of the Dirac spinor representation we have:

To make the formulas uniformly, we make a similarity transformation 7 4 :

From the equivalent theorem of 7a, we obtain

Substituting Eq. (9.59) into it, we determine the constant c :

This is the reason why the matrix 7 4 should be inserted between two spinors $t and 1(, to construct an invariant quantity of the Lorentz transformation in the quantum field theory:

18. Discuss the classes of the proper Lorentz group L,.

Solution. In the self-representation of L, we write the pure rotation R ( M , P, 7 ) and the Lorentz boost A(&, iw) with the relative velocity along the z axis in terms of Eq. (9.43)

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Real Orthogonal Groups 425

Thus, any element A(cp, 8, w, a, P, r) in L, can be expressed as

where

The matrix M is a two-dimensional matrix with determinant +1, which belongs to the two-dimensional unimodular complex matrix group SL( 2 ,C) . We are going to show the homomorphism between SL(2,C) and L, with a 2:l correspondence. First, we have known from Eq. (9.60) that an arbitrary Lorentz transformation A = A(cp, 8, w , a, P, 7) corresponds to a matrix M E

SL(2,C). Conversely, any M given in Eq. (9.61) determines a Lorentz transformation A. We will show that any element P in SL(2,C) can be expressed in the form (9.61) later. If two matrices M and M' correspond to the same Lorentz transformation A, then

Thus, M - l M ' = c l . Since detM = 1 and detM' = 1, c2 = 1 and c = fl . There is a 2:l correspondence between fM given in Eq. (9.61) and the Lorentz transformation A. This correspondence keeps invariant in the multiplication of elements. In fact, if

we have

AA' = N { M M ' x [ 0 2 ( M M ' ) * ~ 2 ] } N - l .

Now, we have proved that the two-dimensional unimodular complex matrix group SL(2,C) is homomorphic onto the proper Lorentz group L, with a 2:l correspondence. Thus, we can determine the class of L, by SL(2,C). Recall that two matrices fM eSL(2,C) correspond to the same Lorentz transfor- mation A E L,. Two classes in SL(2,C) different by a sign correspond to the same class in L,. If two eigenvalues of M eSL(2,C) are different, M can be diagonalized through a similarity transformation X eSL(2,C)

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426 Problems and Solutions in Group Theory

Thus, we have

-n- I cp I r1 0 0 coshw 0 0 isinhw coshw

coscp - s in9 0 sincp coscp 0 A = (

The class is described by two parameters w and cp. If two eigenvalues of M are equal, the eigenvalues have to be fl. fl

correspond to the identity element of L,, which constitutes one class in L,. If M is not equal to fl, M can be transformed into the Jordan form by a similarity transformation Y eSL(2,C) (see Problem 16 in Chap. 1):

Y - % Y = f ( , 1 - 2 1>. Since

we obtain 01 = -22, 0 2 = 2, 0 3 = 0. In order to calculate the six parameters of A, we make a transformation

The solutions are cp = -7r, 6' = r/4, coshu = 3, a = T , p = 37r/4, y = 0, and

1 0 2 - - e-i2T31 -2Tl4

-2 0 -1 -2i -2i 0 - 2i 3

It is easy to check that A is a proper Lorentz transformation with all four eigenvalues to be 1. A has only two linearly independent eigenvectors for the eigenvalue 1. A is transformed into the Jordan form by the similarity

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Real Orthogonal Groups

transformation 2:

427

0 0 2 - 1 0 1 0 0 0 0 0 i/4

0 - 4 0 1 0 - 4 i o 0 O O l i

1 0 0 0

0 0 1 1 *

0 0 0 1

Z - l A Z = ( 0 1 1 0 ) Finally, we are going to show that, any two-dimensional unimodual

complex matrix P ESL(2,C) can be expressed into the form of Eq. (9.61), namely

Solving the equation, we have

where c,g = cos(O/2), direct calculation leads to

= sin(O/2) and cp = cos(P/2) and so on. The

IAI2 = - (sin 8 sin p cos a) /2 + c$cgew + sis$e-*, IB12 = (sin8 sinp cosa) /2 + cis$ew + sic$e-w, ICI2 = (sin8 s inp cosa) /2 + s;4cgew + czsge-", 1OI2 = - (sin 8 sin p cos a ) /2 + sis$ew + cic$e-w, AB = e-iCP { - cos p sin 8 + sin p ( -ciewe-ia + sie-w ia > > / 2 ,

AC = e-iy {case sinp + sin8 (c$ewe-ia - s$e-weia)} /2, AD = sis$ +tic$ - secespcp (ewe-ia - e-+eia),

[A12 + IBI2 + [C12 + = ew + e-w,

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428 Problems and Solutions in Group Theory

w , 8 and p can be respectively calculated from the last three formulas. Then, a is determined by the formula for AD. At last, y and 9 can be calculated from the formulas for AC and AB. Here, the calculated method is not unique due to detP = 1. This completes the proof.

For example, we calculate the parameters of a matrix P :

P = - ( 1 4 [a - 1 - i (a + l)] --a + 1 - i 3 (a + 1) 4 a+ 1 - i 3 (A- 1) d [ A + 1 + i (A- I ) ]

coshw = 3, sinhw = a, ew/2 = Jz+ 1, e-w/2 = Jz- 1 case = 112, e = 7r/3, C O S ~ = -1/2, p = 21~/3 .

Then, from AD = 3(1 - i & ) / S , we obtain

+ e-weiff = 6cosa - i 4 a s i n a =

The solution is a = -7r/2. Finally, from

AC = a [ - 1 - 2i(3 - &)] - - e-iy 8

we have e-zv = -i and e e i Y = -1, and determine cp = 7r/2 and y = 7r. At last, we have

19. Express an arbitrary element in the proper Lorentz group L, in the

Solution. In Problem 18, we have shown that L, - SL(2,C) with the

form of exponential matrix function.

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Real Orthogonal Groups

correspondence (9.60) :

429

An arbitrary element A in L, can be written in the form of exponential matrix function if its corresponding elements f M in SL(2,C) can. Since detM = 1, its two eigenvalues have the same sign. If its eigenvalues are positive, M can be transformed into the form of exponential matrix function by a unimodular similarity transformation. If its eigenvalues are negative, -M can be transformed into the form of exponential matrix function by a unimodular similarity transformation. Ignoring the irrelevant sign, we can express an arbitrary element M in SL(2, C) as

M = f exp (46 - $12) , 02 M*u2 = f exp (46* . $12) .

By Eq. (9.60), the proper Lorentz transformation A corresponding to M can be expressed as

a<b a=l

where Ti*), Tab and Ta4 are the generators in the self-representation of S0(4), and the parameters Wab and wc4 are the real part and the imaginary part of O,, respectively:

1 2i

3 1 2 Wab = - &bc (0, + n:), Wc4 = - (a, - n:).

c= 1

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430 Problems and Solutions in Group Theory

Thus, an arbitrary Lorentz transformation A can be expressed in the form of exponential matrix function, where the parameters u a b are real, and ma4

are imaginary. The set of parameters up, is mainly for the theoretical study, and the set of parameters (cp,8, u, a, P, 7) is convenient for calculation. The relation between two sets of parameters are given by Eq. (9.60), namely

20. Let

x1 + ix2 -23 - ix4 ) ’ 23 - ix4 x1 - ixa 3

x = -ix41+ C xaga = a=l

where xa is real, 24 = ict is imaginary, and X is a hermitian matrix. Let M E S L ( ~ , C ) ,

3

M X M ~ = X I = - i x & l + C X L u a .

a=l

(a) Calculate Tr X and det X . (b) Prove xh = C, Ap,x,, where A is a proper Lorentz transformation. (c) Prove that L, -SL(2,C). (d) Calculate the corresponding matrix M , when A is equal to B(G3, cp ) , R(&, 8), and A(&, i w ) , respectively. (e) Calculate the ma- trix M corresponding to an arbitrary proper Lorentz transformation 4% 6, u, a, P, 7 ) .

4

Solution. (a) TrX = -i2x4, det X = - x:. p=l

(b) Since XI = M X M t is hermitian, it can be expanded with respect to the Pauli matrices and 1, where the coefficients xb, are real, and x i is imaginary. Further, x; can be expressed as the linear combinations of x p ,

4

u=l

where A a b and A 4 4 are real, and A 4 a and A a 4 are imaginary. det X’ = det X ,

Due to

4 4

p=l p=l

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Real Orthogonal Groups 43 1

Therefore, A is a Lorentz transformation matrix, which is an orthogonal matrix with the matrix entries depending upon M . When M = 1, A is the identity. Since the group space of SL(2,C) is connected, the set of A can be continuously extended from the identity, namely, A is a proper Lorentz transformation matrix.

(c) From a given matrix M , we can calculate X' = MXMt, and then, determine a proper Lorentz transformation matrix A uniquely. If the ma- trices X ' calculated from M and M' are the same, so are the matrices A , then M-lM' is commutable with three Pauli matrices ua. Thus, it is a constant matrix. Due to detM=detM' = 1, M' = & M . Namely, there is a 2:l correspondence between &M and A. This mapping is obviously invari- ant in the multiplication of elements. Therefore, SL(2,C) is homomorphic onto L,, L, - SL(2,C).

(d) Ignoring the sign in front of the matrix M , we can express the matrix M and A by the same parameters. If A ESO(3) is a pure rotation, M obviously belongs to SU( 2),

M(&, cp) = u(e'3, cp) = 1 cos (cp/2) - i 0 3 sin (cp/2),

M(e'2,O) = u(e'2, 8) = 1 cos (8/2) - iaa sin (e/2),

For a boost A = A(&,iw), we have

M(&,iw)XM(e'3,iw)t = -i l( ix3 sinhw + 2 4 coshw) + 03(x3 coshw - ix4 sinhw) + 01x1 + 02x2.

If only one component xp is nonvanishing, we have

M(G3, iU)OlM(e'3, i w ) t = 01,

M(e'3, iw)a2hf(e'3,iw)t = 0 2 ,

x1 = 1,

x2 = 1,

2 3 = 1, 2 4 = i.

Ad(&, iw)a3M(Z3, iw)t = 1 sinhw + 0 3 coshw, M(G3,iw)M(G3,iw)t = 1 coshw + 0 3 sinhw,

Let M(e'3,iw) = 1Mo+alM1 +a2M2+a3M3. Extracting the Pauli matrix from the left-hand side of the first two formulas, and then, moving it to

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432 Problems and Solutions an Group Theory

their right-hand side, we have

(1Mo + a lM1- 02M2 - 03M3) M t = 1, (1Mo - a& + a2M2 - a3M3) M+ = 1.

Adding and subtracting two equations, we obtain MI = M2 = 0, IMo12 - lM3I2 = 1, and MOM,* are real. Thus, the last two formulas become the same, and turn to lMOl2 + lM3I2 = coshw and 2MoM,* = sinhw. Recall that det M = 1. The solution is MO = coshw/2 and M3 = sinhw/2, namely

M(e"3,iw) = ( ewoi2 e - o w / 2 ) = u(Z3,iu).

(e) For an arbitrary proper Lorentz transformation A(p, 8, u, a, p,?), we have

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Chapter 10

THE SYMPLECTIC GROUPS

10.1 The Groups Sp(24 R) and Usp(21)

j r Take the subscript a of a vector in a (2l)-dimensional space to be p or p, 1 5 p 5 i?, in the following order: -

a = l , i , 2 , 2 , - , e , Z . (10.1)

Define a (2l)-dimensional antisymmetric matrix J :

1 when a = p, b = p J a b = -1 when a = p , b = p { 0 otherwise, J = l c x (ia2) = -J-l = - JT, det J = 1 .

(10.2)

The set of all (2t) x ( 2 4 real matrices R satisfying

RTJR = J, R* = R, (10.3)

in the multiplication rule of matrices, constitutes the real symplectic group Sp(2l, R). Both the inverse R-' and the transpose RT of any element R of Sp(2l, R) satisfy Eq. (10.3):

R-1 = -JRTJ,

-J ( - J R T J ) ~ J (-JRTJ) J = RJRT = J. ( ~ - 1 ) ~ JR-1 = J ,

(10.4)

Sp(2[, R) is not a compact Lie group, because the general form for a diag- onal matrix & E Sp(2l, R) contains the parameters up without an upper limit:

& = diag {ewl, e-wl, ew2, e -W2, . . . , eWt , e - w t } . (10.5)

433

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434 Problems and Solutions in Group Theory

* The set of all ( 2 4 x ( 2 4 unitary matrices u satisfying

(10.6) T u J u = J , ut = u , -' in the multiplication rule of matrices, constitutes the unitary symplectic group USp(2.t). Similarly, both the inverse u-l and the transpose uT of any element u of USp(2l) satisfy Eq. (10.6). Since any matrix entry of a unitary matrix is finite, USp(2.t) is a compact Lie group. As an example, the general form of a diagonal matrix uo EUSp(2.t) is:

It can be shown that the determinant of R ~Sp(2 . t ,R) or u ~ U S p ( 2 t ) is +l. The group spaces of both Sp(2.t, R) and USp(2.t) are connected.

* For the infinitesimal elements R ~Sp(2 [ , R) and u E U S P ( ~ ~ ? ) ,

R = 1 - i a X , u = 1 - ipY,

X T = J X J , YT = J Y J ,

X * = - X , Y t = Y.

(10.8)

An imaginary matrix X of 2.t dimensions contains 4.t2 real parameters, so does a hermitian matrix Y . The component form of the condition XT = J X J is

which gives .t2 + .t(.t - 1) real constraints. The first equation gives T r X = 0, which means detR = 1. The component form of YT = J Y J is

y =-y- y- =y- V P P" 7 " P P"'

Due to Y t = Y , the first equation contains .t2 real constraints, including T r y = 0, and the second equation contains l ( 1 - 1) complex constraints. Therefore, there are l (2t + 1) real parameters both for R and for u . The order of Sp(2t, R) and the order of USp(2.t) are both t(2.t+ 1).

* We choose the generators TA in the self-representation of USp(2l)

where 7':;) and T::) are the generators in the self-representation of the

su(1) group, and (Ti:)) = &d&b. The generators TA are hermitian and bd

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The Syrnplectic Groups 435

normalized:

Thus, the structure constants of USp(2l) are totally antisymmetric with respect to all three indices.

The diagonal generators span the Cartan subalgebra of USp(2E):

H p = Ti;) x 03/&, 1 5 /L 5 l?. (10.11)

Their simultaneous eigenvectors in the Lie algebra, [H,, E,] = apE,, are

a < b, respectively. The simple roots are

The first ( l - 1) simple roots are shorter, d, = (rp - rp) /2 = 1/2, and the last one is longer, de = (re a re) /2 = 1. The remaining positive roots are

b - 1

/A=,

b - 1 e- 1

p=, p = b e- 1

Thus, the Lie algebra of USp(2t) is Ce. The largest root, which is the highest weight of the adjoint representation of Ce, is

e-i W‘ = h e 1 = 2 C rp +re = 2 ~ 1 . (10.13)

DL=1

where a < b and the eigenvalues (roots) areand

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436 Problems and Solutions an Group Theory

(10.14)

* The generators in the self-representation of the Sp(24 R) group are pure imaginary, X * = - X . The explicit forms of the generators can be obtained from Eq. (10.9) by replacing 0 a with Ta:

r1 = iu1, 72 = 0 2 , 7 3 = ia3. (10.15)

Namely, some generators change by a factor i so that the Sp(2!, R) group is a non-compact Lie group. USp(2l) and Sp(24 R) have the same complex Lie algebra, but different real Lie algebras. In another viewpoint, one may change the relevant parameters of Sp(2l, R) to be imaginary such that the generators in the corresponding irreducible representations of both USp(2l) and Sp(2e,R) are the same.

1. Prove that the determinant of R in Sp(2l, R) and the determinant of u

Solution. In the following proof for the determinant of R in Sp(2t, R) , we did not use the conjugate operation so that the proof is also effective for u in USp(2t).

Let R ~ S p ( 2 t , R ) be the transformation matrix for a vector x a in the 21-dimensional real space:

in USp(2l) are both +l.

Define a pseudo-product for two real vectors x and y :

Obviously, the pseudo-product is

e = c (X,Y,- X D P ) *

invariant in the transformation R:

= {Rx, RY1, * (10.18)

(10.17) p=1

* The Chevalley bases of USp(2l) are

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The Symplectic Groups 437

The self-pseudo-product of one vector is vanishing, {x, x } ~ = 0.

the following identity by Eq. (10.2) For the totally antisymmetric tensor of rank 2t, Eala..aze, we can prove

In fact, the nonvanishing terms in the sum come from the permutations among J's (l! terms) and the transpositions between two subscripts of any J (2' terms), and all are equal to one. Moving the column index of a 2G dimensional matrix X to be a superscript, we rewrite the determinant of X as

Permutating the order of X:, and changing the summing indices to restore the order of the superscripts, we obtain

Thus, we have

det X = (2'l!)

= (2el!)-I a1 ... a 2 ~

Since the right-hand side is invariant in we have

the transformation

det (RX) = det X, det R = 1 .

In the same reason, det u = 1. Both Sp(2l, R) and USp(2l) are the simply- connected Lie groups.

2. Count the number of the independent real parameters of R in Sp(2l, R)

Solution. A (2l)-dimensional real matrix contains (2t)2 real parameters. Equations (10.3) and (10.17) show that the pseudo-product of two column

and u in USp(21) directly from their definitions (10.3) and (10.6).

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438 Problems and Solutions in Group Theory

matrices of R ESp(2l,R) is equal to Jab:

It is an identity when a = b. It gives l (2 l - 1) independent real constraints when a # b, so that the order of Sp(2l, R) is (21)2 - l ( 2 l - 1) = l ( 2 l + 1).

A (2l)-dimensional complex matrix contains 2( 21)2 real parameters. From u-l = - J u T J , we have u * = - J u J ,

Each gives 212 independent real constraints. From the unitary condition Ed U;,Udb = Sab, we have the following constraints. The constraints for a = b = p are the same as those for a = b = 1.. There are l real independent constraints :

e e

V = l c

v= I

When p # u, the constraints for a = p and b = u are the same as those for a = V , b = j!Z as well as those for exchanging a and b. There are l( l - 1)/2 complex independent constraints:

When p # u, the constraints for a = p, b = i7 are the same as those for a = u, b = as well as those for exchanging a and b. There are l( l - 1)/2 complex independent constraints:

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The Symplectic Groups 439

One complex constraint is equivalent to two real constraints. Altogether there are ( 2 t ) 2 + e + 2 t ( ! - 1) = 6e2 - i! real constraints. The number of the independent real parameters of u in USp(2l) is 812 - (612 - l ) = l(2C + l), which is the order of the USp(C) group.

3. Express the simple roots of USp(2C) by the vectors V, given in Eq. (8.3) for the SU(e + 1) group, and then, write their Cartan-Weyl bases of generators in the self-representation of USp(2l)

Solution. From the Cartan-Weyl bases (10.11)) we have

If we choose another set of the normalized bases HL in the Cartan subal- gebra:

p= 1

H: = Ti!:+, x 03 (10.20) e - r + 1

He-T+2,

the component (rL)u of the simple root can be calculated from the formula Hh by replacing H p with (dpp - d p ( p + l l ) when p < e and &dpe when p = e.

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440 Problems and Solutions in Group Theory

Namely,

(10.21) The remaining positive roots are

b- 1

p = a

10.2 Irreducible Representations of Sp( 2C)

* Since the generators in the corresponding irreducible representations of USp(2e) and Sp(2l, R) , as well as the orthonormal bases in the representa- tion space are the same, we will only discuss the irreducible representations of Sp(2l, R ) . In the following, Sp(2l) stands for both USp(2t) and Sp(24 R) for simplicity.

* Let R E Sp(2l,R) be the coordinate transformation in a real (2t)- dimensional space:

(10.22)

From Eq. (10.22) we can define the tensors for Sp(2t, R) transformations:

Similar to the tensors of SU(N) and S O ( N ) , the Weyl reciprocity holds for the tensors of Sp(2l,R). The tensor space can be reduced by the Young operators. In addition, there are two invariant tensors for Sp(2l,R). One is the antisymmetric tensor Jab of rank two

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The Symplectic Groups 441

The contraction cab JabTabc... decreases the rank of the tensor Tabc ... by two:

ab

ab a'b'c' ... (z Ja'b'Ta'b'c' ... * = C R o t . . . (10.25)

Sometimes, this operation is also called the trace. The trace subspace is invariant in Sp(2&,R). In order to reduce the tensor space, we have to decompose the tensor space into the direct sum of a series of traceless tensor subspaces with decreasing rank two by two. Applying the Young operator

yrl 7, corresponding to an irreducible representation [A] of Sp(2l, R ) .

The other invariant tensor of Sp(2l, R) is the totally antisymmetric tensor € a l . . .az t of rank 2&. However, this tensor violates the new traceless condition. It is irrelevant to the reduction of the tensor space for Sp(24 R).

) C' ...

yp 1 XI to a traceless tensor space 7, we obtain the minimal tensor subspace

* The traceless tensor subspace denoted by a Young pattern [A] with the row number larger than & is empty. The irreducible representation of the Sp(2l,R) group is described by a Young pattern with at most l rows. The standard tensor Young tableau is still the common eigenstate of the Chevalley bases H p . The problem is that it is not necessary traceless. Fortunately, the contraction occurs only when a pair subscripts p and F appears. It is convenient to rearrange the bases @ a in the vector space:

@, = @,, @ e + p = @-, 1 5 P 5 e. (10.26)

From Eq. (10.14) the nonvanishing matrix entries of the Chevalley bases in the bases @a are

(10.27)

In the bases @a, 1 5 a 5 2&, the standard tensor Young tableau

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442 Problems and Solutions in Group Theory

Y, [XI @al.. ,an in the representation [A] of Sp(2l, R) corresponds to the high-

est weight if each box in its j t h row, j < L, is filled with the digit j. The standard tensor Young tableau with the highest weight is a traceless tensor because all subscripts take the values less than l + 1, but the trace occurs only when a pair of subscripts p and 2 l - p + 1 appears. The relation between the highest weight M and the Young pattern [A] for Sp(2l,R) is

The remaining traceless tensor bases in an irreducible representation space [A] of Sp(2L, R) can be calculated from the basis with the highest weight by the lowering operators F,.

* The dimension of an irreducible representation [A] of Sp(2l) can be calculated by the hook rule. In this rule, the dimension is expressed as a quotient, where the numerator and the denominator are denoted by the symbols Ypl and YiA1, respectively:

(10.29)

We still use the concept of the hook path (i, j ) in the Young pattern [A], which enters the Young pattern at the rightmost of the ith row, goes leftwards in the i row, turns downwards at the j column, goes downwards in the j column, and leaves from the Young pattern at the bottom of the j column. The inverse path (i, j) is the same path as the hook path (i, j) except for the opposite direction. The number of boxes contained in the hook path (i, j) is the hook number hij of the box in the j t h column of the ith row. YiA1 is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the hook number hij. Define a series of the tableaux Yktl recursively by the rule given below. YF1 is a tableau of the Young pattern [A] where each box is filled with the sum of the digits which are respectively filled in the same box of each tableau Ygl in the series. The symbol Ypl means the product of the filled digits in it, so does the symbol YiA1.

The tableaux Ygl are defined by the following rule:

(a) Ygl is a tableau of the Young pattern [A] where the box in the j t h column of the ith row is filled with the digit (2& + j - i).

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The Symplectic Groups 443

(b) Let [A(1)] = [A]. Beginning with [A(1)], we define recur- sively the Young pattern [A(")] by removing the first row and the first column of the Young pattern [A("-')] until [A(")] contains less than two rows.

(c) If [A(")] contains more than one row, define YE1 to be a tableau of the Young pattern [A] where the boxes in the first (a - 1) rows and in the first (a - 1) columns are filled with 0, and the remaining part of the Young pattern is nothing but [A(")]. Let [A(")] have r rows. Fill the first ( r - 1) boxes along the hook path (1, 1) of the Young pat- tern [A(")], beginning with the box on the rightmost, with the digits A?), A t ) , - - ., A?), box by box, and fill the first A:") boxes in each inverse path (i, 1) of the Young pattern [A(")], 2 5 i 5 r , with -1. The remaining boxes are filled with 0. If a few -1 are filled in the same box, the digits are summed. The sum of all filled digits in the pattern YAtl is zero.

4. Calculate the dimensions of the irreducible representations of the Sp(6) group denoted by the following Young patterns:

(1) [4,2], (2) [ 3 J l , (3) [4,417 (4) [3,3,21, ( 5 ) [4,4, 31.

Solution. 6 7 8 9 0 0 0 2 6 7 8 1 1

5 6 -t 5 4 2 1 -1 - 1 } = { %} =924.

2 1

(1): d[4,2] (SP(6)) =

6 7 8 0 0 2 6 7 10 + -1 -1

6 7 8 9 0 0 0 4 6 7 8 1 3 5 6 7 8 + - 1 - 1 - 1 - 1 } = { 5 4 3 2

4 5 6 5 4 3 2 7 ) = 1274.

4 3 2 1 4 3 2 1

(3): d[4,4] (SP(6)) =

6 7 8 0 2 3 0 0 0 5 6 7 + -1-1 0 + 0 0 1 4 5 -2 - 1 0 - 1

5 4 2 4 3 1 2 1

(4): d[3,3,2] (SP(6)) =

6 9 11 4 5 8

5 4 2 4 3 1 2 1

(2):

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444 Problems and Solutions in Group Theory

6 7 8 9 0 0 3 4 0 0 0 0 6 7 11 13 5 6 7 8 + - 1 - 1 - 1 0 + 0 0 0 2 4 5 6 10

2 3 4 6 5 4 2 5 4 3 1 3 2 1

4 5 6 - 2 - 1 - 1 6 5 4 2 5 4 3 1 3 2 1

(5): d[4,4,3](SP(6)) =

= 2002.

5 . Calculate the orthonormal bases in the irreducible representation [1,1,0] of the Sp(6) group by the method of the block weight diagram, and then, express the orthonormal bases by the standard tensor Young tableaux in the traceless tensor space of rank two for Sp(6).

Solution. The Lie algebra of Sp(6) is C3. The relations between the simple roots rp and the fundamental dominant weights wp are

rl = 2wl - w2, r 2 = -w1+ 2w2 - w3, r3 = -2w2 + 2w3, 3 2

w3 = rl + 2r2 + -r3. 1 2

w1 = rl + 1-2 + -r3, w2 = rl + 2r2 + r3,

The highest weight of the representation [l, 1,0] of Sp(6) is M = (0, 1 , O ) . The dimension of [l, 1,0] is d[l,l,o~ = (7 - 4)/(2 - 1) = 14.

The weights equivalent to the highest weight are

Therefore, the representation [ 1, 1, 01 contains one simple dominant weight (0, 1 , O ) and one double dominant weight (O,O,O) .

From the highest weight (0,1,0) we have a doublet of d 2 with (1, T, 1). From (1, i, 1) we have a doublet of A1 with ( i , O , 1) and a doublet of d 3

with (1,l)i). From (l,l ,i) we have a doublet of .A1 with ( i , 2 , i ) and a doublet of d2 with ( 2 , i , 0). From @,O, 1) we have a doublet of ds with (i, 2, i). Although (i, 2, i) comes from two paths, it is still a single weight because it is equivalent to (0, 1 , O ) . Now, we meet the double weight (O,O, 0) by applying Fl to 1(2,i,O)) and applying F 2 to l ( i , 2 , i ) ) . Letting 1(2,i,O)), I(O,O, O),) and 1(2,1,0)) constitute a triplet of d1, and the other basis state I(O,O, 0)2) is a singlet of d1, we have

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The Symplectic Groups 445

where we neglect the first index (0, 1, 0), which denotes the representation, in the basis state I ( O , l , 0), ( m l , m2, m3)) for simplicity. Applying ElF2 = F2El to the basis state I(%, 2, i)), we have

Thus, u = 0. Choosing the phase of the state basis (0 ,0 ,0)2) such that b is real positive, we have b = d m = m. Applying E2F2 = F2E2 + H2 to 1(0,0,0)1) we have

Choosing the phase of the basis state 1(1,2,1)) such that c is real positive, we have c = and d = m. The remaining basis states and the matrix entries of the lowering operators F, can be calculated similarly. The results are given in Fig. 10.1, where only those matrix entries of F, which are not equal to 1 are indicated.

I I

Fig. 10.1 The block weight diagram and the state bases for [ l , 1,0] of Sp(6).

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446 Problems and Solutions in Group Theory

Now, we expand the orthonormal basis states in the standard tensor Young tableaux. The Young tableau y[ljljO] is . A standard tensor Young tableau is expanded as follows:

There is only one trace tensor in the tensor space of rank two projected by y[ljl*O]. It belongs to the representation [O,O,O] of Sp(6):

Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (10.27) and the block weight diagram. Due to the Wigner-Eckart theorem, the basis state belonging to the representa- tion [1,1,0] of Sp(6) is orthogonal to the basis state belonging to [O,O,O], so it is a traceless tensor.

m{-B - 1 + 2

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The Symplectic Groups 44 7

U

l(o,i ,o)) = p21(i, i,i)) = -

6. Calculate the orthonormal bases in the irreducible representation [I, 1,1] of the Sp(6) group by the method of the block weight diagram, and then, express the orthonormal bases by the standard tensor Young tableaux in the traceless tensor space of rank two for Sp(6).

Solution. The Lie algebra of Sp(6) is C3. The relations between the simple roots rp and the fundamental dominant weights wp was given in Problem 5. The dimension of [l, 1,1] is

131 = 14.

The highest weight of the representation [l, 1,1] of Sp(6) is M = (O,O, 1). The weights equivalent to the dominant weight (O,O, 1) are

From the highest weight state (0, 0 , l ) we have a doublet of d 3 with (0,2, i), From (0 ,2 , i ) we have a triplet of d 2 with (1,0,0) and (2,2,1). Both the matrix entries of F2 are A. The weight (1,0,0) is a single dominant weight. The weights equivalent to the dominant weight (1,0,0) are

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448 Problems and Solutions in Group Theory

Therefore, all weights in the representation [I, 1,1] are single. It is easy to draw the block weight diagram for the representation [l, 1,1]. In Fig. 10.2 we give the block weight diagram for the representation [l, 1,1] where only those matrix entries of Fp which are not equal to 1 are indicated.

11,090 I I IIJZ

Fig. 10.2 The block weight diagram and the state bases for [l, 1,1] of Sp(6).

Now, we expand the orthonormal basis states in the standard tensor . Young tableaux. The Young tableau y[lilsll is . The standard tensor

Young tableau is

The trace tensors in the tensor space of rank three projected by Y['v1i1] belong to the representation [l, 0, 01 of Sp(6):

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The Symplectic Groups 449

Beginning with the highest weight state, we calculate the expansions of the basis states in terms of Eq. (10.27) and the block weight diagram. Due to the Wigner-Eckart theorem, the basis state belonging to the representa- tion [l, 1,1] of Sp(6) is orthogonal to the basis state belonging to [l, O,O], so it is a traceless tensor.

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450 Problems and Solutions in Group Theory

- m { i

m { I - +i} [}

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The Symplectic Groups 45 1

7. Calculate the Clebsch-Gordan series for the reduction of the direct prod- uct representation [l, 1,0] x [l, l , O ] of the Sp(6) group and the high- est weight states for the representations contained in the series by the method of the standard tensor Young tableau.

Solution. There are a few methods to calculate the Clebsch-Gordan series of a reducible representation. The method of the standard tensor Young tableau is one of them. The main point in the calculation is to count the multiplicities of the dominant weights in the relevant representations. The difficulty in Sp(2l) group, as well as S O ( N ) group, is how to remove the trace tensor subspace. In fact, the dimension of the tensor space projected by a Young operator [A] is nothing but the dimension d[A] (SU(2l)) of SU(2l) group. Subtracting the dimension 4x1 (Sp(2t)) of Sp(2l) group, we obtain the dimension of the trace tensor subspace. Conversely, we can calculate the multiplicities of the dominant weights in the representation [A] of Sp(2l) by counting the multiplicities both in the tensor space projected by the Young operator Y[’] and in the trace tensor subspace. As far as the highest weight state is concerned, the condition that each raising operator Ep annihilates the highest weight state is useful to calculate the basis state.

First, we count the multiplicities of the dominant weights in the space of the direct product representation [l, l , O ] x [I, 1,0] of the Sp(6) group. Since the block weight diagram for the representation [I, l , O ] of Sp(6) was given in Fig. 10.1 of Problem 5, it is easy to count the multiplicities of the dominant weights. In the direct product space, there are one basis state with the dominant weight (0,2,0), two basis states with the dominant weight ( l , O , l), four basis states with the dominant weight (2,0,0), eight basis states with the dominant weight (0, 1,0), and 16 basis states with the dominant weight (0, 0,O).

Second, since there is one basis state with the dominant weight (0,2,0) in the direct product space, one representation [2,2,0] is contained in the Clebsch-Gordan series. Its highest weight state is

The decomposition of the tensor subspace projected by y[2)2)o] is

[2,2,0] -+ [a, 2,0] CB [I, 1 , O ] CB [O,O, 01, 105 = 90 + 14 + 1.

The tensor subspace contains the basis states with the dominant weights

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Problems and Solutions in Group Theory

No. State bases

We have to subtract the basis states in the representations [I, l , 01 and [O,O, 01: one state with (0, 1,O) and three states with (O,O, 0).

Third, since the difference between the multiplicities of the dominant weight (1,0,1) in the direct product space and in the representation space [2,2,0] is 2 - 1 = 1, one representation [2,1,1] is contained in the Clebsch- Gordan series. Its highest weight state is

The coefficients are calculated by the condition that each raising operator E, annihilates the highest weight state and the normalization condition. The decomposition of the tensor subspace projected by Y[’i1r1] is

[Z, 1,1] -+ [2,1,1] @ [2,0,0] @ [l, 1,0], 105 = 70 + 21 + 14.

The tensor subspace contains the basis states with the dominant weights

452

.as follows

Weight

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The Symplectic Groups 453

We have to

No.

1

2

5

9

State bases

subtract the basis states in the representations [2,0,0] and [I, 1,0]: one state with (2,0,0), two states with (0,1,0) and five states with (0, 0,O). Note that the tensor space for [2,0,0] is traceless and contains the basis states with the dominant weights as follows.

Weight No. State bases

( 2 ) O ) O ) : 1 I 1 I 1 I ) (0 ,1 ,0) : 1 I 1 1 2 1 ,

Fourth, since the difference between the multiplicities of the dominant weight (2,0,0) in the direct product space and in the representation spaces [2,2,0] and [2,1,1) is 4 - 2 - (2 - 1) = 1, one representation [2,0,0] is

Weight

as follows.

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454 Problems and Solutions in Group Theory

contained in the Clebsch-Gordan series. Its highest weight state is

The coefficients are calculated by the condition that each raising operator E, annihilates the highest weight state and the normalization condition. The basis states with the dominant weights contained in the traceless tensor space of [2,0,O] was given.

Fifth, since the difference between the multiplicities of the dominant weight (0,1,0) in the direct product space and in the representation spaces [2,2,0], [2,1,1], and [2,0,0] is 8-(4-1)-(5-2)-1 = 1, onerepresentation [I, 1,0] is contained in the Clebsch-Gordan series. Its highest weight state is

The coefficients are calculated by the condition that each raising operator E, annihilates the highest weight state and the normalization condition. We have known from Fig. 10.1 of Problem 5 that the traceless tensor space for [l, 1,0] contains one basis state with the dominant weights ( O , l , 0) and two basis states with the dominant weight (O,O,O) .

Finally, counting the multiplicities of the dominant weight (0, 0,O) in the direct product space and in the representation spaces [2,2,0], [2,1,1], [2,0,0], and [I, 1,0], we have

16 - (9 - 3) - (9 - 5) - 3 - 2 = 1.

One representation [0, 0, 01 is contained in the Clebsch-Gordan series. The expression for its highest weight state is quite long. We write it in the form of the basis states instead of the standard tensor Young tableaux. The

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The Symplectic Groups 455

1 4 ~ 1 4 = 1 9 6 = 90 + 70 + 21 + 14 + 1

Fig. 10.3 The dominant weight diagram for [l, 1,0] x [l, 1,0] of Sp(6).

In the permutation of two factors in the direct product representation space [l, l , O ] x [l, l , O ] , the basis states in the representations [2,2, 01, [l, 1,0] and [O,O, 01 are symmetric, and the representations [2,1,1] and [2,0,0] are antisymmetric.

latter form can be calculated from Fig. 10.1.

In summary, the dominant weight diagram for the direct product rep-resentation [1,1,0] ´ [1,1,0] is given in Fig. 10.3.

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Index

l?N matrix group, 403 ya matrix, 403

anti-commutator, 319 associative law, 27

basis, 7 orthogonal, 212 standard, 211

block weight diagram, 282, 285, 289, 292, 298, 303, 309, 336, 345, 348, 355, 378, 386, 394, 412, 445, 448

Bravais lattices, 174, 180

Cartan criteria, 270 Cartan matrix, 272 Cartan subalgebra, 271, 376, 435 Cartan-Weyl bases, 271, 318, 439 character, 43

character table, 48 Chevalley bases, 280, 321, 378, 409,

class, 30, 48, 54, 139, 422, 424

graphic method, 213

436, 441

self-reciprocal, 54 mutually reciprocal, 30 self-reciprocal, 30

Clebsch-Gordan coefficients, 80, 101, 106, 149, 154, 215, 243

Clebsch-Gordan series, 80, 240, 299 composition functions, 140 coset, 29

left, 29 right, 29

crystal, 173 crystal lattice, 173

basis vector of the reciprocal, 175 besis vector, 173 vector, 173

crystal system, 178 crystallographic point group, 174 cycle, 193 cycle structure, 194

determinant, 1 dominant weight diagram, 299, 301,

double-vector form, 175 Dynkin diagram, 272

356

eigenequation, 1 eigenvalue, 1 eigenvector, 1 element, 27

equivalent theorem, 404 Euler angles, 123, 125, 127

conjugate, 30

filling parity, 214 Fock condition, 199, 323 F’robenius theorem, 67, 239

Gelfand bases, 337 generator, 27, 269

461

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462 Problems and Solutions in Group Theory

gliding plane, 178 gliding vector, 177, 178 group, 27

C4, 32 c8, 35 D2n+1, 68 D2n, 68 D4, 33 DN, 32 v4, 32 direct product, 55 homomorphism, 33 I, 73, 75, 106, 121, 134 0, 59, 70, 85, 101 T, 81 Abelian, 27 c10, 37 C2n, 38 C4h, 35 c6, 32 CN, 65, 68, 69 Dn, 38 SO(N) , 375 S0(3), 115 SU(N), 317 SU(2), 117 Sp(2C, R), 433 USp(2C), 434 v2, 34 covering, 117, 140 direct product, 51 improper point, 51 proper point, 51 quaternion, 35 quotient, 31 T, 40

group space, 52, 140, 269

height, 282 homomorphism, 121

kernel of, 33 hook number, 198 hook rule, 198, 325, 380, 405, 442 horizontal operator, 199

ideal, 205

idempotent, 205 equivalent, 205 orthogonal, 205 primitive, 205

identity, 27 infinitesimal element, 141 international notation, 178 inverse, 27 irreducible tensor operator, 146 isomorphic, 27

Jordan forms, 18

Killing form, 270, 320

Lie algebra, 270 compact, 270 rank, 271 simisimple, 270 classical, 272 exceptional, 272

Lie group, 115, 140, 269 compact, 140, 269 order, 269 semisimple, 269 simple, 269

Lie product, 270 Lie Theorem, 141, 142 Littlewood-Richardson rule, 239, 326,

356, 364, 370, 402 Lorentz group, 415

homogeneous, 418 proper, 418

lowering operator, 280

matrix, 2 hermitian, 2, 5 negative definite, 2, 270 positive definite, 2 positive semi-definite real orthogonal, 2, 5 real symmetric, 2, 6 unitary, 2, 5

multiplication rule, 27 multiplication table, 27

2

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Index 463

multiplicity, 49, 53, 67, 77, 80, 97, 280, 282, 289, 299, 304, 308, 313, 355, 356, 451, 454

order, 140 of the element, 29 of the group, 27

Pauli matrices, 3, 33 period, 29 permutation, 193

even, 194 horizontal, 199 odd, 194 vertical, 199

permutation parity, 194 planar weight diagram, 283, 288, 289,

292, 298, 306, 340, 345, 348, 355, 360, 372

projection operator, 81

raising operator, 280 rank, 2, 142, 146, 317, 321, 326, 327,

376, 379, 380, 440, 441 rearrangement theorem, 27 reduced matrix entry, 147, 149 regular application, 214 representation, 43

regular, 52, 81, 85 subduced, 130 adjoint, 142, 283 conjugate, 43 equivalent, 47 faithful, 43 identical, 43 induced, 66, 237 inner product, 237 irreducible, 48, 123, 211, 279, 440 outer product, 238 real orthogonal, 43 reduced, 47 reducible, 47 subduced, 66, 237, 362 unitary, 43 unitary of infinite dimension, 166

root, 271

positive, 271 simple, 271, 319, 435

root chain, 271 rotation, 45, 115

improper, 51 proper, 51

Schur theorem, 47, 143, 245, 321 screw axis, 177 secular equation, 1 Serre relations, 280 similarity transformation, 8 space group, 174, 186

symmorphic, 174 spherical harmonic function, 399 spin- tensor, 405 spinor, 405 spinor operators, 407 spinor representation, 404 structure constants, 141, 269, 319 subgroup, 29

cyclic, 29 index, 30 invariant, 30 normal, 30

symmetric center, 177 symmetric operation, 173 symmetric plane, 177 symmetric straight line, 177 symmetry transformation, 44 symplectic group, 433

real, 433 unitary, 434

tabular method, 212 trace, 1 Transformation Operators for a

Scalar Function, 43 translation group, 173 transposition, 194

vertical operator, 199

weight, 279 dominant, 282, 336 equivalent, 280, 281

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464 Problems and Solutions an Group Theory

fundamental dominant, 281 highest, 282, 337, 379, 442 multiple, 279 single, 279

Weyl orbit, 280 Weyl orbital size, 280 Weyl reciprocity, 323, 377, 440 Weyl reflection, 280 Wigner-Eckart theorem, 51, 147

Young operator, 323, 377, 440 Young pattern, 198, 211, 323, 379,

441, 442 associated, 214

Young tableau, 198 standard, 198 standard tensor, 324, 337, 362, 441 tensor, 323