GROUP THEORY AND PHYSICS - Infis-Ufugerson/grupos/3 - Great books/Teoria de...4 symmetries in physics symmetry arguments. This is not a course in group theory and/or differen-tial

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G R O U P T H E O RYA N DP H Y S I C S

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N O TA

Queste note sono pensate come supporto didattico per il corso di Fisica Teorica(Parte A) del Corso di Laurea Magistrale in Fisica dell’Universita degli Studi diPadova. E’ vietata la loro distribuzione non autorizzata.

C O N T E N T S

i introduction 1

1 symmetries in physics 3

1.1 Introduction 4

1.2 Symmetries and Groups 4

1.3 Examples of Symmetries 5

1.3.1 Symmetry of Geometric Objects 5

1.3.2 Spacetime Symmetries 6

1.3.3 Gauge Symmetries 7

1.3.4 Approximate Symmetries 9

1.4 Symmetry Breaking 10

2 principle of least action 13

2.1 Euler–Lagrange Equations 13

2.2 Examples 16

2.2.1 Free relativistic particle 16

2.2.2 Electromagnetic field and its coupling to a charged parti-cle 18

2.3 Noether’s Theorem 19

ii finite groups 23

3 groups : basic elements 25

3.1 Definitions 25

3.2 Maps between groups 27

3.3 Finite groups 29

3.3.1 The Symmetric Group Sn 31

4 subgroups , quotients and products 37

4.1 Basic definitions 37

4.2 Quotients 38

4.3 Products 42

4.4 Classifying finite groups. 44

5 representations 47

5.1 Basic definitions 48

5.2 Equivalent and reducible representations 51

i

ii Contents

5.3 Unitary representations 54

5.4 Comments 57

6 properties of irreducible representations 61

6.1 The Great Orthogonality Theorem 63

7 properties of irreducible representations 67

7.1 Conjugacy classes 67

7.2 Characters and their properties 68

7.3 Character tables 71

8 irreducible representations of Sn 77

8.1 Young Tableaux and conjugacy classes 78

8.2 Young Tableaux and irreducible representations of Sn 80

iii elements of topology and differential geometry 87

9 homotopy groups 89

9.1 Homotopy 90

9.2 Homotopy groups 92

9.3 Universal covering 95

10 differentiable manifolds 97

10.1 Introduction 97

10.1.1 Differentiable manifolds 97

10.1.2 Properties 99

10.1.3 Examples 100

10.2 Tangent and cotangent space 102

10.3 Flows and exponential map 106

10.4 Lie brackets 108

iv lie groups 111

11 lie groups 113

11.1 Compact and non-compact groups 114

11.2 Invariant measure 114

11.3 Groups of matrices 115

11.3.1 Linear constraints 116

11.3.2 Quadratic constraints 117

12 lie algebras 123

12.1 Algebra as tangent space 124

12.1.1 Algebras for groups of matrices 126

12.2 From the algebra g to the group G 128

12.2.1 Surjectivity 128

Contents iii

12.2.2 Injectivity and global properties 129

12.2.3 The algebra structure and the Baker–Campbell–Hausdorffformula 130

12.2.4 Lie algebras 131

13 lie algebras – properties 135

13.1 Subalgebras, ideals 137

13.2 Simplicity, compactness and connectedness 138

13.3 Cartan’s criteria 139

13.4 Casimir operator 141

14 algebra representations 143

14.1 Morphisms and representations 143

14.2 Matrix representations 144

14.3 Differential representations 145

14.4 Bosonic representations 147

14.4.1 u(1)C ' gl(1, C) 148

14.4.2 u(2)C ' gl(2, C) 149

14.5 Fermionic representations 150

15 representations of su(2) and so(3) 153

15.1 Rotation group and its algebra 153

15.2 Isomorphism su(2) ' so(3) and homomorphism SU(2) 7→ SO(3) 154

15.3 SO(3) topology 157

15.4 Irreducible representations of SU(2) and SO(3) 159

15.5 Irreducible representations of su(2) 160

15.6 Matrix representations 163

16 su(n) irreps and products 165

16.1 Products of irreducible representations 165

16.1.1 Tensor products 165

16.1.2 Clebsch–Gordan decomposition 166

16.1.3 Decomposition in irrepses of a subgroup 167

16.2 SU(N) irrepses and Young Tableaux 167

16.2.1 Fundamental, anti-fundamental and adjoint representa-tions 167

16.2.2 Irrepses and Young Tableaux 170

16.2.3 Products using Young Tableaux 176

16.2.4 Again on SU(2) 178

17 poincare group 183

17.1 Topological properties 184

iv Contents

17.1.1 Topology of SL(2,C) and SO+(3,1) 188

17.2 The so+(3, 1) algebra 189

18 poincare irrepses 193

18.1 Unitary representations 195

18.2 Finite dimensional representations 199

18.3 Reducibility of spinor representations 201

18.4 Casimir operators 204

18.4.1 Casimir for the Lorentz group 204

18.4.2 Casimir for the Poincare group 205

19 symmetries of molecules and solids 209

19.1 Classic vibrations of molecules 209

19.1.1 Symmetries 211

19.1.2 Diatomic molecule 213

19.1.3 Diatomic molecule in 2 dimensions 215

19.1.4 Triatomic molecule 217

a solutions 219

Part I

I N T R O D U C T I O N

1S Y M M E T R I E S I N P H Y S I C S

2012 has been a remarkable year for physics. The discovery of the Higgs boson-like particle at the Large Hadron Collider in Geneva provides a spectacularconfirmation of our understanding of the fundamental interactions governingour Universe, embodied in the Standard Model of elementary particles. Thisdiscovery marks the pinnacle of the journey started with the unification of elec-tricity and magnetism in terms of a single symmetry principle and completedwith the further unification of the weak nuclear force, via the electro-weak sym-metry breaking mechanism.

Symmetry principles guide theoretical physics since its birth. Geometricsymmetries can explain many physical properties of materials and character-ize diffraction patterns found in crystallography. Dynamical symmetries arewidely used in Nuclear Physics to classify the energy of rotational and vibra-tional modes of nuclei. Symmetry properties of spacetime itself and symme-try transformations relating different observers are at the base of Special andGeneral Relativity. Broken symmetries are used to explain a wide variety ofphysical phenomena, from Superconductivity to electro-weak interactions.

In this lectures we want to explore the concept of symmetry in physics. In par-ticular we want to give a theoretical description of this concept, by construct-ing mathematical models of the symmetries appearing in physical systems. Weare mainly going to focus on the applications of symmetry principles in thecontext of Quantum Mechanics, but we are also going to discuss spacetimesymmetries, which are at the base of Relativity and Quantum Field Theory,and internal symmetries (global and local gauge symmetries), which guide ourcurrent understanding of the physics of fundamental interactions.

The objective is to provide students with instruments and techniques thatare useful to extract information on the physics of a definite system by using

3

4 symmetries in physics

symmetry arguments. This is not a course in group theory and/or differen-tial geometry and as such we do not seek completeness nor rigor at the levelof a course on these topics. We will introduce only the ingredients that arenecessary for physical applications and try to provide as many examples andexercises derived from physics as possible.

1.1 introduction

Man is naturally attracted by symmetric objects. Actually, the word symmetryderives from συν− µετρoν and its adjective is usually associated to proportion-ate objects, displaying equal measures. In everyday life we often identify theword symmetry with bilateral symmetry. This is a symmetry under reflectionwith respect to a plane, i.e. invariance under exchange of its left and right sides.The fact that such a symmetry is so common when we look at men and ani-mals has to do with the existence of 2 preferred axis, a vertical one determinedby gravity and a horizontal one, determined by the direction of motion. Theircombination singles out a preferred plane in space and explains the consequentsymmetry of our bodies with respect to such a plane. Plants do not move andin fact they usually display a rotational symmetry around the vertical axis. Aswe know, these are not exact symmetries, but rather approximate symmetries. Still,we are able to somehow predict their existence because of the preferred planeexplained above and their existence explains our appearance.

How do we recognize symmetries? An object is symmetric if we can act onit with an operation that does not change it with respect to our observation.We can therefore argue that a symmetry transformation is a map that leaves an“object” invariant. In general, if we apply a certain transformation rule to aphysical system, we put the system in a new and different state. If we are notable to distinguish it from the previous one, we say that the transformation weused is a symmetry.

1.2 symmetries and groups

We will use a great part of these lectures to discuss group theory. We will doso because symmetry transformations are described by groups.

Consider three different states of a physical system: A, B and C. If state Ais similar to state B and state B is similar to state C, then state A is similarto state C. Hence, the product of two symmetry transformations is again a

1.3 examples of symmetries 5

symmetry transformation. Also, each state is mapped onto itself by a trivialtransformation that leaves everything the same (the identity) and the inverse ofa symmetry transformation is still a symmetry. Thus, the set of all symmetrytransformations that characterize the symmetry of a system, are elements of agroup.

A B CS1 S2

id id id

S1 S2

Figure 1.: Example of symmetry transformations between three different physicalstates A, B and C.

For example, reflections with respect to a plane form a group that containstwo elements: the reflection operation and the identity. Rotations in three-dimensional space, on the other hand, form a group that has an infinite numberof elements. Groups with a finite (or numerable) number of elements are calleddiscrete groups, while groups that depend on continuous parameters are calledcontinuous groups.

Symmetries of a system imply relations between observable quantities, whichmay be obeyed with great precision, independently of the nature of the forcesacting in the system. For example, the energies of several different states ofthe Hydrogen atom are exactly equal, as a consequence of the rotational invari-ance of the system. In other instances symmetries of physical systems are onlyapproximately realized or broken. Still, an approximate or a broken symmetrycan teach us as many important lessons as an exact one, like in the case ofsuperconductivity or of the Higgs particle.

1.3 examples of symmetries

1.3.1 Symmetry of Geometric Objects

The first obvious example is given by symmetries of geometric objects. Ageneric geometric object may display symmetry under rotations and/or reflec-tions and these transformations may generate a discrete or a continuous group.


For instance, we know that an equilateral triangle is mapped to itself wheneverwe perform a rotation of 120 degrees around its center, or a reflection aroundone of three axes connecting the center to one of its vertices. A snowflake lookslike itself when we perform a 60 degrees rotation around its center or if we per-form a mirror reflection. A circle, on the other hand, is mapped to itself by anyrotation around its center. This means that in this case we have a continuousset of symmetry transformations (or isometries).

Figure 2.: Symmetry axes of geometric objects.

Geometric symmetries affect physical properties of molecules and materials,like the appearance of conducting and isolating bands in crystals.

1.3.2 Spacetime Symmetries

As mentioned in the introduction, very important symmetry laws in physics arethe transformations of spacetime coordinates that map intertial frames amongthem. These laws constrain both kinematic and dynamic properties of physicalsystems and can be derived from general properties of spacetime. In Newto-nian mechanics, such transformations change the coordinate system (t,~r) by

• spatial translations:

~r →~r +~a; (1.1)

• time translations:

t→ t + b; (1.2)

• constant rotations:

~r → R~r; (1.3)


• relative motion at constant velocity ~v (Galilean boosts):

~r →~r−~v t. (1.4)

The first two symmetries derive from space and time homogeneity, while thethird one is related to space isotropy. Altogether they form a 10-dimensionalcontinuous group called the Galilei group, which constrains newtonian mechan-ics. In Special Relativity Galilean boosts are replaced by Lorentz boosts andtogether with the other transformations form the Poincare Group, which wewill discuss in detail in chapter 17.

This type of symmetries is extremely important because they severely restrictthe form that physical laws can take. Also, as we will soon discuss, continuoussymmetries give rise to conserved quantities via Noether’s theorem and in thecase of spacetime symmetries we can obtain energy-momentum and angularmomentum conservation laws.

1.3.3 Gauge Symmetries

A special kind of symmetry transformations we encountered when studyingelectrodynamics are gauge symmetries. They are coordinate-dependent trans-formations that leave the equations of motion invariant. In fact, Maxwell’sequations can be written as

∂µFµν = −4π

cjν, εµνρσ∂νFρσ = 0, (1.5)

where the electric and magnetic fields are described by the rank 2 antisymmet-ric tensor Fµν. The second equation in (1.5) can be solved in terms of a potentialAµ, by

Fµν = ∂µ Aν − ∂ν Aµ. (1.6)

However this solution is not unique because gauge transformations by an arbi-trary function Λ(x) change the Aµ potential as

Aµ(x)→ Aµ(x) + ∂µΛ(x), (1.7)

but leave Fµν invariant and hence do not change the value of the electric andmagnetic fields. As we will now see, this function is actually specifying anelement of a group of symmetries of the Schrodinger equation for a chargedparticle.


An electrically charged particle is described by a complex wave functionψ(t,~x) whose Schrodinger equation remains valid if we perform a local rotationin the complex plane

ψ(x)→ eiΛ(x) ψ(x), (1.8)

provided the wavefunction is minimally coupled to the Aµ potential. Thismeans that each derivative acting on ψ has to be replaced by a covariant one

∂µ → Dµ = ∂µ − i Aµ. (1.9)

In fact, if we do so, the Schrodinger equation is mapped to itself under gaugetransformations because

Dµψ 7→ eiΛ Dµψ. (1.10)

The set of all possible phases

U = exp(iΛ(x)) (1.11)

forms a group of 1 by 1 unitary matrices called U(1):

U†U = 1 ⇔ U ∈ U(1). (1.12)

Since a symmetry group of this type has an infinite number of elements,whose value may differ at each spacetime event, the couplings and the dy-namics of particles charged with respect to it are extremely restricted. Gaugesymmetries are therefore extremely powerful symmetry principles.

Fundamental interactions are nowadays described by generalizations of thequantum field theory describing electrodynamics, where the internal symme-try group U(1) has been replaced by larger ones. The Standard Model of ele-mentary particles is based on the gauge group SU(3) × SU(2) × U(1). Greatunification theories use larger groups like SU(5) and SO(10) and String Theorymakes use of even larger groups like E8× E8.

Gravitational interactions, on the other hand, are described by General Rela-tivity. Although this theory is rather different from ordinary gauge theories, itstill shows a local invariance that determines its dynamics.

Physics should be described by the same laws in any reference frame. Forthis reason we should be able to describe them in a way that preserves theirform under general coordinate transformations:

xµ → ξµ(x). (1.13)


This general covariance is at the base of Einstein’s theory of Relativity andthe role of “gauge field” for such transformations is played by the metric fieldgµν(x).

1.3.4 Approximate Symmetries

We already discussed the fact that often approximate symmetries are useful toconstrain the theoretical description of physical systems. These are symmetriesvalid in a certain regime of energy or length and may arise dynamically.

An example is given by crystals. An infinite crystal is invariant under trans-lations for which the displacement is an integral multiple of the distance be-tween two adjacent atoms. Real crystal, however, have a definite size and theirsurface perturbs the symmetry under translations. Nevertheless, such pertur-bations have little effects on the properties at the interior if the crystal containsa sufficiently large number of atoms.

Another example of a symmetry that is only approximately realized is theisospin symmetry. One of the baryons, called ∆+, decays into a nucleon and apion in two ways:

∆+ → n + π+, or ∆+ → p+ + π0. (1.14)

Although the masses of the neutron and the proton are very close and the sameis true for the pions, so that the only distinctive element is the electric chargedistribution, the second process occurs twice as much as the first. The explana-tion for this discrepancy comes from a symmetry argument. In fact nucleonsand pions sit in different representations of the isospin symmetry group. Thisinformation is actually relevant for any symmetry principle. The way physicalquantities transform under symmetry transformations is not determined justby the symmetry group, but also by its representation. For instance, we knowthat we cannot sum mass and velocity as they are quantities that transformdifferently under the group of rotations in 3-dimensional space. The mass isan invariant quantity. It is a scalar under rotations and hence transforms inthe trivial representation. The velocity transforms as a vector and hence thegroup of rotation is represented in its fundamental representation. Comingback to the example above, the two decays mode differ because nucleons forma doublet of the isospin symmetry, pions a triplet and ∆ particles a quadrupletand hence they are constrained by the rules governing products of representa-tions and their decomposition in irreducible components. We stress that thisargument is valid because particles inside the same representation have nearly


equal masses, but different electric charges. Symmetry arguments are thereforevalid in the approximation of equal masses, which we know is violated only byfew percents.

We conclude this discussion of approximate symmetries by recalling thatwe have a very simple example of a discrete symmetry that is omnipresent inphysics: the reflection symmetry Z2: x 7→ −x. Any degree of freedom canbe approximately described by the 1-dimensional harmonic oscillator whenlooking at a minimum of its potential

V(x) ' V(x0) +12

m2 δx2 +O(δx3) (1.15)

and obviously this is invariant for δx 7→ −δx.

1.4 symmetry breaking

Consider now a situation where we have a symmetry principle that is valid ina certain energy regime, but which is (spontaneously) broken in another. Wecan still have essential information on the description of such physical systemin the regime where the symmetry is broken by knowing that it was valid inthe other.

Let us consider for example a system described by 2 complex degrees offreedom zi, i = 1, 2, whose potential is given by

V(zi) = −µ2Z†Z + λ(

Z†Z)2

, (1.16)

where λ > 0 and Z =

(z1z2

). It is easy to check that such a potential is

invariant under unitary transformations of the Z doublet

Z 7→ UZ, where U†U = 12 (U ∈ U(2)) (1.17)

and therefore the resulting physics should keep trace of this fact. We can alsosee that the potential (1.16) has two critical points, satisfying

∂iZ =(−µ2 + 2λ|Z|2

)z∗i = 0, (1.18)

1.4 symmetry breaking 11

at zi = 0, or at |Z|2 = µ2/(2λ). Actually, zi = 0 is a maximum and hence anunstable point, while |Z|2 = µ2/(2λ) is a minimum. Physics at the maximumis still invariant under the whole symmetry group because(

a bc d

)(00

)=

(00

)(1.19)

for any value of a, b, c and d and hence for any U ∈U(2).

Figure 3.: The “mexican hat” potential (1.16). The maximum has U(2) symmetry, whilethe family of minima have only a residual U(1).

On the other hand, the family of minima is invariant only with respect to asubset of such matrices, forming a U(1) subgroup. In fact, chosen for instancethe point z1 = 0 and z2 = σ =

√µ2/(2λ) we have that(

a bc d

)(0σ

)=

(b σd σ

)⇒ b = 0, d = 1. (1.20)

In addition we have to remember that U†U = 1 and hence(a∗ c∗

0 1

)(a 0c 1

)=

(|a|2 + |c|2 c∗

c 1

)=

(1 00 1

), (1.21)

which implies c = 0 and a = eiα. We therefore have a spontaneous breaking of thesymmetry group

U(2)→ U(1). (1.22)

This means that if we study the system around the stable minima, we explic-itly see only a U(1) invariance, but we still have constraints on the possible


interaction terms due to the original larger U(2) symmetry group. For instance,expanding the potential (1.16) around the minima we see that the symmetrybreaking terms are not generic, but fixed by the original U(2)-invariant poten-tial.

2T H E P R I N C I P L E O F L E A S T A C T I O N A N D N O E T H E R ’ ST H E O R E M

In this chapter we are going to derive the Euler–Lagrange equations for con-tinuous systems and examine the relation between conservation laws and theexistence of continuous symmetries. After a brief general discussion we willconcentrate on relativistic systems.

2.1 euler–lagrange equations

The principle of least action is a very elegant formalism that can be used toderive the equations of motion of a physical system, starting from a functional,the action, which is manifestly invariant under a definite set of symmetries.

Imagine we throw an object in the Earth’s gravitational field. We know that inthe presence of gravity the trajectory will be a parabola connecting two extremaA and B. What is the distinctive feature of such a trajectory? We could imaginean infinite number of different paths connecting A and B, but if we sum thedifference between kinetic and potential energy at each of the points of thesepaths the result will be a number that is always greater than the one obtainedfor the route followed by an actual object. In fact, while K(A), K(B), U(A)and U(B) are the same for any trajectory, their value along the path changesand crucially depends on the chosen trajectory. If we simplify the problemand consider simply an object in vertical motion, whose height is described bythe function y(t), the law of motion is the one associated to an extremum (notnecessarily a minimum) of the action functional

S[y] =∫ tB

tA

[12

m(

dy(t)dt

)2

−mg y(t)

]dt. (2.1)

13

14 principle of least action

The result is the same as the one we would obtain by directly solving theequation of motion

y(t) = −g, (2.2)

with boundary conditions specified by y(tA) and y(tB). In fact, computing theextrema of (2.1)

δS[y]δy

= limc→0

S[y + cδy]− S[y]c

= 0 (2.3)

one reproduces (2.2).This line of reasoning works in general for any classical system. We can

recover its equations of motion as the result of a variational problem for anaction functional constructed in terms of a Lagrangian L given by the differenceof kinetic and potential energy of the system. For a generic classical physicalsystem with N degrees of freedom qI(t), where I = 1, . . . , N, the state at a timet is determined by qI(t), qI(t) and the action giving the evolution of suchsystem from state A at time tA to state B at time tB is

S[q] =∫ tB

tA

L(t, qI , qI) dt. (2.4)

The evolution of the system is then described by the functions qI(t) extremizingthe action S. This means that if we change path

q′I(t) = qI(t) + δqI(t), (2.5)

with the assumption that the extrema are fixed δqI(tA) = δqI(tB) = 0, werequire that

δS = S[q′]− S[q] = 0. (2.6)

This leads to (assuming synchronous variations δt = 0)

δS =∫ tB

tA

δL dt =∫ tB

tA

[δLδqI

δqI +δLδqI

δqI

]dt

=∫ tB

tA

[δLδqI

δqI +δLδqI

ddt

δqI

]dt

=∫ tB

tA

[δLδqI

δqI +ddt

(δLδqI

δqI

)− d

dt

(δLδqI

)δqI

]dt

=∫ tB

tA

[δLδqI− d

dt

(δLδqI

)]δqI dt +

[δLδqI

δqI

]B

A.

(2.7)

2.1 euler–lagrange equations 15

The last term is vanishing because extrema are fixed and, since we considerarbitrary variations δqI , the integral vanishes if and only if

δLδqI− d

dt

(δLδqI

)= 0. (2.8)

These are the Euler–Lagrange equations, which must be satisfied by physicaltrajectories.

These equations have a vast application in physics. Although not all systemsnecessarily admit a Lagrangian description, whenever we have such a descrip-tion we can derive its equations of motion by using (2.8). In addition, evenif we restricted our discussion to systems with a finite number of degrees offreedom, we can extend it and repeat the derivation above for continuous sys-tems: N → ∞. Field theories are prominent examples of this type. For instance,the electromagnetic field can have different values at different points of spaceand at different times Aµ = Aµ(t,~r). We can therefore think of it as a vari-able whose index I runs over an infinite number of possible values labellingdifferent points in space:

qI(t) −→ q~r(t) = q(t,~r). (2.9)

Consider now a system that requires several fields φa(x), a = 1, . . . , n, tospecify it. The index a may label components of the same field (like in thecase of the electromagnetic potential Aµ) or it may refer to different fields. TheLagrangian should depend on φa(t,~r), φa(t,~r), but also on ∂~rφa(t,~r) and forrelativistic systems we can use the covariant notation

L = L(φa(x), ∂µφa(x)

). (2.10)

It is also obvious that now we need to replace the sum over I by an integrationon the~r variable

∑I→∫

d3~r (2.11)

and define a Lagrangian density

S =∫ tB

tA

L dt =∫ B

AL d4x, (2.12)

with A and B two events in spacetime. We note that while t depends on theobserver A and B do not. In the following we will often omit the extrema toconsider general equations of motion.


The Euler–Lagrange equations for fields follow as before (sum over repeatedindices a understood):

δS =∫ B

AδL d4x =

∫ B

A

[δL

δφaδφa +

δL

δ∂µφaδ∂µφa

]d4x

=∫ B

A

[δL

δφaδφa +

δL

δ∂µφa∂µδφa

]d4x

=∫ B

A

[δL

δφaδφa + ∂µ

(δL

δ∂µφaδφa

)− ∂µ

(δL

δ∂µφa

)δφa

]d4x

=∫ B

A

[δL

δφa− ∂µ

(δL

δ∂µφa

)]δφa d4x +

[δL

δ∂µφaδφa

]B

A.

(2.13)

Once again the last term vanishes because we assume that the extrema are fixedand therefore δS vanishes if and only if

δL

δφa− ∂µ

(δL

δ∂µφa

)= 0. (2.14)

We observe that S is a scalar (L is a scalar density) and that provided thefields φa have definite transformation properties under the action of (groups of)symmetries the Euler–Lagrange equations are covariant.

2.2 examples

We now discuss some significant examples of actions for classical relativisticparticles immersed in generic electromagnetic fields and compute their equa-tions of motion.

2.2.1 Free relativistic particle

The worldline of a relativistic particle can be parameterized by

xµ = xµ(τ), (2.15)

where τ is its proper time. Physical worldlines are selected by solving therelativistic equations of motion, which, as we will now show, can be derivedusing the principle of least action. In the case of a free particle such action must

2.2 examples 17

be a scalar function constructed solely from worldline data and the simplestpossibility is to use proper time τ for such purpose. An invariant definition ofthe proper time is given by the square root of the line element

ds2 = dxµ ⊗ dxνηµν = −(dx0)2 +3

∑i=1

(dxi)2. (2.16)

In order to determine the action, in the non-relativistic limit it should repro-duce the kinetic energy of a free particle. When β 1

dτ =cγ

dt ' cdt(

1− 12

v2

c2 + . . .)

. (2.17)

This implies that if we multiply proper time by mc we get an action

S0[xµ] = −mc∫

dσ, (2.18)

which is a functional in terms of all possible worldlines xµ(τ). The non-relativistic limit of (2.18) gives

S0 ' −mc∫

cdt(

1− 12

v2

c2

)= −mc2

∫dt +

∫ 12

mv2 dt (2.19)

and the first term does not depend on the position nor on the velocity andtherefore does not contribute to the equations of motion.

The equations of motion then follow from the variation of S0 with respect toall possible paths xµ(τ) connecting the events A and B:

δS = −mc∫ B

Aδdτ = −mc

∫ B

A

δ

δxµ

(√−dxρdxρ

)δxµ

= −mc∫ B

A

[12

1√−ds2

(−2dxµdδxµ

)]

= mc∫ B

A

dxµ

dτdδxµ = mc

∫ B

Auµdδxµ

=∫ B

Ad(mcuµδxµ

)−∫ B

Aδxµd(mcuµ).

(2.20)

Once more the last term is vanishing because δxµ(A) = δxµ(B) = 0 and there-fore the equation of motion is

dpµ

dτ= 0, (2.21)


with pµ = mcuµ, which is the correct equation of motion for a free relativisticparticle.

2.2.2 Electromagnetic field and its coupling to a charged particle

In an analogous fashion, we can compute free Maxwell’s equations from anaction involving only the electromagnetic tensor Fµν = ∂µ Aν − ∂ν Aµ:

S1 = − 116πc

∫d4x FµνFµν. (2.22)

The Euler–Lagrange equations for this case are

δL

δAµ− ∂ν

(δL

δ∂ν Aµ

)= 0 (2.23)

and, using (2.22), we get that

δ(

FρσFρσ)

δ∂ν Aµ= 2

δFρσ

δ∂ν AµFρσ = 2 (Fνµ − Fµν) = −4 Fµν (2.24)

and hence

∂νFµν = 0. (2.25)

There is another quadratic tensor constructed from F, which may define aLagrangian density:

L = εµνρσFµνFρσ. (2.26)

However, we can see that this is a total derivative and therefore it does notcontribute to the equations of motion:

L = εµνρσFµνFρσ = 4 εµνρσ∂µ Aν∂ρ Aσ = 4 ∂µ

(εµνρσ Aν∂ρ Aσ

). (2.27)

Its contribution may become relevant when multiplied by a function of otherfields or in the case that boundary effects have to be taken into account, but wewill not consider it for the time being.

If we introduce the current jµ = cq∫

dxµq δ4(x− xq), we can deduce Maxwell’s

equations in the presence of matter from the interaction term

S2 =1c2

∫d4x jµ Aµ. (2.28)

2.3 noether’s theorem 19

We first prove that such an action is gauge invariant provided the continuityequations ∂µ jµ = 0 are valid. Gauge transformations map Aµ → Aµ + ∂µΛ.Hence, the action S2 changes accordingly

δgaugeS2 =1c2

∫d4xjµ∂µΛ = S2 +

1c2

∫d4x ∂µ (jµΛ)− 1

c2

∫d4x ∂µ jµ Λ. (2.29)

Discarding boundary terms, the action is gauge invariant if ∂µ jµ = 0. Finally,we can put together S0 + S1 + S2 and compute both the equations of motion fora field in the presence of charges and for the motion of the charges in this field.

We already computed δS1δAµ

. If we add δS2δAµ

= 1c2

∫jµδAµ d4x we get

∂µFµν = −4π

cjν. (2.30)

Using the explicit expression of the 4-current for a single charged particle in S2we can also write

S2 =qc

∫dxµ Aµ (2.31)

and varying with respect to δxµ (up to boundary terms)

δx(S0 + S2) = −∫

δxµdpµ

dτdτ +

qc

∫ (dδxµ Aµ + dxµδx Aµ

)= −

∫δxµ

dpµ

dτdτ +

qc

∫d(δxµ Aµ

)− q

c

∫δxµdAµ +

qc

∫dxν∂µ Aνδxµ

= −∫

δxµdpµ

dτdτ +

qc

∫δxµ

(∂µ Aν − ∂ν Aµ

) dxν

dτdτ

(2.32)

leading to the equations of motion

dpµ

dτ=

qc

Fµνuν, (2.33)

which is the relativistic form of Lorentz force.

2.3 noether’s theorem

The action formalism is also extremely useful in deriving conservation lawsby looking for symmetry transformations that leave it invariant. The formal


representation of this idea is given by Noether’s theorem. This theorem showshow in classical mechanics and in (classical and quantum) field theories theexistence of a group of continuous symmetries implies the existence of conservedquantities.

Consider a system with N degrees of freedom, described by a Lagrangianindependent on time L(qI , qI) (obviously we can extend the reasoning to thetime-dependent case) and assume that there is a symmetry transformation

qI → qI , qI → ˙qI , (2.34)

which leaves the Lagrangian invariant without using the equations of motion. For-mally

δL(q, q) = L(q, ˙q)−L(q, q) = 0. (2.35)

Since

δL(q, q) =[

δLδqI− d

dt

(δLδqI

)]δqI +

ddt

[δLδqI

δqI

](2.36)

is vanishing at each instant because of the invariance of the Lagrangian underthe symmetry transformation we are considering, we see that

Q =δLδqI

δqI (2.37)

must be conserved along the motion, where the first term in (2.36) is vanish-ing because of the Euler–Lagrange equations. Note that in this derivation theassumption that the symmetry is continuous is crucial, otherwise the infinites-imal variation of L does not make sense.

Also in Lagrangian field theory, the invariance of the action gets translatedin the existence of conserved quantities, but now this is expressed by a localconservation of a current. In fact, following the same steps performed above

δL =

[δL

δφ− ∂µ

(δL

δ∂µφ

)]δφ + ∂µ

[δL

δ∂µφδφ

], (2.38)

and therefore we have

∂µ jµ = 0, for jµ =δL

δ∂µφδφ. (2.39)

2.3 noether’s theorem 21

Charge conservation follows by defining

Q =∫

d3x j0(x). (2.40)

In fact

ddt

Q =∫

d3x ∂0 j0(x) = −∫

d3x ∂i ji(x) = 0, (2.41)

where the last equality follows from the vanishing of ji at the boundary.

exercises

1. Compute the equations of motion following from the Lagrangian

L = −12

∂µ Ai∂µ Ai −

12

∂µBi∂µBi −m2(A1B2 − B1 A2),

where Ai and Bi are real fields and i = 1, 2.

2. Compute the equations of motion for the two real scalar fields φi, i = 1, 2,following from

L = −12

δij DµφiDµφj +16

Cijkφiφjφk,

where Dµφi ≡ ∂µφi + Aµεikφk, with ε12 = −ε21 = 1, and Aµ is an externalfield. Cijk are constant coefficients.

3. Compute the equations of motion for

L = −∂µΦ†∂µΦ + Φ†ηΦ− 12(Φ†Φ)2,

where Φ is a complex field with 4 components Φ(x) =

φ1(x)φ2(x)φ3(x)φ4(x)

and

η = diag−1,−1, 1, 1. Compute also the extrema of the potential.

Part II

F I N I T E G R O U P S

3G R O U P S : B A S I C E L E M E N T S

We have seen in our introductory lecture why symmetries are important tophysics and why symmetries are related to groups. In this chapter we begin tointroduce some basic elements of abstract Group Theory, which allow us to for-malize the concept of symmetry. Our aim is to apply group theory to situationsrelevant to physics, but the power of abstract algebra is to see the structures un-derlying concrete cases and generalize them. We therefore concentrate on theabstract concepts here, without following an axiomatic approach, but rather il-lustrating them with simple examples. We will then explore the consequencesof these algebraic structures for applications in physics in some later lectures.

3.1 definitions

Definition 3.1. A Group (G, ) is a set G with a composition law, called multi-plication,

: G× G → G(g1, g2) 7→ g1 g2

satisfying the following properties:

i) associativity: a (b c) = (a b) c, ∀a, b, c ∈ G;

ii) existence of neutral element (or identity) 1: a 1 = 1 a = a, ∀a ∈ G;

iii) existence of the inverse: ∀a ∈ G, ∃ a−1 | a−1 a = a a−1 = 1 .

The closure property ensures that the composition law does not generate ele-ments outside G. Associativity implies that the computation of an n-fold prod-uct does not depend on how the elements are grouped together.

25

26 groups : basic elements

The remaining assumptions can be replaced by the existence of a neutral el-ement and an inverse for left multiplication only (or right). The other relationsfollow. For instance, if we assume the existence of a neutral element for left-multiplication, 1 a = a, and the existence of the left inverse a−1 a = 1, wecan prove that a−1 is also an inverse under right multiplication:

a a−1 = 1 a a−1 = (a−1)−1 a−1 a a−1 = (a−1)−1 a−1 = 1, (3.1)

where we used the neutral element from the left and that a−1 should have aninverse from the left, called (a−1)−1. Using this result, we can also prove thatthe identity is an identity also when we compose it from the right:

a 1 = a (a−1 a) = (a a−1) a = 1 a = a. (3.2)

We can also prove that

Theorem 3.1. A group has a unique identity element and each element has a uniqueinverse.

Proof. If 1 and I are both identities, then I 1 = 1, but also I 1 = I and henceI = 1.

If the element a has two inverses, namely a−1 and b, from b a = 1 we get thatb a a−1 = (b a) a−1 = 1 a−1 = a−1, but also b a a−1 = b (a a−1) =b 1 = b.

The neutral element is its own inverse, but this may not be the only one. Wemay have other elements that are their own inverse. For instance in (R∗, ·) wehave that 1 · 1 = 1, but also (−1) · (−1) = 1.

We should also note that in general a group may have a multiplication lawthat is not commutative. In fact we distinguish Abelian and non-Abelian groups

Definition 3.2. A group G is said to be Abelian (commutative) if the composi-tion law is commutative: a b = b a, ∀a, b ∈ G.

Conversely

Definition 3.3. A group G is said to be Non-Abelian (non-commutative) if atleast two elements a, b ∈ G satisfy a b 6= b a.

Example 3.1. A simple example of an Abelian group is (R,+), where the neutralelement is the element 0. Also (Z,+) has a group structure, while (N,+)does not admit an inverse. If we use standard multiplication (R∗+, ·) has agroup structure, while (N, ·) does not. In fact if n ∈ N, the inverse undermultiplication is 1/n /∈N.

3.2 maps between groups 27

All these examples have an infinite number of elements, but there are alsogroups with a finite number of elements. We will call

Definition 3.4. Order of a group G: #(G) = number of elements in G.

Groups with a finite order are called finite groups. An example of a finitegroup is

Definition 3.5. Cyclic group of order n: Zn ≡ gk, 1 ≤ k ≤ n|gn = 1.A simple way to construct this group is by means of the n-th root of unity

g = e2πi/n. Let ζ1, . . . , ζn be the n distinct roots of the equation zn = 1, sothat ζk = gk. This set forms a group under multiplication, with compositionlaw ζk ζm = ζm ζk = e2πi/n(k+m) = ζ(k+m mod n). From this compositionwe can also derive an alternative definition of Zn as (Z,+) with equivalencerelation g + n ∼ g, ∀ g ∈ Z.

From the examples above we can also distinguish discrete groups like Zn or(Z,+) and continuous groups, like (R,+).Example 3.2. Other examples we will in the following use are the sets of realand complex square matrices Mat(n, R) and Mat(n, C). These are semigroups(not all their elements have an inverse) using matrix multiplication.

The sets of real or complex square invertible matrices form a group, instead.They are denoted by GL(n, R) and GL(n, C), respectively, where GL stands forgeneral linear.

We also note that while Zn, (Z,+) and (R,+) are Abelian groups, GL(n, R)and GL(n, C) are non-Abelian.

3.2 maps between groups

If we have two groups with the same number of elements and analogous prop-erties, how do we distinguish them? When can we consider the two groups tobe the same? In order to answer to these questions we need to discuss mapsbetween groups.

As we explained above, a group is defined by a set, but also by a multiplica-tion. The map between the elements of the two groups may or may not respectsuch multiplication law.

Definition 3.6. A Homomorphism is a map φ : G1 → G2 such that the composi-tion in G1 is mapped into the composition in G2:

φ(g1 1 g2) = φ(g1) 2 φ(g2).


Homomorphisms map the identities of the two groups among themselvesφ(1G1) = 1G2 . Also, the map of the inverse of an element is the inverse elementof the map of the original one: φ(a−1) = [φ(a)]−1. In fact, if we respect thegroup structure:

φ(a) = φ(1 G1 a) = φ(1) G2 φ(a), (3.3)

and

1G2 = φ(1G1) = φ(a−1 G1 a) = φ(a−1) G2 φ(a). (3.4)

However, homomorphic maps may loose some of the structure of the originalgroup and relate groups with different properties. For instance we will see thathomomorphisms can map non-Abelian groups to Abelian ones.

If we want to construct a map that fully respects the structure of a group weneed to require that it is also 1 to 1:

Definition 3.7. A homomorphism φ : G1 → G2 is called isomorphism if it issurjective and invertible:

∃ φ−1 : G2 → G1 | φ−1(φ(g)) = g, ∀g ∈ G1.

If the map φ is an isomorphism each element of G1 is uniquely mapped intoan element of G2 and therefore the two groups are essentially the same, up toa relabeling of their elements. Symbolically G2 ' G1.

Finally, it is useful to introduce a name for the maps from a group to itself:

Definition 3.8. An isomorphism between a group and itself is an automorphism.

Example 3.3. The first example is a homomorphism between the non-Abeliangroup of invertible matrices and the set of reals (without zero), which form anAbelian group under multiplication. The map is given by the determinant:

ϕ : GL(n, R) −→ R∗

M 7−→ det(M).(3.5)

This map preserves the group multiplication thanks to the properties of thedeterminant: ϕ(AB) = det(AB) = det(A)det(B) = ϕ(A)ϕ(B).

Example 3.4. The second example is a homomorphism between a continuousand a discrete group:

ϕ : R∗ −→ 1,−1a 7−→ a

|a| .(3.6)

3.3 finite groups 29

The map ϕ satisfies ϕ(ab) = ab|ab| = a

|a|b|b| = ϕ(A)ϕ(B), as expected for a

homomorphism.

Example 3.5. Finally, we present an isomorphism between two continuous groups:

exp : (R,+) −→ (R∗+, ·). (3.7)

The exponential map correctly satisfies exp(a + b) = exp(a) · exp(b). In addi-tion the identity of the real numbers with addition as the group compositionlaw 1(R,+) = 0 is mapped to the identity of the real numbers with multiplica-tion as composition law: 1(R∗+ ,·) = 1. The inverse map is the logarithm.

3.3 finite groups

Finite groups have properties that are not shared by infinite (or continuous)groups. For example, if an element g of a finite group is multiplied by itselfenough times, the identity of the group is recovered. In fact, multiplying anyelement of a finite group by itself more than #G times must lead to a recurrenceof the product, because there are at most #G distinct elements. Take g ∈ G,where G is a finite group with #G = n. We can easily prove that there must bea m ≤ n such that gm = 1. Consider the set g, g2, . . . , gn+1, which has oneelement more than those of G. Since each of the elements of this set is in G byclosure, at least two of them must coincide. This means that ∃ p = q + m suchthat gp = gq. We then deduce that gp = gqgm = gq and hence gm = 1, togetherwith m ≤ n.

We then call

Definition 3.9. The order of an element g ∈ G is the smallest integer n suchthat gn = 1.

We can also prove that multiplying a group by any of its element we recoverall its elements.

Theorem 3.2. Rearrangement theorem.

g G = G, ∀g ∈ G.

Proof. g G contains #G elements and hence they are all distinct and thereforeit is identified with G, or at least two of them are equal. If G = g1 , . . . , gnand g = gj, two elements of gG are equal if gjgk = gjgm, but this implies thatgk = gm and therefore gG = G.


An important application of this theorem is given by the fact that we canrepresent the composition law for a finite group via a multiplicative table. Thisis a square array whose raws and columns are labelled by the elements of thegroup and whose entries correspond to the products. The element in the i-thraw and j-th column corresponds to the element gi gj, where gi and gj are theelements labeling that raw and column.

A simple instance is the table for the group Z2. Since the requirement thatZ2 is a group implies that 1 1 = 1 and a 1 = 1 a = a, we only have to fixa a, which is determined by the rearrangement theorem to be 1:

1 a1 1 aa a 1

(3.8)

Note that the table is symmetric about the diagonal, which means that thegroup is Abelian.

A different and efficient way to present a discrete group is via its generatorsand their relations.

Definition 3.10. A group presentation 〈S|R〉 contains a set of generators S and aset of relations R between elements constructed in terms of the generators in S.

The group defined by such presentation is the result of all the inequivalent(with respect to relations in R) words constructed by using the generators in S(and their inverse) as letters. Note that for the same group we may have severalpresentations!

Example 3.6. Cyclic groups have the following presentation:

Zn = 〈g | gn = 1〉. (3.9)

We have only one generator and the inequivalent words are g, g2, . . . , gn = 1.

Using group presentations we can introduce another group we will exten-sively use during our lectures, the Dihedral group. Its definition is given by

Definition 3.11. Dihedral group: Dn = 〈ρ, σ | ρn = 1 = σ2, σρ = ρ−1σ〉

and from such presentation we can deduce all the elements of the group. Wesee from the relations in the presentation that both ρ and σ have finite order(n and 2). We can also use the last relation in the presentation to move allρ’s appearing in the definition of a group element to the left of any σ. We


therefore conclude that inequivalent words are ρAσB, with A = 0, 1, . . . , n− 1and B = 0, 1. This means that #Dn = 2n.

This group has a simple geometric origin: it is the group of symmetries ofa regular polygon with n sides. ρ is the group element related to a rotation of2π/n radians and σ is the reflection around a vertical axis.

We conclude this lecture by introducing two more groups that will be widelyused in this course: the Symmetric group and the Alternating group.

3.3.1 The Symmetric Group Sn

The symmetric or permutation group of n elements (Sn) has a special role ingroup theory because of a theorem (Cayley’s theorem) that shows that anygroup of order n is isomorphic to some subgroup of Sn. It also has an obviousimportant relevance in physics whenever one discusses identical particles.

Definition 3.12. The Permutation group of n elements (Sn) is the group of bijec-tive maps from the set X = 1, 2, . . . , n to itself.

Note that its definition depends only on the order of the elements and noton the labels.

1

2

3

2

3

1

3

2

1

2

1

3

Figure 4.: Two identical elements of the permutation group of three objects.

First of all we can see that such maps admit a group structure. The compo-sition of two maps f , g : X → X defined as f g : x 7→ f (g(x)) is obviouslyagain a map f g : X → X and hence the group is closed under multiplication. This product is also clearly associative and the identity is the map 1 : x 7→ x,∀x ∈ X. If we restrict our attention to bijective maps then there is an inversemap for each one of them and hence we have an inverse element for each ofthe group elements.

Note that if we have two sets X and Y that have the same number of elements,the associated permutation groups are isomorphic.


We can also determine the order of Sn rather easily. Since permutations aremaps σ : X → X, we just have to check how many inequivalent ways we haveto map X to itself. The first element can be mapped into each of the n elementsof X. Hence σ(1) can have n different values. At this point σ(2) can take atmost n− 1 different values, because we have to exclude σ(1), and so on. Thismeans that

#Sn = n! . (3.10)

There are several different ways to describe these maps, but, instead of givingadditional abstract definitions, we will introduce them by examples.

Example 3.7. Consider S3, the group of permutations of 3 elements. This grouphas order 3! = 6. One way to describe these elements is to use matrices describ-ing how each element is mapped: 1 2 3

σ(1) σ(2) σ(3)

. (3.11)

In this notation the top line represents the initial order of the objects and thelower line represents the final order, after the permutation. The 6 inequivalentmaps are then

1 =

1 2 3

1 2 3

, p1 =

1 2 3

2 3 1

, p2 =

1 2 3

3 1 2

,

d1 =

1 2 3

1 3 2

, d2 =

1 2 3

3 2 1

, d3 =

1 2 3

2 1 3

.

(3.12)

An alternative representation is given by cycles. This is actually more conciseand much more efficient for large n. A permutation that shuffles k < n objectsinto themselves, leaving the others untouched is called a k-cycle. For instance,we see that d1 permutes 2 and 3, but leaves 1 untouched, so we represent it by(23), or by (23)(1) if we want to keep track of the fixed element. The permuta-tion p1 on the other hand maps 1 → 2 → 3 → 1. We therefore represent it bythe 3-cycle (123). The whole group is then represented by:

1 = (1)(2)(3), p1 = (123), p2 = (132),

d1 = (1)(23), d2 = (2)(13), d3 = (3)(12).(3.13)


The composition is trivial and follows by applying the maps in series. Forinstance:

(123) (13) :1 → 3 → 12 → 2 → 33 → 1 → 2

= (23). (3.14)

Another important property of Sn (which we will not prove) is that everypermutation can be uniquely resolved into cycles which operate on mutually esclusivesets. These are called disjoint cycles, like in d1 = (1)(23). Another generalproperty is that each element can be obtained as a product of 2-cycles (in thiscase they may not act on mutually esclusive sets). Also, the decomposition in2-cycles is not generically unique. For instance,

(123) = (32) (31) =1 → 3 → 22 → 2 → 33 → 1 → 1

(3.15)

but also (13) (12) = (123). For each element however there is a minimumnumber of 2-cycles needed to describe it and one can also see that if suchnumber is even all the decompositions of that group element will be givenin terms of an even number of 2-cycles and if such number is odd all thedecompositions will be given in terms of an odd number of 2-cycles.

We therefore define

Definition 3.13. The parity of a permutation g ∈ Sn is π(g) = (−1)n(g), wheren(g) is the number of 2-cycles needed in an arbitrary decomposition of g.

We can then split our symmetric group in terms of the even and odd elementswith respect to parity. It is interesting to note that the subset of elements witheven parity define also a group:

Definition 3.14. The Alternating group An is defined by the subset of elementsof Sn with even parity and form a group under the multiplication rules of Sn.

In the case of S3 we see that A3 = 1, p1, p2. Note also that in this case A3is Abelian, while S3 is not.


exercises

1. Consider the following sets of elements and composition laws. Determinewhether they are groups and, if not, identify which group property isviolated:

• Q+, with a b = a/b;

• Even integers with respect to ordinary addition;

• Z with a b = a− b;

• Q∗, with a b = ab;

• Z∗n ≡ a ∈ Zn | gcd(a, n) = 1, with a b = ab. (Tricky. Try firstwith Z∗5 = 1, 2, 3, 4)

2. Show that a group is Abelian if and only if (ab)−1 = a−1b−1.

3. Write the multiplicative table for a group of order 3 and show that it isunique.

4. Determine all groups of order 4 (up to isomorphisms) and discuss if theyare Abelian or not.

5. The group of quaternions is the group generated by the elements 1,−1, I, J, Ksatisfying the composition laws

I2 = J2 = K2 = −1, I J = K, (−1)2 = 1.

Compute its multiplication table.

6. Consider the following couples of groups:

Z4, S3,Z6, D3,S3, D3.

Which lines contains isomorphic groups?

7. Show that the inverse of a k-cycle is the same k-cycle written in reverseorder.

8. Compute (34) (623174) (34), (12) (13)(45) (12) and (43) (13)(452) (34). What do we learn from this?


9. A3 is isomorphic to another group we already know, which one?

10. A presentation for A4 is A4 = 〈a, b|a2 = 1 = b3 = (ab)3〉. Find twopermutations in S4 representing the elements a and b of the given presen-tation.

4S U B G R O U P S , Q U O T I E N T S A N D P R O D U C T S .

Now that we have introduced the basic notions of group theory, we would liketo understand when a subset of the elements of a group form themselves agroup under the same composition law, how to take products and quotients ofgroups.

4.1 basic definitions

Definition 4.1. A subset of a group H ⊂ G is a subgroup H < G if and only ifit is closed under the composition law of G (and contains the inverse of eachelement):

H < G ⇔ h1 G h−12 ∈ H, ∀h1, h2 ∈ H. (4.1)

For finite groups there is no need of the requirement of having the inverse,because an = 1 for some n for any a ∈ H.

According to this definition, the unit element 1 forms a subgroup of G aswell as the full G. These are improper subgroups.

In physics subgroups will be extremely useful, especially in the context ofsymmetry breaking. We often encounter systems whose Lagrangian (or Hamil-tonian) is invariant under a certain symmetry group, which is broken by theaddition of additional terms to a subgroup H < G.

Example 4.1. Dn contains a subgroup isomorphic to Zn. This is obtained as theresult of the repeated application of the generator ρ in the presentation (3.11).

Example 4.2. Zn < Sn. If we name ρ = (12 . . . n), we see that ρn = 1 and ρk 6= 1

for any k 6= n.

37

38 subgroups , quotients and products

Example 4.3. Dn < Sn: ρ has been identified above. σ = (1 n)(2 n − 1) . . . =(1 2 . . . n− 1 nn n− 1 . . . 2 1

).

Once again all these examples let us sense that all finite groups can be real-ized as subgroups of Sn for n sufficiently large (which is the content of Cayley’stheorem). If we only had finite groups we could then forget all the theoreticalconstructions we introduced so far and directly study Sn. However it is stilluseful to use finite groups as a laboratory to develop the necessary intuition tostudy the properties of continuous groups.

A special subgroup in every group is

Definition 4.2. The Center of a group is the subgroup ZG < G defined as

ZG = a ∈ G | a g = g a ∀g ∈ G.

This subgroup contains all the elements of a group that commute with all theothers. By definition, the center of a group is Abelian and it is straightforwardto see that if G is Abelian than the ZG is the whole group.

Example 4.4. Consider GL(2,R), i.e. the group of 2 by 2 invertible matrices. Thecenter is given by all the matrices proportional to the identity: s1, with s ∈ R∗.

4.2 quotients

When the symmetry group G of a given physical system is broken to a sub-group H, some of the elements of the original invariance will have a non-trivialaction. In detail, while H will still leave the system invariant, elements of Gthat are not in H will not. It is therefore interesting to ask what is the structureemerging for the trasformations under G when we identify all the transfor-mations in H < G. For this reason we now discuss the quotients of groups.Quotients essentially emerge by identifying elements in a group via equiva-lence relations. In order to define them we therefore need first to recall theproperties of

Definition 4.3. An equivalence relation ∼ is a binary relation ∼ on a set if andonly if it is

1. reflexive: a ∼ a;

2. symmetric: a ∼ b ⇒ b ∼ a;

4.2 quotients 39

3. transitive: a ∼ b and b ∼ c ⇒ a ∼ c.

Once we have an equivalence relation we can split the elements according totheir properties under such a relation. In particular, we will group together allthe equivalent elements in so-called equivalence classes:

Definition 4.4. An equivalence class for an element a ∈ A is the set of all ele-ments in A equivalent to a:

[a] ≡ b ∈ A | b ∼ a.

It can be seen that equivalence relations split a set into disjoint subsets givenby equivalence classes. This splitting is going to be crucial in the definition ofgroup quotients.

If we now want to identify elements in G that are related by elements in H,we need to define appropriate equivalence classes by multiplying each elementof G by all the elements in H. Since for a generic group the product is notcommutative, the result will depend on the order. Given H < G we thereforedefine distinct equivalence classes using left and right multiplication.

Definition 4.5. Right cosets are the sets obtained by multiplying H < G byelements g ∈ G from the right: Hg. Analogously, Left coset ≡ gH, with H < Gand g ∈ G.

Note that in general gH and Hg are not in H unless also g ∈ H and they arenot even subgroups of G.

Example 4.5. Consider H = 1, d1 < S3. This group is isomorphic to Z2. Rightcosets are obtained by multiplying H from the right with the elements of S3.The group structure shows that we have only 3 distinct cosets:

H1 = Hd1 = 1, d1, Hd2 = Hp1 = d2, p1, Hd3 = Hp2 = d3, p2. (4.2)

Similarly, left cosets are

1H = d1H = 1, d1, d2H = p2H = d2, p2, d3H = p1H = d3, p1. (4.3)

As expected, left and right cosets can be different and indeed in this case theyare different. We can also note that the intersection between left cosets (orbetween right cosets) is empty and that their union is the original group S3.

In fact, this is a consequence of a general theorem:


Theorem 4.1. Two cosets of a subgroup have the same elements or they do not shareany element in common.

Proof. Let Hg1 and Hg2 be two different (right) cosets. If there is at least oneelement in common, this means that for hi, hj ∈ H, we have hig1 = hjg2. Fromthis assumption we derive that g2(g1)

−1 = (hj)−1hi ∈ H. Using the rearrange-

ment theorem, we then conclude that Hg2(g1)−1 = H and hence Hg2 = Hg1,

so all their elements are in the same.

We can finally discuss quotients of groups. Identifying elements in the samecoset we get

Definition 4.6. The Right quotient G/H of a group G by its subgroup H is theset of all equivalence classes of elements of G obtained by right multiplicationswith H:

g1 ∼ g2 if ∃ h ∈ H | g1 = g2h.

The left quotient H\G is obviously defined by the relation g1 ∼ g2 if ∃ h ∈H | g1 = hg2. From these definitions we see that quotients are collections ofequivalence classes defined by cosets.

In general G/H 6= H\G and they do not form a group with respect to thegroup multiplication in G, unless H satisfies a special condition: being self-conjugate. This is a special condition under which the group multiplication inG can be transferred without inconsistencies to G/H.

First of all we need to define what we mean by conjugation. This is anoperation by which we relate different elements of a group. In detail:

Definition 4.7. a, b ∈ G are conjugate if ∃g ∈ G | a = gbg−1.

This definition implies that the elements a and b give the same element whenmultiplied by g, the first from the right and the second from the left. We thensee that we can use such a definition to establish a relation between left andright cosets. Actually, we can extend this definition to whole subgroups of agroup:

Definition 4.8. N < G is Normal (and we write NC G) if it is selfconjugate:

gNg−1 = N, ∀g ∈ G.

This definition implies that the conjugate of any element in N is also an ele-ment in N for any element in G defining the conjugation: gng−1 ∈ N, ∀g ∈

4.2 quotients 41

G and ∀n ∈ N. The straightforward consequence is that right and left cosets de-fined in terms of N coincide: gN = Ng. Since left and right cosets are the same,also left and right quotients coincide. This eventually leads to the fact that G/Nquotients inherit the property of being a group under the multiplication of theoriginal group G:

Theorem 4.2. If NC G then G/N is a group.

Proof. The elements of the quotient are defined as the equivalence classes givenby the cosets. Hence all the elements in gN for a given g ∈ G have to beidentified. The product in G/N is the product in G, where again we identifythe elements which are part of the same coset. We then have to verify that allgroup properties are respected for G/N. Closure follows from the compositionin G:

g1N g2N = g1(g2N(g2)−1)g2N = g1g2N(g2)

−1g2N = g1g2N,

where we first used the fact that N is normal and then the fact that N is agroup (N N = N). The associative property is straightforward. The existenceof an inverse for each element follows from the proof of closure above. Givenan element gN, its inverse is defined by g−1N.

Example 4.6. We first give an instance of a subgroup that is not normal. Con-sider Z2 < Dn, defined by the σ generator in the presentation (3.11). Fromthe equivalence relations given in the presentation we can see that ρσρ−1 =σρ−2/∈Z2. As expected, this implies that the product of elements in the quo-tient using the multiplication in Dn does not close consistently. In fact, Dn/Z2implies the identification ρk ∼ ρkσ, but if we multiply elements in the sameequivalence class, we get inconsistent results:

ρaρb = ρa+b 6= (ρaσ)(ρbσ) = ρa−b. (4.4)

Example 4.7. An example of a normal subgroup is given by Zn < Dn, definedby ρ. Since we identify all the elements ρk ∼ 1 and σρk ∼ σ, the quotient isgiven by two elements, which define the group Z2 = 1, σ. The multiplicationof elements in the same equivalence class now gives consistent results. Forinstance, we see that

σρaσρb = σ2ρb−a = ρb−a ∼ 1 = σσ. (4.5)

Example 4.8. Although for n = 3 D3 ' S3, in general Zn < Sn is not normalfor n > 3. This can be seen right away by checking that (12)(12345)(12) =(21345) 6= (12345)a for any a.


4.3 products

In Physics we can imagine systems where we have symmetries of differentnature that are simultaneously present. Both should be represented by groups,with independent actions. We could also have instances where one symmetryinterferes with the way the other acts. In this case the two actions will not beindependent and the resulting group will be a non-trivial combination of theoriginal ones.

For these reasons we now discuss group products.

Definition 4.9. Given two groups G and K their (direct) product is

G× K = (g, k), g ∈ G, k ∈ K,

with composition

(g1, k1) (g2, k2) = (g1 g2, k1 k2).

For finite groups we can deduce immediately the order of the product as

#(G× K) = #G #K. (4.6)

In general, there are two obvious normal subgroups: G× 1K and 1G × K. Thequotients with respect to these normal subgroups are K and G, respectively.

Using the presentations of the two groups forming the product, we can definethe product group via a presentation that uses all their generators and relationswith an additional one, which we now explain. If we call S1 = gi the set ofgenerators of G and S2 = ki the set of generators of K, the generators ofG× K are S = S1 ∪ S2. Moreover, the fact that we have a direct product G× Kis equivalent to the fact that all the elements gi ∈ G× 1K and k j ∈ 1G × K (weloose the identity factors in the following) should commute: gik j = k jgi. Wecan then write

G× K = 〈S1 ∪ S2 | R1 ∪ R2 ∪ gkg−1k−1 = 1, ∀g ∈ G, k ∈ K〉. (4.7)

This means that we need to take all the words constructed from the genera-tors of G and K with their equivalence relations plus the commutation relationbetween elements of G and K.

As we discussed in the introduction to this section, we may also need tospecify groups that are products where the action of one of the factors depends

4.3 products 43

on which of the elements of the other factor is taken in the product. For thisreason we can introduce a new definition, where the last relation in (4.7) isreplaced by a new one.

Definition 4.10. Given two groups G and K their semidirect product is definedby

G n K ≡ 〈S1 ∪ S2 | R1 ∪ R2 ∪ gφg(k)g−1k−1 = 1, ∀g ∈ G, k ∈ K〉,

where φg : K → K is a g-dependent automorphism.

Note that the symbol defining the product contains an asymmetry, which isreflected in the different ways the two groups enter in the definition: G n K =K o G. In fact, for semidirect products elements in G × 1K and in 1G × K donot necessarily commute.

From the definition we also see that K is a normal subgroup, i.e. KC (GnK),while G is not, i.e. G < (G n K).

Example 4.9. Dn ' Z2 n Zn, where Z2 = 〈σ | σ2 = 1〉, Zn = 〈ρ | ρn = 1〉 andφσ(ρ) = ρ−1, so that σρ = ρ−1σ. Zn is a normal subgroup, Z2 is not.

Example 4.10. The group of orthogonal matrices in N dimensions O(N) is de-fined as the set of matrices O ∈ GL(N, R), which satisfy OTO = 1N = OOT .Since det (OTO) = (det O)2 = det 1N = 1, the matrices of O(N) satisfy detO = ±1. Those with positive determinant form a subgroup, called special or-thogonal group SO(N) < O(N), which obviously contains the identity. Thequotient with respect to this subgroup generates two classes of equivalence,the one of matrices with positive determinant (which contains the identity)and the set of matrices with negative determinant. This is actually a Z2 groupunder the original multiplication. In fact we leave as an exercise to the readerto check that generically O(N) ' SO(N)o Z2, which, for odd N reduces to adirect product.

Example 4.11. The Euclidean group E(N) ' ISO(N). It is defined as the combi-nation of rotations SO(N) and translations in RN :

E(N) = (O, a) | OTO = 1N , det O = 1, a ∈ RN. (4.8)

Given v ∈ RN the action is v→ Ov + a. The composition gives

(O2, a2) (O1, a1) = (O2O1, O2a1 + a2), (4.9)


in fact v(O1,a1)−→ O1v + a1

(O2,a2)−→ O2O1v + O2a1 + a2. We have two subgroups(O, 0) and (1, a), but

(A, 0)(1, a)(A−1, 0) = (1, Aa) = φA−1(a). (4.10)

Hence

ISO(N) ' SO(N)n RN . (4.11)

4.4 classifying finite groups .

Using direct and semidirect products we can construct groups from othergroups, but obviously these are not really new, because they can be classifiedas products. So, how do we distinguish genuinely new groups?

Definition 4.11. A group is called simple if it does not admit proper normalsubgroups.

Definition 4.12. A group is called semisimple if it does not admit nontrivialnormal Abelian subgroups.

Non-simple groups can be obtained as (semi)direct products.The classification of all finite groups (simple ones) required computers. Most

of them are in infinite families (like Zn with prime n, An, etc...) There are somefinite simple exceptional groups (for instance the Monster Group has about 8

·1053 elements)The enormous theorem shows that there are a few infinite families and a finite

number of exceptional cases (sporadic groups).

exercises

1. Prove Lagrange’s theorem, which states that the order of a subgroup Hof a finite group G is a divisor of the order of G, i.e. #H divides #G.

2. Compute the center of Dn.

3. Find subgroups of order 2,3 and 4 of S4.

4. Show that for Abelian groups left and right cosets coincide.

4.4 classifying finite groups . 45

5. Consider the following permutations in S8:

a =

(1 2 3 4 5 6 7 83 6 1 5 4 2 8 7

),

b =

(1 2 3 4 5 6 7 82 3 4 5 6 7 1 8

),

c =(

1 2 3 4 5 6 7 82 4 6 1 3 5 7 8

).

a) Are these conjugate to each other?

b) a, b, c generate a group G < S8. Is G < A8?

c) Take now H < G generated from a e b. Is HC G?

(Attention! #G = 168 and #H = 56. Find meaningful products to solvethe exercise)

6. Find all proper subgroups of the group of quaternions Q and discuss theright and left quotients of Q with respect to them.

7. Is Z2 ×Z2 isomorphic to Z4? And what about Z2 ×Z3 and Z6?

8. Take A = Z2 and B = Zk. Discuss all possible products C = A n B fork = 2, 3, 4 and check if the resulting group is isomorphic to Zn, Dn, An orSn.

5R E P R E S E N TAT I O N S

So far we discussed groups using their abstract definition. To really start ap-plying them (especially to physics), we must be able to represent them in a waythat could be used for practical calculations. For this reason we need to mapabstract groups into groups of mathematical objects we know how to deal with,like matrices or differential operators. We therefore introduce in this lecture theconcept of representations of a group.

If a group codifies the symmetries of a physical system, i.e. the relationsbetween the transformations we can apply to a system without changing thephysics, we want to be able to represent this action on the system.

Beware that the existence of a symmetry does not mean that nothing changeswhen applying it. For instance the discrete group D3 rotates or reflects a trian-gle, leaving the image fixed, but not the vertices:

σ

refl.

ρ

rot.

Figure 5.: Example of some of the actions of D3 on a triangle.

In the same way, when we discuss a physical system, its invariance under acertain symmetry does not imply that all its quantities remain unchanged. Onthe other hand, the existence of a symmetry transformation tells us how thevariables describing a certain physical state change. For instance, Special Rela-tivity dictates invariance of physics under Lorentz transformations. This doesnot mean that two inertial frames, related by a Lorentz transformation, share

47

48 representations

exactly the same value for all their physical quantities, but rather that theirtransformation is such that physical laws do not change in form. For instance,the value of pµ from an inertial system to another changes by pµ → Λµ

ν pν, butfree particles are described by the same law dpµ

dσ in both systems. In fact, Λ pro-vides a concrete representation of how the group of Lorentz symmetries actson the various components of the momentum, which are grouped in a multipletof the symmetry group. In a similar fashion, when we consider a quantum-mechanical system with a certain symmetry, we know that there must exist aunitary operator commuting with the Hamiltonian, which gives a representa-tion to the action of such symmetry. The wavefunctions, however, do not sharethe symmetry of the Hamiltonian and they will transform in a way determinedby such operator and, in particular, by the representation of the symmetrygroup provided by the unitary operator mentioned above. This, in turn, willprovide a way to classify the eigenfunctions of the Hamiltonian, as we willdiscuss shortly.

5.1 basic definitions

As understood from our introduction, there may be different ways to repre-sent a group. However, we will be mainly interested in linear representations.These are representations by means of matrices. We will therefore dedicate aconsiderable amount of time to the discussion of groups of matrices (both finiteand continuous) and to their properties.

Definition 5.1. A linear representation of a group G is a homomorphism (iso-morphism) D : G → GL(N, C).

In a linear representation, each element of the group is represented by aN× N complex matrix D(g), whose composition law respects the compositionlaw in G:

D(g1 g2) = D(g1)D(g2),

D(1) = 1N ,

D(g−1) = [D(g)]−1.

(5.1)

From a different point of view, we can describe the linear representation D asa map relating elements of a linear vector space D : G × V → V, which wegenerically take to be a complex space V = CN . N fixes the size of the matricesby which we represent the group G and therefore we call

5.1 basic definitions 49

Definition 5.2. The dimension of the vector space on which the matrices D(g)act is the dimension of the representation.

Since we are representing an abstract group G by a group of matrices, themappings between the two groups may not necessarily preserve all the infor-mation of the original group G. For this reason we distinguish between faithfulrepresentations, obtained by means of isomorphisms, and unfaithful representa-tions, obtained by homomorphisms. In this last case, we will have that differentelements of the group are represented in the same way. If more than one el-ement is represented by the same matrix, the group element connecting themmust be represented by the identity. This further implies that more than onegroup element must be represented by the identity matrix and therefore thatthe kernel of the map is going to be non-trivial. A faithful representation thenis such if D(g) = D(g′)⇔ g = g′.

It is also easy to convince ourselves that the elements in the kernel of therepresentation are going to be a normal subgroup of the original group G:H = ker DC G. In fact, if D(g) = D(g′) = 1, we also have that D(g g′) =D(g)D(g′) = 1. Obviously D(1) = 1, the inverses are again represented bythe identity and finally D(g h g−1) = D(g)1D(g)−1 = 1 for any g ∈ G andh ∈ H. We can then conclude that if our representation is unfaithful for G,the same matrices are providing a representation of G/H that is faithful! Wetherefore conclude that:

Theorem 5.1. All non-trivial representations of a simple group are faithful.

Once we choose a basis for V, ei with i = 1, . . . , N, the matrix action isdefined as

D(g)ei = ejDji(g), (5.2)

where sum over repeated indices is understood. This is needed to preserve thecomposition rule

D(g g) = D(g)D(g). (5.3)

In fact:

D(g g)ei = ejDji(g g) = D(g)[D(g)ei] =

D(g)ekDki(g) = ejDjk(g)Dki(g)(5.4)

and hence

Dji(g g) = Djk(g)Dki(g), (5.5)

50 representations

which provides the standard matrix composition law. Be careful with the indexposition when passing to an explicit representation.

We can now ask ourselves whether we can always represent a group in sucha way and also what kind of matrix representations are really useful. A repre-sentation that always exists is

Definition 5.3. The trivial representation is defined by

D(g) = 1, ∀g ∈ G.

Another representation that is easy to construct is

Definition 5.4. The regular representation DR acts on a basis of vectors |g〉 ∈ C#G

asDR(g)|g′〉 = |g = g g′〉.

We directly used g as an index for such a representation, because we areassuming that the linear space on which it acts has one basis vector for eachelement of G. The matrix representation can then be given explicitly by theformula:

DRg1g2

(g3) = δg1,g3g2 . (5.6)

Example 5.1. Take Z2 = 1, σ, V = span|1〉, |σ〉 and then

DR(1) =

(1 00 1

), DR(σ) =

(0 11 0

). (5.7)

So that DR(1)|1〉 = |1〉, DR(σ)|1〉 = |σ〉, DR(1)|σ〉 = |σ〉 and finally DR(σ)|σ〉 =|1〉.

Representations of G induce in an obvious way representations of any H < G.We just need to restrict the maps D to the elements h ∈ H.

Example 5.2. The representations of D3 induce representations of Z3 by D(ρ)and also of Z2 < D3 by using D(σ).

Since representations are specified by matrices describing each of the groupelements, we may ask ourselves whether we really need to give explicitly eachof them or not. Actually, it turns out that it is often sufficient to provide theexplicit representation of a meaningful subset. For instence, once we have apresentation for a group, we can provide a representation for the same groupby choosing D(s) for each s ∈ S in a way that satisfies the presentation relations:D(r) = 1 ∀r ∈ R.

5.2 equivalent and reducible representations 51

Example 5.3. Zn = 〈ρ | ρn = 1〉 can be represented trivially by D(ρ) = 1. ThenD(ρn) = D(ρ) . . . D(ρ) = 1 = D(1). However, we can also provide a faithfulrepresentation by choosing D(ρ) = e2πi/n.

Example 5.4. D3 can be represented by its geometric action on a plane. This canbe represented by two matrices acting on the x and y coordinates giving the lo-cation of the vertices of an equilateral triangle. The first one, D2(ρ), representsa rotation by 2/3π around the origin and the second one, D2(σ), is a reflectionon the x axis:

D2(ρ) =12

(−1

√3

−√

3 −1

), D2(σ) =

(−1 00 1

). (5.8)

It is easy to check that the relations in the presentation are identically satisfied:

[D2(ρ)]3 = [D(σ)]2 = 12, (5.9)

D(ρ)D(σ) =12

(1√

3√3 −1

)= D(σ)[D(ρ)]2. (5.10)

In general, we may have an infinite number of different representations forthe same group. To our advantage, we can often classify them, and this is thesubject of the next part of this lecture. In fact, knowing all the representationsof a given group allows us to specify all its (linear) actions on a set of quantitiesand it is therefore of primary importance for applications to physics.

5.2 equivalent and reducible representations

We already identified the conditions under which two groups are the same. Inthis section we would like to identify the conditions under which two represen-tations are the same.

Definition 5.5. Two representations D1 and D2 of the same group G are equiv-alent if we can relate them by a change of basis, i.e.

∃S | D1(g) = S−1D2(g)S, ∀g ∈ G.

The matrix S is called a similarity matrix and its application to V changes thebasis of the representation space.

From the definition above, we see that S preserves the trace

trD1 = tr(S−1D2S) = trD2 (5.11)

52 representations

as well as the determinant

detD1 = det(S−1D2S) = det(S)−1det(D2)det(S) = trD2. (5.12)

Hence, if detD1 6= detD2 or trD1 6= trD2 the two representations D1 and D2cannot be equivalent.

Once we have constructed a map D that provides a representation for G, wecan easily construct other representations of the same group by consideringD = [DT ]−1 and D∗. We should stress that, although we can construct themin terms of D, they are not necessarily equivalent representations! When theyare equivalent, if we have a basis such that D(g) = D∗(g) then we call D a realrepresentation and if D(g) = S−1D∗(g)S, but we do not have a basis where thetwo coincide, then D is pseudoreal.

This brief discussion already shows that for a given group we can have morerepresentations. We could now ask which is the most efficient way to find themand classify them. For instance, starting from two non-trivial representationsD1 and D2 we can construct a new representation D1 ⊗ D2 acting on V1 ⊗ V2,but we could also construct another one by

D1(g)⊕ D2(g) =(

D1(g) 00 D2(g)

), ∀g ∈ G. (5.13)

It is obviously not useful to study all equivalent representations. We just needone element in each equivalence class. How do we decide whether it is possibleto reduce a representation to more fundamental ones? First we try to separateall possible different cases and then give criteria to understand which is theone we face.

Definition 5.6. A representation is reducible (or totally reducible) if we can writeit as direct sum of other representations up to a change of basis.

Alternatively, D : G×V → V is reducible is V = V1⊕V2 with V1 and V2 bothinvariant under the action of D: D(g)V1 ⊆ V1 and D(g)V2 ⊆ V2, ∀ g ∈ G.

Definition 5.7. A representation is decomposable (or reducible) if it is equivalentto a block triangular representation.

Alternatively, D : G×V → V is decomposable if there is a V1 ⊂ V invariantunder the action of D:

D(g) =(

D1(g) B(g)0 D2(g)

), ∀g ∈ G. (5.14)

This representation maps (v10) vectors to themselves.

5.2 equivalent and reducible representations 53

Definition 5.8. A representation is irreducible if it is non-decomposable.

Following the same arguments as above, a representation is irreducible ifthere are no invariant subspaces under its action.

Example 5.5. S3 ' D3. We already saw the 2-dimensional representation D2 inequation (5.8). Using the fact that S3 is the group of permutations of 3 objects,we can now construct a 3-dimensional representation, where the basis of thevector space on which it acts is given by the three elements acted upon by thepermutations. A simple computation gives

D(1) =

11

1

, D(d1) =

11

1

,

D(d2) =

11

1

, D(d3) =

11

1

,

D(p1) =

11

1

, D(p2) =

11

1

.

(5.15)

1 ρ ρ2 σ σρ σρ2

Figure 6.: Graphic representation of the 3-dimensional representation of equation (5.15)as the action on the vertices of a triangle: (1, 0, 0) is the red one, (0, 1, 0) isthe blu one and (0, 0, 1) is the green one.

It is straightforward to check that such a representation is reducible. We canimmediately see that the three vertices are always mapped among themselves

54 representations

and therefore e3 = 1/√

3

111

is invariant. Also the subspace generated by its

orthogonal complement

e1 =1√2

1−10

, e2 =1√6

11−2

(5.16)

is invariant. This becomes obvious if we use as basis for the vector spaceon which these matrices act the set e1, e2, e3. In this new basis the samerepresentation matrices become

D(ρ = p1) =

−1/2√

3/2 0−√

3/2 −1/2 00 0 1

, D(σ = d3) =

−1 0 00 1 00 0 1

. (5.17)

These are block-diagonal matrices and so will be all the matrices obtained bytheir multiplications. We therefore see that this representation is the direct sumof the 2-dimensional representation D2 and of the trivial representation.

Before proceeding let us stress the importance of the field used for the vectorspace V in order to decide whether a given representation is reducible or not.If we consider SO(2), the group of rotations on a plane, a generic element isrepresented by

D(θ) =

(cos θ sin θ− sin θ cos θ

). (5.18)

This representation is irreducible if we act on R2, but becomes reducible on C2:

D(θ) = S−1D(θ)S =

(eiθ 00 e−iθ

), (5.19)

where S = 1√2

(1 1i −i

).

5.3 unitary representations

If V is a complex vector space with a positive definite norm then V is a pre-Hilbertian space (Hilbert if complete). It is then obvious the importance ofdiscussing unitary representations of groups.

5.3 unitary representations 55

Definition 5.9. A unitary representation of a group G is a representation D suchthat ∀ g ∈ G and x, y ∈ V

〈x|y〉 = 〈D(g)x|D(g)y〉 ⇔ D(g)†D(g) = 1.

Hence D(g−1) = [D(g)]−1 = [D(g)]†.

A very interesting fact comes from the following theorem, which proves thatfor finite groups unitary representations are all we need.

Theorem 5.2. Any representation of a finite group is equivalent to a unitary represen-tation.

Proof. Take D(g) a non-unitary representation of g ∈ G. Define

M ≡ ∑g∈G

D†(g)D(g). (5.20)

This satisfies

D†(g)MD(g) = ∑g∈G D†(g)D†(g)D(g)D(g)

= ∑g∈G D†(g g)D(g g) = ∑g′ D†(g′)D(g′) = M,(5.21)

where we used the rearrangement theorem in the second line. We also have thatM = M† and hence M is a matrix with positive real eigenvalues (v† Mv > 0).We can diagonalize it by means of a unitary matrix U

M = U†mU, with m = diagm1, . . . , mN (5.22)

and we can define s = diag√m1, . . . ,√

mN. In this way

M = S2 = U†sUU†sU = U†s2U = U†mU, (5.23)

where S = U†sU and S† = S. We can now prove that SD(g)S−1 is unitary forany g ∈ G:

(SDS−1)†(SDS−1) = (S−1)†D†S†SDS−1

= S−1D† MDS−1 = S−1MS−1 = S−1SSS−1 = 1.(5.24)

56 representations

Although we proved this theorem for finite groups, we could prove an anal-ogous one for continuous compact groups. For continuous compact groups, how-ever, we need to sum over infinite elements. This means that we can still per-form the same proof provided we replace the sum with an integral over thegroup, with an appropriate invariant measure (called Haar measure). Only theexistence of such a measure allows us to repeat the demonstration with

M ≡∫

dµgD†(g)D(g), (5.25)

otherwise we cannot use the rearrangement theorem.For finite groups we can also prove that

Theorem 5.3. Every unitary decomposable representation is reducible.

Proof. If D is decomposable we know that there is a subspace of the total vectorspace such that D(g)V ⊆ V, ∀ g ∈ G. To show that it is reducible we shouldalso show that D(g)V⊥ ⊆ V⊥, ∀ g ∈ G. This follows because for any |v〉 ∈ Vand |w〉 ∈ V⊥ we have that 〈w|v〉 = 0, but also

D(g−1)|v〉 = |v′〉 ∈ V, ∀g ∈ G. (5.26)

These facts imply that D(g)|w〉 = |w′〉 ∈ V⊥, because

〈w′|v〉 = 〈w|D†(g)|v〉 = 〈w|D(g−1)|v〉 = 0. (5.27)

This means that for finite groups and for continuous compact groups we canalways construct representations that are unitary and block diagonal. Hence wecan constrain ourselves to studying the irreducible representations composingthem.

We stress, however, that if the group is not finite nor compact, we may haverepresentations that are decomposable, but not reducible, like

D(x) =(

1 x0 1

), (5.28)

representing (R,+), in fact D(x)D(y) = D(x + y).

5.4 comments 57

5.4 comments

Unitary representations play a fundamental role in QM. If we prepare a systemin a state |1〉, the probability to find it in a state |2〉 is

P12 =|〈2|1〉|2

|| |1〉 ||2 || |2〉 ||2 .

The change of state given by the action of the symmetry group G (for instancethe rotation group or Sn for n identical particles) is represented by a unitaryoperator and therefore its action on the Hilbert space is such that the transitionprobabilities are not affected. We see then that knowledge of group theoryimplies knowledge of QM systems.

For instance, energy levels are given by eigenstates of the Hamiltonian: H|〉 =E|〉. If G is a group of transformations leaving H invariant, there is a uni-tary representation of this group that commutes with H and hence UH|E〉 =EU|E〉 = H(U|E〉). This means that the unitary operator U maps eigenstateswith eigenvalue E into states with the same eigenvalue and that thereforethe portion of the Hilbert space containing the eigenvectors with energy E,HE ⊂ H, is invariant. If such a space is invariant under the action of U, this op-erator must be in an irreducible representation of the symmetry group. Hencethe level degeneracy, which is equivalent to the dimension of HE, must also bethe dimension of an irreducible representation.

exercises

1. Given two matrices A ∈ Mat(p× q, R) and B ∈ Mat(r× s, R), their directsum is the matrix

A⊕ B =

Aij i ≤ p and j ≤ q,

B(i−p)(j−q) i > p and j > q,

0 otherwise.

Their direct product is defined by

(A⊗ B)(ik)(jl) = AijBkl .

Show that if the dimensions are the same (i.e. p = r and q = s) then(A⊕ B)(C⊕ D) = AC⊕ BD and (A⊗ B)(C⊗ D) = AC⊗ BD.

58 representations

If A ∈ GL(n, R) and B ∈ GL(m, R), show that

tr(A⊗ B) = (trA)(trB).

Consider now Pauli matrices:

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

).

Compute

γ0 = −i σ1 ⊗ 1, γ1 = σ2 ⊗ σ1, γ2 = σ2 ⊗ σ2, γ3 = σ2 ⊗ σ3.

Verify that σiσj = iεijk σk + δij 1 and, using what you just learned, computeγ5 = −iγ0γ1γ2γ3.

2. Consider again exercise 8 of chapter 4. Construct the regular representa-tion for all cases with k = 2 and show their decomposition in irreduciblerepresentations.

3. Verify that the matrices

D(1) =

(1 00 1

), D(a) =

12

(−1

√3√

3 1

)form a representation for the group Z2. Is this representation irreducible?If not, determine the 1-dimensional representations which form the directsum of this representation.

4. The group A4 < S4 is generated by all possible products of the followingelements of S4: a = (12)(34) and b = (123). A linear representation forthese permutations is given by

D(a) =

−1 0 0 00 1 0 00 0 −1 00 0 0 1

, D(b) =

0 1 0 00 0 1 01 0 0 00 0 0 e2πi/3

.

a) Discuss whether this representation is irreducible or not.

b) Knowing that A4 admits 3 irreducible 1-dimensional representationsand 1 irreducible 3-dimensional representation, provide the explicitform of the matrices D(a) and D(b) for such representation.

5.4 comments 59

c) Discuss the quotient of A4 with H = 1, a, b2ab, bab2.

5. In this exercise we want to discuss a representation of the group of per-mutations of two objects S2 = Z2 = 1, ρ, when the two objects are acouple of quantum-mechanical particles. The state of each particle is anelement of a Hilbert space H and the state of the couple is an element ofH⊗H. The action of the S2 group on vectors of H⊗H is

D(ρ)[ψ⊗ χ] = χ⊗ ψ, ∀ψ, χ ∈ H.

If the two particles are identical, only states satisfying D(ρ)Ψ = Ψ orD(ρ)Ψ = −Ψ are physical. Let us analyze these states as representationsof the group S2.

Once constructed the operators P± = (I±D(ρ))/2, where I is the identityon H⊗H

a) show that P2± = P±, P+ + P− = I and that D(ρ)P± = ±P±;

b) find the invariant subspaces under the action of D(ρ);

c) if H is finite-dimensional, compute the dimension of P±(H⊗H).

6. Consider a particle in the potential well

V(x, y) =

0 if x ∈ [−1, 1] and y ∈ [−1, 1],∞ for other values of x, y.

The Hamiltonian is

H =1

2m

[p2

x + p2y + V(x, y)

].

The group D4 acts on the wavefunction Ψ(x, y) as (U(g)Ψ)(x, y) ≡ Ψ(D(g)(x, y)),where g ∈ D4 and D(g) is the 2-dimensional representation of D4.

a) Show that H commutes with the representation U.

b) For n positive integer and kn = nπ/2 define the functions

fn(z) =

cos(knz) if n is odd,sin(knz) if n is even.

Show that Ψn,m(x, y) = fn(x) fm(y) inside the square (and zero out-side) is an eigenfunction of H and compute its eigenvalue. Discusstheir degeneracy.

60 representations

c) Compute the action of U on the eigenfunctions Ψ2,1 and Ψ1,2. Dotheir linear combinations span an invariant subspace? If so, is theinduced representation on this subspace irreducible?

d) Compute the action of U on Ψ1,1. Is this generating an invariantsubspace? If so, is the induced representation on this subspace irre-ducible?

6P R O P E RT I E S O F I R R E D U C I B L E R E P R E S E N TAT I O N S

Reducible representations of a group can be put in a block diagonal form bymeans of similarity transformations and in each block one has irreducible rep-resentations of the same group. We can then use irreducible representations asthe fundamental building blocks of more general representations. It is thereforeimportant to find criteria telling us when a given representation is reducible ornot and how to compare irreducible representations. We will see various waysto do so, but all of them are based on a fundamental theorem that we are goingto discuss in this lecture: the Great Orthogonality Theorem.

In order to prove this theorem we need first a few technical results collectedin Schur’s lemmas and their corollaries. The idea behind these lemmas is verysimple: a matrix commuting with all those of a given irreducible representationof a group must be proportional to the identity and irreducible representationsof different dimensions cannot be connected by a non-trivial similarity trans-formation.

Theorem 6.1. Schur’s lemma: Given D1 and D2 two irreducible representationsof G (D1 : G × V1 → V1 and D2 : G × V2 → V2), if T is a linear map (a matrix)T : V2 → V1 such that

D1(g)T = TD2(g), ∀g ∈ G,

then we have one of these two cases:

1. T = 0;

2. T is a bijection and D1 and D2 are equivalent.

Proof. We first show that ker T is an invariant subspace of V2, namely thatD2(ker T) ⊆ ker T. A vector v is in the kernel of the map T, v ∈ ker T, iff

61

62 properties of irreducible representations

Tv = 0. If we act on a null vector with the irreducible representation D1 we arebound to obtain again a null vector. However, by using the main assumptionof the theorem, we see that

0 = D1(Tv) = T(D2v)

and therefore D2v ∈ ker T. Since D2 is an irreducible representation we con-clude that either ker T = V2 or that ker T = 0. If ker T = V2, T maps everyelement of V2 to zero and therefore T = 0, closing our proof. Otherwise, weproved that the map T is injective. We can now prove that it is bijective byshowing that im T ⊆ V1 is an invariant subspace for D1. Take a vector w in theimage of the map T: w ∈ im T ⇔ w = Tv. Using again the main assumption ofthe theorem, we see that the action of D1 on this vector gives again a vector inthe image of T:

D1w = D1(Tv) = T(D2v) ∈ im T.

Since also D1 is irreducible we conclude that either im T = V1 or im T = 0.If im T = 0 we would have again that T = 0, but we already excluded that,because we assumed that the map was injective. We therefore conclude thatim T = V1. This means that our map T satisfies both im T = V1 and ker T = 0,from which we conclude that dim V2 = dim V1 and that T is an invertiblesquare matrix. Summarizing, if V1 6= V2 then T = 0. If V1 ' V2, then eitherT = 0 or D1 and D2 are equivalent and T is the similarity transformation.

A direct consequence of this theorem is the following corollary:

Theorem 6.2. An invertible matrix T commuting with all the matrices of an irre-ducible representation is proportional to the identity:

[D(g), T] = 0, ∀g ∈ G ⇒ T = c 1.

Proof. This is just a special case of the previous theorem where D = D1 = D2and we have TD = DT, for T = T − c 1 (if a matrix T commutes with D alsoT commutes with it). In this case, however, we just need to show that ker Tis an invariant subspace for D. This follows in the same way as before usingthe assumption: 0 = D(Tv) = T(Dv). The consequence is once again thateither ker T = 0 or ker T = V. Now, however, T is invertible and thereforeker T 6= 0 because there is at least one eigenvector of T with eigenvalue λ,which means that Tv = λv for this vector. Choosing c = λ, we conclude thatTv = 0. Hence ker T = V and then T = T − c 1 = 0.

6.1 the great orthogonality theorem 63

We also have another corollary:

Theorem 6.3. Irreducible representations on C for an Abelian group are 1-dimensional.

Proof. If G is Abelian we must have that [D(g), D(g′)] = 0, ∀g, g′ ∈ G. Thisimplies that D(g) = cg 1, again ∀g ∈ G, but this would be reducible, unless itis 1-dimensional.

Let us stress that this corollary uses the fact that the vector space on whichthe representation acts is V = CN . As we have seen before we can have irre-ducible 2-dimensional representations for SO(2) over the reals.

6.1 the great orthogonality theorem

Now that we established these technical results, we can finally prove the greatorthogonality theorem between matrix elements.

In order to introduce an orthogonality criterion, we need first to introducean inner product between matrix elements. This is easily identified if we real-ize that matrix elements can be viewed as complex functions defined over thegroup G:

Dij : G → C.

They are therefore part of the vector space C[G] of complex functions definedover G and we can use as inner product the ordinary inner product for a com-plex vector space. Note that this vector space has complex dimension equal tothe order of the group, because C[G] ' C#G, where each factor C is the set ofall possible values on which one can map each element g ∈ G. We can thendefine a scalar product on this space by summing the values of the functionsf , h ∈ C[G] (vectors in this space) over their components, labelled by the groupelements:

( f , h) ≡ 1#G ∑

g∈Gf (g)h(g). (6.1)

Using this scalar product we can finally introduce the concept of orthogonalitybetween matrix elements.

Theorem 6.4. Great Orthogonality Theorem. Given D1 and D2 irreducible repre-sentations of dimensions N1 and N2 of a finite group G. We have that either D1 /'D2and then

(D1ij, D2

kl) = 0,


or D1 ' D2 (which means that N = N1 = N1 and that there is a basis whereD = D1 = D2) and then

(Dij, Dkl) =1N

δikδjl .

Proof. The proof is tedious, but straightforward. We apply Schur’s lemmas,building an appropriate map interpolating between D1 and D2. We start bydefining a basis for the N2 × N1 matrices:

E(kl)ij ≡ δk

i δlj ,

where kl label different matrices and ij are the row and column indices. Wethen build T such that D1T = TD2. There are actually N1N2 matrices that playthis role and they are given by

T(E(kl)) = ∑g∈G

D1(g)E(kl)D†2(g). (6.2)

We can indeed check that

D1(h)T(E(kl)) = ∑g D1(h g)E(kl)D2(g−1) = ∑g D1(g)E(kl)D2(g−1 h)

= ∑g D1(g)E(kl)D2(g−1)D2(h) = T(E(kl))D2(h),(6.3)

where we used the rearrangement theorem (g = h g) as well as the fact that fora finite group all representations are equivalent to unitary ones (which meansthat we can assume D2(g−1) = D−1

2 (g) = D†2(g)). Because of Schur’s lemmas

we conclude that if the two representations have different dimensions N1 6= N2the interpolating matrix T must vanish, which means that

Tij(E(kl)) = ∑g

D1im(g)δk

mδlnD2

jn(g) = #G (D2jl , D1

ik) = 0. (6.4)

If, on the other hand, N1 = N2 = N, then we have that either the two represen-tations are inequivalent, and then again T = 0, or they are equivalent D1 ' D2,in which case we can assume they coincide. If D1 = D2 = D then we can repeatthe same argument, using the corollary of Schur’s lemmas to argue that

T(E(kl)) = c(kl) 1. (6.5)

This means that Tij(E(kl)) = c(kl) δij, whose trace determines the constants c(kl):

Tii(E(kl)) = Nc(kl). (6.6)

6.1 the great orthogonality theorem 65

For a finite group we can always choose for D a unitary representation (D†D =1) and therefore

Tii(E(kl)) = ∑g

Dim(g)δkmδl

nD†ni(g) = ∑

gδkl = #G δlk. (6.7)

We are then able to completely fix the proportionality coefficient

c(kl) =#GN

δkl (6.8)

and close our proof.

We can also combine the two conditions in a unique condition on two genericirreducible representations a and b saying that

(Daij, Db

kl) =δab

Nδikδjl . (6.9)

From this theorem we also get an important limit on the number of inequiv-alent irreducible representations for a finite group. In fact we said we can con-sider the matrix elements of a given representation as elements in the vectorspace C[G]. Since dim C[G] = #G we can have at most #G mutually orthogonalvectors in this space. On the other hand, for each representation a of dimensionNa we can have at most N2

a different matrix elements, from which we deducethat the total of orthogonal elements for a given representation is N2

a . Summingover al irreducible representations we obtain the inequality

∑a

N2a ≤ #G. (6.10)

We will actually see in the next lecture that such inequality should be an exactequality.

exercises

1. Show that the matrices

D(1) = D(a) = D(b) = 12,

D(c) = D(d) = D(e) =12

(−1 −

√3

−√

3 1

),


provide a representation for the 6 group elements of D3. Use Schur’slemma to determine if this representation is reducible or not and if re-ducible show which irreducible representations compose the one given.

2. Verify the Great Orthogonality Theorem for the following irreducible rep-resentation of S3:

D(ρ) =12

(−1

√3

−√

3 −1

), D(σ) =

(−1 00 1

).

3. Verify the Great Orthogonality Theorem for the following representationof Z3:

D(ρ) =12

(−1

√3

−√

3 −1

).

7P R O P E RT I E S O F I R R E D U C I B L E R E P R E S E N TAT I O N S

We saw that the Great Orthogonality Theorem (GOT) gives us a chance todistinguish inequivalent representations, by providing a statement about theorthogonality of matrix elements corresponding to different irreducible repre-sentations of a given group. Unfortunately, the procedure needed to prove thattwo representations are inequivalent by means of the GOT can be rather te-dious. In this lecture we are going to see how we can obtain analogous results(derived from the GOT) by introducing the concept of character for a representa-tion. Characters, as we will see, allow us to obtain a lot of useful information onthe irreducible representations of a given group as well as on their properties.

The character of a representation is a particular function in C[G], which hasdifferent values only on certain classes of elements of G. For this reason, in or-der to introduce it in an efficient way, we first need to consider further relationsbetween the elements of G.

7.1 conjugacy classes

Recall that if g1 = gg2g−1 for some g ∈ G, then g1 and g2 are conjugate. Alsorecall that conjugation is an equivalence relation. We can therefore split a givengroup in disjoint subsets defined by equivalence classes under conjugation asfollows:

Definition 7.1. All the elements of a group G that are conjugate to g ∈ G definethe class of conjugation of g.

We can immediately see that the identity is always alone in its conjugacyclass:

1 = g 1 g−1, ∀g ∈ G. (7.1)

67


We can also see that all the equivalence classes in an Abelian group have asingle element:

g g g−1 = g g g−1 = g, ∀g, g ∈ G. (7.2)

This implies that an Abelian group has #G conjugacy classes.For a generic non-Abelian group the number of conjugacy classes is less than

the number of elements and their union will give back the full set of groupelements.

Example 7.1. As an example we discuss here the conjugacy classes of D3 ' S3.One equivalence class is given by the identity, as expected: C1 = 1. The otherelements are either of the form ρa, for a = 1, 2, with inverse ρ−a, or of the formσρb, where b = 0, 1, 2, whose inverse is again σρb (in fact σρbσρb = σσρ−bρa =1). The conjugacy class of ρ is then determined by

ρaρρ−a = ρ, σρaρaσρa = σ2ρ−a−1ρa = ρ−1 = ρ2. (7.3)

Since conjugation is an equivalence relation, if ρ2 is conjugate to ρ, we alsohave that ρ is conjugate to ρ2. The conjugacy class of σρ on the other hand isdetermined by

ρaσρρ−a = σρ1−2a, σρaσρσρa = σρ2a−1. (7.4)

The group elements split therefore into three different conjugacy classes

C1 = 1, C2 = ρ, ρ2, C3 = σ, σρ, σρ2. (7.5)

Once we group the elements of G in conjugacy classes we can study themcollectively. In particular we can study the functions defined on these classes.

Definition 7.2. A function that is constant on each conjugacy class is calledcentral:

f (g) = f (ggg−1), ∀g ∈ G.

As we will now see, the character of a representation is a special centralfunction.

7.2 characters and their properties

Definition 7.3. The character of a representation is the function χD : G → C

defined asχD(g) = tr D(g), for g ∈ G.

7.2 characters and their properties 69

We can see immediately that the character is a central function:

χD(ggg−1) = tr(D(ggg−1)) = tr(

D(g)D(g)D(g−1))

= tr(

D(g)D(g)[D(g)]−1) = trD(g) = χD(g),(7.6)

where we used the cyclic property of the trace. Because of this fact we can seethat the value of χD is the same on all the elements of a conjugacy class. Usingthe cyclic property of the trace we can also conclude that all equivalent repre-sentations have the same character. If two representations are equivalent thenD1 = SD2S−1 and therefore χD1(g) = trD1(g) = tr(SD2(g)S−1) = tr(D2(g)) =χD2(g).

As we did for the single matrix elements of a representation, we can intro-duce the concept of orthogonal characters, as follows from their interpretationas elements of the vector space C[G], where we introduced the inner product(6.1). Actually, it is an immediate consequence of the GOT that

Theorem 7.1. Characters associated to inequivalent irreducible representations areorthogonal:

(χDa , χDb) = δab.

Proof. It follows straightforwardly from the definition:

(Daii, Db

kk) =δab

Nδikδik = δab δii

N= δab. (7.7)

However we can say more. For finite groups all decomposable represen-tations are also reducible. This means that they are always direct sums ofirreducible representations. We then see that

Theorem 7.2. Representations with the same set of characters are equivalent.

Proof. If D is reducible then D = ∑a caDa = Da ⊕ Da ⊕ . . .⊕ Da︸︷︷︸ca

⊕ . . . From this

we get that

(χD, χa) = ca (7.8)

and hence D is fully determined by χD.


Since we just saw that characters fully specify a representation we can limitour study to the irreducible ones. The first thing we should determine thenis how many inequivalent irreducible representations we can construct for agiven group. This is determined by a theorem in the following way:

Theorem 7.3. The number of inequivalent irreducible representations of a finite groupis equal to the number of conjugacy classes.

Proof. The proof has three steps.

First we show that matrix elements are a good basis for C[G].

Given a representation Da, the matrix elements Daij are orthogonal among

them and they are elements of C[G]. In order to form a basis of C[G], we mustshow that they span C[G], i.e. that any arbitrary function f ∈ C[G] can beexpanded in terms of matrix elements of irreducible representations of G. Forany given g ∈ G we can write

f (g) = fg = ∑g′

fg′ δg′g = ∑g′

fg′DRg′1(g), (7.9)

where the first equality is really just a definition of an element of C#G andthe last one follows from the definition of regular representation. The regularrepresentation is reducible and using previous results we can always writeDR = ∑a caDa, which implies

f (g) = ∑a,i,j

f aijD

aij(g). (7.10)

Hence any element of C[G] can be written by using Daij as a basis.

The second step is the proof that if f ∈ C[G] is a central function then wecan expand it using characters as a basis:

f (g) = ∑ fa ij Daij(g) = f (ggg−1) = ∑ fa ij Da

ij(ggg−1)

= ∑ fa ij Daik(g)Da

km(g)Damj(g−1)

= ∑ fa ij Daik(g)Da

km(g)Dajm(g)

(7.11)

7.3 character tables 71

and being central f (g) = f (ggg−1) = 1#G ∑g f (ggg−1), which means that

f (g) = 1#G ∑g ∑a fa ijDa

ik(g)Dajm(g)Da

km(g)

= ∑a fa ij (Dajm, Da

ik)Dakm(g)

= ∑a fa ijδijδkm

NaDa

km(g)

= ∑afa iiNa

χDa(g).

(7.12)

Hence characters form a basis for central functions.The third step is simply a correspondence between characters and central

functions. Since inequivalent irreducible representations have orthogonal char-acters, they are in 1 to 1 correspondence with the basis of characters, whichare also a basis for central functions. This also means that characters are in 1

to 1 correspondence with conjugacy classes and that in turn these are in 1 to 1

correspondence with irreducible representations.

7.3 character tables

As we said before, characters are central functions and therefore they have thesame value on each element of the same conjugacy class. This means that whenwe evaluate the inner product between two characters we can actually restrictthe sum to elements of different conjugacy classes:

(χa, χb) =1

#G ∑c

nc χa(gc)χb(gc) = δab, (7.13)

where c runs on the different conjugacy classes and nc is the number of ele-ments in the c conjugacy class. Using this rewriting we can now produce someinteresting definitions and relations that will completely specify irreducible rep-resentations of a given group. The first thing we can introduce is the matrix ofcharacters. For any group we have the same number of conjugacy classes andirreducible representations and this means that to fully specify all irreduciblerepresentations of a given group we need a square matrix encoding the valueof the character of a given irreducible representation on an element of a givenconjugacy class, for each one of them.

Definition 7.4. The matrix of characters of a representation is

Uab ≡√

na

#Gχb(ga).


This matrix has some very interesting properties that restrict its possibleentries.

Theorem 7.4. The character matrix is unitary

Proof.

U†abUbc = UbaUbc = ∑

b

(√nb#G

)2

χa(gb)χc(gb) = δac. (7.14)

This property means that different rows or different columns are orthogonaland therefore we can fix their entries. Using it in the opposite way we can alsosay more and actually prove a theorem that allows us to fix the number anddimensions of the irreducible representations of a finite group.

Theorem 7.5.

∑a

N2a = #G.

Proof. We use unitarity of the character matrix, only in a way different from(7.14):

UabU†bc = UabUcb = ∑

b

√nanc

#Gχb(gc)χb(ga) = δac. (7.15)

If we specify this formula for ga = gc = 1, we can also conclude that

∑b

χb(1)χb(1) = #G, (7.16)

where we used that na = nc = 1, and since now χa(1) = tr(1Na) = Na weconclude.

Another noteworthy relation follows from what we just proved and tells uswhen a representation is irreducible:

Theorem 7.6. If ∑c ncχD(gc)χD(gc) = #G then D is irreducible.

Proof. It follows from the orthogonality relation given above UabU†bc, for the

choice gc = ga.


Often we do not want to keep track of the normalization factors in definingthe unitary character matrix and therefore we introduce

Definition 7.5. The character table is the matrix collecting all the values of thecharacters for each irreducible representations and for every conjugacy class.

We conclude this lecture by computing the character table for a couple ofexamples.

Example 7.2. The first example is given by the group Z3.Each Zn is Abelian and therefore we have n conjugacy classes and n irre-

ducible representations. As we already discussed, all these representationsmust be 1-dimensional and in fact they fulfill the condition ∑n

a=1 N2a = n = 12 +

. . . + 12. All these 1-dimensional representations can be defined by ωk = e2πin k,

where k = 1, . . . , n.For Z3 the conjugacy classes are then C1 = 1, C2 = ρ, C3 = ρ2 and

the representations are generated by D1(ρ) = 1, D1′(ρ) = e2πi/3 = ω andD1”(ρ) = ω2. We eventually deduce the character table, which is

C1 C2 C31 1 1 11′ 1 ω ω2

1” 1 ω2 ω

(7.17)

Example 7.3. We now discuss D3 ' S3.We already know that the conjugacy classes are C1 = 1, C2 = ρ, ρ2 and

C3 = σ, σρ, σρ2. Since irreducible representations must be in 1 to 1 corre-spondence with the conjugacy classes, we conclude that we have 3 irreduciblerepresentations and from theorem 7.5 we can deduce that their dimension is 1,1 and 2. In fact

#G = 6 = 12 + 12 + 22. (7.18)

The characters of the trivial representation are 1 and that of the dimension 2

representation can be computed from their explicit expression, we gave previ-ously. The rest follows by orthogonality

C1 C2 C31 1 1 11′ 1 1 −12 2 −1 0

(7.19)


exercises

1. Consider the group of geometric symmetries of a square (D4). Split itselements in conjugacy classes, find all its subgroups and discuss whichones are normal. Compute the quotient of D4 with its normal subgroupsand identify the resulting group.

2. Find all irreducible representations of D4.

3. Take the representations of D4 generated by

D1(ρ) =

(0 1−1 0

), D1(σ) =

(1 00 −1

),

D2(ρ) =

(83 130−53 −83

), D2(σ) =

(43 66−28 −43

).

Using characters, discuss if they are equivalent.

4. Prove that given two representations D1 and D2, χD1⊗D2 = χD1 · χD2 .

5. Prove that for unitary representations χD(g−1) = (χD(g))∗.

6. Consider once more the group of quaternions Q.

a) Provide a matrix representation for each of its irreducible represen-tations.

Hint: note that Q can be constructed as a discrete subset of the SU(2)

matrices U(θ, α, β) =

(cos θ eiα sin θ eiβ

− sin θ e−iβ cos θ e−iα

)b) Give its character table.

7. The dicyclic group Γ(Q6) is defined by Γ(Q6) = 〈a, b|a6 = 1, b2 = a3, aba =b〉. Its conjugacy classes are

C1 = 1, C2 = a, a5, C3 = a2, a4,

C4 = a3, C5 = b, a2b, a4b, C6 = ab, a3b, a5b.

a) Are the following subgroups normal?

Z2 = 1, a3, Z3 = 1, a2, a4, Z4 = 1, b, b2, b3.

Hint: use the information contained in the conjugacy classes.


b) Compute Γ(Q6)/Z3, where Z3 is the one considered in the previousquestion.

c) Discuss which of the following representations are equivalent:

D1(a) =(

eiπ/3 00 e−iπ/3

), D1(b) =

(0 1−1 0

),

D2(a) =(

ei2π/3 00 e−i2π/3

), D2(b) =

(0 11 0

),

D3(a) =(

e−i2π/3 00 e2iπ/3

), D3(b) =

(0 −1−1 0

),

D4(a) =(

cos π3 i sin π

3i sin π

3 cos π3

), D4(b) =

(0 −11 0

).

8I R R E D U C I B L E R E P R E S E N TAT I O N S O F S n

All finite groups are subgroups of some group of permutations Sn. Also, ir-reducible representations of any H < G can be obtained by restricting therepresentations of G to the elements of H. Combining these two facts we seethat we can obtain the representations of any finite group by studying andbuilding the irreducible representations of an appropriate permutation group.It is therefore extremely useful and interesting to try to provide a classificationof the irreducible representations of Sn.

In order to do so, we will use a graphical description that is very efficient:Young Tableaux. This type of diagrams not only is useful to discuss and classifyboth conjugacy classes as well as irreducible representations of permutationgroups, but, as we will see in a subsequent lecture, this graphical descriptionwill also be very useful for the construction and classification of representationsof continuous groups.

Definition 8.1. A standard Young Tableau of order k is a diagram with k boxes,where the (j + 1)th row contains a number of boxes that is less or equal to thatof the jth row and the (j + 1)th column contains a number of boxes that is lessor equal to that of the jth column.

For instance, good Young Tableaux are

, , ,

while the following are not good Young Tableaux:

, , , .

77

78 irreducible representations of sn

8.1 young tableaux and conjugacy classes

In this first part we discuss the relation between Young Tableaux and conjugacyclasses. We will see that there is a 1-to-1 correspondence between regular YoungTableaux and conjugacy classes. In order to establish such a correspondence wefirst need to classify the conjugacy classes of Sn.

Recall that any permutation associated to an irreducible j-cycle can be writtenas a product of 2-cycles. For instance

(12 . . . m) = (1 m)(1 m− 1) . . . (13)(12).

We also established that the inverse of a given j-cycle is a j-cycle with the sameelements written in the opposite order, i.e.

(12 . . . m)−1 = (1 m . . . 2),

and this in turn can be written as the product of the same 2-cycles definingthe original j-cycle, but in the opposite order, (1m . . . 2) = (12)(13) . . . (1 m −1)(1 m). This implies that we can constrain our analysis of the conjugacy classesof Sn using 2-cycles.

It is easy to see that the conjugates of a j-cycle by 2-cycles is again a j-cycle,where the elements of the 2-cycle in the j-cycle are swapped:

(12) · (a1b . . . c2d . . .) · (12) = (a2b . . . c1d . . .). (8.1)

An analogous demonstration works for composite cycles (like, for instance,(12)(345) ∈ S5).

Since conjugation by 2-cycles can change any element in any cycle, we seethat all the cycles with the same length must be in the same conjugacy class.We can then put all conjugacy classes in 1-to-1 correspondence with YoungTableaux where each column represents one of the cycles defining the conju-gacy class. For instance, the cycle (1352)(4) in S5 is composed by a 4-cycle anda 1-cycle. The corresponding representation via a Young Tableaux is therefore

. The same thing is true for the cycle (361)(24)(5) ∈ S6, which is an element

in the conjugacy class represented by .

Example 8.1. Conjugacy classes of S3. We already know that we have 3 conjugacyclasses C1 = 1 = (1)(2)(3), C2 = p1 = (123), p2 = (132) and C3 = d1 =(1)(23), d2 = (2)(13), d3 = (3)(12). As expected, they can be distinguished

8.1 young tableaux and conjugacy classes 79

by the length of the cycles defining them. In terms of Young Tableaux we canrepresent them as

C1 ∼ , C2 ∼ , C3 ∼ .

From this analysis we can deduce a general lesson: we can find all the conju-gacy classes of Sn by considering all Young Tableaux with n boxes arranged inall possible admissible ways. We also learn that the number of classes dependson the number of ways we can partition n in terms of integer numbers. Forinstance, for S3, we split n = 3 in 3 possible ways

3 = 1 + 1 + 1 = 2 + 1 = 3, (8.2)

corresponding to its three conjugacy classes and to its 3 inequivalent YoungTableaux.

The shape of the tableau also tells us how many elements are present ineach conjugacy class. This is determined by the number of different waysone can arrange the n elements one is permuting in the boxes of the tableau.For instance, the Young Tableau corresponds to a triple 1-cycle and since(1)(2)(3) = (1)(3)(2) = (2)(1)(3) = . . . = (3)(2)(1) we see that there is onlyone element in this class:

1 2 3 = 1 3 2 = 2 1 3 = . . . = 3 2 1 .

The Young Tableau on the other hand has three different elements in itsclass

1 32

= (12)(3), 2 31

= (23)(1), 1 32

= (13)(2). (8.3)

We see then that computing the number of elements in the conjugacy classdefined by the Young Tableau under consideration follows by a combinatorialanalysis. In detail, if we have k j j-cycles, the corresponding Tableau represents

nr. of permutations =n!

∏nj=1 jkj k j!

. (8.4)

Example 8.2. The Tableau has n = 3, k1 = 3, k2 = k3 = 0 and hence it

contains 1 element. The Tableau has n = 3, k1 = k2 = 1 and k3 = 0, hence it

contains 3 elements. The Tableau has n = 3, k1 = k2 = 0 and k3 = 1, hence itcontains 2 elements.


8.2 young tableaux and irreducible representations of Sn

We established that Young Tableaux are in 1-to-1 correspondence with conju-gacy classes of the permutation group. On the other hand, we also know thatirreducible representations are in 1 to 1 correspondence with conjugacy classes.This means that we can use Young Tableaux of rank n to represent irreduciblerepresentations of the group of permutations Sn, too.

Each Young Tableau of rank k is associated to an irreducible representationof dimension

d =n!

∏ki=1 hi

, (8.5)

where hi is the hook number associated to each single box in the diagram. Thehook number is the sum of all the boxes to the right and below a given boxplus one (see the picture 7 for an example).

1

4

5

9 6 3 1

7 4 1

5 2

4 1

2

1

Figure 7.: Example of hook number computation.

The proof of this formula is very long, tedious and not particularly illumi-nating. For this reason, we will not see here a full proof of this formula, butwe will discuss a significative example that is also useful to understand how toconstruct all the irreducible representations of a given permutation group.

The starting point is given by the regular representation of Sn, which isreducible and decomposes in the sum of all the irreducible representationsof a given permutation group. For the sake of simplicity we will focus ouranalysis to the group S3, but we will mention the consequences of consider-ing general n whenever possible. The regular representation of S3 is a map

8.2 young tableaux and irreducible representations of Sn 81

DR : S3 ×C6 → C6, where the natural basis for the vector space C6 is given byvectors associated to the elements of S3:

|g〉 = |1〉, |(12)〉, |(13)〉, |(23)〉, |(123)〉, |(132)〉 (8.6)

and the explicit action is DR(g)|g〉 = |g g〉. We can then decompose thematrices DR(g) by looking at its action on the subspaces associated to eachYoung Tableau as follows from the theorem:

Theorem 8.1. Given a Young Tableau T, the vectors |v〉 ≡ aTsT |1〉 generate invariantsubspaces under the action of the regular representation.

Here sT and aT are the operators that symmetrize all the elements in the rowsof a given Young Tableau and anti-symmetrize the elements in the columns asfollows. Given a Tableau T, we can fill its boxes with numbers from 1 to n in allpossible ways and define HT , which contains all permutations of the numbersin the rows, and VT , which contains all the permutations of the numbers in thecolumns. The symmetrizing and anti-symmetrizing operators are then definedby

sT ≡ ∑σ∈HT

DR(σ), aT = ∑σ∈VT

π(σ)DR(σ), (8.7)

where π is the parity of the element σ ∈ Sn, defined in Def. 3.13. For in-

stance, the tableau1 32 corresponds to the operators sT = DR(1) + DR(13),

aT = DR(1)− DR(12).

We will now explicitly show that for S3 the regular representation admitsinvariant subspaces associated to the vector spaces V , V and V , defined

by the corresponding Young Tableaux.Take first V . We can fill the corresponding diagram in 3! different ways:

1 2 3 , 1 3 2 , 2 1 3 , 2 3 1 , 3 1 2 , 3 2 1 (8.8)

and we can see that they are related by all the permutations in S3. Conse-quently sT is given by the sum of all the elements in S3 and aT = DR(1). Thecorresponding vector, generating V , is

|v 〉 = DR(1)[DR(1) + DR(12) + DR(13) + DR(23)

+DR(123) + DR(132)]|1〉

= |1〉+ |(12)〉+ |(13)〉+ |(23)〉+ |(123)〉+ |(132)〉.

(8.9)


This means that V is a 1-dimensional subspace and, since it is generatedby the sum of all the vectors corresponding to the different elements of S3,the action of the regular representation on |v 〉 is trivial: each element ismapped to a different element in S3 and the sum of all the elements is mappedto itself, because of the rearrangement theorem

DR(g)|v 〉 = |v 〉, ∀g ∈ S3. (8.10)

The second invariant subspace V is again 1-dimensional and is spanned by

the vector

|v 〉 =[DR(1)− DR(12)− DR(13)− DR(23)

+DR(123) + DR(132)]

DR(1)|1〉

= |1〉 − |(12)〉 − |(13)〉 − |(23)〉+ |(123)〉+ |(132)〉.

(8.11)

In this case the regular representation reduces to a 1-dimensional representa-tion determined by the parity of the element which is representing. For in-stance:

DR(12)|v 〉 = |(12)〉− |1〉− |(123)〉− |(132)〉+ |(23)〉+ |(13)〉 = −|v 〉. (8.12)

We are left with the space V . In this case sT and aT are not uniquely

determined, but they depend on the arrangement of the elements of Sn in the

tableau. For instance, for1 23 we have sT = DR(1) + DR(12) and aT = DR(1)−

DR(13), while for1 32 we have sT = DR(1) + DR(13) and aT = DR(1)−DR(12).

The corresponding vectors are

|v 1 23〉 = |1〉+ |12〉 − |13〉 − |123〉, (8.13)

|v 1 32〉 = |1〉+ |13〉 − |12〉 − |132〉. (8.14)


These vectors however are not invariant under the action of the elements of S3,but rather generate two orthogonal invariant subspaces of dimension 2. In fact,the action of the S3 elements on the first gives

DR(1)|v 1 23〉 = |v 1 2

3〉,

DR(12)|v 1 23〉 = |(12)〉+ |1〉 − |(132)〉 − |(23)〉 ≡ |w〉,

DR(13)|v 1 23〉 = |(13)〉+ |(123)〉 − |1〉 − |(12)〉 = −|v 1 2

3, 〉 (8.15)

DR(23)|v 1 23〉 = |(23)〉+ |(132)〉 − |(123)〉 − |(13)〉 = −|w〉+ |v 1 2

3〉,

DR(123)|v 1 23〉 = |(123)〉+ |(13)〉 − |(23)〉 − |(132)〉 = |w〉 − |v 1 2

3〉,

DR(132)|v 1 23〉 = |(132)〉+ |(23)〉 − |(12)〉 − |1〉 = −|w〉.

By consistency, the application of the same representation matrices to |w〉 re-mains also in the same 2-dimensional subspace, generated by |w〉 and |v 1 2

3〉

and hence, we can reconstruct the 2-dimensional representation of S3 by look-ing at DR restricted to this space.

The same computation on the second element gives, on the other hand:

DR(1)|v 1 32〉 = |v 1 3

2〉,

DR(12)|v 1 32〉 = |(12)〉+ |(132)〉 − |1〉 − |(13)〉 = −|v 1 3

2〉,

DR(13)|v 1 32〉 = |(13)〉+ |1〉 − |(123)〉 − |(23)〉 ≡ |u〉, (8.16)

DR(23)|v 1 32〉 = |(23)〉+ |(123)〉 − |(132)〉 − |(12)〉 = −|u〉+ |v 1 3

2〉,

DR(123)|v 1 32〉 = |(123)〉+ |(23)〉 − |(13)〉 − |1〉 = −|u〉,

DR(132)|v 1 32〉 = |(132)〉+ |(12)〉 − |(23)〉 − |(123)〉 = −|v 1 3

2〉+ |u〉.


Also in this case, by consistency, the application of DR to |u〉 remains in thesame 2-dimensional subspace and its restriction to this subspace gives a 2-dimensional irreducible representation that is equivalent to the previous one.

We conclude that we have two orthogonal invariant subspaces of C6 corre-sponding to the two Young Tableaux we started from and by a more accuratecomputation we can also check that the vectors |w〉 and |u〉 can be associatedto linear combinations of vectors corresponding to other Young Tableaux of thesame type. We have then determined 4 invariant subspaces of dimensions 1, 1,2 and 2 and therefore we exhausted the rewriting of the original basis of vectorsgenerating C6 in terms of new ones associated to the same invariant subspaces.We conclude that

DR = D1 + D1′ + D2 + D2, (8.17)

where the 2-dimensional representations are obviously equivalent, because weknow that there is only one such irreducible representation for S3. This isactually a result that applies in general for the decomposition of the regularrepresentation of Sn, where each irreducible representation appears a numberof times equal to its dimension. The important lesson we learn from this is that

among all possible arrangements in the tableaux , only 2 of them generateorthogonal subspaces and these are in correspondence with the tableaux wherethe numbers are always growing in any row and column of the diagram. Byrepeating this decomposition for a generic Sn we can then deduce the generalformula for the dimension of its irreducible representations, but also constructthem explicitly.

The general formula for the dimension of the irreducible representations ofSn associated to a given Young Tableau can also be obtained by induction, con-sidering the Tableaux of Sn−1 < Sn. In fact, each Young Tableau of Sn containsonly a subset of the diagrams associated to Sn−1 and hence its dimension isthe sum of the dimensions of the corresponding diagrams. In this way we candetermine the resulting dimension by an iterative procedure, like in figure 8.


Figure 8.: Dimension of Sn representations determined by induction from the dimen-sion of the representations of Sn−1.

exercises

1. Find the character table of S4. To do so, we can use that the charac-ters of the symmetric group are always integers and that the characterof the product of two representations is the product of the characters,i.e. χDa⊗Db = χa · χb.

a) Draw all Young Tableaux with 4 boxes and then determine the con-jugacy classes, the number of their elements and the dimension ofthe corresponding irreducible representations.

b) Fill the column related to χ(1) and the raw associated to the trivialrepresentation.

c) By using the results of this chapter, fill the raw χ1A , where 1A refersto the 1-dimensional alternating representation defined by the parityof the elements of S4.

d) Using χDa⊗Db = χa · χb deduce that 2 ⊗ 1A ' 2. What does thisimply for the conjugacy classes containing elements constructed byodd numbers of 2-cycles?

e) Use orthogonality of the characters of irreducible representations tocomplete χ2.


f) Argue that for the 3-dimensional representations we must have 3′ =3⊗ 1A.

g) Complete the table using again that (χa, χb) = δab for irreduciblerepresentations.

Part III

E L E M E N T S O F T O P O L O G Y A N D D I F F E R E N T I A LG E O M E T RY

9H O M O T O P Y G R O U P S

So far we discussed general properties of groups and some of their applica-tions to finite groups. On the other hand, in physics we have to deal veryoften with symmetry transformations that are continuous (e.g. rotations) andhence we must expand our discussion to include symmetry groups with aninfinite number of elements. In order to do so, we need to introduce and re-call some concepts in topology and differential geometry that are relevant forunderstanding their properties. In fact, continuous groups are also topological(and differential) spaces where each point is identified with an element of thegroup.

The first important task is to classify and distinguish them according to theirtopological and differential properties. In this lecture we will discuss one partic-ular topological property that allows to separate spaces (and hence continuousgroups) in equivalence classes: homotopy. Actually, homotopy is not only rel-evant in the context of group theory, but it has more interesting applicationsboth in condensed matter physics (where it allows to distinguish materials interms of their defects) and in theoretical high energy physics (both in gauge the-ories and in higher dimensional field theories). For this reason we will give ageneral introduction to this concept, with some final remarks on its applicationto group theory.

First we need to introduce maps between topological spaces:

Definition 9.1. Let X and Y be two topological spaces. The map φ : X → Y is ahomeomorphism if φ is a continuous bijection and φ−1 is continuous. Two spacesrelated by such a map are homeomorphic to each other.

The existence of homeomorphic maps between two spaces implies that wecan continuously deform them into each other. In this sense we can consider

89

90 homotopy groups

them to be topologically equivalent and they will be characterized by the sametopological invariants (like the number of connected components, ...). We canactually group all spaces with the same topological invariants in equivalenceclasses, whose equivalence relations are homeomorphisms. A complete charac-terization in terms of invariants is rather difficult, but one important thing weknow is that

Theorem 9.1. Topological spaces having different topological invariants are not home-omorphic.

9.1 homotopy

One way to construct invariants is the use of homotopy. This is a weaker notionthan that coming from homeomorphisms. In fact, in the case of homotopicspaces we consider continuous functions between them that need not have aninverse.

Homotopy also creates equivalence classes of continuous maps.

Definition 9.2. Let X and Y be two topological spaces. Consider two contin-uous maps f0, f1 : X → Y, then f0 is homotopic to f1 ( f0 can be continuouslydeformed into f1):

f0 ∼ f1 (9.1)

if there is a continuous function

F : X× [0, 1]→ Y (9.2)

such that

F(x, 0) = f0(x), and F(x, 1) = f1(x). (9.3)

Definition 9.3. F is called a homotopy.

Homotopies define homotopy classes for functions, where two functions liein the same class if they are homotopic.

We can also apply the same idea to topological spaces and define homotopicequivalent spaces.

Definition 9.4. Two topological spaces X and Y are homotopic equivalent ifthere are continuous maps f : X → Y and g : Y → X such that

f g ∼ 1Y and g f ∼ 1X . (9.4)

9.1 homotopy 91

Note that we did not require that g was the inverse of the function f , butonly that their combination is homotopically equivalent to the identity.

Example 9.1. Here is an example of homotopic spaces. Let X = Rn r 0,Y = Sn−1. We can show that these two spaces are homotopic, which meansthat we can continuously deform one into the other. The proof is simple if oneconsider the maps

f : Rn r 0 3 ~x 7→ ~x|~x| ∈ Sn−1, (9.5)

g : Sn−1 3 ~n 7→ ~n ∈ Rn r 0 where |~n| = 1. (9.6)

We see that f g = 1Y, but g is not the inverse of f , because g f 6= 1X .However, we can construct a continuous map F : X × [0, 1] → X that showsthat g f is homotopically equivalent to the identity map in Y. This is built bydefining F such that F(~x, 0) = 1X(~x) = ~x, and F(~x, 1) = g f (~x) = ~x

|~x| :

F(~x, t) = (1− t)~x + t~x|~x| . (9.7)

We see then that f g ∼ 1Y.

S1

Figure 9.: The space Rn r 0 can be squeezed to a sphere Sn−1.

Example 9.2. We now discuss an interesting example of non-trivial homotopicmaps between X = S1, the unit circle described by an angle θ, and Y =U(1),represented by the unimodular complex numbers eiϕ. Consider the maps fα :X → Y defined by

fα(θ) = ei(nθ+α). (9.8)

92 homotopy groups

These maps generate homotopy classes for different α and fixed integer n. Ex-plicitly, for a fixed n, fα0 ∼ fα1 by means of the continuous family of functions

F(θ, t) = exp(i[nθ + (1− t)α0 + tα1]), (9.9)

so that F(θ, 0) = fα0(θ) and F(θ, 1) = fα1(θ). An important point, which willbecome of primary importance later, is that different integers n and m definedifferent homotopy classes:

einθ /∼eimθ , if n 6= m. (9.10)

In fact, a map generalizing (9.9) and interpolating between the two would be

exp(i n(1− t) θ + i t m θ), (9.11)

but this is not continuous around θ = 0, 2π for t ∈]0, 1[ and generic m and n.We actually have a multivalued function.

We can then differentiate equivalence classes in terms of the integer definingthe maps fα. This integer is called winding number, because we can see that itrepresents how many times the circle is covered when θ goes from 0 to 2π.

n = 0 n = 1

Figure 10.: Different homotopy classes of functions covering the circle.

We then see that we cannot unwind in a continuous fashion functions withdifferent winding numbers (see figure 15).

9.2 homotopy groups

Introducing loops, like those mentioned above, in a topological space is usefulto classify the same spaces in terms of their connectedness properties.

9.2 homotopy groups 93

Definition 9.5. A continuous map α : [0, 1] → X, where X is a topologicalspace, with α(0) = α(1) = x0 ∈ X is called a loop with base point x0.

Once again we can continuously deform loops among themselves by meansof homotopies. A homotopy between two loops α and β can be constructed bydefining

H : [0, 1]× [0, 1]→ X (9.12)

such that

H(s, 0) = α(s), H(s, 1) = β(s), ∀s ∈ [0, 1], (9.13)

H(0, t) = H(1, t) = x0, ∀t ∈ [0, 1]. (9.14)

x0

H(

s, 12

)H(

s, 14

)

Figure 11.: Example of homotopy shrinking a loop with base point x0.

Using the homotopy H we can then define equivalence classes of loops.There are two kinds of loops, those containing a hole in them, which cannot beshrunk to a point, and those that do not contain a hole and can be shrunk. Wecan then use loops to classify topological spaces:

Definition 9.6. If any loop in X can be continuously shrunk to a point then Xis simply connected.

94 homotopy groups

x0

α

β

Figure 12.: Space X with a hole. Loop α cannot be shrunk to a point. Loop β can.

The interesting point for us is that the equivalence classes defined by loophomotopies have a group structure.

Definition 9.7. The fundamental group or first homotopy group of a topologicalspace X at a point x0, Π1(X, x0) is the group of equivalence classes of homo-topic loops on X with base point x0.

We can check the group properties rather easily by composing loops havingthe same base point. In fact we define the product of loops by

β α =

α(2s), for s ∈ [0, 1/2],

β(2s− 1), for s ∈ [1/2, 1].(9.15)

This means that we first follow the loop α, from x0 to x0 and then the loop βfrom the same point to x0. The result is a special loop with basepoint x0, whichpasses in x0 for s = 1/2. The inverse is trivially defined by

α−1(s) = α(1− s), (9.16)

and the identity map is 1(s) = x0, ∀s. It must be noted that α−1 α 6= 1, butobviously it is homotopic to the identity map α−1 α ∼ 1.

It is very interesting that in the case of arcwise connected topological spaces,the fundamental group does not depend on the base point x0 anymore:

Theorem 9.2. If X is an arcwise connected topological space and x0, x1 ∈ X, then

Π1(X, x0) ' Π1(X, x1) (9.17)

and we generically talk about Π1(X).

This is understood by looking at figure 14, where the arcwise connectednessassures the existence of the maps β and β−1 connecting x0 with x1.

9.3 universal covering 95

x0

α

α−1

x0

α α−1

Figure 13.: A loop α times its inverse α−1 is homotopic to the identity.

x0 x1

αβ

β−1

Figure 14.: Loops with base-point x0 can be extended to loops with base-point x1 if thespace is arcwise connected.

Example 9.3. From our previous discussion we immediately see that Π1(U(1)) 'Π1(S1) = Z, because inequivalent loops are classified by n ∈ Z. Combiningloops in the equivalence classes n and m we obtain a loop in the equivalenceclass defined by n + m.

Example 9.4. Π1(Sn≥2) = 0. In this case all loops can be shrunk to a point.

9.3 universal covering

Whenever Π1(X) = 0 we see that X is simply connected.If, on the other hand X is not simply connected, then we can define its univer-

sal covering X, which is simply connected and which is locally homeomorphicto X. Actually, it can be proved that X is unique up to homeomorphisms.

In the special case that X = G is a group, it can also be proved that G is agroup, which is homomorphic to G (i.e. there is a map ϕ : G → G, called thecovering map, respecting the products). Also, the kernel of the homomorphismϕ is Π1(G), so that Π1(G) / G and therefore

G/Π1(G) = G. (9.18)

For instance, R is the universal covering group of U(1) ' R/Z.

96 homotopy groups

In the following lectures we will also need that Π1(SO(n)) = Z2, for n > 2.We will define then the spin group

Spin(n) ' SO(n). (9.19)

exercises

1. Consider the disk D2 = ~x ∈ R2||~x| ≤ 1 where we make two holes atpoints ~x1 and ~x2. Show that the first homotopy group Π1(D2 r ~x1r~x2) is non Abelian.

10D I F F E R E N T I A B L E M A N I F O L D S

10.1 introduction

In the rest of the course we want to analyze infinite-dimensional continuousgroups. In order to do so, we need to introduce some notions of derivability onthese spaces and for this reason we will discuss now differentiable manifolds.These are one of the most fundamental concepts in mathematics and physics.

We already learned in previous courses how to perform derivatives and in-tegrals of functions defined on the n-dimensional Euclidean space Rn. In thislecture we will learn how to extend these concepts to spaces that cannot beidentified directly with Rn, but which look like Rn if considered in appropriateregions. Generally, we can think of a differentiable manifold M as a smoothsurface that is locally homeomorphic to Rn. We can then associate to each pointof M some coordinates in Rn and require that the transition maps from a set ofcoordinates to another be regular. With this approach we can analyze functionson such spaces by converting them (locally) to functions in a Euclidean space.

10.1.1 Differentiable manifolds

We will introduce the concept of differentiable manifold by introducing one byone the ingredients needed for its definition. The starting point is obviously atopological space M, but then we need to add some structure to it. The firststep is to parameterize locally this space by coordinates in Rn.

Definition 10.1. Given a topological space M, a chart (Uα, ϕα) is a homeomor-phism ϕα from an open set Uα ⊂ M to an open set Rα ⊂ Rn:

ϕα : M ⊃ Uα → Rα ⊂ Rn.

97

98 differentiable manifolds

U1

U2

ϕ2

ϕ1

ϕ1 ϕ−12

ϕ2 ϕ−11

M

Rn

Rn

Figure 15.: Example of a manifold with arrows exemplifying its local maps ϕ1, ϕ2 toRn and the maps ϕ1 ϕ−1

2 , ϕ2 ϕ−11 giving the coordinate changes in the

overlapping region.

Then we would like that different sets of coordinates on overlapping chartscould be exchanged.

Definition 10.2. Two charts (U1, ϕ1) and (U2, ϕ2) are compatible if ϕ1 ϕ−12 ∈

C∞ and ϕ2 ϕ−11 ∈ C∞ (when U1 ∩U2 6= ∅).

Finally, we want to cover the whole topological space M with charts, so thatwe can give a coordinate in Rn to all its points.

Definition 10.3. The set of (compatible) charts covering M is called atlas.

This may not be a unique procedure and in this case we will say that twoatlases are compatible if all their charts are compatible.

We are now in a position that lets us introduce the concept of differentiablemanifold.

Definition 10.4. A smooth differentiable manifold M satisfies:

1. M is a topological space;

2. there is a family of charts Uα, ϕαα covering M, i.e. ∪αUα = M;

10.1 introduction 99

3. for any couple Uα, Uβ ⊂ M such that Uα ∩Uβ 6= ∅, ϕ1 ϕ−12 ∈ C∞ and

ϕ2 ϕ−11 ∈ C∞.

If the last assumption is not satisfied we have a manifold that is not smooth.In conclusion a manifold is constructed by smoothly sewing together regions

which are homeomorphic to a Euclidean space. A crucial point it that thedimensionality of the Euclidean spaces being used must be the same in everypatch.

Definition 10.5. The dimension of a manifold M is given by the dimension ofthe vector space Rm to which it is locally homeomorphic.

It should be noted that this definition does not rely on an embedding of themanifold in some higher-dimensional Euclidean space. While any n-dimensionalmanifold can be embedded in R2n by Whitney’s embedding theorem, it is im-portant to recognize the existence of a manifold independently of any embed-ding.

10.1.2 Properties

Now that we have introduced the notion of a manifold, we can start looking atits properties and at the possibility of deforming different manifolds onto eachother. Since we are now interested also in the differential structure and notjust to the topological structure, we better introduce a new definition of mapbetween topological spaces that preserves also the differentiable properties.

Definition 10.6. A map f : M1 → M2 continuous and differentiable with in-verse continuous and differentiable is a diffeomorphism.

Two manifolds related by a diffeomorphism are called diffeomorphic.Other properties of manifolds should be decided looking at the underlying

topological space, but, thanks to their atlases, we can often analyze them bylooking at their image in Rn (with some care). This is the case for their com-pactness or orientability.

Definition 10.7. A manifold is compact if for any atlas Uαα=1,...,∞ there is afinite sub-collection Uαα=1,...,n covering the whole manifold.

For instance, in R we can always have a single open subset covering thewhole manifold, R itself, but if we choose the atlas given by Uj =]j, j + 2[j∈Z

there is no finite sub-collection covering the whole manifold.


An interesting concept, although with limited applications in our course, isthe orientability of a manifold. For a sphere it is obvious that one can dis-tinguish an interior and an exterior, but this is not always the case for anymanifold. Mathematically, we can use the determinant of the Jacobian of thecoordinate change to decide whether a manifold is orientable or not.

Definition 10.8. A manifold is orientable if there is an atlas such that det (ϕα ϕ−1

β ) > 0 for any couple of charts (Uα, ϕα), (Uβ, ϕβ) with non-trivial intersec-tion Uα ∩Uβ 6= ∅.

It is interesting to know that compact orientable manifolds have been classi-fied up to dimension 3.

10.1.3 Examples

We now give some examples to illustrate the concepts introduced so far. Firstof all let us introduce some obvious examples of manifolds: Euclidean spaces.

Example 10.1. Rn is a manifold with an obvious atlas given by a unique chartwith homeomorphism the identity map.

On such manifolds it is easier to see why the concept of compatible charts isimportant.

Example 10.2. Let M = R. We can construct charts covering the whole manifoldby using the maps ϕ1 : R → R and ϕ2 : R → R such that ϕ1(x) = x andϕ2(x) = x3. These are not compatible, because ϕ1 ϕ−1

2 = x1/3, which is notC∞ at x = 0.

It is very easy to produce examples of manifolds that are not simply Rn.For instance, the n-sphere Sn, defined as the locus of all points at some fixeddistance from the origin of Rn+1, is a manifold. The n-torus Tn, resulting fromtaking an n-dimensional cube and identifying points on opposite sides, is amanifold. And we can easily continue.

It is therefore also interesting to understand what is not a manifold and forthis reason we now give a few simple examples.

Example 10.3. , , are not manifolds. In the intersectionsthey are not locally R. If we remove the points at the intersections of the firsttwo examples in the picture we get 3 or 4 disconnected components, while inR we always get 2 disconnected components if we remove a point. A similarargument applies to the third picture.

10.1 introduction 101

When introducing the notion of manifold, we insisted on charts, atlases andtheir compatibility. The reason is that for a generic manifold one cannot coverthe whole space by a single chart and we therefore need the discussion pre-sented above. This is already clear for one of the simplest manifolds one canthink of: the circle.

Example 10.4. The circle S1 ≡ (x, y) ∈ R2 | x2 + y2 = 1 is a compact 1-dimensional manifold and therefore we can locally construct homeomorphismsfrom open subsets of S1 to open sets of R. It is easy to see that the minimal atlasis composed of two charts. The first one associates points of a maximal opensubset of the circle to points in the open set ]0, 2π[⊂ R via the homeomorphismϕ1 : S1 \ (1, 0) →]0, 2π[⊂ R, whose inverse gives the embedding of the circlein R2:

ϕ−11 : θ 7→ (cos θ, sin θ). (10.1)

The second one does the same, but in a way that covers the point we excludedin the previous map. This means that now we map the open set U2 = S1 \(−1, 0) to ]− π, π[⊂ R, so that also the homeomorphism ϕ2 : U2 →]− π, π[has an inverse map that has the same form as before:

ϕ−12 : θ 7→ (cos θ, sin θ). (10.2)

The two maps are represented in figure 16.

x

y

(1,0)θ

ϕ1

0

2π

x

y

(-1,0)θ

ϕ2 −π

π

Figure 16.: Atlas for S1 containing two charts.

With this procedure we are assigning two different coordinates sets to twodifferent patches of the manifold. We should then require that the two corre-sponding charts are compatible in the overlapping regions. The points that arenot covered by one of the two charts are at y = 0 in the R2 embedding. We


then have two regions of overlap: one given by all the points with y > 0 andone with all the points with y < 0.

When y > 0 the coordinate changes are trivial because ϕ1 ϕ−12 (θ) = θ and

ϕ2 ϕ−11 (θ) = θ. This effectively means that we do not really need to change

the coordinate assigned to the points of S1 in this region when we pass fromone chart to the other. However, when y < 0 we have ϕ1 ϕ−1

2 (θ) = 2π + θ andϕ2 ϕ−1

1 (θ) = θ− 2π, which means that we have assigned different coordinatesto the same points and we need to perform a coordinate change when we gofrom one chart to the other. It is straightforward to see that both change ofcoordinates are C∞.

Example 10.5. The sphere Sn = ~r ∈ Rn+1 | |~r| = 1 also needs two charts,constructed generalizing the previous example. They come from the so-calledstereographic projection. Using once again the embedding of the sphere in ahigher-dimensional space, we can define the two charts as follows. From a pole,calling ~x ∈ Rn the local coordinates we can use

ϕ−11 (~x) =

(2~x

1 + |~x|2 ,1− |~x|21 + |~x|2

),

which covers the whole Sn except the point 0, 0, . . . , 1, and

ϕ−12 (~x) =

(2~x

1 + |~x|2 ,−1 + |~x|21 + |~x|2

),

which covers the whole Sn except the point 0, 0, . . . ,−1.

10.2 tangent and cotangent space

Useful structures on manifolds are vector fields and tangent spaces and wenow discuss how to define and construct them.

We are already familiar with the concept of tangent vectors to curves in Eu-clidean spaces. We know that they can be constructed by evaluating derivativesof the functions defining such curves. When we deal with a generic topologicalspace M that is also a differentiable manifold, we can construct the tangentspace to one of the points of M by generalizing this approach, describing thevectors tangent to the curves passing through this point via its image in Rn.

A curve on M is a function

γ : R ⊃ I → M. (10.3)

10.2 tangent and cotangent space 103

p γ(t) xi

xi(p)

Figure 17.: We can define tangent vectors to a manifold by looking at tangent vectorsto curves defined on the same manifold using their image in Rn.

When its image has a non-trivial intersection with the open subset of M thatdefines a chart x : M ⊃ U → Rn, we can associate to the curve an image in Rn

with coordinates xi(γ(t)), i = 1, . . . , n (see fig. 17 for a representation of this).The vector tangent to the curve γ at the point p = γ(t∗) ∈ M can be definedvia the tangent vector to its image:

ddt

xi(γ(t))∣∣∣∣t=t∗≡ Vi(p). (10.4)

The functions Vi are maps from points of M to Rn, associating an element ofthe vector space Rn to each of the points of γ(I) ⊂ M. Unfortunately, thisdefinition of tangent vector heavily depends on the homeomorphism definingthe chart. For two overlapping charts we may have different values for thefunctions Vi(p) defining the vector tangent to the same point of the same curve.

In order to have an intrinsic definition of tangent vectors, independent of thecoordinates, it is useful to introduce real functions defined on a manifold:

f : M→ R. (10.5)

Once again, using coordinates defined by an atlas of M, we can write theirvalue in each chart as f (ϕ−1

α (xi)), or, with an abuse of notation, as f (xi). Thevariation of the function f along the curve γ is (in terms of local coordinates)

ddt

f (γ(t)) =∂

∂xi f (ϕ−1(x))ddt

xi(γ(t)), (10.6)

where we evaluated f as a function of Rn via f ϕ−1 : Rn → R. This allows usto define the differential operator on the set of real functions on a manifold M

V ≡ Vi ∂

∂xi , (10.7)


where Vi = ddt xi(γ(t)), so that the directional derivative of the test function f

along the curve γ can be obtained by application of the differential operator V:

V[ f ] ≡ ddt

f (γ(t)). (10.8)

Using these differential operators we can eventually give an intrinsic definitionof vectors tangent to a curve.

Definition 10.9. The differential operator V ≡ Vi ∂∂xi computed in p ∈ M is the

vector tangent to M at p in the direction of the curve γ.

Now V is defined in an intrinsic way. In fact, if the same point is covered bytwo different charts (U, x) and (U, x), i.e. p ∈ U ∩ U, we have

V(p) = Vi(p)∂i = Vi(p) ∂i = Vi(p)∂xj

∂xi ∂j ⇒ Vi(p) =∂xi

∂xj V j(p). (10.9)

We can also generalize this concept in order to obtain a differential operatordefined on the whole manifold and not just on a curve. This is called a vectorfield.

Definition 10.10. A vector field V on M is a linear operator from C∞(M) toC∞(M), which, in each chart, is defined via the directional derivative V =Vi(x)∂i.

V(x)

M

Figure 18.: Integral curves following from the definition of the vector field V on themanifold M.

We can imagine a vector field as a set of differential operators defined ateach point of the manifold, whose action is to take differentiable functions andproduce directional derivatives at the point where they are evaluated.

10.2 tangent and cotangent space 105

The action of these vector fields is linear on the functions

V[ f + g] = V[ f ] + V[g] (10.10)

and there is no obstruction to adding them or scaling them by a real numberin order to obtain new vector fields:

(a V + b W)[ f ] = a V[ f ] + b W[ f ], for a, b ∈ R. (10.11)

This implies that directional derivatives form a vector space and we can usethis vector space to introduce the notion of tangent space to a manifold at apoint p in a coordinate-independent way.

Tp M

pM

Figure 19.: The tangent space Tp M to a point p ∈ M can be defined by the equivalenceclasses of directional derivatives along all the curves passing through p.

The tangent space to a point p ∈ M, whose symbol is Tp M, is the vectorspace of linearly independent directional derivatives evaluated at that point.Since we associate these derivatives to curves passing through p we can alsodefine Tp M as the collection of curves generating different directional deriva-tive operators.

Definition 10.11. The tangent space to the manifold M at the point p (in short Tp M)is the vector space of dimension N = dim M, defined by the equivalence classes

Tp M ≡ γ : R→ M | γ(0) = p ∈ Mγ1 ∼ γ2 ⇔ d

dt γ1(0) = ddt γ2(0)

. (10.12)

The collection of all the tangent vector spaces at each of the points of M isthen the tangent space of the manifold.

Definition 10.12. The Tangent Space to a manifold M is defined as

TM =⋃

p∈MTp M. (10.13)


Although we will not make much use of it in this course, starting from Tp Mwe can define its dual space T∗p M, called the cotangent space.

Definition 10.13. The cotangent space to a manifold M at the point p ∈ M (inshort T∗p M) is defined as the dual space of Tp M via the duality operation(

dxi,∂

∂xj

)= δi

j, (10.14)

so that the set of differentials

dxi define a basis on this space.

10.3 flows and exponential map

A vector field on a manifold describes quite naturally a flow in M. These flowsare the curves that that gave origin to the vector field V and which can bereconstructed by using the exponential map we are going to introduce in thefollowing.

Definition 10.14. Given a vector field V on a manifold M, the curve x(t, x0)solving the differential equation

ddt

xi(t, x0) = Vi(x(t, x0)), and xi(0, x0) = xi0, (10.15)

where x0 ∈ M and xi are the local coordinates, is called the integral curve orflow generated by the vector field V, passing through x0 at the time t = 0.

For simplicity, we use x to indicate both points in M as well as their localcoordinates in Rn.

Theorem 10.1. Let V ∈ TM be a vector field on the manifold M. For any pointx0 ∈ M there is an integral curve of V, a flow x : R×M → M such that x(t, x0) isa solution of the differential equation (10.15).

It is interesting to note that flows satisfy the properties required to be agroup. In fact, for fixed t, xt(x0) ≡ x(t, x0) is a diffeomorphism xt : M → Mand represents a 1-parameter Abelian group:

1. the product closes (and is Abelian): xt xs = xs xt = xt+s (or, explicitlyxt(xs(x0)) = xs(xt(x0)) = xt+s(x0));

2. there is an identity: x0 = 1;

10.3 flows and exponential map 107

3. each element has an inverse: x−1t = x−t.

Also, flows can be expressed in terms of the exponential map of the vectorfields that generate them:

Definition 10.15. The exponential map of a vector field V (denoted by exp(tV))is defined as the family of diffeomorphisms mapping the point x0 ∈ M tox(t, x0):

exp(tV)[x0] = x(t, x0). (10.16)

We can verify this by the analysis of the expansion of the coordinate functionsdefining the image of the flow around x0:

xi(t, x0) =∞

∑n=0

[dn

dtn xi(t, x0)

]t=0

tn

n!

= exp(

tddt

)[xi(t, x0)]

∣∣∣∣t=0

(10.17)

= exp(tV)[xi0],

where in the last step we used the definition (10.15).We give here only a simple example of the construction of the tangent space

to a manifold and the reconstruction of the curves on the manifold via theexponential map.

Example 10.6. Take M = R. The tangent space Tx0R can be constructed byconsidering the diffeomorphisms γ(x0, t) = x0 + t. In fact

ddt

f (γ(x0, t)) = limt→0

f (γ(x0, t))− f (γ(x0, t))t

= limt→0

f (x0 + t)− f (x0)

t=

ddx

f (x0).

(10.18)

This means that ddt γ(x0, t) = 1 = Vx and the vector field is V = d

dx .We can get back the integral curve that originated V by the exponential map

of V:

exp(tV)[x0] = exp(

tddx

)[x0]

=

(1 + t

ddx

+ . . .)

x∣∣∣∣x=x0

= x0 + t = γ(x0, t).(10.19)


10.4 lie brackets

p

p′

p′

V(p) W(p′)

W(p) V(p′′)

[V, W]

Figure 20.: Geometrically the Lie bracket [V, W] describes the non-commutativity ofthe flows generated by the vector fields V and W.

The last concept we need to introduce is the Lie bracket. This is an operationthat associates a new vector field to the combination of two vector fields. Itexpresses how a vector field changes along the flow generated by another vectorfield. In detail, we have seen that there is a direct relation between a vector fieldand its integral curves by means of the exponential map. If, starting from thesame point p ∈ M, we compose the result of integral curves coming fromdifferent vector fields, the result will depend on the order. At the linearizedlevel, this should be represented once again by a vector associated to anothervector field and this is represented by the Lie brackets.

Definition 10.16. The Lie brackets are the commutator of two vector fields onM:

[V, W] = Vi∂i(W j∂j

)−Wi∂i

(V j∂j

)=

=(Vi∂iW j −Wi∂iV j) ∂j

= Zj∂j ≡ Z.

(10.20)

We do not discuss further this new ingredient at this stage, but we will comeback to it when we will discuss the relation between a Lie group and its tangentspace, defined by a Lie algebra.

exercises

1. The maps φi : R → R defined as φ1(x) = x and φ2(x) = x3 are homeo-morphisms from R to R.

10.4 lie brackets 109

Are they diffeomorphisms?

If we consider φi as homeomorphisms in two different charts coveringMi = R (the manifolds covered by such charts), are M1 and M2 diffeo-morphic?

2. On the real projective space RPn we can introduce homogeneous coordi-nates

[x] = [x0 : . . . : xn] ∈ RPn.

Define also the open sets Ui = [x0 : . . . : xn]|xi 6= 0 and the charts

φ−1i : Rn → RPn

(x0, . . . , xi, . . . , xn) 7→ [x0 : . . . : 1︸︷︷︸i

: . . . : xn],

where xi signals the fact that we remove the coordinate xi from the set.Compute the transition functions. Recall the stereographic projection ofthe spheres and build an homeomorphism between RP1 and S1. (Onecan proceed in a similar way for CP1 and S2 or for HP1, the projectivespace on quaternions, and S4).

3. Show that if a smooth differentiable manifold M ⊂ Rn of dimension k isdefined by the zeros of f : U ⊂ M→ Rn−k (or U ∩M = f−1(0)) we havethat

Tp M = ker [d f (p)].

Use this fact to find the tangent space to the unit sphere of dimension k.

Part IV

L I E G R O U P S

11L I E G R O U P S .

In the first part of these lectures we mainly discussed discrete and finite groups.We now want to consider the theory of continuous groups, where the groupelements depend continuously on a set of parameters.

The theory of continuous groups is much more complicated than the the-ory of discrete groups. In this case the elements cannot be enumerated andin addition to algebraic concepts we need topological notions like continuity,connectedness and compactness and this is why we reviewed in the previouslectures some of these notions. However, we will not discuss all possible typesof continuous group, but rather focus on a particular class of continuous groupsrelevant for physical applications: Lie groups.

Definition 11.1. A Lie group is a smooth differentiable manifold with a groupstructure such that the following group operations are differentiable (smooth):

• the composition: : G× G → G;

• the inversion map: G → G given by g 7→ g−1.

The dimension of a Lie group G is determined by the dimension of G as amanifold.

A Lie group is a special kind of differentiable manifold, where each pointis an element of a group and where different points may be related by groupoperations.

Remember that our goal is not to give a full mathematical treatment of Liegroups and their associated Lie algebras, but to show the results that are mostrelevant for physical applications. For this reason we will often mention theo-rems without demonstrations, just to inform the student of their existence, andprovide examples that are instructive to understand more general mathematicalresults.

113

114 lie groups

11.1 compact and non-compact groups

Before discussing examples of Lie groups, we present some general statementsabout their general properties that are going to be useful in the following.

Lie groups are also differentiable manifolds and this implies that we canassociate topological properties to Lie groups by means of the same propertiesfor their underlying manifold. For instance, we can separate compact andnon-compact groups using the corresponding definition 10.7 for differentiablemanifolds.

Since a generic element of a Lie group is determined by the values of dimG parameters defined via local charts, we call compact groups the ones havinga compact parameter domain and non-compact those having a non-compactparameter domain. For instance, the rotation group in 3 dimensions, SO(3), iscompact and its elements are defined via angles. The group of translations onthe other hand is non-compact.

11.2 invariant measure

Most of the theorems we proved in the first part of this course relied on carryingout sums over the group elements, often in conjuction with the RearrangementTheorem 3.2. For instance, thanks to the rearrangement theorem, we oftenreplaced the sum over the group elements of a function of the elements ofthe group, with the sum of the values of the same function evaluated on theelements multiplied by the same g′ from the left

∑g∈G

f (g′g) = ∑h∈G

f (h = g′g) = ∑g∈G

f (g), (11.1)

or from the right

∑g∈G

f (gg′) = ∑g∈G

f (g), (11.2)

or summing over the inverses

∑g∈G

f (g−1) = ∑g∈G

f (g). (11.3)

When we deal with continuous groups, discrete sums have to be replacedwith continuous ones∫

Gdµg f (g) (11.4)

11.3 groups of matrices 115

and in order to do so, we need to introduce a measure on the group dµg,possibly in a way that the integral gives finite results for continuous functionsf . Obviously, different choices of measures may lead to different results, evenif we sum over the same elements. For this reason we would like to havea measure that remains invariant when the group elements in the sum arerearranged. This means that we need an invariant measure over G, namely ameasure dµg satisfying

dµgg′ = dµg′g = dµg−1 = dµg. (11.5)

Unfortunately such a measure does not always exist:

Theorem 11.1. If G is a compact group there is a measure satisfying dµgg′ = dµg′g =dµg−1 = dµg and it is unique up to a normalization factor.

This measure is called the Haar measure.

This theorem implies that only for compact groups we can safely extend ourproofs of orthogonality for the matrix elements of their representations, for theequivalence of the group representations to unitary ones, etc. One needs morecare when discussing non-compact groups. For instance, one may still haveleft-invariant or right-invariant measures, but their integral on the group maydiverge.

11.3 groups of matrices

The groups of invertible matrices play a special role among Lie groups for tworeasons:

• Invertible matrices provide linear representations of groups;

• Groups of matrices can be systematically classified using constraints.

We start by repeating a definition that we already encountered, but whichprovides the starting point of our discussion.

Definition 11.2. The set of N × N invertible matrices with entries in F = R orC forms a group called General Linear group, or GL(N, F).

If A, B ∈ GL(n, C) it is obvious that AB ∈ GL(n, C) and also (AB)−1 =B−1 A−1 ∈ GL(n, C), where the composition is given by the usual matrix prod-uct. Also, we have an obvious identity element, which is the n× n unit matrix

116 lie groups

1n. The only constraint needed to have a group is that the matrices in GL(n, C)should have non-vanishing determinant, so that we can invert them. This doesnot constrain single entries of the matrix and therefore

dim GL(n, R) = n2 (11.6)

and

dim GL(n, C) = 2n2. (11.7)

Recall that we discussed real manifolds and therefore we are giving here theirreal dimension.

If we restrict ourselves to the group of matrices with unit determinant wecan define

Definition 11.3. The Special Linear group

SL(n, F) = M ∈ GL(n, F) | detM = 1.

The proof that this is still a group is straightforward because det (AB) =det A · det B, det 1n = 1 and det (A−1) = 1/det A.

Since one of the entries of the matrices in SL(n, F) is fixed in terms of theothers because of the constraint on the determinant we have that

dim SL(n, R) = n2 − 1 (11.8)

and

dim SL(n, C) = 2(n2 − 1). (11.9)

Starting from these groups we will now see a series of other groups definedas subsets satisfying linear or quadratic constraint. The matrices resulting fromthese constraints provide the so-called fundamental representation of a group.As we will see in the following, this is not the only irreducible representationexisting for such groups, but it is fundamental in the sense that the others couldbe obtained by decomposing its products (in general a Lie group has an infinitenumber of irreducible representations).

11.3.1 Linear constraints

Definition 11.4. The upper triangular matrices form a group

UT(p, q) = M ∈ GL(p + q, F) | Mia = 0, ∀a ≤ p and i ≥ p.


The form of a generic matrix in UT(p, q) is

m =

(Ap×p Bp×q

Oq×p Cq×q

). (11.10)

Composition of matrices of this form close on matrices of the same form. Theidentity is part of UT(p, q) and the inverse is also in the group

m−1 =

(A−1

p×p −A−1p×pBp×qC−1

q×q

Oq×p C−1q×q

). (11.11)

This definition can be generalized straightforwardly to UT(p, q, r) = UT(p, q+r)∩UT(p + q, r) and this can be repeated, up to UT(1, . . . , 1). In this case thegroup gets a new name:

Definition 11.5. The group of n× n strictly triangular matrices is called Solvablegroup:

Solv(n) = UT(1, 1, . . . , 1).

Definition 11.6. The subgroup of matrices of Solv(n) having all diagonal ele-ments equal to 1 is called Nilpotent group, Nil(n).

11.3.2 Quadratic constraints

We can also restrict invertible matrices by requiring that they preserve anothermatrix (a metric), that is not necessarily part of the group. We then introduce aquadratic constraint, where matrices M ∈ GL(n, F) satisfy also a constraint ofthe form MTηM = η, or M†ηM = η, or M∗ηM = η, for η a matrix that may besingular or non-singular, antisymmetric or symmetric and, if symmetric, posi-tive definite or indefinite. All these distinct cases will generate distinct groups,provided that the constraint allows for group composition to be closed and forthe inverse to satisfy the same constraint (M = 1 always satisfies quadraticconstraints like those mentioned above).

Compact groups

Definition 11.7. The Unitary group

U(n) ≡ M ∈ GL(n, C) | M†1n M = 1n.

118 lie groups

This group has n2 real constraints because (M† M)† = M† M and hence

dim U(n) = n2. (11.12)

Definition 11.8. The Special Unitary group

SU(n) ≡ M ∈ U(n) | detM = 1.

Obviously SU(n) < U(n), but, contrary to naive expectations, U(n) is notsimply the product of SU(n) with a phase. In fact the map

SU(n)×U(1) −→ U(n)

(M, eiα) 7→ M = eiα M(11.13)

is a homomorphism, but not an isomorphism. If det(M) = 1, det(M) = eiαn ∈C, but obviously this map is not 1 to 1. If we choose M = e2πi/n1n, det(M) = 1,but we have n such matrices that we can map to the same element in U(n):

gm =(

e2πi m/n1, e−2πi m/n)7→ 1n, (11.14)

where m = 1, . . . , n. Since the elements gm generate a Zn < SU(n)×U(1), wehave that

U(n) ' (SU(n)×U(1))/Zn. (11.15)

If we restrict ourselves to real matrices, the same quadratic constraint givesrise to another notable group.

Definition 11.9. The Orthogonal group

O(n) ≡ M ∈ GL(n, R) | MT M = 1n.

We now have a real symmetric constraint (MT M)T = MT M. We thus haven(n + 1)/2 constraints on n2 real entries leaving

dim O(n) =n(n− 1)

2. (11.16)

We already saw that the constraint defining O(n) implies that its matrices havedeterminant equal to 1 or to −1. Those with unit determinant include theidentity and form a subgroup:

Definition 11.10. The Special Orthogonal group

SO(n) ≡ M ∈ GL(n, R) | MT M = 1n, detM = 1.


Non-compact groups

If we consider the same type of constraints as those given above, but now foran indefinite metric of the form

η =

(1p

−1q

)(11.17)

we get some non-compact groups:

Definition 11.11.

U(p, q) ≡ M ∈ GL(p + q, C) | M†ηM = η.

Definition 11.12.

O(p, q) ≡ M ∈ GL(p + q, R) | MTηM = η.

It is very easy to see that the set of parameters labeling the elements of thesegroups becomes non-compact. For instance,(

cosh ψ sinh ψ

sinh ψ cosh ψ

)= m ∈ O(1, 1), ∀ψ ∈ R. (11.18)

Another non-compact group important for physical applications is the

Definition 11.13. Symplectic group

Sp(2n, F) = M ∈ GL(2n, F) | MTΩM = Ω,

where Ω =

(On 1n

−1n On

).

Other groups may be obtained using singular metrics, like

η =

1

−1qOr

,

generating U(p, q, r), O(p, q, r), etc.

120 lie groups

We will not discuss all possibilities here. We close this lecture mentioningthe fact that some of these groups with different definitions are actually iso-morphic. For instance, a matrix M ∈ Sp(2,R) has unit determinant. In detail

M =

(a bc d

)∈ Sp(2, R) if and only if

(a cb d

)(0 1−1 0

)(a bc d

)=

(0 1−1 0

)(11.19)

and this gives ad− bc = 1. But this is the definition of a matrix in the speciallinear group of 2-dimensional real matrices and therefore

Sp(2, R) ' SL(2, R). (11.20)

exercises

1. Take the fundamental representation of SO(2) and show that Zn < SO(2).The matrices you obtained provide a representation of Zn. Is it reducible?

2. Show that Dn < O(2) by using appropriate matrices in the fundamentalrepresentation of O(2).

3. Consider the matrices in the fundamental representation of SO(3) andshow that S4 < SO(3).

Hint: look for a transformation corresponding to a 3-cycle and one correspondingto a 4-cycle.

4. Let (qi, pi), i = 1, . . . , n, be the coordinates and momenta of a classical me-chanical system with n degrees of freedom. Poisson brackets are definedas

f , g =n

∑i=1

(∂ f∂qi

∂g∂pi− ∂g

∂qi∂ f∂pi

).

a) Show that qi, qj = pi, pj = 0 e qi, pj = δij.

b) Show that if we redefine the coordinates with a linear transformation(QP

)= S

(qp

),

the relations Qi, Qj = Pi, Pj = 0 e Qi, Pj = δij are still true if

and only if S ∈ Sp(2n, R).


5. Write the matrices R1, R2 ∈ SO(3) corresponding to a rotation of π/2around the z axis and of π/4 around the x axis respectively. ComputeR3 = R2R1 and show that the corresponding element gives a rotation ofan angle φ around an axis determined by ~n = (nx, ny, nz). Show that tr(R3) = 1 + 2 cos φ.

6. Using the results of the previous exercise show that conjugate elementsof SO(3) are characterized by the angle specifying the rotation.

7. This is just for fun. Use what you just learned to find the conjugacy classesof S4.

8. Find the center of O(n) and of SO(n).

9. Find the maximal Abelian subgroup of U(2).

10. Given M ∈ Sp(2n, R), show that M−1 ∈ Sp(2n, R) and also MT ∈Sp(2n, R), i.e. prove that if MTΩM = Ω, then M−TΩM−1 = Ω as well asMΩMT = Ω.

11. Discuss the isomorphism between Sp(2, R) and SU(1,1).

Hint: Use the map µ(g) = τgτ−1, where τ = 1√2

(1 ii 1

)and check that it

defines an isomorphism.

12. Consider again the Lagrangian density

L = −∂µΦ†∂µΦ + Φ†ηΦ− 12(Φ†Φ)2,

where Φ is a complex scalar field with 4 components Φ(x) =

φ1(x)φ2(x)φ3(x)φ4(x)

e η = diag−1,−1, 1, 1.

a) What is the group of linear transformations Φ(x) → SΦ(x), withS ∈ Mat(4, C), preserving the kinetic term and the potential, respec-tively?

122 lie groups

b) For which constant values Φ0 =

c1c2c3c4

the Euler–Lagrange equa-

tions are satisfied and what is the residual symmetry group?

(Otherwise, which matrices S satisfy Φ0 → SΦ0 = Φ0)

12L I E A L G E B R A S .

As already done for finite groups, also for Lie groups we would like to have arapid way to understand their structure and their possible relations. Given theirdouble nature of groups and manifolds, in order to specify an isomorphism weneed to have their equivalence as topological spaces as well as groups. Wethen need to understand when we can deform two groups onto each other in acontinuous way, so that the composition laws are equivalent up to a change ofvariables. This last requirement may be very complicated because compositionlaws are not necessarily linear. Also, we would like to compare only a finitenumber of elements if possible, rather than the infinite number that is defininga continuous group. For these reasons we now introduce the notion of Liealgebra, which is going to help us in this tasks.

A Lie algebra is the linearization of the corresponding group around theidentity element. It contains the information that is needed to reconstruct thegroup in a neighborhood of the neutral element. Actually, the first useful thingto note is that linearizing the group G at any of its elements gives the sameresult. In fact, since G is a manifold endowed with a group structure, we canmap any element g to any other g′ using the composition of the first withh = g′g−1 and in the same way we can map any open neighborhood of g, Ug,to an open neighborhood of g′, Ug′ , by multiplying points of Ug with h fromthe left. Since the composition map on a Lie group is a continuous function,we can do the same for the curves in Ug and eventually for the directionalderivatives defining the tangent space in g. The result is that we can focus onthe tangent space to the group G at the identity without losing in generality,but simplifying some of the calculations because of the properties of the neutralelement.

123

124 lie algebras

g

Ug

TgG

g′Ug′Tg′G

g′ g−1

GFigure 21.: Tangent spaces at different points on a group manifold can be related by

group multiplication.

12.1 algebra as tangent space

The general procedure to construct the tangent space requires the identificationof curves M(t) ∈ G, passing through the element around which we want toconstruct the tangent space, which is the identity. Since we do not want to enterin a mathematical discussion that would be useless for our purpose, we willfocus on groups of matrices, though obvious generalizations apply to genericLie groups. We therefore choose M(t) as a 1-dimensional subgroup of thegroup of matrices G, so that

M(0) = 1 (12.1)

and

M(t1)M(t2) = M(t1 + t2), (12.2)

which implies

M(t)−1 = M(−t). (12.3)

Locally this group is isomorphic to R. This means that we can define in anobvious way its derivatives, following the construction explained in chapter 10.For each of the inequivalent curves M(t) we define the corresponding tangentvector to the identity as

g ≡ T1G 3 X =ddt

M(0). (12.4)

12.1 algebra as tangent space 125

The subgroup defined by M(t) can actually be seen as an integral curve of avector field V as follows from

ddt

M(t) = limδt→0

M(t + δt)−M(t)δt

= M(t) limδt→0

M(δt)−M(0)δt

= M(T)ddt

M(0) = M(t)X, (12.5)

and hence

ddt

M(t) = V(t)[M(0)] = M(t)X, V(0) = X. (12.6)

From the general procedure of section 10.3 we also know that we can recon-struct the 1-parameter group M(t) < G via the exponential map built fromX ∈ g:

M(t) = exp(t X)[M(0) = 1]. (12.7)

We then provided explicit maps between a group and its algebra in both direc-tions:

ddt−→

G g

←−exp

When G = GL(n, C), both maps are easy to compute. In particular, for M ∈GL(n, C), we can easily obtain the tangent space at the identity by consideringthe expansion

M = 1 + ε m + . . . , ε 1, (12.8)

where the condition det M 6= 0 is trivially satisfied for ε 1 for any

m ∈ Mat(n, C), (12.9)

the space of square complex matrices. The inverse operation gives back theoriginal group close to the identity via the exponential map. By exponentiationwe can indeed obtain n2 different 1-parameter subgroups of GL(n, C):

M(t) = exp(tm) =∞

∑k=0

tk mk

k!, (12.10)

126 lie algebras

where mk is defined by the usual matrix product. This series is always conver-gent in the space of matrices and we can easily check that M(t) is in fact inGL(n, C). Its inverse is simply M(t)−1 = M(−t) = exp(−tm). We thereforeconclude that

Definition 12.1. The Lie algebra of the general linear group is

gl(n, F) = Mat(n, F).

We are obviously interested in understanding if and how we can reconstructthe group G and all its properties from its algebra g = T1G, but first we pauseand analyze some other specific instances of groups of matrices.

12.1.1 Algebras for groups of matrices

Also for the Special Linear group we can obtain its algebra by expanding theconstraint defining its fundamental representation around the identity:

1 = det M = det(1 + ε A + . . .) = 1 + ε trA + . . . . (12.11)

Definition 12.2.sl(n, F) = A ∈ Mat(n, F) | trA = 0.

In the case of Unitary groups an analogous expansion gives

(1 + ε A + . . .)†(1 + ε A + . . .) = 1 + ε (A + A†) + . . . , (12.12)

for ε 1 and real (any phase can be reabsorbed in the definition of A).

Definition 12.3.u(n) = A ∈ Mat(n, C) | A† = −A.

Obviously dim U(n) = dim u(n) = n2. This is easy to check for the algebra,where the n diagonal entries are purely imaginary and the n(n− 1)/2 entriesabove the diagonal are complex conjugate of those below the diagonal. We thenhave only 2× n(n−1)

2 + n = n2 real degrees of freedom.Since we obtained the conditions that matrices in the algebra should satisfy

by looking at the linear expansion around the identity, we may ask whethertheir exponential, which goes beyond linear level, is really in the group. Thisis straightforward to verify because of matrix properties:

[exp(tA)]† exp(tA) = exp(tA†) exp(tA) = exp(−tA) exp(tA) = 1, (12.13)

12.1 algebra as tangent space 127

where we used the definition of the exponential of a matrix in the first equalityand the fact that A ∈ u(n) in the second equality.

Note also that in Quantum Mechanics the relation between elements of theUnitary group and the algebra is usually done by considering the exponen-tial of Hermitian operators, which are related to our group generators by animaginary unit.

For the orthogonal group and the special orthogonal group we get that

(1 + ε A + . . .)T(1 + ε A + . . .) = 1 + ε (A + AT) + . . . , (12.14)

which means that matrices in their algebras are antisymmetric and thereforewith n(n − 1)/2 degrees of freedom. In addition, for the special orthogonalgroup we should add the constraint giving unit determinant matrices, but thisis given by the vanishing of the trace, which is guaranteed by their antisymme-try.

Definition 12.4.

o(n) ' so(n) = A ∈ Mat(n, R) | AT = −A.

Note that, as expected, the algebras described above have the structure of avector space, but differently from the corresponding groups, they do not closeunder matrix multiplication. For instance, if A, B ∈ so(n), AB /∈ so(n), because

(AB)T = BT AT = (−B)(−A) = BA 6= −AB. (12.15)

Using the same trick we can also establish that:

Definition 12.5.

sp(2n, R) =

A ∈ Mat(2n, R) | ATΩ = −ΩA

,

where Ω =

(0 1

−1 0

).

Since ΩT = −Ω, we can also rewrite the constraint as (ΩA)T = ΩA, andthis means that the number of parameters defining the matrices A is equal tothat of 2n× 2n symmetric matrices. Hence

dimRsp(2n, R) = n(2n + 1). (12.16)

128 lie algebras

12.2 from the algebra g to the group G

Given g ≡ T1G, we have a description of G in a neighborhood of the identityas follows from:

Theorem 12.1. The exponential map is bijective in a neighborhood of the identity.

However, using g 3 X 7→ exp X ∈ G can we reconstruct all the informationneeded to specify G? In particular, we are interested in the following questions:

• Is the exponential map surjective?

• Is the exponential map injective?

• If we have isomorphic algebras (in a sense that we will specify soon), arethe corresponding groups isomorphic?

We will see that unfortunately the answer to all these questions is NO. However,the exponential map becomes surjective if the group is connected and compact andit becomes injective if G is simply connected. Hence we conclude that using theexponential map we can fully reconstruct a group that is simply connected tothe identity and compact.

12.2.1 Surjectivity

The counterexample has been found by Cartan analyzing the relation betweenSL(2, R) and sl(2, R). If A is a matrix in sl(2, R), it has vanishing trace and canbe parameterized by three real parameters a, b, c:

A =

(a b + c

b− c −a

). (12.17)

The generic matrix in SL(2, R) that can be obtained from an element of thealgebra by exponentiation is going to be exp(A), for an arbitrary choice ofa, b, c. We will now show that such elements, however, do not cover the wholegroup. The reason is that they satisfy a constraint on the trace that some of theelements in the group do not fulfill. In fact, any matrix A of the form (12.17)can be diagonalized

SAS−1 =

(λ−λ

), (12.18)

12.2 from the algebra g to the group G 129

by means of the similarity matrix S. Here ±λ = ±√

a2 + b2 − c2 are the eigen-values of the matrix A. They are real if a2 + b2 ≥ c2 and purely imaginaryotherwise. Using once more the properties of the exponential map for matri-ces, we see that

tr(exp(A)) = tr(S exp(A)S−1) = tr(exp(SAS−1)) = eλ + e−λ. (12.19)

This implies that

tr[exp(A)] ≥ −2. (12.20)

In fact, if λ is real tr(exp A) = 2 cosh λ ≥ 2 and if λ is complex, λ = iµ andtr(exp A) = 2 cos µ ∈ [−2, 2]. On the other hand we can easily construct amatrix in SL(2, R) that does not fulfill this constraint. This is

X =

(−x

− 1x

), for x ∈ R∗+. (12.21)

We have that det X = 1 and therefore X ∈ SL(2, R), but also

trX = −(

x +1x

)≤ −2, (12.22)

which does not fulfill the constraint (12.20) for x 6= 1.Note that, although the exponential map is clearly not surjective, we can still

obtain the element X by multiplying elements obtained from the exponentialof the algebra:

X = exp[

π

(0 1−1 0

)]exp

[log x

(1 00 −1

)]. (12.23)

This, on the other hand, is a consequence of the fact that SL(2, R) is connectedand cannot be extended to other groups (for instance in O(n) we have twodisconnected components and we will never get matrices with determinant −1by exponentiating matrices in the algebra, nor by taking their products).

12.2.2 Injectivity and global properties

We already saw that so(n) = o(n), but we can also construct identical algebrasthat generate both compact and non-compact groups. This is the case of o(2) 'u(1) ' R. All these algebras have a unique element, which we can choose to

130 lie algebras

be X =

(0 1−1 0

)for o(2), X = i for u(1) and X = 1 for R. The group

elements are obtained by taking exp(θX). For U(1) and SO(2) we get that forany θ = c + 2π Z we get the same element of the group, while for (R+, ·) themap is injective.

12.2.3 The algebra structure and the Baker–Campbell–Hausdorff formula

As long as we consider g = T1G just as a vector space we only recover thenotion of G as a differentiable manifold, but not its group structure. However,once we introduce an additional operation (the commutator), we can relateproperties of g with those of G.

In order to see how this structure emerges we analyze the simple case ofSO(n). Starting from A, B ∈ so(n) we know that exp(tA), exp(tB) ∈ SO(n),for any t. We also know that the closure of the product operation within thegroup ensures that also exp(tA) · exp(tB) ∈ SO(n). How can we characterizethis from the point of view of the algebra? Take the expansion of the productfor small t:

exp(tA) · exp(tB) =

(1 + tA +

t2

2A2 + . . .

)(1 + tB +

t2

2B2 + . . .

)

= 1 + t(A + B) +t2

2(A2 + B2 + 2AB) +O(t3)

= 1 + t(A + B) +t2

2(A + B)2 +

t2

2[A, B] +O(t3)

= exp(

t(A + B) +t2

2[A, B] + . . .

), (12.24)

where (A + B)2 = A2 + B2 + AB + BA and

[A, B] ≡ AB− BA. (12.25)

We see that in order to determine the new element of the group from the prod-uct of exponentials of elements of the algebra we need to know also their com-mutator, no matter how close we are to the identity.

Note also that

[A, B]T = (AB)T − (BA)T = BT AT − AT BT = BA− AB = −[A, B] (12.26)


and therefore [A, B] ∈ so(n), so the commutator associates a new element of thealgebra to two other elements of the algebra. Hence we can use the commutatoras a product operation that closes in the algebra.

We conclude that close to the identity, the structure of a (matrix) group isdetermined by the vector space given by T1G with the additional informationof the commutators of the elements of the algebra.

We could continue order by order, but the result is just a particular case of ageneral formula we are not going to prove in detail:

Theorem 12.2. Baker–Campbell–Hausdorff formula: eXeY = eZ, where

Z = X +∫

dt ψ(exp(adX) exp(t adY))Y,

where ψ(u) = u log uu−1 = 1 + (u− 1) + 1

6 (u− 1)2 + . . . is regular in u = 1 and

adXY ≡ [X, Y]. (12.27)

In the following we will also need the

Theorem 12.3. Hadamard Lemma

eABe−A = exp(adA)B.

This is rather easy to check explicitly for matrices:(1 + A +

A2

2+ . . .

)B(

1− A +A2

2+ . . .

)

= B + AB− BA +12

(A2B− 2ABA + BA2

)+ . . . (12.28)

= (B + [A, B] + [A, [A, B]] + . . .) = exp([A, ·])B.

12.2.4 Lie algebras

Definition 12.6. A Lie algebra is a vector space g on a field F with a map

[·, ·] : g× g→ g

which is

132 lie algebras

1. bilinear: ∀x, y, z ∈ g and a, b ∈ F

[ax + by, z] = a[x, z] + b[y, z],

[z, ax + by] = a[z, x] + b[z, y];

2. antisymmetric[x, y] = −[y, x], ∀x, y ∈ g;

3. such that the map [x, ·] : g→ g satisfies the Leibnitz rule

[x, [y, z]] = [[x, y], z] + [y, [x, z]], ∀x, y, z ∈ g.

The last condition also goes under another name

Definition 12.7. Jacobi identity. Given any X, Y, Z ∈ g:

[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0.

The map we just introduced is simply the commutator for matrices, but inthe general case it is an operation that we can define starting from Lie brackets.In full generality we can define g from G as its tangent space at the identity T1G.This is a vector space on which we can introduce an operation associating a newvector of the tangent space [X, Y] to any other couple of vectors X, Y ∈ T1G bytaking the generator of the vector field that is obtained from the Lie brackets ofthe vector fields generated by the other two. Operatively, we introduce exp(tX)and exp(tY), compute the Lie bracket [exp(tX), exp(tY)] and expand for t→ 0.The element we obtain is [X, Y].

exercises

1. Matrix properties and exponential of a matrix.

a) Assume that A and B are generic N × N complex matrices. Checkthat tr [A, B] = 0, eUAU†

= UeAU†, det eA = etrA.

b) Compute eαM and eαN , where α ∈ C and

M =

0 0 10 0 01 0 0

, N =

0 0 10 0 0−1 0 0

.


2. The Heisenberg algebra, generating the Heisenberg group, can be repre-sented by the following matrices:

tx =

0 1 00 0 00 0 0

, tp =

0 0 00 0 10 0 0

, th =

0 0 10 0 00 0 0

.

Compute their commutators, compute their exponentials Ux, Up e Uh andverify the Baker–Campbell–Hausdorff formula for the product UxUp.

3. Build the fundamental representation of SU(1,1) and derive su(1, 1) =T1SU(1,1) by taking derivatives of appropriate 1-parameter subgroups.

13P R O P E RT I E S O F L I E A L G E B R A S .

Lie algebras are first of all vector spaces. As such, for any Lie algebra g wecan introduce a basis of n = dim g = dim G vectors tII=1,...,n, so that anyelement of g can be written as their linear combination. Lie algebras, however,are also closed under the product defined by the map in 12.6 (which we willcall commutator in the following). Because of its vector space nature, all theinformation contained in the commutator of two generic elements is fixed bythat of the basis vectors:

[tI , tJ ] = f I JKtK. (13.1)

Definition 13.1. The coefficients f I JK in (13.1) are called structure constants.

Using the linearity properties of the commutator, we can always express theresult of the commutator of any couple of elements of the algebra X, Y ∈ g interms of the structure constants. In detail, we can expand the generic vectorsX and Y as X = xI tI , Y = yJtJ , where xI , yJ ∈ F, the field on which the algebrais defined, and do the same for the vector Z, associated to the commutator:

[X, Y] = xIyJ [tI , tJ ] = xIyJ f I JK tK = Z ∈ g. (13.2)

From the relation (13.1), defining the structure constants, we can also deducesome of their properties. First of all, they must be antisymmetric in the firsttwo indices

f I JK = − f J I

K. (13.3)

Then, expanding the Jacobi identity 12.7 in terms of the basis vectors, we canalso deduce that the structure constants must satisfy

0 = f[I JL fK]L

M =13

(f I J

L fKLM + fKI

L f JLM + f JK

L f ILM)

. (13.4)

135

136 lie algebras – properties

Example 13.1. The algebras u(1) ' R are generated by a unique element t1 andall their structure constants are vanishing.

Example 13.2. The algebra of the Heisenberg group is defined by 3 generators,with a unique non-trivial commutator:

[t1, t2] = t3. (13.5)

The only non-vanishing structure constants are f123 = − f21

3 = 1.

Example 13.3. The algebra of angular momenta so(3) has 3 generators tI , I =1, 2, 3, satisfying the commutator relations:

[tI , tJ ] = εI JK tK. (13.6)

From this introduction, we see that a Lie algebra is specified by the dimen-sion of the vector space and by the commutator relations. Hence a classificationof inequivalent structure constants allows to classify inequivalent algebras. Ifwe want to compute two algebras, though, we should also remember that wecan always choose different basis for the same algebra. This means that we canalways change the values of the structure constants by taking linear combina-tions of the generators:

tI = ∑K

cIKtK. (13.7)

Since this is just a redefinition of the basis of vectors generating the vector spaceg, it does not change the algebra. For this reason, if two algebras g and g areconnected by a change of basis like (13.7), we say that they are isomorphic (wewill give a more appropriate definition in 14.2). One should be careful on thechoice of the coefficients cI

K, though. The algebra g is defined a vector spaceon a specific field F, which is usually taken to be C or R. We should thenonly take linear combinations where the coefficients cI

K are in the same field,otherwise we are effectively changing the nature and dimension of the algebra.

Later we will see that sometimes establishing an isomorphism between dif-ferent real Lie algebras using complex change of coordinates can be useful tostudy their representations. For this reason we introduce another useful con-cept: the complexification of a real Lie algebra.

Definition 13.2. If gR = span(tI) is a real vector space, its complexification gC

is the real span of the basis tI , i tI ∈ gC.Note that dimR gC = 2 dimR gR.

13.1 subalgebras , ideals 137

The process of complexification of an algebra is useful to connect algebrasthat are not isomorphic. For instance,

so(3) = span(t1, t2, t3) , [t1, t2] = t3, [t2, t3] = t1, [t3, t1] = t2 (13.8)

and

so(2, 1) = span(t1, t2, t3) , [t1, t2] = t3, [t2, t3] = t1, [t3, t1] = −t2 (13.9)

are not isomorphic as real algebras, but we can map them onto each other ifwe allow the use of complex coefficients:

t1 = it1, t2 = t2, t3 = it3. (13.10)

We then say that so(3) and so(2, 1) are different real forms of the same complexalgebra so(3)C ' so(2, 1)C ' sl(2, C).

Definition 13.3. A real form gR of the complex algebra gC is obtained by con-sidering a complex basis tI ∈ gC, I = 1, . . . , dimC gC and restricting thevector space to the one generated by their real combinations.

13.1 subalgebras , ideals

Given an algebra g, we can always consider some subspace that is again a vectorspace and we can ask whether the result is also an algebra.

Definition 13.4. A subspace h ⊂ g is a subalgebra of g if it is closed with respectto the bracket operation in g, i.e.

∀x, y ∈ h, [x, y] ∈ h.

An invariant subalgebra is called ideal.

Definition 13.5. An ideal n ⊂ g is a subspace of an algebra that is stable undermultiplication, i.e.

[n, g] ∈ n.

It is clear that such a definition derives from the linearization of the conditionfor a subgroup to be normal. In the same fashion, we can distinguish Abelianand non-Abelian algebras by their commutators.

Definition 13.6. An algebra a is Abelian if and only if all its structure constantsare vanishing.


13.2 simplicity, compactness and connectedness

We can now analyze how various group properties are translated into prop-erties of their algebras. For instance, in 4.11 and 4.12 we introduced simpleand semi-simple groups. The corresponding definitions for the algebras arethe following:

Definition 13.7. A Lie algebra g is simple is it does not contain any ideal.

Definition 13.8. A Lie algebra g is semi-simple if it does not admit any Abelianideal.

Note that while it is obvious that

G simple ⇒ g simple, (13.11)

G semi-simple ⇒ g semi-simple, (13.12)

the inverse is not guaranteed! A simple counterexample is given by the su(2)and so(3) algebras. As we will see in detail in section 15.2, although the alge-bras are isomorphic, the corresponding groups have different properties: SO(3)is simple, while SU(2) has an invariant Abelian subgroup1: Z2 (which is alsonot a Lie group).

By extension of meaning, we will also call compact the Lie algebra of a com-pact Lie group.

Putting together these definitions one can come to an interesting result thatexplains several properties of many algebras and algebra representations thatare useful in Physics.

Theorem 13.1. Every semi-simple complex Lie algebra admits a unique compact realform.

Let us stress that since algebras come from groups by linearizing around theidentity, we cannot deduce global properties of G starting from g. In particular,if G is not connected and K < G is the part connected to the identity

T1G = g = k = T1K. (13.13)

In addition, if a group is not simply connected and G is its universal covering,we have that

g = g. (13.14)

1 Pay attention, because some texts define semi-simple groups as those that do not have invariantAbelian continuous groups, and in this sense also SU(2) would be semi-simple.

13.3 cartan’s criteria 139

This is what happens for instance for u(1) = R, so(3) = su(2), so(1, 3) =

sl(2, C), where R = U(1), SU(2) = SO(3), SL(2, C) = ˜SO(1, 3), but obviouslyR 6= U(1), etc.

Summarizing, for any group G there is an algebra g that is obtained as itstangent space at the identity g = T1G, while for any algebra g there is a uniquesimply connected group G that has g as its tangent space at the identity T1G.All the other connected groups K that have g as algebra are of the form

K/H = G, (13.15)

where HC K (maybe discrete).For non-connected groups, G is the component connected to the identity.

13.3 cartan’s criteria

We can understand even more properties of algebras and the relations withproperties of the groups that generated them by introducing a new object: theCartan–Killing form.

Definition 13.9. The Cartan–Killing form is defined by

(X, Y) = (Y, X) = tr (adXadY(·)) ,

where X, Y ∈ g.

If we recall formula (12.27), we see that adX is an operator acting on theelements of the vector space spanned by the basis of generators of the algebra.Effectively, we can define it as a matrix operator linearly mapping elements ofg into other elements of g:

adX(tI) = tJ D(adX)J

I , (13.16)

where

D(adX)J

I ≡ xL fLIJ . (13.17)

In this way adX(adY(·)) can be interpreted as a product of matrices, whosetrace is easy to compute. If we then expand X and Y on the basis of the algebraX = xI tI , Y = yI tI , we can rewrite the definition 13.9 as

(X, Y) = kI J xIyJ , (13.18)


where

kI J = k J I = f IML f JL

M. (13.19)

Note also that

kI J = (tI , tJ). (13.20)

Using this form one can prove the following very useful theorem:

Theorem 13.2. Cartan’s criteria:

1. A Lie algebra is semi-simple if and only if det k 6= 0;

2. g semi-simple, real and compact if and only if k is negative definite.

Example 13.4. The algebra so(3) has non-vanishing structure constants f123 =

f312 = f23

1 = 1 and therefore k11 = f123 f13

2 + f132 f12

3, etc... so that

k =

−2−2

−2

. (13.21)

We see immediately that det k = −8 6= 0 and that it is negative definite. In factso(3) is semi-simple and compact.Example 13.5. In heis(3) the only non-vanishing structure constants are f12

3 =− f21

3 = 1. The Cartan–Killing matrix is then

k =

00

0

, (13.22)

which is obviously degenerate. The algebra is not semi-simple. It is straight-forward to see that a = span(t3) is a 1-dimensional vector space that is also anAbelian subalgebra of g:

[a, g] = 0 ⊂ a ⊂ g. (13.23)

Example 13.6. For so(2, 1) the structure constants are f123 = f13

2 = f231 = 1,

which gives

k =

2−2

2

. (13.24)

This is non-degenerate, but it is not definite. The algebra is semi-simple, butnot compact.

13.4 casimir operator 141

13.4 casimir operator

When k is non-degenerate, we can construct its inverse k−1, which we denotewith upper indices. In this case there is another interesting object that we canintroduce: the quadratic Casimir operator.

Definition 13.10. Quadratic Casimir operator

C2 ≡ kI J tI tJ .

This operator2 plays an important role in the analysis of representations andhas the property that

Theorem 13.3.[C2, X] = 0, ∀X ∈ g.

Proof. The theorem follows if we prove that C2 commutes with all the basisvectors tI .

[C2, tk] = kI J [tI tJ , tK] = kI JtI [tJ , tK] + kI J [tI , tK]tJ

= kI J f JKL tI tL + kJ I f IK

L tLtJ

= kI J f IKL (tJtL + tLtJ

)= kI J kLP f IKP(tJtL + tLtJ),

(13.25)

where we have introduced the completely antisymmetric tensor

f I JL ≡ f I JK kKL. (13.26)

This is completely antisymmetric, because f I JK = f[I J]K by definition, but alsof I JK = f JKI , as follows from

([tI , tJ ], tK) = tr(tI tJtK − tJtI tK) = tr(tI tJtK − tI tKtJ) = (tI , [tJ , tK]), (13.27)

where we used the definition of commutator as well as the cyclic property ofthe trace. The conclusion is that we are contracting a symmetric tensor with anantisymmetric one and hence the result is zero.

2 Formally C2 is a combination that does not live in the algebra g, but in the so-called universalenveloping algebra, defined as the associative algebra of polynomials in the elements in g. Sincewe will be interested mainly in matrix representations, also C2 will be a matrix obtained from theproduct of the matrices representing the generators.


exercises

1. Verify that (13.4) follows from the Jacobi Identity for three vectors defin-ing a basis of the algebra.

2. Compute the Cartan–Killing form for su(1, 1) and su(2) and show thatthey are different real forms of sl(2, C).

3. The algebra su(3) in the fundamental representation is determined byGell-Mann matrices

λ1 =

0 1 01 0 00 0 0

, λ2 =

0 −i 0i 0 00 0 0

, λ3 =

1 0 00 −1 00 0 0

,

λ4 =

0 0 10 0 01 0 0

, λ5 =

0 0 −i0 0 0i 0 0

, λ6 =

0 0 00 0 10 1 0

,

λ7 =

0 0 00 0 −i0 i 0

, λ8 =1√3

1 0 00 1 00 0 −2

,

via Ta =i2 λa so that T†

a = −Ta and tr(TaTb) = − 12 δab.

a) Compute fabc.

b) Show that su(3) has a proper su(2)⊕ u(1) subalgebra.

4. Let G = SO(2,2) and its algebra g = so(2, 2).

a) Provide dim g matrices for the basis of elements of g in the funda-mental representation.

b) Show that H = SO(2) × SO(2) < G and that the two elements a1, a2 ∈g corresponding to the generators of the two SO(2) commute.

c) Are there other elements of g that are not in the span of a1 and a2and that commute with them?

5. Let A = Sp(4, R) and B = SO(4).

a) Give their dimension and define their algebras in terms of matrices.

b) Discuss C = A ∩ B by analyzing a and b.

c) Compute the Cartan–Killing metric of the corresponding algebra c

and discuss if C can be simple, semi-simple and/or compact.

14R E P R E S E N TAT I O N S O F L I E A L G E B R A S

We have discussed at length representations of finite and of continuous groups.We would like now to discuss how to provide representations of Lie algebrasand how these representations may be connected to those of the underlyinggroup.

The first kind of representation we are going to discuss is provided by ma-trices, but obviously in physics we are often interested in other types of repre-sentations, either by differential operators or by quantum bosonic or fermionicoperators. For this reason we will also briefly discuss these instances, mainlyproviding examples for some particularly interesting cases.

14.1 morphisms and representations

When we want to provide a representation of an algebra g we need to map itselements to some matrices or operators, which fulfill the same commutationrelations that define g. More in general, we may be interested in maps betweenalgebras and see when we can map an algebra into another, preserving thecomposition operation.

Definition 14.1. An algebra morphism between the algebras g1 and g2 is a linearmap φ : g1 → g2 preserving the bracket operation

[φ(t1), φ(t2)]g2 = φ ([t1, t2]g1) , ∀t1, t2 ∈ g1.

Definition 14.2. An invertible morphism is an isomorphism.

Since Lie algebras are first of all vector spaces, invertible linear maps respect-ing the structure of the commutators are changes of basis.

143

144 algebra representations

As expected, we get isomorphic algebras from isomorphic groups, but notthe opposite:

G1 ' G2 ⇒ g1 ' g2, (14.1)

but

g1 ' g2 /⇒ G1 ' G2. (14.2)

Representations of an algebra are obtained when we consider morphisms ofan abstract algebra to an algebra on which we know how to perform explicitcomputations.

14.2 matrix representations

Definition 14.3. A matrix representation of an algebra g is a morphism

D : g→ gl(n, C).

Definition 14.4. We call the representation D : g → gl(n, C) faithful if it is aninjective morphism.

When dealing with matrices it is easy to relate group representations to al-gebra representations. We already saw in chapter 11 how to provide a matrixrepresentation (the fundamental representation) to many groups and in chapter12 how this generates a matrix representation for the algebra, by differentiating1-parameter subgroups of the original group. The two representations can alsobe related in the opposite direction by the exponential map:

exp [D(t1)] = D(exp[t1]), (14.3)

where t1 ∈ g, exp(t1) ∈ G and exp [D(t1)] is the standard matrix exponentia-tion.

An important point for our interest in physical applications is the followingtheorem:

Theorem 14.1. (Ado’s theorem) Every finite-dimensional Lie algebra admits a faith-ful matrix representation.

This implies that we can always represent finite-dimensional Lie algebrasusing matrices and therefore we can always map other types of representationsto matrix representations.

14.3 differential representations 145

As in the case of groups, we may have many different irreducible represen-tations for the same algebras. Most of them depend on the algebra one isstudying, but some can be constructed in the same way for any algebra. Onesuch representation is the adjoint representation. We already saw in section12.2.3 the definition of the adX operator, for X ∈ g. This is an operator actingon the vector space defined by the algebra g itself. We now show that it actuallyprovides a representation of the algebra (though not always faithful!).

Definition 14.5. The adjoint representation of an algebra g is provided by

adtI (·) ≡ [tI , ·], tI ∈ g.

This is a representation of dimension dim g because the vector space onwhich it acts is g itself and it is indeed a representation because

[adX , adY] = ad[X,Y], (14.4)

as follows straightforwardly from the Jacobi identity.It is also interesting to note that we can lift this definition to a representation

of the corresponding group by defining

Adg(tI) = gtI g−1, ∀g ∈ G, tI ∈ g. (14.5)

This is also a representation of dimension dim g and its tangent space is pre-cisely given by the adjoint representation of the algebra:

ddt

Adg(t)(tI)

∣∣∣∣t=0

=ddt

g(t)∣∣∣∣t=0

tI g(0)−1 + g(0)tIddt

g(t)−1∣∣∣∣t=0

=

= [X, tI ] = adX(tI),

(14.6)

where g(0) = 1 and X = ddt g(t)

∣∣∣t=0∈ g.

The adjoint representation of a Lie algebra has some resemblance with theregular representation of a finite group. In that case group elements wereused as a basis of the vector space on which the representation was actingand in this case we use the Lie algebra itself as the vector space on which therepresentation acts.

14.3 differential representations

We can often provide representations for Lie algebras in terms of differentialoperators, both in classical and quantum mechanics (with obvious limitations


on the spaces of functions on which they act). In the following we will notbe careful about the definition of a proper space of functions on which theseoperators act (be it L2(R) or C∞ or tempered distributions or else), but alwaysassume that the operations are legitimate. We are only interested in showinghow some of the most common groups and algebras in physics can be repre-sented in this way.

We already met the first example, which is the group of translations, isomor-phic to (R,+). A representation for its 1-dimensional algebra is given by thederivative of real functions:

D(t1) =ddx

. (14.7)

A representation for the group follows from the exponential map

D(exp(a t1)) = exp(

addx

)(14.8)

and its action on test functions realizes the group action:

exp(

addx

)f (x) = f (x + a). (14.9)

A group that is extremely useful in physics is the group of rotations in n-dimensional Euclidean space. The corresponding algebra is so(n) and is givenby the generators of such rotations. In general we can construct infinitesimalgenerators of rotations by means of the differential operators

Tij ≡ xi ∂

∂xj − xj ∂

∂xi . (14.10)

The antisymmetric double index ij = −ji is useful to create a 1 to 1 correspon-dence with the n(n − 1)/2 generators of the group and therefore provide afaithful representation. The commutator relations are

[Tij, Tkl ] = −δikTjl + δjkTil − δjlTik + δilTjk, (14.11)

as it can be easily verified by computing it on a test function f (~x) and recallingthat ∂ixj = δ

ji and noting that the terms with two derivatives acting on the

function f (~x) disappear because of antisymmetrizations.

14.4 bosonic representations 147

In the simplest case, namely so(2), we have only one such generator T =x ∂y − y ∂x and this can be simplified using polar coordinates x = r cos θ, y =r sin θ, leading to

T =∂

∂θ=

∂x(r, θ)

∂θ∂x +

∂y(r, θ)

∂θ∂y = −r sin θ ∂x + r cos θ ∂y. (14.12)

For so(3) we get the expected commutation rules if we define

Tij = −εijk lk, (14.13)

so that

[li, lj] = εijk lk. (14.14)

The same type of operators could be easily introduced for so(p, q), defining

Tij ≡ ηik xk ∂j − ηjk xk ∂i, (14.15)

where

η =

(1p

−1q

). (14.16)

The commutator of the generators (14.15) can also be computed rather easilyand gives

[Tij, Tkl ] = ηjkTil + ηilTjk − ηikTjl − ηjlTik. (14.17)

14.4 bosonic representations

The realization of Lie algebras using bosonic operators is very useful in Physicsin the resolution of problems like the harmonic oscillator in Quantum Mechan-ics and in Quantum models of rotational and vibrational models of nuclei andmolecules.

Assume that we have n bosonic operators bα, α = 1, . . . , n, satisfying thefollowing commutation relations

[bα, b†β] = δαβ, [bα, bβ] = [b†

α, b†β] = 0. (14.18)

We can think of these operators as creation and destruction operators for quan-tum states, starting from a vacuum satisfying bα|0〉 = 0.


By means of the defining relations (14.18) we can easily prove that

Tαβ ≡ b†αbβ (14.19)

provide a representation of the gl(n, C) generators, satisfying

[Tαβ, Tγδ] = δβγTαδ − δδαTγβ. (14.20)

This is clear if one notes that they are in 1-to-1 correspondence with the matricesrepresenting the same group. For n = 2 for instance

T11 =

(1 00 0

), T12 =

(0 10 0

),

T21 =

(0 01 0

), T22 =

(0 00 1

) (14.21)

satisfy exactly the same commutation relations and therefore we can establishan isomorphism between the two algebras. Note also that these generatorscan be further split into those of sl(n, C), whose matrix representation hasvanishing trace, plus the identity matrix, that commutes with all the others,providing a “linear Casimir” C1 = ∑i Tii.

Recall that sl(n, C) is the complexification of su(n) and gl(n, C) is the com-plexification of u(n), which are their real sections. This implies that we canrelate their representations by taking appropriate real linear combinations.

We now provide some examples that hopefully will clarify the role of thesesymmetry generators in Physics.

14.4.1 u(1)C ' gl(1, C)

Consider the 1-dimensional harmonic oscillator, whose Hamiltonian is

H =12

p2 +12

x2. (14.22)

Using the operator

b ≡ 1√2(x + i p) (14.23)

we can rewrite (14.22) as

H = b†b +12= T11 +

12

, (14.24)

14.4 bosonic representations 149

where, according to the discussion above, T11 is the generator of gl(1, C). Thisalgebra is obviously Abelian and hence all its irreducible representations are1-dimensional and labeled by the value of the action of the linear Casimir T11on the corresponding invariant subspaces of the Hilbert space. A basis of statesgenerating these invariant spaces is given by

|N〉 = 1√N!

(b†)N |0〉. (14.25)

Since the Hamiltonian is given in terms of the linear Casimir, we can immedi-ately deduce its eigenvalues and their degeneracies, computing its values onthe invariant spaces of the Hilbert space:

H|N〉 =(

N +12

)|N〉. (14.26)

14.4.2 u(2)C ' gl(2, C)

We now move to the 2-dimensional harmonic oscillator. In this case the Hamil-tonian is

H =12( p2

x + p2y) +

12(x2 + y2) = b†

1b1 + b†2b2 + 1, (14.27)

where

b1 ≡1√2(x + i px), b2 ≡

1√2(y + i py). (14.28)

Using bα and b†α operators we can construct the 4 generators of the gl(2, C)

algebra:

T11 = b†1b1, T12 = b†

1b2, T21 = b†2b1, and T22 = b†

2b2. (14.29)

A basis for the states of the Hilbert space is

|N〉 = 1N b†

α1. . . b†

αN|0〉, (14.30)

where N is some normalization factor, and they form an irreducible and com-pletely symmetric representation of dimension N + 1 (We will come back onthis in section 16.2.3). Once again we see that the Hamiltonian is a function of


the linear Casimir N = b†1b1 + b†

2b2. This implies that its eigenvalues and theirdegeneracies are also determined by such Casimir.

Note that usually one speaks of the symmetry generators of the harmonicoscillator as generators of su(n) rather than sl(n, C). The reason is that weare really interested in unitary transformations that preserve the orthonormal-ity of the state vectors and this poses obvious restrictions on the actual linearcombinations of the Tij operators that generate admissible symmetries.

14.5 fermionic representations

In certain instances creation and annihilation operators have to be taken withthe opposite statistic, i.e. they have to be (complex) fermionic operators, satis-fying

aα, a†β = aαa†

β + a†βaα = δαβ, (14.31)

and

aα, aβ = 0 = a†α, a†

β. (14.32)

Also in this instance

Tαβ = a†αaβ (14.33)

generate gl(n, C), but now the states forming a basis of the Hilbert space onwhich they act are in fully antisymmetric representations

|N〉 = 1N a†

α1. . . a†

αN|0〉. (14.34)

The check is straightforward also in this case, but pay attention to the commu-tators:

[Tαβ, Tγδ] = a†αaβa†

γaδ − a†γaδa†

αaβ

= a†αaβ, a†

γaδ − a†γaδ, a†

αaβ − a†αa†

γaβaδ + a†γa†

αaδaβ (14.35)

= δβγTαδ − δδαTγβ − a†α, a†

γaβaδ + a†γa†

αaδ, aβ

= δβγTαδ − δδαTγβ. (14.36)

14.5 fermionic representations 151

exercises

1. Prove that the Adjoint representation Dad(tK)I

J = fKJI for tI ∈ g is indeed

a representation of g, i.e. verify that

[Dad(tI), Dad(tJ)] = f I JKDad(tK).

2. What about (TI)JK = − f I J

K?

Explicitly: consider the matrices TI , with non-zero elements as given above andcheck the commutator [TI , TJ ].

3. What is the algebra defined by the operators Li = − i2 xα(σi)α

β∂β actingon functions f (x1, x2)?

4. LetD(Jµν) = −ηµρxρ ∂ν + ηνρxρ ∂µ, D(Pµ) = ∂µ,

where µ, ν, ρ = 0, 1, 2, 3. Find the structure constants of the resulting alge-bra.

5. Casimir operators for gl(n, C).

a) Show that the matrices (Tij)ab = δiaδjb, form a basis of gl(n, C) andcompute their structure constants.

Again: Tij are matrices with row and column indices a and b.

b) Compute the Cartan–Killing form in this basis and verify the algebraproperties using Cartan’s criteria.

c) Show that the elements of the enveloping algebra

Cr ≡ ∑1≤i1,i2,...,ir≤n

Ti1i2 Ti2i3 . . . Tir i1

commute with all the elements Tij and therefore they are Casimiroperators of degree r.

Try this at least for r = 1, 2, 3.

6. Important! Recall the definition of so(n) as given in (14.10). Show thatTij, where i, j = 1, . . . , n− 1 generate a so(n− 1) ⊂ so(n).

Focus now on the n = 4 case. In this case we can prove that so(4) =su(2)⊕ su(2).


Split the so(4) generators in L(1)i = T23, T31, T12 and L(2)

i = T14, T24, T34.

Compute [L(A)i , L(B)

j ].

Define L±i ≡12

(L(1)

i ± L(2)i

)and close the argument.

7. An exceptional group. (very long)

Let Jab, with a, b = 1, . . . , 7, be the 21 generators of so(7). Introduce thetotally antisymmetric tensor φabc, whose non-zero entries follow from

φ123 = φ516 = φ624 = φ435 = φ471 = φ673 = φ572 = 1.

Show that the generators surviving the projection ∑j,k φijk Jjk = 0 define asubalgebra of so(7) (i.e. consider the vector space generated by the linearcombinations of the Jij generators orthogonal to the 7 vectors defined by∑j,k φijk Jjk). Compute its structure constants and the Cartan–Killing form.Is this one of the algebras we studied so far (so(n), su(n), sp(n))?

15R E P R E S E N TAT I O N S O F S U ( 2 ) A N D S O ( 3 )

15.1 rotation group and its algebra

The relevance of the group associated to the invariance of a physical system un-der spatial rotations in 3 dimensions is obvious. This group can be determinedas the set of transformations preserving the norm of 3-dimensional vectors

~V → R~V, |~V|2 → ~VT RT R~V = |~V|2 (15.1)

and is isomorphic to SO(3), defined in 11.10. Its algebra, so(3), is defined bythe commutators

[li, lj] = εijk lk, (15.2)

where li, i = 1, 2, 3 form a basis of the so(3) vector space. In the fundamentalrepresentation, the generators li are represented by 3 × 3 antisymmetric realmatrices, fulfilling commutation relations isomorphic to (15.2):

D3(l1) =

00 −11 0

, D3(l2) =

0 10

−1 0

, D3(l3) =

0 −11 0

0

.

(15.3)

Using an explicit index notation for the matrix elements

D3(li)jk = −εijk. (15.4)

153

154 representations of su(2) and so(3)

15.2 isomorphism su(2) ' so(3) and homomorphism su(2) 7→ so(3)

The algebra of angular momenta, as already mentioned in previous lectures,is isomorphic to the su(2) algebra. We can easily show this by explicitly con-structing the latter and relating it to the one provided in (15.2).

The starting point is the fundamental matrix representation of SU(2). A 2 ×2 matrix U provides such a representation if it satisfies U†U = 1 and det U = 1(see definition 11.8). This means that we can parameterize U in terms of twocomplex numbers z, w ∈ C as follows:

U =

(z w−w∗ z∗

), with |z|2 + |w|2 = 1. (15.5)

As a byproduct, this parameterization shows that we can identify the manifoldunderlying the SU(2) Lie group with the 3-sphere:

SU(2) ' S3. (15.6)

In fact, rewriting z = x4 + i x3 and w = x2 + i x1, for xi ∈ R, we have that theconstraint in (15.5) becomes

∑i(xi)2 = 1, (15.7)

which is the defining equation of S3 as a hyper-surface embedded in R4.Starting from general U matrices as in (15.5), by differentiation we obtain

the su(2) algebra in the fundamental representation. This is the space of 2-dimensional matrices A satisfying A† = −A and tr A = 0. Pauli matrices canbe used to provide a basis for such a space:

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

). (15.8)

In fact these matrices satisfy

σ†i = σi and σiσj = i εijk σk + δij 12 (15.9)

and we can see that the combinations

D2(li) = −i2

σi (15.10)

15.2 isomorphism su(2) ' so(3) and homomorphism su(2) 7→ so(3) 155

are anti-hermitian D2(li)† = −D2(li) and have vanishing trace. We then con-clude that D2(li) is a 2-dimensional representation of su(2) whose commutatorsdefine the commutators of the algebra. It is also easy to see that the resultingstructure constants are isomorphic to the ones of so(3):

[D2(li), D2(lj)] = εijk D2(lk). (15.11)

We have then established that

su(2) ' so(3). (15.12)

We can now prove that by taking the exponential map of this 2-dimensionalrepresentation of su(2) we can generate back the 2-dimensional representationof SU(2) that covers the whole group.

A generic element of the algebra can be given in terms of the D2 representa-tion by A = − i

2 θi σi ∈ su(2). Its exponential gives

U = exp(A) =∞

∑n=0

(− i

2θi σi

)n 1n!

= 1− i2

θi σi −12

14

θiθ j σiσj + . . .

(15.13)

and it can be simplified using that

θiθ j σiσj =12

θiθ j σi, σj=

12

θiθ j δij 12, (15.14)

which implies

U = 12 −i2

θi σi −18

θiθ j δij 12 +13!

(− i

2θk σk

)(−1

4θiθ j δij

)+ . . .

= 12

(1− 1

2|~θ|2

4+ . . .

)− i

2θk σk

(1− 1

3!|~θ|2

4+ . . .

)

= 12 cos|~θ|2− i

θi

|~θ|σi sin

|~θ|2

.

(15.15)

Matching this expression with the one for U given in (15.5), we see that

x4 = cos|~θ|2

and ~x = −~θ

|~θ|sin|~θ|2

, (15.16)


so that ∑i(xi)2 = 1 and we cover the whole S3.Note that su(2) also has a 3-dimensional irreducible representation: the ad-

joint representation. This acts on the su(2) elements as

ad− i2~θ·~σ[σa] =

[− i

2~θ ·~σ, σa

]= θc εcab σb = σb (−θc εcba) = σb~θ ·D3(~l)ba, (15.17)

where D3(~l) provides its matrix representation. It is straightforward to checkthat this representation coincides with the fundamental representation of so(3).

As presented in (14.5), the adjoint representation of an algebra induces theadjoint representation for the corresponding group:

AdU [σa] = UσaU† = σb D3(U)ba. (15.18)

Explicitly

exp(− i

2~θ ·~σ

)σa exp

(i2~θ ·~σ

)= exp

(ad− i

2~θ·~σ

)[σa] =

= σa + ad(− i2~θ·~σ)σa +

12 ad(− i

2~θ·~σ)ad(− i

2~θ·~σ)σa + . . .

= σa + σb ~θ · D3

(~l)

ba+ σc

12~θ · D3

(~l)

cb~θ · D3

(~l)

ba+ . . .

= σb exp(~θ · D3

(~l))

ba.

(15.19)

Since SU(2) is locally parameterized by the coordinates θi, each element of SU(2)corresponds to a distinct value of θi. By using this fact we can now build a mapthat relates the 2-dimensional and the 3-dimensional representations of SU(2)and in turn elements of SU(2) with elements of SO(3):

exp(− i

2

(θ3 θ1 − i θ2

θ1 + i θ2 −θ3

))7→ exp

0 −θ3 θ2

θ3 0 −θ1

−θ2 θ1 0

. (15.20)

This map shows two important things:

• the adjoint representation of SU(2) is not faithful;

• SU(2) and SO(3) are homomorphic, but not isomorphic.

15.3 so(3) topology 157

In fact we can see that the map in (15.20) is not a bijection, because it is notinvertible. Let us consider the simplified case where θ1 = θ2 = 0. The maprelates elements of SU(2) and SO(3) as

(e−iθ3/2

eiθ3/2

)7→

cos θ3 − sin θ3

sin θ3 cos θ3

1

(15.21)

and this means that different elements in SU(2) may have the same adjointrepresentation and therefore be mapped to the same element in SO(3). A simpleinstance is obtained by looking at the elements associated to the values θ3 =0 and θ3 = 2π. While in SU(2) these two elements are distinct and the 2-dimensional faithful representation shows this explicitly, their 3-dimensionalrepresentation is the same. In fact in SO(3) there is a unique element associatedto such values of the parameters. Explicitly, at θ3 = 0 we have

D2(0, 0, 0) = 12 7→ D3(0, 0, 0) = 13, (15.22)

while at θ3 = 2π we have

D2(0, 0, 2π) = −12 7→ D3(0, 0, 2π) = 13. (15.23)

We conclude that the map (15.20) is not a bijection, but a 2 to 1 homomorphism,whose kernel is a subgroup of SU(2) given by Z2 = 12,−12.

As usual, we can restore an isomorphism by identifying group elements inthe kernel:

SO(3) ' SU(2)/Z2. (15.24)

15.3 so(3) topology

From (15.24) we see that topologically SO(3) is equivalent to the 3-sphere wherepoints are identified by a Z2 projection.


Z2

x4 < 0 x4 > 0

Figure 22.: Z2 map identifying points on the 3-sphere. Contractible and non-contractible cycles.

The 3-sphere can be described by its embedding in R4 as the set of pointssatisfying ∑i(xi)

2 = 1. Thus, for each value of |x4| the coordinates x1, x2and x3 span two 2-spheres of radius

√1− |x4|2. While each of these points

corresponds to a different element in SU(2), only half of them are distinct inSO(3). For instance, the map (15.20) shows that the identity matrix 12 appearswhen x1 = x2 = x3 = 0 and x4 = 1 and that −12 corresponds to the choicex1 = x2 = x3 = 0 and x4 = −1. On the other hand, both points are representingthe identity of SO(3). We then conclude that the action of Z2 identifies pointsof the 2-sphere with x4 > 0 to those of the 2-sphere with x4 < 0 (see figure 22

for a description of this fact). The resulting manifold is SO(3) and it is multiplyconnected, with first homotopy group

π1(SO(3)) = Z2. (15.25)

This implies that we have 2 classes of equivalence for loops in SO(3): the classof contractible loops and the class of loops that cannot be contracted to a point.

A pictorial explanation of this fact is the following. The SO(3) manifold is theset of rotations in 3 dimensions. Any element g ∈ SO(3) can be described by arotation of an angle between −π and π around an axis, oriented in a specificdirection in R3. Obviously rotations by π must be identified with rotations by−π. This means that we can describe SO(3) as a closed ball in 3 dimensionsof π radius, where opposite points on the surface must be identified. We cannow easily see that a loop that starts from a point on the boundary and runsin the interior up to the opposite point on the surface of the ball is a closedloop in SO(3) that can never be contracted to a point. On the other hand if wehave a loop that starts from a point on the boundary, runs in the interior up

15.4 irreducible representations of su(2) and so(3) 159

to another point on the boundary, different from the first, and does the samething on the opposite side, we can continuously deform it to a closed loop thatcan be contracted to a point by moving different points on the boundary on topof each other (see fig. 23).

Figure 23.: Contraction of a loop that passes twice through the boundary of the ballrepresenting SO(3). Red and green dots are points on the boundary, whichare identified. First we move the red dot and its image on top of the greenones and then we can contract the loop.

This is just a special instance of what happens for any SO(n) group. AllSO(n) spaces are not simply connected and their universal coverings, whichare simply connected, are the Spin groups Spin(n). We conclude that

Spin(3) ' SU(2). (15.26)

15.4 irreducible representations of su(2) and so(3)

What we just presented shows that SU(2) and SO(3) representations are relatedand that SU(2) representations provide representations for the universal cover-ing group of SO(3).

In Quantum Mechanics we are interested in unitary representations for thegroup elements and hermitian representations for the algebra. The homomor-phism between SU(2) and SO(3) shows that irreducible representations of SU(2)are identical to irreducible representations of SO(3) up to a sign. In fact, if westill want to insist and think of D2 as a representation of SO(3), we can thinkof it as a projective representation, whose composition law depends on a phase(which is invisible in the ray defining a state):

D2(g1)D2(g2) = exp(iφ(g1, g2)) D2(g1 g2), g1, g2 ∈ SO(3). (15.27)


What we can infer from this is that SO(3) admits genuinely projective represen-tations and that its covering group automatically takes them into account. Infact, for all practical applications we can actually study the representations ofthe latter. For this reason we will now discuss representations of SU(2) and ofits algebra rather than those of SO(3).

In general, a series of theorems by Bargmann1 shows that

Theorem 15.1. A simply connected group without continuous Abelian normal sub-groups does not admit genuinely projective irreducible representations.

This means that whenever the group we want to represent is not simplyconnected, we can use its covering group.

15.5 irreducible representations of su(2)

In order to study representations of su(2) it is useful to realize that su(2)C 'sl(2, C). In fact it is easier to construct representations for this algebra andthen deduce the form of the su(2) representations by taking appropriate linearcombinations sections.

We start by showing that su(2)C ' sl(2, C), indeed. If li, i = 1, 2, 3, are thesu(2) generators, we can construct a complex basis for its complexification bydefining the linear combinations

J± ≡ i (l1 ± i l2), J3 ≡ i l3. (15.28)

These generators satisfy

[J3, J±] = ±J±, [J+, J−] = 2 J3 (15.29)

and the isomorphism of this algebra with sl(2, C) is obvious in the 2-dimensionalrepresentation

D2(J3) =

( 12 00 − 1

2

), D2(J+) =

(0 10 0

), D2(J−) =

(0 01 0

), (15.30)

which is equivalent to the fundamental representation of sl(2, C), i.e. 2 × 2matrices with vanishing trace.

1 Again, we admit finite Abelian subgroups and therefore we do not use the word semi-simple,consistently with our definition.

15.5 irreducible representations of su(2) 161

In order to build a generic finite-dimensional irreducible representation, weneed a basis of vectors |m〉 spanning the corresponding invariant space V. Wecan always choose a basis in which one of the generators is diagonal. It is cus-tomary to use J3 for this purpose, so that the index m denotes its eigenvalues:

D(J3)|m〉 = m|m〉. (15.31)

We then note that J±|m〉 are still eigenstates of J3 with eigenvalues increased orreduced by 1 with respect to |m〉:

D(J3)D(J±)|m〉 = [D(J3), D(J±)]|m〉+ D(J±)D(J3)|m〉

= ±D(J±)|m〉+ m D(J±)|m〉

= (m± 1) D(J±)|m〉.

(15.32)

If the representation is finite-dimensional there must be a j such that D(J+)|j〉 =0 and a j− such that D(J−)|j−〉 = 0. The basis of the invariant subspace isthen constructed acting with D(J−) on |j〉 until we reach the state j−. In factD(J−)|j〉 ∼ |j− 1〉, D(J−)|j− 1〉 ∼ |j− 2〉, etc., up to the moment that we finda j− such that D(J−)|j−〉 = 0.

Another important point that fixes the representation is that tr D(J3) = 0,which follows from

tr D(J3) =12

tr ([D(J+), D(J−)]) = 0, (15.33)

where we used (15.32) in the first equality and the cyclic property of the tracein the second. We then conclude that

j + (j− 1) + (j− 2) + . . . + j− = 0, (15.34)

which implies that j− = −j and that the vector space has dimension 2j + 1,with basis

|j〉, |j− 1〉, . . . , | − j〉. (15.35)

Since the dimension of the vector space V must be an integer number, we alsoconclude that

j ∈N/2. (15.36)


The number j is called the spin of the representation and the vector space onwhich it acts is given by

V = span(|j, m〉), where m = j, j− 1, . . . ,−j. (15.37)

It is interesting to note that we can always find the spin of the representationof sl(2, C) by using an operator that commutes with the entire set of matricesproviding the representation. By means of Schur’s lemmas we know that onthe common invariant space V this must be proportional to the identity andtherefore have a unique value when acting on any of the vectors of V. Anoperator that has this property is the quadratic Casimir. In the sl(2, C) basisdefined by J3, J± satisfying (15.29), the Cartan–Killing form is

k =

44

2

(15.38)

and the quadratic Casimir is

C2 =12

(J23 +

12

J+ J− +12

J− J+

). (15.39)

Its representation D(C2) must be diagonal on V, i.e.

D(C2)|j, m〉 = λ(j) |j, m〉 (15.40)

for any m. Since D(J−)|j,−j〉 = 0, we can simplify the expression of C2 usingthe fact that J− J+ = J+ J− − 2J3 and therefore

D(C2)|j,−j〉 = 12

(D(J3)

2 − D(J3) + D(J+)D(J−))|j,−j〉 = 1

2j(j + 1)|j,−j〉.

(15.41)

We conclude that the Casimir operator on the 2j+ 1-dimensional representationof sl(2, C) has always the same value on V, which is

λ(j) =12

j(j + 1). (15.42)

Irreducible representations of sl(2, C) are then distinguished by the eigenvaluesof D(C2) and the basis of vectors in each of the invariant spaces is labelled bythe eigenvalues of D(J3).

15.6 matrix representations 163

15.6 matrix representations

The procedure we just outlined let us also find a (2j + 1)-dimensional matrixrepresentation of su(2) explicitly. We start from the basis of normalized vec-tors providing a basis for the invariant space of the corresponding irreduciblerepresentation of sl(2, C) and then take the linear combinations leading to theappropriate real form.

Define |j, j〉 as a norm one state:

||D2j+1(J−)|j, j〉 ||2 = 〈j, j|D2j+1(J+)D2j+1(J−)|j, j〉

= 〈j, j|[D2j+1(J+), D2j+1(J−)]|j, j〉

= 〈j, j|2 D2j+1(J3)|j, j〉 = 2j.

(15.43)

We can then define

|j, j− 1〉 ≡ 1√2j

D2j+1(J−)|j, j〉. (15.44)

We can then proceed in the same way for the other states and eventually definethe form of D2j+1(la) as

D2j+1(l3) = −i D2j+1(J3), (15.45)

D2j+1(l1) = −i2(

D2j+1(J+) + D2j+1(J−))

(15.46)

and

D2j+1(l2) = −12(

D2j+1(J+)− D2j+1(J−))

(15.47)

Note that when j ∈N we also have a representation for so(3).

Example 15.1. The j = 1/2 representation has highest state | 12 , 12 〉. We start by

D2(J3)| 12 ,± 12 〉 = ±

12 |

12 ,± 1

2 〉. (15.48)

We then have that

| 12 ,− 12 〉 ≡ D2(J−)| 12 , 1

2 〉 (15.49)


and

D2(J−)| 12 ,− 12 〉 = 0. (15.50)

Analogously

D2(J+)| 12 , 12 〉 = 0, (15.51)

and

D2(J+)| 12 ,− 12 〉 = D2(J+)D2(J−)| 12 , 1

2 〉

= [D2(J+), D2(J−)]| 12 , 12 〉

= 2D2(J3)| 12 , 12 〉 = |

12 , 1

2 〉.

(15.52)

If the generic vector in Vj= 12

has two components and | 12 , 12 〉 is the upper com-

ponent and | 12 ,− 12 〉 is the lower component, the corresponding representation

2 is given by

D2(J3) =

( 12 00 − 1

2

), D2(J+) =

(0 10 0

), D2(J−) =

(0 01 0

), (15.53)

which in fact is the same as

D2(li) = −i2

σi. (15.54)

exercises

1. Build the 8-dimensional adjoint representation of su(3). Take the 3 ma-trices related to su(2) ⊂ su(3) and show that they provide a reduciblerepresentation of su(2).

2. Build the irreducible representation of dimension 5 for so(3).

a) Start from the basis of eigenstates of its complexified algebra |2j +1 = 5, m〉, with m = −2, . . . , 2 and build D5(J3).

b) Using the properties of J± and their action on the same states buildD5(J±) and D5(J1), D5(J2).

c) Build the corresponding representation for the quadratic Casimir op-erator D5(C2).

16I R R E D U C I B L E R E P R E S E N TAT I O N S O F S U ( N ) A N D T H E I RP R O D U C T S

There are many groups relevant to physics, but a special role is played by SU(N)groups. These groups appear naturally in Quantum Mechanics, for instance inthe analysis of the harmonic oscillator and of rotational symmetries, but theyare also at the core of the so-called Standard Model of Particle Physics. Thismodel is in fact a gauge theory coupled to matter, with gauge group SU(3)×SU(2)×U(1) and fundamental particles are sitting in specific representationsof these groups. In addition, there are approximate global symmetries that alsoplay a role in the analysis of fundamental processes, which are again describedby groups isomorphic to SU(N).

We are not just interested in constructing representations of SU(N) groups,but also in understanding their products, because it is often useful to under-stand how a composite system transforms once the transformation propertiesof its components are known. Thus, after a brief general discussion, we presentin detail SU(N) irrepses and their products using Young Tableaux. We stressthat, although some details of the techniques we use here are specific to SU(N)groups, most of the discussion can be repeated and extended to other groups.

16.1 products of irreducible representations

16.1.1 Tensor products

A method that allows to build all irreducible representations (irrepses) of agroup starting from its fundamental representation is that of considering theirproducts and their decompositions according to their invariant subspaces.

165

166 su(n) irreps and products

The product of irrepses of a given group or algebra provides a new represen-tation for the same group or algebra. The resulting representation, however, isin general reducible, unless one of the two irrepses has dimension 1. In orderto obtain a reduction of the product according to its invariant subspaces it isuseful to work at the algebra level and then exponentiate the fragments result-ing from the decomposition. For SU(N) groups we are sure to cover the wholegroup by doing so, because SU(N) is compact, simply connected and has nocontinuous Abelian normal subgroups.

Definition 16.1. Let Da be a representation of a group (algebra) acting on thevector space Va and Db a representation acting on Vb, their product Da⊗Db actson Va ⊗Vb as

(Da ⊗ Db)(g)[va ⊗ vb] = Da(g)[va]⊗ Db(g)[vb],

for g ∈ G (or g ∈ g) and va,b ∈ Va,b.

16.1.2 Clebsch–Gordan decomposition

The result of a product of representations is not irreducible in general. When itis also reducible (and not just decomposable), we can decompose it in irrepses.This is always the case for compact Lie groups.

The decomposition of the product takes the name of Clebsch–Gordan decom-position:

Definition 16.2. Clebsch–Gordan decomposition

D⊗ D′ = ⊕jDj.

If G is finite or if it is a compact Lie group and its irrepses have been classified,we can use the multiplicity ma of each irreducible fragment to write

D⊗ D′ = ⊕a ma Da. (16.1)

Obviously we have that their characters are also related by

χD⊗D′(g) = χD(g) χD′(g) = ∑a

ma χDa(g) (16.2)

and their dimensions (take the previous formula for g = 1) by

ND⊗D′ = ND · ND′ = ∑a

maNa. (16.3)

16.2 su(n) irrepses and young tableaux 167

This decomposition is in 1 to 1 correspondence with the decomposition ofthe vector space on which the same linear representations act. In detail, if|a, m〉 ∈ Va and |b, n〉 ∈ Vb, with m = 1, . . . , dim Va and n = 1, . . . , dim Vb aretheir bases, we can decompose the vectors in the product space as follows:

|a, m; b, n〉 ≡ |a, m〉 ⊗ |b, n〉 = ∑c,i,p|ci, p〉〈p, ci|a, m; b, n〉, (16.4)

where c is a label for the invariant subspace for the related irreps, i runs overthe number of times the same representation is repeated and p is the indexlabeling the basis vectors. The coefficients Cci ,p|a,m,b,n = 〈p, ci|a, m; b, n〉 arecalled the Clebsch–Gordan coefficients.

16.1.3 Decomposition in irrepses of a subgroup

Although it is not going to be addressed here, it is useful to note that thereis another case of interest where a given representation can be decomposedin irreducible fragments. This happens in the case of a group H < G, whereirrepses of H are constructed by restricting those of G to the elements in H.In this case D(h) is not in general irreducible, even if D(g) is irreducible. Forinstance, the dimension 2 irreps of the group D3 induces a representation ofZ2 < D3 which is reducible, because Z2 is Abelian.

16.2 su(n) irrepses and young tableaux

Without giving too many technical details, we now discuss irrepses of the Spe-cial Unitary group SU(N).

16.2.1 Fundamental, anti-fundamental and adjoint representations

The fundamental representation of SU(N) is defined by matrices U ∈ GL(N, C)that fulfill U†U = 1N and det U = 1. These linear operators act on a vectorspace V = CN , mapping vectors v = viei ∈ V (where ei is a basis for V) as

vi 7→ Uij vj, (16.5)


where i, j = 1, . . . , N label rows and columns of the matrix U. It is customaryto label irreducible representations with their dimension and therefore we havethat

Fundamental Representation ⇔ N. (16.6)

Once we proved that the matrix U(g) provides a good representation of theelement g ∈ SU(N), we see that also U∗(g) provides a representation for thesame element:

(U∗)†U∗ = (U†U)∗ = 1∗N = 1N (16.7)

and

det U∗ = (det U)∗ = 1. (16.8)

For N > 2 this is an inequivalent representation called anti-fundamental, whichwe can represent by its action on w ∈ CN as

w∗ ı 7→ U∗ ı w∗ . (16.9)

Once more, it is customary to refer also to this representation using its dimen-sion, though in this case we also use a bar to distinguish it from the fundamen-tal representation and stress that it can be obtained from the former by complexconjugation:

Anti-Fundamental Representation ⇔ N. (16.10)

Using the scalar product in CN we also see that

w†v = w∗ıvi δiı = w∗i vi (16.11)

and we can characterize the N representation by the action on covariant vectors,whose index has been lowered by means of δiı:

w∗i 7→ U∗ ij w∗j . (16.12)

We now move to the adjoint representation. The product of the two irrepsesN ⊗ N can be described by the set of operators acting on the vector spacedefined by rank 2 tensors as

vi ⊗ w∗j ≡ Mij 7→ Ui

kU∗ jl Mk

l . (16.13)


As expected, this is not an irreducible representation. In fact the scalar productof the elements is invariant (i.e. it transform in the trivial representation)

w†v 7→ w∗k U∗ ik Ui

l vl = w† U†U v = w†v (16.14)

and thus we can decompose the tensor Mij into its trace and its traceless parts:

Mij =

1N

δij tr M+

M i

j, with

Mkk = 0. (16.15)

In summary

N⊗N = 1⊕ (N2 − 1), (16.16)

where the first one is the trivial representation and the second one is the adjoint.We recognize that this is really the adjoint representation by the fact that thedimension is the correct one (dimRSU(N) = N2 − 1) and by the fact that the

action on

M is precisely the one expected by the Adjoint representation, whichtakes vectors in the algebra and acts on them by elements of the group fromthe left and inverse elements from the right:

AdU(

M) = U

M U−1 = U

M U†. (16.17)

As we said in the introduction, we would like to see how all irrepses canbe generated by taking products of the fundamental representation. We startby giving a simple example, where we consider rank 2 tensors built from 2

fundamental representations. These tensors transform as

Mij ≡ vi ⊗ wj 7→ UikU j

l Mkl . (16.18)

They are obviously always decomposable in the symmetric and antisymmetricparts

Mij = Aij + Sij = −Aji + Sji, (16.19)

but what is interesting to note is that these fragments define invariant sub-spaces under the action of SU(N):

Aij ′ = UikU j

l Akl = −UikU j

l Alk = −U jlUi

k Alk = −Aji′, (16.20)

Sij ′ = UikU j

l Skl = UikU j

l Slk = U jlUi

k Slk = Sji′. (16.21)


We conclude that

N⊗N =N(N− 1)

2⊕ N(N + 1)

2. (16.22)

Using this tensorial method we can extract all irreducible components in aproduct, but it is generically too elaborate and complicated. However, a muchmore effective method can be put in place by using Young Tableaux.

16.2.2 Irrepses and Young Tableaux

Tensorial representations can be decomposed in invariant parts using theirproperties under permutations of their indices. Since conjugacy classes andirreducible representations of the group of permutations are effectively repre-sented by Young Tableaux, we can apply this technique to decompose productsof SU(N) irrepses. Each irreps is obtained as a product of k fundamental ir-repses, projected on the irreducible part that is selected by the elements of thepermutation group represented by a specific Young tableau.

The fundamental representation acts on a single vector and can be repre-sented by the unique element of S1:

N ⇔ . (16.23)

Other irrepses follow from its k-fold product, with symmetrization and anti-symmetrization rules determined by the Young tableau we use to represent it.For instance the completely symmetric representation is

· · ·︸︷︷︸k

⇔ Ti1 ...ik = T(i1 ...ik) ⇔ (N + k− 1)!(N− 1)!k!

. (16.24)

and the fully antisymmetric one is

· · ·

k ⇔ Ti1 ...ik = T[i1 ...ik ] ⇔ N!

(N− k)!k!. (16.25)

There is an obvious limit on the number of boxes one can pile in vertical,determined by the fact that in CN we have at most N different indices. This


case is unique and it is represented by the totally antisymmetric tensor, whichtransforms trivially:

εi1 ...iN 7→ Ui1 k1 . . . UiN kN εk1 ...kN = det U εi1 ...iN = εi1 ...iN . (16.26)

Using this same tensor we can also construct an object transforming in the anti-fundamental representation, starting from N − 1 vectors in the fundamentalrepresentation. In detail, we can define

w∗i1 ≡ εi1 ...iN vi2 . . . viN , (16.27)

which transforms as

w∗i1 7→ U∗ i1k1 εk1 ...kN U∗ i2

k2 . . . U∗ iNkN Ui2

l2 . . . UiN lN vl2 . . . vlN . (16.28)

Using U†U = 1 the right hand side is the same as

U∗i1k1 w∗k1

≡ U∗i1k1 εk1 ...kN vk2 . . . vkN . (16.29)

We therefore see that N − 1 vertical boxes correspond to the N representation.Other tensors will be related to mixed symmetry diagrams with k boxes,

whose permutations can be applied to rank k tensors. In detail, when consider-ing such diagrams we construct irrepses by defining tensors whose symmetryproperties are determined by the projector

P = aTsT , (16.30)

where aT and sT are the antisymmetric and symmetric permutation operatorsdefined in Theorem 8.1. Obviously the result may depend on the order of theindices, but it is easy to see that because of the symmetry properties of theoperator we acted with, we can always exchange indices in the same row andremove diagrams where in one of its columns we have the same index repeated.For instance, SU(2) tensors have indices with i = 1, 2 and applied to a tensorwith 2 indices has only 3 inequivalent combinations because

1 2 = 2 1 . (16.31)

The other tableau with two boxes, applied to the same tensor gives theremaining combination

12

= − 21

, 11

= 0 , 22

= 0 . (16.32)


The number of inequivalent possibilities is then represented by standard tableaux.A standard tableau is a tableau where the index numbers do not decrease whengoing from left to right in rows and always increase from top to bottom incolumns. Non-standard tableaux give tensors that, by symmetrization or anti-symmetrization either vanish or are not independent of the standard tableaux.By using some easy combinatorics, we can argue that the dimension of eachof the irrepses can be computed using the formula that gives the number ofstandard tableaux associated to a given configuration:

D = ∏boxes i,j

N − i + jh(i, j)

, (16.33)

where i labels the row, j labels the column and h(i, j) is the corresponding hooknumber.

Example 16.1. The SU(4) Young Tableau has dimension D = 140. ForN = 4 the numerator is given by the product of the coefficients in

4 5 6 73 4 52 3

(16.34)

and the denominator is given by the products of the coefficients in

6 5 3 14 3 12 1

. (16.35)

Example 16.2. We apply this decomposition technique to rank 3 tensors. Ageneric rank 3 tensor has 3 indices without any definite symmetry property.Once we have a basis for the vector space VN where the fundamental represen-tation acts, we can construct a basis of the vector space where rank 3 tensorsare defined, namely V⊗3N = VN ⊗VN ⊗VN, by taking their products, so that

T = Tijk ei ⊗ ej ⊗ ek. (16.36)

In order to split T into its irreducible fragments we build the projector operatorsassociated to the different representations of S3. There are a total of 6 possible


states and 4 different projectors: P , P , P 1 23

and P 1 32

, associated to the

different standard tableaux. Explicitly:

P = 1 + (12) + (13) + (23) + (123) + (132), (16.37)

P = 1− (12)− (13)− (23) + (123) + (132), (16.38)

P 1 23

= 1 + (12)− (13)− (123), (16.39)

P 1 32

= 1 + (13)− (12)− (132). (16.40)

Their action on the basis of the vector space where rank 3 tensors are defined,splits V⊗3N into invariant subspaces whose basis elements are linear combina-tions of the original ones (we omit the tensor product symbol ⊗ for the sake ofhaving more compact expressions):

Sijk =16(eiejek + eiekej + ejekei + ejeiek + ekeiej + ekejei

),

Aijk =16(eiejek − eiekej + ejekei − ejeiek + ekeiej − ekejei

),

M1,1 ijk =12(eiejek + ejeiek − ekeiej − ekejei

), (16.41)

M1,2 ijk =1

2√

3

(eiejek − 2 eiekej − 2 ejekei + ejeiek + ekeiej + ekejei

),

M2,1 ijk =12(eiejek − ejeiek − ejekei + ekejei

),

M2,2 ijk =1

2√

3

(eiejek + 2 eiekej − 2 ejekei − ejeiek + ekeiej + ekejei

).

The labels S, A and M stand for symmetric, antisymmetric and mixed sym-metry, which are the symmetry properties of the corresponding basis. Theprojectors in (16.37)–(16.40) select these invariant spaces: P projects into S,P projects into the 1-dimensional space spanned by A and finally, P 1 2

3and

P 1 32

project into the 2-dimensional spaces spanned by M1 and M2, respectively.


A generic tensor can be decomposed by considering all the fragments that re-main when contracting its Tijk components with the various basis elements.For instance, the fully symmetric and fully antisymmetric combinations are

TijkSijk =16

(Tijk + Tikj + Tkij + T jik + T jki + Tkji

)ei ⊗ ej ⊗ ek,

Tijk Aijk =16

(Tijk − Tikj + Tkij − T jik + T jki − Tkji

)ei ⊗ ej ⊗ ek, (16.42)

while the mixed symmetry combinations give

Tijk M1,1 ijk =12

(Tijk + T jik − T jki − Tkji

)ei ⊗ ej ⊗ ek,

Tijk M1,2 ijk =1

2√

3

(Tijk − 2Tikj − 2Tkij + T jik + T jki + Tkji

)ei ⊗ ej ⊗ ek,

Tijk M2,1 ijk =12

(Tijk − T jik − Tkij + Tkji

)ei ⊗ ej ⊗ ek, (16.43)

Tijk M2,2 ijk =1

2√

3

(Tijk + 2Tikj − 2Tkij − T jik + T jki + Tkji

)ei ⊗ ej ⊗ ek.

One thing, however, that needs to be stressed here is that while M1 and M2belong to different invariant subspaces of the S3 basis space, they give tensorsassociated to the same representation of the permutation group on their indicesand therefore they provide equivalent representations of SU(N). In fact they allfulfill the same conditions

TijkMa

+ T jkiMa

+ TkijMa

= 0, a = 1, 2, (16.44)

which remove the fully symmetric TS and fully antisymmetric TA representa-tions, and have the symmetry property

TijkM1

= −TkjiM1

, TijkM2

= −T jikM2

. (16.45)

Clearly the fact that the antisymmetry is on the first two indices or in the firstand third, or, for that matter, in the second and third, gives obviously the sameSU(N) representation.

Example 16.3. The same kind of decomposition is also useful to analyze invari-ant subspaces of the Hilbert space of k identical particles transforming under


the action of the fundamental representation of the SU(N) group. The Hilbertspace for each of these particles is HN = CN and the total Hilbert space isH = (HN)

k, which we now decompose using Sk irrepses.For the sake of simplicity we focus on the decomposition of the 8-dimensional

Hilbert space of k = 3 identical particles, transforming as a doublet of SU(2),i.e. N = 2. We denote the basis of the 2-dimensional space on which the fun-damental representation of SU(2) acts by |+〉 and |−〉. The generic element ofthe 8-dimensional space H = H2⊗H2⊗H2 is then a linear combination of thestates labelled by | ± ±±〉 = |±〉 ⊗ |±〉 ⊗ |±〉, where each of the signs refersto the corresponding sign in the choice of the basis elements for each of the2-dimensional Hilbert spaces of the 3 identical particles.

Since k = 3, the SU(2) irrepses acting on H are in correspondence with the

Young Tableaux with 3 boxes: , and . According to (16.33), theirdimensions are 4, 0 and 2, respectively. The decomposition of H into invariantsubspaces follows now from the application of the projectors associated to theseYoung Tableaux as in the previous example.

The fully symmetric tensor is unique in S3 (dS3 = 1) and the application ofthis operator to H selects 4 different invariant states, classified according totheir total spin:

|+++〉, 1√3(|++−〉+ |+−+〉+ | −++〉) ,

1√3(|+−−〉+ | −+−〉+ | − −+〉) , | − −−〉.

(16.46)

The totally antisymmetric tensor has no states.The mixed tensor has dS3 = 2 (there are 2 independent permutations) and

has dimension 2, because we have 2 different states in each invariant space.The corresponding states in H are (again according to their total spin)

1√2(|++−〉 − | −++〉) , (16.47)

12(|++−〉+ | −++〉 − 2|+−+〉) , (16.48)

1√2(| − −+〉 − |+−−〉) , (16.49)

12(| − −+〉+ |+−−〉 − 2| −+−〉) . (16.50)


Note that the states with the same spin transform in one of the irreduciblerepresentations ψMa of S3. In conclusion dimH = 8 = d · D + d ·D = 1 · 4 + 2 · 2.

As a general rule one can prove that

Theorem 16.1. SU(N) irrepses are in 1 to 1 correspondence with Young Tableauxwith a number of rows lower than N.

The dimension of the irreps is determined by the formula (16.33) and thenumber of times it appears is determined by the dimension as a representationof the corresponding Symmetric group.

This classifies completely irreducible representations. We can now move onand analyze how to explicitly compute products of representations for the samegroup.

16.2.3 Products using Young Tableaux

The analysis given in the previous section for the construction of the irreduciblerepresentations of SU(N) suggests that Young Tableaux should be used effec-tively to compute products of representations. In this case we do not discussthe derivation of the rules that determine how to construct this product andobtain the Clebsch–Gordan decomposition, but rather explain the result andprovide some examples that can convince us of the effectiveness of the adoptedtechnique.

Theorem 16.2. The Clebsch–Gordan decomposition for products of representations inSU(N) can be represented by means of Young Tableaux using the following rule:

1. Label the second tableau in the product with the letters a, b, c, . . . in the 1st, 2nd,3rd, etc. row:

a a a ab b bc c c

. . .

(16.51)

2. Attach all the boxes with an a to the first tableau in all possible different ways, butsuch that there are never two a in the same column and such that the resultingdiagram is still a legitimate Young tableau.


3. Repeat the procedure for the boxes with b, c etc...

4. Read all tableaux in the outcome from right to left and from top to bottom anddelete all those associated to strings that have more b than a to the left of anyletter or more c than b etc.

Example 16.4. In SU(2) the fundamental and antifundamental representationscoincide

2 ∼ 2⇔ . (16.52)

The product gives

⊗ = ⊕ ⇔ 2× 2 = 1 + 3. (16.53)

Since corresponds to the trivial representation, the only irreducible represen-tations that can be obtained are those with k boxes in a row, whose dimensionis k + 1. This is easily seen in the product of two 3-dimensional representations:

⊗ a a = a a + aa

+a a

= + + · , (16.54)

which can be represented by

3× 3 = 5 + 3 + 1. (16.55)

Example 16.5. In SU(3) the fundamental representation 3 and the anti-fundamentalrepresentation 3 are represented by two different tableaux:

3⇔ and 3⇔ . (16.56)

Using the first 3 rules in 16.2, their product gives

⊗ ab

=

(a +

a

)b = a b + a

b+ b

a+ a

b. (16.57)

Using the fourth rule in 16.2, the 4 resulting tableaux correspond to the stringsba, ab, ba and ab respectively. This means that the first and third tableau haveto be discarded and we are left with

⊗ = + ·, (16.58)


which corresponds to

3× 3 = 8 + 1, (16.59)

in agreement with the result discussed in (16.16).We can also compute the product of two fundamentals, which is the same

in terms of Young Tableaux as the one for SU(2), but corresponds to differentdimensionality of the corresponding irreducible representations:

⊗ = ⊕ ⇔ 3⊗ 3 = 6⊕ 3. (16.60)

Another interesting lesson we can take home from these examples is thatYoung Tableaux corresponding to conjugate irrepses have complementary shapes.This means that complex conjugation of a given representation is representedby the complementary tableau, rotated by 180 degrees.

8 = = = 8 = 8

6 = = = 6

Figure 24.: Conjugate representations in SU(3). The adjoint representation 8 is real andits conjugate Young Tableaux has the same shape as the original one. Onthe other hand, the representation 6 is complex and its complex conjugateis represented by a different tableau.

16.2.4 Again on SU(2)

We conclude this lecture, by reminding that you already encountered the Clebsch–Gordan decomposition for SU(2) in physical problems related to spin and an-gular momentum. In that case it is also useful to it may be useful to obtainthe exact multiplicative coefficients in the decomposition and this was done asfollows.


Let Dj1 and Dj2 two irrepses of SU(2)

Dj1(g)⊗Dj2(g) = e~ψ·~J(1) ⊗ evecψ·~J(2) = 1+ ~ψ

(~J(1) ⊗ 1 + 1⊗~J(2)

)+ . . . (16.61)

We can define the product representation of the algebra generators as

~J = ~J(1) ⊗ 1 + 1⊗~J(2). (16.62)

The spaces on which the irrepses are defined, namely V1 and V2, admit a or-thonormal basis of eigenstates of J3

(1) and J3(2):

|j1, m1〉 ∈ V1, |j2, m2〉 ∈ V2. (16.63)

Obviously J3 is diagonal in this basis:

D(J3)(|j1, m1〉 ⊗ |j2, m2〉) = (m1 + m2)(|j1, m1〉 ⊗ |j2, m2〉). (16.64)

We can then discuss J±. Now the states are mixed by the operators and, startingfrom the highest spin state, whose spin is j1 + j2, one generates 2 states withspin j1 + j2 − 1:

|j1, j1〉 ⊗ |j2, j2 − 1〉 (16.65)

and

|j1, j1 − 1〉 ⊗ |j2, j2〉. (16.66)

This implies that we have a non-trivial degeneracy in the interval [−|j1 −j2|, |j1 − j2|]. In particular we have 2 min(j1, j2) + 1 states.

The natural basis of the product is given in terms of the basis vectors byunitary matrices whose coefficients are Clebsch–Gordan coefficients:

|j, m〉 = ∑j1,m1,j2,m2

Cj,m,j1,m1,j2,m2 |j1, m1〉 ⊗ |j2, m2〉. (16.67)

Example 16.6. We show that

D1 ⊗ D1/2 = D3/2 ⊕ D1/2. (16.68)

The spin 3/2 state is unique

| 32 , 32 〉 = |1, 1〉 ⊕ | 12 , 1

2 〉. (16.69)


We apply J− to this:

D(J−)| 32 , 32 〉 =

√3| 32 , 1

2 〉 =√

2(|1, 0〉 ⊗ | 12 , 12 〉) + |1, 1〉 ⊗ | 12 ,− 1

2 〉. (16.70)

We deduce that

| 32 , 12 〉 =

√23(|1, 0〉 ⊗ | 12 , 1

2 〉) +1√3(|1, 1〉 ⊗ | 12 ,− 1

2 〉). (16.71)

Applying once more D(J−) we get

| 32 ,− 12 〉 =

1√3(|1,−1〉 ⊗ | 12 , 1

2 〉) +√

23(|1, 0〉 ⊗ | 12 ,− 1

2 〉), (16.72)

| 32 ,− 32 〉 = |1,−1〉 ⊗ | 12 ,− 1

2 〉. (16.73)

In addition, to complete the expected degeneracy, we should also consider thestates orthogonal to | 32 , m〉:

| 12 , 12 〉 =

1√3(|1, 0〉 ⊗ | 12 , 1

2 〉)−√

23(|1, 1〉 ⊗ | 12 ,− 1

2 〉), (16.74)

| 12 ,− 12 〉 =

√23(|1,−1〉 ⊗ | 12 , 1

2 〉)−1√3(|1, 0〉 ⊗ | 12 ,− 1

2 〉), (16.75)

where obviously | 12 ,− 12 〉 = D(J−)| 12 , 1

2 〉.

exercises

1. Product of two fundamental representations of SO(4).

a) The first step requires to build the tensor Mij = Vi ⊗W j where i, j =1, 2, 3, 4. If V and W transform in the fundamental representationVi 7→ Ri

jV j and Wi 7→ RijW j, we have that Mij 7→ Ri

kRjl Mkl . Build

the corresponding matrix representation for the product (this is a 16

× 16 matrix).

b) We now show that this representation is not irreducible.

Show that the subspaces generated by symmetric matrices Sij = Sji

and by antisymmetric matrices Aij = −Aji are invariant subspacesof dimensions 10 and 6.


c) Show that the space of symmetric tensors can be further reducedinto the sum of two irreducible representations of dimensions 9 and1.

d) Show that also the space of antisymmetric tensors can be furtherreduced as follows:

Aij+ = 1

2 εijkl Akl+;

Aij− = − 1

2 εijkl Akl−.

2. We can repeat the analysis of the previous problem by using the isomor-phism so(4) ' su(2)L ⊕ su(2)R established in an exercise of previouschapter. (We use here L, R to distinguish the two copies of su(2) in corre-spondence with L+

i and L−i ). Use the symbol (m, n) = Dm(gL)⊗ Dn(gR)for the representation of the elements (gL, gR) ∈ SU(2)L × SU(2)R bymeans of matrices of dimensions m and n.

Show that the fundamental representation of SO(4) is a couple of dou-blets of SU(2) and compute their products using the properties of SU(2)products.

a) Compute L(1)i on (2, 2) and note that it contains a triplet and a scalar

(the fourth component).

b) Compute (2, 2)⊗ (2, 2) and decompose the result in its irreducibleparts.

c) Relate this result to the one obtained in the previous exercise.

3. Describe the decomposition of (H2)5 in irreducible representations of S5

and SU(2).

4. Describe the decomposition of (H3)4 in irreducible representations of S4

and SU(3).

5. Using Young Tableaux describe the 6 representations of SU(4) of dimen-sions lower or equal to 10 and consider their products.

17T H E P O I N C A R E G R O U P

We already discussed some direct applications of group theory to Physics,mainly focusing on Quantum Mechanics. So far, however, we concentratedon simple instances where the only spacetime symmetry was related to thegroup of rotations or to its (discrete) subgroups. On the other hand, we are ob-viously interested in descriptions of systems where one of the basic principlesof modern Physics is taken into account: the principle of Relativity. By thisprinciple, observers in different inertial frames must have equivalent physicaldescriptions of the same phenomena.

Inertial frames measure the same spacetime interval between any two events

ds2 = −(dx0)2 + (dx1)2 + (dx2)2 + (dx3)2 = dxµ ηµν dxν , (17.1)

where

η =

−1

11

1

(17.2)

is the Minkowski metric. Coordinate transformations relating different inertialframes should then preserve ds2 and have the form

xµ′ = Λµνxν + aµ, where a ∈ R4, ΛTηΛ = η. (17.3)

The point of interest for us is that coordinate changes of the form (17.3) forma group: the Poincare group. This fact tells us that all relevant physical quan-tities in a relativistic theory must transform in definite representations of thisgroup and that therefore we can classify quantum relativistic states by quantumnumbers that specify irreducible representations of the Poincare group.

183

184 poincare group

Definition 17.1. The Poincare group is the set(Λ, a) ∈ Mat(4, R)×R4 | ΛTηΛ = η

with composition law

(Λ2, a2) (Λ1, a1) = (Λ2Λ1, Λ2a1 + a2),

where matrix product and the action of matrices on vectors is understood.

We leave to the reader the check that this composition law satisfies all groupaxioms.

The Poincare group contains several interesting subgroups. The first andmost interesting one is the subset of elements of the form (Λ, 0), which is calledthe Lorentz group. Other interesting subgroups are the group of translations,whose matrices have the special form (14, a), and the group of rotations, whosematrices are of the form (R, 0), with R0

0 = 1, R0i = Ri

0 = 0 and RijRk

j = δik.Any rotation matrix can actually be parameterized by the unit norm vector nfixing the axis of rotation and by the rotation angle θ:

Rij = δi

j cos θ + εijk nk sin θ + ni nj (1− cos θ). (17.4)

17.1 topological properties

Before proceeding with the analysis of the representations of the Poincaregroup and of its Lorentz subgroup, we analyze its topological properties. Weactually focus on the Lorentz subgroup first, because some interesting facts arealready clear when we analyze elements of this subgroup.

As explained above, the Lorentz group is specified by 4 by 4 matrices Λ thatpreserve the mixed signature metric η. These matrices provide the fundamentalrepresentation of the Lorentz group, which is trivially isomorphic to

O(3, 1) = Λ ∈ Mat(4, R) | ΛTηΛ = η,

which describes “rotations” in a 4-dimensional non-Euclidean space with 3

isotropic dimensions. Just like any O(N) group has two disconnected compo-nents, characterized by the determinant of the matrices representing its ele-ments, also O(3,1) has two disconnected components. In detail:

−1 = det η = det(ΛTηΛ) = det η (det Λ)2 (17.5)

17.1 topological properties 185

implies

det Λ = ±1. (17.6)

The full Lorentz group has therefore at least 2 disconnected components, cor-responding to the set of matrices having determinant equal to plus or minus1.

The subset of matrices with unit determinant contains also the identity ma-trix and actually forms a subgroup of the original one called

SO(3, 1) = Λ ∈ O(3, 1) | det Λ = 1. (17.7)

This is also analogous to the SO(N) < O(N) groups. However, we now provethat SO(3,1) is not connected.

If we focus on the constraint that characterizes the matrices of the groupSO(3,1) we see that the 00 component gives

−1 = η00 = Λµ0Λν

0 ηµν = −(

Λ00

)2+ ∑

i

(Λi

0

)2, (17.8)

which implies(Λ0

0

)2= 1 + ∑

i

(Λi

0

)2≥ 1. (17.9)

Using this inequality we can separate elements of SO(3,1) in two disconnectedcomponents. One component, connected to the identity, is represented by ma-trices with

Λ00 ≥ 1, (17.10)

while the other component has

Λ00 ≤ −1. (17.11)

Obviously we cannot change continuously the value of Λ00 to go from one

component to the other.The same conclusion can also be drawn for matrices with det Λ = −1. We

then see that the Lorentz group has 4 disconnected components, specified bythe value of det Λ and by the sign of Λ0

0 for the matrices Λ providing its fun-damental representation. We distinguish these sets by calling proper all trans-formations with positive determinant and orthochronous all transformationspreserving the direction of time.

186 poincare group

Definition 17.2. The set of matrices Λ ∈ O(3, 1) with det Λ = 1 and Λ00 ≥ 1

forms a subgroup called the proper and orthochronous Lorentz group:

SO+(3, 1).

The other components are subsets but not subgroups. One can cover them bytaking the product of elements in SO+(3, 1) with appropriate elements defininga discrete subgroup of SO(3,1). In detail, the subset with det Λ = −1 andΛ0

0 ≥ 1 contains the space parity operator

P =

1−1

−1−1

, (17.12)

which satisfies P2 = 1, while the subset with det Λ = −1 and Λ00 ≤ 1 contains

the temporal inversion operator

T =

−1

11

1

, (17.13)

which also satisfies T 2 = 1. Finally, the subset with det Λ = 1 and Λ00 ≤ −1

contains the combination of these two discrete operators: −14 = PT = T P .We therefore see that a generic O(3,1) transformation is the combination of aSO+(3,1) transformation with those of the Z2 ×Z2 group generated by P andT .

There are many more considerations that are interesting from a physicalpoint of view, like the fact that boosts in the same direction form a subgroupwhile boosts in different directions close only up to rotations, that physics isinvariant only under SO+(3,1) etc., but these are for the courses introducingspecial relativity.

In the following we focus on the component connected to the identity andwe now prove that it is homomorphic to the simply connected group SL(2,C).We will then discuss the physical implications of this fact.

Theorem 17.1. There is a 2 to 1 homomorphism between SL(2, C) and SO+(3,1).

Proof. The fundamental representation of SO+(3,1) acts linearly on spacetimecoordinates: x 7→ Λx. In order to obtain an action of SL(2,C), we think of

17.1 topological properties 187

the coordinates xµ as parameters specifying a generic 2 by 2 hermitian matrixA†(x) = A(x). A generic 2 by 2 hermitian matrix is specified by 4 real parame-ters, which determine its expansion in terms of the basis1

σµ = (12, σi). (17.14)

In fact, σ†µ = σµ and any 2 by 2 hermitian matrix can be written as A(x) = xµσµ,

where xµ are 4 real parameters. We can actually extract the xµ parameters fromA(x) by taking

xµ =12

tr (A(x)σµ) , (17.15)

where σµ = (12, σi), and tr (σµσν) = δνµ. Note that

det A(x) = det(

x0 + x3 x1 − i x2

x1 + i x2 x0 − x3

)= −xµηµνxν. (17.16)

We can then map hermitian matrices into themselves by

A 7→ A = SAS†, (17.17)

so that A† = A, which means that A = xµσµ, with xµ = ΛSµ

νxν. This in turnimplies that we have a map from SL(2, C) to SO+(3, 1). We can see that themap is linear because

xµ = 12 tr(

A(x)σµ)= 1

2 tr(SA(x)S†σµ

)= 1

2 tr(SσνS†σµ

)xν = ΛS

µνxν.

(17.18)

If, in addition, we require that S ∈ SL(2, C) (and hence that det S = 1), then

−xµ xνηµν = det A = det (SAS†) = det S(det S)∗ det A = −xµxνηµν. (17.19)

This implies that

xµxνηµν = (ΛSx)µ(ΛSx)νηµν (17.20)

and hence that

ΛS ∈ O(3, 1). (17.21)

1 If you are interested in a proof that (17.14) provides an orthonormal basis, check the exercises atthe end of this chapter.

188 poincare group

In addition, SL(2, C) is connected, and the same will be true by the image ofthe continuous map given by ΛS. We therefore conclude that

S 7→ ΛS (17.22)

maps elements of SL(2, C) into elements of SO+(3,1). Finally, since S and −Sare mapped to the same element ΛS we see that the map is 2 to 1.

The fact that such map is a homomorphism follows by the proof that

S1S2 7→ ΛS1S2 = ΛS1 ΛS2 . (17.23)

This is tedious, but easy to obtain by using the explicit expression for ΛS:

ΛS1S2µ

ν = 12 tr(S1S2σνS†

2S†1 σµ)= 1

2 tr(S2σνS†

2S†1 σµS1

)=

= 12(S2σνS†

2)

ab (S†

1 σµS1)

ba,

(17.24)

where we used the cyclic property of the trace, and using that (see the exercisesfor a proof of this identity)

(σρ)ab(σρ)c

d = 2 δda δb

c . (17.25)

Inserting this identity in the expression above:

ΛS1S2µ

ν = 14(S2σνS†

2)

dc (2 δd

a δbc )(S†

1 σµS1)

ba

= 14(S2σνS†

2)

dc(σρ)c

d(σρ)ab (S†

1 σµS1)

ba

= 12 tr

(S2σνS†

2 σρ) 1

2 tr(σρS†

1 σµS1)= ΛS1

µρΛS2

ρν.

(17.26)

From this theorem we also have a hint of the fact that SL(2,C) provides theuniversal covering of SO+(3,1) and that

SO+(3, 1) ' SL(2, C)/Z2. (17.27)

17.1.1 Topology of SL(2,C) and SO+(3,1)

Given that SL(2,C) and SO+(3, 1) are homomorphic, they are also related astopological spaces.

17.2 the so+(3, 1) algebra 189

We can actually determine their topologies by studying them as spaces ofmatrices. SL(2,C) is the space of 2 by 2 complex matrices with unit determinantand SO+(3,1) is the same space, but where couples of matrices are identifiedby means of the map (17.22). A generic complex matrix S can be written as theproduct of a unitary matrix and the exponential of a hermitian one

S = u eh, u†u = 1, h† = h. (17.28)

Since we are interested in SL(2,C) matrices, we compute the determinant of(17.28), which is

det S = (det u)(exp(tr h)), (17.29)

where det u = eiα ∈ U(1) and exp(tr h) ≥ 0. The request of having a unitdeterminant for S forces

det S = 1 ⇔ det u = 1, tr h = 0. (17.30)

Thus any SL(2,C) matrix can be written as the product of a matrix u ∈ SU(2)' S3 and the exponential of a traceless, hermitian, 2 by 2 matrix h, which isparameterized by three real numbers. We conclude that topologically

SL(2, C) ' R3 × S3. (17.31)

Also, since the 2 to 1 map (17.22) identifies matrices S with opposite signs andonly u can change the overall sign, we argue that

SO+(3, 1) ' R3 × S3/Z2. (17.32)

17.2 the so+(3, 1) algebra

As is often the case, also for the SO+(3, 1) group, it is convenient to determinerepresentations of the algebra rather than those of the group. For this reasonwe now discuss the structure of the algebra generators and their fundamentalrepresentation. The group SO+(3, 1) can be thought of as a group of matrices.We can then follow the same derivation of the algebra as in section 12.2, byexpanding matrices Λ ∈ SO(3,1) close to the identity. In detail, we can linearizethe condition ΛTηΛ = η by choosing Λµ

ν = δµν + ωµ

ν, assuming a small ω:

ηµν = Λρµ Λσ

ν ηρσ = (δρµ + ωρ

µ) (δσν + ωσ

ν) ηρσ

= ηµν + ηµσωσν + ηνσωσ

µ + . . .(17.33)

190 poincare group

The so(3, 1) algebra is then determined by the set of matrices ω fulfilling thecondition above, namely

so(3, 1) =

ω ∈ Mat(4, R) | ηω = −(ηω)T

. (17.34)

This also tells us immediately that

dim so(3, 1) = 6, (17.35)

because this is the number of independent conditions in a 4 by 4 antisymmetricmatrix (which is the set of ηω matrices, otherwise, if one considers ω, we wouldhave 3 symmetric and 3 antisymmetric matrices).

A basis for its fundamental representation is then given by

D(l1) =

0

00 −11 0

, D(l2) =

0

0 10

−1 0

,

D(l3) =

0

0 −11 0

0

,

(17.36)

which represent generators of so(3) ⊂ so(3, 1), and by

D(k1) =

0 11 0

00

, D(k2) =

0 1

01 0

0

,

D(k3) =

0 1

00

1 0

,

(17.37)

which make the algebra non-compact and whose exponential generates boosts:

exp(ψD(k1)) =

cosh ψ sinh ψsinh ψ cosh ψ

11

, (17.38)

17.2 the so+(3, 1) algebra 191

where ψ is the rapidity tanh ψ = −β. Note that since the group SO+(3,1) isnon-compact, we cannot generate all its elements by exponentiation. However,being SO+(3,1) connected, we can always cover the whole underlying manifoldby taking products of boosts and rotations:

Λ = exp(~ψ · D(~k)

)exp

(~θ · D(~l)

). (17.39)

A description of the same algebra, independent on the representation, isgiven by the commutators. Let us define

J0i ≡ ki, Jij ≡ εijklk, (17.40)

whose fundamental representation, provided in (17.36)–(17.37), is given by

D(Jµν)αβ = −ηµα δ

βν + ηνα δ

βµ . (17.41)

From the commutators of the faithful representation given above, we can nowdefine the algebra so(3, 1) by means of its structure constants, as follows fromthe brackets

[Jµν, Jρσ] = 4 δ[µ[ρ Jν]

σ], (17.42)

where indices have been raised by means of Minkowski metric η and squarebrackets on the indices denote antisymmetrizations with weight one. Alterna-tively

[li, lj] = εijk lk, [li, k j] = εijk kk, [ki, k j] = −εijk lk. (17.43)

As for the group of rotations, it is going to be extremely useful to complexifythis algebra in order to study its representations.

Theorem 17.2. so(3, 1)C ' sl(2, C)⊕ sl(2, C).

Proof. Define

l±i ≡12(li + i ki). (17.44)

These fulfill

[l±i , l±j ] = εijk l±k , [l+i , l−j ] = 0, (17.45)

which show that we have two copies of su(2)C ' sl(2, C).

This simple proof shows that also in this case we can build representationsof so(3, 1) by means of representations of sl(2, C), just like we did for su(2) 'so(3).

192 poincare group

exercises

1. Check that (17.4) satisfies RTηR = η.

2. Show that the space of 2 by 2 hermitean matrices is spanned by real linearcombinations of the σµ, µ = 0, 1, 2, 3, matrices, where σ0 = 12 and σi arethe Pauli matrices. For this purpose we introduce the inner product

〈A|B〉 = Tr (A†B).

a) Show that 〈σµ|σν〉 = N δµν, for some N ∈ R.

b) We can therefore think of σµ as orthogonal vectors in R4, whosecomponents are (σµ)ν. Show that

∑ρ

(σρ)∗µ(σρ)

ν = N δµν,

which leads to∑µ

(σµ)∗

αβ(σµ)γ

δ = N δαγδβδ

and finally to

∑i(σi)α

β(σi)γδ = −δ

βα δδ

γ + 2 δδαδ

βγ .

18I R R E D U C I B L E R E P R E S E N TAT I O N S O F T H E P O I N C A R EG R O U P

Now that we have determined the structure of the Poincare and Lorentz groupsand of their algebras we can discuss their representations. As we will see inthe following, we will be interested in two different kinds of representationsfor these groups, because we will need to employ different strategies in thedescription of states and operators of a relativistic quantum mechanical theory.Let me anticipate that we will need unitary infinite-dimensional representa-tions to describe physical states, while we will make use of non-unitary, butfinite-dimensional representations for operators.

We restrict our discussion to elements of the Poincare group that are con-nected with the identity, i.e.

ISO+(3, 1) = SO+(3, 1)n R3,1, (18.1)

where ISO stands for inhomogeneous special orthogonal transformations. Weknow how the corresponding algebra can be represented in terms of differentialoperators. Using anti-hermitian operators:

D(Jµν) = −ηµρ xρ∂ν + ηνρ xρ∂µ, (18.2)

D(Pµ) = ∂µ. (18.3)

The corresponding commutators are those of the Lorentz algebra (17.42) for Jµν

and

[Pµ, Pν] = 0, [Jµν, Pρ] = 2 ηρ[µPν], (18.4)

193

194 poincare irrepses

for the others. The last commutator tells us that translation generators trans-form in the fundamental representation of the Lorentz rotations (and in factthey carry a vector index µ):

[Jµν, Pρ] = PσD(Jµν)σρ, (18.5)

with D(Jµν)ρσ = −ηµρδσ

ν + ηνρδσµ , as in (17.42).

Starting from the basis of generators of the Poincare algebra we can con-struct by exponentiation a representation for the Poincare group acting on anappropriate vector space, which we want to be the Hilbert space of a quantumrelativistic theory. In particular, if we want the Poincare group to be the set ofspacetime symmetries that leave physics invariant, we are interested in a rep-resentation of its elements by unitary operators. However, as noted before, thePoincare group is not compact and therefore unitary irreducible representationsare infinite dimensional. This implies that the vector space on which they act isinfinite dimensional and a basis of states for this space has labels that vary overan infinite number of values. Although this may sound scary, it is a rather rea-sonable fact that one should expect, given that among the quantum numberscharacterizing a generic physical state there is the momentum pµ, which hascontinuous values.

However, these are not going to be the only representations of the Poincaregroup we are interested in. Today we know that quantum relativistic theoriesare field theories. Fields (continuous functions of spacetime coordinates xµ

at the classical level) contain creation and destruction operators for quantumstates. In order to be able to work with them and in order to have a sensibleclassical limit it is therefore useful to introduce also finite-dimensional representa-tions of the Poincare group, which are necessarily non unitary. For instance, theelectromagnetic field is described by the vector potential Aµ(x). Under Lorentztransformations Aµ 7→ Λµ

ν Aν and Λ is generically not unitary, as can be seenimmediately from the matrix representation given in the previous chapter.

Summarizing, we will use unitary representations for the Poincare group ac-tion on the Hilbert space of physical states of quantum field theories and wewill use finite-dimensional representations for the operators acting on the sameHilbert space. In this lecture we therefore first analyze in detail the constructionof unitary representations and then discuss finite-dimensional ones. We con-clude by discussing the role of Casimir operators to determine the appropriatequantum numbers defining these irreducible representations.


18.1 unitary representations

For the sake of simplicity, we are going to consider single particle states, trans-forming under the action of irreducible representations of the Poincare group.The classification of these representations has been worked out first by Wigner,but one can look for a more detailed discussion in Weinberg’s book ”Quantumtheory of Fields: I”.

As mentioned in the introduction, an irreducible unitary representation ofthe Poincare group must be infinite dimensional. The states providing a ba-sis for the vector space on which this representation acts form an infinite-dimensional multiplet and transform onto each other when acted upon byPoincare generators. As is always the case, we can decide that a set of com-muting operators are diagonal on this basis and start labeling the correspond-ing states by using their eigenvalues. Recall that irreducible representations ofabelian groups are always 1-dimensional.

In the Poincare algebra, Pµ operators commute among themselves and wecan therefore classify quantum states by using their eigenvalues

Pµ|pµ, σ〉 = i pµ|pµ, σ〉, (18.6)

where σ are other possible quantum numbers. As expected, their exponentialprovides a unitary representation for translations

U(1, a) = exp(aµPµ) (18.7)

and the action on this basis of states is also diagonal:

U(1, a)|p, σ〉 = eiaµ pµ |p, σ〉. (18.8)

In order to complete the basis of states of the vector space on which a unitaryrepresentation of the Poincare group acts, we need to provide a representationfor the proper Lorentz group elements. The corresponding operators U(Λ) ≡U(Λ, 0) should map Pµ eigenstates into states with momentum Λp, because Pµ

transform in the fundamental representation of the Lorentz group and so musttransform its eigenstates. Actually, using Hadamard’s lemma 12.3 and (18.5),we can see this explicitly by taking the action of

U(Λ) = exp(

12

ωµν Jµν

)(18.9)


on the momentum operators

e−12 ωµν Jµν

Pρe12 ωµν Jµν

= Pσ exp(−1

2ωµνD(Jµν)

)σ

ρ, (18.10)

which also implies that

U†(Λ)PρU(Λ) = U(Λ−1)PρU(Λ) = Pσ(Λ−1)σρ = Λρ

σPσ, (18.11)

where we used that ΛµρηµνΛν

σ = ηρσ and hence (Λ−1)ρσ = ηρα(Λ−1)α

βηβσ =Λσ

ρ. This in turn implies that an eigenstate of Pµ with momentum pµ trans-forms into an eigenstate with momentum Λµ

ν pν, when acted upon by U(Λ):

Pµ (U(Λ)|p, σ〉) = U(Λ)U†(Λ)PµU(Λ)|p, σ〉 = U(Λ)ΛµνPν|p, σ〉

= i (Λµν pν) U(Λ)|p, σ〉.

(18.12)

Does this imply that the only result of the action of a Lorentz transformationon a state |p, σ〉 is to map it to

|Λp, σ〉 = U(Λ)|p, σ〉? (18.13)

Actually, we can see that this cannot be correct, because we did not take intoaccount the possibility that there may different states whose momenta are in-variant under the action of Λ and which therefore may mix between then whenacted upon by U(Λ). In order to solve this issue we introduce the so-called littlegroup and show that irreducible representations of Poincare group are relatedto irreducible representations of this group.

Definition 18.1. The little group of a vector v ∈ V (where V is the space onwhich the representation D acts) is the subgroup Gv < G that leaves invariantv:

IsoG(v) = g ∈ G | D(g)v = v.

Since we are interested in the possible degeneracy between momentum eigen-states acted upon by Lorentz transformations, we are interested in the littlegroups of generic 4-vectors pµ with respect to G = SO+(3, 1). We can provethat irreducible representations of the Poincare group must then be determinedin terms of irreducible representations of the little groups associated to physicalmomenta pµ.


Theorem 18.1. Unitary irreducible representations of the Poincare group are fixed byirreducible representations of Iso(k) for fixed k.

Proof. Start from a fixed momentum k, whose little group Iso(k) is determinedby elements Λ ∈ Iso(k) fulfilling

Λk = k. (18.14)

All eigenstates of the momentum operator with eigenvalue k are then mappedonto each other by the action of Lorentz transformations

U(Λ)|k, σ〉 = Dσ′σ(Λ)|k, σ′〉, (18.15)

where D provides an irreducible representation of Iso(k), with σ = 1, . . . , dim D,which we want to be unitary in order to preserve orthonormality of states

〈p, σ|p′σ′〉 = δ3(~p− ~p′)δσσ′ . (18.16)

Also D may not be finite dimensional if Iso(k) is not compact. Fortunately thiswill not be necessary if one is not interested in non-physical cases.

We can now determine the result of the action of a unitary Lorentz transfor-mation on a generic state of momentum p. First we find a Lorentz transforma-tion relating p to k

pµ = Λpµ

ν kν, (18.17)

which we assume does not mix states with the same momentum k:

|p, σ〉 ≡ U(Λp)|k, σ〉. (18.18)

Then we compute the action of a generic Lorentz transformation on |p, σ〉 andnote that any such transformation can be rewritten as the product of a boostmapping k to Λp (which again does not mix states with the same momentumk) and a Lorentz transformation in the little group of k. In fact, if we computeU(Λ)|p, σ〉, we see that we can rewrite it as

U(Λ)|p, σ〉 = U(ΛΛp)U†(ΛΛp)U(Λ)|p, σ〉

= U(ΛΛp)U†(ΛΛp)U(Λ)U(Λp)|k, σ〉

= U(ΛΛp)U(Λ−1Λp

ΛΛp)|k, σ〉,

(18.19)


where

Λ−1Λp

ΛΛpk = Λ−1Λp

Λp = k (18.20)

and therefore

Λ−1Λp

ΛΛp ∈ Iso(k). (18.21)

We can use this result to see that the generic unitary representation of a Lorentztransformation acts on states |p, σ〉 mixing them according to the little group ofthe associated momentum k

U(Λ)|p, σ〉 = U(ΛΛp)Dσ′σ(Λ)|k, σ′〉 = Dσ′σ(Λ)|Λp, σ′〉 (18.22)

where Λ = Λ−1ΛpΛΛp ∈ Iso(k). This provides a representation for Λ on each

state |p, σ〉.

While the final expression may not be particularly elegant, we reduced theproblem of classifying representations of the Poincare group to the problem ofclassifying representations of the little groups associated to physical momenta,which can be differentiated by the value of their invariant scalar combinations.

Since the physical interpretation of pµ does not allow for the case p0 < 0 andsince p2 > 0 implies tachyonic states, we can distinguish three physical orbits1:

• kµ = 0. Iso(k) = SO(3, 1) and its orbit is given by a unique state. It hasthe property that all generators of the Poincare group are representedby 1-dimensional null matrices and, consequently, all group elements arerepresented by identity matrices. This state has momentum and angularmomentum equal to zero and is invariant under Poincare transformations.We interpret it as the vacuum.

• k2 < 0. There is a frame where kµ = (k, 0, 0, 0). In this case the little groupconsists of space rotations Iso(k) = SO(3), whose covering group is SU(2).These states describe massive particle and are classified by well knownspin representations D2j+1 and σ = (j, j3). However, since [l3, Pi] = ε3ijPj,j3 should not be directly interpreted as the eigenvalue of the state withrespect to l3, but rather with respect W3, where Wµ is the Pauli–Lubanskioperator we will introduce later, when discussing Casimir operators forthe Poincare group.

1 Also, since we are interested in proper and orthochronous transformations, we should distinguishp0 > 0 from p0 < 0. However, once more, physics tells us we should only concentrate on p0 > 0.

18.2 finite dimensional representations 199

• k2 = 0. There is a frame where k = (1, 1, 0, 0), or k = (1,−1, 0, 0). Theseare massless particles. At the infinitesimal level the elements of so(3, 1)that leave invariant k follow from Λk = (1 + ω)k = k and hence ωk = 0:

θ1l1 + a(k2 − l3) + b(k3 + l2) =

0 0 a b0 0 a ba −a 0 −θ1b −b θ1 0

. (18.23)

The resulting group is ISO(2) ' E2, the Euclidean group in 2 dimen-sions, where l1 is the rotation generator and k2 − l3 and k3 + l2 commuteand transform onto each other by the action of exp(αl1). This is a non-compact group and most of its irreducible representations are infinitedimensional. The corresponding quantum numbers are, however, not ob-served in nature. We conclude that physical states are classified only bythe representations of l1 ∈ u(1), which are labeled by an integer number,called helicity.

18.2 finite dimensional representations

Since we are interested also in finite-dimensional non unitary representationsof the Lorentz group, because they are directly relevant to determine the fieldsof quantum field theories, we now discuss how to find them in a constructivemanner. We proceed in the same way as for the su(2) algebra, by consideringfirst the complexification of the Lorentz algebra and then taking the appropriatereal section. We therefore first build representations for so(3, 1)C ' sl(2, C)⊕sl(2, C) and then consider the appropriate linear combinations. We know thatirrepses of each of the two sl(2, C) factors are characterized by a half-integerlabel j ∈N/2, called spin, and have dimension 2j + 1.

Theorem 18.2. Finite-dimensional representations of so(3, 1) are characterized by twosemi-integer numbers (j1, j2), with j1, j2 ∈N/2.

We therefore proceed by effectively generalizing the spinorial representationsof 3-dimensional rotations to 4 dimensions, taking into account the Lorentziansignature. For 3 dimensional rotations we introduced Pauli matrices σi, i =1, 2, 3, satisfying

σi, σj = 2 δij, (18.24)


where δij is the Euclidean metric in 3 dimensions, and we saw that by means ofthem we could provide an irreducible 2-dimensional spinorial representationfor the so(3)C algebra by using

DS(Jij) = −12

σij ≡ −14[σi, σj] = εijk

[− i

2σk

]= εijkDS(lk) (18.25)

and then, starting from that, build all the other representations.For so(3, 1) we introduce 4 by 4 matrices γµ, µ = 0, 1, 2, 3, with analogous

anti-commutation properties:

Definition 18.2. The Clifford algebra Cl(3, 1) is the algebra generated by γµ,µ = 0, 1, 2, 3, satisfying

γµ, γν = 2 ηµν.

The minimal matrix representation one can give to γµ is 4 by 4 (see the ex-ercises) and there are obviously various legitimate equivalent choices for theirexpression. One of them is the following:

γ0 = i σ1 ⊗ 12, γi = σ2 ⊗ σi. (18.26)

We can now provide the spinorial representation of the Lorentz generators byconsidering

DS(

Jµν

)= −1

2γµν, (18.27)

where

γµν ≡12[γµ, γν]. (18.28)

We leave to the reader the check of the commutation relations2 (17.42).

Definition 18.3. The spinorial representation of the Lorentz group is provided byDS(

Jµν

)= − 1

2 γµν.

States transforming under the action of such a representation of the Lorentzgroup are called spinors.

The group representation is obtained by exponentiation:

DS

(exp

(12

ωµν Jµν

))= exp

(12

ωµνDS(Jµν)

)= exp

(−1

4ωµν γµν

). (18.29)

2 It may be useful to realize that [γµν, γρσ ] = −8 δρσµν .

18.3 reducibility of spinor representations 201

From the discussion above, this is actually a representation of the coveringgroup of the Lorentz group

Spin(3, 1) =

exp(−1

4ωµνγµν

)| ωµν = −ωνµ

(18.30)

and again, just like between SU(2) and SO(3) we have a 2 to 1 map, we alsohave a 2 to 1 map between Spin(3, 1) and SO+(3, 1). We can actually identify

Spin(3, 1) ' SL(2, C). (18.31)

Differently from what happened for SU(2) we can now see that this representa-tion is not irreducible, though.

18.3 reducibility of spinor representations

Starting from the 4 γ-matrices defining Cl(3, 1) we can always introduce a newmatrix3

γ5 ≡ i γ0γ1γ2γ3, (18.32)

satisfying

γ5, γµ = 0 (18.33)

and

(γ5)2 = 1. (18.34)

We can then introduce two projectors

P± ≡12(1± γ5) , (18.35)

which split the vector space on which the spinorial representation acts in twocomplementary subspaces. This is actually straightforward to see using therepresentation (18.26) for the γ-matrices, which gives

P+ =

(12

O2

), P− =

(O2

12

). (18.36)

3 The index 5 is not related to spacetime properties and therefore γ5 = γ5.


We now show that the subspaces im(P±) are invariant with respect to the actionof the spinorial representation of the Lorentz group DS.

Take an element in any of the two subspaces: η± ∈ Im(P±). By definitionP±η± = η± and P∓η± = 0. Hence

γ5η± = ± η±. (18.37)

If we apply a Lorentz transformation to this spinor, we get a new spinor

η± = exp(−1

4ωµνγµν

)η±, (18.38)

which is again in the same subspace because

γ5η± = γ5 exp(−1

4ωµνγµν

)η± = exp

(−1

4ωµνγµν

)γ5η± = ±η±. (18.39)

We will call spinors in im(P+) spinors with positive chirality and those inim(P−) spinors with negative chirality.

Actually im(P±) are vector spaces where irreducible representations of thesl(2, C)⊕ sl(2, C) algebra act. In fact we can write

12

ωµν Jµν = ω0i J0i +12

ωij Jij = −ω0iki +12

εijkωij lk = l+i ω+i + l−i ω−i , (18.40)

where

l±i ≡12(li ± i ki), ω±i ≡

12

(εijkωjk + 2i ω0i

). (18.41)

In the Weyl basis (18.26)

−14

ωµνγµν =

ω+i

(− i

2 σi

)ω−i

(− i

2 σi

) . (18.42)

We then see that Vs = V+ ⊕ V− on which we have a split action of l+i and l−i .We also conclude that spinors with positive chirality are objects in the ( 1

2 , 0)representation and those with negative chirality in the (0, 1

2 ). Alternatively,using the dimensions of the irrepses, we denote them by (2, 1) and (1, 2).

18.3 reducibility of spinor representations 203

Finally, we can show that Spin(3, 1) ' SL(2, C) and that we have a 2 to 1

map between the Spin group and SO+(3, 1). A generic element of Spin(3, 1) isrepresented by4

exp(−1

4ωµνγµν

)=

(S 00 (S−1)†

), (18.43)

where

S = exp(− i

2√

2ω+

i σi

)(18.44)

and S ∈ SL(2, C), with (S−1)† = exp(− i

2 ω−i σi

), because (ω+

i )∗ = ω−i . Asfor SU(2), we can build the adjoint representation on γµxν and produce the2 to 1 map between SL(2, C) and SO+(3, 1). This map translates elements ofSL(2, C) into ΛS matrices giving the action xµ 7→ Λµ

Sνxν and again S and −Sare mapped to the same element ΛS.

The generic representation, labeled by (j1, j2), is built from the tensor productof these representations and the elements of the corresponding vector space onwhich they act can be interpreted as tensors with 2j1 + 2j2 indices

. . .︸︷︷︸2j1

⊗ . . .︸︷︷︸2j2

⇔ Sα1 ...α2j1,α1 ...α2j2

, (18.45)

whose transformation properties are obvious.Before closing this discussion, it is interesting to point out that each of the ten-

sors Sα1 ...α2j1,α1 ...α2j2

also contain pieces with well defined transformation prop-erties under the group of spatial rotations. In fact, SO(3) < SO(3,1) and theLorentz algebra so(3, 1) contains the so(3) rotation generators li as special lin-ear combinations of its generators: li = l+i + l−i . This immediately implies thatthe (j1, j2) irreps of so(3, 1) decomposes into irrepses of so(3) according to thebranching of the product j1 ⊗ j2. For instance, the vector transforming in thefundamental representation of SO(3,1) Vµ corresponds to an object in the ( 1

2 , 12 ).

In fact12⊗ 1

2= 0⊕ 1, (18.46)

where the scalar under SO(3) is V0 and the vector is Vi.

4 We can now also come back to the issue of covering the whole group from the algebra. Theelement S = − exp(γ0γ1 + γ1γ2) = −12 − γ0γ1 − γ1γ2 is in Spin(3, 1) ' SL(2, C), but it can neverbe obtained as the exponential of any combination of the form (18.43).


18.4 casimir operators

During these lectures we stressed several times that an effective way of classi-fying group representations is given by Casimir operators. Casimir operatorscommute with all the generators of a given algebra and therefore allow us todistinguish irreducible representations by their eigenvalues. This is true alsofor the Lorentz and Poincare group and for this reason we now revisit the dis-cussion of their irreducible representations in the light of the existence of suchCasimir operators.

18.4.1 Casimir for the Lorentz group

The Lorentz algebra has two quadratic Casimir operators

C1 =12

Jµν Jµν (18.47)

and

C2 =14

εµνρσ Jµν Jρσ. (18.48)

The existence of these two quadratic Casimir operators is obviously relatedto the fact that its complexification is isomorphic to the direct sum of twoother algebras and therefore we should be able to use their quadratic Casimiroperators to classify irrepses:

so(3, 1)C = sl(2, C)⊕ sl(2, C). (18.49)

Recalling that J0i = ki, Jij = εijklk and l±i = 1√2(li ± i ki) we have that

C1 = −J20i +

12

Jij Jij = −|~k|2 + |~l|2 = (~l+)2 + (~l−)2 (18.50)

and

C2 = εijk J0i Jjk = 2~k ·~l = −i((~l+)2 − (~l−)2

). (18.51)

We then conclude that we can use two labels (j1, j2) to differentiate the irre-ducible representations.

18.4 casimir operators 205

18.4.2 Casimir for the Poincare group

In the case of the Poincare group [C1, Pµ] 6= 0 and also [C2, Pµ] 6= 0. We thenneed to find new Casimir operators. The first one is simply the operator defin-ing the mass of the state:

P2|m〉 = −m2|m〉. (18.52)

In fact [P2, Pµ] = 0 trivially and [P2, Jµν] = 0 because P2 is a Lorentz scalar.This allows us to distinguish between the two orbits of massive and masslessstates.

The second invariant must reflect the discussion in section 18.1, namely thatphysical states are classified according to their transformation properties underthe little group. The correct operator is called the Pauli–Lubanski operator and isbuilt using the square of the quadratic operator

Wµ =12

εµνρσ JνρPσ. (18.53)

Before proving that W2 is a good Casimir operator, we note that

W0 =12

εijkεijkll Pk =~l · ~P, (18.54)

Wi = −εijk J0iPk −12

εijk JjkP0 =(−~k ∧ ~P + P0~l

)i, (18.55)

which tells us that W0 is proportional to the helicity

η =~l · ~P|~P|

, (18.56)

which defines the spin in the direction of the motion, and that Wi becomesproportional to the rotation generators li whenever we consider the rest frameof massive particles. We also see that

WµPµ =12

εµνρσ JνρPσPµ = 0 (18.57)

and that Wµ commutes with the momentum operators Pν:

[Wµ, Pν] = 12 εµαβγ[JαβPγ, Pν] = 1

2 εµαβγ[Jαβ, Pν]Pγ

= 12 εµαβγ 2 δν

αPβPγ = εµνβγPβPγ = 0.(18.58)


This means that Wµ are in fact combinations of the Lorentz generators thatleave the momentum invariant and therefore generate the little group.

We then easily prove that W2 is a good Casimir operator, because Wµ is avector under Lorentz transformations

[Jµν, Wρ] = 2 ηρ[µWν], (18.59)

and

W2 =12

J2P2 + Pµ Jµν JνρPρ (18.60)

is therefore a scalar:

[Jµν, W2] = 0. (18.61)

When acting on massive particle states, whose rest frame momentum is pµ =(m, 0, 0, 0), it reduces to

W2 = −m2(−|~k|2 + |~l|2)−m2|~k|2 = −m2 |~l|2, (18.62)

which means that irreducible representations are classified by the so(3) Casimir,as expected.

When acting on massless particle states we have once more the problem withthe infinite quantum numbers not visible in nature. Choose a frame wherepµ = (p, 0, 0, p) and decompose Wµ accordingly

Wµ = η Pµ + N1 nµ1 + N2 nµ

2 + Ws sµ, (18.63)

where

n1 = (0, 1, 0, 0), n2 = (0, 0, 1, 0) (18.64)

define the directions orthogonal to pµ and

sµ = (s0, 0, 0, s3), with s0 6= |s3|. (18.65)

Since WµPµ = 0 and WµPµ ∼ Ws we have that Ws = 0. We then have thatW2 = N2

1 + N22 , but there are no physical states with such continuous quantum

numbers and therefore we set them to zero, so that W2 = 0. As expected, weare left with η to classify the remaining states. In fact, since W2 = 0, the helicitydetermines fully W because

W0 = η|~P| and ~W = η ~P. (18.66)

18.4 casimir operators 207

exercises

1. Prove that the minimal representation for γ-matrices satisfying γµ, γν =2ηµν 1 with η = diag−1,+1,+1,+1 is 4 by 4.

a) Compute the trace of γ0 and of γi using the definition of the Cliffordalgebra.

b) Show that the only admissible eigenvalues for iγ0 and γi are ±1.

c) Show that 2 by 2 matrices are not enough to provide a faithful repre-sentation.

2. In relativistic theories we often use two different representations for thegamma matrices. One representation is called the Weyl representationand was given in (18.26). The other representation is called the Diracrepresentation and is related to the previous one by a similarity transfor-mation:

γµD = Sγ

µWS−1, where S =

1√2

(12 12−12 12

).

Build it explicitly.

3. Show that the 16 matrices Γn given by all possible products of different γmatrices are independent (Γ0 = 1, Γ1 = γ0, Γ5 = γ0γ1 = −γ1γ0, etc...).

a) Compute (Γn)2.

b) Show that tr Γn = 0, but for n = 0.

c) Show that ΓmΓn is equal to one of the other Γp up to a sign, andp = 0 only when n = m.

d) Assuming that ∑n anΓn = 0 and taking traces of this relation aftermultiplication with Γm one finally shows that all an coefficients mustbe vanishing.

4. Compute ( 12 , 1

2 )⊗ [( 12 , 0)⊕ (0, 1

2 )].

5. What kind of tensor corresponds to the (1, 0) ⊕ (0, 1) representation ofthe Lorentz group?

6. Fields described by tensors with definite symmetry properties like gµν(x) =gνµ(x), with gµ

µ = 0, and Bµνρ(x) = B[µνρ](x) transform under the actionof finite-dimensional representations of the Lorentz group. What are the


labels of sl(2, C)⊕ sl(2, C) describing them? Which spin states they con-tain with respect to SO(3) < SO(1,3)?

7. A Dirac spinor ψ transforms under the action of the proper and orthochronousLorentz group in the spinorial representation D(ω) = exp

(− 1

4 ωµνγµν)

,i.e. ψ→ D(ω)ψ.

Using [γρ, γµν] = 4ηρ[µγν] show that D−1γµD = Λµνγν (prove it at the

infinitesimal level).

Using also γµ† = γ0γµγ0 show that ψψ ≡ i ψ†γ0ψ is a scalar quantity.

8. Prove that [Jµν, Wρ] = 2 ηρ[µWν].

19S Y M M E T R I E S O F M O L E C U L E S A N D S O L I D S .

In this lecture we discuss some applications of the theory of group represen-tations to Physics. In particular we will try to see what are the consequencesof the existence of symmetry properties of a physical systems and how to usethem to compute relevant physical quantities.

We will focus on systems consisting of identical particles (atoms) in an or-dered configuration in space. The symmetries one has for these configurationsare described by discrete groups and are also subgroups of the groups of trans-lations, rotations and reflections in space: the 3-dimensional Euclidean group

E3 = O(3)n R3. (19.1)

Let ~xI ∈ R3 denote the position of the constituents of the molecule

M = ~xI , I = 1, . . . , N. (19.2)

Definition 19.1. The symmetry group of a molecule M is the subgroup of E3 thatleaves M invariant:

G = g ∈ E3 | D(g)M = M.

19.1 classic vibrations of molecules

Imagine a molecule where the atoms are in a fixed position. The electroniclevels can be obtained as a function of their configuration and their energy de-pends on the relative positions of the nuclei. These levels will change even if wemove the nuclei by a little amount. Fix then the average positions 〈~xI〉 of the N

209

210 symmetries of molecules and solids

units forming a molecule and call qA, where A = 1, . . . , 3N, the displacements,weighted with their mass:

mI(~xI − 〈~xI〉) =

q3I−2q3I−1

q3I

. (19.3)

The potential energy of the system depends on the values of qA. Although theresulting function is strictly valid only for fixed qA, we can still use V(qA) tocompute the molecule vibrations, because qA is going to be much smaller thanthe average speed of the electrons.

The total energy of the system is the sum

H = K + V, (19.4)

where K = 12m |~pI |2 and V = V(~xI).

If we use as coordinates the displacements qA, we can simplify the potentialand expand it around the equilibrium positions:

K =12 ∑

Aq2

A, V ' 12

FABqAqB, (19.5)

where we subtracted the constant part V(0) and we stopped at second order inthe expansion of the potential. The equations of motion of this system are

qA + FABqB = 0. (19.6)

Whenever FAB = ω2A δAB we have the obvious solution

qA = cA cos(ωAt + φA). (19.7)

However, in general F is not diagonal. We assume it is real, symmetric andpositive definite, as a consequence of the expansion around a local minimumconfiguration. In this case

F = ST∆S, (19.8)

where

∆ =

λ1. . .

λ3N

, λA = ω2A > 0 (19.9)

19.1 classic vibrations of molecules 211

and the columns of S are given by the 3N eigenvectors c(A), so that the generalsolution can be expressed in terms of the so-called Normal Modes

qA = ∑B

c(B)A αB cos(ωBt + φB), (19.10)

where α and φ are integration constants, fixed by initial conditions.

Definition 19.2. The coordinates Q = Sq are called Normal Coordinates.

Using normal coordinates, the Hamiltonian simplifies to

H =12

PAPA +12

ω2A QAQA. (19.11)

We now understand that the problem of studying molecule vibrations is equiv-alent to the analysis of its normal modes. This problem becomes purely grouptheoretical, once we use the symmetry properties of the molecule.

19.1.1 Symmetries

The ensemble of discrete points providing the equilibrium positions of a moleculeare invariant (or better, are mapped among themselves in a way that the finalresult is indistinguishable from the original configuration) under the action ofthe symmetry group G. For instance, a molecule described by a regular poly-gon in 2 dimensions has symmetry group G = Dn.

The same group, however, also has a non-trivial action on the displacements.In fact, we have to combine the vertex permutations with the rotations of therelative reference frames. In figure 25 the two configurations are obtained byreflection.

1 2

3

x1

y1

x2

y2

x3

y3

2 1

3

x2

y2

x1

y1

x3

y3

Figure 25.: Caption


Since the atom 2 is displaced (by the pink arrow in the picture), the finalconfiguration is obtained from the original one by permuting the vertices 1 and2, but also acting on the displacement q2, which is mapped by means of thesame element of D3 by a reflection around the y axis.

The generic action of G is therefore realized on the qA coordinates by the realand unitary matrix

D(g) = p(g)⊗ R(g), (19.12)

where p acts by permuting the I indices and R is a rotation acting on (xI , yI , zI),in the same fashion for any value of I.

Since the matrix D(g) comes from a product of different representations ofthe same group, it is often reducible

D(g) = ∑a

naDa(g), (19.13)

where a labels different irreducible fragments. Since D should represent asymmetry of the system, whenever

q 7→ D(g)q, (19.14)

K and V should be left invariant. Invariance of the kinetic term is straightfor-ward because D is real and unitary. Invariance of the potential gives

DT(g)FD(g) = F, (19.15)

or, equivalently

FD(g) = D(g)F, ∀g ∈ G. (19.16)

If we decompose D in irreducible fragments we get

FmnDn(g) = Dm(g)Fmn, (19.17)

where m and n are the dimensions of the irreducible blocks. Using Schur’slemma we can deduce that Fmn = 0 whenever m 6= n and is proportional tothe identity on the same irreducible fragments. We conclude that F is blockdiagonal on invariant subspaces of the space of displacements and thereforewe can reduce the computation of normal modes to the analysis of irreduciblerepresentations of discrete groups.


19.1.2 Diatomic molecule

Let us start by considering a diatomic molecule, like H2, O2 or N2. Assume thatdisplacements are allowed only in 1 dimension, we can fix the reference frameso that

〈x1〉 = −d/2, 〈x2〉 = d/2, (19.18)

where d is the average distance between the two nuclei.

Figure 26.: Caption

We call q1 and q2 the displacements, so that the distance between the twonuclei is

δ = |(−d/2 + q1)− (d/2 + q2)| = |d + q2 − q1|. (19.19)

Following Hooke’s law, for small displacements

V =12

k (δ− d)2 =12

k (q2 − q1)2, (19.20)

so that

F = k(

1 −1−1 1

). (19.21)

The symmetry group is the permutation group of two objects S2 ' Z2, repre-sented on (q1, q2) by

D2(1) =

(1 00 1

), D2(σ) =

(0 11 0

). (19.22)

This representation is clearly reducible, because Z2 is abelian and its irreduciblerepresentations are all 1-dimensional. The characters of the representationgiven above are χ2(1) = 2 and χ2(σ) = 0 and they must be the sum of the char-acters of Z2 irrepses. Since χ1(1) = χ1(σ) = 1 and χ1′(1) = 1, χ1′(σ) = −1,we have that D2 = D1 ⊕ D1′ . As we proved above, F must be diagonal in thisbasis and hence

tr(D(g)F) = λ1 χ1(g) + λ2 χ1′(g). (19.23)


We then deduce that

tr(D(1)F) = tr[

k(

1 −1−1 1

)]= 2k = λ1 + λ2, (19.24)

tr(D(σ)F) = tr[

k(−1 11 −1

)]= −2k = λ1 − λ2, (19.25)

which implies that

λ1 = 0, λ2 = 2k. (19.26)

This is exactly what we should get from the split of the displacements intothe center of mass displacement X = 1√

2(q1 + q2) and the displacement with

respect to the center of mass x = 1√2(q2 − q1). The Hamiltonian becomes

H =12

(X2 + x2

)+

12(2k) x2. (19.27)

We see that the null eigenvalue is associated to the center of mass. In fact, itsmotion is that of a translation and hence has no vibrational energy. On theother hand, x describes precisely a vibrational mode with frequency ω =

√2k.

We can actually understand what is described by the normal modes by pro-jecting the displacement vector into the subspace left invariant by the a-th irre-ducible representation. This can be done by means of the projector

Pa ≡Na

#G ∑g

χa(g)∗D(g). (19.28)

If one evaluates this projector as a matrix one sees that it is proportional to theidentity only in the block corresponding to the a irreducible representation:

(Pa)ij = Na (Dakk, Dij) = ∑

bδab δij, (19.29)

where b runs over the various irrepses contained in D. For the case at hand wehave that

P1 =12

(1 11 1

)=

1√2

(11

)1√2(1, 1) (19.30)

and

P1′ =12

(1 −1−1 1

)=

1√2

(1−1

)1√2(1, −1) . (19.31)

Their projection on (q1, q2) gives QCM = q1 + q2 and√

2 x = q2 − q1.


19.1.3 Diatomic molecule in 2 dimensions

If we now consider the same diatomic system in a 2-dimensional space, wehave 4 coordinates for the displacements and the symmetry group is G = D2 =Z2 ×Z2, i.e. the set of reflections around the two axes.

x

y

1 2

y1

x1

y2

x2

Figure 27.: Caption

The symmetry group G has to be realized as a set of permutations of thetwo centers and as the reflections of the x1, y1 and x2, y2 axes. We call ρ thegenerator associated to the reflection symmetry around the x axis and σ thegenerator of the reflection symmetry around the y axis. The representation ofthe group elements as permutations of the index I = 1, 2 is

p(1) = p(ρ) =(

1 00 1

), p(σ) = p(σρ) =

(0 11 0

). (19.32)

This realizes the fact that the equilibrium positions of the nuclei is not affectedby a reflection around the x axis.

Considering the embedding in O(2), the same elements are represented as

R(1) =(

1 00 1

), R(σ) =

(−1 00 1

),

R(ρ) =(

1 00 −1

), R(σρ) =

(−1 00 −1

).

(19.33)


We can now deduce the form of the representation matrices for G = D2 onthe displacements qA, A = 1, . . . , 4. They are

D4(1) =

1

11

1

, D4(σ) =

−1

1−1

1

,

D4(ρ) =

1−1

1−1

, D4(ρσ) =

−1

−1−1

−1

.

(19.34)

We can now show that D4 = D1 + D1′ + D1′′ + D1′′′ , by computing χ4(1) = 4,χ4(σ) = χ4(ρ) = χ4(ρσ) and using the character table of Z2 ×Z2:

1 σ ρ ρσ1 1 1 1 11′ 1 −1 1 −11′′ 1 1 −1 −11′′′ 1 −1 −1 1

(19.35)

In this basis

F = k

1 −1

0−1 1

0

. (19.36)

We therefore deduce that

tr(D4(1)F) = 2 k = λ1 + λ1′ + λ1′′ + λ1′′′ , (19.37)

tr(D4(ρ)F) = 2 k = λ1 + λ1′ − λ1′′ − λ1′′′ , (19.38)

tr(D4(σ)F) = 2 k = λ1 − λ1′ + λ1′′ − λ1′′′ , (19.39)

tr(D4(ρσ)F) = 2 k = λ1 − λ1′ − λ1′′ + λ1′′′ . (19.40)


We solve this by λ1 = 2k and λ1′ = λ1′′ = λ1′′′ = 0. The correspondingprojectors are

P1 =12

1 −1

0−1 1

0

, (19.41)

which correctly gives the vibration mode in x, with frequency√

2k,

P1′ =12

1 1

01 1

0

, (19.42)

which gives the zero mode related to translations in x,

P1′′ =12

0

1 10

1 1

, (19.43)

which gives the zero mode related to translations in y, and

P1′′′ =12

0

1 −10

−1 1

, (19.44)

which produces anti-clockwise rotations around the origin.

19.1.4 Triatomic molecule

If we have three nuclei at the vertices of an equilateral triangle, the symmetrygroup is D3 ' S3, whose 3-dimensional representation as vertex permutationsshould be multiplied by its 2-dimensional representation as rotations on theplane. The resulting representation is reducible as

D6 = D1 = D1′ + D2 + D2. (19.45)

The corresponding subspaces give the non-trivial dilatation mode, with fre-quency ω =

√3k, the zero energy, rotational mode, the 2-dimensional space of


translations in x and y (again with zero energy) and the 2-dimensional spaceof vibrations, with frequency ω =

√3k/2. We leave this straightforward, but

tedious analysis to the reader.

AS O L U T I O N S

Here are some possible very rough outlines for the solutions of the exercisesprovided in the text.

chapter 2

1. The lagrangian can also be rewritten as

L = −12

∂µ Ai∂µ Ai −

12

∂µBi∂µBi −m2 εij AiBj, (A.1)

where ε12 = −ε21 = 1. The equations of motion follow from the straigth-forward application of the Euler–Lagrange equations:

Ai −m2εijBj = 0, (A.2)

Bi + m2εij Aj = 0. (A.3)

2. Euler–Lagrange equations follow from

δL

δφi − ∂µδL

δ∂µφi = 0. (A.4)

One has to pay attention to the fact that Dµφi contains a piece without

derivatives, so that δDµφa

δφi = Aµεai and δDνφa

δ∂µφi = δai δ

µν . We then have

δLδφi = −δkj

δDµφk

δφi Dµφj + 16

(Cijk + Cjik + Cjki

)φjφk

= −δkj AµεkiDµφj + 12 C(ijk)φ

jφk.(A.5)

219

220 solutions

When varying the last term in the Lagrangian one has to take into accountthe fact that only the fully symmetrized terms in the C tensor survive,because they are contracted with three identical fields. Also

δL

δ∂µφi = −δajδDνφa

δ∂µφi Dνφj = −δijDµφj. (A.6)

We then get that

δijDµ(Dµφj) =12

C(ijk)φjφk, (A.7)

where

Dµ(Dµφj) = (∂µδjk + Aµεjk)[(∂µδka + Aµεka)φa]. (A.8)

3. Euler–Lagrange equations:

Φ + ηΦ− (Φ†Φ)Φ = 0. (A.9)

When Φ is constant we have ηΦ− (Φ†Φ)Φ = 0, which can be solved onlyfor Φ = 0 or when (Φ†Φ) = 1, which implies c1 = c2 = 0, |c3|2 + |c4|2 =1.

chapter 3

1. One by one:

• Q+ with a b = a/b is not a group because the associative propertyfails. Let a, b, c ∈ Q+, we have that

(a b) c =( a

b

) c =

abc=

abc

. (A.10)

On the other hand

a (b c) = a (

bc

)=

abc

=cab

. (A.11)

• Yes. The sum of two even integers 2m and 2n, where m, n ∈ Z, is2(m + n), which is even. Addition is associative. The unit is zero,which is an even integer, and the inverse of 2n is −2n.

solutions 221

• Z with a b = a− b is not a group because we do not have a neutralelement, nor multiplication is associative. For instance

(n−m)− p 6= n− (m− p). (A.12)

Also, while 0 could be a good identity from the right:

n− 0 = n, (A.13)

it is not a good identity from the left

0− n = −n 6= n (A.14)

and we know that any identity from the right in a group must alsobe an identity from the left.

• Yes. A generic element of Q∗ can be written asmn

, where m and nare coprime integers. Closure:(m

n

)(

kp

)=

mknp

, (A.15)

which reduces to the formxy

with x and y coprimes, after simplifi-

cations. The product is obviously associative. Neutral element is the

number 1. The inverse ofmn

isnm

.

• Associativity is clear and 1 is the obvious neutral element undermultiplication. The existence of the inverse is tricky. Let us try firstwith Z∗5 . Multiplying 2 with the other numbers

2 · 2 = 4, 2 · 3 = 6 ∼ 1, 2 · 4 = 8 ∼ 3 (A.16)

we see that 2−1 = 3 and 3−1 = 2. Multiplying 4 with the othernumbers

4 · 2 = 8 ∼ 3, 4 · 3 = 12 ∼ 2, 4 · 4 = 16 ∼ 1 (A.17)

and therefore 4−1 = 4. So there is always an inverse.

In general, take a ∈ Zn, we need to prove that there is always anelement b ∈ Zn such that ab = 1 mod n. First we show that for agiven a the map

Z∗n → Z∗n

b 7→ ab mod n(A.18)

222 solutions

is injective. If it was not, we should have two elements x, y ∈ Z∗nso that ax mod n = ay mod n. This, however, implies that a(x −y) mod n = 0 and therefore that x− y can be divided by n, but thisis true in Z∗n only if x = y. Since it is injective and maps all elementsof Z∗n into elements of Z∗n it must also be surjective. This means thatthere is an element b ∈ Z∗n for any a ∈ Z∗n such that ab mod n = 1.

2. For any group

(ab)−1 = b−1a−1. (A.19)

If G is Abelian any two elements commute and therefore we also havethat b−1a−1 = a−1b−1, which implies (ab)−1 = a−1b−1. On the otherhand if

(ab)−1 = a−1b−1 (A.20)

for all a, b ∈ G we can multiply from the right by ba getting

(ab)−1ba = 1 (A.21)

and multiplying from the left by ab we get

ba = ab. (A.22)

3. Assume we have 3 elements 1, a, b. We can fill immediately the linesand columns where the identity appears

1 a b1 1 a ba ab b

(A.23)

The product ab cannot be equal to a or b otherwise we would imply thateither b = 1 or a = 1. Hence ab = 1. At this point the rearrangementtheorem fixes the other entries, which must be different in each row andin each column

1 a b1 1 a ba a b 1

b b 1 a

(A.24)

solutions 223

4. Take G = 1, a, b, c. We start as usual by filling rows and columns asso-ciated to the identity:

1 a b c1 1 a b ca ab bc c

(A.25)

We then consider the product aa. This cannot be equal to a otherwiseit would imply a = 1. We then have two possibilities, either aa = 1, oraa = b (the case aa = c is just a relabeling of the elements). In the firstcase, the table is

1 a b c1 1 a b ca a 1

b bc c

(A.26)

and from the rearrangement theorem we immediately fill the rest of therow:

1 a b c1 1 a b ca a 1 c bb bc c

(A.27)

Also, ba 6= b, otherwise a = 1 and from the rearrangement theorem weget

1 a b c1 1 a b ca a 1 c bb b cc c b

(A.28)

224 solutions

The remaining entries could be completed in two possible ways:

1 a b c1 1 a b ca a 1 c bb b c 1 ac c b a 1

1 a b c1 1 a b ca a 1 c bb b c a 1

c c b 1 a

(A.29)

These are two distinct groups. We now come back to the other optionabove, aa = b. By the rearrangement theorem we complete immediatelythe second row and the second column

1 a b c1 1 a b ca a b c 1

b b cc c 1

(A.30)

Actually, at this point also the remaining entries are fixed:

1 a b c1 1 a b ca a b c 1

b b c 1 ac c 1 a b

(A.31)

This table is identical to the second in (A.29) if we relabel a 7→ b and b 7→ a.We then have only two possible groups. In the first one all the elementshave order 2 and in fact it is Z2 ×Z2, where the first is generated by aand the second by b. In the second we have an element of order 4, it isZ4, generated by b: b2 = a, b3 = ab = ba = c, b4 = a2 = 1.

5. Let us compute the multiplication table. From the data we have somemultiplications, but not all. First of all we introduce −I = (−1)I, −J =(−1)J and −K = (−1)K and we show that they cannot be any of theelements we already have. It is tedious, but straightforward. For instance,−I cannot be I or −1 otherwise we would have that I = 1 or −1 = 1.Then we prove that −I 6= 1, otherwise I = (−I)I = (−1)I2 = (−1)2 = 1.Finally we prove that −I 6= K, otherwise −IK = −1 and hence IK = 1 =KI, which implies J = KI J = K2 = −1. The same goes though for theother elements.

solutions 225

So we assume that we have at least 8 elements ±1,±I,±J,±K, whoseproducts are 1g = g1 = g, for any g, I2 = J2 = K2 = −1, (−1)2 = 1,(−1)I = −I, (−1)J = −J, (−1)K = −K and I J = K.

The rest of the table is computed by using the properties above. Forinstance, we have that I(−1) = I I I = (−1)I = −I and the same for Jand K. Also −I I = (−1)2 = 1 and hence also I(−I) = 1 (and the samefor J and K). From I J = K we also get JK = I by multiplying by I fromthe left and −K from the right. In the same fashion we also get KI = Jby multiplying by K from the left and −J from the right. From I J = Kwe also get KJ = −I by multiplying by J from the right and we can dosimilar manipulations to get IK = −J and J I = −K. The rest follows.

1 I J K −1 −I −J −K1 1 I J K −1 −I −J −KI I −1 K −J −I 1 −K JJ J −K −1 I −J K 1 −IK K J −I −1 −K −J I 1−1 −1 −I −J −K 1 I J K−I −I 1 −K J I −1 K −J−J −J K 1 −I J −K −1 I−K −K −J I 1 K J −I −1

(A.32)

6. Case by case:

• Z4 and S3 have different order, hence they cannot be isomorphic.#(Z4) = 4, #(S3) = 3! = 6.

• Z6 and S3 have the same order, but the first is Abelian, while thesecond is not. We conclude that they cannot be isomorphic.

• S3 and D3 are isomorphic. In general one can easily prove theisomorphism by mapping generators between them and showingthat they respect the relations in the presentations. In this case wecan associate a 3-cycle permutation to ρ and a 2-cycle to σ, so thatσρ = ρ−1σ. An instance that works is σ = (12)(3) ρ = (123).

7. A generic p-cycle of Sn (p ≤ n) has the form

a = (i1, i2, . . . , ip), (A.33)

226 solutions

where ij = 1, . . . , n, for j = 1, . . . , p, and must all be different betweenthemselves. The same p-cycle reversed is

b = (ip, ip−1, . . . , i1). (A.34)

Their product gives the identity

ab = (i1, i2, . . . , ip) (ip, ip−1, . . . , i2, i1) = 1 (A.35)

as follows fromip 7→ ip−1 7→ ip...i2 7→ i1 7→ i2i1 7→ ik 7→ i1

(A.36)

The same argument can be applied to the generic cycle, which is a disjointproduct of cycles.

8. Explicitly:

(34) (623174) (34) =

1 7→ 1 7→ 7 7→ 72 7→ 2 7→ 3 7→ 43 7→ 4 7→ 6 7→ 64 7→ 3 7→ 1 7→ 15 7→ 5 7→ 5 7→ 56 7→ 6 7→ 2 7→ 27 7→ 7 7→ 4 7→ 3

= (624173), (A.37)

(12) (2)(13)(45) (12) =

1 7→ 2 7→ 2 7→ 12 7→ 1 7→ 3 7→ 33 7→ 3 7→ 1 7→ 24 7→ 4 7→ 5 7→ 55 7→ 5 7→ 4 7→ 4

= (1)(23)(45), (A.38)

solutions 227

(34) (13)(452) (34) =

1 7→ 1 7→ 3 7→ 42 7→ 2 7→ 4 7→ 33 7→ 4 7→ 5 7→ 54 7→ 3 7→ 1 7→ 15 7→ 5 7→ 2 7→ 2

= (14)(352). (A.39)

We see that the result of the application of a 2-cycle (and its inverse) to a genericpermutation p, exchanges in p the position of the two numbers defining such2-cycle:

(ab) (. . . a . . . b . . .) (ab) = (. . . b . . . a . . .). (A.40)

9. A3 is given by all the elements in S3 with positive parity. These are 1,(123) = (13) (12) and (132) = (12) (13). The only group of order 3 isZ3 = 〈ρ | ρ3 = 1〉 and indeed the two are isomorphic as follows from theidentification ρ = (123).

10. We should identify two permutations of S4 with a and b satisfying a2 =1 = b3 and (ab)3 = 1. The order of a generic permutation is the smallestcommon multiple between the order of the disjoint cycles composing it.This means that a should be given by a 2-cycle or by the product of 2-cycles, while b should be a 3-cycle. All 3-cycles are in A4, because theyarise as the product of connected 2-cycles. On the other hand single 2-cycles are not in A4, while the product of two 2-cycles is in A4. Hence ashold be one among

(12)(34), (13)(24), (14)(23), (A.41)

and b one among

(1)(234), (1)(324), (2)(413), (2)(431),

(3)(412), (3)(421), (4)(123), (4)(132).(A.42)

We just need to check which couple satisfy the relation (ab)3 = 1. Aninstance is

a = (12)(34), b = (4)(123). (A.43)

228 solutions

chapter 4

1. Cosets have all their elements in common or they are distinct (see The-orem 4.1). This means that every element of the group must appear inexactly one distinct coset. Thus, since each coset has the same numberof elements, the number of distinct cosets multiplied by the number ofelements in the cosets is equal to the order of the group.

2. The center of a group G is C = g ∈ G | gh = hg, ∀h ∈ G. Let nowDn = 〈ρ, σ | ρn = σ2 = 1, ρσ = σρn−1〉. From this presentation wesee that any h ∈ Dn can be written as h = σaρb, where a = 0, 1 andb = 1, ..., n. In order to check what is the center we can then see for whichvalues of a and b the element h commutes with the generators of Dn. Thecommutator with σ gives

σaρbσ = σa+1ρn−b = σσaρb ⇔ b = n− b⇔ b = n/2, (A.44)

which is satisfied only for even n, or when b = 0. The commutator withρ gives

ρσaρb =

σaρb−1 6= σaρbρ, a = 1, n 6= 2ρb+1, a = 0

(A.45)

We then conclude that

CDn = 1, ρn/2, (A.46)

when n is even, and

CDn = 1, (A.47)

when n is odd. Explicitly:

σaρbρn/2 = ρn−n/2σaρb = ρn/2σaρb. (A.48)

3. We have 9 subgroups of order 2: 1, (12), 1, (13), 1, (14), 1, (23),1, (24), 1, (34), 1, (12)(34), 1, (13)(24), 1, (14)(23). 4 subgroupsof order 3: 1, (123), (132), 1, (124), (142), 1, (134), (143), 1, (234),(243). 7 subgroups of order 4, 3 of them are isomorphic to Z4, namely 1,(1234), (13)(24), (1432), 1, (1243), (14)(23), (1342), 1, (1324), (12)(34),(1423), and 4 are isomorphic to Z2×Z2: 1, (12)(34), (13)(24), (14)(23),1, (12), (34), (12)(34), 1, (13), (24), (13)(24), 1, (14), (23), (14)(23).Other subgroups of higher order are S3, D4 and A4.

solutions 229

4. Any two elements of an Abelian group G commute, i.e. ab = ba, ∀ a, b ∈ G.This implies that gH = Hg for any H < G and hence that left and rightcosets are identical.

5. As a first step we write the same permutations using disjoint cycles:

a = (13)(26)(45)(78), b = (1234567), c = (124)(365). (A.49)

a) From this expression it is obvious that they sit in different conju-gacy classes because the disjoint cycles describing them have differ-ent length.

b) Rewriting also b and c as product of 2-cycles, namely

b = (12)(23)(34)(45)(56)(67), c = (12)(24)(36)(56), (A.50)

we conclude that a, b, c ∈ A8.

c) H is generated by a e b, while the elements of G come from productsof a, b and c. It is straightforward to see that a and c commute

ca = ac = (1, 6, 4, 3, 2, 5)(78), (A.51)

while we can move any b to the left of any c obtaining a power of b:

cb = b2c = (1, 3, 5, 7, 2, 4, 6). (A.52)

We deduce that any g ∈ G can be written as the product g = cαh,with h ∈ H and α = 0, 1, 2. From this we have that n ∈ H impliesgng−1 ∈ H. In detail

gng−1 = cαhnh−1c3−α = cαnc3−α, (A.53)

where n ∈ H. Using what we just found we can then move cα to theright and we are left with an element in H.

6. The identity must be in any subgroup. We can then take the other ele-ments one by one and see if they form a subgroup. An obvious subgroupis H0 = 1,−1. For any of the other elements the set 1, Ji does notclose and, for any given i, needs also −1 = Ji Ji and −Ji = (−1)Ji. Theresult is a subgroup Hi = ±1,±Ji, where i = 1, 2, 3. Starting from1,−Ji we end up with the same subgroups Hi. Adding a third element

230 solutions

to H0 we end up again with Hi and if we add another element to Hi we re-cover the full Q. From the multiplication table found before one sees thatall such subgroups are normal and hence left and right quotients coincide.In detail Q/H0 = 1 ∼ −1, J1 ∼ −J1, J2 ∼ −J2, J3 ∼ −J3 ∼ Z2 ×Z2 andQ/H1 = 1 ∼ −1 ∼ J1 ∼ −J1, J2 ∼ −J2 ∼ J3 ∼ −J3 ∼ Z2 (and similarresults follow for Q/H2 and Q/H3).

7. Z2 × Z2 is not isomorphic to Z4 bacause the latter has an element oforder 4, while all the elements in the first have at most order 2. On theother hand Z2 ×Z3 ' Z6. In fact if a generates Z2 and b generates Z3,i.e. a2 = 1 and b3 = 1, we can construct an isomorphism between the twogroups by mapping φ : (a, b) 7→ ρ, where ρ is the Z6 generator. Explicitly:

φ(a, b) = ρ, φ(a2, b2) = φ(1, b2) = ρ2,

φ(a3, b3) = φ(a, 1) = ρ3, φ(a4, b4) = φ(1, b) = ρ4,

φ(a5, b5) = φ(a, b2) = ρ5, φ(a6, b6) = φ(1, 1) = 1.

(A.54)

8. We have A = 〈a | a2 = 1〉 and B = 〈b | bk = 1〉. The presentation for thesemidirect product A n B is C = 〈a, b | a2 = bk = 1, aφ(b)a = b〉, where φis a B automorphism.

Whenever φ(b) = b we have the direct products, which are Z2×Z2 ' D2,Z2 ×Z3 = Z6 and Z2 ×Z4, for k = 2, 3, 4. When k = 3 also φ(b) = b2 isa good automorphism, and the result is the group D3 = S3. When k = 4also φ(b) = b3 is an automorphism and the result is the group D4.

Other choices for φ(b) do not give an automorphism. We can never mapφ(b) = 1, because any homomorphism already has φ(1) = 1 and we wantφ to be 1 to 1. When k = 4 we also cannot choose φ(b) = b2, because it isnot surjective: φ(bn) = b2n will never cover b and b3.

chapter 5

1. Using an explicit matrix form

A⊕ B =

(A O

O B

). (A.55)

solutions 231

We then have that when A and C have the sane dimensions and so do Band D

(A⊕ B)(C⊕D) =

(A O

O B

)(C O

O D

)=

(AC O

O BD

)= AC⊕ BD.

(A.56)

For the direct product, using double indices as in the text, we have that

(A⊗ B)(ij)(kl)(C⊗ D)(kl)(mn) = AikBjlCkmDln

= (AC)im(BD)jn = (AC⊗ BD)(ij)(mn).(A.57)

From the same formulae we also have that

tr(A⊗ B) = (A⊗ B)(ij)(ij) = AiiBjj = (tr A)(tr B). (A.58)

We see that

γ0 =

−i

−i−i

−i

, γ1 =

−i

−ii

i

, (A.59)

γ2 =

−1

11

−1

, γ3 =

−i

ii−i

. (A.60)

The product

γ5 = −i γ0γ1γ2γ3 (A.61)

follows by taking the products in each factor of the tensor product defin-ing the matrices:

γ5 = −i (−iσ1σ2σ2σ2)⊗ (1σ1σ2σ3) = σ3 ⊗ 1. (A.62)

2. When k = 2 we had only one case Z2×Z2. Its elements are 1, a, b, ab = ba,where a and b are the generators of the two Z2 factors. We want to

232 solutions

represent these elements on a vector space of dimension #(Z2 ×Z2) = 4,whose basis is

|g〉 = |1〉, |a〉, |b〉, |ab〉 . (A.63)

The regular representation is then given by matrices DR(g) such thatDR(g)|g〉 = |gg〉. We then have

DR(1) = 14, DR(a) =

1

11

1

, (A.64)

DR(b) =

1

11

1

, DR(ab) =

1

11

1

. (A.65)

Since the group is abelian all its irreducible representations are 1-dimensional.This implies that we should reduce this 4-dimensional representation intothe sum of 4 1-dimensional representations. In other words, we can putall its elements in a diagonal form. This is done by computing the eigen-values and eigenspaces of the generators a and b. From this computationwe get that

1√4

1111

,1√4

1−1−11

,1√4

1−11−1

,1√4

11−1−1

(A.66)

are a basis for the 4 invariant subspaces of C4 so that

DR(a) =

1−1

1−1

, DR(b) =

1−1

−11

. (A.67)

3. D(a)D(a) = 12 and therefore they represent Z2. This representation mustbe reducible because all irreducible representations of abelian groups are

solutions 233

1-dimensional. Since D(1) and D(a) commute, we can diagonalize themsimultaneously and the diagonal form of a is simply(

1 00 −1

).

This had to be expected from the trace, which is invariant. trD(a) = 0implies that the two 1-dimensional representations of a that appear in the2-dimensional one provided in the text must have opposite signs.

4. a) The representation given in the text for the generators of the groupis block diagonal, with the first block given by a 3 by 3 matrix andthe second one 1-dimensional. Since all the other elements of A4 aregiven by products of these matrices, they all have the same block-diagonal structure.

b) From the decomposition of the matrices D(a) and D(b) we alreadyhave two representations, a 3-dimensional one and a 1-dimensionalone. We also know that we always have the trivial representationD(a) = D(b) = 1, which has also dimension 1. The last one is obvi-ously the complex conjugate of the 1-dimensional complex represen-tation D(a) = 1, D(b) = e2πi/3, namely D(a) = 1, D(b) = e4πi/3.

c) The elements of H have order 2, hence it is isomorphic to Z2 ×Z2. This is a normal subgroup of A4. Its elements are (12)(34),(13)(24), (14)(23) and the identity, which we know are closed underconjugation in S4 and hence also in A4. (Note that in general conjugacyclasses of Sn and An differ, because elements with the same length of cyclesmay be connected by odd cycles, which are not in An. However the onlyresult may be to split a conjugacy class in multiple conjugacy classes andhence the argument here goes through) The quotient gives Z3, generatedby b.

5. Let H = CN , then (ψ ⊗ χ) ∈ CN ⊗ CN and therefore we can representit by elements of CN2

, namely N × N matrices, M = ψχT . The opera-tion generated by ρ is the transposition D(ρ)[M] = MT . Obviously anyrepresentation D is such that D2 = 1, being a representation of S2.

Let us now verify the properties given in the text. From the action

P±(ψ⊗ χ) =12(ψ⊗ χ± χ⊗ ψ) (A.68)

234 solutions

we get

P±P±(ψ⊗ χ) = P±[

12 (ψ⊗ χ± χ⊗ ψ)

]= 1

4 [(ψ⊗ χ± χ⊗ ψ)± (χ⊗ ψ± ψ⊗ χ)]

= 12 (ψ⊗ χ± χ⊗ ψ) = P±(ψ⊗ χ).

(A.69)

It is then trivial that P+ + P− = 12 (1 + D + 1 − D) = 1 and P+P− =

P−P+ = 0.

We also see that DP± = D 1±D2 = 1

2 (D ± D2) = 12 (D ± 1) = ±P±.

This means that the spaces H± ≡ P±(H ⊗H) are invariant under theaction of D. Obviously one could further reduce the representation to1-dimensional invariant subspaces, because S2 is Abelian. Note also thatthe invariant subspacesH+ andH− are each other’s complement becauseP+P− = 0. The dimension of P+(H⊗H) equals that of N×N symmetricmatrices, and therefore is N(N+1)

2 . The dimension of P−(H⊗H) equals

that of N × N antisymmetric matrices: N(N−1)2 .

6. To be filled in...

chapter 6

1. The group D3 has presentation D3 = 〈ρ, σ | ρ3 = σ2 = 1, ρσ = σρ2〉. Notethat D(c)2 = D(d)2 = D(e)2 = D(1). We therefore have a representationof D3 where c, d, e are identified with σ, σρ and σρ2, while a and b areidentified with ρ and ρ2.

The representation is reducible. In fact the matrix D(c) commutes withall matrices of the representation provided in the text, but it is not pro-portional to the unit matrix.

By diagonalization: D(c) =(

1 00 −1

).

solutions 235

2. We can explicitly check (summing in the order the elements 1, ρ, ρ2, σ, σρand σρ2):

(D11, D11) =16

(1 +

14+

14+ 1 +

14+

14

)=

12

, (A.70)

(D11, D12) =16

(0− 1

4

√3 +

14

√3 + 0− 1

4

√3 +

14

√3)= 0,(A.71)

(D11, D21) =16

(0− 1

4

√3 +

14

√3 + 0 +

14

√3− 1

4

√3)= 0,(A.72)

(D11, D22) =16

(1− 1

4− 1

4− 1 +

14+

14

)= 0, (A.73)

(D12, D12) =16

(0 +

34+

34+ 0 +

34+

34

)=

12

, (A.74)

(D12, D21) =16

(0 +

34+

34+ 0− 3

4− 3

4

)= 0, (A.75)

(D12, D22) =16

(0 +

14

√3− 1

4

√3 + 0− 1

4

√3 +

14

√3)= 0,(A.76)

(D21, D21) =16

(0 +

34+

34+ 0 +

34+

34

)=

12

, (A.77)

(D21, D22) =16

(0 +

14

√3− 1

4

√3 + 0 +

14

√3− 1

4

√3)= 0,(A.78)

(D22, D22) =16

(1 +

14+

14+ 1 +

14+

14

)=

12

. (A.79)

They agree with the GOT, which, for S3 requires

(Dij, Dkl) =12

δikδjl (A.80)

for any irreducible representation.

236 solutions

3. Again, we can check:

(D11, D11) =13

(1 +

14+

14

)=

12

, (A.81)

(D11, D12) =13

(0−√

34

+

√3

4

)= 0, (A.82)

(D11, D21) =13

(0 +

√3

4−√

34

)= 0, (A.83)

(D11, D22) =13

(1 +

14+

14

)=

12

, (A.84)

(D12, D12) =13

(0 +

34+

34

)=

12

, (A.85)

(D12, D21) =13

(0− 3

4− 3

4

)= −1

2, (A.86)

(D12, D22) =13

(0−√

34

+

√3

4

)= 0, (A.87)

(D21, D21) =13

(1 +

14+

14

)=

12

, (A.88)

(D21, D22) =13

(0 +

√3

4−√

34

)= 0, (A.89)

(D22, D22) =13

(1 +

14+

14

)=

12

. (A.90)

In this case the theorem fails because of (A.84) and (A.86).

chapter 7

1. A presentation for D4 is 〈ρ, σ | ρ4 = σ2 = 1, σρ = ρ3σ〉. Its elements are

ρ, ρ2, ρ3, ρ4 = 1 = σ2, σ, σρ = ρ3σ, σρ2 = ρ2σ, σρ3 = ρσ. (A.91)

solutions 237

We can then write the generic element as σmρn, with m = 0, 1 and n =0, . . . , 3. Conjugacy classes are determined by taking all the elements ofthe form

σmρngρ−nσm (A.92)

for a given g ∈ D4 and m = 0, 1, n = 0, . . . , 3.

One obvious conjugacy class is

C1 = 1, (A.93)

because the identity commutes with all other elements.

When g = ρp we get

σρnρpρ−nσ = σρpσ = σ2ρ−p = ρ−p. (A.94)

This means that p = 1 is conjugate to p = 4 − 1 = 3 and p = 2 top = 4− 2 = 2:

C2 = ρ, ρ3, C3 = ρ2. (A.95)

For g = σρn we get that

σρmσρnρ−mσ = σρmσ2ρm−n = σρ2m−n. (A.96)

This means that when n = 0 we have conjugate elements σρ2m, for m =0, 1, 2, 3:

C4 = σ, σρ2. (A.97)

When n = 1 we have conjugate elements σρ2m−1 for m = 0, 1, 2, 3:

C5 = σρ, σρ3. (A.98)

Obvious subgroups are Z4 = 〈ρ | ρ4 = 1〉, which is normal, and Z2 =1, σ, which is not normal. Other Z2 subgroups are 1, σρ (whichis not normal, because σ(σρ)σ = σρ3 is not in the subgroup), 1, σρ3(again not normal, because σ(σρ3)σ = σρ is not in the subgroup), 1, σρ2(ρ(σρ2)ρ3 = σ is not in the subgroup) and Z2 = 1, ρ2, which is normal(it is given by the union of conjugacy classes).

We also have Z2×Z2 generated by 1, σρ, σρ3, ρ2 or 1, σ, ρ2, σρ2. Theseare again union of conjugacy classes and hence normal.

Quotients: D4/Z4 = Z2 = 1, σ, D4/Z2 = Z2 ×Z2 = 1 ∼ ρ2, σ ∼σρ2, ρ ∼ ρ3, σρ ∼ σρ3, D4/Z2 ×Z2 = Z2.

238 solutions

2. From the results of previous exercise we have 5 conjugacy classes andhence 5 irreducible representations. We know from the lectures that thereis a 2-dimensional faithful representation

D2(ρ) =

(0 1−1 0

), D2(σ) =

(−1 00 1

). (A.99)

We also know that we always have the trivial representation D1(ρ) =D1(σ) = 1. We are then left with 3 more irreducible representations, sothat

∑ N2a = 1 + 22 + 1 + 1 + 1 = 8 = #G. (A.100)

These follow from the determinant of the 2-dimensional representation:D1′(ρ) = 1, D1′(σ) = −1, while the other two are determined by thefact that D(σ)2 = 1, which means that D(σ) = ±1 and D(ρ)D(σ) =D(σ)D(ρ)3, which means that also D(ρ)2 = 1. We find D1”(ρ) = −1,D1”(σ) = 1 e D1′′′(ρ) = −1, D1′′′(σ) = −1.

3. The two representations are equivalent because

(χD1 , χD2) = 1. (A.101)

The explicit change of basis is given by the unitary matrix

U = eiπ/3(

11 −3−7 2

). (A.102)

4.

χD1⊗D2(g) = tr(D1 ⊗ D2)(g) = (D1 ⊗ D2)(g)(ij)(ij) = D1(g)iiD2(g)jj

= trD1(g) trD2(g) = χD1 · χD2 .(A.103)

5. If D is unitary D†D = 1. Then

χD(g−1) = tr D(g−1) = tr D−1(g) = tr D†(g) = tr D∗(g) = (χD(g))∗.

(A.104)

solutions 239

6. a) Irreducible representations are in 1 to 1 correspondence with conju-gacy classes. One obvious conjugacy class is given by C1 = 1. Itis straightforward to see that also −1 commutes with all the otherelements and hence C2−1. By using (Ji)−1 = −Ji and −J j Ji J j =2δij Ji − Ji (without summation over i), we also deduce that C2+i =±Ji. Hence we have 5 conjugacy classes.

Since #Q = 8 = ∑a N2a and a = 1, . . . , 5, we deduce that 8 = 1 + 1 +

1 + 1 + 22. The first 1-dimensional representation is the trivial oneD1(±1) = D1(±Ji) = 1. All other representations must preserve therelations in the presentation of the group and therefore D(Ji)D(J j) =δijD(−1) + εijkD(Jk). This implies that D(J1)2D(J2)2 = D(−1)2 =D(1) = 1 and D(J1)D(J2)D(J1) = D(J2), which gives D(J1)2 = 1and therefore D(J1) = ±1 and D(J2) = ±1. The three possibilitiesare then

D(±1) = D(±J1) = 1, D(±J2) = D(±J3) = −1, (A.105)

D(±1) = D(±J2) = 1, D(±J1) = D(±J3) = −1, (A.106)

D(±1) = D(±J3) = 1, D(±J2) = D(±J1) = −1. (A.107)

The 2-dimensional representation follows from the suggested matrix,by choosing

D(1) = U(0, 0, 0) =(

1 00 1

), D(−1) = U(π, 0, 0) =

(−1 00 −1

),

(A.108)

D(J1) = U(

0,π

2, 0)=

(i 00 −i

), D(−J1) = U

(0,−π

2, 0)=

(−i 00 i

),

(A.109)

D(J2) = U(π

2, 0, 0

)=

(0 1−1 0

), D(−J2) = U

(−π

2, 0, 0

)=

(0 −11 0

),

240 solutions

(A.110)

D(J3) = U(π

2, 0,

π

2

)=

(0 ii 0

), D(−J3) = U

(π

2, 0,−π

2

)=

(0 −i−i 0

).

(A.111)

b) The character table is straightforward from the traces of the matricesgiven above:

C1 C2 C3 C4 C51 1 1 1 1 11′ 1 1 1 −1 −11′′ 1 1 −1 1 −11′′′ 1 1 −1 −1 12 2 −2 0 0 0

(A.112)

7. a) A group N < G is a normal subroup iff gNg−1 ∈ N, ∀g ∈ G. Fromthe conjugacy classes we have that ga3g−1 = a3 and hence Z2 isnormal. Also Z3 is given by the union of conjugacy classes andhence is normal. Z4 is not normal: gbg−1 gives a2b and e a4b, whichare not part of the group

b) Since Z3 is normal, left and right quotients coincide. In this case thequotient is

Γ(Q6)/Z3 = Z4 = 1 ∼ a2 ∼ a4, b ∼ a2b ∼ a4b, b2 = a3 ∼ a5 ∼ a,

b3 = a3b ∼ a5b ∼ ab. (A.113)

c) All elements of the form anb are off-diagonal and hence χ1(anb) =χ2(anb) = χ3(anb) = χ4(anb) = 0. We then need only to com-pute the characters of an element for each of the conjugacy classesC2, C3 and C4. We have that χ1(a) = χ4(a) and χ2(a) = χ3(a),χ1(b2 = a3) = χ4(b2 = a3) = −2 and χ2(b2 = a3) = χ3(b2 = a3) = 2,χ1(a2) = χ4(a2) = 2 cos(2π/3) and χ2(a2) = χ3(a2) = 2 cos(4π/3).This implies that the characters of D1 and D4 are identical and there-fore they are equivalent representations. The same is true for D2 andD3. We can also check that (χ1, χ2) = 0:

(χ1, χ2) = 112 (1 · 2 · 2 + 2 · 2 cos π/3 · 2 cos(2/3π)

+ 2 · 4 cos(2/3π)2 + (−2)(2))= 0.

(A.114)

solutions 241

chapter 8

1. a) The tableaux are:

• which corresponds to the conjugacy class C1 of theidentity, with a unique element. The corresponding representa-tion has dimension 1;

• , representing the conjugacy class C2 of 4-cycles. It contains6 elements and the corresponding representation has dimension1′;

• , representing double 2-cycles. There are 3 such elementsin C3. The dimension of the corresponding irreps is 2;

• , representing the 8 3-cycles in C4. The dimension of thecorresponding irreps is 3;

• , representing 6 2-cycles in C5. The dimension of thecorresponding irreps is 3′.

We check easily that #G = 4! = 24 = 1 + 1 + 22 + 32 + 32.

b) The column C1 and the first row, given by χ(1), are straightforwardto fill:

C1 C2 C3 C4 C51 1 1 1 1 11′ 12 23 33′ 3

242 solutions

c) The alternating group selects elements that can be written with aneven number of 2-cycles. 4-cycles come from multiplication of 3 2-cycles (for instance (1234) = (12)(23)(34)) and therefore elementsin C2 are odd. The same is true for C5, which contains single 2-cycles. The other classes contain elements which appear from aneven number of 2-cycles.

We then fill the table accordingly:

C1 C2 C3 C4 C51 1 1 1 1 11′ 1 −1 1 1 −12 23 33′ 3

d) We can show that (χ2⊗1′ , χ2⊗1′) = 1 and it must be an irreduciblerepresentation of dimension 2. Since there is only one such represen-tation it must be equivalent to the one we already have. The resultfollows from χ2⊗1′(g) = χ2(g)χ1′(g):

(χ2⊗1′ , χ2⊗1′) =1

12 ∑g(χ2⊗1′)∗(g)(χ2⊗1′)(g))

= 112 ∑g |χ2(g)|2 · |χ1′(g)|2 = 1

12 ∑g |χ2(g)|2 = (χ2, χ2),(A.115)

because χ1′(g) = ±1, for any g ∈ S4.

If 2⊗ 1′ ' 2, then χ2χ1′ = χ2 and therefore it must be vanishing forC2 and for C5:

C1 C2 C3 C4 C51 1 1 1 1 11′ 1 −1 1 1 −12 2 0 03 33′ 3

e) Orthonormality tells us that (χ1, χ2) = 0 = 124 (χ2(1) + 2 χ2(C3) +

8 χ2(C5)) and (χ2, χ2) = 1 = 124 (χ2(1)2 + 3 χ2(C3)

2 + 8 χ2(C4)2). If

solutions 243

only solutions with integer numbers can be accepted we fix χ2(C4) =−1 and χ2(1) = χ2(C3) = 2:

C1 C2 C3 C4 C51 1 1 1 1 11′ 1 −1 1 1 −12 2 0 2 −1 03 33′ 3

f) By the same line of reasoning as the one above 3⊗ 1′ is an irreduciblerepresentation and therefore it must be either equivalent to 3 or to3′. If we assume that it is equivalent to 3 then we should againconclude that we have a 0 in correspondence with the characters ofthe elements in C2 and C5, but then it would be impossible to fulfillat the same time the condition following from (χ3, χ3) = 1 and thosefrom (χ3, χ1) = (χ3, χ2) = 0. We then conclude that 3⊗ 1′ ' 3′.

g) Writing down the system of orthonormality equations:

C1 C2 C3 C4 C51 1 1 1 1 11′ 1 −1 1 1 −12 2 0 2 −1 03 3 1 −1 0 −13′ 3 −1 −1 0 1

We could exchange 3 and 3′ obtaining again a consistent matrix.

244 solutions

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GROUP THEORY AND PHYSICS - Infis-Ufugerson/grupos/3 - Great books/Teoria de...4 symmetries in physics symmetry arguments. This is not a course in group theory and/or differen-tial