Roots and Weights

8/10/2019 Roots and Weights

1/19

Particles I, Tutorial notesSessions I-III: Roots & Weights

Kfir Blum

June 3, 2008

Comments/corrections regarding these notes will be appreciated.My Email address is: [email protected]

Contents

1 Groups and representations 2

1.1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Representations: equivalence, unitarity, (ir)reducibility . . . . . . . . . . . . . . . 21.3 Schurs lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Lie groups and algebras 3

2.1 Lie groups and algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Jacobi identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 The adjoint representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.3.1 States in the adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 U(1) factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Roots & weights: Classifying irreps 7

3.1 Cartan subalgebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Roots & weights . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.3 SU(2) subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.4 Example: SU(3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.4.1 The fundamental representation ofS U(3) . . . . . . . . . . . . . . . . . . 83.4.2 The adjoint representation ofSU(3) . . . . . . . . . . . . . . . . . . . . . 9

3.5 Weighteology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.5.1 The master formula of roots and weights . . . . . . . . . . . . . . . . . . . 103.5.2 Simple roots & Highest weight . . . . . . . . . . . . . . . . . . . . . . . . 11

3.6 Building all the irreps, and fundamental weights . . . . . . . . . . . . . . . . . . 123.6.1 Weyl reflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.6.2 Another example: The 6 ofS U(3) . . . . . . . . . . . . . . . . . . . . . . 14

3.7 Product decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

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1 Groups and representations

1.1 Groups

A group is a set G with a multiplicationsuch that:1. x,y G x y G2.e G such thatx G : e x= x e= x3. x,y,z G x (y z) = (x y) z4. x G there is an inverse x1 G such that x x1 =x1 x= e

In what follows, I will denote xy x y.

A group is finite if it has afinitenumber of elements. O.w. it is infinite.Theorderof a finite group Gis the number of elements in G.

A group is abelian if g1, g2 G : g1g2= g2g1.A subgroup ofGis a group Hwhich is contained in G.All elements of an invariant subgroup H of G transform among themselves under the adjointaction of any element ofG

g G, h1 H,h2 H s.t. gh1 = h2g

A simple group contains no nontrivial invariant subgroup.A semisimple group contains no nontrivial abelian invariant subgroup.

1.2 Representations: equivalence, unitarity, (ir)reducibility

A representation ofG is a mapping, D ofG, into linear operators such that:

1. D(e) =I

2. D(g1)D(g2) =D(g1g2)

Thedimensionof a representation is the dimension of the vector space on which it acts.

Let D be a representation of dimension N. We span the states in the vector space of thisrepresentation by the set {|i} , i 1,...,N. Thematrix elementsof the linear operatorsD(g) inthis representation are given by [D(g)]ij = i| D(g) |j.One may choose to represent the states in a more convenient way, by making linear transforma-tions |i |i= S |i. An invertible transformationS induces a similarity transformation onthe linear operators D D= S DS1. D and D areequivalent representations.

A representation is unitary ifD (g) is unitary for every g : g G : D(g) = D(g)1.All of the representations of a finite group are equivalent to unitary representations.

An invariant subspace is a subspace which transforms within itself under the action of D(g)for any g G.A representation is reducible if it contains an invariant subspace.

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The restriction of the representation to the invariant subspace is itself a representation.A representation is completely reducibleif it is equivalent to a block diagonal form:

D1(g) 0 . . .

0 D2(g) . . ....

... . . .

(1)where all of the Di(g) are not reducible. (i.e. irreducible or irrep.)Eq.(1) means that a completely reducible rep. can be decomposed into a direct sum of irreps.

1.3 Schurs lemma

Schurs lemma: Let D(g) be a finite dimensional irrep. of G. Then an operator A whichcommutes with D (g) for every g must be proportional to the identity:

[D(g), A] = 0 g G A I (2)

Schurs lemma has crucial implications for QM, where the common practice is precisely to classifythe eigenspaces of hermitian operators. In QM a symmetry is typically realized by putting thestates of the Hilbert space in unitary representations. The transformation law is

| D(g) | , | | D(g)

for states, andO D(g)OD(g)

for operators. An invariant observable satisfies O = D(g)OD(g), which, since D is unitary,means that we may use Schurs lemma. Eq.(2) then says that the eigenstates of O can be

arranged in irreps Dk ofG, where each irrep corresponds to some fixed eigenvalue ok .

2 Lie groups and algebras

2.1 Lie groups and algebras

The elements of a Lie group depend smoothly on some continuous parameter: g = g(). Wemay label the operators in a representation of such group by D() = D(g()).We will mainly deal with compact groups, i.e. groups for which is confined to some finitevolume parameter space.Any representation of a compact lie group is equivalent to a unitary representation.

It is convenient to set g(= 0) = e

Now we can use an exponential parametrization in order to represent a group element in thevicinity of the identity:

D() =eiaXa (3)

The Xa, a= 1,...,Nare called the generators:

Xa= i

D

a

=0

(4)

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TakingN as the minimal number of independent real parameters required in order to representany element in the group, the generators Xa are linearly independent. Due to the form of (3),and as long as we stick to real , the Xa of a unitary representation are hermitian. They span

a linear vector space of hermitian operators.

Since a representation maintains the closure property of the group, we must have

eiXeiX =eiX (5)

Take log of both sides and expand in the vicinity of the unity

iX= ln

1 +i(X+ X) 1

2(X+ X)2 1

2[X,X] +...

=

i(+)X 12

[X,X] +... (6)

We can now define = 2(+

), thereby (5) becomes

X= i [X,X]

This tells us that the vector space of the X is closed under commutation:

[Xa, Xb] =ifabcXc (7)

with indices now explicit. is given by c= abfabc.

The relation (7) is called the Lie algebra.Thestructure constants fabc obey

fabc= fbac

They are determined by the multiplication law of the group. One can verify that the ellipses onthe right hand side of (6), i.e. terms of higher order in the arbitrary parameters, , dependagain only onfabc. This means that the fabcalsodeterminethe multiplication law of the group.The structure constants can be computed in any nontrivial representation. The result will bethe same every time. We may thus think of representations of the algebra as representationsof the group. Typically, when people talk about groups in physics, they actually think of somerepresentation of the group. Therefore it turns out that you may typically just swap the wordgroup by the word algebra. Notions like reducibility, invariance or compactness can be carriedto the algebra in obvious ways.

An invariant subalgebra Y in the algebra X is some subset of the generators which transformwithin itself upon commutation with any element of the algebra

x X, y Y : [x, y] YAn invariant subalgebra generates an invariant subgroup.A simple algebracontains no nontrivial invariant subalgebra.A semisimple algebracontains no nontrivial abelian invariant subalgebra.

Exercise: show that if a group has a unitary representation, then the structure constantsare real.

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2.2 Jacobi identity

The generators Xobey the Jacobi identity

[Xa, [Xb, Xc]] + [Xc, [Xa, Xb]] + [Xb, [Xc, Xa]] = 0 (8)

Which is easy to prove for matrices, but true also in a more abstract sense for linear operators.A similar identity occurs for the structure constants

fbcdfade+fabdfcde+fcadfbde= 0 (9)

2.3 The adjoint representation

Define the set of matrices(Ta)bc = ifabc (10)

Using the Jacobi identity in the form (9), one finds

[Ta, Tb] =ifabcTc

This means that (10) is a representation of the algebra, with dimension N. This representationis called the adjoint representation.

We should define a scalar product on the linear space of the generators. Consider

tr (TaTb) (11)

This is a real symmetric matrix. We can diagonalize it by performing a linear transformation onthe generators:

Xa Xa = SabXb (12)with some orthogonal matrix S. Eq.(12) induces the following transformation of the structure

constantsfabc fabc= SadSbefdegS1gc

The matrices Ta of the adjoint representation transform by

Ta Ta= Sab(STbS1) (13)combining a similarity transformation simultaneously with a linear transformation of the form(12). In the trace, however, only the latter transformation matters and we obtain

tr (T T) tr (TT) = S tr (T T) ST

Selecting a diagonal basis, we write

tr (TaTb) =aab (no sum!)

However, we still have freedom to re-scale our generators. We shall assume that all of the a arepositive. This will be the case if the algebra is compact1 and semisimple2. Then we can just goahead and write

tr (TaTb) =ab (14)

1Non-compact Lie algebras, in which some of the a are negative, do not have non-trivial, finite dimensionalunitary representations. They still, however, play an important role in physics: The lorentz algebra is an example.

2The requirement for the algebra to be semisimple is discussed in section 2.4.

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with some positive .

The basis in which (14) holds is particularly nice: In this basis, the structure constants are

completely antisymmetric fabc= fbac= facb= ...This property follows because we can write

fabc= i

tr ([Ta, Tb]Tc)

The Ta matrices are therefore antisymmetric and pure imaginary, and so hermitian. Thus, inthis basis, the adjoint representation is unitary.

2.3.1 States in the adjoint

The states in the adjoint correspond to the generators themselves

Xa |Xa

We can find the action of the generators on the states by inserting a complete set of states:

Xa |Xb = |Xc Xc| Xa |Xb = |Xc (Ta)cb = |ifacbXc = |[Xa, Xb] (15)

The scalar product is defined on this space by

Xa|Xb = 1

tr

TaTb

(16)

The dagger is there because we will sometimes work with complex linear combinations of gener-ators.

2.4 U(1) factors

In order to write (14), we had to assume that our algebra was compact and semisimple. We nowdiscuss the notion of a semisimple algebra.

An algebra is semisimple if it does not contain a non-trivial abelian invariant subalgebra. Anabelian subalgebra follows from an abelian subgroup. Such invariant subgroup is called a U(1)

factor. In the algebra, it corresponds to a specific generatorXwhich commutes with all of theother generators. The structure constants give us no information about this subalgebra.However, Schurs lemma tells us that the existence of such a U(1) factor is not really much trou-ble. The representation of the group will split into irreps of the semisimple part of the algebra,each irrep characterized by some fixed eigenvalue under the generatorXof the U(1) factor.In order to classify the algebra representations, we may therefore keep to semisimple algebras.These are built by factoring simple algebras to each other.For example:

SU(2) : simple. SU(2) SU(3) : semisimple. U(2) U(1) SU(2) : neither simple nor semisimple.

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3 Roots & weights: Classifying irreps

The useful thing about the adjoint representation is that it is a representation of the algebra onitself. This means that every structure we expose in the adjoint representation is in fact inherentto the algebra: studying the adjoint representation is studying the algebra. This notion getsexploited to a great extent in the analysis of Roots & Weights.

3.1 Cartan subalgebra

Suppose theXa,a = 1,...,Nare the generators of a given irrep of dimension3 D. The maximum

subset of generators which can be diagonalized simultaneously is called the Cartan subalgebra.Denote these generators by Hi, i= 1,...,M. They satisfy

Hi= Hi , [Hi, Hj ] = 0 i, j (17)

Mis called the rankof the algebra.We can choose a basis in which

tr (HiHj) = Dij

After diagonalizing all of the Hi, we have for the states of the representation

Hi | =i | is a real vector of length M; We have D of those.The vector is called the weight vector.The entries i are the weights.In different irreps we will get different weight vectors.

3.2 Roots & weights

We now return for a moment to the adjoint representation.The weights of the adjoint representation are called roots.Using the conventions of section 3.1 we have, in the adjoint, D = N. Due to (15,17) we knowthat

Hi |Hj = 0 (18)for any Hi, Hj in the cartan. We can label the other N M states by|E, where

Hi |E =i |E (19)This implies

[Hi, E] =iE (20)

It follows that[Hi, E

] =

iE

(21)

And, thereforeE= E (22)

The E are not hermitian. They should remind you of the raising and lowering operators in theanalysis of the irreps ofSU(2). Indeed, one sees

HiE | = ([Hi, E] +EHi) | = (i i) E | (23)3Make sure you understand the difference between the dimension of the group (N) and the dimension of the

representation (D).

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The convention (16) ensures us of the normalization

Hi|Hj =ij

E

|E

= (24)

You can easily check that this is so, by noting that the similarity transformation connecting theEs to the Xas is unitary: it is the diagonalizing transformation for the hermitian operatorsHi.Notice that the state

E |E (25)has weight zero. This implies that the operator [E, E] is in the cartan:

[E, E] = iHi

We can compute by projecting the state (25) onto a cartan state:

i= Hi| E |E = 1

tr (Hi[E, E]) = 1

tr (E[Hi, E]) =i (26)

so[E, E] =iHi (27)

Now comes an important point.Equations (20-22) and (27) that we have found are operator equations. These equations, togetherwith the conclusion (23), also hold in an arbitrary representation!

3.3 SU(2) subalgebras

Putting equations (22,27) together, we can make explicit the relation between the Es andSU(2) irreps.For every non-zero root vector there exists an SU(2) subalgebra, with the generators:

E3= iHi

2 , E=

E|| (28)

such that[E3, E] = E , [E+, E] =E3

3.4 Example: SU(3)

3.4.1 The fundamental representation ofSU(3)

The fundamental representation of the group SU(3) consists of 33 unitary matrices withdeterminant 1, acting on 3-dimensional complex vectors. It is also calledthe3. These matricesare generated by 3 3 hermitian traceless matrices, which form an 8-dimensional algebra. Abasis for the algebra is

1= 12

0 1 01 0 00 0 0

2= 12 0i 0i 0 0

0 0 0

3= 12 1 0 001 0

0 0 0

4= 12 0 0 10 0 0

1 0 0

5=

12

0 0i0 0 0i 0 0

6= 12 0 0 00 0 1

0 1 0

7= 12 0 0 00 0i

0 i 0

8= 123 1 0 00 1 0

0 02

(29)

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This basis is normalized such that

tr (ij) = 1

2ij

The cartan consists ofH1= 3, H2= 8

The weights are: 100

12

, 1

2

3

,

010

12

, 1

2

3

,

001

0, 13

(30)

Figure 1 shows a geometrical diagram of the weights. Since the cartan is two dimensional, wecan draw the weights on a two dimensional plain. There is no degeneracy: Each weight vectorcorresponds to a unique state.

3.4.2 The adjoint representation of SU(3)

We now continue to construct the adjoint representation ofSU(3). The vector space on whichthis representation acts is eight dimensional - this representation is also simply called the 8.However, using our weight diagram from the previous section, we dont even need to write downexplicitly the eight-by-eight matrices which act on this space.

We know that the six generators which are not in the cartan can be rearranged as raisingand lowering operators ofSU(2) subalgebras. These subalgebras appear as SU(2) irreps whichare contained in the SU(3) representation we have found. The arrows connecting the states infigure 1 reveal these S U(2) irreps: We have three of those, each of them spin half. The weightvectors we have found are separated by the vectors: (I chose the labels for later convenience)

(3) = (1, 0) , (1) = 12

, 32 , (2) = 1

2, 3

2 (31)

The (i)s are precisely the root vectors of the non-cartan generators in the adjoint. Indeed, theEs are given by

E(3) = 1

2(1+i2) , E(1) =

12

(4+i5) , E(2) = 1

2(6 i7) (32)

The two cartan generators will correspond to two vanishing root vectors: All together, we haveeight roots corresponding to eight states. We can now go ahead and draw a weight diagramfor the adjoint representation. This diagram is depicted in figure 2. The non vanishing rootsform a regular hexagon. The states corresponding to the cartan generators are degenerate:These are two states with the same root. I have marked with arrows 3 specific SU(2) irreps (all

corresponding to the same S U(2) subalgebra) which are hiding in this S U(3) representation.

3.5 Weighteology

It may be the case that several states in a representation will have the same weight vector. Theadjoint representation is an example: All of the states corresponding to cartan generators havezero weight vectors. We would like to have a systematic method to label all of the states in agiven representation by unique labels.In order to do this, lets start working with the tools of section 3.2.

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3.5.1 The master formula of roots and weights

Define:E

|

=N

,

|

(33)

Now compute:

| [E, E] | = | iHi | = =| EE | | EE | = |N,|2 |N,|2 (34)

However, notice that

N,= | E | = | E | = | E | = N,so that (34) implies

= |N,|2 |N,|2 (35)Because theEare raising and lowering operators ofS U(2) subalgebras, we know that, for anystate|, there exists p, q N such that

E |+p =E | q = 0In other words, the N, vanish for high enough or low enough :

N,+p= 0 , N,q = N,(q+1)= 0

We can stack evaluations of (35) to write the following tower of equations:

|N,+(p1)|2 0 = (+p)|N,+(p2)|2 |N,+(p1)|2 = (+ (p 1))

...

|N,q|2 |N,(q1)|2 = ( (q 1))

0 |N,q|2

= ( q)

(36)

Summing both sides of the tower, we obtain

0 = (p+q+ 1)

+

2

2 (p q)

We conclude:

2

= qp

2 =

integer

2 (37)

This is a surprising constraint on the weight vectors! We will call (37) the master formula.

Specifically, in the adjoint

2 =

m

2but, just the same

2

= m

2

We can write:( )2

22 = cos2 =

mm

4 (38)

There are only very limited choices to the relative orientation of root vectors. I wrote them intable 1.

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Table 1: relative root orientations

mm

0 90

1 60, 120

2 45, 135

3 30, 150

4 180

3.5.2 Simple roots & Highest weight

A weight vector is positiveif its first non-zero entry is positive.

We say > if is positive.A highest weightwithin an irrep is larger than any other weight in the irrep.The highest weight of an irrep is unique.In the adjoint, asimple rootis a positive root which cannot be written as a sum of positive roots.

Some facts concerning simple roots:

1. If, are simple, than is not a root. The master formula (37) then implies that

E |E = 0 2

= 12

p , (i.e., q= 0) (39)

Similarly,

E |

E = 0

2 = 1

2p ,

(i.e., q

= 0) (40)This restricts the length and the relative orientation of any two simple roots:

2

2 =

p

p , cos =

pp

2

Note also that

2 <

This follows because, by (39-40), the cosine cannot be positive.

2. It follows that the simple roots are linearly independent.In fact, they are also a complete set: Any positive root can be decomposed as a sum of

simple roots: = ri(i), with ri = integer 0.3. The number of simple roots is equal to the rank of the group, and so to the number of

generators in the cartan.

Knowing the ps of all the simple roots, we can construct the whole algebra.This is done by finding all the roots, through recursive application of the master formula.

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Example: Take the adjoint representation ofS U(3).The two simple roots are

(1) = 12 ,

32 , (2) = 12 ,

32 They give (1) = (2) = 1 , (1) (2) = 1

2 p= p= 1

Since both (1), (2) are simple roots, we know that the qs are zero.Thus (1) can raise (2) once (and vice verse), giving the third positive root (3) =(1) +(2). Note that (3) cannot be raised further, because p = p = 1. So (1) + 2(2), forexample, is nota root.Since roots come in pairs(k), we now know all of the negative roots, too.

3.6 Building all the irreps, and fundamental weightsLet (k), k = 1,...,m be the simple roots. Consider a weight in some arbitrary irrep. is the highest weight if and only if+(k) is not a weight for any k.

This means that E(k) | = 0 k, so that

(k) (k)2 = qk2 0 , (i.e. pk = 0) (41)Since the (k) are independent and since m is the rank of the group, the qks determine thehighest weight completely.

Because the highest weight of an irrep is unique, we arrive at an important result, instruct-ing us how to build all of the irreps of a Lie algebra:

Given a set of integers qk, k = 1,...,m, we have an irrep with highest weight , defineduniquely through (41).

The qks are called the Dynkin indices.

We define the fundamental weightsby the set of weights (i) such that

(l) : qk = kl

Example: SU(3)...The two simple roots are

(1) =

1

2,

3

2

, (2) =

1

2,

3

2

We have two fundamental weights, corresponding to the two different choices:

q= (1, 0) or q= (0, 1)

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The first choice, which we will label by (1), gives us

(1) (1) = 12(2)

(1)

= 0 (1) =

1

2

, 1

23The fundamental weight (1) is just the highest weight of the fundamental representation3, which we already know. We will also call this representation (1, 0).

In a similar way the second choice, labeled (2), gives us

(1) (2) = 0(2) (2) = 1

2

(2) =

1

2, 1

2

3

The fundamental weight (2) is the highest weight of another three-dimensional irrep ofSU(3), called the anti-fundamentalrepresentation, or 3. We will also call this representa-tion (0, 1). We can continue to construct this representation:Since q1= 0, we can not lower (2) by(1). We can, however, lower it once with (2). Sothe weight

(2) (2) =

0, 1

3

does belong in the irrep. The master formula dictates:

(1) (2) (2)(1)2 =12 = q p

2 (42)

However, we know that[E(1) , E(2) ] = 0

because (1)

(2)

isnt a root. (HW!) Therefore we know that

E(1)E(2)(2) =E(2)E(1) (2) = 0

telling us that p = 0. Eq.(42) than implies q = 1. We have found the last missing stateof the irrep: It is

(2) (2) (1) =

12

, 12

3

Comparing the three weights that we have found to the weights (30) of the 3, we see thatthe weight diagram for the 3 is obtained by reflecting the diagram of figure 1 around the1 axis.

It is true in general that if the set of matrices Xa form a representation D of the algebra, than

the matricesXa obtained from the Xa by Xa = Xaalso form a representation of the algebra, called D .The 3 and the 3 are an example for a situation in which the D and the D are not equivalent:There is no similarity transformation which maps the set of matrices a of (29) intoa.

IfD is equivalent to D, than D is called a real representation.

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3.6.1 Weyl reflections

The fact that roots come in pairs which generate SU(2) subalgebras (together with the car-tan element

H) is manifest in weight diagrams through Weyl reflection symmetry.

A Weyl reflection operation on the weight diagram corresponds to a reflection of the weightsaround an axis perpendicular to one of the roots. Weyl reflection symmetry means that theweight diagram of any representation is invariant under such operation. Multiple reflections arealso a symmetry.

Weyl reflection symmetry is easily verified for the SU(3) irreps we have studied so far. Allare invariant for reflection around the axes perpendicular to any of (1), (2) or (3) of (31).This property is responsible for the regular triangular and hexagonal patterns in the irreps ofSU(3).Weyl reflection symmetry is very useful when constructing irreps from highest weights, savingmuch computing time.

3.6.2 Another example: The6 of SU(3)Lets construct another irrep ofS U(3): The one with highest weight defined by

q= (2, 0)

Once again, we will not need to write explicitly the matrix realization of the generators. All wewill need is Weighteology.

The highest weight is just

= 2(1) =

1,

13

Two states we immediately recognize are those corresponding to the weights

(1) = 12 , 123 2(1) =

0, 2

3

We start by checking the possibility to raise or lower the first of these weights by (2), using themaster formula:

(2)

(1)

=

1

2,

3

2

1

2, 1

2

3

=

1

2 =

q p2

I leave it as an exercise to show that, in fact4, p= 0. This gives q= 1, and so a forth weight:

(1)

(2) = 0,

1

3Obviously, this weight cannot be raised by (1), since this would contradict being highestweight. So p= 0. It can, however, be lowered by (1):

(1)

(1) (2)

= 1

2 =

q

24The proof follows from a two-step logic: A. Assume that there is a state with the weight (1) + (2). B.

Weyl reflection around the vector perpendicular to (1) then implies that there exists a weight which is largerthan . This contradicts with being highest weight.

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So we have a fifth weight

2(1) (2) =

12

, 12

3

We proceed by checking the possibility to raise or lower the weight 2(1) by(2), using themaster formula:

(2)

2(1)

=

1

2,

3

2

0, 23

= 1 =

q p2

You can show that the consistent solution is q= 2, p= 0.The weight corresponding to lowering once by (2) is just the fifth weight weve found above.The weight corresponding to lowering twice by (2) is the sixth and last of this irrep:

2(1) 2(2) =

1, 13

It is indeed the last weight, since it cannot be neither lowered nor raised by (1):

(1)

2(1) 2(2)

= 0 = q p

2

and you can show that the correct solution is q = p= 0.

The weight diagram we have found is depicted in figure 3.

3.7 Product decomposition

The product of irreps is usually not an irrep.

However, theres a very painless way to decompose a product of irreps into a direct sum of irreps.It consists of writing down all of the combinations of weight sums, arranging these new (summed)weights in degenerate bunches, and then removing the revealed irreps one by one until theresnothing left. I will demonstrate the method with S U(3).

By now, we are familiar with five inequivalent irreps of SU(3): These are the 8, 6, 3, 3 andthe trivial rep 1. Let us find how the product 3 3 is decomposed into a direct sum.

Putting together what weve learned in sections 3.4-3.6, we construct new weights by addingall the weights of the 3 to each weight of the 3, keeping track of the multiplicity. I have collectedthe result in table 2. There are nine states, as the sum over multiplicities confirms. Substitutingfor the vectors (k), (k) enables us to plot the weight diagram. The three-fold degenerate statehas weight zero.

Subtracting the diagram of the 8, we are left with a simple irrep consisting of one zero weight.This is of course just the trivial representation.We conclude that:

3 3 = 8 + 1 (43)Its convenient to do what we have just done pictorially, with weight diagrams. To re-derive theresult above, take the following steps:

1. Draw the weight diagram of the 3, the triangle.

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Table 2: States of 3 3

weight multiplicity

(1) +(2) 1

(1) +(2) (1) 1(1) +(2) (1) (2) 3

(1) +(2) (2) 1(1) +(2) (1) 2(2) 1(1) +(2) 2(1) (2) 1

(1) +(2) 2(1) 2(2) 1

2. For each vertex of the triangle, draw the (reversed triangle) weight diagram of the 3,centered around it. The result is the diagram of (43). I tried to illustrate this process infigure 4.

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Figure 1: SU(3) fundamental

Figure 2: SU(3) adjoint

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Figure 3: SU(3) 6

Figure 4: 3 3 = 8 + 1

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References

[1] H. Georgi, Lie Algebras in Particle Physics

[2] B. C. Hall, Lie Groups, Lie Algebras, and representations

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Roots and Weights