Finite dimensional group representationsfolk.ntnu.no/myrheim/FY8104-09/notes09part02.pdfFinite...

Chapter 3

Finite dimensional group

representations

In the theory of linear representations of groups it is useful to distinguish between repre-sentations on finite and on infinite dimensional vector spaces. Another useful distinction isbetween finite and infinite groups, although the fundamental difference is rather between thefinite groups and the compact infinite groups, on the one hand, and the non-compact infinitegroups, on the other hand.

In this chapter and the next we will consider finite dimensional representations of groups,beginning with results that do not depend on whether the group is finite or infinite. In thenext chapter we will restrict ourselves mostly to the case of representations of finite groupson finite dimensional complex vector spaces.

It is possible to study representations on vector spaces based on a general number fieldF, and many results are valid in such a general setting. The cases of physical interest aremainly F = R, the real numbers, and F = C, the complex numbers. We will specialize hereto complex representations, which are of special interest both from the mathematical and thephysical point of view. It will usually be obvious which results depend on special propertiesof the complex numbers, and which results hold also for other number fields.

The great mathematical advantage of the complex numbers is the basic fact that theyare algebraically complete, that every polynomial of degree n with complex coefficients hasexactly n complex roots. This result, known as the fundamental theorem of algebra, simplifiesparts of the theory. Complex representations are of physical interest because states of aphysical system are represented in quantum mechanics as vectors in a complex Hilbert space.Real group representations also occur in physics, in order to understand them we will haveto relate them to complex representations.

Definition 3.1 A “linear representation” (or “representation”) ρ of the group G on thecomplex vector space V is a realization of G as a group of linear transformations on V . Thegroup element g is represented as the linear transformation ρ(g), and the group product isrepresented as composition of linear transformations, ρ(gh) = ρ(g) ρ(h).

The representation is “faithful” if the function ρ : g 7→ ρ(g) is one to one.The function ρ should be continuous if G is a topological group, and differentiable (smooth)

if G is a Lie group.Two representations ρ on V and σ on W are “equivalent” if there exists an invertible

linear transformation S : V → W such that σ(g) = Sρ(g)S−1 for all g ∈ G.

41

42 CHAPTER 3. FINITE DIMENSIONAL GROUP REPRESENTATIONS

A “matrix representation” D represents every group element g as an n × n matrix D(g),in such a way that the group product becomes matrix multiplication, D(gh) = D(g)D(h).

A matrix representation is just a special case of a linear representation, since a complexn × n matrix D(g) may be regarded as a linear transformation on C

n.Since a representation ρ is a group homomorphism from G into the group of invertible

linear transformations on V , it follows from the general group theory that ρ(e) = I, theunit element e ∈ G is represented by the identity transformation I on V . It also follows thatρ(g−1) = ρ(g)−1, where g and g−1 are inverse group elements, and ρ(g) and ρ(g)−1 are inverselinear transformations.

A representation ρ on V and an invertible linear transformation S : V → W always definean equivalent representation SρS−1 on W . In particular, since an n dimensional complexvector space V is isomorphic to C

n, a linear representation ρ on V is always equivalent tosome representation by n × n matrices with complex matrix elements.

Equivalent representations are identical for classification purposes, and can all be identifiedwith some standard matrix representation through the proper choice of basis. Thus, theproblem of classifying all linear representations of a group G reduces to the problem of findingall the inequivalent matrix representations.

3.1 Matrix representations

To demonstrate explicitly how a general linear representation ρ is equivalent to a matrixrepresentation D, choose a basis [e] = (e1, e2, . . . , en) in V . Every vector v ∈ V may bewritten in a unique way as a linear combination of the basis vectors,

v =

n∑

i=1

aiei , (3.1)

with complex coefficients ai. The correspondence

v ↔ a =

a1

a2...

an

(3.2)

is an isomorphism between the vector spaces V and Cn. Next, introduce n × n matrices

D(g) =

D11(g) D12(g) . . . D1n(g)D21(g) D22(g) . . . D2n(g)

......

. . ....

Dn1(g) Dn2(g) . . . Dnn(g)

, (3.3)

with complex matrix elements Dij(g) defined by the relation

ρ(g) ei =

n∑

j=1

Dji(g) ej . (3.4)

Note that the summation index j here is the first index of the matrix.

3.1. MATRIX REPRESENTATIONS 43

The correspondence v ↔ a, eq. (3.2), implies that ρ(g)v ↔ D(g)a, where D(g)a is thematrix product of the n × n matrix D(g) and the n × 1 matrix a. The proof is simple,

ρ(g)v =

n∑

i=1

ai ρ(g) ei =

n∑

i=1

ai

n∑

j=1

Dji(g) ej =

n∑

j=1

(n∑

i=1

Dji(g) ai

)ej . (3.5)

It follows that

D(gh)a ↔ ρ(gh)v = (ρ(g) ρ(h))v = ρ(g) (ρ(h)v) ↔ D(g) (D(h)a) = (D(g)D(h))a . (3.6)

The last equality holds because matrix multiplication is associative. Since the above relationholds for every a ∈ C

n, we conclude that

D(gh) = D(g)D(h) . (3.7)

We may give a second proof of this matrix equation using directly the definition (3.4),

∑

j

Dji(gh) ej = ρ(gh) ei = ρ(g) ρ(h) ei = ρ(g)∑

k

Dki(h) ek =∑

k

Dki(h) ρ(g) ek

=∑

k

Dki(h)∑

j

Djk(g) ej =∑

j

(∑

k

Djk(g)Dki(h)

)ej . (3.8)

We have omitted the summation limits here in order to save writing, and to make the formulaemore easily readable. Eq. (3.8) proves eq. (3.7) as written explicitly in terms of matrixelements,

Dji(gh) =∑

k

Djk(g)Dki(h) . (3.9)

A linear transformation S on V has matrix elements Sij defined by the expansion

Sei =∑

j

Sji ej . (3.10)

If S is invertible, then the n vectors ei = Sei are linearly independent and constitute a newbasis in V . Relative to the new basis the representation ρ on V corresponds to a new matrixrepresentation D such that

ρ(g) ei =∑

j

Dji(g) ej . (3.11)

The left hand side of this equation is

ρ(g) ei = ρ(g)∑

j

Sji ej =∑

j

Sji ρ(g) ej =∑

j

Sji

∑

k

Dkj(g) ek

=∑

k

∑

j

Dkj(g)Sji ek . (3.12)

The right hand side is∑

j

Dji(g) ej =∑

j

Dji(g)∑

k

Skj ek =∑

k

∑

j

Skj Dji(g) ek . (3.13)


We see that the two matrix representations D and D corresponding to the same representationρ are related by the equation

∑

j

Dkj(g)Sji =∑

j

Skj Dji(g) , (3.14)

which we write in matrix notation as

D(g)S = S D(g) . (3.15)

Since the matrix S relating the two bases must be invertible, we have that

D(g) = S−1 D(g)S . (3.16)

A matrix transformation of the type D(g) → D(g) = S−1 D(g)S, resulting from a change ofbasis, is called a similarity transformation.

3.2 Reducible and irreducible representations

Definition 3.2 A subspace U ⊂ V is “invariant” if ρ(g)u ∈ U for all g ∈ G and u ∈ U . Aninvariant subspace U is “irreducible” if it contains no invariant subspace W apart from thetrivial ones, W = U and W = {0}.

The representation ρ is “irreducible” if the whole vector space V is irreducible, otherwiseρ is “reducible”.

ρ is “fully reducible” if it has two or more non-trivial complementary irreducible subspaces.

By definition, subspaces V1, V2, . . . , Vk ⊂ V are complementary if every vector v ∈ V hasa unique decomposition as

v = v1 + v2 + · · · + vk with vi ∈ Vi . (3.17)

Note that the definition of invariant or irreducible subspaces has to be modified slightly ifV is an infinite dimensional real or complex vector space. In that case the invariant subspaceswe consider are further required to be closed with respect to some topology. If V is finitedimensional it has a standard topology, and every subspace is closed in this topology, so thatthe condition of closure need not be stated explicitly.

Note also that the reducibility or irreducibility of a group representation may depend onthe number field F. Thus, a representation by real matrices may well be irreducible whenregarded as a real representation, but reducible when regarded as a complex representation.This is one of the reasons for restricting our discussion to the complex case.

There always exist the trivial invariant subspaces V and {0}. More generally, each vectorv ∈ V generates an invariant subspace, spanned by the vectors ρ(g)v which are the transformsof v by elements g ∈ G. This subspace is finite dimensional if G is finite. It need not beirreducible, if the representation ρ is reducible. It may be all of V , in which case v is calleda “generator”, or “cyclic vector”, and ρ is a “cyclic representation”.

Theorem 3.3 A group representation is irreducible if and only if every vector v 6= 0 is cyclic.Hence, every irreducible representation of a finite group must be finite dimensional.

3.3. DIRECT SUM AND DIRECT PRODUCT 45

Unitary or real orthogonal representations are always either irreducible or fully reducible,and the same is true for all representations of a group G which is either finite or compact.However, there are very simple counterexamples for infinite, non-compact groups. Take forexample G = Z, the additive group of integers, and let a ∈ Z be represented by the 2 × 2matrix

D(a) =

(1 a0 1

). (3.18)

It is easy to verify that D is a representation, that D(a)D(b) = D(a + b), and that there isone single non-trivial invariant subspace of C

2,

U = {

(c0

)| c ∈ C } . (3.19)

Since there is no invariant subspace complementary to U , the representation D is reduciblebut not fully reducible.

3.3 Direct sum and direct product

Direct sum and product of vector spaces

Let V and W be vector spaces of dimensions n and p, respectively.

Definition 3.4 Given basis vectors e1, e2, . . . , en ∈ V and f1, f2, . . . , fp ∈ W .The “direct sum” V ⊕ W is an n + p dimensional vector space with basis vectors

e1, e2, . . . , en, f1, f2, . . . , fp .

The “direct product” (or “tensor product”) V ⊗ W is an np dimensional vector spacewith basis vectors

e1 ⊗ f1, . . . , e1 ⊗ fp, e2 ⊗ f1, . . . , e2 ⊗ fp, . . . , en ⊗ f1, . . . , en ⊗ fp .

Every vector u ∈ V ⊕ W is the sum u = v + w = v ⊕ w of two uniquely determinedcomponent vectors

v =n∑

i=1

ai ei ∈ V , w =

p∑

j=1

bj fj ∈ W . (3.20)

But not every vector in V ⊗ W is a direct product vector of the form

v⊗ w =n∑

i=1

p∑

j=1

aibj ei ⊗ fj . (3.21)

In fact, the most general vector u ∈ V ⊗ W can be expanded as

u =n∑

i=1

p∑

j=1

cij ei ⊗ fj , (3.22)


with np complex coefficients cij . The equation u = v ⊗ w means that cij = aibj for all i, j,and solutions exist for ai and bj only if

cij ckl = cil cjk for i, k = 1, . . . , n ; j, l = 1, . . . , p . (3.23)

The above bases in V and W introduce the following isomorphisms,

v ∈ V ↔ a =

a1...

an

∈ C

n , w ∈ W ↔ b =

b1...bp

∈ C

p , (3.24)

next,

v ⊕ w ∈ V ⊕ W ↔ a ⊕ b =

(ab

)=

a1...

an

b1...bp

∈ Cn+p , (3.25)

and finally,

v ⊗ w ∈ V ⊗ W ↔ a ⊗ b =

a1b...

anb

=

a1b1

a1b2...

anbp

∈ C

np . (3.26)

Direct product Hilbert spaces are frequently encountered in quantum theory. A physicalsystem is very often a composite system made up of two or more subsystems, for exampletwo or more interacting particles. If there are two subsystems, and if the quantum statesof the two subsystems are represented as vectors in the complex Hilbert spaces V and W ,respectively, then the Hilbert space of the total system is V ⊗ W .

In such a composite system any product state of the form v⊗w has the special propertythat there is no correlation between the results of measurements performed separately on thetwo subsystems. In all other states such correlations can be observed. If a state is not aproduct state, it is said to be entangled, because some measurements on the subsystems willshow stronger correlations than what is possible according to classical physics.

Direct sum and product of operators

Given two linear operators A on V and B on W , the direct sum operator A ⊕ B and thedirect product operator A ⊗ B act on V ⊕ W and on V ⊗ W , respectively, so that

(A ⊕ B)(v ⊕w) = (Av) ⊕ (Bw) ,

(A ⊗ B)(v ⊗w) = (Av) ⊗ (Bw) . (3.27)

3.3. DIRECT SUM AND DIRECT PRODUCT 47

Relative to the bases defined above, the operator A is represented by an n × n matrix A, Bby a p × p matrix B, A ⊕ B by the (n + p) × (n + p) matrix

A⊕ B =

(A 00 B

)=

A11 . . . A1n 0 . . . 0...

. . ....

.... . .

...An1 . . . Ann 0 . . . 00 . . . 0 B11 . . . B1p

.... . .

......

. . ....

0 . . . 0 Bp1 . . . Bpp

, (3.28)

and A ⊗ B by the (np) × (np) matrix

A⊗ B =

A11B . . . A1nB...

. . ....

An1B . . . AnnB

=

A11B11 A11B12 . . . A1nB1p

A11B11 A11B12 . . . A1nB1p

......

. . ....

An1Bp1 An1Bp2 . . . AnnBpp

. (3.29)

Direct sum and product of representations

Definition 3.5 Given two linear representations of the group G, ρ on the vector space V andσ on W . The “direct sum” τ = ρ⊕ σ and “direct product” π = ρ⊗ σ are representations onV ⊕ W and V ⊗ W , respectively, such that

τ(g) = ρ(g) ⊕ σ(g), π(g) = ρ(g) ⊗ σ(g) ∀g ∈ G . (3.30)

A fully reducible representation is the direct sum of its irreducible components, and itis therefore completely described by an enumeration of the irreducible representations itcontains. Obviously, it may contain the same irreducible representation several times. Instatements of this kind equivalent representations are always counted as identical.

More precisely, given a representation ρ of the group G on the vector space V , and givena set of complementary invariant subspaces V1, V2, . . . , Vk ⊂ V . Each subspace Vi is assumedto be invariant, but not necessarily irreducible. Then the restriction of ρ to one subspace Vi

is a representation, reducible or irreducible, which we will call ρi. That is, we define

ρi(g)v = ρ(g)v ∀g ∈ G,v ∈ Vi , (3.31)

but ρi(g)v is undefined for v /∈ Vi. With this definition ρ is the direct sum,

ρ = ρ1 ⊕ ρ2 ⊕ . . . ⊕ ρk . (3.32)

Let ni be the dimension of Vi, and choose a basis for V such that the first n1 basis vectorsare in V1, the next n2 in V2, and so on. Relative to this basis any element g ∈ G will have amatrix representation D(g) which is “block diagonal”, that is,

D(g) =

D1(g) 0 . . . 00 D2(g) . . . 0...

.... . .

...0 0 . . . Dk(g)

, (3.33)


with zeros everywhere outside the matrix blocks of dimension ni×ni on the diagonal. If everyrepresentation ρi is irreducible, then this is as near as it is possible to come to a simultaneousdiagonalization of all the representation matrices D(g). The basis may always be chosen sothat two submatrices Di(g) and Dj(g) are identical whenever the two representations areequivalent.

Recall the definition, that V1, V2, . . . , Vk are complementary subspaces of V when everyvector v ∈ V has a unique decomposition as

v = v1 + v2 + · · · + vk with vi ∈ Vi . (3.34)

Every subspace Vi must contain the zero vector 0, and the unique decomposition of 0 is0 = 0 + 0 + · · · + 0. Assume that u ∈ Vi ∩ Vj with i 6= j. Because 0 = u + (−u) with u ∈ Vi

and −u ∈ Vj, and because the decomposition of 0 is unique, we must have u = 0. This showsthat Vi ∩ Vj = {0}.

By definition, a projection operator, or projection, P , is idempotent, P 2 = P . DefiningPiv = vi, we get linear projection operators P1, P2, . . . , Pk such that PiV = Vi. The projec-tions onto the complementary subspaces V1, V2, . . . , Vk form a decomposition (or resolution)of the identity. That is, P 2

i = Pi for i = 1, 2, . . . , k, PiPj = 0 when i 6= j, and

I = P1 + P2 + · · · + Pk . (3.35)

To summarize, complementary subspaces V1, V2, . . . , Vk define projections P1, P2, . . . , Pk thatform a decomposition of the identity. And conversely, a decomposition of the identity definecomplementary subspaces Vi = PiV .

Theorem 3.6 A representation ρ of the group G is a direct sum as in eq. (3.32) if and onlyif there exists a decomposition of the identity as in eq. (3.35), such that every projection Pi

commutes with ρ. That is, Pi ρ(g) = ρ(g)Pi ∀g ∈ G, i = 1, 2, . . . , k.

Proof. First the “if” part. Given a decomposition of the identity such that every Pi

commutes with ρ. For g ∈ G define the following operator on each of the complementarysubspaces Vi = PiV ,

ρi(g) = Pi ρ(g)Pi = P 2i ρ(g) = Pi ρ(g) = ρ(g)Pi . (3.36)

We see that ρ is a direct sum, because

ρ(g) = ρ(g) I = ρ(g)

k∑

i=1

Pi =

k∑

i=1

ρi(g) . (3.37)

And then the “only if” part. Given complementary invariant subspaces V1, . . . , Vk ⊂ V ,not necessarily irreducible, and the corresponding decomposition of the unity, I = P1+· · ·+Pk,with Vi = PiV . If i 6= j, we must have for g ∈ G that

Pi ρ(g)Pj = 0 . (3.38)

3.4. SCHUR’S LEMMA 49

In fact, for every v ∈ V we have Pj v ∈ Vj , ρ(g)Pj v ∈ Vj , and Pi ρ(g)Pj v = 0, sincePi vj = 0 whenever vj ∈ Vj with i 6= j. It follows that every Pi commutes with ρ, in fact,

Pi ρ(g) = Pi ρ(g) I = Pi ρ(g)

k∑

j=1

Pj = Pi ρ(g)Pi

=

k∑

j=1

Pj ρ(g)Pi = I ρ(g)Pi = ρ(g)Pi . (3.39)

QED

Note that as soon as we have one nontrivial projection P (P 6= 0 and P 6= I) thatcommutes with a representation ρ, then we know that ρ is a direct sum, because the definitionQ = I − P gives us a second projection commuting with ρ. We verify directly that

Q2 = (I − P )2 = I − 2P + P 2 = I − P = Q , (3.40)

and furthermore that P + Q = I and PQ = QP = P − P 2 = 0.

3.4 Schur’s lemma

Schur’s lemma is very useful both in the general theory of group representations and inphysical applications. A quantum mechanical problem where there is a non-trivial symmetrygroup can often be very much simplified by means of Schur’s lemma, or by means of theWigner–Eckart theorem, which is closely related.

A division algebra is an algebra with multiplicative unit element I in which every nonzeroelement A has a multiplicative inverse A−1 such that A−1A = AA−1 = I. In particular, theonly complex division algebra is the one consisting of the multiples of the identity, cI withc ∈ C. See Appendix A.

Theorem 3.7 (Schur’s lemma) Let ρ and σ be irreducible linear representations of thegroup G over the (finite dimensional) vector spaces V and W .

If A : V → W is a linear transformation such that Aρ(g) = σ(g)A ∀g ∈ G, then eitherA = 0, or A is invertible so that ρ and σ are equivalent, σ(g) = Aρ(g)A−1.

The commutant ρ(G)′ of ρ, defined as the set of all linear transformations on V commutingwith ρ,

ρ(G)′ = {B : V → V | Bρ(g) = ρ(g)B ∀g ∈ G } , (3.41)

is a division algebra. In particular, when ρ is an irreducible complex representation, then

ρ(G)′ = { cI | c ∈ C } . (3.42)

In other words, the only linear transformations commuting with ρ(g) for every g ∈ G are themultiples of the identity I.

Proof. In the first part of the theorem, the image of V under A,

ImgA = {Av | v ∈ V } , (3.43)


is an invariant subspace of W , since σ(g)Av = Aρ(g)v ∈ ImgA for all g ∈ G. Since σ isirreducible, either ImgA = {0} or ImgA = W . Similarly, the kernel of A,

Ker A = {v ∈ V | Av = 0 } , (3.44)

is an invariant subspace of V . For if Av = 0 and g ∈ G, then Aρ(g)v = σ(g)Av = 0. Since ρis irreducible, either Ker A = {0} or KerA = V .

Therefore, either Ker A = V and ImgA = {0} so that A = 0, or Ker A = {0} andImgA = W so that A is invertible.

The last part of the theorem follows when we take σ = ρ. For example, it is obvious thatρ(G)′ is an algebra containing the identity I.

QED

An operator A : V → W such that Aρ = σA is called an intertwining operator of therepresentations ρ and σ. We say that it commutes with the action of the group G on V andon W .

It follows from Schur’s lemma that if A,B : V → W are both invertible and commute withthe action of G, so that Aρ(g) = σ(g)A and Bρ(g) = σ(g)B for all g ∈ G, then A−1B ∈ ρ(G)′,hence in the complex case A−1B = cI and B = cA for some complex number c.

Another immediate consequence is the following.

Theorem 3.8 Every finite dimensional irreducible complex representation ρ of an Abeliangroup G is one dimensional.

In fact, in this case every ρ(g) belongs to the commutant ρ(G)′, and so ρ(g) = c(g) I withc(g) ∈ C. Then every one dimensional subspace of V is invariant.

Note that the group structure plays no part at all in the proof of Schur’s lemma, therole of the group G is only that the group element g defines a correspondence between thelinear transformations ρ(g) on V and σ(g) on W . Thus we have actually proved the moregeneral result that any set X of linear transformations on V which leaves no proper subspaceinvariant, has a division algebra as its commutant, and in the complex case only multiples ofthe identity commute with every transformation in X.

Theorem 3.9 The commutant of a direct sum of two (equivalent or inequivalent) group rep-resentations is an algebra of dimension at least two, and it is not a division algebra.

Proof. The direct sum (aI)⊕ (bI) where a, b ∈ C commutes with every ρ(g)⊕ σ(g) whereρ and σ are representations of the group G, and g ∈ G. In particular, 0 ⊕ I and I ⊕ 0 arenon-invertible non-zero elements of the commutant.

QED

This result is as close as we can get to a converse of Schur’s lemma, that a group represen-tation is irreducible if its commutant is a division algebra. As we shall see, all representationsof finite groups and compact topological groups, as well as all unitary representations of anygroup, are either irreducible or fully reducible, and then the converse of Schur’s lemma holds.

Theorem 3.10 Assume that ρ is a complex representation of the group G, and that G isfinite or compact, or G is non-compact but ρ is unitary.

Then ρ is irreducible if the commutant ρ(G)′ consists of multiples of the identity.

3.5. UNITARY AND ORTHOGONAL REPRESENTATIONS 51

A simple counterexample is the infinite and non-compact group of triangular 2×2 matricesof the form

D(a, b, c) =

(a b0 c

), (3.45)

with complex parameters a, b, c such that

detD(a, b, c) = ac 6= 0 . (3.46)

This matrix representation is reducible but not fully reducible, and its commutant consistsof the multiples of the identity matrix.

3.5 Unitary and orthogonal representations

Unitary and orthogonal operators

A Hermitean scalar product on the complex vector space V assigns to every pair of vectorsu,v ∈ V a complex number (u,v) = (v,u)∗. It is a linear function of its second argument,

(u, av + bw) = a (u,v) + b (u,v) ∀ u,v,w ∈ V, a, b ∈ C . (3.47)

Hence, it is antilinear in its first argument,

(au + bv,w) = a∗ (u,w) + b∗ (v,w) . (3.48)

We will assume here that the scalar product is positive definite, that (u,u) > 0 for all u 6= 0.A complex vector space with a Hermitean positive definite scalar product is called a complexHilbert space.

If A is an operator (a linear transformation) on the complex Hilbert space V , the adjoint,or Hermitean conjugate, operator A† is defined by the condition that

(u, A†v) = (Au,v) ∀ u,v ∈ V . (3.49)

We say that A is self-adjoint, or Hermitean, if A† = A (in the infinite dimensional case there isa subtle distinction between self-adjoint and Hermitean operators). We say that A is unitaryif it is invertible and A† = A−1. A unitary operator A preserves the scalar product,

(Au, Av) = (A†Au,v) = (A−1Au,v) = (u,v) . (3.50)

A real Hilbert space is a real vector space with a positive definite scalar product. In thiscase the scalar product is real valued and symmetric, (u,v) = (v,u). The condition

(u, A⊤v) = (Au,v) ∀ u,v ∈ V (3.51)

defines the transpose A⊤ of an operator A on the real Hilbert space V . We say that A issymmetric if A⊤ = A, and that A is orthogonal if it is invertible and A⊤ = A−1. Like aunitary operator on a complex Hilbert space, an orthogonal operator on a real Hilbert spacepreserves the scalar product.

Note that the term “orthogonal” is sometimes used also for an operator on a complexvector space if it preserves a scalar product which is not positive definite, but is symmetric,such that (u,v) = (v,u), and is complex linear in both factors. To avoid confusion one mayspeak of “real orthogonal” and “complex orthogonal” operators.


Orthonormal bases

A basis [e] = (e1, e2, . . . , en) for the n dimensional complex Hilbert space V is orthonormal if

(ei, ej) = δij ∀ i, j = 1, 2, . . . , n . (3.52)

The matrix elements of an operator A relative to the basis [e] is defined in general by therelation

Aei =

n∑

j=1

Ajiej . (3.53)

An equivalent definition for an orthonormal basis is that

Aji = (ej, Aei) . (3.54)

Then the Hermitean conjugate operator A† has the matrix elements

(A†)ji = (ej , A†ei) = (Aej, ei) = (ei, Aej)

∗ = (Aij)∗ . (3.55)

Thus, when the operator A on V is represented, relative to an orthonormal basis, by then × n matrix A, then the Hermitean conjugate operator A† is represented by the Hermiteanconjugate matrix A†, obtained from A by transposition and complex conjugation. In partic-ular, a Hermitean operator A, with A† = A, is represented by a Hermitean matrix A, withA† = A, and a unitary operator A, with A† = A−1, is represented by a unitary matrix A,with A† = A−1.

Similarly, when we define matrix elements relative to an orthonormal basis in a real Hilbertspace, the transposition of operators, A → A⊤, corresponds to the transposition of matrices,A → A⊤. In this case, a symmetric operator A, with A⊤ = A, is represented by a symmetricmatrix A, with A⊤ = A, and an orthogonal operator A, with A⊤ = A−1, is represented byan orthogonal matrix A, with A⊤ = A−1.

The Gram–Schmidt orthogonalization procedure

It is always possible to choose an orthonormal basis in a complex or real Hilbert space, forexample, starting from an arbitrary basis [f] and constructing an orthonormal basis [e] by theGram–Schmidt procedure.

Let e1 = a1f1, where a1 is a (complex or real) normalization factor chosen in such a waythat

(e1, e1) = |a1|2 (f1, f1) = 1 . (3.56)

Then for k = 2, 3, . . . , n define

ek = ak

(fk −

k−1∑

i=1

(ei, fk) ei

), (3.57)

where ak is a normalization factor chosen such that (ek, ek) = 1.

3.6. CHARACTERS 53

Unitary and orthogonal representations

Definition 3.11 A complex linear representation ρ of the group G is “unitary” if the rep-resentation space V is a complex Hilbert space (either finite or infinite dimensional), and ifρ(g) is unitary for every g ∈ G. Then

(ρ(g)u, ρ(g)v) = (u,v) ∀ u,v ∈ V, g ∈ G . (3.58)

A real linear representation is “orthogonal” if the representation space is a real Hilbertspace, and if ρ(g) is orthogonal for every g ∈ G.

Theorem 3.12 Let ρ be a unitary representation of G on V , and let U be an invariantsubspace. The orthogonal complement

U⊥ = {v | (u,v) = 0 ∀u ∈ U} (3.59)

is then a complementary invariant subspace.

Proof. Let g ∈ G and v ∈ U⊥. We want to prove that ρ(g)v ∈ U⊥. For every u ∈ U wehave that ρ(g−1)u ∈ U , and

(u, ρ(g)v) = ((ρ(g))†u,v) = ((ρ(g))−1u,v) = (ρ(g−1)u,v) = 0 . (3.60)

QED

Theorem 3.13 Every (finite dimensional) reducible unitary representation is fully reducible.

Proof. Let ρ be a unitary representation of G on V . If it is reducible, there exists a non-trivial invariant subspace U , and U and U⊥ are complementary invariant subspaces. If Uand/or U⊥ are reducible, they can be split further in a similar way. In the finite dimensionalcase irreducible components will be obtained in a finite number of steps.

QED

These two theorems hold also for real orthogonal representations, by the same proofs.

3.6 Characters

Definition 3.14 The “character” of a linear representation ρ of the group G is the functionχ : G → C such that

χ(g) = Tr ρ(g) ∀g ∈ G . (3.61)

The character of the unit element e is just the dimension n of the representation, χ(e) =Tr ρ(e) = Tr I = n.

In order to compute the trace we need a basis [e] = (e1, e2, . . . , en) for the vector spaceV , turning ρ(g) into an n × n matrix D(g), by eq. (3.4). Then

Tr ρ(g) = TrD(g) =

n∑

i=1

Dii(g) . (3.62)


The trace is actually basis independent, even though we use a basis to compute it. In fact,we have seen that a change of basis corresponds to a similarity transformation, D(g) →S−1 D(g)S, and this leaves the trace invariant. Using the property of the trace that

Tr(AB) =n∑

i=1

n∑

j=1

AijBji = Tr(BA) , (3.63)

we find that

Tr(S−1AS) = Tr((S−1A)S) = Tr(S(S−1A)) = Tr((SS−1)A) = Tr(A) . (3.64)

Given an invertible linear transformation S : V → W we have a representation σ = SρS−1

on W , equivalent to ρ on V . We have then that

Tr σ(g) = Tr(Sρ(g)S−1) = Tr(S−1Sρ(g)) = Tr ρ(g) . (3.65)

Theorem 3.15 Equivalent representations have identical characters.

The converse statement, that equality of characters implies equivalence, again is not truein general, although it is true for irreducible or fully reducible representations, such as complexrepresentations of finite and compact groups. As a counterexample we need an infinite group,let us consider the additive group of the integers, Z = { 0,±1,±2, . . . }. It has for examplethe two matrix representations

D1(a) =

(1 a0 1

), D2(a) =

(1 00 1

), where a ∈ Z . (3.66)

We may verify that D1(a)D1(b) = D1(a + b) and D2(a)D2(b) = D2(a + b) . These tworepresentations have the same character, χ1(a) = χ2(a) = 2, but they are not equivalent.

Theorem 3.16 The character is a class function. That is, it is a function only of the con-jugation class to which a group element belongs.

Proof. If g and h are conjugate group elements, h = fgf−1, then

χ(h) = Tr ρ(h) = Tr(ρ(f)ρ(g)ρ(f−1)) = Tr(ρ(f−1)ρ(f)ρ(g)) = Tr ρ(g) = χ(g) . (3.67)

QED

We have seen two ways of building new representations from old. Given two represen-tations ρ1 and ρ2, we may construct the direct sum ρs = ρ1 ⊕ ρ2 and the direct productρp = ρ1 ⊗ ρ2. The characters of these new representations are the sum and product of thecharacters of ρ1 and ρ2,

χs(g) = χ1(g) + χ2(g) χp(g) = χ1(g)χ2(g) . (3.68)

3.7. THE CONTRAGREDIENT REPRESENTATION 55

3.7 The contragredient representation

From a matrix representation D we may construct the contragredient, or dual, representationD, defined by the relation D(g) = (D(g−1))⊤, or explicitly in terms of matrix elements,

Dij(g) = Dji(g−1) . (3.69)

Because of the general relation (AB)⊤ = B⊤A⊤, we see that D is a representation,

D(gh) = (D((gh)−1))⊤ = (D(h−1)D(g−1))⊤ = (D(g−1))⊤(D(h−1))⊤ = D(g)D(h) . (3.70)

When D is a unitary matrix representation, so that D(g−1) = (D(g))−1 = (D(g))†, then Dis just the complex conjugate of D, in fact, D(g) = (D(g−1))⊤ = ((D(g))†)⊤ = (D(g))∗.

The two representations D and D are often equivalent. A necessary condition for equiv-alence is that they have the same character, that is,

TrD(g) = TrD(g) = TrD(g−1) ∀g ∈ G . (3.71)

If D is irreducible or fully reducible, as is always the case for example if G is a finite group,then this condition is also sufficient. If for example G = Sn, the symmetric group, where g−1

is always conjugate to g, D and D always have the same character and are equivalent.

Chapter 4

Complex representations of finite

groups

In the linear representation theory for finite groups only representations on finite dimensionalvector spaces are of interest. Every cyclic and therefore every irreducible representation ofa group G of finite order |G| = N is finite dimensional, of dimension at most equal to N .As we shall see immediately below, every representation of a finite group, at least if it isfinite dimensional, is a direct sum of irreducible representations. Hence, in order to classifyall possible finite dimensional representations, it is sufficient to classify all the irreduciblerepresentations. This is our purpose in the present chapter.

Most of the time we will consider matrix representations, since we do not distinguishbetween equivalent representations, and since every representation is equivalent to a matrixrepresentation. When D is an n dimensional matrix representation, the group element g isrepresented by the n × n matrix D(g), having n2 matrix elements Dij(g) ∈ C.

We will choose more or less arbitrarily a complete set of standard irreducible matrixrepresentations, which we will denote by D(µ), with µ = 1, 2, . . . ,M . Completeness meansthat every irreducible representation is equivalent to some D(µ), and it is to be understoodthat D(µ) and D(ν) are inequivalent when µ 6= ν. The dimension of D(µ) we denote by nµ.As we shall see, the number M of inequivalent irreducible representations is finite and equalto K, the number of conjugation classes in G.

4.1 Unitarity and reducibility

Theorem 4.1 Every complex representation of a finite group is unitary with respect to somepositive definite scalar product. Hence it is either irreducible or fully reducible.

Proof. Given the group G and a representation ρ over V . There always exists a positivedefinite scalar product ((·, ·)) on V , invariant or not. For example, choose an arbitrary basise1, e2, . . . , en and define ((ei, ej)) = δij . Then define

(u,v) =∑

g∈G

((ρ(g)u, ρ(g)v)) ∀ u,v ∈ V . (4.1)

This is now our invariant positive definite scalar product. It is invariant under h ∈ G because

57

58 CHAPTER 4. COMPLEX REPRESENTATIONS OF FINITE GROUPS

h just permutes the terms in the defining sum,

(ρ(h)u, ρ(h)v) =∑

g∈G

((ρ(g)ρ(h)u, ρ(g)ρ(h)v))

=∑

g

((ρ(gh)u, ρ(gh)v)) =∑

f

((ρ(f)u, ρ(f)v)) = (u,v) , (4.2)

where f = gh. To prove that it is a positive definite scalar product, we have to prove forexample that (v,v) > 0 when v 6= 0. We have that

(v,v) =∑

g∈G

((ρ(g)v, ρ(g)v)) ≥ ((ρ(e)v, ρ(e)v)) = ((v,v)) > 0 . (4.3)

Other required properties are equally easily demonstrated.QED

This theorem implies that we know all complex representations of a finite group as soonas we know the finite dimensional irreducible representations. The theorem holds also forcontinuous representations of infinite compact topological groups, by a similar proof. Thedifference is that the sum over group elements must be replaced by an integral over thegroup, called the invariant Haar integral.

Since we may always choose an orthonormal basis in the Hilbert space, we have nowactually proved the following result.

Theorem 4.2 Every complex representation of a finite or compact group is equivalent to aunitary matrix representation.

In the same way, every real representation of a finite or compact group is equivalent to areal orthogonal matrix representation.

We will make use of this theorem to assume that our standard irreducible representationsD(µ) are unitary, that (D(µ)(g))† = (D(µ)(g))−1 = D(µ)(g−1). Or, when we write explicitlythe matrix indices,

((D(µ)(g))†)ij = (D(µ)ji (g))∗ = D

(µ)ij (g−1) . (4.4)

4.2 The group algebra

Definition 4.3 The “group algebra” A(G) of the group G over the complex numbers C is acomplex vector space with the group elements as basis vectors, and with multiplication definedby the group product. That is, A(G) consists of all linear combinations

a =∑

g∈G

a(g) g (4.5)

with complex coefficients a(g).The “natural scalar product” (·, ·) on the group algebra A(G) is defined such that the

natural basis vectors (the group elements) are orthonormal,

(g, h) = δg,h ∀ g, h ∈ G . (4.6)

4.3. THE REGULAR REPRESENTATIONS 59

The vector space dimension of the group algebra A(G) is the order |G| = N of the group.It is easily verified that the group product turns A(G) into an associative linear algebra,

with a multiplicative unit equal to the unit element e ∈ G. Explicitly, the product of twoelements a, b ∈ A(G) is

ab =∑

g∈G

∑

h∈G

a(g) b(h) gh =∑

f∈G

∑

g∈G

a(g) b(g−1f)

f =

∑

f∈G

(∑

h∈G

a(fh−1) b(h)

)f , (4.7)

where f = gh, g = fh−1, h = g−1f .Alternatively, the element a ∈ A(G) may be regarded as a complex valued function on G,

a : g 7→ a(g) ∈ C. The set of all such function is denoted by CG, thus A(G) = C

G. Since bydefinition

ab =∑

f∈G

ab(f) f , (4.8)

we see that

ab(f) =∑

g∈G

a(g) b(g−1f) =∑

h∈G

a(fh−1) b(h) . (4.9)

This kind of product of two functions a and b is called a convolution product, and is thendenoted by a ∗ b. Here we will most often use the convolution notation and write a ∗ b insteadof ab, in an attempt to make some formulae more readable.

The scalar product of two general elements a, b ∈ A(G) is

(a, b) =∑

g∈G

∑

h∈G

a(g)∗ b(h) (g, h) =∑

g∈G

a(g)∗ b(g) . (4.10)

This kind of scalar product of functions is called an L2 scalar product.

4.3 The regular representations

The left and right regular realizations L and R, by which each g ∈ G acts as permutationsLg and Rg of the group elements,

Lg(h) = gh , Rg(h) = hg−1 , ∀h ∈ G , (4.11)

have another straightforward interpretation as linear representations of G, called the left andright regular representations, over the N dimensional vector space A(G). These are bothcyclic representations, in fact, every h ∈ G is a cyclic vector for the representation L as wellas for R.

In order to distinguish between Lg as a permutation over G and as a linear transformationover A(G), we may write L(g) for the latter. If

a =∑

h∈G

a(h)h , (4.12)


then by definition

L(g) a =∑

h∈G

a(h)L(g)h =∑

h∈G

a(h)Lg(h) =∑

h∈G

a(h) gh =∑

f∈G

a(g−1f) f . (4.13)

Alternatively, if we regard a ∈ A(G) as a function a : G → C, we have that

[L(g) a](f) = a(g−1f) = a(L −1g (f)) ∀f ∈ G , (4.14)

or,

L(g) a = aL −1g . (4.15)

The left hand side in this equation is a linear transformation L(g) acting on a vector a ∈ A(G),whereas the right hand side is the composite of two functions L −1

g : G → G and a : G → C.In a similar way we have that

R(g) a =∑

h∈G

a(h)R(g)h =∑

h∈G

a(h)Rg(h) =∑

h∈G

a(h)hg−1 =∑

f∈G

a(fg) f . (4.16)

In the alternative notation, with a as a function, we have that

[R(g) a](f) = a(fg) = a(R −1g (f)) , (4.17)

or,

R(g) a = aR −1g . (4.18)

Exercise 4.4 Show that the left and right regular representations are equivalent. (Hint:Rg(h) = (Lg(h

−1))−1 ∀g, h ∈ G.)

Theorem 4.5 The left and right regular representations L and R of the group G are bothfaithful representations.

Proof. We have to show that L(g) and L(h) are different linear transformations on A(G)when g 6= h. Take for example the unit element e ∈ G, regarded as a vector in A(G). ThenL(g) e = ge = g 6= L(h) e = he = h.

QED

It is easy to verify that the two regular representations are unitary with respect to thenatural scalar product on the group algebra. Indeed, it follows from the relation

(L(f) g, L(f)h) = (fg, fh) = δfg,fh = δg,h = (g, h) , (4.19)

valid for every f, g, h ∈ G, that L(f) is unitary, thus the left regular representation L isunitary. Similarly, the right regular representation R is unitary.

We may use the natural basis for the group algebra, the group elements, in order to definematrix representations L and R equivalent to L and R. The relation

L(g)h = gh =∑

f∈G

Lf,h(g) f , (4.20)

4.4. REPRESENTATION MATRICES IN THE GROUP ALGEBRA 61

which corresponds to eq. (3.4), defines the matrix elements Lf,h(g) of the matrix L(g) rep-resenting the linear transformation L(g). We see that the matrix elements are Kroneckerδ’s,

Lf,h(g) = δf,gh . (4.21)

Similarly, the relation

R(g)h = hg−1 =∑

f∈G

Rf,h(g) f (4.22)

defines the matrix elements of the matrix R(g),

Rf,h(g) = δfg,h . (4.23)

4.4 Representation matrices in the group algebra

Let D be an n dimensional matrix representation of G. This means that D(gh) = D(g)D(h)for all g, h ∈ G, or in terms of matrix elements,

Dik(gh) =n∑

j=1

Dij(g)Djk(h) . (4.24)

Then the matrix elements Dij are complex valued functions on G and may be regarded aselements of the group algebra A(G),

Dij =∑

h∈G

Dij(h)h . (4.25)

They span a subspace of A(G),

A(D) = {∑

i,j

aijDij | aij ∈ C } . (4.26)

Theorem 4.6 Equivalent matrix representations D and D define the same subspace, A(D) =A(D).

Proof. Equivalence of D and D means that there exists an invertible matrix S such that

D(g) = SD(g)S−1 ∀g ∈ G . (4.27)

It follows that A(D) ⊂ A(D), because

Dij =∑

k,m

SikDkm(S−1)mj ∈ A(D) , (4.28)

Similarly, A(D) ⊂ A(D).QED


Theorem 4.7 To every (finite dimensional) representation ρ of the group G there corre-sponds a unique subspace of A(G), A(ρ) = A(D), where D is an arbitrary matrix represen-tation equivalent to ρ.

If ρ is equivalent to a direct sum

D = D1 ⊕ D2 ⊕ . . . ⊕ Dr , (4.29)

where each Di is a matrix representation of dimension ni, then A(ρ) is spanned by the n12 +

n22 + . . . + nr

2 matrix elements [Di]jk.In particular, A(ρ) is spanned by the matrix elements of the irreducible matrix represen-

tations contained in D.

4.5 The irreducible matrix representations

Theorem 4.8 (Completeness) The matrix elements D(µ)ij of the irreducible matrix repre-

sentations span the whole group algebra.

Proof. By Theorem 4.7, it is enough to find one representation ρ such that A(ρ) = A(G),then A(G) is spanned by the matrix elements of the irreducible matrix representations con-tained in ρ. One possible choice is ρ = L, the left regular representation. In fact, eq. (4.21)gives that

Lf,h =∑

g∈G

Lf,h(g) g =∑

g∈G

δf,gh g = fh−1 . (4.30)

In order to make the elements Lf,h ∈ A(G) run through all elements of G, which form thenatural basis in A(G), we may for example fix h = e and let f vary freely.

QED

Assume now that D(µ) and D(ν) are unitary irreducible representations, inequivalent if

µ 6= ν. We regard the matrix elements D(µ)ij and D

(ν)kl as elements of the group algebra, and

compute their group algebra product,

D(µ)ij ∗ D

(ν)kl =

∑

f∈G

∑

g∈G

D(µ)ij (g)D

(ν)kl (g−1f) f =

∑

f∈G

∑

h∈G

D(µ)ij (fh−1)D

(ν)kl (h) f . (4.31)

Because D(ν) is a representation, we have that

D(ν)kl (g−1f) =

∑

p

D(ν)kp (g−1)D

(ν)pl (f) , (4.32)

and consequently,

D(µ)ij ∗ D

(ν)kl =

∑

p

∑

g∈G

D(µ)ij (g)D

(ν)kp (g−1)

∑

f∈G

D(ν)pl (f) f . (4.33)

Similarly, because D(µ) is a representation, we have that

D(µ)ij (fh−1) =

∑

q

D(µ)iq (f)D

(µ)qj (h−1) , (4.34)

4.5. THE IRREDUCIBLE MATRIX REPRESENTATIONS 63

and,

D(µ)ij ∗ D

(ν)kl =

∑

q

∑

h∈G

D(µ)qj (h−1)D

(ν)kl (h)

∑

f∈G

D(µ)iq (f) f . (4.35)

By substituting g ↔ h−1 we see that∑

g∈G

D(µ)ij (g)D

(ν)kl (g−1) =

∑

h∈G

D(µ)ij (h−1)D

(ν)kl (h) . (4.36)

Since we assumed the representations D(µ) and D(ν) to be unitary, this quantity is just thescalar product of the matrix elements as vectors in the group algebra. Let us introduce thenotation

C(µ,ν)ijkl =

∑

h∈G

D(µ)ij (h−1)D

(ν)kl (h) =

∑

h∈G

(D(µ)ji (h))∗ D

(ν)kl (h) = (D

(µ)ji ,D

(ν)kl ) . (4.37)

What we have shown is that

D(µ)ij ∗ D

(ν)kl =

∑

p

C(µ,ν)ijkp D

(ν)pl =

∑

q

C(µ,ν)qjkl D

(µ)iq . (4.38)

Here C(µ,ν)ijkl ∈ C, whereas D

(µ)ij ∈ A(G) and D

(ν)kl ∈ A(G). Thus, the second equality in the

above equation may be written as∑

p

C(µ,ν)ijkp D

(ν)pl (f) =

∑

q

D(µ)iq (f)C

(µ,ν)qjkl ∀f ∈ G . (4.39)

Since we assumed the representations D(µ) and D(ν) to be irreducible, we may now applySchur’s lemma. We first conclude that if µ 6= ν, so that the two irreducible representations

are inequivalent, we must have C(µ,ν)ijkl = 0. If µ = ν, then the dependence of C

(µ,ν)ijkl on the

two indices i and l must be the Kronecker delta δil. Thus we may write

C(µ,ν)ijkl = B

(µ)jk δµν δil . (4.40)

The coefficients B(µ)jk are so far unknown, but are easily calculated. When we set µ = ν and

i = l, and sum over i, we get the equation∑

i

C(µ,µ)ijki = B

(µ)jk nµ , (4.41)

where nµ is the dimension of the irreducible representation D(µ). The left hand side of thisequation is

∑

i

C(µ,µ)ijki =

∑

i

∑

h∈G

D(µ)ij (h−1)D

(µ)ki (h) =

∑

h∈G

∑

i

D(µ)ki (h)D

(µ)ij (h−1)

=∑

h∈G

D(µ)kj (hh−1) =

∑

h∈G

D(µ)kj (e) =

∑

h∈G

δkj = N δkj . (4.42)

Hence,

B(µ)jk =

N

nµδjk . (4.43)


To summarize, by a somewhat lengthy computation we have obtained two importantresults. One is the group algebra product of the matrix elements,

D(µ)ij ∗ D

(ν)kl =

N

nµδµν δjk D

(µ)il . (4.44)

And as a byproduct we have proved an orthogonality relation.

Theorem 4.9 (Orthogonality) The natural scalar products in the group algebra betweenmatrix elements of the irreducible unitary representations are

(D(µ)ij ,D

(ν)kl ) =

N

nµδµν δik δjl . (4.45)

It follows that these matrix elements, as vectors in the group algebra, are nonzero andlinearly independent. Since they span the whole group algebra, they form a basis.

We will show at the end of this chapter that the ratio N/nµ is always an integer. In otherwords, the dimension of an irreducible representation is always a divisor of the order of thegroup.

The scalar product of one matrix element with itself is

(D(µ)ij ,D

(µ)ij ) =

∑

g∈G

|D(µ)ij (g)|2 =

N

nµ. (4.46)

This shows that whatever matrix element we pick in an irreducible representation, there isalways at least one group element for which it is nonzero.

Any two bases in a finite dimensional vector space must have the same number of basisvectors. Thus, the finite group G can have only a finite number M of inequivalent irreduciblerepresentations, and by counting basis vectors we find a very useful relation between the orderof the group, |G| = N , and the dimensions of the irreducible representations,

N =

M∑

µ=1

n 2µ . (4.47)

We will prove below that the number of inequivalent irreducible representations, M , isequal to the number of conjugation classes, K.

The group algebra as a direct sum of matrix algebras

For an element a =∑

g a(g) g ∈ A(G) we now define

a(µ)ij = (D

(µ)ij , a) =

∑

g

(D(µ)ij (g))∗ a(g) =

∑

g

D(µ)ji (g−1) a(g) . (4.48)

Then the expansion of a in the orthogonal basis of the irreducible matrix elements is

a =1

N

M∑

µ=1

nµ∑

i=1

nµ∑

j=1

nµ a(µ)ij D

(µ)ij . (4.49)

4.6. THE CENTRE OF THE GROUP ALGEBRA 65

The product of a with another element b ∈ A(G), expanded as

b =1

N

M∑

ν=1

nν∑

k=1

nν∑

l=1

nν b(ν)kl D

(ν)kl , (4.50)

is

a ∗ b =1

N2

∑

µ

∑

i

∑

j

∑

ν

∑

k

∑

l

nµ nν a(µ)ij b

(ν)kl D

(µ)ij ∗ D

(ν)kl

=1

N

∑

µ

∑

i

∑

l

nµ c(µ)il D

(µ)il . (4.51)

with

c(µ)il =

∑

j

a(µ)ij b

(µ)jl . (4.52)

This shows that the group algebra product a ∗ b = c corresponds to M matrix prod-ucts A(µ)B(µ) = C(µ), where the matrix elements of the matrices A(µ),B(µ),C(µ) are the

components a(µ)ij , b

(µ)ij , c

(µ)ij of a, b, c ∈ A(G).

Thus, we may picture the general element a ∈ A(G) as a block diagonal matrix

A =

A(1) 0 . . . 00 A(2) . . . 0...

.... . .

...0 0 . . . A(M)

, (4.53)

where each diagonal block A(µ) is a general nµ × nµ matrix. And the group algebra productis the same as the matrix product of such matrices.

4.6 The centre of the group algebra

An important subalgebra of the group algebra is its centre, or commutant, defined as

A(G)′ = { c ∈ A(G) | c ∗ a = a ∗ c ∀ a ∈ A(G) } . (4.54)

The centre is commutative, by definition.We will use two different approaches for analysing the centre of the group algebra. Let

c ∈ A(G)′. One approach will give c as a linear combination of the conjugation classes, andthe other approach will give c as a linear combination of the irreducible characters.

The conjugation classes as a basis

Obviously, an element c ∈ A(G) belongs to A(G)′ if and only if it commutes with every h ∈ G.Recall that the conjugation class of a group element g ∈ G is the subset

C = {hgh−1 | h ∈ G} ⊂ G . (4.55)


Here we introduce the conjugation class C also as an element of the group algebra,

C =∑

g∈C

g . (4.56)

Then C commutes with every h ∈ G, that is, h ∗ C = C ∗ h, because

h ∗ C ∗ h−1 =∑

g∈C

hgh−1 = C . (4.57)

The effect of the mapping g → hgh−1 is just to permute the elements of C.Let K be the number of conjugation classes of the group G, and enumerate them as

C1, C2, . . . , CK . They are linearly independent vectors in A(G), and they are orthogonal,

(Ci, Cj) =∑

g∈Ci

∑

h∈Cj

(g, h) =∑

g∈Ci

∑

h∈Cj

δg,h = Ni δij , (4.58)

where Ni = |Ci| is the number of elements of Ci. The unit element e ∈ G is a conjugationclass by itself, and it is natural to define C1 = {e}, with N1 = 1.

As we have seen, the conjugation classes belong to the centre A(G)′. Moreover, if

c =∑

g∈G

c(g) g ∈ A(G)′ , (4.59)

then c is a linear combination of conjugation classes, because

c =1

N

∑

h∈G

h ∗ c ∗ h−1 =1

N

∑

g∈G

c(g)∑

h∈G

hgh−1 =

K∑

i=1

1

Ni

∑

g∈Ci

c(g)

Ci . (4.60)

The last equality follows because when Ci is the conjugation class of g, then

1

N

∑

h∈G

hgh−1 =1

NiCi . (4.61)

This completes the proof of the following theorem.

Theorem 4.10 The conjugation classes C1, C2, . . . , CK are basis vectors of the centre A(G)′.They are orthogonal in the natural scalar product,

(Ci, Cj) = Ni δij . (4.62)

The irreducible characters as a basis

The character χ of any linear representation ρ is a class function, χ(g) = χi for g ∈ Ci.Therefore, when χ is regarded as a member of the group algebra A(G) it is a linear combinationof conjugation classes,

χ =∑

g∈G

χ(g) g =

K∑

i=1

χi Ci . (4.63)

4.6. THE CENTRE OF THE GROUP ALGEBRA 67

From what we have just learned, it follows that χ ∈ A(G)′.We now introduce the characters of the irreducible representations,

χ(µ)(g) = TrD(µ)(g) =

nµ∑

i=1

D(µ)ii (g) , (4.64)

and treat them as members of the group algebra,

χ(µ) =∑

g∈G

χ(µ)(g) g =

nµ∑

i=1

D(µ)ii =

nµ∑

i=1

∑

g∈G

D(µ)ii (g) g . (4.65)

The orthogonality relation (4.45) implies directly that

(χ(µ), χ(ν)) =∑

i

∑

k

(D(µ)ii ,D

(ν)kk ) =

∑

i

∑

k

N

nµδµν δik δik = N δµν . (4.66)

Only the identity matrix and multiples of it commute with every matrix in a full matrixalgebra. For example, the 2 × 2 matrix

(a bc d

)(4.67)

commutes with the two matrices(

1 00 0

),

(0 10 0

)(4.68)

if and only if b = c = 0 and a = d.Hence, when we represent c ∈ A(G)′ by a block diagonal matrix C, like in eq. (4.53),

C =

C(1) 0 . . . 00 C(2) . . . 0...

.... . .

...0 0 . . . C(M)

, (4.69)

it follows that the blocks on the diagonal must be multiples of the identity, C(µ) = c(µ)I(µ),where c(µ) ∈ C and I(µ) is the nµ × nµ identity matrix. This means that c is a linearcombination of the irreducible characters, in fact,

c =1

N

∑

µ

∑

i

∑

l

nµ c(µ) δil D(µ)il =

1

N

∑

µ

∑

i

nµ c(µ) D(µ)ii =

1

N

∑

µ

nµ c(µ) χ(µ) . (4.70)

We have now established the conjugation classes Ci and the irreducible characters χ(µ)

as two different orthogonal bases for A(G)′. Since two bases of a finite dimensional vectorspace must have the same number of basis vectors, it follows that the number of irreduciblerepresentations is equal to the number of conjugation classes.

We may write the irreducible characters as linear combinations of the conjugation classes,

χ(µ) =∑

g∈G

χ(µ)(g) g =

K∑

i=1

χ(µ)i Ci . (4.71)


The scalar product of χ(µ) with Ci is

(χ(µ), Ci) =∑

g∈C

(χ(µ)(g))∗ = Ni (χ(µ)i )∗ . (4.72)

And we may write the conjugation classes are linear combinations of the irreducible characters,

Ci =1

N

K∑

µ=1

(χ(µ), Ci)χ(µ) =Ni

N

K∑

µ=1

(χ(µ)i )∗ χ(µ) . (4.73)

Let us sum up in a theorem.

Theorem 4.11 The number of irreducible representations of the finite group G is K, thenumber of conjugation classes in G.

The irreducible characters χ(1), χ(2), . . . , χ(K) are basis vectors of the centre A(G)′ of thegroup algebra A(G). They are orthogonal in the natural scalar product,

(χ(µ), χ(ν)) =∑

g∈G

(χ(µ)(g))∗ χ(ν)(g) =K∑

i=1

Ni (χ(µ)i )∗ χ

(ν)i = N δµν . (4.74)

An equivalent orthogonality relation is the following,

K∑

µ=1

(χ(µ)i )∗ χ

(µ)j =

N

Niδij . (4.75)

The second orthogonality relation may also be called a completeness relation, because itis what we need to derive eq. (4.73) from eq. (4.71).

To see that the relations (4.74) and (4.75) are equivalent, consider the K × K matrix Mwith matrix elements

Mµi =

√Ni

Nχ

(µ)i . (4.76)

Eq. (4.74) says that MM† = I, whereas eq. (4.75) says that M†M = I. In the case of finitedimensional square matrices any one of the two equations MM† = I and M†M = I impliesthat M is invertible, with M−1 = M†, and then the other equation follows.

4.7 More about characters

If a linear representation ρ is a direct sum,

ρ = ρ1 ⊕ ρ2 ⊕ · · · ⊕ ρn , (4.77)

then its character χ is a sum,

χ = χ1 + χ2 + · · · + χn . (4.78)

4.7. MORE ABOUT CHARACTERS 69

Since ρ can always be decomposed as a direct sum of irreducible representations, its characteris a sum over the irreducible characters,

χ =∑

µ

mµ χ(µ) , (4.79)

where mµ = 0, 1, 2, . . . is the multiplicity with which the irreducible representation µ occurs inthe direct sum. Given the character χ, we may use the orthogonality relation of the irreduciblecharacters to compute this multiplicity,

mµ =1

N

∑

g∈G

(χ(µ)(g))∗ χ(g) =1

N

K∑

i=1

Ni (χ(µ)i )∗ χi . (4.80)

The irreducible characters as projection operators

The group algebra product of a character and a matrix element is

χ(µ) ∗ D(ν)kl =

∑

i

D(µ)ii ∗ D

(ν)kl =

∑

i

N

nµ

δµν δik D(µ)il =

N

nµ

δµν D(ν)kl . (4.81)

The product in the opposite order is the same,

D(ν)kl ∗ χ(µ) =

∑

i

D(ν)kl ∗ D

(µ)ii =

∑

i

N

nνδνµ δli D

(ν)ki =

N

nµδµν D

(ν)kl . (4.82)

Thus, the result of multiplying any a ∈ A(G) by χ(µ), from the left or from the right, is toproject out the component of a lying in the subspace A(D(µ)). Defining

eµ =nµ

Nχ(µ) , (4.83)

and using the expansion introduced in eq. (4.49), we have the identity

∑

µ

eµ ∗ a =1

N

∑

µ

∑

i

∑

j

nµ a(µ)ij D

(µ)ij = a . (4.84)

The special choice a = e, the unit element of the group G and of the group algebra A(G),gives that

∑

µ

eµ =∑

µ

eµ ∗ e = e . (4.85)

The algebra product of two irreducible characters is

χ(µ) ∗ χ(ν) =N

nµδµν χ(µ) . (4.86)

Thus, the elements eµ of the group algebra, defined above, are projections with the propertythat

eµ ∗ eν = δµν eµ . (4.87)


Eq. (4.86) is an equality between N dimensional vectors. The component of the vectorequation relative to the basis vector e is

∑

g∈G

χ(µ)(g−1)χ(ν)(g) =N

nµ

δµν χ(µ)(e) . (4.88)

The unitarity of the representation D(µ) implies that

χ(µ)(g−1) = TrD(µ)(g−1) = Tr(D(µ)(g))† = (χ(µ)(g))∗ . (4.89)

Note that it is a more or less arbitrary convention when we choose the matrix representationD(µ) to be unitary, but the relation χ(µ)(g−1) = (χ(µ)(g))∗ does not depend on any particularconvention, because the character is invariant under similarity transformations. Since

χ(µ)(e) = TrD(µ)(e) = Tr I = nµ , (4.90)

we arrive once more at the orthogonality relation of the irreducible characters,

(χ(µ), χ(ν)) =∑

g∈G

(χ(µ)(g))∗ χ(ν)(g) = N δµν . (4.91)

Number theoretical properties of characters

Every element g of a finite group G has a finite order n ≥ 1, such that gn = e, and gk 6= e if1 ≤ k < n. In any matrix representation D we must have that

(D(g))n = I . (4.92)

An eigenvalue λ of D(g) is then an n-th root of unity, it is one of the roots

λk = e i 2kπn , k = 0, 1, . . . , n − 1 (4.93)

of the characteristic equation

λn = 1 . (4.94)

The matrix D(g) can be completely diagonalized, because it is either unitary or thesimilarity transform of a unitary matrix, and every unitary matrix can be diagonalized. Bythe way, the fact that D(g) is diagonalizable also follows from Theorem B.38, because thepolynomial equation

(D(g))n − I =n−1∏

k=0

(D(g) − λkI) = 0 (4.95)

has n different complex roots λk.Anyway, the character value χ(g) = TrD(g) is the sum of the eigenvalues of D(g), which

are n-th roots of unity. Therefore it is an algebraic integer, as defined in Appendix A.

Theorem 4.12 A character value χ(g) for an element g of a finite group G is an algebraicinteger. In particular, if χ(g) is a rational number, it is an integer.

The ratio N/nµ of the group order N and the dimension nµ of an irreducible representationis also an algebraic integer. Since it is a rational number, it is an integer.

4.7. MORE ABOUT CHARACTERS 71

That N/nµ is an algebraic integer, and hence an integer, is a rather deep result. The proofgoes as follows. We read eq. (4.86) with µ = ν as an eigenvalue equation for the irreduciblecharacter χ(µ) as a linear operator,

χ(µ) ∗ χ(µ) =N

nµχ(µ) . (4.96)

In words: the operator χ(µ) has χ(µ) as an eigenvector with eigenvalue N/nµ.Now use the conjugation classes C1, C2, . . . , CK as a basis for the centre A(G)′. The

multiplication table of the conjugation classes is given by a set of constants ckij that we may

call structure constants of the commutative algebra A(G)′,

Ci ∗ Cj =∑

k

ckij Ck . (4.97)

Every coefficient ckij is either zero or a positive integer. To prove this, use the definition

Ci ∗ Cj =∑

g∈Ci

∑

h∈Cj

gh . (4.98)

Given one element g1 ∈ Ck, every element of Ck is a conjugate g′1 = fg1f−1 for some f ∈ G.

If g1 = gh with g ∈ Ci, h ∈ Cj , then g′1 = g′h′ with g′ = fgf−1 ∈ Ci, h′ = fhf−1 ∈ Cj . Thisshows that the two equations g1 = gh and g′1 = g′h′ have the same number of solutions. Thecoefficient ck

ij is just that number of solutions.We choose to read eq. (4.97) in the following way. We regard Ci as a linear transformation

of the basis vector Cj into a linear combination of basis vectors Ck with k = 1, 2, . . . ,K. Inanalogy with eq. (3.4), this gives us a matrix representation of the linear algebra of conjugationclasses, Ci → Ci, such that the matrix elements of Ci are

(Ci)kj = ckij . (4.99)

Since the character χ(µ) is a linear combination of the conjugation classes,

χ(µ) =

K∑

i=1

χ(µ)i Ci , (4.100)

where the character values χ(µ)i ∈ C are algebraic integers, we have a representation of χ(µ)

as a matrix

A(µ) =

K∑

i=1

χ(µ)i Ci , (4.101)

with matrix elements that are algebraic integers. The characteristic polynomial of A(µ),

det(λI − A(µ)) = λnµ + anµ−1λnµ−1 + · · · a1λ + a0 , (4.102)

has coefficients ai that are algebraic integers. Hence, every eigenvalue λ, including λ = N/nµ,is an algebraic integer.

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