NOTES ON REPRESENTATIONS OF FINITE GROUPS Contents 1 ...

NOTES ON REPRESENTATIONS OF FINITE GROUPSBEN MCKAYContents1. Definitions 12. Examples 23. Invariants 34. Subrepresentations 35. Irreducible representations 56. Schur’s lemma 57. Characters 78. Characters and class functions 101. DefinitionsFrom here on, all vector spaces will be finite dimensional complex vector spaces.All groups will be finite groups.Definition 1. Let V be a vector space. Then GL (V ) is the the group of all isomorphismsfrom V to V .Definition 2. A representation of a group G is a group morphism ρ: G → GL (V ).So to each abstract group eleemnt g ∈ G, we associate a linear map ρ(g): V → V ,so that for any g, h ∈ G, ρ(gh) = ρ(g)ρ(h). In particular ρ(1) = I andρ ( g −1) = ρ(g) −1 .When the particular morphism ρ is clear, we will often say that V is a representationof G.Definition 3. A isomorphism of two representations ρ 1 : G → GL (V 1 ) and ρ 2 : G →GL (V 2 ) is an isomorphism of vector spaces (just one linear map, not depending onthe group element) say φ: V 1 → V 1 so thatfor all g ∈ G.ρ 2 (g) ◦ φ = φ ◦ ρ 1 (g),Date: December 9, 2011.1

2 BEN MCKAY2. Examples(1) Let G be the group of permutations of the numbers 1, 2, . . . , n. Let V bean n-dimensional vector space with a basis v 1 , v 2 , . . . , v n . Associate to eachpermutation p the linear map ρ(p) given byρ(p)v i = v p(i) .(2) Let G be the group Z n and V = C. Each linear map V → V is justmultiplication by some complex number. Letρ(m) = e 2πim/n ∈ C × .(3) For any group G, let V = C, we can just let ρ(g) = I for every g ∈ G. Thisρ is called the trivial representation.(4) Suppose that G is a finite group, and that V is a vector space whose dimensionequals the number of elements in G. Suppose that V has a basiswhich is indexed by the elements of G, say{v i } i∈G.Then for each g ∈ G, define ρ(g) byρ(g)v i = v gi .This representation is called the regular representation.(5) We can generalize the notion of representation.Definition 4. If G is a group and X is a set, an action of G on X is a choice,for each g ∈ G, of permutation p(g): X → X, so that p(gh) = p(g)p(h). Ifthe choice of permutation p(g) is understood, we will write p(g)x as gx.(6) For example, the group of permutations G of the numbers 1, 2, . . . , n actsin the obvious way onX = {1, 2, . . . , n} .(7) If G acts on X, take V to be any vector space with a basis{v i } i∈Xindexed by the elements of the set X. Then letρ(g)v x = v gx ,for each x ∈ X. We call ρ the permutation representation associated to theaction on X.(8) Recall that if V 1 and V 2 are two vector spaces, they have a direct sumV 1 ⊕ V 2 consisting of the pairs (v 1 , v 2 ) for v 1 ∈ V 1 and v 2 ∈ V 2 , withpairwise addition and scaling. If ρ 1 : G → GL (V 1 ) and ρ 1 : G → GL (V 2 )are two representations, then V 1 ⊕ V 2 is a representation, via the ruleρ(g) (v 1 , v 2 ) = (ρ 1 (g)v 1 , ρ 2 (g)v 2 ) .(9) Recall that if V 1 and V 2 are two vector spaces, they have a tensor productV 1 ⊗ V 2 . It is tricky to say what the elements of V 1 ⊗ V 2 are, but we cansay that ifu 1 , u 2 , . . . , u pis a basis of V 1 andw 1 , w 2 , . . . , w q

NOTES ON REPRESENTATIONS OF FINITE GROUPS 3is a basis of V 2 , then there are some elementsu 1 ⊗ w 1 , u 1 ⊗ w 2 , . . . , u 1 ⊗ w q ,u 2 ⊗ w 1 , u 2 ⊗ w 2 , . . . , u 2 ⊗ w q ,.,u p ⊗ w 1 , u p ⊗ w 2 , . . . , u p ⊗ w q ,forming a basis of V 1 ⊗V 2 (called the product basis). If ρ 1 : G → GL (V 1 ) andρ 1 : G → GL (V 2 ) are two representations, then V 1 ⊗ V 2 is a representation,via the ruleρ(g) (u i ⊗ w j ) = ρ 1 (g)u i ⊗ ρ 2 (g)w j .3. InvariantsDefinition 5. If ρ: G → GL (V ) is a representation, a fixed vector of the representationis a vector v ∈ V so that ρ(g)v = v for every g ∈ G. We denote by V G theset of all fixed vectors. For any vector v ∈ V , we denote by p(v) the vectorp(v) = 1|G|∑ρ(g)v,the average of all of the things that G can do to v. We call p the averaging operator,and sometimes write p(v) as v G .Lemma 1. The averaging operator is linear, and G-invariant, and is a projectionto V G , i.e. p ◦ p = p, and p is the identity on V G .Proof. Linearity is clear. Clearly since we averaged over all of the different elementsof G, p is G-invariant. If v is fixed, each ρ(g) drops out, and we get p(v) = v. Clearlythe p(v) is G-invariant, i.e. the image of p is contained in V G so p ◦ p = p. □g∈G4. SubrepresentationsLemma 2. Suppose that ρ: G → GL (V ) is a representation of a finite group G.Take any inner productv, w ∈ V ↦→ 〈v, w〉 ∈ C.From this inner product, define a new operation, which we will write asand is defined byv, w ∈ V ↦→ 〈v, w〉 G ∈ C,〈v, w〉 G = 1|G|∑〈ρ(g)v, ρ(g)w〉 .g∈G(Average over all ways of getting G to act on the pair v, w of vectors). Then thisnew operation is a G-invariant inner product, i.e.〈ρ(g)v, ρ(g)w〉 G = 〈v, w〉 G ,for any g ∈ G and v, w ∈ V .In particular, V has a G-invariant inner product.

4 BEN MCKAYProof. Because we averaged over all the things that G can do, the result is clearlyG-invariant. It is clearly linear in v and conjugate linear in w. We only needto check that 〈v, v〉 G ≥ 0 and equals 0 only if v = 0. But 〈v, v〉 G is a sum ofnonnegative terms, vanishing only when ρ(g)v = 0 for all g. But ρ(g) is linearisomorphism, so ρ(g)v = 0 just when v = 0.□Example 1. If we let G be the group of permutations of X = {1, 2, . . . , n}, with thepermutation representation on V = C n , then the usual inner product〈z, w〉 = ∑ iz i ¯w iis G-invariant.Lemma 3. If ρ: G → GL (V ) is a representation of a finite group, then everyelement ρ(g) is diagonalizable, and its eigenvalues are complex numbers of unitlength.Proof. By lemma 2 on the preceding page, we can pick an invariant inner product.By definition, the inner product is invariant under every ρ(g), so every ρ(g)is unitary for that inner product. Unitary linear maps are normal, so unitarilydiagonalizable.□Warning: we can diagonalize each individual ρ(g) in some basis, but maybe notall of them at once.Definition 6. If V is a representation of G, then a subrepresentation of V is aG-invariant subspace of V .Lemma 4. If ρ: G → GL (V ) is a representation of a finite group, then everysubrepresentation has an invariant complement.Proof. If W ⊂ V is a G-invariant subspace, then take any G-invariant inner product,and clearly W ⊥ is also G-invariant, and V = W ⊕ W ⊥ .□Exercise 1. Suppose that G is a finite abelian group and ρ: G → GL (V ) is arepresentation. Prove that there is a basis, sayv 1 , v 2 , . . . , v n ,so that all of the ρ(g) are diagonal this one basis, i.e. for every g,for some functions λ i : G → C × .ρ(g)v i = λ i (g)v i ,Example 2. Let G be the symmetric group on 3 letters.(1) One obvious representation is the trivial one, on V = C.(2) Consider the permutation representation of G on V = C 3 . Take the subspaceU ⊂ C 3 to be the vectors⎛ ⎞z 1z = ⎝z 2⎠z 3for which z 1 + z 2 + z 3 = 0. Clearly U is G-invariant.(3) Define a representation of G on V = C by ρ(g)z = sgn(g)z (where sgn isthe usual sign of a permutation).

NOTES ON REPRESENTATIONS OF FINITE GROUPS 5Lemma 5. Suppose that V is a representation of a finite group G, with averagingoperator p: V → V . Then tr p = dim V G .Proof. The map p is a projection map (i.e. p is the identity on its image), andtherefore we can split V into the kernel and image of p. On the kernel, p = 0, whileon the image, p = I.□5. Irreducible representationsDefinition 7. A representation V of a group G is irreducible if it admits no G-invariant subspaces other than 0 and V .Theorem 1. Every representation of a finite group is a direct sum of irreduciblerepresentations.The proof is clearly by induction: take any subrepresentation W , and then Vsplits V = W ⊕ W ⊥ . Then try to split W and W ⊥ , etc.Exercise 2. Prove that every representation of any finite abelian group is a directsum of dimension 1 representations.Exercise 3. Prove that an irreducible representation of a finite group G has at mostone G-invariant inner product.Exercise 4. Prove that the 2-dimensional representation U of example 2 on thepreceding page is irreducible.6. Schur’s lemmaTheorem 2. Given two irreducible representations ρ 1 : G → GL (V 1 ) and ρ 2 : G →GL (V 2 ), any G-invariant linear map φ: V 1 → V 2 is either an isomorphism ofrepresentations or is 0.Either there is no isomorphism of representations from V 1 to V 2 , or else there isan isomorphism and any two isomorphisms, say φ and ψ, satisfy ψ = cφ for somenonzero complex number c.Proof. The kernel of φ is a G-invariant subspace of V 1 , since φ is G-invariant.Therefore the kernel of φ is 0 or is V 1 . The kernel is V 1 just when φ = 0, so we canassume that the kernel of φ is 0. So φ is 1-1.The image of φ is a G-invariant subspace of V 2 , so either is 0 or is V 2 . If theimage of φ is 0 then φ = 0, so we can assume that the image if V 2 . But then φ isan isomorphism.If ψ is another isomorphism, then let α = ψ −1 ◦ φ. So α is a G-invariant linearisomorphism V 1 → V 1 . The eigenspaces of α are G-invariant, so each eigenspaceis either V 1 or 0. So there can only be at most one eigenspace. Every linear mapV 1 → V 1 on any finite dimensional complex vector space has an eigenvalue, whoseeigenspace must be nonzero, so must be V 1 ; let a be the corresponding eigenvalue.Then α = aI, i.e. φ = aψ.□If V is a vector space, it is traditional in representation theory to write 2V tomean V ⊕ V , etc.Theorem 3. If V is a representation of a finite group G, then V splits into a sumof multiples of irreducible representationsV = n 1 V 1 ⊕ n 2 V 2 ⊕ · · · ⊕ n k V k ,

6 BEN MCKAYso that each V i is not isomorphic to any V j for i ≠ j. The summands n i V i areuniquely determined, as are the multiplicities n i , up to ordering. The various V iare uniquely determined up to isomorphism.Proof. We know that we can split V into irreducible representations by theorem 1on the preceding page. Suppose that we have two splittings as above, sayV = n 1 V 1 ⊕ n 2 V 2 ⊕ · · · ⊕ n k V k , = p 1 W 1 ⊕ p 2 W 2 ⊕ · · · ⊕ p l W l .The identity map V → V yields a map on each factor V i , and either identifies itwith a factor W j , or maps it to that factor by the 0 linear map.□Example 3. Careful: take any group G and the trivial G representation on V = C 2 .Then we can split C 2 = C ⊕ C. But we could also (for example) let V 1 be the spanof the vector ( 11)and let V 2 be the span of the vector( ) 1.−1Then V = V 1 ⊕ V 2 . So V has many decompositions into G-invariant subspaces, but(by the theorem) only one into maximal sums of irreducible subspaces.Definition 8. If ρ 1 : G → GL (V 1 ) and ρ 2 : G → GL (V 2 ) are two representations ofa finite group, then W = Hom (V 1 , V 2 ) (the set of all linear maps V 1 → V 2 ) hasG-representationρ: G → GL (W ) ,given by taking ρ(g)φ to be the linear map(ρ(g)φ) (v) = ρ 2 (g)φ ( ρ 1 (g) −1 v ) .Exercise 5. Prove that this makes Hom (V 1 , V 2 ) into a G-representation.Definition 9. A morphism of representations is a linear map φ: V 1 → V 2 so thatφ (ρ 1 (g)v) = ρ 2 φ(v),for all vinV 1 and g ∈ G. Let Hom G (V 1 , V 2 ) be the set of all morphisms of representationsV 1 → V 2 .Exercise 6. Prove that Hom G (V 1 , V 2 ) is precisely the set of vectors in Hom (V 1 , V 2 )which are invariant under all elements of G, i.e.Hom G (V 1 , V 2 ) = Hom (V 1 , V 2 ) G .Lemma 6. If V and W are two representations of a finite group G, say with eachsplitting into irreducibles asandV = p 1 V 1 ⊕ p 2 V 2 ⊕ . . . p k V k ,W = q 1 W 1 ⊕ q 2 W 2 ⊕ . . . q l W l ,then Hom G (V, W ) splits into irreducibles asHom G (V, W ) = ⊕ ijp i q j Hom (V i , W j ) ,

NOTES ON REPRESENTATIONS OF FINITE GROUPS 7where the direct sum is a sum over pairs i and j for which V i and W j are isomorphic.Moreover, Hom (V i , W j ) is then 1-dimensional.Proof. From theorem 3 on page 5, we have unique splittings into irreducibles. Fromtheorem 2 on page 5, any linear map φ ∈ Hom (V, W ) must split into a sum ofisomorphisms and 0 maps. Moreover, each isomorphism is unique up to scaling, a1-dimensional representation Hom (V i , W j ).□7. CharactersDefinition 10. The trace of a square (say n × n) matrix A isthe sum of the diagonal entries.tr A = A 11 + A 22 + · · · + A nn ,Exercise 7. Prove that tr(AB) = tr(BA) for any n × n matrices A and B.Definition 11. The traceof a linear map φ: V → V on a finite dimensional vector space V is defined bytr φtr φ = tr Awhere A is the matrix of φ in some basis.Exercise 8. Prove that tr φ does not depend on the choice of basis.Choosing a basis in which A is in Jordan normal form, we see that tr φ is thesum of the eigenvalues, counted with multiplicities.Definition 12. If ρ: G → GL (V ) is a representation, its character is the functionχ (also written as χ ρ ) given byχ(g) = tr ρ(g).Exercise 9. Let G be the group of permutations of 1, 2, 3. Write out each permutation,and find the value of the character of the permutation representation on eachpermutation.Exercise 10. Prove that if G is the group of permutations of 1, 2, . . . , n then thecharacter of the permutation representation is given by χ(g) equal to the numberof numbers from among 1, 2, 3, . . . , n which are fixed by g.Exercise 11. If χ is the character of an n-dimensional representation of a finitegroup G, then χ(1) = n.Lemma 7. The average of the character of a representation is the dimension ofthe set of fixed vectors. In other words, if ρ: G → GLV is a representation withcharacter χ, then1 ∑χ(g) = dim V G .|G|g∈GProof. Let p: V → V be the averaging operator. By lemma 5 on page 5, dim V G =tr p. Butp(v) = 1 ∑ρ(g)v,|G|g∈G

8 BEN MCKAYi.e.Thusp = 1|G|∑ρ(g).g∈Gdim V G = tr p= 1 ∑tr ρ(g)|G|= 1|G|g∈G∑χ(g).Example 4. Let ρ: G → GL (V ) be the regular representation of a finite group G,and χ its character. Then χ(1) = |G|. Moreover, exercise 10 on the preceding pagetells us that χ(g) is the number of elements of G fixed by g. But if g ≠ 1, thenthe action takes each h ∈ G to gh ∈ G, and has no fixed points: if gh = h thenmultiply on the right by h to find g = 1. So χ(g) = 0 for g ≠ 1. The average ofχ is therefore 1. So there is a one dimensional space V G of fixed vectors, which isprecisely the span of the vector∑v h .h∈GLemma 8. If χ is the character of an n-dimensional representation of a finitegroup G, then(1) χ ( g −1) = ¯χ(g) (complex conjugation), and(2) χ (gh) = χ (hg),for any g, h ∈ G.Proof. The second property is exercise 7 on the previous page. The first is justthat, for each g, the linear map ρ(g) is unitarily and so is unitarily diagonalizablewith eigenvalues being unit length complex numbers, say λ j = e iθj . Sog∈Gχ(g) = tr ρ(g)□= ∑ j= ∑ jλ je iθj .But thenχ ( g −1) = tr ρ ( g −1)= tr (ρ(g)) −1= ∑ e −iθj .jLemma 9. If ρ 1 : G → GL (V 1 ) and ρ 2 : G → GL (V 2 ) are two representations of afinite group, with characters χ 1 and χ 2 , then the character of V 1 ⊕ V 2 is χ 1 + χ 2 ,and the character of V 1 ⊗ V 2 is χ 1 χ 2 .□

NOTES ON REPRESENTATIONS OF FINITE GROUPS 9Proof. Fix one element g ∈ G. Pick a basis of eigenvectorsu 1 , u 2 , . . . , u p ∈ V 1of ρ 1 (g), sayρ 1 (g)u i = λ i u i .So these λ i are the eigenvalues, andχ 1 (g) = ∑ iλ i .Similarly, pick a basis of eigenvectorsw 1 , w 2 , . . . , w q ∈ V 2of ρ 2 (g), sayandρ 2 (g)w j = µ j w j ,χ 2 (g) = ∑ jµ j .Let ρ(g) be the direct sum representation. Make the basis(u 1 , 0) , (u 2 , 0) , . . . , (u p , 0) , (0, w 1 ) , (0, w 2 ) , . . . , (0, w q ) ∈ V 1 ⊕ V 2 .These are eigenvectors of the representation on the direct sum, with eigenvaluesλ 1 , λ 2 , . . . , λ p , µ 1 , µ 2 , . . . , µ q .So the trace is the sum of the traces.The proof for tensor products is similar. Clearlyρ(g)u i ⊗ w j = λ i µ j u i ⊗ w j ,so diagonal in the product basis, with characterχ(g) = ∑ λ i µ jij( ) ⎛ ∑= λ j⎝ ∑ijµ j⎞⎠= χ 1 (g)χ 2 (g).Exercise 12. If ρ: G → GL (V ) is the permutation representation of some finitegroup G acting on some finite set X, and χ = χ ρ , prove that, for any g ∈ G, χ(g)is the number of elements of X fixed by g.Lemma 10. If ρ 1 : G → GL (V 1 ) and ρ 2 : G → GL (V 2 ) are two representations ofa finite group, with characters χ 1 and χ 2 , then the character of the representationon Hom (V 1 , V 2 ) isχ = χ 2 ¯χ 1 .□

10 BEN MCKAYProof. For each g ∈ G, again pick a basis {v i } ⊂ V 1 of eigenvectors of ρ 1 (g),ρ 1 (g)v i = λ i v i ,and a basis {w j } ⊂ V 2 of eigenvectors of ρ 2 (g),ρ 2 (g)w j = µ j w j .Then using a G-invariant inner product on V 1 , we can define linear mapsφ ij ∈ Hom (V 1 , V 2 )byφ ij (v) = 〈v, v i 〉 w j .We then check thatρ(g)φ ij = µ j ¯λi φ ij .Exercise 13. Check the last step in the last proof.□8. Characters and class functionsDefinition 13. A class function for a finite group G is a function f : G → C so thatf(gh) = f(hg) for any g, h ∈ G.For example, the character of any representation is a class function.Exercise 14. Let C G be the vector space of all class functions on G, with usualaddition and scaling of functions. For G = Z n , what is the dimension of C G ?Definition 14. We define an inner product on class functions. If f 1 and f 2 are twoclass functions, define the inner product〈f 1 , f 2 〉 = 1 ∑f 1 (g)|G|¯f 2 (g) ,(where the bar is complex conjugate).g∈GExercise 15. Prove that the space C G of class functions is a finite dimensional innerproduct space, with this inner product.Theorem 4. If V 1 and V 2 are irreducible representations of a finite group G, saywith characters χ 1 and χ 2 , then{1 if V 1 is isomorphic to V 2 ,〈χ 1 , χ 2 〉 =0 otherwise.Proof. Let χ be the character of the representation Hom (V 1 , V 2 ). Recall χ = χ 2 ¯χ 1from lemma 10 on the preceding page.〈χ 1 , χ 2 〉 = 1 ∑χ 1 (g) ¯χ 2 (g)|G|so by lemma 7,= 1|G|g∈G∑χ (g)g∈G= dim Hom (V 1 , V 2 ) G ,

NOTES ON REPRESENTATIONS OF FINITE GROUPS 11and by exercise 6 this is= dim Hom G (V 1 , V 2 ) ,and because V 1 and V 2 are irreducible this is{1 if V 1 is isomorphic to V 2 ,=0 otherwise.□Theorem 5. Suppose that V is a representation of a finite group G, with characterχ, and V splits into a sum of multiples of irreducibles, sayand V i has character χ i , thenV = n 1 V 1 ⊕ n 2 V 2 ⊕ · · · ⊕ n k V k ,n i = 〈χ, χ i 〉 .In particular, two representations with the same character are isomorphic.Corollary 1. There are only finitely many irreducible representations of any finitegroup up to isomorphism.Proof. Each one gives us a unit vector in C G , and all of the unit vectors given thisway are orthonormal. But C G has finite dimension.□Theorem 6. The character χ of a representation has 〈χ, χ〉 an integer, and thisinteger is 1 just when the representation is irreducible.Proof. Write out our representation as a sum of multiples of irreducibles, sayV = n 1 V 1 ⊕ n 2 V 2 ⊕ · · · ⊕ n k V k ,and soχ = n 1 χ 1 + n 2 χ 2 + . . . n k χ k ,if V i has character χ i . Then〈χ, χ〉 = ∑ n 2 i .□Theorem 7. Every irreducible representation of a finite group appears as a summandin the regular representation. Its multiplicity equals its dimension.Proof. By exercise 12 on page 9, the character of the regular representation V hasχ(g) equal to the number of fixed points of g acting on X = G. But g acts bytaking h ↦→ gh. There are no fixed points unless g = 1, when every point is fixed.So{|G| if g = 1,χ(g) =0 otherwise.Take any character η of any irreducible representation W . Then the number n oftimes that W occurs in the regular representation V isn = 〈χ, η〉 = 1 ∑χ(g)¯η(g).|G|g∈G

12 BEN MCKAYBut χ(g) = 0 except at g = 1, son = 1|G| χ(1)¯η(1)= 1|G| |G|¯η(1)= ¯η(1)= dim W.□