NOTES ON REPRESENTATIONS OF FINITE GROUPS Contents 1 ...

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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
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