NOTES ON REPRESENTATIONS OF FINITE GROUPS Contents 1 ...

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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 .□
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