By Kevin McGerty
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Extra resources for Group Representation Theory [Lecture notes]
Example text
K∈G Thus z ∈ Z(A) if and only if λk = λgkg−1 for all g ∈ G, that is, if and only if z is a class function. 9. Let (V, ρ) be an irreducible representation of G, and let f ∈ Ck (G). Then we have |G| ρ(f ) = f, χ∗V idV dim(V ) Proof. idV for some λ ∈ k. idV ) = λ. dim(V ), hence we see that 1 λ= f (g)tr(ρ(g)) dim(V ) g∈G |G| = f, χ∗V . 10. For V be an irreducible representation let cV = dim(V ) |G| χV (g −1 )eg ∈ k[G]. g∈G Since χV is a class function, so is cV , indeed cV = dim(V ) ∗ |G| χV . 36 KEVIN MCGERTY Notice that if V is the trivial representation, then cV is just the element IG we studied earlier.
We have already seen that given ρ : G → GL(V ), we may construct an algebra map from k[G] → End(V ). Thus we need only check the converse. Suppose that ρ : k[G] → End(V ). For g ∈ G set ρG (g) = ρ(eg ). h) = ρ(egh ) = ρ(eg ) ◦ ρ(eh ) = ρG (g) ◦ ρG (h). But now since ρ(ee ) = idV , it follows that ρG (g) is invertible for every g ∈ G with inverse ρG (g −1 ), and hence ρG is a homomorphism from G to GL(V ) as required. We now use the group algebra to reformulate the averaging trick we used in the proof of Maschke’s theorem.
It is clear that C, the group generated by g, is a finite cyclic subgroup of G, hence we see that as a Crepresentation, V splits as a direct sum of lines. Since g ∈ C we immediately see that g is diagonalizable as required. Picking a basis {e1 , e2 , . . , en } of eigenvectors n of g, so that g(ei ) = λi ei say, we have χV (g) = i=1 λi . Since ρ(g −1 ) = ρ(g)−1 is n −1 diagonal on the same basis, with eigenvalues λ−1 ) = i=1 λ−1 i , it follows χV (g i . If k = C, then since ρ(g)|G| = id for any g ∈ G, the eigenvalues of ρ(g) are roots of unity, and hence they have modulus 1.