1 Homework 8 Solutions

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1 1 Homework 8 Solutions (1) Let G be a finite group acting on a finite set S. Let V be the vector space of functions f : S C and let ρ be the homomorphism G GL C (V ) such that (ρ(g)f)(s) = f(g 1 s) g G, s S. (a) Show ρ is a homomorphism. SOLUTION: We must check that ρ(gh) = ρ(g)ρ(h); to check this it is enough to show their actions on V is the same. From the definition of the action of ρ, one has: (ρ(gh)f)(s) = f((gh) 1 s) = f(h 1 g 1 s) = f(h 1 (g 1 s)) = (ρ(h)f)(g 1 s) = ρ(g)(ρ(h)f)(s) = (ρ(g)ρ(h)f)(s) (b) Show χ ρ (g) = Card({s S gs = s}) SOLUTION: Using the hint, we know the δ-functions form a basis for V ; furthermore, one checks that ρ(g)δ s = δ gs. Therefore, writing the explicit matrix of ρ(g) relative to the basis of δ-functions, the trace of the matrix ρ(g) is the number of δ-functions fixed by the action of g, i.e. Card({δ s V δ gs = δ s }) whereupon the desired result follows. (c) Letting 1 G denote the trivial representation of G, show χ ρ, 1 G equals the number of G-orbits on S. SOLUTION: χ ρ, 1 G is the multiplicity of the trivial representation in χ ρ. Let O s = {s 1, s 2,...} be the G-orbit of s; then the δ-function δ s1 + δ s is invariant under the G-action on V ; hence forms a trivial subrepresentation of ρ on V (i.e a vector in V fixed by all elements of G); there is clearly one such function for each G-orbit, so χ ρ, 1 G is at least as big as the number of orbits. On the other hand, take some v V such that gv = v g and write v relative to the δ-basis. If the coefficient of δ s is c s, then since for any t O s there is g G such that gs = t, the coefficient of δ t must also be c s ; therefore the coefficient of δ si is equal for all s i O s (note that since the orbits are disjoint, there is no issue with coefficients from different orbits interfering with one another). But if the coefficients of v are constant on orbits, then v is clearly a linear combination of the functions t O s δ t, hence the dimension of such v is at most equal to the number of G-orbits which proves the equality. 1

2 (2) Let ρ 1,ρ 2,ρ 3 be the three irreducible representations of S 3 as determined in the previous homework and let χ i be the character of ρ i. (a) Write down the elements e ρi = χi(1) 6 g S 3 χ i (g 1 )[g] SOLUTION: One has: e 1 = 1 6 [(1)(2)(3)] [(12)(3)] [(1)(23)] [(13)(2)] [(123)] [(132)] e 2 = 1 6 [(1)(2)(3)] 1 6 [(12)(3)] 1 6 [(1)(23)] 1 6 [(13)(2)] [(123)] [(132)] e 3 = 2 6 [(1)(2)(3)] 1 6 [(123)] 1 6 [(132)] (b) Verify the elements {e i } belong to Z(C[S 3 ]). SOLUTION: In the previous homework, we showed C[S 3 ] = C C Mat 2 2 (C) under the isomorphism h(x) = (α(x), β(x), γ(x)). Under this isomorphism, one has: ( ( )) h(e 1 ) = 1, 0, ( ( )) h(e 2 ) = 0, 1, h(e 3 ) = ( 0, 0, ( )) Since C is commutative, α(e i ) and β(e i ) obviously lie in Z(C) for all i. Furthermore, it is clear that the matrices γ(e i ) lie in Z(Mat 2 2 (C)) so h(e i ) commutes with any h(x) and so the e i are central in C[S 3 ]. (c) Verify the e i are idempotent and orthogonal to one another. This is clear using the isomorphism h from above. 2

3 (3) Let ρ be a representation of G acting on a C-vector space V. (a) Prove: χ ρ (z) = d χ ρ (x 1 )χ ρ (xz) SOLUTION: Note that we have defined d := χ ρ (1) = dim C (V ). Letting ξ i,j (x) be the i, j-entry of ρ(x), and following the hint, the RHS of the above expression can be rewritten: d χ ρ (x 1 )χ ρ (xz) = d 1 i,j,k d ξ i,i (x 1 )ξ j,k (x)ξ k,j (z) (1) The RHS of this expression is simply writing out the definition of matrix multiplication. Now consider the coefficient of ξ k,j (z) in this sum, it is given by: Coeff(ξ k,j (z)) = 1 i,j,k d ξ i,i (x 1 )ξ j,k (x) For i, j, k fixed, this sum is simply the orthogonality relation for the matrix entries ξ i,i and ξ j,k ; it is therefore equal to 0 if i j or i k. If i = j = k it is equal to d. Thus the RHS of (1) becomes: d 1 i,j,k d ξ i,i (x 1 )ξ j,k (x)ξ k,j (z) = d = d d = χ ρ (z) 1 i d 1 i d ξ i,i (x 1 )ξ i,i (x)ξ i,i (z) ξ i,i (z) 3

4 (b) Deduce from (a) that e ρ as defined before is a central idempotent in C[G]. SOLUTION: We will show that the e ρ commute with the basis of δ-functions, hence with all of C[G]. One has: ( ) d e ρ δ z = χ ρ (x 1 )[x] [z] = d = d χ ρ (x 1 )[xz] χ ρ (x 1 )[zz 1 xz] = d χ ρ (zy 1 z 1 )[zy] y G y := z 1 xz (2) = δ z d χ ρ (y 1 )[y] (3) = δ z e ρ y G (3) follows from (2) upon noting that χ ρ is constant on conjugacy classes, hence χ ρ (zy 1 z 1 ) = χ ρ (y 1 ). To see that e ρ is idempotent we first expand: e ρ e ρ = ( d ) χ ρ (x 1 )[x] d χ ρ (y 1 )[y] y G Now we focus on the coefficient of [z] on the right side: Coeff([z]) = d2 2 = d2 2 = d2 2 x,y G xy=z = d χ ρ(z) χ ρ (x 1 )χ ρ (y 1 ) χ ρ (x 1 )χ ρ (z 1 x) ( d χ ρ(x 1 zx) ) y = x 1 z Since the coefficient of [z] in e 2 ρ is equal to the coefficient of [z] in e ρ for all z G, it follows e ρ is idempotent. 4

5 (c) Suppose ρ 1 and ρ 2 are non-isomorphic irreducible representations of G. Show e ρ1 e ρ2 = 0. SOLUTION: Expanding out as in (b), we see that this is exactly the orthogonality condition on irreducible characters which follows from the orthogonality condition on the matrix coefficients. 5

6 (4) Let σ : C C be a ring homomorphism. (a) Let ζ n denote an n th root of 1, so ζ n = e 2πim n for some m. Show that!a Z/nZ such that σ(ζ n ) = ζn. a SOLUTION: Given a homomorphism σ, we must determine how it moves an arbitrary element of C. Since the ζ n are precisely the elements of C of finite order, σ must preserve this order in order to be a homomorphism. Thus, since ζn n = 1, σ(ζ n ) n = 1 as well whereupon we see that σ(ζ n ) is another n th root of unity. Assuming ζ n is primitive, i.e. gcd(m, n) = 1 so that the smallest power of ζ n equal to 1 is ζn, n one has that all n th roots of unity (primitive or otherwise) are powers of ζ n ; hence σ(ζ n ) = ζn a and thus σ(ζn m ) = ζn am m Z/nZ. (b) Suppose x G has order n. For ρ a representation of G, show χ ρ (x a ) = σ(χ ρ (x)). SOLUTION: χ ρ (x) is the sum of the eigenvalues of ρ(x). Since x has order n in G, the eigenvalues of ρ(x) must also be n th roots of unity, otherwise ρ(x) n = Id will not be satisfied. Therefore the eigenvalues of ρ(x a ) are the eigenvalues of ρ(x) raised to the a th power whereupon χ ρ (x a ) = σ(χ ρ (x)) as σ is a ring homomorphism. Alternatively, to use the hint, we define N σ : GL C (V ) GL C (V ) by sending each matrix entry m i,j to σ(m i,j ) giving a new matrix. Since σ is a ring homomorphism, it follows that N σ is a group homomorphism. Now show N σ ρ(x) is conjugate to x a. 6