Solutions to Example Sheet 3 1. Let G = A5. For each pair of irreducible representations ρ, ρ0, (i) decompose ρ ρ0 into a sum of irreducible representations, ⊗ (ii) decompose 2 ρ and S2 ρ into irreducible representations, V (iii) decompose n ρ into irreducible representations, for n 3, V ≥ (iv) check that if ρ is a non-trivial representation of A5 (hence faithful), every representation of A5 2 occurs with non-zero multiplicity in at least one of ; ρ, ρ⊗ ;::: . Solution. Straightforward but tedious. Use Problem 3(iii) for n ρ. The last part follows from V Problem 11. 2. (i) Let CG be the space of class functions on G, and δ be the basis of \delta functions on conjugacy O classes". show that the functions δ are orthogonal idempotents, and hence CG is isomorphic as an algebra to a direct sum of copiesO of the one-dimensional algebra C. (ii) Let G = Z=nZ, and λ be the one-dimensional representation of G which sends 1 G to 2πi=n n 2 multiplication by e . show CG = C[λ]=(λ 1). Check this is consistent with part (i). − 3 2 (iii) Let G = S3, and ρ be the two-dimensional representation of G. Show CG = C[ρ]=(ρ ρ 2ρ). − − Solution. (i) Let 1;:::; n be the conjugacy classes of G. We define δ : G C by O O O ! 1 if x ; δ (x) = ( 2 O O 0 if x = : 2 O 2 Clearly δ = δ and so δ is an idempotent element in the ring CG (with respect to addition and multiplicationO O of classO functions). We have 1 = G if i = j; ( i δ i ; δ j = δ i (x)δ j (x) = jO j j j (2.1) h O O i G X O O 0 if i = j x G j j 2 6 since every x G lies in a unique k for some k 1; : : : ; n . Note that i = G = 0 and so (2.1) 2 O 2 f g jO j j j 6 shows that δ 1 ; : : : ; δ n are orthogonal to each other and in particular, linearly independent over O O C. Given any f CG, f constant on conjugacy classes means that there exists c1; : : : ; cn C 2 2 such that f(x) = ci for every x i and we have 2 O f = c1δ 1 + + cnδ n : O ··· O C Hence δ 1 ; : : : ; δ n is a basis for CG. So as vector spaces over , f O O g ★ C C C C Cn CG = δ 1 δ n = = : O ⊕ · · · ⊕ O ∼ ⊕ · · · ⊕ n copies | {z } n n Recall that C may be made into an algebra by defining product of (a1; : : : ; an); (b1; : : : ; bn) C by 2 (a1; : : : ; an)(b1; : : : ; bn) = (a1b1; : : : ; anbn): (2.2) 1 Since δ 1 ; : : : ; δ n are idempotent and furthermore O O δ if i = j; ( i δ i δ j = O O O 0 if i = j: 6 So if f = aiδ i ; g = bjδ j CG, we have P O P O 2 n n n fg = aiδ i bjδ j = aibjδ i δ j = aibiδ i : (2.3) X O X O X O O X O i=1 j=1 i;j i=1 Comparing (2.2) and (2.3), we see that ★ is indeed an isomorphism of algebras. 1On second thoughts, this might be what the lecturer meant by `orthogonal' idempotents. So maybe (2.1) is superfluous. But since it took me a while to TEX it, I'll leave it. 1 2πi=n 2πi=n (ii) λ : Z=nZ C×, 1 e is an irreducible character. Since e is a primitive nth root of ! n 17! \ unity, ; λ, : : : ; λ − forms a complete set of irreducible characters of Z=nZ (recall that Z=nZ = λ from the solution of Example Sheet 2, Problem 7(b)) and hence a basis of CZ Z. Now h i =n consider the map ' : C[x] CZ=nZ, f(x) f(λ). ' is clearly a homomorphism of algebras. n 1 ! 7! n 1 Since 1; x; : : : ; x − would be mapped to ; λ, : : : ; λ − , so rank(') = dimC(CZ=nZ) and ' is C n n surjective. We have [x]= ker(') ∼= im(') = CZ=nZ. Since λ = , we have that x 1 ker(') and so the (xn 1) ker('). But − 2 − ⊆ n dimC(C[x]=(x 1)) = n = dimC(CZ Z) = dimC(C[x]= ker(')) − =n and so we must have equality (xn 1) = ker('). Therefore − n C[x]=(x 1) = CZ Z: − ∼ =n (iii) It is easy to construct the character table of S3 | it has three conjugacy classes and its three irreducible characters are just the obvious ones: , "(σ) = sgn(σ) and χ(σ) = 1; 2; 3 σ 1. χ is the character of the two-dimensional representation ρ. The character table togetherjf g withj − the 2 3 characters corresponding to ρ⊗ and ρ⊗ is: σS3 1 (1 2)S3 (1 2 3)S3 1 1 1 " 1 1 1 χ 2− 0 1 − χ2 4 0 1 χ3 8 0 1 − from which we can easily see that χ2 = + " + χ and that χ3 χ2 2χ = . Since ; "; χ is 2 − − f g a basis for CS3 , then so is ; χ, χ by the first relation. Arguing as in the previous part, the C f g 3 2 map ' : [x] CS3 , f(x) f(χ) is surjective and x x 2x ker(') by the second relation. Again since ! 7! − − 2 3 2 dimC(C[x]=(x x 2x)) = 3 = dimC(CS ) = dimC(C[x]= ker(')); − − 3 we have ker(') = (x3 x2 2x) and − − 3 2 C[x]=(x x 2x) = CS : − − ∼ 3 Remark. Note that in Part (ii) and (iii) of this problem, all we need to arrive at the conclusion is the fact that the powers of λ and χ can be used to replace the irreducible characters as a basis for CG. By Problem 11, every faithful character of G would have this property. So given such a character, we can follow the same procedure above to get the structure of CG explicitly as a quotient of the polynomial algebra C[x]. 3. Let V be a representation of G. (i) Compute dim Sn V and dim n V for all n. V n (ii) Let g G. Suppose g has eigenvalues λ1; : : : ; λd on V . What are the eigenvalues of g on S V and 2n V ? V (iii) Let f(x) = det(g xI) be the characteristic polynomial of g on V . Describe how to read tr(g; n V ) from the− coefficients of f(x). V (iv) ✳ Find a relation between tr(g; Sn V ) and the polynomial f(x). (Hint: first do the case when dim V = 1). Solution. (i) Let d = dimC V . Then n n + d 1 n d dimC S V = − and dimC V = n V n 2 i n where we adopt the convention that = 0 when j > i (so dimC V = 0 when n > dimC V ). j V n n 2 Define the linear operators S, A : V ⊗ V ⊗ by ! 1 1 S = σ and A = sgn(σ)σ: n! X n! X σ Sn σ Sn 2 2 Note that σS = Sσ = S and σA = Aσ = sgn(σ)A: n n n So for any w V ⊗ , σ(S(w)) = S(w) and σ(A(w)) = sgn(σ)A(w), which implies S(V ⊗ ) S V n 2 n n ⊆ and A(V ⊗ ) V . Conversely, if w S V , then σ(w) = w and so ⊆ V 2 1 1 S(w) = σ(w) = (w) = w n! X n! X σ Sn σ Sn 2 2 n n implying that w S(V ⊗ ); likewise, if w V , then σ(w) = sgn(σ)w and so 2 2 V 1 1 1 A(w) = sgn(σ)σ(w) = sgn(σ)2w = w = w n! X n! X n! X σ Sn σ Sn σ Sn 2 2 2 n n n n n implying that w A(V ⊗ ). Hence we have shown that S V = S(V ⊗ ) and V = A(V ⊗ ). 2 V Let e1; : : : ; ed be a basis of V . We adopt the following standard shorthand: f g ei ei := S(ei ei ) and ei ei := A(ei ei ): 1 · ··· · n 1 ⊗ · · · ⊗ n 1 ^ · · · ^ n 1 ⊗ · · · ⊗ n Since Sσ = S, the term ei1 ein depends only on the number of times each ei enters this product and we can write · ··· · k1 kd ei ei = e e 1 · ··· · n 1 · ··· · d where ki is the multiplicity of occurrence of ei in ei ei . Since Aσ = sgn(σ)A, the 1 · ··· · n term ei ei changes sign under the permutation of any two factors ei ei and so 1 ^ · · · ^ n α $ β ei ei = 0 if ei = ei . In particular, ei ei = 0 whenever n > d. 1 ^ · · · ^ n α β 1 ^ · · · ^ n The above discussion shows that k1 kd B1 = ei ei 1 i1 in d = e e k1 + + kd = n f 1 · ··· · n j ≤ ≤ · · · ≤ ≤ g f 1 · ··· · d j ··· g spans Sn V and B2 = ei ei 1 i1 < < in d : f 1 ^ · · · ^ n j ≤ ··· ≤ g spans n V . V k1 kd Vectors in B1 are linearly independent: if (k1; : : : ; kd) = (l1; : : : ; ld), then the tensors e1 ed l1 ld 6 ····· n and e1 ed are linear combinations of two non-intersecting subsets of basis elements of V ⊗ . ····· n Likewise, vectors in B2 are linearly independent. Hence B1 and B2 are bases for S V and n V respectively. V n+d 1 The cardinality of B1 is n− = number of partitions of d into a sum of n non-negative d integers.