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 ∈ 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) = ( ∈ O O 0 if x / . ∈ 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) = |O | | | (2.1) h O O i G X O O 0 if i = j x G | | ∈ 6 since every x G lies in a unique k for some k 1, . . . , n . Note that i / G = 0 and so (2.1) ∈ O ∈ { } |O | | | 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 ∈ ∈ such that f(x) = ci for every x i and we have ∈ O

f = c1δ 1 + + cnδ n . O ··· O C Hence δ 1 , . . . , δ n is a basis for CG. So as vector spaces over , { O O } ★ 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 ∈ (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 ∈ 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 ϕ : C[x] CZ/nZ, f(x) f(λ). ϕ is clearly a homomorphism of algebras. n 1 → 7→ n 1 Since 1, x, . . . , x − would be mapped to , λ, . . . , λ − , so rank(ϕ) = dimC(CZ/nZ) and ϕ is C n n surjective. We have [x]/ ker(ϕ) ∼= im(ϕ) = CZ/nZ. Since λ = , we have that x 1 ker(ϕ) and so the (xn 1) ker(ϕ). But − ∈ − ⊆ n dimC(C[x]/(x 1)) = n = dimC(CZ Z) = dimC(C[x]/ ker(ϕ)) − /n and so we must have equality (xn 1) = ker(ϕ). 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: , ε(σ) = sgn(σ) and χ(σ) = 1, 2, 3 σ 1. χ is the character of the two-dimensional representation ρ. The character table together|{ } with| − the 2 3 characters corresponding to ρ⊗ and ρ⊗ is: σS3 1 (1 2)S3 (1 2 3)S3 1 1 1 ε 1 1 1 χ 2− 0 1 − χ2 4 0 1 χ3 8 0 1 − from which we can easily see that χ2 = + ε + χ and that χ3 χ2 2χ = . Since , ε, χ is 2 − − { } a basis for CS3 , then so is , χ, χ by the ﬁrst relation. Arguing as in the previous part, the C { } 3 2 map ϕ : [x] CS3 , f(x) f(χ) is surjective and x x 2x ker(ϕ) by the second relation. Again since → 7→ − − ∈

3 2 dimC(C[x]/(x x 2x)) = 3 = dimC(CS ) = dimC(C[x]/ ker(ϕ)), − − 3 we have ker(ϕ) = (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 ∈n 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 ∈ ∈ 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 ∈ n n ⊆ and A(V ⊗ ) V . Conversely, if w S V , then σ(w) = w and so ⊆ V ∈ 1 1 S(w) = σ(w) = (w) = w n! X n! X σ Sn σ Sn ∈ ∈ n n implying that w S(V ⊗ ); likewise, if w V , then σ(w) = sgn(σ)w and so ∈ ∈ 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 ∈ ∈ ∈ n n n n n implying that w A(V ⊗ ). Hence we have shown that S V = S(V ⊗ ) and V = A(V ⊗ ). ∈ V Let e1, . . . , ed be a basis of V . We adopt the following standard shorthand: { }

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 { 1 · ··· · n | ≤ ≤ · · · ≤ ≤ } { 1 · ··· · d | ··· } spans Sn V and B2 = ei ei 1 i1 < < in d . { 1 ∧ · · · ∧ n | ≤ ··· ≤ } 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. The cardinality of B2 is clearly . n n (ii) Recall that if G acts linearly on a vector space V , then the natural action on V ⊗ is given by

g(ei ei ei ) = gei gei gei . 1 ⊗ 2 ⊗ · · · ⊗ n 1 ⊗ 2 ⊗ · · · ⊗ n This action restricts to the G-invariant subspaces Sn V and n V as follows: V 1 g(ei ei ) = ge ge , 1 · ··· · n n! X σ(i1) ⊗ · · · ⊗ σ(in) σ Sn ∈ 1 g(ei ei ) = sgn(σ)ge ge . 1 ∧ · · · ∧ n n! X σ(i1) ⊗ · · · ⊗ σ(in) σ Sn ∈ 2The 1/n! term is optional. It is there so that S and A are projections onto Sn V and n V respectively; also, S may 1 V then be interpreted as the ‘average’ of Sn, ie. G g G g. | | P ∈

3 When gei = λiei the expressions simplify as follows: 1 g(ei ei ) = λ λ e e 1 · ··· · n n! X σ(i1) ··· σ(in) σ(i1) ⊗ · · · ⊗ σ(in) σ Sn ∈ = λi λi ei ei , 1 ··· n 1 · ··· · n 1 g(ei ei ) = sgn(σ)λ λ e e 1 ∧ · · · ∧ n n! X σ(i1) ··· σ(in) σ(i1) ⊗ · · · ⊗ σ(in) σ Sn ∈ = λi λi ei ei 1 ··· n 1 ∧ · · · ∧ n n since λ λ = λi λi for all σ Sn. Hence, the set of eigenvalues of g on S V is σ(i1) ··· σ(in) 1 ··· n ∈

k1 kd λi λi 1 i1 in d = λ λ k1 + + kd = n { 1 ··· n | ≤ ≤ · · · ≤ ≤ } { 1 ··· d | ··· } and the set of eigenvalues of g on n V is V

λi λi 1 i1 < < in d . { 1 ··· n | ≤ ··· ≤ } (iii) Let g G. By (ii), ∈ n tr(g, V ) = λi λi V X 1 ··· n 1 i1<

d f(x) = (λ x) (λ x) = ( x)d + tr(g, n V )( x)d n. 1 d X − − ··· − − n=1 V − So n d n d n tr(g, V ) = ( 1) − coeﬃcient of x − in f(x) V − × for n = 1, . . . , d. (iv) Let g G. By (ii), ∈ tr(g, Sn V ) = λk1 λkd X 1 ··· d k1+ +kd=n ··· and we have

∞ (1 + λ x + λ2x2 + ) (1 + λ x + λ2x2 + ) = 1 + tr(g, Sn V )xn. 1 1 d d X ··· ··· ··· n=1

1 2 2 Noting that (1 λix)− = 1 + λix + λ x + , we have − i ··· ∞ 1 1 + tr(g, Sn V )xn = X (1 λ1x) (1 λdx) n=1 − ··· − 1 = d ( x) (λ1 1/x) (λd 1/x) − − ··· − 1 = . ( x)df(1/x) − So 1 tr(g, Sn V ) = ( 1)d coeﬃcient of xd+n in power series expansion of − × f(1/x) for n = 1, 2,... .

Remark. Note that the two boxed formulas would have to be changed accordingly if you deﬁned the characteristic polynomial as f(x) = det(xI g). − 4. Let G act on a ﬁnite set X, and C[X] be the permutation representation.

4 (i) Show S2 C[X] = C[S2 X], 2 2 where S X is the set of orbits of S2 on X X. (Hence S X = A X A = 1 or 2 .) × { ⊆ | | | } (ii) Compute the character of S2 C[X]. (iii) Show that the trivial representation occurs in S2 C[X] with multiplicity at least two. (Do not use Part (ii)!) (iv) Show Sn C[X] = C[Sn X]. (v) ✳ Can you ﬁnd a description of 2 C[X] analogous to the description of Part (i)? (You should V use a non-trivial S2-equivariant vector bundle on X X to replace the use of a permutation representation). ×

Solution. Let X = x1, x2, . . . , xd (note that we have implicitly ordered the elements in X by { } n giving them integer indices; this is important in our solution). Sn acts on X = X X n × · · · × by permuting coordinates. If (xj , xj , . . . , xj ) X , we can certainly ﬁnd a σ Sn such that 1 2 n ∈ ∈ σ(xj1 , xj2 , . . . , xjn ) = (xi1 , xi2 , . . . , xin ) where the resulting indices satisfy 1 i1 i2 in d. Hence the set of orbits Sn X may be represented as ≤ ≤ ≤ · · · ≤ ≤

n Sn S X = (xi , . . . , xi ) 1 i1 in d . { 1 n | ≤ ≤ · · · ≤ ≤ }

For notational simplicity, we write xi = i for i = 1, . . . , d. When n = 2, we write:

2 S2 C[X] = span e1, . . . , ed S X = (i, j) 1 i j d C{ } { | ≤ ≤ ≤ } 2 2 S C[X] = span ei ej 1 i j d C[S X] = span ei,j 1 i j d C{ · | ≤ ≤ ≤ } C{ | ≤ ≤ ≤ } (i) It is clear that S2 C[X] and C[S2 X] have the same dimension and are thus isomorphic as vector spaces. What we need to show is that they are isomorphic as G-modules. This follows from the G-equivariance of the linear map

ϕ : S2 C[X] C[S2 X] → ei ej ei,j · 7→

ie. ϕ(gei ej) = ϕ(egi egj) = egi,gj = gei,j = gϕ(ei ej). · · · 2 g (ii) χ 2 C (g) = χC 2 (g) = (S X) . S [X] [S X] | | (iii) This is obvious since S2 C[X] = C[S2 X]. Note that for any G-set S, the permutation representation C[S] always contains the trivial representation (so in this case, we just take S = S2 X). (iv) This follows from the G-equivariance of the map

ϕ : Sn C[X] C[Sn X] → ei ei ei ei ,i ,...,i 1 · 2 · ··· · n 7→ 1 2 n as above.

5. For each irreducible representation ρ of A , determine the character of IndS4 ρ, and decompose this 4 A4 into irreducible representations. For each irreducible representation ρ of S , decompose ResS4 ρ into 4 A4 irreducible representations. Check your answer is compatible with Frobeniuse reciprocity. Now doe this for S3 S4. ⊆ Solution. Straightforward but tedious.

6. Determine the character table of the dihedral group D2n of symmetries of the n-gon by inducing up representations of the rotation group Z/nZ , D2n. → Solution. Straightforward.

5 7. Let G be the Heisenberg group of order p3. This is the subgroup

1 a x F G = n 0 1 b a, b, x po 0 0 1 ∈

of matrices over the ﬁnite ﬁeld Fp, p a prime. Let H be the subgroup of G of matrices with a = 0 and Z be the subgroup of G of matrices with a = b = 0.

2 (i) Show Z is the center of G, and that G/Z = Fp. Note that this implies that the commutator subgroup [G, G] is contained in Z. You can check by explicit computation that it equals Z, or if you are lazy you can deduce this from the list of irreducible representations, below. (ii) Find all one dimensional representations of G.

(iii) Let ψ : Fp = Z/pZ C× be a non-trivial one-dimensional representation of Fp, and deﬁne a → one-dimensional representation ρψ of H by

1 0 x ρψ 0 1 b = ψ(x). 0 0 1

G Show that IndH ρψ is an irreducible representation of G. (iv) Prove that the collection of representations constructed in Part (ii) and (iii) complete the list of all irreducible representations. G (v) Determine the character of the representations IndH ρψ. Solution. Note that

1 a1 x1 1 a2 x2 1 a1+a2 x1+x2+a1b2 g1g2 = 0 1 b1 0 1 b2 = 0 1 b1+b2 . 0 0 1 0 0 1 0 0 1

(i) Let g1 Z. Then g1g2 = g2g1 for all g2 G iﬀ b1a2 = a1b2 for all a2, b2 Fp iﬀ a1 = b1 = 0. So Z is∈ the center of G. ∈ ∈ 1 a 0 F F2 G/Z = nh 0 1 b i a, b po ∼= p. 0 0 1 ∈

Since G/Z is abelian, G0 = [G, G] must be contained in Z. Now [G : G0] = [G : Z][Z : G0] and 2 2 3 [G : Z] = p . So [G : G0] = p ([G : G0] = p is not possible since G is non-abelian) and thus [Z : G0] = 1, ie. Z = G0. (ii) By Example Sheet 2, Problem 7, every one-dimensional representation of G is lifted from a 2 2 representation of G/G0 = Fp. By Example Sheet 1, Problem 9, the representations of Fp are precisely 2(la+mb)πi/p χl,m : Fp Fp C×, (a, b) e × → 7→ for l, m = 0, . . . , p 1, (ie. we let λ = e2lπi/p and µ = e2mπi/p in Example Sheet 1, Problem 9). So the lifted one-dimensional− representations of G are

1 0 x 2(la+mb)πi/p χl,m : G C×, 0 1 b e → 0 0 1 7→ e for l, m = 0, . . . , p 1. − (iii) Let 1 1 0 1 0 0 1 0 1 α = 0 1 0 , β = 0 1 1 , ζ = 0 1 0 . 0 0 1 0 0 1 0 0 1 Then 1 a x a b x 0 1 b = α β ζ . 0 0 1 In particular Z = ζx x = 0, . . . , p 1 and H = βbζx b, x = 0, . . . , p 1 . { | − } { | − } 2xnπi/p Recall that ψ has the form ψn : Fp C×, x e for n = 0, . . . , p 1. We will now G → 7→ − compute the character of IndH ρψ. Case: a = b = 0. Let x 0, . . . , p 1 . Then ζx Z. So ∈ { − } ∈ x x x χ G (ζ ) = pχρ (ζ ) = pψ(ζ ). IndH ρψ ψ

6 Case: a = 0, b = 0. Let b 1, . . . , p 1 . Suppose βb is conjugate to some g G, then βbZ 6 ∈ { − } ∈ is conjugate to gZ in the abelian group G/Z, so βbZ = gZ, and therefore g = βbζx for some x 0, . . . , p 1 . Since βb / Z, the conjugacy class (βb)G does not have size 1, and hence ∈ { − } ∈ (βb)G = βbζx x = 0, . . . , p 1 . { | − } By the formula for character of induced representations, we have

b x b b b p 1 G χInd ρ (β ζ ) = χρψ (β ) + χρψ (β ζ) + + χρψ (β ζ − ) H ψ ··· b b b p 1 = ψ(β ) + ψ(β ζ) + + ψ(β ζ − ) ··· p 1 − = ψ(βb) (ψ(ζ))i = 0 X i=0

since ψ(ζ) is a pth root of unity. Case: a = 0. In this case, αaβbζx / H and (αaβbζx)G H = ∅, so 6 ∈ ∩ a b x χ G (α β ζ ) = 0 IndH ρψ by the deﬁnition of induced characters.

Summarizing, if ψ = ψn, then

2xnπi/p a b x pe if a = b = 0, χIndG ρ (α β ζ ) = ( H ψ 0 otherwise,

G for n = 0, . . . , p 1. Ind ρψ is irreducible since − H

1 2 1 2 1 2 χIndG ρ , χIndG ρ = χIndG ρ (g) = χIndG ρ (g) = p = 1. h H ψ H ψ i G X| H ψ | p3 X| H ψ | p3 X g G g Z g Z | | ∈ ∈ ∈

2 (iv) The p representations of dimension 1 in (ii), χl,m l, m = 0, . . . , p 1 , and the p 1 G { | − } − representations of dimension p in (iii), IndH ρψn n = 1, . . . , p 1 , are all distinct, and the sum of the squares of their degrees is { | − }

p2 12 + (p 1) p2 = p3 = G × − × | | and so these complete the list of all irreducible representations. (v) See (iii).

8. Suppose that V is a representation of G, and the character χ of V satisfies χ(g) R for all g G. ∈ 2 ∈ show that the trivial representation always occurs in V V . Conclude that if dim V > 2, V ⊗ is the sum of at least three irreducible representations. ⊗ Solution. Since χ(g)2 0, ≥ 1 , χ2 = χ(g)2 0; h i G X ≥ g G | | ∈ it cannot be 0 as χ(1) = 0 and so the multiplicity of in χ2 must be non-zero. 6 2 2 If dimC V 3, then dimC S V 6 and dimC V 3. We know that the trivial representation must ≥ 2 ≥ ≥ occur in either S2 V or V (or both). SinceV the trivial representation has dimension 1, either S2 V or 2 V (or both) mustV decompose further into two or more irreducible representations. In all cases, V we would have at least three irreducible representations. 9. Let (ρ, V ) be an irreducible representation of a finite group G. How unique is the positive definite hermitian form on V ?

7 10. Show that the number of irreducible characters of G taking only real values equals the number of 1 conjugacy classes of elements g such that g− is conjugate to g. Solution. The character table of G may be viewed as an n n matrix A GL(n, C) where n is the number of conjugacy classes of G (A is invertible by row or column× orthogonality).∈ Observe that χ is an irreducible character if and only if χ is. So A = χ(g) is simply a permutation of the rows of A = χ(g) . If we let P denote the permutation matrix, then PA = A. (10.1)

Since χ = χ precisely when χ(g) R for all g G, the number of R-valued characters equals the number of rows in A that are ﬁxed∈ under A A∈which in turn equals the trace of P . 7→ Now observe that χ(g 1) = χ(g), so we may also view A = χ(g 1) as a permutation of the columns − − of A = χ(g) . Let Q denote the permutation matrix, then AQ = A. (10.2)

G 1 G G Since g = (g− ) precisely when the column corresponding to g equals that corresponding to 1 G (g− ) (distinct conjugacy classes must assume diﬀerent values on at least one irreducible character G 1 G by column orthogonality), the number of conjugacy classes satisfying g = (g− ) equals the number of columns that are ﬁxed under A A which in turn equals the trace of Q. 7→ By (10.1) and (10.2), AP = QA, and the required result follows from

1 tr(P ) = tr(AP A− ) = tr(Q).

11. ✳ Let ρ be a faithful irreducible representation of G, with character χ, and suppose that χ takes t distinct values. Show that every irreducible representation of G occurs with non-zero multiplicity in at least one of 2 (t 1) , ρ, ρ⊗ , . . . , ρ⊗ − . Hint: Let µ be an irreducible character, and suppose that (χj, µ) = 0 for j = 0, . . . , t 1. Consider this an equation − (t t-matrix of character of powers of ρ)(vector of µ) = 0 × and invert the matrix to get a contradiction. 1 Solution. Let α0, . . . , αt 1 C be the t distinct values in im(χ) and let Gi := χ− (αj) = x G − ∈ { ∈ | χ(x) = αi for i = 0, . . . , t 1. We may assume α0 = χ(1). Since χ is faithful, G0 = ker(χ) = 1 . } − { } If µ is an irreducible character satisfying (χj, µ) = 0 for j = 0, . . . , t 1, then − t 1 − j j αi µ(x) = G (χ , µ) = 0 Xi=0 Xx Gi ∈ | | · for j = 0, . . . , t 1. Let β = µ(x) (note that β = µ(1)). Rewriting the system of linear i x Gi 0 equations in matrix− form, we getP ∈

1 1 1 β0 0  ···      α0 α1 αt 1 β1 0 ··· − = .  . . .. .   .  .  . . . .   .  .  t 1 t 1 t 1     α0− α1− αt−1 βt 1 0 ··· − −

Since the αi’s are all distinct and the Vandermonde matrix is invertible, we have βi = 0 for all i = 0, . . . , t 1. In particular, 0 = β0 = µ(1) = 0 gives the required contradiction. − 6 Remark. The rows v0,..., vt 1 of the Vandermonde matrix are linearly independent since a linear combination − a0v0 + + at 1vt 1 = 0 ··· − − t 1 gives a polynomial a0 + a1X + + at 1X − that has t distinct zeroes 1, α0, . . . , αt 1 and so must ··· − − be the zero polynomial, ie. a0 = = at 1 = 0. ··· −

8 12. ✳ Prove the following theorem of Burnside. If ρ is an irreducible representation of G with character χ and dim ρ > 1 then there exists a g G such that χ(g) = 0. ∈ Solution. (Sketch) Let ψ be a character (not necessarily irreducible) of the cyclic group Cn ∼= Z Z /n . Let X = g Cn g = Cn . We could show that g X ψ(g) is an integer. Hence, if { ∈ 2| h i } 2 Q ∈ ψ(g) = 0, then g X ψ(g) 1 and so g X ψ(g) X (by the fact that Arithmetic Mean Geometric6 Mean).Q ∈ | | ≥ P ∈ | | ≥ | | ≥

Now define an equivalence relation on the elements of G by g h iff g = h . Let Xi = g ∼ h i h i { ∈ G g = gi , i = 1, . . . , m, be the equivalence classes. If χ(g) = χρ(g) = 0 for all g G, then | h i h i} 6 ∈ χ G (g) = 0 for all g G and we may apply the above result to Xi and ψ = χ G to get Res g ρ Res g ρ h ii 6 ∈ h ii 2 ψ(g) Xi . X | | ≥ | | g Xi ∈ Let g = 1. Then X = 1 . So ψ(g) 2 = ψ(1) 2 > 1 = X since dim ρ > 1. 1 1 g X1 1 { } P ∈ | | | | | | From ψ, ψ = 1, we get h i m m 2 2 G = ψ(g) = ψ(g) > Xi = G , | | X| | X X | | X| | | | g G i=1 g Xi i=1 ∈ ∈ which gives a contradiction. Hence χ(g) = 0 for some g G. ∈ 13. Show that an abelian group has a faithful irreducible representation if and only if it is cyclic. Solution. We assume that the group G is finite and the representation ρ is over C. Recall that complex irreducible representations of abelian groups are one-dimensional (Example Sheet 1, Problem 9) and that each ρ(x) C× is an Nth root of unity where N = G (Example Sheet 1, Problem 7). So ρ(x) ∂D = z C∈ z = 1 for all x G. | | ∈ { ∈ | | | } ∈ C D 1 If G is abelian and ρ : G × is faithful, then G ∼= ρ(G). But ρ(G) ∂ ∼= S and since the only finite subgroups of S1 are→ either dihedral or cyclic, we must have G is≤ cyclic as it is abelian. n 2nπi/N If G = x is a cyclic group, then ρ : G C×, x e is clearly a faithful representation. h i → 7→ 14. Let Γr = A 1, . . . , n A = r , and χr be the character of the permutation representation C[Γr] { ⊆ { } | | | } of Sn. Compute (χr, χs) and hence decompose C[Γr] as a representation of Sn. Solution. (Impetus for this solution provided by Dominic Pinto, Caius College) By Problem 9 in Example Sheet 2, (χr, χs) = #Sn-orbits on Γr Γs. × Let ∆ := (x, x) x X . Recall that Sn acts transitively on the Sn-stable subsets ∆ and X X ∆. { | ∈ } × \ It follows that the Sn-orbits on Γr Γs are × i = A B Γr Γs A B = i . (14.1) O { × ∈ × | | ∩ | } for i = max(0, s + r n),..., min(s, r) (a moment’s thought would show that these are the only possible values i could− take). Hence

(χr, χs) = min(s, r) max(0, s + r n) + 1. (14.2) − − Note that (14.1) also follows from a direct argument: if A B, C D Γr Γs and A B = i = C D , let × × ∈ × | ∩ | | ∩ |

A B = x1, . . . , xi ,A B = xi+1, . . . , xr ,B A = xi0+1, . . . , xs0 ,X (A B) = xr00+s i+1, . . . , xn00 , ∩ { } \ { } \ { } \ ∪ { − } C D = y1, . . . , yi ,C D = yi+1, . . . , yr ,D C = yi0+1, . . . , ys0 ,X (C D) = yr00+s i+1, . . . , yn00 . ∩ { } \ { } \ { } \ ∪ { − } Then the permutation in Sn deﬁned by

x1 xi xi+1 xr xi0+1 xs0 xr00+s i+1 xn00  ··· ··· ··· − ···  ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ y1 yi yi+1 yr yi0+1 ys0 yr00+s i+1 yn00 ··· ··· ··· − ··· would send A B to C D, ie. they lie in the same Sn-orbit. Since Sn permutes elements in ∆, × × A B and C D with A B = C D cannot lie in the same Sn-orbit. × × | ∩ | 6 | ∩ |

To decompose C[Γr] into irreducibles, observe that

9 ➀ Γ1 = 1, 2, . . . n , so C[Γ1] is just the usual permutation representation of Sn which we know breaks{ up into exactly} two representations: the trivial representation and an (n 1)-dimensional − irreducible representation V1 (see your lecture notes).

➁ Γn is a singleton, so C[Γn] is the trivial representation. Hence χn = .

➂ C[Γr] and C[Γn r] are equivalent representations of Sn. The obvious G-isomorphism being − C[Γr] C[Γn r], A X A for A Γr. G-equivariance follows from σ(X A) = X σA for → − 7→ \ ∈ \ \ σ Sn. Hence χr = χn r. ∈ − By ➁, (14.2) with s = n gives (χr, ) = (χr, χn) = 1. By ➂, we may assume that s r n/2, so s + r n and (14.2) gives ≤ ≤ ≤

(χr, χs) = s + 1.

For r = 2, (χ2, ) = 1 and (χ2, χ1) = 2 imply and V1 both occur with multiplicity 1 in C[Γ2]. Since n n (χ2, χ2) = 3, there is a third irreducible representation V2 of dimension deg(χ2) deg(χ1) = . − 2−1 In general, since (χr, χr 1) = r and (χr, χr) = r + 1, C[Γr] contains an irreducible representation Vr − n n C of degree deg(χr) deg(χr 1) = r r 1 that is not in [Γr 1]. Inductively we get − − − − −

C[Γr] = V1 Vr. ⊕ ⊕ · · · ⊕

m n 15. Using Problems 14 and 4, decompose S C as representation of Sn for all m 1. ≥ n Solution. Let X = 1, 2, . . . , n . Note that C[X] = C as Sn-modules. { } ∼ Sm Cn = Sm C[X] = C[Sm X]

= C[Γm] C[Γm 1] C[Γ1] ⊕ − ⊕ · · · ⊕ = ( V1 Vm) ( V1 Vm 1) ( V1) ⊕ ⊕ · · · ⊕ ⊕ ⊕ ⊕ · · · ⊕ − ⊕ · · · ⊕ ⊕ = m (m 1)V1 Vn ⊕ − ⊕ · · · ⊕ 1 2 16. Let V be an irreducible representation of G, with character χ. Show that G − g G χ(g ) is either 0, 1 or 1, and that this zero precisely when χ is not real valued, ie. when| | V Pis not∈ isomorphic to − 2 2 V ∗. (Hint: consider V V = S V V ). ⊗ ⊕ V Solution. Since χ is irreducible, it occurs with multiplicity one in itself and if it is not real valued, the χ is a diﬀerent irreducible character, ie.

1 if χ is real valued, χ, χ = ( h i 0 if χ is not real valued.

Since 1 2 2 χ, χ = χ (g) = χ , = χS2 V + χ 2 V , = χS2 V , + χ 2 V , , h i G X h i h ∧ i h i h ∧ i g G | | ∈ we have the following possibilities ➀ χS2 V , = 1 and χ 2 V , = 0 when χ is real valued, h i h ∧ i ➁ χS2 V , = 0 and χ 2 V , = 0 when χ is not real valued, h i h ∧ i ➂ χS2 V , = 0 and χ 2 V , = 1 when χ is real valued. h i h ∧ i Hence this gives 1 if ➀ holds, 1 2  χ(g ) = χS2 V , χ 2 V , = 0 if ➁ holds, G X h i − h ∧ i  g G ➂ | | ∈ 1 if holds. − Note that ➁ happens precisely when χ is not real valued.

10 C 17. Let G GLn( ), G a ﬁnite group. Show that g G tr(g) = 0 implies g G g = 0. ≤ P ∈ P ∈ Solution. We may regard G, GLn(C), g g as a representation of G. Recall that every representation of a ﬁnite group is→ equivalent to a7→ unitary representation (the proof is similar to that in Example Sheet 1, Problem 8(i)). In this case, it means that we may assume that G U(n), ie. T 1 C ≤ g∗ := g = g− for all g G. Let A = g G g Mn( ). Observe that ∈ P ∈ ∈ ➀ A is self-adjoint (a.k.a. hermitian) since

A = g = g 1 = g = A. ∗ X ∗ X − X g G g G g G ∈ ∈ ∈ ➁ For all x G, ∈ xA = xg = g = A. X X g G g G ∈ ∈ 2 Suppose A = 0. Then by ➀, A = AA∗ is positive deﬁnite and so has eigenvalues λ1, . . . , λn (0, ). Since 6 ∈ ∞ ➁ A2 = gA = gA = A = G A, X X X | | g G g G g G ∈ ∈ ∈ 2 2 we get tr(A ) = G tr(A) and tr(A ) = λ1 + + λn > 0 would then imply that tr(A) > 0. | | ··· Remark. Given any G GLn(C), G, GLn(C) is often known as the natural representation of G. ≤ →

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