Solutions to Example Sheet 3

Transcription

1 Solutios to Example Sheet 3 1. Let G = A 5. For each pair of irreducible represetatios ρ, ρ, (i decompose ρ ρ ito a sum of irreducible represetatios, (ii decompose 2 ρ ad S 2 ρ ito irreducible represetatios, (iii decompose ρ ito irreducible represetatios, for 3, (iv check that if ρ is a o-trivial represetatio of A 5 (hece faithful, every represetatio of A 5 occurs with o-zero multiplicity i at least oe of, ρ, ρ 2,.... Solutio. Straightforward but tedious. Use Problem 3(iii for ρ. The last part follows from Problem (i Let C G be the space of class fuctios o G, ad δ O be the basis of delta fuctios o cojugacy classes. show that the fuctios δ O are orthogoal idempotets, ad hece C G is isomorphic as a algebra to a direct sum of copies of the oe-dimesioal algebra C. (ii Let G = Z/Z, ad λ be the oe-dimesioal represetatio of G which seds 1 G to multiplicatio by e 2πi/. show C G = C[λ]/(λ 1. Check this is cosistet with part (i. (iii Let G = S 3, ad ρ be the two-dimesioal represetatio of G. Show C G = C[ρ]/(ρ 3 ρ 2 2ρ. Solutio. (i Let O 1,..., O be the cojugacy classes of G. We defie δ O : G C by { 1 if x O, δ O (x = 0 if x / O. Clearly δo 2 = δ O ad so δ O is a idempotet elemet i the rig C G (with respect to additio ad multiplicatio of class fuctios. We have { δ Oi, δ Oj = 1 O i / G if i = j, δ Oi (xδ Oj (x = (2.1 G 0 if i j x G sice every x G lies i a uique O k for some k {1,..., }. Note that O i / G 0 ad so (2.1 shows that δ O1,..., δ O are orthogoal to each other ad i particular, liearly idepedet over C. Give ay f C G, f costat o cojugacy classes meas that there exists c 1,..., c C such that f(x = c i for every x O i ad we have f = c 1 δ O1 + + c δ O. Hece {δ O1,..., δ O } is a basis for C G. So as vector spaces over C, C G = Cδ O1 Cδ O = C C = C }{{}. copies Recall that C may be made ito a algebra by defiig product of (a 1,..., a, (b 1,..., b C by (a 1,..., a (b 1,..., b = (a 1 b 1,..., a b. (2.2 Sice δ O1,..., δ O are idempotet ad furthermore 1 δ Oi δ Oj = { δ Oi if i = j, 0 if i j. So if f = a i δ Oi, g = b j δ Oj C G, we have ( ( fg = a i δ Oi b j δ Oj = i=1 j=1 i,j a i b j δ Oi δ Oj = a i b i δ Oi. (2.3 i=1 Comparig (2.2 ad (2.3, we see that is ideed a isomorphism of algebras. 1 O secod thoughts, this might be what the lecturer meat by orthogoal idempotets. So maybe (2.1 is superfluous. But sice it took me a while to TEX it, I ll leave it. 1

2 (ii λ : Z/Z C, 1 e 2πi/ is a irreducible character. Sice e 2πi/ is a primitive th root of uity,, λ,..., λ 1 forms a complete set of irreducible characters of Z/Z (recall that Ẑ/Z = λ from the solutio of Example Sheet 2, Problem 7(b ad hece a basis of C Z/Z. Now cosider the map ϕ : C[x] C Z/Z, f(x f(λ. ϕ is clearly a homomorphism of algebras. Sice 1, x,..., x 1 would be mapped to, λ,..., λ 1, so rak(ϕ = dim C (C Z/Z ad ϕ is surjective. We have C[x]/ ker(ϕ = im(ϕ = C Z/Z. Sice λ =, we have that x 1 ker(ϕ ad so the (x 1 ker(ϕ. But dim C (C[x]/(x 1 = = dim C (C Z/Z = dim C (C[x]/ ker(ϕ ad so we must have equality (x 1 = ker(ϕ. Therefore C[x]/(x 1 = C Z/Z. (iii It is easy to costruct the character table of S 3 it has three cojugacy classes ad its three irreducible characters are just the obvious oes:, ε(σ = sg(σ ad χ(σ = {1, 2, 3} σ 1. χ is the character of the two-dimesioal represetatio ρ. The character table together with the characters correspodig to ρ 2 ad ρ 3 is: σ S3 1 (1 2 S3 (1 2 3 S ε χ χ χ from which we ca easily see that χ 2 = + ε + χ ad that χ 3 χ 2 2χ =. Sice {, ε, χ} is a basis for C S3, the so is {, χ, χ 2 } by the first relatio. Arguig as i the previous part, the map ϕ : C[x] C S3, f(x f(χ is surjective ad x 3 x 2 2x ker(ϕ by the secod relatio. Agai sice dim C (C[x]/(x 3 x 2 2x = 3 = dim C (C S3 = dim C (C[x]/ ker(ϕ, we have ker(ϕ = (x 3 x 2 2x ad C[x]/(x 3 x 2 2x = C S3. Remark. Note that i Part (ii ad (iii of this problem, all we eed to arrive at the coclusio is the fact that the powers of λ ad χ ca be used to replace the irreducible characters as a basis for C G. By Problem 11, every faithful character of G would have this property. So give such a character, we ca follow the same procedure above to get the structure of C G explicitly as a quotiet of the polyomial algebra C[x]. 3. Let V be a represetatio of G. (i Compute dim S V ad dim V for all. (ii Let g G. Suppose g has eigevalues λ 1,..., λ d o V. What are the eigevalues of g o S V ad V? (iii Let f(x = det(g xi be the characteristic polyomial of g o V. tr(g, V from the coefficiets of f(x. Describe how to read (iv Fid a relatio betwee tr(g, S V ad the polyomial f(x. (Hit: first do the case whe dim V = 1. Solutio. (i Let d = dim C V. The ( + d 1 dim C S V = ad dim C V = ( d 2

3 where we adopt the covetio that ( i j = 0 whe j > i (so dimc V = 0 whe > dim C V. Defie the liear operators S, A : V V by 2 Note that S = 1 σ ad A = 1 sg(σσ. σ S σ S σs = Sσ = S ad σa = Aσ = sg(σa. So for ay w V, σ(s(w = S(w ad σ(a(w = sg(σa(w, which implies S(V S V ad A(V V. Coversely, if w S V, the σ(w = w ad so S(w = 1 σ(w = 1 (w = w σ S σ S implyig that w S(V ; likewise, if w V, the σ(w = sg(σw ad so A(w = 1 sg(σσ(w = 1 sg(σ 2 w = 1 w = w σ S σ S σ S implyig that w A(V. Hece we have show that S V = S(V ad V = A(V. Let {e 1,..., e d } be a basis of V. We adopt the followig stadard shorthad: e i1 e i := S(e i1 e i ad e i1 e i := A(e i1 e i. Sice Sσ = S, the term e i1 e i depeds oly o the umber of times each e i eters this product ad we ca write e i1 e i = e k1 1 ek d d where k i is the multiplicity of occurrece of e i i e i1 e i. Sice Aσ = sg(σa, the term e i1 e i chages sig uder the permutatio of ay two factors e iα e iβ ad so e i1 e i = 0 if e iα = e iβ. I particular, e i1 e i = 0 wheever > d. The above discussio shows that B 1 = {e i1 e i spas S V ad 1 i 1 i d} = {e k1 1 ek d d k k d = } B 2 = {e i1 e i 1 i 1 < < i d}. spas V. Vectors i B 1 are liearly idepedet: if (k 1,..., k d (l 1,..., l d, the the tesors e k1 1 ek d d ad e l1 1 el d d are liear combiatios of two o-itersectig subsets of basis elemets of V. Likewise, vectors i B 2 are liearly idepedet. Hece B 1 ad B 2 are bases for S V ad V respectively. The cardiality of B 1 is ( +d 1 = umber of partitios of d ito a sum of o-egative itegers. The cardiality of B 2 is clearly ( d. (ii Recall that if G acts liearly o a vector space V, the the atural actio o V is give by g(e i1 e i2 e i = ge i1 ge i2 ge i. This actio restricts to the G-ivariat subspaces S V ad V as follows: g(e i1 e i = 1 ge σ(i1 ge σ(i, σ S g(e i1 e i = 1 sg(σge σ(i1 ge σ(i. σ S 2 The 1/ term is optioal. It is there so that S ad A are projectios oto S V ad V respectively; also, S may 1 the be iterpreted as the average of S, ie. G g G g. 3

4 Whe ge i = λ i e i the expressios simplify as follows: g(e i1 e i = 1 λ σ(i1 λ σ(ie σ(i1 e σ(i σ S = λ i1 λ i e i1 e i, g(e i1 e i = 1 sg(σλ σ(i1 λ σ(ie σ(i1 e σ(i σ S = λ i1 λ i e i1 e i sice λ σ(i1 λ σ(i = λ i1 λ i for all σ S. Hece, the set of eigevalues of g o S V is {λ i1 λ i 1 i 1 i d} = {λ k1 1 λk d d k k d = } ad the set of eigevalues of g o V is {λ i1 λ i 1 i 1 < < i d}. (iii Let g G. By (ii, ad we have tr(g, V = λ i1 λ i 1 i 1< <i d d f(x = (λ 1 x (λ d x = ( x d + tr(g, V ( x d. =1 So tr(g, V = ( 1 d coefficiet of x d i f(x for = 1,..., d. (iv Let g G. By (ii, ad we have tr(g, S V = k 1+ +k d = λ k1 1 λk d d (1 + λ 1 x + λ 2 1x 2 + (1 + λ d x + λ 2 dx 2 + = 1 + Notig that (1 λ i x 1 = 1 + λ i x + λ 2 i x2 +, we have 1 + tr(g, S V x 1 = (1 λ 1 x (1 λ d x =1 tr(g, S V x. =1 1 = ( x d (λ 1 1/x (λ d 1/x 1 = ( x d f(1/x. So tr(g, S V = ( 1 d coefficiet of x d+ i power series expasio of 1 f(1/x for = 1, 2,.... Remark. Note that the two boxed formulas would have to be chaged accordigly if you defied the characteristic polyomial as f(x = det(xi g. 4. Let G act o a fiite set X, ad C[X] be the permutatio represetatio. 4

5 (i Show S 2 C[X] = C[S 2 X], where S 2 X is the set of orbits of S 2 o X X. (Hece S 2 X = {A X A = 1 or 2}. (ii Compute the character of S 2 C[X]. (iii Show that the trivial represetatio occurs i S 2 C[X] with multiplicity at least two. (Do ot use Part (ii! (iv Show S C[X] = C[S X]. (v Ca you fid a descriptio of 2 C[X] aalogous to the descriptio of Part (i? (You should use a o-trivial S 2 -equivariat vector budle o X X to replace the use of a permutatio represetatio. Solutio. Let X = {x 1, x 2,..., x d } (ote that we have implicitly ordered the elemets i X by givig them iteger idices; this is importat i our solutio. S acts o X = X X by permutig coordiates. If (x j1, x j2,..., x j X, we ca certaily fid a σ S such that σ(x j1, x j2,..., x j = (x i1, x i2,..., x i where the resultig idices satisfy 1 i 1 i 2 i d. Hece the set of orbits S X may be represeted as S X = {(x i1,..., x i S 1 i 1 i d}. For otatioal simplicity, we write x i = i for i = 1,..., d. Whe = 2, we write: C[X] = spa C {e 1,..., e d } S 2 X = {(i, j S2 1 i j d} S 2 C[X] = spa C {e i e j 1 i j d} C[S 2 X] = spa C {e i,j 1 i j d} (i It is clear that S 2 C[X] ad C[S 2 X] have the same dimesio ad are thus isomorphic as vector spaces. What we eed to show is that they are isomorphic as G-modules. This follows from the G-equivariace of the liear map ϕ : S 2 C[X] C[S 2 X] e i e j e i,j ie. ϕ(ge i e j = ϕ(e gi e gj = e gi,gj = ge i,j = gϕ(e i e j. (ii χ S 2 C[X](g = χ C[S 2 X](g = (S 2 X g. (iii This is obvious sice S 2 C[X] = C[S 2 X]. Note that for ay G-set S, the permutatio represetatio C[S] always cotais the trivial represetatio (so i this case, we just take S = S 2 X. (iv This follows from the G-equivariace of the map as above. ϕ : S C[X] C[S X] e i1 e i2 e i e i1,i 2,...,i 5. For each irreducible represetatio ρ of A 4, determie the character of Id S4 A 4 ρ, ad decompose this ito irreducible represetatios. For each irreducible represetatio ρ of S 4, decompose Res S4 A 4 ρ ito irreducible represetatios. Check your aswer is compatible with Frobeius reciprocity. Now do this for S 3 S 4. Solutio. Straightforward but tedious. 6. Determie the character table of the dihedral group D 2 of symmetries of the -go by iducig up represetatios of the rotatio group Z/Z D 2. Solutio. Straightforward. 5

6 7. Let G be the Heiseberg group of order p 3. This is the subgroup } G = a, b, x Fp {( 1 a x 0 1 b of matrices over the fiite field F p, p a prime. Let H be the subgroup of G of matrices with a = 0 ad Z be the subgroup of G of matrices with a = b = 0. (i Show Z is the ceter of G, ad that G/Z = F 2 p. Note that this implies that the commutator subgroup [G, G] is cotaied i Z. You ca check by explicit computatio that it equals Z, or if you are lazy you ca deduce this from the list of irreducible represetatios, below. (ii Fid all oe dimesioal represetatios of G. (iii Let ψ : F p = Z/pZ C be a o-trivial oe-dimesioal represetatio of F p, ad defie a oe-dimesioal represetatio ρ ψ of H by (( 1 0 x ρ ψ 0 1 b = ψ(x. Show that Id G is a irreducible represetatio of G. (iv Prove that the collectio of represetatios costructed i Part (ii ad (iii complete the list of all irreducible represetatios. (v Determie the character of the represetatios Id G. Solutio. Note that g 1 g 2 = ( 1 a1 x b 1 ( 1 a2 x b 2 = ( 1 a1+a 2 x 1+x 2+a 1b b 1+b 2 (i Let g 1 Z. The g 1 g 2 = g 2 g 1 for all g 2 G iff b 1 a 2 = a 1 b 2 for all a 2, b 2 F p iff a 1 = b 1 = 0. So Z is the ceter of G. ] } G/Z = a, b Fp = F 2 p. {[ 1 a b Sice G/Z is abelia, G = [G, G] must be cotaied i Z. Now [G : G ] = [G : Z][Z : G ] ad [G : Z] = p 2. So [G : G ] = p 2 ([G : G ] = p 3 is ot possible sice G is o-abelia ad thus [Z : G ] = 1, ie. Z = G. (ii By Example Sheet 2, Problem 7, every oe-dimesioal represetatio of G is lifted from a represetatio of G/G = F 2 p. By Example Sheet 1, Problem 9, the represetatios of F 2 p are precisely χ l,m : F p F p C, (a, b e 2(la+mbπi/p (iii Let for l, m = 0,..., p 1, (ie. we let λ = e 2lπi/p ad µ = e 2mπi/p i Example Sheet 1, Problem 9. So the lifted oe-dimesioal represetatios of G are χ l,m : G C, e 2(la+mbπi/p for l, m = 0,..., p 1. The α = ( , β = ( 1 0 x 0 1 b ( , ζ = ( 1 a x 0 1 b = α a β b ζ x.. ( I particular Z = {ζ x x = 0,..., p 1} ad H = {β b ζ x b, x = 0,..., p 1}. Recall that ψ has the form ψ : F p C, x e 2xπi/p for = 0,..., p 1. We will ow compute the character of Id G. Case: a = b = 0. Let x {0,..., p 1}. The ζ x Z. So χ Id G (ζ x = pχ ρψ (ζ x = pψ(ζ x. 6

7 Case: a = 0, b 0. Let b {1,..., p 1}. Suppose β b is cojugate to some g G, the β b Z is cojugate to gz i the abelia group G/Z, so β b Z = gz, ad therefore g = β b ζ x for some x {0,..., p 1}. Sice β b / Z, the cojugacy class (β b G does ot have size 1, ad hece (β b G = {β b ζ x x = 0,..., p 1}. By the formula for character of iduced represetatios, we have sice ψ(ζ is a pth root of uity. χ Id G (β b ζ x = χ ρψ (β b + χ ρψ (β b ζ + + χ ρψ (β b ζ p 1 = ψ(β b + ψ(β b ζ + + ψ(β b ζ p 1 p 1 = ψ(β b (ψ(ζ i = 0 Case: a 0. I this case, α a β b ζ x / H ad (α a β b ζ x G H =, so by the defiitio of iduced characters. Summarizig, if ψ = ψ, the i=0 χ Id G (α a β b ζ x = χ Id G (α a β b ζ x = 0 for = 0,..., p 1. Id G is irreducible sice χ Id G, χ Id G = 1 G { pe 2xπi/p if a = b = 0, 0 otherwise, χ Id G (g 2 = 1 p 3 χ Id G (g 2 = 1 p 3 p 2 = 1. g G (iv The p 2 represetatios of dimesio 1 i (ii, {χ l,m l, m = 0,..., p 1}, ad the p 1 represetatios of dimesio p i (iii, {Id G = 1,..., p 1}, are all distict, ad the sum of the squares of their degrees is g Z p (p 1 p 2 = p 3 = G ad so these complete the list of all irreducible represetatios. (v See (iii. 8. Suppose that V is a represetatio of G, ad the character χ of V satisfies χ(g R for all g G. show that the trivial represetatio always occurs i V V. Coclude that if dim V > 2, V 2 is the sum of at least three irreducible represetatios. Solutio. Sice χ(g 2 0,, χ 2 = 1 G it caot be 0 as χ(1 0 ad so the multiplicity of χ(g 2 0; g G i χ 2 must be o-zero. If dim C V 3, the dim C S 2 2 V 6 ad dim C V 3. We kow that the trivial represetatio must occur i either S 2 V or 2 V (or both. Sice the trivial represetatio has dimesio 1, either S 2 V or 2 V (or both must decompose further ito two or more irreducible represetatios. I all cases, we would have at least three irreducible represetatios. 9. Let (ρ, V be a irreducible represetatio of a fiite group G. How uique is the positive defiite hermitia form o V? g Z 7

8 10. Show that the umber of irreducible characters of G takig oly real values equals the umber of cojugacy classes of elemets g such that g 1 is cojugate to g. Solutio. The character table of G may be viewed as a matrix A GL(, C where is the umber of cojugacy classes of G (A is ivertible by row or colum orthogoality. Observe that χ is a irreducible character if ad oly if χ is. So A = ( χ(g is simply a permutatio of the rows of A = ( χ(g. If we let P deote the permutatio matrix, the P A = A. (10.1 Sice χ = χ precisely whe χ(g R for all g G, the umber of R-valued characters equals the umber of rows i A that are fixed uder A A which i tur equals the trace of P. Now observe that χ(g 1 = χ(g, so we may also view A = ( χ(g 1 as a permutatio of the colums of A = ( χ(g. Let Q deote the permutatio matrix, the AQ = A. (10.2 Sice g G = (g 1 G precisely whe the colum correspodig to g G equals that correspodig to (g 1 G (distict cojugacy classes must assume differet values o at least oe irreducible character by colum orthogoality, the umber of cojugacy classes satisfyig g G = (g 1 G equals the umber of colums that are fixed uder A A which i tur equals the trace of Q. By (10.1 ad (10.2, AP = QA, ad the required result follows from tr(p = tr(ap A 1 = tr(q. 11. Let ρ be a faithful irreducible represetatio of G, with character χ, ad suppose that χ takes t distict values. Show that every irreducible represetatio of G occurs with o-zero multiplicity i at least oe of, ρ, ρ 2,..., ρ (t 1. Hit: Let µ be a irreducible character, ad suppose that (χ j, µ = 0 for j = 0,..., t 1. Cosider this a equatio (t t-matrix of character of powers of ρ(vector of µ = 0 ad ivert the matrix to get a cotradictio. Solutio. Let α 0,..., α t 1 C be the t distict values i im(χ ad let G i := χ 1 (α j = {x G χ(x = α i } for i = 0,..., t 1. We may assume α 0 = χ(1. Sice χ is faithful, G 0 = ker(χ = {1}. If µ is a irreducible character satisfyig (χ j, µ = 0 for j = 0,..., t 1, the t 1 i=0 αj i ( µ(x = G (χ j, µ = 0 x G i for j = 0,..., t 1. Let β i = x G i µ(x (ote that β 0 = µ(1. Rewritig the system of liear equatios i matrix form, we get β 0 0 α 0 α 1 α t 1 β = 0.. α0 t 1 α1 t 1 αt 1 t 1 β t 1 0 Sice the α i s are all distict ad the Vadermode matrix is ivertible, we have β i = 0 for all i = 0,..., t 1. I particular, 0 = β 0 = µ(1 0 gives the required cotradictio. Remark. The rows v 0,..., v t 1 of the Vadermode matrix are liearly idepedet sice a liear combiatio a 0 v a t 1 v t 1 = 0 gives a polyomial a 0 + a 1 X + + a t 1 X t 1 that has t distict zeroes 1, α 0,..., α t 1 ad so must be the zero polyomial, ie. a 0 = = a t 1 = 0. 8

9 12. Prove the followig theorem of Burside. If ρ is a irreducible represetatio of G with character χ ad dim ρ > 1 the there exists a g G such that χ(g = 0. Solutio. (Sketch Let ψ be a character (ot ecessarily irreducible of the cyclic group C = Z/Z. Let X = {g C g = C }. We could show that g X ψ(g is a iteger. Hece, if ψ(g 0, the g X ψ(g 2 1 ad so g X ψ(g 2 X (by the fact that Arithmetic Mea Geometric Mea. Now defie a equivalece relatio o the elemets of G by g h iff g = h. Let X i = {g G g = g i }, i = 1,..., m, be the equivalece classes. If χ(g = χ ρ (g 0 for all g G, the χ Res G g i ρ (g 0 for all g G ad we may apply the above result to X i ad ψ = χ Res G g i ρ to get ψ(g 2 X i. g X i Let g 1 = 1. The X 1 = {1}. So g X 1 ψ(g 2 = ψ(1 2 > 1 = X 1 sice dim ρ > 1. From ψ, ψ = 1, we get which gives a cotradictio. Hece χ(g = 0 for some g G. G = g G ψ(g 2 = m i=1 g X i ψ(g 2 > m X i = G, 13. Show that a abelia group has a faithful irreducible represetatio if ad oly if it is cyclic. Solutio. We assume that the group G is fiite ad the represetatio ρ is over C. Recall that complex irreducible represetatios of abelia groups are oe-dimesioal (Example Sheet 1, Problem 9 ad that each ρ(x C is a Nth root of uity where N = G (Example Sheet 1, Problem 7. So ρ(x D = {z C z = 1} for all x G. If G is abelia ad ρ : G C is faithful, the G = ρ(g. But ρ(g D = S 1 ad sice the oly fiite subgroups of S 1 are either dihedral or cyclic, we must have G is cyclic as it is abelia. If G = x is a cyclic group, the ρ : G C, x e 2πi/N is clearly a faithful represetatio. 14. Let Γ r = {A {1,..., } A = r}, ad χ r be the character of the permutatio represetatio C[Γ r ] of S. Compute (χ r, χ s ad hece decompose C[Γ r ] as a represetatio of S. Solutio. (Impetus for this solutio provided by Domiic Pito, Caius College By Problem 9 i Example Sheet 2, (χ r, χ s = #S -orbits o Γ r Γ s. Let := {(x, x x X}. Recall that S acts trasitively o the S -stable subsets ad X X \. It follows that the S -orbits o Γ r Γ s are i=1 O i = {A B Γ r Γ s A B = i}. (14.1 for i = max(0, s + r,..., mi(s, r (a momet s thought would show that these are the oly possible values i could take. Hece (χ r, χ s = mi(s, r max(0, s + r + 1. (14.2 Note that (14.1 also follows from a direct argumet: if A B, C D Γ r Γ s ad A B = i = C D, let A B = {x 1,..., x i }, A \ B = {x i+1,..., x r}, B \ A = {x i+1,..., x s }, X \ (A B = {x r+s i+1,..., x }, C D = {y 1,..., y i }, C \ D = {y i+1,..., y r}, D \ C = {y i+1,..., y s }, X \ (C D = {y r+s i+1,..., y }. The the permutatio i S defied by x 1 x i x i+1 x r x i+1 x s x r+s i+1 x y 1 y i y i+1 y r y i+1 y s y r+s i+1 y would sed A B to C D, ie. they lie i the same S -orbit. Sice S permutes elemets i, A B ad C D with A B C D caot lie i the same S -orbit. To decompose C[Γ r ] ito irreducibles, observe that 9

10 ➀ Γ 1 = {1, 2,... }, so C[Γ 1 ] is just the usual permutatio represetatio of S which we kow breaks up ito exactly two represetatios: the trivial represetatio ad a ( 1-dimesioal irreducible represetatio V 1 (see your lecture otes. ➁ Γ is a sigleto, so C[Γ ] is the trivial represetatio. Hece χ =. ➂ C[Γ r ] ad C[Γ r ] are equivalet represetatios of S. The obvious G-isomorphism beig C[Γ r ] C[Γ r ], A X \ A for A Γ r. G-equivariace follows from σ(x \ A = X \ σa for σ S. Hece χ r = χ r. By ➁, (14.2 with s = gives (χ r, = (χ r, χ = 1. By ➂, we may assume that s r /2, so s + r ad (14.2 gives (χ r, χ s = s + 1. For r = 2, (χ 2, = 1 ad (χ 2, χ 1 = 2 imply ad V 1 both occur with multiplicity 1 i C[Γ 2 ]. Sice (χ 2, χ 2 = 3, there is a third irreducible represetatio V 2 of dimesio deg(χ 2 deg(χ 1 = ( ( 2 1. I geeral, sice (χ r, χ r 1 = r ad (χ r, χ r = r + 1, C[Γ r ] cotais a irreducible represetatio V r of degree deg(χ r deg(χ r 1 = ( ( r r 1 that is ot i C[Γr 1 ]. Iductively we get C[Γ r ] = V 1 V r. 15. Usig Problems 14 ad 4, decompose S m C as represetatio of S for all m 1. Solutio. Let X = {1, 2,..., }. Note that C[X] = C as S -modules. S m C = S m C[X] = C[S m X] = C[Γ m ] C[Γ m 1 ] C[Γ 1 ] = ( V 1 V m ( V 1 V m 1 ( V 1 = m (m 1V 1 V 16. Let V be a irreducible represetatio of G, with character χ. Show that G 1 g G χ(g2 is either 0, 1 or 1, ad that this zero precisely whe χ is ot real valued, ie. whe V is ot isomorphic to V. (Hit: cosider V V = S 2 V 2 V. Solutio. Sice χ is irreducible, it occurs with multiplicity oe i itself ad if it is ot real valued, the χ is a differet irreducible character, ie. { 1 if χ is real valued, χ, χ = 0 if χ is ot real valued. Sice χ, χ = 1 G χ 2 (g = χ 2, = χ S 2 V + χ 2 V, = χ S 2 V, + χ 2 V,, g G we have the followig possibilities ➀ χ S 2 V, = 1 ad χ 2 V, = 0 whe χ is real valued, ➁ χ S 2 V, = 0 ad χ 2 V, = 0 whe χ is ot real valued, ➂ χ S 2 V, = 0 ad χ 2 V, = 1 whe χ is real valued. Hece this gives 1 G 1 if ➀ holds, χ(g 2 = χ S 2 V, χ 2 V, = 0 if ➁ holds, g G 1 if ➂ holds. Note that ➁ happes precisely whe χ is ot real valued. 10

11 17. Let G GL (C, G a fiite group. Show that g G tr(g = 0 implies g G g = 0. Solutio. We may regard G GL (C, g g as a represetatio of G. Recall that every represetatio of a fiite group is equivalet to a uitary represetatio (the proof is similar to that i Example Sheet 1, Problem 8(i. I this case, it meas that we may assume that G U(, ie. g := g T = g 1 for all g G. Let A = g G g M (C. Observe that ➀ A is self-adjoit (a.k.a. hermitia sice A = g G g = g G g 1 = g G g = A. ➁ For all x G, xa = g G xg = g G g = A. Suppose A 0. The by ➀, A 2 = AA is positive defiite ad so has eigevalues λ 1,..., λ (0,. Sice ( A 2 = g A = ga = ➁ A = G A, g G g G g G we get tr(a 2 = G tr(a ad tr(a 2 = λ λ > 0 would the imply that tr(a > 0. Remark. Give ay G GL (C, G GL (C is ofte kow as the atural represetatio of G. Bug report: L.H.Lim@dpmms.cam.ac.uk Versio 0.2, beta release (Jue 1,