24 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS

Transcription

1 24 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS Corollary Suppose that the semisimple decompositio of the G- module V is V = i S i. The i = χ V,χ i Proof. Sice χ V W = χ V + χ W, we have: χ V = j χ j. So, χ V,χ i = j χ j,χ i = i proof of the mai theorem. The theorem will follow from three lemmas. The first lemma calculates the dimesio of the fixed poit set of V. Defiitio If V is a G-module the the fixed poit set of the actio of G is give by V G := {v V σv = v σ G} Lemma The dimesio of the fixed poit set is equal to the average value of the correspodig character: dim C V G = 1 χ V (σ) Proof. The projectio map is give by σ G π : V V G π(v) = 1 σv It is clear that (1) π(v) V G sice multiplicatio by ay τ G will just permute the summads. (2) π(v) =v if v V G because, i that case, each σv = v ad there are terms. Therefore, π is a projectio map, i.e., a liear retractio oto V G. Lookig at the formula we see that π is multiplicatio by the idempotet e 1 = 1 σ G σ. (This is the idempotet correspodig to the trivial represetatio.) So: dim V G = Tr(π) =χ V (e 1 )=χ V ( 1 Explaatios: ) σ = 1 χ V (σ) σ G σ G

2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 25 (1) dim V G = Tr(π) because V = V G W (W = ker π). So, the matrix of π is: ( ) 1V G 0 π = 0 0 W makig Tr(π) = Tr(1 V G) = dim C V G. (2) Tr(π) =χ V (e 1 ) by defiitio of the character: χ V (e 1 ) := Tr(e 1 : V V ) This is the trace of the mappig V V give by multiplicatio by e 1. But we are callig that mappig π. Lemma If V, W are represetatios of G the Hom G (V, W ) = Hom C (V, W ) G where G acts o Hom C (V, W ) by cojugatio, i.e., σf = σ f σ 1 which meas that (σf)(v) =σf(σ 1 v) Proof. This is trivial. Give ay liear map f : V W, f is a G- homomorphism iff σ f = f σ σ f σ 1 = f σf = f iff f Hom C (V, W ) G. Lemma Hom C (V, W ) = V W as G-modules. Proof. Let φ : V W Hom C (V, W ) be give by φ(f w)(v) =f(v)w To check that this is a G-homomorphism we eed to show that φσ = σφ for ay σ G. So, we compute both sides: which seds v V to φσ(f w) =φ(σf σw) =φ(f σ 1 σw) O the other side we have: which also seds v V to φ(f σ 1 σw)(v) =f(σ 1 v)σw σφ(f w) =σ φ(f w) σ 1 σ φ(f w) σ 1 v = σ(f(σ 1 v)w) =f(σ 1 v)σw This shows that φ commutes with the actio of G. The fact that φ is a isomorphism is well-kow: If v i,v i form a basis-dual basis pair for V ad w j form a basis for W the v j w i form a basis for V W ad φ(v j w i ):v = a j v j v j (v)w i = a j w i

3 26 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS is the mappig whose matrix has ij-etry equal to 1 ad all other etries 0. So, these homomorphisms form a basis for Hom C (V, W ) ad φ is a isomorphism. Proof of mai theorem Usig the three lemmas we get: dim C Hom G (V, W )= 2.33 dim C Hom C (V, W ) G = 2.34 dim C (V W ) G 1 = 2.32 χ V W (σ) σ G = 1 χ V (σ)χ W (σ) σ = 1 χ V (σ 1 )χ W (σ) = χ V,χ W σ character table of S 4. Usig these formulas we ca calculate the character table for S 4. First ote that there are five cojugacy classes represeted by 1, (12), (123), (12)(34), (1234) The elemets of cycle form (12)(34) form (with 1) a ormal subgroup K = {1, (12)(34), (13)(24), (14)(23)} S 4 called the Klei 4-group. The quotiet S 4 /K is isomorphic to the symmetric group o 3 letters. Imitatig the case of D 4, this allows us to costruct the followig portio of the character table for S 4 : c j (12) (123) (12)(34) (1234) χ χ χ χ 4 3 χ 5 3 Explaatios: (1) Sice (12)(34) K, the value of the first three characters o this cojugacy class is d i, the same as i the first colum. (2) Sice (1234)K = (12)K, these two colums have the same values of χ 1,χ 2,χ 3.

4 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 27 (3) Fially, the two ukow characters χ 4,χ 5 must be 3-dimesioal sice 24 = d 2 i = d d 2 5 has oly oe solutio: d 4 = d 5 = 3. To figure out the ukow characters we eed aother represetatio. The permutatio represetatio P is the 4-dimesioal represetatio of S 4 i which the elemets of S 4 act by permutig the uit coordiate vectors. For example ρ P (12) = Note that the trace of ρ P (σ) is equal to the umber of letters left fixed by σ. So, χ P takes values 4, 2, 1, 0, 0 as show: c j (12) (123) (12)(34) (1234) χ χ χ χ P χ V = χ P χ The represetatio P cotais oe copy of the trivial represetatio ad o copies of the other two: χ P,χ 1 = 1 (4 + 6(2) + 8(1)) = 1 24 χ P,χ 2 = 1 (4 + 6( 1)(2) + 8(1)(1)) = 0 24 χ P,χ 3 = 1 ((2)(4) + 8( 1)(1)) = 0 24 So, P = S 1 V where V is a 3-dimesioal module which does ot cotai S 1,S 2 or S 3. So, V = S 4 ms 5. But S 4,S 5 are both 3- dimesioal. So, V = S 4 (or S 5 ). Usig the fact that χ 1 + χ 2 +2χ 3 +3χ 4 +3χ 5 = χ reg

5 28 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS we ca ow complete the character table of S 4 : c j (12) (123) (12)(34) (1234) χ χ χ χ χ From the character table of S 4 we ca fid all ormal subgroups. First, the kerels of the 5 irreducible represetatios are: (1) ker ρ 1 = S 4. (2) ker ρ 2 = A 4 cotaiig the cojugacy classes of 1, (123), (12)(34). (3) ker ρ 3 = K cotaiig 1, (12)(34) ad cojugates. (4) ker ρ 4 = 1. I.e., ρ 4 is a faithful represetatio. (5) ker ρ 5 = 1. So, ρ 5 is also faithful. Sice these subgroups cotai each other: 1 < K < A 4 <S 4 itersectig them will ot give ay other subgroups. So, these are the oly ormal subgroups of S 4.

6 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS colum orthogoality. The colums of the character table also satisfy a orthogoality coditio. To see it we first have to write the row orthogoality coditio χ i,χ j = ad write it i matrix form: T b k=1 c k χ i(c k )χ j (c k )=δ ij c c b T t = I b where T is the character table T =(χ i (c j )). This equatio shows that the character table T is a ivertible matrix with iverse T 1 = DT t where D is the diagoal matrix with diagoal etries c i. Multiplyig both sides of this equatio o the right by T ad o the left with D 1 ad we get: T t T = D 1 = c c b Lookig at the etries of these matrices we get the colum orthogoality relatio: Theorem If σ, τ G the b χ i (σ)χ i (τ) = i=1 { c if σ, τ are cojugate 0 if ot Here c is the umber of cojugates of σ i G. (So, / c is the order of the cetralizer C(σ) ={τ G στ = τσ} of σ.) Corollary The character table T =(χ i (c j )) determies the size of each cojugacy class c j. Proof. Takig σ = τ i the above theorem we get C(σ) = i χ i (σ) 2 The size of the cojugacy class c of σ is the idex of its cetralizer: c = G : C(σ) = / C(σ).

7 30 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS As a example, look at the character table for S 3 : 1 (12) (123) χ χ χ Colum orthogoality meas that the usual Hermitia dot product of the colums is zero. For example, the dot product of the first ad third colum is (1)(1) + (1)(1) + (2)( 1) = 0 Also the dot product of the jth vector with itself (its legth squared) is equal to / c j. For example, the legth squared of the third colum vector is = 3 Makig the umber of cojugates of (123) equal to 6/3 = 2.

8 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS Iductio If H is a subgroup of G the ay represetatio of G will restrict to a represetatio of H by compositio: H G ρ Aut C (V ) Iductio is a more complicated process which goes the other way: It starts with a represetatio of H ad produces a represetatio of G. Followig Lag, I will costruct the same object i several differet ways startig with a elemetary equatio for the iduced character iduced characters. Defiitio 3.1. Suppose that H G (H is a subgroup of G) ad χ : H C is a character (or ay class fuctio). The the iduced character Id G H χ : G C is the class fuctio o G defied by Id G H χ(σ) = 1 χ(τστ 1 ) H where χ(σ) = 0 if σ/ H. τ G The mai theorem about the iduced character is the followig. Theorem 3.2. If V is ay represetatio of H the there exists a represetatio W of G so that χ W = Id G H χ V Furthermore, W is uique up to isomorphism. The represetatio W is writte W = Id G H V ad is called the iduced represetatio. We will study that tomorrow. Before provig this theorem let me give two examples example 1. Here is a trivial observatio. Propositio 3.3. If G is abelia the Id G H χ(σ) = G : H χ(σ) Now suppose that G = Z/4 ={1, σ, σ 2,σ 3 } ad H = {1,τ} with τ = σ 2. The the character table of H = Z/2 is H = Z/2 1 τ χ χ 1 1