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

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1 2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 1. The group rig k[g] The mai idea is that represetatios of a group G over a field k are the same as modules over the group rig k[g]. First I defied both terms Represetatios of groups. Defiitio 1.1. A represetatio of a group G over a field k is defied to be a group homomorphism where V is a vector space over k. ρ : G Aut k (V ) Here Aut k (V ) is the group of k-liear automorphisms of V. This also writte as GL k (V ). This is the group of uits of the rig Ed k (V )= Hom k (V, V ) which, as I explaied before, is a rig with additio defied poitwise ad multiplicatio give by compositio. If dim k (V )=d the Aut k (V ) = Aut k (k d )=GL d (k) which ca also be described as the group of uits of the rig Mat d (k) or as: GL d (k) ={A Mat d (k) det(a) 0} d = dim k (V ) is called the dimesio of the represetatio ρ examples. Example 1.2. The first example I gave was the trivial represetatio. This is usually defied to be the oe dimesioal represetatio V = k with trivial actio of the group G (which ca be arbitrary). Trivial actio meas that ρ(σ) = 1 = id V for all σ G. I the ext example, I poited out that the group G eeds to be writte multiplicatively o matter what. Example 1.3. Let G = Z/3. Writte multiplicatively, the elemets are 1, σ, σ 2. Let k = R ad let V = R 2 with ρ(σ) defied to be rotatio by 120 =2π/3. I.e., ( ) 1/2 3/2 ρ(σ) = 3/2 1/2 Example 1.4. Suppose that E is a field extesio of k ad G = Gal(E/k). The G acts o E by k-liear trasformatios. This gives a represetatio: ρ : G Aut k (E) Note that this map is a iclusio by defiitio of Galois group.

2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS axioms. I a elemetary discussio of group represetatios I would write a list of axioms as a defiitio. However, they are just logwided explaatios of what it meas for ρ : G Aut k (V ) to be a group homomorphism. The oly advatage is that you do t eed to assume that ρ(σ) is a automorphism. Here are the axioms. (I switched the order of (2) ad (3) i the lecture.) (1) ρ(1) = 1 1v = v v V (2) ρ(στ) =ρ(σ)ρ(τ) σ, τ G (στ)v = σ(τv) v V (3) ρ(σ) is k-liear σ G σ(av + bw) =aσv + bσw v, w V, a, b k The first two coditios say that ρ is a actio of G o V. Actios are usually writte by juxtapositio: σv := ρ(σ)(v) The third coditio says that the actio is k-liear. So, together, the axioms say that a represetatio of G is a k-liear actio of G o a vector space V Modules over k[g]. The group rig k[g] is defied to be the set of all fiite k liear combiatios of elemets of G: a σ σ where α σ k for all σ G ad a σ = 0 for almost all σ. For example, R[Z/3] is the set of all liear combiatios x + yσ + zσ 2 where x, y, z R. I.e., R[Z/3] = R 3. I geeral k[g] is a vector space over k with G as a basis. Multiplicatio i k[g] is give by ( )( ( ) aσ σ bτ τ) = cλ λ where c λ G ca be give i three differet ways: c λ = στ=λ a σ b τ = Propositio 1.5. k[g] is a k-algebra. a σ b σ 1 λ = τ G a λτ 1b τ This is straightforward ad tedious. So, I did t prove it. But I did explai what it meas ad why it is importat. Recall that a algebra over k is a rig which cotais k i its ceter. The ceter Z(R) of a (ocommutative) rig R is defied to be the set of elemets of R which commute with all the other elemets: Z(R) is a subrig of R. Z(R) := {x R xy = yx y R}

3 4 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS The ceter is importat for the followig reaso. Suppose that M is a (left) R-module. The each elemet r R acts o M by left multiplicatio λ r λ r : M M, λ r (x) =rx This is a homomorphism of Z(R)-modules sice: λ r (ax) =rax = arx = aλ r (x) a Z(R) Thus the actio of R o M gives a rig homomorphism: ρ : R Ed Z(R) (M) Gettig back to k[g], suppose that M is a k[g]-module. The the actio of k[g] o M is k-liear sice k is i the ceter of k[g]. So, we get a rig homomorphism ρ : k[g] Ed k (M) This restricts to a group homomorphism ρ G : G Aut k (M) I poited out that, i geeral, ay rig homomorphism φ : R S will iduce a group homomorphism U(R) U(S) where U(R) is the group of uits of R. Ad I poited out earlier that Aut k (M) is the group of uits of Ed k (M). G is cotaied i the group of uits of k[g]. (A iterestig related questio is: Which fiite groups occur as groups of uits of rigs?) This discussio shows that a k[g]-module M gives, by restrictio, a represetatio of the group G o the k-vector space M. Coversely, suppose that ρ : G Aut k (V ) is a group represetatio. The we ca exted ρ to a rig homomorphism ρ : k[g] Ed k (V ) by the simple formula ( ) ρ aσ σ = a σ ρ(σ) Whe we say that a represetatio of a group G is the same as a k[g]-module we are talkig about this correspodece. The vector space V is also called a G-module. So, it would be more accurate to say that a G-module is the same as a k[g]-module. Corollary 1.6. (1) Ay group represetatio ρ : G Aut k (V ) exteds uiquely to a rig homomorphism ρ : k[g] Ed k (V ) makig V ito a k[g]-module.

4 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 5 (2) For ay k[g]-module M, the actio of k[g] o M restricts to give a group represetatio G Aut k (M). (3) These two operatios are iverse to each other i the sese that ρ is the restrictio of ρ ad a actio of the rig k[g] is the uique extesio of its restrictio to G. There are some coceptual differeces betwee the group represetatio ad the correspodig k[g]-module. For example, the module might ot be faithful eve if the group represetatio is: Defiitio 1.7. A group represetatio ρ : G Aut k (V ) is called faithful if oly the trivial elemet of G acts as the idetity o V. I.e., if the kerel of ρ is trivial. A R-module M is called faithful if the aihilator of M is zero. (a(m) ={r R rx =0 x M}). These two defiitios do ot agree. For example, take the represetatio ρ : Z/3 GL 2 (R) which we discussed earlier. This is faithful. But the extesio to a rig homomorphism ρ : R[Z/3] Mat 2 (R) is ot a moomorphism sice 1 + σ + σ 2 is i its kerel Semisimplicity of k[g]. The mai theorem about k[g] is the followig. Theorem 1.8 (Maschke). If G is a fiite group of order G = ad k is a field with char k (or char k =0) the k[g] is semisimple. Istead of sayig char k is either 0 or a prime ot dividig, I will say that 1/ k. By the Wedderbur structure theorem we get the followig. Corollary 1.9. If 1/ G k the k[g] = Mat d1 (D i ) Mat db (D b ) where D i are fiite dimesioal divisio algebras over k. Example R[Z/3] = R C I geeral, if G is abelia, the the umbers d i must all be 1 ad D i must be fiite field extesios of k.

5 6 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS homomorphisms. I order to prove Maschke s theorem, we eed to talk about homomorphisms of G-modules. We ca defie these to be the same as homomorphisms of k[g]-modules. The the followig is a propositio. (Or, we ca take the followig as the defiitio of a G- module homomorphism, i which case the propositio is that G-module homomorphisms are the same as homomorphisms of k[g]-modules.) Propositio Suppose that V, W are k[g]-modules. The a k- liear mappig φ : V W is a homomorphism of k[g]-modules if ad oly if it commutes with the actio of G. I.e., if for all σ G. σ(φ(v)) = φ(σv) Proof. Ay homomorphism of k[g]-modules will commute with the actio of k[g] ad therefore with the actio of G k[g]. Coversely, if φ : V W commutes with the actio of G the, for ay a σ σ k[g], we have ( ) φ a σ σv = a σ φ(σv) = ( ) a σ σφ(v) = a σ σ φ(v) So, φ is a homomorphism of k[g]-modules. We also have the followig Propositio/Defiitio of a G-submodule. Propositio A subset W of a G-module V over k is a k[g]- submodule (ad we call it a G-submodule) if ad oly if (1) W is a vector subspace of V ad (2) W is ivariat uder the actio of G. I.e., σw W for all σ G. Proof of Maschke s Theorem. Suppose that V is a fiitely geerated G-module ad W is ay G-submodule of V. The we wat to show that W is a direct summad of V. This is oe of the characterizatios of semisimple modules. This will prove that all f.g. k[g]-modules are semisimple ad therefore k[g] is a semisimple rig. Sice W is a submodule of V, it is i particular a vector subspace of V. So, there is a liear projectio map φ : V W so that φ W = id W. If φ is a homomorphism of G-modules, the V = W ker φ ad W would split from V. So, we would be doe. If φ is ot a G- homomorphism, we ca make it ito a G-homomorphism by averagig over the group, i.e., by replacig it with ψ = 1 λσ 1 φ λ σ.

6 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 7 First, I claim that ψ W = id W. To see this take ay w W. The σw W. So, φ(σw) =σw ad ψ(w) = 1 σ 1 φ(σw) = 1 σ 1 (σw) =w Next I claim that ψ is a homomorphism of G-modules. To show this take ay τ G ad v V. The ψ(τv) = 1 σ 1 φ(στv) = 1 αφ(βv) αβ=τ = 1 τσ 1 φ(σv) =τψ(v) So, ψ gives a splittig of V as required R[Z/3]. I gave a logwided explaatio of Example 1.10 usig the uiversal property of the group rig k[g]. I these otes, I will just summarize this property i oe equatio. If R is ay k-algebra ad U(R) is the group of uits of R, the: Hom k-alg (k[g],r) = Hom grp (G, U(R)) The isomorphism is give by restrictio ad liear extesio. The isomorphism R[Z/3] = R C is give by the mappig: φ : Z/3 R C which seds the geerator σ to (1,ω) where ω is a primitive third root of uity. Sice (1, 0), (1,ω), (1, ω) are liearly idepedet over R, the liear extesio φ of φ is a isomorphism of R-algebras group rigs over C. We will specialize to the case k = C. I that case, there are o fiite dimesioal divisio algebras over C (Part C, Theorem 3.12). So, we get oly matrix algebras: Corollary If G is ay fiite group the I particular, = G = d 2 i. C[G] = Mat d1 (C) Mat db (C) Example If G is a fiite abelia group of order the C[G] = C. Example Take G = S 3, the symmetric group o 3 letters. Sice this group is oabelia, the umbers d i caot all be equal to 1. But

7 8 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS the oly way that 6 ca be writte as a sum of squares, ot all 1, is 6 = Therefore, C[S 3 ] = C C Mat 2 (C) This ca be viewed as a subalgebra of Mat 4 (C) give by Each star ( ) represets a idepedet complex variable. I this descriptio, it is easy to visualize what are the simple factors Mat di (C) give by the Wedderbur structure theorem. But what are the correspodig factors of the group rig C[G]? 1.4. idempotets. Suppose that R = R 1 R 2 R 3 is a product of three subrigs. The the uity of R decomposes as 1 = (1, 1, 1). This ca be writte as a sum of uit vectors: 1 = (1, 0, 0) + (0, 1, 0) + (0, 0, 1) = e 1 + e 2 + e 3 This is a decompositio of uity (1) as a sum of cetral, orthogoal idempotets e i. Recall that idempotet meas that e 2 i = e i for all i. Also, 0 is ot cosidered to be a idempotet. Orthogoal meas that e i e j = 0 if i j. Cetral meas that e i Z(R). Theorem A rig R ca be writte as a product of b subrigs R 1,R 2,,R b iff 1 R ca be writte as a sum of b cetral, orthogoal idempotets ad, i that case, R i = e i R. A cetral idempotet e is called primitive if it caot be writte as a sum of two cetral orthogoal idempotets. Corollary The umber of factors R i = e i R is maximal iff each e i is primitive. So, the problem is to write uity 1 C[G] as a sum of primitive, cetral ( orthogoal) idempotets. We will derive a formula for this decompositio usig characters.

8 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS Ceter of C[G]. Before I move o to characters, I wat to prove oe last thig about the group rig C[G]. Theorem The umber of factors b i the decompositio C[G] b = Mat di (C) is equal to the umber of cojugacy classes of elemets of G. i=1 For, example, the group S 3 has three cojugacy classes: the idetity {1}, the traspositios {(12), (23), (13)} ad the 3-cycles {(123), (132)}. I order to prove this we ote that b is the dimesio of the ceter of the right had side. Ay cetral elemet of Mat di (C) is a scalar multiple of the uit matrix which we are callig e i (the ith primitive cetral idempotet). Therefore: Lemma The ceter of b i=1 Mat d i (C) is the vector subspace spaed by the primitive cetral idempotets e 1,,e b. I particular it is b-dimesioal. So, it suffices to show that the dimesio of the ceter of C[G] is equal to the umber of cojugacy classes of elemets of G. (If G is abelia, this is clearly true.) Defiitio A class fuctio o G is a fuctio f : G X so that f takes the same value o cojugate elemets. I.e., for all σ, τ G. Usually, X = C. f(τστ 1 )=f(σ) For example, ay fuctio o a abelia group is a class fuctio. Lemma For ay field k, the ceter of k[g] is the set of all a σσ so that a σ is a class fuctio o G. So, Z(k[G]) = k c where c is the umber of cojugacy classes of elemets of G. Proof. If a σσ is cetral the a σ σ = τ a σ στ 1 = a σ τστ 1 The coefficiet of τστ 1 o both sides must agree. So a τστ 1 = a σ I.e., a σ is a class fuctio. The coverse is also clear. These two lemmas clearly imply Theorem 1.18 (which ca ow be stated as: b = c if k = C).

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