REPRESENTATION THEORY.WEEK 4VERA SERGANOVA1. Induced modulesLet B ⊂ A be rings and M be a B-module. Then one can construct inducedmodule IndABM = A ⊗BM as the quotient of a free abelian group with generatorsfrom A × M by relations(a1+ a2) × m − a1× m − a2× m, a × (m1+ m2) − a × m1− a × m2, ab × m − a × bm,and A acts on A ⊗BM by left multiplication. Note that j : M → A ⊗BM defined byj (m) = 1 ⊗ mis a homomorphism of B-modules.Lemma 1.1. Let N be an A-module, then for ϕ ∈ HomB(M, N) there exists aunique ψ ∈ HomA(A ⊗BM, N) such that ψ ◦ j = ϕ.Proof. Clearly, ψ must satisfy the relationψ (a ⊗ m) = aψ (1 ⊗ m) = aϕ (m) .It is trivial to check that ψ is well defined. Exercise. Prove that for any B-modul e M there exists a unique A-module satisfyingthe conditions of Lemma 1.1.Corollary 1.2. (Frobenius recipro city.) For any B-module M and A-modul e Nthere is an isomorphism of abelian groupsHomB(M, N)∼=HomA(A ⊗BM, N) .Example. Let k ⊂ F be a field exte nsion. Then induction IndFkis an exactfunctor from the category of vector spaces over k to the category of vector spacesover F , i n the se nse that the short exact sequence0 → V1→ V2→ V3→ 0becomes an e xact sequence0 → F ⊗kV1⊗ → F ⊗kV2→ F ⊗kV3→ 0.Date: September 27, 2005.12 VERA SERGANOVAIn general, t he latter property is not true. It is not difficult to see that induction isright exact, i.e. an exact sequence of B-module sM → N → 0induces an exact sequence of A-modulesA ⊗BM → A ⊗BN → 0.But an e xact sequence0 → M → Nis not necessarily ex act after induction.Later we discuss general properties of induction but now we are going to studyinduction for the case of groups.2. Induced representations for groups.Let H be a subgroup of G and ρ : H → GL ( V ) be a representation. Then theinduced representation IndGHρ is by definition a k (G)-modulek (G) ⊗k(H)V.Lemma 2.1. The dimension of IndGHρ equals the product of dim ρ and the index[G : H] of H. More precisely, let S is a set of r epresentatives of left cosets, i.e.G =as∈SsH,then(2.1) k (G) ⊗k(H)V = ⊕s∈Ss ⊗ V.For any t ∈ G, s ∈ S there exist unique s′∈ S, h ∈ H such that ts = s′h and theaction of t is given by(2.2) t (s ⊗ v) = s′⊗ ρhv.Proof. Straightforward check. Lemma 2.2. Let χ = χρand IndGHχ denote the character of IndGHρ. Then(2.3) IndGHχ (t) =Xs∈S,s−1ts∈Hχs−1ts=1|H|Xs∈G,s−1ts∈Hχs−1ts.Proof. (2.1) and (2.2) implyIndGHχ (t) =Xs∈S,s′=sχ (h) .Since s = s′implies h = s−1ts ∈ H, we obtai n the formula for the induced character.Note also that χ (s−1ts) does not depend on a choice of s in a left coset. REPRESENTATION THEORY. WEEK 4 3Corollary 2.3. Let H be a normal subgroup in G. Then IndGHχ (t) = 0 for anyt /∈ H.Theorem 2.4. For any ρ : G → GL (V ), σ : H → GL (W ), we have the identity(2.4)IndGHχσ, χρG= (χσ, ResHχρ)H.Here a subindex indicates the group where we take inner product.Proof. It f ol lows from Frobenius reciprocity, sincedim HomGIndGHW, V= dim HomH(W, V ) . Note that (2.4) can be proved directly from (2.3). Define two mapsResH: C (G) → C (H) , IndGH: C (H) → C (G) ,the former is the restric tion on a subgroup, the latter is defined by (2.3). Then forany f ∈ C (G) , g ∈ C (H)(2.5)IndGHg, fG= (g, ResHf)H.Example 1. Let ρ be a trivial representation of H. Then IndGHρ is the permutationrepresentation of G obtained from t he natural l eft action of G on G/H (the set ofleft cosets).Example 2. Let G = S3, H = A3, ρ be a non-trivial one dimensional representa-tion of H (one of two possible). ThenIndGHχρ(1) = 2, IndGHχρ(12) = 0, IndGHχρ(123) = −1.Thus, by induction we obtain an irreducible two-dimensional representation of G.Now consider another subgroup K of G = S3generated by the tr ansposition (12),and let σ be the (unique) non-trivial one-dimensional representation of K. ThenIndGKχσ(1) = 3, IndGKχσ(12) = −1, IndGHχρ(123) = 0.3. Double cosets and restriction to a subgroupIf K and H are subgroups of G one can define the equivalence relation on G : s ∼ tiff s ∈ KtH. The equivalence classes are called double cosets. We can choose a set ofrepresentative T ⊂ G such thatG =as∈TK tH .We define the set of double cosets by K\G/H. One can identify K\G/H with K-orbits on S = G/H in the obvious way and with G-orbits on G/K × G/H by theformulaKtH → G (K, tH) .4 VERA SERGANOVAExample. Let Fqbe a field of q elements and G = GL2(Fq)def= GLF2q. Let B bethe subgroup of upper-triangular matrices in G. Check that |G| = (q2− 1) (q2− q),|B| = (q − 1)2q and therefore [G : B] = q + 1. Identify G/B with the set of lines P1in F2qand B\G/B with G-orbits on P1× P1. Check that G has only two orbits onP1× P1: the diagonal and its complement. Thus, |B\G/B| = 2 andG = B ∪ BsB,wheres =0 11 0Theorem 3.1. Let T ⊂ G such that G =`s∈TKtH. ThenResKIndGHρ = ⊕s∈TIndKK∩sHs−1ρs,whereρshdef= ρs−1hs,for any h ∈ sHs−1.Proof. Let s ∈ T and Ws= k (K) (s ⊗ V ). Then by construction, Wsis K-invariantandk (G) ⊗k(H)V = ⊕s∈TWs.Thus, we need t o check that the representation of K in Wsis isomorphic to IndKK∩sHs−1ρs.We define a homomorphismα : IndKK∩sHs−1V → Wsby α (t ⊗ v) = ts ⊗ v for any t ∈ K, v ∈ V . It is well definedα (th ⊗ v − t ⊗ ρshv) = ths ⊗ v − ts ⊗ ρs−1hsv = tss−1hs⊗ v − ts ⊗ ρs−1hsv = 0and obviously surjective. Injectivity can be proved by counting dimensions. Example. Let us go back to our example B ⊂ SL2(Fq). Theorem 3.1 tells us thatfor any representation ρ of BIndGBρ = ρ ⊕ IndGHρ′,where H = B ∩ sBs−1is a subgroup of diagonal matrices andρ′a 00 b= ρb 00 aCorollary 3.2. If H is a normal subgroup of G, thenResHIndGHρ = ⊕s∈G/Hρs.REPRESENTATION THEORY. WEEK 4 54. Mackey’s criterionTo findIndGHχ, IndGHχwe can use Frobenius reciprocity and Theorem 3.1.IndGHχ, IndGHχG=ResHIndGHχ, χH=Xs∈TIndHH∩sHs−1χs, χH==Xs∈T(χs, ResH∩sHs−1χ)H∩sHs−1= (χ, χ)H+Xs∈T \{1}(χs, ResH∩sHs−1χ)H∩sHs−1.We call two representation disjoint if they do not have the same irreducible com-ponent, i.e. their characters are orthogonal.Theorem 4.1. (Mackey’ s c r iterion) IndGHρ is irreducible iff ρ is irreducible and ρsand ρ are disjoint representations of H ∩ sHs−1for any s ∈ T \