RATIONALITY QUESTIONS AND HECKE ALGEBRAVERA SERGANOVA1. Splitting fieldWe assume that k is a field of characteristic 0,¯k is its algebraic closure and G is afinite group.Let χ1, . . . , χrbe the characters of all non-isomorphic irreducible representationsof G over¯k and ψ1, . . . , ψsbe the characters of all non-isomorphic representations ofG over k. Recall that if ρ : G → GL(V ) and σ : G → GL(W ) are two representationsof G over any field of characteristic zerodim Ho mG(V, W ) = (χρ, χσ).Therefore (ψi, ψj) = 0 if i 6= j.LetRk(G) = {sXj=1mjψj|mj∈ Z}andR(G) = {rXi=1miχi|mi∈ Z}.Note that Rk(G) ⊂ R(G), and both Rk(G) and R(G) have natural structures ofcommutative rings.A representation ρ of G over¯k is rational over k if ρ = σ¯k:=¯k ⊗kσ for somerepresentation σ of G over k. This condition is equivalent to χρ∈ Rk(G).Lemma 1.1. The f ollowing conditions on a field k are equivalent(a) every irreducible r epresentation of G over k is absolutely irreducible;(b) every representation over¯k is rational over k;(c) Rk(G) = R(G);(d) the group algebra k(G) is isomorphic to a direct sum of matrix algebras overk.Exercise. Prove the lemma.A field k is a spli tting field for G if it satisfies the conditions of Lemma 1 .1 .Theorem 1.2. For any field k and a group G there exists a finite extension F ⊃ ksuch that F is a splitting field for G.Date: March 31, 2011.12 VERA SERGANOVAProof. Recall t hat¯k(G) = ⊕ri=1End¯k(Vi) is a direct sum of matrix algebras. Let Epijbe the standard basis consisting of elementary matrices. Then Epij=Pg∈Gcpij(g) gand g =Pi,j,pdijp(g) Epij. Then a finite extension F of k conta ining cpij(g) and dijp(g)is a splitting field for G. In fact, the stronger result is true.Theorem 1.3. Let m be the least common multiple of orders of all elements in G,and F contains a primitive m-th root of 1. Then F is a splitting field for G.2. The number of irreducible representationsLemma 2.1. Assume that F is such that χi(g) ∈ F for all i ≤ r and g ∈ G. ThenRF(G) is a subgroup of finite index in R(G).Proof. Let τ be an irreducible representation of G over F . The operator of projectionon isotypic component pi=ni|G|Pχi(g−1)τgis well defined. By irreducibility of τ it iseither zero or non-degenerate. Hence χτis a multiple of χifor one i. The statementfollows. Let us assume now that F is a splitting field of G and F is a finite Galois extensionof F . By Γ denote the corresponding Galois group. Note tha t if ρ is a representationof G over F , then for each γ ∈ Γ one can define the representation ργby twisting thematrix coefficients by action of γ. It is also clear that if ρ is irreducible, then ργisalso irreducible. Thus, Γ acts on R(G) by permuting χ1, . . . , χr.Lemma 2.2. Let ρ : G → GL(V ) be an irreducible representation of G over k. Thenthere exist an integer nρand i ≤ r such thatχρ= nρXγ∈Γχγi.This nρis called the Schur index of ρ.Proof. Write χρ=Paiχi. Note that χρis Γ-invariant. Therefore if χi= χγjfor someγ ∈ Γ, then ai= aj. Let ai6= 0 and ψ =Pχjover the Γ-orbit of χi. Then ψ isΓ-invariant and hence ψ(g) ∈ k f or all g ∈ G. Thus, the operator p =Pg∈Gψ(g−1)ρgis an intertwiner. It is not zero as can be easily shown by calculating the trace,therefore it is non-degenerate. Hence it is non-degenerate on VF. The latter impliesthat the char acters of all irreducible components of ρFare not orthogonal to ψ. Thestatement follows. Theorem 2.3. The number of irreducible representation of G over k equals thenumber of Γ-orbits in {χ1, . . . , χr}.Proof. From the previous lemma we obtain that the number of irreducible repre-sentation is not larger t han the number of orbits. On the other hand let ρ be anRATIONALITY QUESTIONS AND HECKE ALGEBRA 3irreducible representation of G over F , and ρkbe the same representation consideredas a representation over k. Thenχ(ρk) =Xγ∈Γχ.This shows that the number of irreducible representation of G over k is not less thanthe number of o rbits. Let us make an additional assumption that F conta ins all roots of xm−1. The set ofall roots is a cyclic group of order m and and Γ acts o n this group by automorphisms.More precisely, if ε is a primitive roo t then fo r each γ ∈ Γ t here exists r(γ) relativelyprime to m such that γ(ε) = εr(γ). Introduce the action of Γ on G by settinggγ= gr(γ). If ρ is a representation of G, then all eigenvalues of ρgare roots of xm− 1.Thereforeχγρ(g) = χρ(gγ).Now we consider the action of Γ × G on G, where G acts by conjugation. The o r bitsunder this action are called Γ-classes.Theorem 2.4. The number of Γ classes equals the number of irreducible represen-tations of G over k.Proof. Let R(G)Γbe the abelian subgroup of R(G) fixed by the Γ- action. Thenrk R(G)Γcoincides with dim F ⊗ZR(G)Γ. The representation of Γ in F ⊗ZR(G) isa permutation representation. Hence dim F ⊗ZR(G)Γequals the number of Γ-orbitsin {χ1, . . . , χr}. On the other hand, F ⊗ZR(G) is the subspace of functions on Gconsisting of function constant on Γ classes. That implies the theorem. Example 2.5. Let k = R, F = C, Γ = Z2with generator σ which is conjugation.It is easy to see that σ(g) = g−1for any g ∈ G. Therefore the number of irreduciblerepresentation of G over R equals the number of conjugacy classes stable under σplus half the number of classes unstable under σ.3. Hecke algebra and double cosetsLet G be a finite group. The group algebra k(G) is isomorphic to the algebra offunction on G with the operation of convolutionf1∗ f2(g)Xs∈Gf1(gs−1)f2(s).The isomorphism can be constructed by the map g → δg, where δg(x) = δg,x. Theproof is based on the the simple identity δg∗ δh= δgh.A certain generalization of this construction is so called Hecke algebra. Let H bea subgroup of G, then the induced representation IndGH(triv) can be identified withthe space of k(X) of functions X = G/H. The action of G is given by the formulaρgf(x) = f(g−1(x)).4 VERA SERGANOVAThe Hecke algebra A(G, H) is by definition EndG(k(X)). Note that the ringEndk(k(X)) is isomorphic to the algebra k(X × X) with operation of convolutionF1∗ F2(x, z) =Xy ∈XF1(x, y)F2(y, z).The action of k(X × X) on k(X) is given by the formulaF f(x) =Xy ∈XF (x, y)f(y).The condition that F (x, y) commutes with ρgis equivalent to the followingρg(F f)(x) =Xy ∈XF (g−1x, y)f (y) = F ρgf(x) =Xy ∈XF