CHARACTERS AS CENTRAL IDEMPOTENTS
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1 CHARACTERS AS CENTRAL IDEMPOTENTS CİHAN BAHRAN I have recently noticed (while thinking about the skewed orthogonality business Theo has mentioned) that the irreducible characters of a finite group G are nothing but the central idempotents (up to a scalar multiple) of the group algebra CG. The purpose of this document is to make sense of and explain this. Actually, I will show how characters can be defined in this way and the fact that they are scaled central idempotents immediately imply the usual and the skew orthogonality relations. 1. Endomorphisms Induced by Central Elements In this section, I will work with a more general setup than the group algebra. The main results are Theorem 7, which is stated in a form that doesn t refer to previous notation; so the reader may jump to there. Let k be an algebraically closed field of characteristic 0 (can be taken as C for all intents and purposes) and let A be a finite-dimensional k-algebra. Definition 1. For a left A-module M, we write ρ M : A End k M for the ring homomorphism which defines the A-action on M as a k-vector space. Definition 2. Let S be a simple left A-module. For any A-module M, define M S to be the sum of submodules of M which are isomorphic to S. If M has no submodule isomorphic to S, then M S = 0. Note that M S can be written as a direct sum of copies of S. If M is semisimple, then we have M = M S in which case we may refer to M S as the S-constituent of M. Note that a finitedimensional algebra has finitely many simple (left) modules up to isomorphism. But of course each individual M S might be a direct sum of infinitely many S s. Remark. In the remaining of this section, we assume M to be a fixed finitely generated semisimple left A-module. Therefore each M S consists of finitely many copies of S, say M S = S n S. In fact n S is determined by M: it is the number of copies of S occurring in M, the well-definition of n S can be checked by Jordan-Hölder, for instance. Proposition 3. Writing ι S : M S M for the inclusion and π S : M M S for the projection maps, the map Ξ : End A M End A M S is a k-algebra isomorphism. ϕ (π S ϕ ι S ) S 1
2 CHARACTERS AS CENTRAL IDEMPOTENTS 2 Proof. This is because Hom A (M S, M T ) = 0 if S T. Observe that if a A is central, then ρ M (a) End k M actually lies inside End A M. This is because the subring End A M of End k M is nothing but the centralizer of ρ M (A). And in the notation of Proposition 3, since ι S : M S M is an A-module homomorphism, we have π S ρ M (a) ι S = π S ι S ρ MS (a) = ρ MS (a). Thus the isomorphism Ξ in Proposition 3 sends ρ M (a) to the tuple (ρ MS (a)) S. understand the ρ MS (a) s, we start with a standard fact. Proposition 4. For every simple S, we have Z(End A M S ) = k. Proof. Writing n = n S, we have k-algebra isomorphisms End A M S = EndA S n = Mn (End A S) but End A S = k by Schur s lemma (here we use the assumption that k is algebraically closed). Thus Z(End A M S ) = Z(M n (k)) = k. Definition 5. Given a A, we write tr M (a) for the trace of the k-linear operator ρ M (a) : M M. Corollary 6. If a Z(A) and ρ M (a) Z(End A M), then for every simple S M, the endomorphism ρ MS (a) End A M S is the scalar multiplication by tr MS (a) n S, where tr denotes taking the trace of a k-linear operator on a finite-dimensional k-vector space. Proof. The isomorphism Ξ in Proposition 3 yields an isomorphism Z(End A M) = Z(End A M S ) which sends ρ M (a) to the tuple (ρ MS (a)) S, where each coordinate ρ MS (a) is scalar multiplication by Proposition 4. The scalar can be recovered by dividing the trace by n S. Note that we use the assumption that char k = 0 to be able to divide with n S. Also n S 0 because we are assuming S M. Application: If A is semisimple as a ring, then we can take M to be the regular left module A A, as this is finitely generated and semisimple. Let us write I S for the S-constituent of M (it can be shown that I S A is not just a left ideal, but a two-sided ideal). Moreover, it is a standard fact that every simple S occurs in M with multiplicity dim k S. Moreover, the endomorphism algebra End A M is isomorphic to the opposite algebra A op, via the maps A op = EndA M a σ a ϕ(1) ϕ To
3 CHARACTERS AS CENTRAL IDEMPOTENTS 3 where σ a : A A is the multiply by a from the right map. Composing with the isomorphism in Proposition 3, we get a k-algebra isomorphism A op ( ) = End A I S. S-simple Moreover if a Z(A), then σ a Z(End A M) is also equal to the multiply by a from the left map, which is exactly ρ M (a). Thus the second assumption of Corollary 6 comes for free. In short, we get the following result: Theorem 7. Let A be a finite-dimensional semisimple algebra over an algebraically closed field of characteristic zero. Let S be a simple left A-module and write I S for the S-constituent of the regular left module A A. Then the action of a central element a Z(A) on I S is given by its trace divided by dim k S. Proof. Immediate from above. Really. We may also deduce the following from the technology we ve developed: Theorem 8. Let A be as in Theorem 7. Then for every simple left A-module S, there exists a unique element e S in A such that the action of e S on S is trivial and on any other simple T S is zero. Consider the (finite) set B := {e S : S-simple}. (1) B is the set of central primitive idempotents of A. (2) B is a k-basis of Z(A). (3) Let S, T be simple left A-modules. Writing tr T (a) for the trace of the action of an element a A on T, we have dim k S if S tr T (e S ) = = T, 0 otherwise. = dim k S δ S,T. Thus, because taking trace is k-linear, the map Z(A) B k (a, e T ) tr T (a) uniquely extends to a non-degenerate bilinear form on Z(A) for which B is an orthogonal basis and e S, e S = dim k S. Proof. No kidding. The Group Algebra Next we apply the generalities in the previous section to the group algebra CG where G is a finite group. CG definitely satisfies the conditions of Theorem 7. The general convention to write an element of CG is something like G a g g, that is, formal C-linear combinations of the group elements. This is very useful in calculating products like (1+g)(1+h). But in this document, I consider CG as the set of functions from G to C. This allows to realize a character χ as a genuine element of the group
4 CHARACTERS AS CENTRAL IDEMPOTENTS 4 algebra, instead of writing them like g G χ(g)g. t g CG for the function t g : G C And for every g G, we write h δ g,h The set {t g : g G} forms a C-basis for CG and the multiplication of basis elements is defined by t g t h = t gh. and is extended linearly to all elements. I write for the product instead of a or juxtaposition because for α, β CG, it is easy to make the mistake that their product should send g to α(g)β(g). The multiplication is NOT pointwise. In fact, by definition it is given by (α β)(g) = ÑÑ h G h,k G h,k G hk=g α(h)t h é Ñ k G α(h)β(k)(t h t k )(g) α(h)β(k)t hk (g) α(h)β(k) k G α(gk 1 )β(k). β(k)t k Note the similarity to the convolution product (f g)(y) = f(y x)g(x)dx from real analysis. In fact, putting the counting measure on G, the map α β is precisely the convolution of α and β. By Theorem 8, we already have a non-degenerate C-bilinear form on Z(CG) via traces, for which we write, tr. By using the convolution product (which is really the product on CG), we shall define another bilinear form,. Here it is:, : Z(CG) Z(CG) C éé (α, β) (α β)(1). This is clearly a bilinear form (it is actually symmetric since elements of the center commute with respect to, by definition). What is not clear, for now, is that whether, is non-degenerate. We will show, tr and, are basically the same forms. For this, we make use of the basis {t g : g G} of CG and chase the definition of, tr. For every pair S, T of simple left CG-modules, we have Ñ é e S, e T tr = tr T (e S ) = tr T e S (g)t g g G e S (g) tr T (t g ). (g)
5 CHARACTERS AS CENTRAL IDEMPOTENTS 5 We want to realize this expression as a convolution product of e S with some other function. The expression suggests to define a function χ M : G C g tr M (t g ). for every left CG-module M which is of course the character associated to M! Clearly χ T is a class function, hence lies in Z(CG). Then, writing T for the dual of T, which is also a simple CG-module, we get dim k S δ S,T = e S, e T tr e S (g)χ T (g) e S (g)χ T (g 1 ) e S (g)χ T (g 1 ) = (e S χ T )(1) = e S, χ T. Thus, is non-degenerate, {χ S : S-simple} is also a basis of Z(CG) which the bilinear form, pairs with the basis {e S : S-simple} via e S χ S. But on the other hand because e S s are central primitive idempotents, we have e S, e T = (e S e T )(1) = e S (1) δ S,T. Therefore, also pairs the basis {e S : S-simple} with itself. So e S = e S(1) dim k S χ S. With using extra standard information about the characters, it can be shown that e S (1) = (dim k S) 2 / G, but we don t need this for the skew orthogonality: Given S T simples, then S T are also simples; so e S e T = 0 and χ S, χ T are (nonzero) scalar multiples of e S and e T, respectively; so χ S χ T = 0 as functions from G to C. Thus for every g G, we have 0 = (χ S χ T )(g) h G χ S (gh 1 )χ T (h).
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