A Reciprocity for Symmetric Algebras
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(2) 202. MARKUS LINCKELMANN AND BERNARD STALDER. Proof. Let s ∈ A∗ and t ∈ B ∗ be symmetrizing forms on A and B, respectively. It is well known (see [1], also [3, Appendix]) that there is an isomorphism of B-A-bimodules Hom A (M, A) ∼ = M∗ f. → s ◦ f. which is functorial in M. Moreover, since M and N are finitely generated projective as left and right modules, we have an isomorphism of B-B-bimodules Hom A (M, A) ⊗ A N ∼ = Hom A (M, N ) f ⊗n. → (m → f (m)n). which is functorial in both M and N . Taking B-fixpoints yields (M ∗ ⊗ A N ) B ∼ = (Hom A (M, A) ⊗ A N )B ∼ = (Hom A (M, N )) B = Hom A⊗B 0 (M, N ). Similarly, there is an isomorphism of B-Abimodules Hom B 0 (M, B) ∼ = M∗ g and we have an isomorphism of A-A-bimodules N ⊗ B Hom B 0 (M, B) n⊗g. → t ◦ g ∼ = Hom B 0 (M, N ) → (m → ng(m)).. As before, taking A-fixpoints yields (N ⊗ B M ∗ ) A (Hom B 0 (M, N )) A = Hom A⊗B 0 (M, N ).. ∼ =. (N ⊗ B Hom B 0 (M, B)) A. ∼ =. R EMARK The proof of Theorem 1 shows that the two expressions in the statement of Theorem 1 are isomorphic to Hom A⊗B 0 (M, N ). In particular, for M = N , this induces algebra structures on (M ∗ ⊗ A M) B and (M ⊗ B M ∗ ) A . Taking derived functors of the fixpoint functors in Theorem 1 yields the following consequence on Hochschild cohomology. C OROLLARY 1 With the notation and assumptions of Theorem 1, we have an isomorphism of graded O-modules H H ∗ (B, M ∗ ⊗ A N ) ∼ = H H ∗ (A, N ⊗ B M ∗ ). Proof. Let P be a projective resolution of M as A-B-bimodule. Then P ∗ = HomO (P, O) is an O-injective resolution of M ∗ . Thus N ⊗ B P ∗ and P ∗ ⊗ A N are O-injective resolutions of N ⊗ B M ∗ and M ∗ ⊗ A N , respectively. Using Theorem 1, we have isomorphisms of cochain complexes Hom B⊗O B 0 (B, P ∗ ⊗ A N ) ∼ = (P ∗ ⊗ A N ) B ∼ = (N ⊗ B P ∗ ) A ∼ = Hom A⊗O A0 (A, N ⊗ B P ∗ ). Taking cohomology yields the statement. Let A be an O-algebra. Following the terminology in [2,3] (which generalizes [4]), an interior Aalgebra is an O-algebra B endowed with a unitary algebra homomorphism σ : A → B. If A, B are O-algebras, C is an interior B-algebra and M is an A-B-bimodule, we set Ind M (C) = EndC 0 (M ⊗ B C), considered as an interior A-algebra via the homomorphism A → Ind M (C) sending a to the C 0 -endomorphism given by left multiplication with a on M ⊗ B C..
(3) A RECIPROCITY FOR SYMMETRIC ALGEBRAS. 203. T HEOREM 2 Let A, B be symmetric O-algebras and let M be an A-B-bimodule which is finitely generated projective as left and right module. There is a canonical anti-isomorphism of O-algebras (Ind M (B)) A ∼ = (Ind M ∗ (A)) B . Proof. We have Ind M (B) = End B 0 (M) and Ind M ∗ (A) = End A0 (M ∗ ). Since taking O-duality is a contravariant functor, this algebra is isomorphic to End A (M)0 . Taking fixpoints completes the proof. The group algebra OG of a finite group G is a symmetric algebra. More precisely, OG has a canonical symmetrizing form, namely the form s : OG → O mapping a group element g ∈ G to zero if g = 1 and to 1 if g = 1. Following the terminology of Puig [4], an interior G-algebra is an O-algebra endowed with a group homomorphism σ : G → A× . Such a group homomorphism extends uniquely to an O-algebra homomorphism OG → A, and thus A becomes an interior OGalgebra (and vice versa). If H is a subgroup of G and B an interior H -algebra, the induced algebra IndG H (B) defined in [4] is the O-module OG ⊗O H B ⊗O H OG endowed with the multiplication (x ⊗ b ⊗ y)(x ⊗ b ⊗ y) = (x ⊗ byx b ⊗ y ) provided that yx ∈ H , and 0 otherwise, where x, y, x , y ∈ G and b, b ∈ B. The algebra IndG H (B) is viewed as an interior G-algebra with the structural homomorphism mapping x ∈ G to y∈[G/H ] x y ⊗ 1 B ⊗ y −1 . For B = O H , we have the obvious identification IndG H (O H ) = OG ⊗O H OG, with multiplication given by (x ⊗ y)(x ⊗ y ) = x ⊗ yx y if yx ∈ H and 0 otherwise, where x, y, x , y ∈ G. The previous notion of algebra induction is consistent with this concept. L EMMA 1 Let G be a finite group, H a subgroup of G and let B be an interior H -algebra. Set M = OG H . There is an isomorphism of O-algebras ∼ IndG = Ind M (B) H (B) (x ⊗ b ⊗ y). → (z ⊗ c → x ⊗ byzc if yz ∈ H and 0 otherwise) ,. where x, y, z ∈ G and b, c ∈ B. Proof. Straightforward verification. T HEOREM 3 (Stalder [5]) Let G be a finite group, let H, K be subgroups of G. Consider OG as O H -OK -bimodule via multiplication in OG. Then there is an isomorphism of O-algebras ∼ G K H −→ (IndG (Ind H (O H )) K (OK )) kx ⊗ yk −→ hx ⊗ yh, k∈[K /K (x⊗y) ]. h∈[H/H(x⊗y) ]. where K (x⊗y) is the stabilizer in K of x ⊗ y ∈ IndG H (O H ) under the action of K and H(x⊗y) is the stabilizer in H of x ⊗ y ∈ IndG (OK ) under the action of H . K There are (at least) three ways to go about the proof of Theorem 3: by explicit verification or by interpreting Theorem 3 as special case of either Theorem 1 or Theorem 2. We sketch the three different proofs..
(4) 204. MARKUS LINCKELMANN AND BERNARD STALDER. Proof 1 of Theorem 3. The image of the set G × G in IndG H (O H ) = OG ⊗O H OG is an O-basis K which is permuted under the action of K by conjugation. Thus the subalgebra (IndG H (O H )) of K K -stable elements has as O-basis the set of relative traces Tr K (x⊗y) (x ⊗ y), where x, y ∈ G. If x, x , y, y ∈ G and k ∈ K such that kx ⊗ yk = x ⊗ y in IndG H (O H ), there is a (necessarily unique) h ∈ H such that kx = x h and yk = hy , which in turn is equivalent to the equality hx ⊗ yh = (x ) ⊗ (y ). in IndG K (OK ). Thus the map x ⊗ y → x ⊗ y induces a bijection between the sets of K -orbits G and of H -orbits of the images of G × G in IndG H (O H ) and Ind K (OK ), respectively. In particular, with the notation above, we have k ∈ K (x⊗y) if and only if h ∈ H(x⊗y) , and the correspondence k → h induces a group isomorphism K (x⊗y) ∼ = H(x⊗y) . From this it follows that the map given in Theorem 3 is an O-linear isomorphism. It remains to verify that this is an algebra homomorphism. In IndG H (O H ), multiplication is given by x ⊗ yzt if yz ∈ H, (x ⊗ y)(z ⊗ t) = 0 otherwise, −1 ⊗ t −1 (yz)−1 and z −1 ⊗ where x, y, z, t ∈ G. If yz ∈ H , then in IndG K (OK ), the elements (yz)z G −1 t are in the same H -orbit, and the multiplication in Ind K (OK ) yields (x −1 ⊗ y −1 )((yz)z −1 ⊗ t −1 (yz)−1 ) = x −1 ⊗ t −1 z −1 y −1 , and this corresponds precisely to the bijection between the sets of G K -orbits and H -orbits of the images of the set G × G in IndG K (OK ) and Ind H (O H ), respectively.. Proof 2 of Theorem 3. We are going to apply Theorem 1 to the particular case where A = O H , B = OK , M = N = OG viewed as A-B-bimodule (through the inclusions H ⊆ G, K ⊆ G). This yields an O-linear isomorphism ((OG)∗ ⊗O H OG) K ∼ = (OG ⊗ (OG)∗ ) H . OK. Composing this with the canonical isomorphism (OG)∗ ∼ = OG mapping f ∈ (OG)∗ to −1 x∈G f (x )x yields the isomorphism in Theorem 3. Proof 3 of Theorem 3. Applying Theorem 2 and the above lemma to A = OK , B = O H and G K ∼ H M = OG as A-B-bimodule yields an anti-isomorphism (IndG H (O H )) = (Ind K (OK )) . The map G sending x ⊗ y to y ⊗ x is an anti-automorphism of Ind H (O H ) which induces an anti-automorphism K of (IndG H (O H )) . Composing both maps yields again the isomorphism in Theorem 3. R EMARK The proof 3 of Theorem 3 is essentially the proof given in [5, §11]. References 1. M. Brou´e, On Representations of Symmetric Algebras: an Introduction, Notes by M. Stricker, Mathematik Department, ETH, Z¨urich, 1991. 2. B. K¨ulshammer, T. Okuyama, and A. Watanabe, A lifting theorem with applications to blocks and source algebras, preprint, 1999..
(5) A RECIPROCITY FOR SYMMETRIC ALGEBRAS. 205. 3. M. Linckelmann, Induction for interior algebras, Quart. J. Math. 53 (2002), 195–200. 4. L. Puig, Pointed groups and construction of characters, Math. Z. 176 (1981), 265–292. 5. B. Stalder, Une extension de l’alg`ebre de groupe en th´eorie des repr´esentations modulaires, Th`ese, Universit´e Lausanne, 2000..
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