Some bialgebroids constructed by Kadison and Connes-Moscovici are isomorphic.pdf

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Some bialgebroids constructed by Kadison and Connes-Moscovici are isomorphic.pdf

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Some bialgebroids constructed by Kadison and Connes-Moscovici are isomorphic

a r X i v : m a t h / 0 5 0 8 6 3 8 v 1 [ m a t h .Q A ] 3 1 A u g 2 0 0 5 Some bialgebroids constructed by Kadison and Connes-Moscovici are isomorphic ? Florin Panaite Institute of Mathematics of the Romanian Academy PO-Box 1-764, RO-014700 Bucharest, Romania e-mail: Florin.Panaite@imar.ro Freddy Van Oystaeyen Department of Mathematics and Computer Science University of Antwerp, Middelheimlaan 1 B-2020 Antwerp, Belgium e-mail: Francine.Schoeters@ua.ac.be Abstract We prove that a certain bialgebroid introduced recently by Kadison is isomorphic to a bialgebroid introduced earlier by Connes and Moscovici. At the level of total algebras, the isomorphism is a consequence of the general fact that an L-R-smash product over a Hopf algebra is isomorphic to a diagonal crossed product. 1 Introduction Let H be a Hopf algebra with bijective antipode, A a left H-module algebra and denote as usual Ae = A?Aop. In [10], Kadison constructed a bialgebroid (in the sense of [12]) with base A and having as total algebra a certain algebra structure on Ae ?H, which we denote by Ae ?H. We first note that this algebra structure is actually an L-R-smash product Ae ? H as introduced in [14]. An L-R-smash product is isomorphic to a diagonal crossed product as in [9], [6], hence Ae ? H ? Ae ?? H. Finally, the diagonal crossed product Ae ?? H is isomorphic to a certain algebra, which we denote by A ⊙ H ⊙ A, used by Connes and Moscovici in [8] (see also [11]) and which is also a bialgebroid over A. It turns out that the resulting algebra isomorphism is actually a bialgebroid isomorphism between Ae ?H and A⊙H ⊙A. We establish also a universal property of Ae ?H as a bialgebroid and we give an example (up to isomorphism) of the type A⊙H ⊙A: the Cibils-Rosso algebra from [7]. 2 The isomorphism We work over a ground field k. All algebras, linear spaces etc. will be over k; unadorned ? means ?k. ThroughoutH will be a Hopf algebra with bijective antipode S. We use the following version of Sweedl