Nikita Markarian’s mathblog

Equivariant Hochschild homology

Posted in Mathematics by nikitamarkarian on April 18, 2011

Here is the appendix to the forthcoming paper “Crystalline characteristic classes via the Atiyah gerbe and generalized Chern-Weil construction”.

Equivariant Hochschild homology of an algebra with an algebraic group action is homology of a mixed complex, that is a complex with an extra differential $B$ of degree -1 commuting with the main differential. For a trivial group this is the usual Hochschild complex. I believe that if the group is finite, then the equivariant Hochschild homology is the Hochschild homology of the semidirect product of the algebra and the group algebra. If  the algebra is trivial, then the equivariant Hochschild homology is the complex of functions on the cyclic nerve of the group, that is the cohomology of the group with coefficients in the functions on the group under the adjoint action. The aim is to show that the operator $B$  is up to a coboundary equal to the standard operator that is induced by the cyclic structure on the cyclic nerve. This is a routine consequence of routine facts about cyclic sets. I did not find exact references that is why I devoted first three subsections to a reminder about cyclic stuff.

The only new concept in these three parts is functor $\mathop{Tot}$ that produces a cyclic set from a cocylcic simplicial set. The point is this functor respects “circle action”. To be more precise, when passing to the chain complex of a cyclic set and to the total chain-cochain  complex of a cocylcic simplicial set functor $\mathop{Tot}$ respects (up to a boundary) operators $B$ that are naturally defined on both sides.

This concept implicitly appeared in the classical papers on cyclic homology as follows. Given a cyclic set, on its chain complex an operator of degree -1 naturally acts and it gives an operator of degree -1 on homology. On the other hand, the  geometric realization of the cyclic set is acted by circle. This also gives an operator of degree -1 on homology. The problem is to prove that this is the same operator.  Informally the question is as follows. Consider the set of $Hom$s of the cyclic category $\Lambda$. It is naturally a cyclic and a cocyclic set. Roughly speaking one needs to prove that these two “circle actions” (“from the right” and “from the left”) are homotopy equivalent to each other.

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1. Anonymous said, on September 7, 2012 at 9:20 pm

What is equivariant Hochschild homology for a reductive G acting on an algebra A over the complex numbers?