Monday, 25 July 2011

How exactly is Hochschild homology a monad homology?

This is partly in response to Reid, but also intended as general clarification.



As I understand it, Peter's original question was:



-- here is the Hochschild chain complex for an algebra $A$ and bimodule $M$, as defined in Hochschild's original papers;
-- it is the chain complex associated to a certain simplicial object as defined on the Wikipedia page;
-- one is told that this object comes from the bar construction (or standard resolution) associated to some monad;
-- where/what is the monad?



The last one seems to be Reid's underlying point/question. Tyler says you can get it, up to a dimension-shift, from the adjunction between k-modules and k-algebras (at least when $A=M$). My earlier recollection was that this more naturally leads to cyclic homology a.k.a. additive K-theory as defined by Feigin and Tsygan, but I have yet to check this against a copy of their paper. (The point is that in characteristic zero, the cyclic homology of a free tensor algebra on a given k-module, coincides with the cyclic homology of the ground field, so one can take free resolutions of a given $k$-algebra and then use spectral sequence arguments.) On reflecting a bit more, because the Hochschild homology of a free (=tensor) algebra is confined to degrees 0 and 1, perhaps one can also obtain $H_n(A,M)$ as Tyler suggests, by taking the free algebra resolution of A (in the category of k-algebras) and then hitting the resulting simplicial object with a suitable functor - but this seems trickier than in the commutative case (Andre-Quillen) and I can't get hold of a copy of Quillen's paper at the moment.



Alors. As I understand it, following Weibel's book (and the papers of Barr & Beck et al), the simplicial object (in the category of $k$-modules) that yields the Hochschild chain complex, arises by applying a certain Hom-functor (namely ${}_A{rm Hom}_A( cdot ,X)$ ) to another simplicial object, say $beta(A)$, in the category of $A$-bimodules.



Now $beta(A)$ is not contractible in the category of $A$-bimodules, in general, and doesn't come from a (co)monad on that category. However, $beta(A)$ can be identified with another simplicial object $F(A)$, which lives in the category of $A$-modules.



What is $F(A)$?



Well, take a step back and consider the adjunction between $k$-modules and $A$-modules (maybe you need $k$ to be a field at this point, maybe not). That gives rise to a bar construction in $A$-mod, namely for any given $M$ in $A$-mod one obtains a simplicial object $F(M)$ which is given in each degree by



$$ F_{-1}(M)=Mquad,quad F_n(M) = M otimes A^{otimes n+1} {rm for } n geq 0. $$



Note that this is contractible in $A$-mod by the general machinery of the bar resolution associated to a monad. There was nothing to stop us taking $M=A$, that's a perfectly good $A$-module; and on doing so, lo and behold, we get the same simplicial object $F(A)$.



Thus, Hochschild homology, regardless of the choice of coefficients, can be thought of as "coming from" a comonad - namely, that induced on $A$-mod by the forgetful functor from $A$-mod to $k$-mod. In my opinion, that is probably the (co)monad they are talking about.



It so happens that, since $F(A)$ is contractible in $A$-mod and hence a fotiori in $k$-mod, the "chain-complex-ification" of $beta(A)$ is, as a chain complex in $R$-bimod, a resolution of $R$ by $k$-relatively projective $R$-bimodules -- and hence applying ${}_R{rm Hom}_R( cdot ,X)$ to it and taking homology coincides with taking $k$-relative Tor of $R$ and $X$ as R-bimodules. Hence the point of view that Hochschild homology is a special case of relative Tor.



Finally, I actually agree with Reid that this is not the best example to motivate (co)monad (co)homology. Group cohomology with coefficients in the ground field; or indeed André-Quillen cohomology, which is given by a "free algebra" adjunction but only for commutative algebras, or sheaf cohomology, would be better. (No originality in my choices; I've cribbed them out of Weibel Section 8.6).



(Apologies for the length and the tediousness, by the way.)

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