# Hochschild (co)homology of differential operators

I googled the title on the internet, and arrived at the following result

$$HH_k(D)\cong H_{DR}^{2n-k}(M).$$

Here $M$ is a smooth manifold of dimension $n$, and $D$ is the ring of differential operators on $M$. My first question is

(A): is there a known chain map realizing the above isomorphism?

I think the above isomorphism remains true if we replace $D$ by the ring of differential operators on $M$ with coefficients in a vector bundle $E$. But I am not so sure. If this is indeed the case, my second question is

(B): question (A) with coefficients in $E$?

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If I remember right this was computed in Wodzicki's paper "Cyclic homology of differential operators", but I don't have access right now to know if there is an explicit chain map. The way I think about this for Hochschild cohomology is via deformation quantization for the cotangent bundle, namely the Hochschild cohomology should be (T_poly(T^*(X), [\omega*,) which is Poisson cohomology with a symplectic form. It is then fairly easy to see gives the same thing as cohomology by using the Poisson structure to turn differential forms into polyvector fields. Maybe that can help construct your map? –  Daniel Pomerleano Mar 11 '12 at 6:25
Hi Daniel, I have the paper with me. The argument there was indeed along your comments. But this is exactly where the problem is: to reduce to the cotangent bundle, one uses the spectral sequence associated to the order filtration on D, which does not hint on a chain level map. –  tu_junwu Mar 11 '12 at 8:17
Do you care whether the chain map goes (Hochschild -> de Rham) or (de Rham -> Hochschild)? Or would it be useful to have a third complex with quasi-isomorphisms to (or from) both? The homology isomorphism is natural with respect to open embeddings (in particular diffeomorphisms), but a nontrivial chain map cannot be natural, at least in the direction (de Rham -> Hochschild). You can see this by letting $n=1$, $M=\mathbb R$, and considering the subcomplexes of chains invariant under affine motions of the line. –  Tom Goodwillie Mar 11 '12 at 13:08
Actually I was even expecting maps in both directions, but this is a good example to know. –  tu_junwu Mar 11 '12 at 18:32