# Hochschild Cohomology of Differential Operators in characteristic 0

In Mariusz Wodzicki's paper "Cyclic homology of differential operators," the following result is mentioned: for $D_M$ the algebra of differential operators on a smooth manifold $M$ we have that $HH_n(D_M) \cong H_{DR}^{2m-n}(M)$ where $m=\dim M$. I'm having trouble finding a reference for the Hochschild Cohomology of $D_M$. Does anyone know of a paper with the result?

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Hi,

This is an instance of a more general fact concerning deformation quantization of symplectic varieties. The general theorem is:

`Let $X$ be an symplectic manifold, and let $A_\hbar$ be any quantization of the Poisson algebra $(C^\infty(X), \{, \})$. Then we have $HH^*(A_\hbar)\cong H_{DR}^*(X)$.

See the paper:

Hochschild cohomology and Characteristic Classes by Weinstein and Xu

for an early proof. As Jeremy noted, one proof of this goes through spectral sequences. Morally the point, as I understand it, is that there is a canonical quantization of a symplectic vector space, which is the Weyl algebra, and which has trivial Hochschild cohomology. Since every symplectic variety is locally symplectomorphic to a symplectic vector space, you can compute the Hochschild cohomology of $HH^*(A_\hbar)$ on some covering of $X$ via a Cech complex consisting of constant sheaves at each basic open set (constant sheaves because you choose the covering fine enough so that the symplectic form can be put in canonical form, and then you use the result about the Hochschild cohomology of the Weyl algebra). This is also how you compute DeRham comology, so the answers coincide.

The reason this relates to differential operators is that for a smooth manifold $M$, its cotangent space $T^*M$ is a symplectic manifold, and we can regard $D_M$ as a quantization of $C^\infty(T^*M)$. as above. So we find that $HH^*(D_M)\cong H_{DR}^*(T^*M)\cong H_{DR}^*(M)$, since $T^*M$ contracts onto $M$.

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There are many references for this fact, which is essentially a spectral sequence argument along with the Poincare Lemma. The full proof is in section 5 of "THE HOMOLOGY OF ALGEBRAS OF PSEUDO-DIFFERENTIAL SYMBOLS AND THE NON-COMMUTATIVE RESIDUE" by Brylinski and Getzler.

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Dear Jeremy, I am not finding any discussion of cohomology in this paper, just the homology. –  KReiser May 17 '12 at 4:50
I think about the same argument will work for cohomology but instead you should get something like $HH^*(D_M)\simeq H_{DR}^*(M)$ but this time with no shift in the grading. This is like a Poincare duality between Hochschild cohomology and Hochschild homology. –  Jeremy Pecharich May 17 '12 at 5:14

I'm afraid I don't know the literature well enough to give a proper reference (my default would have been the paper Jeremy suggested) but I wanted to point out that at least in the algebraic setting such a result follows from results describing functors on D-modules as integral transforms (in the spirit of Goncharov's papers on D-modules and Integral Transforms, work of Kashiwara, Schapira, d'Agnolo, Polesello, and others --- a general theorem of this kind and more references can be found in this paper. Namely once you know that functors on D-modules are identified with D-modules on the product, it follows that Hochschild cohomology - ie self-ext of the identity functor - can be calculated by self-Ext of the diagonal, which for smooth schemes is exactly the de Rham cohomology of the scheme itself.

[One interesting consequence however of the theorems on integral transforms is that in the equivariant setting -- ie on stacks -- the Hochschild cohomology is bigger: you get the de Rham cohomology not of your original space but of its inertia groupoid (aka derived loop space) - ie you see "twisted sectors", cohomology of the stabilizer groups of the points in X..]

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Thank you for your comments and alternate perspective! –  KReiser May 19 '12 at 5:32
Do you have any reference for the assertion about Hochschild cohomology on stacks? –  Clemens Koppensteiner Apr 5 '14 at 23:56
Thanks for pointing out I'm really calculating Hochschild homology not cohomology in the parenthetical comment. Please update if you find a good argument for the cohomology! –  David Ben-Zvi Apr 24 '14 at 16:51