Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a given commutative algebra $A$ over a field $\mathbb{K}$(with char=0) the algebra of differential operators on $A$ is the set of endomorphism $D$ of $A$ such for some $n$ we have that for any sequence $\left\lbrace a_i\right\rbrace_{1\leq i\leq n}$ for $a_i\in A$ one has $[\ldots[[D,a_0],a_1],\ldots,a_n]=0$.

By analogy, I think of Hochschild Cohomology as a sort of "algebraic differential forms"(maybe this isn't the right approach, but $C^{n}(A,A)=Hom(A^{\otimes n},A)$ and it gives a cohomology theory etc.), thus it seems like there should be a connection to differential operators on the algebra.

Sorry if the question is a little general, but I am open to accepting a variety of answers.

Thanks in advance!

share|cite|improve this question
The first sentence seems to contain some kind of structural error: I am not so knowledgeable in this field, but I doubt that the alebra of differential operators on $A$ is an endomorphism of $A$... – Pete L. Clark Jun 4 '11 at 16:52
hehe, thanks @Pete L. Clark and @Mariano, the edit was what I intended. :) – BBischof Jun 4 '11 at 17:06

If $X$ is an smooth affine variety over a field of characteristic zero, with coordinate ring $\mathcal O$, the ring of global differential operators $\mathcal{D}$ on $X$ is the same thing as the Grothendieck algebra of differential operators on $A$.

One can show that the Hochschild homology $HH_\bullet(D)$ of the algebra $\mathcal D$ is, according to a theorem of Mariusz Wodzicki, precisely the same thing as the algebraic de Rham cohomology $H_{\mathrm{dR}}(X)$ of $X$, which is the same thing as the de Rham cohomology of $A$ and, if the base field is $\mathbb C$, classical theorems of Grothendieck and Hartshorne then tell you that these two latter cohomologies are precisely the same thing as the topological de Rham cohomology of the analytic variety $X_{\mathrm{an}}$.

(All this can be globalized to schemes, but one needs to be careful. For example, not all schemes are $\mathcal D$-affine, &c.)

share|cite|improve this answer
Sorry, but I have a few more questions, first is $\mathcal{O}$ the same as $A$ here? Second, what do you mean de Rham of $A$? If you mean de Rham on the variety with coordinate ring $A$, then how is that different than the topological de Rham. Sorry if these questions are stupid. – BBischof Jun 4 '11 at 22:41
@BBischof: Indeed: $A$ was supposed to be $\mathcal O$ here. By "de Rham cohomology of $A$" I mean "construct the module of Kähler differential forms on $A$, from it construct its exterior algebra and define a differential in complete analogy of the exterior differential, and tak e the homology of the resulting complex". It is a non-trivial result that this gives the same thing (o0ver $\mathbb C$) as the topological de Rham cohomology of the underlying analytic variety. – Mariano Suárez-Alvarez Jun 6 '11 at 16:10
BBischof: Yes, here in the affine situation, $\mathcal{O}$ is $A$. As for de Rham of $A$, Mariano means the algebraic de Rham cohomology, see e.g. – Kevin H. Lin Jun 6 '11 at 16:13

Like Grigory M says, look up the Hochschild-Kostant-Rosenberg theorem.

This states that for a smooth algebra $A$, the Hochschild chain complex is quasi-isomorphic to the chain complex $(\Omega^\bullet_A, d=0)$ of differential forms with zero differential.

There is another version of HKR which states that, again for smooth $A$, the Hochschild cochain complex is quasi-isomorphic to the cochain complex $(\Lambda^\bullet T_A, d=0)$ of polyvector fields with zero differential. So here you see (poly)vector fields (i.e. derivations), so maybe that gives you one connection to differential operators.

Now in this second version of HKR, these are actually more than just chain complexes but dg Lie algebras. The formality theorem of Kontsevich -- it's in his deformation quantization paper -- says that, as dg Lie algebras (or $L_\infty$ algebras) the two sides are still quasi-isomorphic.

Moreover, if you look at the deformation quantization paper, you'll notice that Kontsevich's definition of Hochschild cohomology is not the "standard" definition (that is, Hochschild's original definition) involving $\operatorname{Hom}(A^{\otimes n}, A)$. Instead, he takes the subcomplex of the Hochschild cochain complex $\operatorname{Hom}(A^{\otimes n}, A)$ consisting of those maps which are polydifferential operators. [However, see e.g. the paper "The Continuous Hochschild Cochain Complex of a Scheme" by Yekutieli for comparison of different definitions of Hochschild cohomology (both for algebras and more generally for schemes).] Then he shows that this thing is quasi-isomorphic to $(\Lambda^\bullet T_A, d=0)$ (as a chain complex and as a dg Lie algebra). So that gives you another connection to differential operators...

share|cite|improve this answer

The usual way to relate Hochschild (co)homology of a commutative algebra to "differential-geometric" concepts is the Hochschild-Kostant-Rosenberg theorem: if A is a smooth commutative k-algebra, there exists a (graded) isomorphism $\Omega^\bullet A\to HH_\bullet(A,A)$ (see e.g. theorem 3.4.4 in Loday's Cyclic homology book or the nLab entry and links there).

Now, I don't know about any direct connection to differential operators. But there is a general way to get differential operators from differential forms: to apply (relative) Koszul duality to de Rham complex (it's described... well, e.g. in Positselski's "Two kinds of derived categories...", although it's an overkill, perhaps; see also "What is Koszul duality?").

Hope, it helps.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.