Background: in deformation quantization one wants to construct formal associative deformations (the star products $\star$) of the algebra of smooth complex-valued functions on a Poisson manifold $M$ in direction of the Poisson bracket. The cochains in order $\hbar^r$, where $\hbar$ is the deformation parameter, are not just any $2$-cochains of $C^\infty(M)$ but typically one requires them to be either continuous with respect to the usual $C^\infty$-topology, or even bi-differential. Beyond the deformation of $C^\infty(M)$ one also is interested in deforming all kinds of modules over $C^\infty(M)$ once a star product $\star$ is fixed in such a way that one still obtains modules. Also here the "classical" modules usually come from geometry (sections of bundles etc) and hence allow for the notion of differential cochains.

The deformation problem is mainly governed by the Hochschild cohomology of $C^\infty(M)$ with values in the corresponding endomorphisms of the "classical" module one wants to deform (or the algebra itself). However, the additional requirement of being differential brings us to a sub-complex. Thus the computation of its cohomology is a new problem which in the cases I know has to be done more or less by hand...

For a general commutative algebra one still can define mutlidifferential operators (with values in modules) a la Grothendieck in a completely algebraic way (reproducing the above differential operators in the case of $C^\infty(M)$). This gives a sub-complex of the Hochschild complex which I would like to understand: for the algebraic situation and also for continuous cochains (under some conditions) one has the usual tools of homological algebra to identify the Hochschild cohomology as certain Ext groups etc.

However, the additional requirement of being differential does not seem to fit into this nice algebraic abstract nonsense theory. I guess it would be very nice to have these sort of tools also available in the differential setting, which, after all, is entirely algebraic in its nature.

So my question is: are there any possibilities to transfer the usual notions of Ext etc to the differential case?

  • $\begingroup$ I don't understand what exactly it is you want to know. Wat do you mean exactly by "the differential case"? What notions of "Ext etc" do you want to transfer? $\endgroup$ – Mariano Suárez-Álvarez Feb 3 '11 at 0:48
  • $\begingroup$ Well, things like: Hochhschild cohomology is an Ext group and hence we can choose the projective resolution we want to compute it. This sometimes is very handy to replace the rather large bar complex by some smaller and easier one (like Koszul...) But since the differential Hochschild complex is a sub-complex, I don't see why this should be possible a priori. IN the case of $C^\infty$ functions one sort of checks by hand that certain other complexes compute the same cohomology. But I wonder if there is a good principle behind this. $\endgroup$ – Stefan Waldmann Feb 3 '11 at 12:44
  • $\begingroup$ Isn't it true that the inclusion of differentiable cochains into continuous cochains is a quasi-isomorphism ? Then working with the category of topological algebras should be OK. $\endgroup$ – DamienC Apr 24 '11 at 23:14
  • $\begingroup$ Hi Damien, nice to meet you this way ;) Well, I have no idea whether this is true in general. I know that it works for $C^\infty(M)$ for a smooth manifold $M$, but beyond? and a´for values in general bimodules? I know that for a infinite-dim vector space $V$ with topology there are different topologies on the symmetric algebra $S^\bullet(V)$ (something like the polynomials on the pre-dual of $V$ if it has some) such that the completed algebras have quite different Hochschild cohomologies. Moreover, not every multidifferential operator on $S^\bullet$ is continuous at all...??? $\endgroup$ – Stefan Waldmann Apr 25 '11 at 12:34

If I understand the question correctly Stefan is asking for an Ext interpretation of the polydifferential Hochschild cochain complex. Elements of this are not just continuous linear maps $C^\infty(M)^{\otimes n} \to C^\infty(M)$, but they have to be polydifferential operators. (This version of Hochschild cohomology is used in Kontsevich's formality theorem).

Anyway, one can understand the polydifferential condition as follows. Consider the jet bundle $J$ on $M$; this is an infinite-rank vector bundle whose fibre at a point $p \in M$ is the algebra of formal power series at $p$. If we choose coordinates $x_1,\dots, x_n$ at $p$, then we can identify the fibre $J_p$ as $\mathbb{R}[[x_1,\dots,x_n]]$.

It's standard that $J$ is a left $D$-module. Further, the obvious product on the fibre of $J$ makes $J$ into a commutative algebra in the symmetric monoidal category of left $D$-modules.

Then, one can take Hochschild cochains of $J$ in the symmetric monoidal category of left $D$-modules.

This is the same as the complex of poly-differential Hochschild cochains. The key point is that $D$-module maps $J^{\otimes n} \to J$ are the same as polydifferential operators.

Of course, this means that you can apply any of the standard interpretations of Hochschild cohomology in this context (e.g. $\operatorname{Ext}_{J \otimes J}(J,J)$).

One needs a little care with these definitions, because $J$ is a topological $D$-module. However, if you take continuous $D$-module maps and appropriately completed tensor products you get the right answer.

  • $\begingroup$ Dear Kevin: thank for this point of view. My question was probably a little bit missleading as the algebras I have in mind are not the smooth functions but something more general. For smooth functions, the differential Hochschild cohomology is fine. In fact, my concerns are algebras like functions (whatever reasonable class) on some infinite-dim spaces/manifolds. I fear that for such examples much of $D$-module theory etc does not apply. $\endgroup$ – Stefan Waldmann Apr 26 '11 at 13:44

Let me try an answer.

  1. It seems to me that the appropriate language to use is the one of ringed spaces. For a given ring space $(X,\mathcal{O}_X)$ one can consider the category of sheaves of right $\mathcal{O}_X$-modules (see e.g. Section 7 of http://alpha.uhasselt.be/Research/Algebra/Publications/hochschild_ab.pdf, where you will find some evidences for such an approach).

  2. Another approach, more in the spirit of David Ben-Zvi answer is to use the notion of Lie algebroid. In http://arxiv.org/abs/0908.2630 we interpret differential Hochschild cohomology as Ext's of $J_L$-modules, where $J_L$ is the jet algebra sheaf associated to the Lie algebroid $L=(\mathcal{O},Der(\mathcal{O}))$ (i.e. the $\mathcal{O}$-dual of its universal enveloping algebra).

  3. Maybe yet another point of view could help. $J_L$ is the algebra of functions on the formal groupoid integrating the Lie algebroid $L$. Now for a groupoid $G=(G_1,G_0)$ we can consider $Ext_{G_1}(e_*(-),e_*(-))$, where $e:G_0\to G_1$ is the unit inclusion. Now it appears that this only depends on the inclusion of $G_0$ into its formal neighbourhood in $G_1$. In other words, it only depends on $J_L$, where $L$ is the Lie algebroid of $G$. SO (up to details) 2 and 3 are equivalent.

In David's answer $G=(X,X\times X)$, and its associated Lie algebroid is $(\mathcal{O}_X,Der(\mathcal{O}_X))$. In this case the relation between 3 and 1 is again explained in Section 7 of http://alpha.uhasselt.be/Research/Algebra/Publications/hochschild_ab.pdf for schemes. For complex analytic and differentiable manifolds the proof is basically the same, but you will probably have to use http://arxiv.org/abs/0908.2630

I hope this can help.

  • $\begingroup$ Dear Damien: thanks again. I will take a look if these ideas apply to the setting I have in mind. I was probably not very specific here. The main concern will be to apply this to function spaces (smoothness?) on some infinite-dim manifolds/spaces. I hope that some of the ideas still apply to this situation, but bad surprises lurk at every corner... $\endgroup$ – Stefan Waldmann Apr 26 '11 at 13:47
  • $\begingroup$ argh... then I am afraid that it will not work (since we use some finitness assumption), unless you can write these spaces as (co)limits of finite dimensional ones. $\endgroup$ – DamienC Apr 26 '11 at 15:11
  • $\begingroup$ Hmm... That might still be an interesting situation. However, at least for the usual distribution/test function space and their symmetric algebras I have no idea whether this can be done. I fear not... :( Anyway, thanks again.. $\endgroup$ – Stefan Waldmann Apr 26 '11 at 15:36

Given any bimodule over a commutative ring (or over a scheme, i.e. coherent sheaf on the product) we can consider its differential part, namely the part supported set-theoretically on the diagonal. Applying this to End(R) we find Grothendieck's definition of differential operators. Can you not repeat the Hochschild story working strictly in the world of differential bimodules? In this case you get a quasiisomorphic complex to the Hochschild complex -- ie in the algebraic setting self-Exts of the identity functor are already supported on the diagonal. (Put another way, if everything in sight is coherent, differential bimodules -- such as the diagonal -- are I think a full subcategory of bimodules, so it doesn't hurt to restrict to them). Is this the kind of picture you're looking for?

Another trick that may be useful is the setting of induced D-modules: there's an equivalence of categories between the category with objects quasicoherent sheaves and morphisms given by differential operators, and the full subcategory of D-modules given by induced D-modules (for right D-modules, these are things of the form $M\otimes_{O_X} D_X$). So working with induced D-modules might be a convenient place to think of differential Hochschild theory..

  • $\begingroup$ @David: Thanks for the long answer. I will have to think about it. The only problem I see is that the algebras I'm intersted are like smooth functions etc for which the usual machinery of algebraic geometry might be more tricky (at least, I'm not so much familiar with these techniques in the context of smooth functions). BUt I will take a closer look at what you suggest. Thanks again. $\endgroup$ – Stefan Waldmann Feb 3 '11 at 12:47

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