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The Hochschild-Kostant-Rosenberg theorem says, that for a commutative algebra $R$ over a field $k$ with certain smoothness and finiteness, we have an identification $\mathrm{HH}^\bullet(R)\cong \wedge_R^\bullet\mathrm{Der}_k(R)$. Namely, the Hochschild cohomology of $R$ is the space of poly-vector fields, whose element is just a bunch of first-order differential operators tensored together.

Question: What would be an analogous statement for differential operators of higher order?

I guess higher Hochschild cohomology may have something to do with them (with the only reason being the word 'higher' is in its name; I know nothing about it). Also, I've heard that $\mathrm{Ker}(\mathrm{mult}\colon R^{\otimes n}\to R)$ is the space of universal $n$th-order differentials, so I expect this should be incorporated into the bar complex of $R$ of some kind.

Any comments are appreciated.

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  • $\begingroup$ Just a remark. Derivations are not the most general first order differential operators (the latter don't always annihilate constants, but derivations always do). So if HH doesn't even capture all first order differential operators, what hope is there for it to capture higher order operators? $\endgroup$
    – Igor Khavkine
    Commented Sep 21 at 22:51
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    $\begingroup$ I would rather say differential operators are a LOWER analog of Hochschild cohomology (or better, Hochschild cochains). The former are generated by (in fact the "universal enveloping algebroid" of) the tangent sheaf T, and quantize the cotangent bundle, while the latter are generated by (and form a "universal enveloping E_2 algebra on") the shift T[-1], and quantize the shifted cotangent bundle T^*[1], and can be considered "categorified differential operators" - see the work of Beraldo, eg arxiv.org/abs/1709.07867 and arxiv.org/abs/1801.03752 $\endgroup$
    – David Ben-Zvi
    Commented Sep 22 at 1:29

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I do not have an analogous statement for higher-order differential operators, and I do not think that "higher Hochschild homology" is relevant here. But these remarks might be of interest:

The composition of two first-order differential operators is a second-order operator, and so on. Composition is not commutative. But if you filter the algebra of all differential operators according to order, then the associated graded algebra is commuative. For example, if $X$ and $Y$ are first-order then $XY-YX$ is first order.

So the quotient of second-order operators by first-order operators receives a map from $S^2Der$, the symmetrized quotient of $Der\otimes Der$. I believe that with smoothness assumptions, and assuming characteristic zero, $S^nDer$ is isomorphic to the quotient of $n$th order operators by $(n-1)$st order operators.

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