Let $X$ be your "favourite" kind of space, and let $\mathcal{D}_X$ be the (sheaf of) ring(s) of differential operators on $X$. What does the ring $\mathcal{D}_X$ tell us about $X$?

I know this might be too broad or naive a question, but the literature seems to be equally vast and scattered. Browsing the canonical references on rings of differential operators (Bjork, McConnell & Robson) and/or $\mathcal{D}$-modules (Bjork, Borel, Hotta et al), quite a bit is known in the forward direction, e.g. if $X$ is smooth over $\mathbb{C}$ then $\mathcal{D}_X$ is $\mathcal{O}_X$-quasicoherent and (left and right) noetherian; $\text{gr}\mathcal{D}_X\simeq \mathcal{O}_{T^*X}$; ...

In fact, one of the points of this question is also to have an idea for what kind of "favourite" spaces this is an approachable problem, and what is known out there in the literature. In particular,

If $\mathcal{D}_X\simeq \mathcal{D}_Y$ as rings, what does this tell us about the relation between $X$ and $Y$? That is, how much the isomorphism class of rings $\mathcal{D}_X$ is an interesting object to understand $X$? And, in a more Tannakian fashion, how much of $X$ can we recover from $\mathcal{D}_X$?

References most welcome. Please, re-tag if appropriate.


1 Answer 1


The paper

  • J. Grabowski: Isomorphisms and Ideals of the Lie Algebras of Vector Fields, Inventiones Math. 50, 13-33 (1978)

shows that any smooth manifold (or real analytic manifold, or Stein manifold) is completely determined by its Lie algebra of vector fields of the appropriate kind. So if you can extract the filtration by degree from your ring of differential operators, you get your hand on the Lie algebra of vector fields and you can use the result.

Moreover, it is well known, that the algebra of smooth functions determines the manifold (this is called Milnor's exercise). See page 296 of here for a short proof. Now, the algebra of smooth functions is the space of global sections of the zero order part of the sheaf of differential operators.

  • $\begingroup$ Good, thank you, that's what I was aiming at first, but wasn't so sure. So, essentially, at least in the smooth setting and in characteristic zero, the algebra of differential operators is a complete invariant of the underlying manifold. The "Milnor exercise" gives a rather explicit construction of the underlying manifold via the evaluation functional on the algebra of smooth functions. Looking at the Grabowski and related papers, it doesn't, however, seem so straightforward to obtain the same manifold from its Lie algebra of vector fields, or is it? $\endgroup$
    – Carlos
    Apr 25, 2018 at 15:59

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