Let $R$ be a regular ring over a field of char 0. Let $X=Spec R$ and $D=\mathcal{D}_X$ the algebra of differential operators over it.

The overall vague question is what kind of algebraic object is $D$ and what kind of category is the category of its modules? Here are some points whose answers could together be considered an answer to this.

  1. When (meaning for what kind of $D$-modules $M,N$ left or right) is there a $D$-module structure on $Hom_R(M,N)$? On $M\otimes_RN$? Does this make $D$-mod into an abelian monoidal category with fiber functor to $R$-mod? If so is it closed monoidal?

  2. When does a $D$-module $M$ admit a dual $M^*$? (in the sense of monoidal categories).

  3. Is the abelian category of $D$-modules isomorphic to $D^{op}$-mod? (before deriving). If not how are they related? Maybe the correct thing to consider is the opposite co-opposite $D^{op}_{cop}$? Specifically Why does the dualizing sheaf $\omega_X$ pop up in this context?

  4. Is the $R$-linear dual $Hom_R(D,R)$ a bialgebra? Is it related to the opposite of $D$?

  5. How do I derive correctly the category of $D$-modules? Suppose $M$ and $N$ are $D$-modules and suppose $Hom_R(M,N)$ is the "correct" internal Hom in the abelian monoidal category of $D$-modules. If we hope to have a fiber functor between the derived categories $Hom_R(Q,N)$ (with $Q \to M$ a resolution as a $D$-module) should go to (something quasi isomorphic) to $Hom_R(P,N)$ (with $P \to M$ a resolution as an $R$-module). It doesn't seem to follow easily from the rest of the structure.

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    $\begingroup$ Not competent enough unfortunately to make this an answer but I gained a lot from learning about the approach through the infinitesimal path groupoid. In a nutshell - from a scheme (in particular affine, as in your question) one obtains a groupoid whose objects are points and morphisms are pairs of infinitesimally nearby points. Then it turns out that sheaves on this groupoid are more or less equivalently described by D-modules. Intuitively this is even clear - such a sheaf is essentially an "ordinary" sheaf with a flat connection. $\endgroup$ Dec 21, 2016 at 18:05
  • $\begingroup$ On reflection, I'm not sure I've understood the bold part of question 5. How do you derive what? Do you mean the category or functors between such categories for different choices of $X$? If the latter, which functors? $\endgroup$ Dec 22, 2016 at 11:10
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    $\begingroup$ Comment about 3: From the analytic point of view the fact that top forms carry a \emph{right} action of $D$ is very natural. Namely, there is a pairing between top forms and functions given by $(\omega,f)\mapsto \int f\omega$ (if the space is not compact, we'd need to say something about behavior at infinity to ensure convergence). If you think of top forms as functionals on the space of functions (i.e., distributions), it is clear where the right action of $D$ comes from. $\endgroup$
    – t3suji
    Dec 23, 2016 at 17:17
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    $\begingroup$ @SaalHardali I mean, it follows from your 1. If you have any right D-module $\ell$ that happens to be a line bundle, then $M\mapsto M\otimes\ell$ and $N\mapsto Hom(\ell,N)$ give mutually inverse equivalences between categories of left and right $D$-modules. The minor miracle here is that any variety carries a naturally defined canonical right $D$-module, namely $\omega$. If you like analytical analogies, think of the following (to be continued) P.S. I apologize if some of the things I say are obvious to you - I don't know your background, and other people might find the comment useful, too. $\endgroup$
    – t3suji
    Dec 23, 2016 at 18:41
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    $\begingroup$ Here is an analytic analogy. Suppose you have a manifold with a given closed 2-form $\omega$. You can then consider two categories: 1. Bundles with flat connections and 2. Bundles with connections whose curvature is $\omega$ (and in particular, scalar). Think of these as counterparts of left and right D-modules, and you'll see what happens to tensor product and hom functor. In particular, if there is somehow a canonical line bundle with a connection whose curvature is exactly $\omega$, it would give an equivalence between categories 1 and 2. $\endgroup$
    – t3suji
    Dec 23, 2016 at 18:44

1 Answer 1

  1. Proposition 1.2.9 of http://math.columbia.edu/~scautis/dmodules/hottaetal.pdf explains that if $M$ and $N$ are both left $D$-modules and $M'$ and $N'$ are both right $D$-modules then

    (a) $M\otimes_{R} N$ is naturally a left $D$-module;

    (b) $M'\otimes_{R} N$ is naturally a right $D$-module;

    (c) $\mathrm{Hom}_{R}(M,N)$ is naturally a left $D$-module;

    (d) $\mathrm{Hom}_{R}(M',N')$ is naturally a left $D$-module;

    (e) $\mathrm{Hom}_{R}(M,N')$ is naturally a right $D$-module.

    This Proposition also gives explicit formulae that explain why each of these is the case. Remark 1.2.10 explains why $M'\otimes_{R}N'$ is not naturally a left or right $D$-module in general.

    Section 2 of https://arxiv.org/abs/dg-ga/9702008 fits these results into a wider framework. The thing about $D$ that makes these things true is that it is the universal enveloping algebra of a $(k,R)$ Lie-Rinehart algebra.

    I think that it is clear from what I've written that $D$-mod (=cat of left $D$-modules) is a (symmetric) monoidal category with $\otimes=\otimes_{R}$ and the forgetful functor $D$-mod to $R$-mod preserves the monoidal product. I'm not sure if a fiber functor requires more than this.

    As noted by t3suji in the comments below the internal Hom $\mathrm{Hom}_R(-,-)$ on $R$-mod which induces an internal Hom on $D$-mod via part (c) above does make $D$-mod into a closed monoidal category. Once again this observation works for the enveloping algebra of any Lie-Rinehart algebra.

  2. I think it is probably the case that $M$ has a dual precisely if it is a projective $R$-module of finite rank (that is also a left $D$-module). In this case $M^\ast$ should just be $\mathrm{Hom}_R(M,R)$ which is a left $D$-module by 1(c).

  3. Yes. The reason that $\omega_X$ crops up is that it is a right $D$-module that is a rank $1$ projective $R$-module. It follows that tensoring with it defines an auto-equivalence of categories on $R$-mod (the inverse is given by $\mathrm{Hom}_R(\omega_X,-)\cong \mathrm{Hom}_R(\omega_X,R)\otimes_R-$. It follows from 1(b) that this auto-equivalence sends left $D$-modules to right $D$-modules and from 1(d) that its inverse sends right $D$-modules to left $D$-modules.

  4. Since $D$ is not a finitely generated $R$-module, $\mathrm{Hom}_R(D,R)$ is pretty badly behaved. In particular it is unlikely that the algebra structure on $D$ will induce a coalgebra structure on its 'dual'. Incidentally I'm not sure what you mean by saying that $D$ is an $R\otimes R$ bialgebra (in particular I don't know what the coalgebra structure is nor the counit).

  5. I don't really understand what this part is asking.

Edited in the light of the comments below by t2suji

  • $\begingroup$ I should note that the equivalence $D$-mod to $D^{op}$-mod is not monoidal and cannot be since $D^{op}$-mod is not a monoidal category as noted in part 1. $\endgroup$ Dec 21, 2016 at 17:21
  • $\begingroup$ Are you talking about rigid monoidal category or about closed monoidal category here? $\endgroup$
    – t3suji
    Dec 21, 2016 at 17:28
  • $\begingroup$ Oops. You are right. I confused the two. I think I stand by my claim that it is not closed. Certainly that $\mathrm{Hom}_R(M,N) is not an internal Hom. I'll need to think further though. $\endgroup$ Dec 21, 2016 at 17:52
  • $\begingroup$ Perhaps pertinent to part 5: any projective resolution of $M$ as a $D$-module is also a projective resolution of $M$ as an $R$-module since $D$ is a projective $R$-module. Indeed $D$ is isomorphic to $\mathrm{Sym}(Der_k(R,R))$ as an $R$-module. $\endgroup$ Dec 21, 2016 at 18:29
  • $\begingroup$ @SimonWadsley This settles a lot of my confusion. One last query: Is there a monoidal structure on right $\mathcal{D}$-modules before deriving? (as I understand there's a shriek tensor product in the derived category). $\endgroup$ Dec 22, 2016 at 11:34

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