4
$\begingroup$

Let $X=\mathbb{C}^n$ be affine n-space (with the Zariski topology), $\mathcal{D}$ its sheaf of differential operators. Let $D$ be the $n$th Weyl algebra, $M$ a right $D$-module, and $N$ a left $D$-module. Write $\tilde M$ for the $\mathcal{D}$-module associated to $M$, and similarly for $N$. Is the $D$-module tensor product $M \otimes_D N$ always isomorphic to the global sections of the $\mathcal{D}$-module tensor product $ \tilde M\otimes_\mathcal{D} \tilde N$? I assume it is, since it's used (without proof or reference) throughout the literature, but I've been unable to prove it.

$\endgroup$
7
  • 1
    $\begingroup$ I think the question may be confusing. - In the category of D-modules, there is an inner tensor product, but it is by definition their product as O-modules equipped with a D-module structure essentially by the Leibniz rule. For this tensor product, the kind of local-to-global correspondence that you are asking about is true because it holds for quasicoherent O-modules. This is the tensor product used `throughout the literature', and it may be that this is what you are interested in. (Continued below.) $\endgroup$
    – t3suji
    Sep 19, 2015 at 17:11
  • 1
    $\begingroup$ -On the other hand, as posed, the question is about a different object. The tensor product of modules over the sheaf of differential operators may be a reasonable operation in the classical topology (used in the Riemann-Hilbert correspondence), but I don't think I ever saw it in the Zariski topology. In either case, I suspect that the claim is false in this formulation: you are composing left-exact functor (global sections) with a right-exact functor (tensor product), both of them could have higher cohomology, and compatibility of the sort you ask should hold in the derived category only. $\endgroup$
    – t3suji
    Sep 19, 2015 at 17:21
  • 1
    $\begingroup$ @t3suji This is definitely something used, for instance in the definitions of the $\mathcal{D}$-module pushforward and pullback. $\endgroup$ Sep 19, 2015 at 17:53
  • 1
    $\begingroup$ True, I wasn't thinking about that. Anyway, the rest of the comment still stands. In fact, that is why push-forward of D-modules is defined directly in the derived category, right? $\endgroup$
    – t3suji
    Sep 19, 2015 at 17:58
  • 2
    $\begingroup$ OK, I'm posting details as an answer (`these comment field is too small for it' :) Let me know if it makes sense, because I am still not sure how detailed it should be. $\endgroup$
    – t3suji
    Sep 20, 2015 at 20:39

1 Answer 1

5
$\begingroup$

As discussed in comments, the claim holds in the derived world; here is a counterexample to the naive statement. As I was writing it, I realized that I looked for a counterexample in the classical topology, but the same idea can be used to produce an easier counterexample that works in both Zariski and classical topology. Let's see:

Let $$n=1,\quad M=vD/v\frac{d}{dx}D,\quad N=Dw/D(x(x-1)\frac{d}{dx})w.$$ ($v$ and $w$ are the generators.) Thus, $M$ corresponds to the right ${\mathcal D}$-module $\mathcal M=\omega_{\mathbb{A}^1}$, and $N$ corresponds to the left $\mathcal{D}$-module $\mathcal N=j_{!}\mathcal{O}_U$ for $U=\mathbb{A}^1-\{0,1\}$ and $j:U\hookrightarrow\mathbb{A}^1$. I will write $\mathcal M$ and $\mathcal N$ for the corresponding $\mathcal D$-modules (rather than $\tilde M$ and $\tilde N$ as in the question).

Put $$\mathcal F:=\mathcal{Tor}_1^{\mathcal{D}}(\mathcal M,\mathcal N)=\ker(\frac{d}{dx}:\mathcal N\to\mathcal N).$$ Because $\mathcal M$ has homological dimension one, we get an exact triangle $$\to\mathcal F[1]\to\mathcal M\otimes^L_{\mathcal D}\mathcal N\to \mathcal M\otimes_{\mathcal D}\mathcal N\to\mathcal F[2];$$ taking cohomology, we get an exact sequence $$0\to H^1(\mathbb{A}^1,\mathcal F)\to H^0(\mathbb A^1,\mathcal M\otimes^L_{\mathcal D}\mathcal N)\to H^0(\mathbb A^1,\mathcal M\otimes_{\mathcal D}\mathcal N).$$ (Of course, this is just a simple case of the corresponding spectral sequence.)

Note that the middle term is exactly $$M\otimes_D N=N/\frac{d}{dx}N=H^1_{dR}(\mathbb{A}^1,\mathcal N),$$ while the term on the right is exactly the space of global sections of $\mathcal M\otimes_{\mathcal D}\mathcal N$. Thus, it suffices to check that $H^1(\mathbb A^1,\mathcal F)\ne 0.$

Indeed, over $U$, $\mathcal N|_U=\mathcal O_U$, and $\mathcal F$ is identified with the constant sheaf $\mathbb{C}$, while its stalks at $0$ and $1$ are easily seen to be zero. Thus, $\mathcal{F}$ is the $j_!$-extension of the constant sheaf by zero. This has non-trivial $H^1$ in either Zariski or classical topology.

Edit. Here's how to compute the stalk. (Say, at 0). Let $R$ be the local ring of $0\in\mathbb A^1$, i.e., the stalk of $\mathcal O$. The stalk of $\mathcal D$ is then $$\mathcal D_0=R[\frac{d}{dx}],$$ the ring of differential operators with coefficients in $R$. The stalk of $\mathcal N$ is therefore $$\mathcal N_0=\mathcal D_0 w/\mathcal D_0(x(x-1)\frac{d}{dx}w)=\mathcal D_0 w/\mathcal D_0(x\frac{d}{dx}w);$$ the last equality holds because $(x-1)$ is invertible in $R$. As a vector space, $$\mathcal N_0\simeq Rw\bigoplus{Span}\left\langle \frac{d^i}{dx^i}(w)\right\rangle_{i>0}$$ (in a more geometric way, the point is that $j_!\mathcal O_U$ is obtained as an extension of $\mathcal O_{\mathbb A^1}=j_{!*}\mathcal O_U$ by $\delta$-functions). At any rate, from this description, $\frac{d}{dx}$ acts injectively on $\mathcal N_0$, and therefore the stalk of its kernel is zero.

Remark. The example mentioned in the comments ($n=2$, $M$ is $\omega_{\mathbb A^2}$, $N$ is the sheaf of $\delta$-functions on the hyperbola) was based on the same idea: choose sheaves so that $\mathcal M\otimes^L_{\mathcal D}\mathcal N$ is concentrated in cohomological degrees $0$ and $1$, and that $\mathcal F=\mathcal{Tor}^1$ has non-trivial cohomology. In this example, $\mathcal F$ was the constant sheaf on the hyperbola; this works fine in the classical topology (hyperbola has higher cohomology), but not in the Zariski topology.

$\endgroup$
3
  • $\begingroup$ Can you elaborate on why the stalks of $\mathcal{F}$ at $0$ and $1$ are zero? $\endgroup$ Sep 21, 2015 at 1:42
  • $\begingroup$ @t3suji By the way, I'm sure that if you ask one of the moderators that they could merge your new account with your old one. $\endgroup$ Sep 21, 2015 at 5:50
  • $\begingroup$ @Peter Samuelson: Thanks! I've got the accounts merged now (maniacal laughter). $\endgroup$
    – t3suji
    Sep 21, 2015 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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