Global sections of D-module tensor product

Question

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.

Answer 1

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.