# Category of $\mathcal{D}$-modules on a singular variety

Take $$X\to V$$ a closed embedding, where $$X$$ is not necessarily smooth, $$V$$ is affine and smooth. Define the category $$\mathcal{C}$$ of $$\mathcal{D}$$ modules on $$X$$ to be the full subcategory of $$\mathcal{D}$$ modules on $$V$$ with support on $$X$$.

I want to ask if the inclusion $$D(\mathcal{C})\to D_\mathcal{C}(\mathcal{D}_V\text{-mod})$$ is an equivalence (where $$D_\mathcal{C}(\mathcal{D}_V\text{-mod})$$ is the full subcategory of $$D(\mathcal{D}_V\text{-mod})$$ consisting complexes with cohomologies in $$\mathcal{C}$$). I believe it is. In fact for what I really need, I only need the inclusion to be fully faithful, but nevertheless it should be an equivalence. The derived category is the one with either quasi-coherent or coherent objects, and should be bounded. Note when $$X$$ is smooth, this is just a version of Kashiwara's Theorem.

What we know: $$\mathcal{C}$$ is thick/Serre, $$\mathcal{C}$$ and $$\mathcal{D}_V\text{-mod}$$ are Grothedieck (so has enough injectives).

There is a potential useful theorem in Kashiwara-Schapira Category and Sheaves, Theorem 13.2.8, but I don't know how to show the conditions. Also, these overflow questions can be useful: Equivalence between a derived subcategory and a subcategory of the derived category

Derived category of $\mathcal{D}_X$ modules

Edited later: Actually there is more to ask: similar to Derived category of $\mathcal{D}_X$ modules, is the same result still true for singular $$X$$, because normally Kashiwara's theorem is stated using 'the other' version of derived category instead of the version I am using (and they turn out to be the same in the smooth case), but I probably don't need this...

• any suggestion will help – FunctionOfX Aug 21 '19 at 17:26

I'm going to assume that we're dealing with bounded-below complexes, as I'm most familiar with those.

Denote by $$\Gamma_X$$ the sections-supported-on-$$X$$ functor, i.e. $$\Gamma(U, \Gamma_X(\mathcal{F})) = \ker\left(\Gamma(U; \mathcal{F}) \xrightarrow{\text{restriction}} \Gamma(U\setminus X; \mathcal{F})\right).$$ Let $$\mathcal{M}^\bullet$$ be a bounded-below complex of $$\mathcal{D}_V$$-modules whose cohomology is supported on $$X$$. It suffices to show that $$\mathcal{M}^\bullet$$ is quasi-isomorphic to a complex of objects in $$\mathcal{C}$$.

I claim that the inclusion $$i^\bullet\colon \Gamma_X(\mathcal{M}^\bullet) \hookrightarrow \mathcal{M}^\bullet$$ is a quasi-isomorphism. I do this by induction on the cohomological degree $$p$$.

When $$p\ll 0$$, both complexes vanish, so $$H^p\Gamma_X(\mathcal{M}^\bullet) = H^p(\mathcal{M}^\bullet) = 0$$, and therefore $$H^p(i^\bullet)$$ is trivially an isomorphism.

Assume we know that $$H^{p-1}(i^\bullet)$$ is an isomorphism. Consider the short exact sequence $$0 \to \Gamma_X(\mathcal{M}^\bullet) \to \mathcal{M}^\bullet \to \mathcal{M}^\bullet/\Gamma_X(\mathcal{M}^\bullet) \to 0.$$ This gives rise to an exact sequence $$H^{p-1}(\mathcal{M}^\bullet) \to H^{p-1}(\mathcal{M}^\bullet/\Gamma_X(\mathcal{M}^\bullet)) \to H^{p-1}(\mathcal{M}^\bullet/\Gamma_X(\mathcal{M}^\bullet)) \to H^p(\Gamma_X(\mathcal{M}^\bullet)) \to H^p(\mathcal{M}^\bullet) \to H^p(\mathcal{M}^\bullet/\Gamma_X(\mathcal{M}^\bullet)).\tag{1}$$ By the induction hypothesis, the first map is an isomorphism, so (1) gives the exact sequence $$0\to H^{p-1}(\mathcal{M}^\bullet/\Gamma_X(\mathcal{M}^\bullet)) \xrightarrow{a} H^p(\Gamma_X(\mathcal{M}^\bullet)) \xrightarrow{H^p(i^{\bullet})} H^p(\mathcal{M}^\bullet) \xrightarrow{b} H^p(\mathcal{M}^\bullet/\Gamma_X(\mathcal{M}^\bullet)).\tag{2}$$ For each $$q$$, no non-zero section of $$\mathcal{M}^q/\Gamma_X(\mathcal{M}^q)$$ vanishes outside of $$X$$ (prove this!), so the same is true of each $$H^q(\mathcal{M}^\bullet/\Gamma_X(\mathcal{M}^\bullet))$$. On the other hand, every non-zero section of $$H^q(\mathcal{I}^\bullet)$$ and $$H^q(\Gamma_X(\mathcal{I}^\bullet))$$ vanishes outside of $$X$$ (the first by hypothesis and the second because by the definition of $$\Gamma_X$$). Therefore, $$a$$ and $$b$$ must be the zero map. Hence, $$H^p(i^{\bullet})$$ is an isomorphism.

Edit 1: To see fully faithfulness: $$\Gamma_X\colon \mathcal D_V\text{-mod} \to \mathcal{C}$$ preserves injectives (because it is right adjoint to the inclusion functor, which is fully faithful). Therefore, the hom sets in $$D^+(\mathcal C)$$ and $$D^+_X(\mathcal D_V)$$ are computed in the same way.

Edit 2 (2019/10/06): Using the following fact, along with the non-crossed-out part of Edit 1, I am going to prove fully faithfulness.

Fact. Although in general, a morphism $$u$$ between bounded below complexes with $$H^n(u)=0$$ for all $$n$$ is not necessarily $$0$$ in the derived category, if such a morphism $$u$$ maps between complexes of injectives, then it is $$0$$. (See, e.g., the proof of Prop. 1.7.10 from Kashiwara and Schapira's Sheaves on Manifolds).

Let $$\mathcal{M}^\bullet, \mathcal{N}^\bullet$$ be bounded below complexes of objects of $$\mathcal{C}$$. Choose $$\mathcal{D}_V$$-injective resolutions $$\mathcal{I}^\bullet,\mathcal{J}^\bullet$$ of $$\mathcal{M}^\bullet, \mathcal{N}^\bullet$$, respectively. Because $$\Gamma_X(\mathcal{M}^\bullet)=\mathcal{M}^\bullet$$ (as $$\mathcal{M}^\bullet$$ is a complex of guys in $$\mathcal{C}$$), the resolving quasi-isomorphism $$\mathcal{M}^\bullet \to \mathcal{I}^\bullet$$ factors through the natural inclusion $$\Gamma_X(\mathcal{I}^\bullet) \hookrightarrow \mathcal{I}^\bullet$$. Hence, because by the above, the natural inclusion $$\Gamma_X(\mathcal{I}^\bullet) \hookrightarrow \mathcal{I}^\bullet$$ is a quasi-isomorphism, we get a quasi-isomorphism (i.e. a resolution) $$\mathcal{M}^\bullet\to \Gamma_X(\mathcal{I}^\bullet)$$. But $$\Gamma_X(\mathcal{I}^\bullet)$$ is a complex of $$\mathcal{C}$$-injectives, so we in fact have a $$\mathcal{C}$$-injective resolution $$\mathcal{M}^\bullet\to \Gamma_X(\mathcal{I}^\bullet)$$ of $$\mathcal{M}^\bullet$$. Similarly, $$\mathcal{N}^\bullet\to \Gamma_X(\mathcal{J}^\bullet)$$ is a $$\mathcal{C}$$-injective resolution of $$\mathcal{N}^\bullet$$.

Replacing $$\mathcal{M}^\bullet$$ and $$\mathcal{N}^\bullet$$ with $$\Gamma_X(\mathcal{I}^\bullet)$$ and $$\Gamma_X(\mathcal{J}^\bullet)$$, resp., it is enough to show that

1. for every morphism $$u^\bullet\colon \Gamma_X(\mathcal{I}^\bullet)\to \Gamma_X(\mathcal{J}^\bullet)$$, there exists a unique (in $$D^+_X(\mathcal{D}_V)$$) morphism $$v^\bullet\colon \mathcal{I}^\bullet\to \mathcal{J}^\bullet$$ extending $$u^\bullet$$; and

2. if $$u^\bullet=0$$ in $$\mathcal{D}^+(\mathcal{C})$$, then $$v^\bullet=0$$ in in $$D^+_X(\mathcal{D}_V)$$.

Let $$i^\bullet,j^\bullet$$ be, resp., the inclusions $$\Gamma_X(\mathcal{I}^\bullet) \hookrightarrow \mathcal{I}^\bullet$$ and $$\Gamma_X(\mathcal{J}^\bullet) \hookrightarrow \mathcal{J}^\bullet$$. The existence of $$v^\bullet$$ is because everything in town is injective. Then 2. and the remaining part of 1. follow from the Fact together with the equation $$v^\bullet\circ i^\bullet = j^\bullet\circ u^\bullet.$$

• how do you show fully faithfulness? is that automatic? – FunctionOfX Aug 27 '19 at 1:26
• @FunctionOfX see my edit – Avi Steiner Aug 28 '19 at 19:43
• Sorry, I am not understanding why $\Gamma_X: \mathcal{D}_X$-mod$\to \mathcal{C}$ preserve injectives. In fact, what is $\mathcal{D}_X$ – FunctionOfX Aug 30 '19 at 1:59
• oh you mean D_V? – FunctionOfX Aug 30 '19 at 13:26
• yeah... i dont understand it, but it is a result we can quote. I only found it out a few days ago. Thank you so much for your answer. – FunctionOfX Oct 7 '19 at 2:56