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...

  • $\begingroup$ any suggestion will help $\endgroup$ Aug 21, 2019 at 17:26

1 Answer 1


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 $$ v^\bullet\circ i^\bullet = j^\bullet\circ u^\bullet.$$

  • $\begingroup$ how do you show fully faithfulness? is that automatic? $\endgroup$ Aug 27, 2019 at 1:26
  • $\begingroup$ @FunctionOfX see my edit $\endgroup$ Aug 28, 2019 at 19:43
  • $\begingroup$ Sorry, I am not understanding why $\Gamma_X: \mathcal{D}_X$-mod$\to \mathcal{C}$ preserve injectives. In fact, what is $\mathcal{D}_X$ $\endgroup$ Aug 30, 2019 at 1:59
  • $\begingroup$ oh you mean D_V? $\endgroup$ Aug 30, 2019 at 13:26
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    $\begingroup$ 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. $\endgroup$ Oct 7, 2019 at 2:56

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