6
$\begingroup$

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...

$\endgroup$
1
  • $\begingroup$ any suggestion will help $\endgroup$ Aug 21, 2019 at 17:26

1 Answer 1

1
$\begingroup$

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

$\endgroup$
18
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.