# Does an exact sequence of holonomic D-modules imply an exact sequence in the solution space?

Here is a probably quite basic question on holonomic D-modules, but I am only a physicist so please bear with me.

If I have the weyl algebra $$D$$ in $$n$$ variables, and an exact sequence of holonomic $$D$$-modules $$$$0\rightarrow M_1 \rightarrow M_2 \rightarrow M_3\rightarrow 0$$$$ I know that the holonomic ranks must satisfy $$rank(M_2)=rank(M_1)+rank(M_3)$$. To me this seems to imply that there is an exact sequence $$$$0\rightarrow Sol(M_3)\rightarrow Sol(M_2) \rightarrow Sol(M_1)\rightarrow 0\;,$$$$ where $$Sol(\underline\;)$$ the solution space of the module, since these are all just vector spaces.

However, I can also consider the solution space as $$Hom(\underline\;,\mathcal{O}_X)$$ for an appropriate function space $$\mathcal{O}_X$$. Applying this functor to the exact sequence above then leads to the exact sequence $$$$0\rightarrow Hom(M_3,\mathcal{O}_X)\rightarrow Hom(M_2,\mathcal{O}_X)\rightarrow Hom(M_1,\mathcal{O}_X)\rightarrow Ext_1(M_3,\mathcal{O}_X)\;.$$$$ It seems to me that these facts combined somehow seem to imply that $$Ext_1(\underline\;,\mathcal{O}_X)=0$$ for all holonomic $$D$$-modules but this seems wrong somehow since I don't see why $$\mathcal{O}_X$$ should be an injective module.

Could someone spot the mistake I made or explain why it is actually correct?
I am mostly interested in if the exact sequence of solution spaces hold so if someone knows this I'd already be very thankful.

Edit: Could it be related to $$Ext^i(M,D)=0$$ for $$i\neq n$$ and $$M$$ a holonomic $$D$$-module?

• Two questions: 1) How do you define exactly $Sol(M)$ if not as $RHom(M,\mathcal{O}_X)$? 2) I don't see how the third sequence is exact. If you apply $RHom(-,\mathcal{O_X})$ you get the exact sequence: $0 \rightarrow Hom(M_3,\mathcal{O}_X) \rightarrow Hom(M_2,\mathcal{O}_X)\rightarrow Hom(M_1,\mathcal{O}_X) \rightarrow R^1Hom(M_3,\mathcal{O}_X) \rightarrow R^1Hom(M_2,\mathcal{O}_X) \rightarrow$....
– MPos
Commented Jan 29 at 13:29
• @MPos thanks for the response! Usually I'd consider $Sol(M/I)$ as the set of functions $f$ with $Pf=0$ for all $P\in I$. However, I think in this case it doesn't matter since the third exact sequence is also an exact sequence of abelian groups so the same argument still holds I think. Regarding your second point. I realize now my notation is a bit wrong, I meant to write $Ext^1(M_3,\mathcal{O}_X)$. Given this, is $R^1 Hom(M_3,\mathcal{O}_X)$ not $Ext_1(M_3,\mathcal{O}_X)$? I realize the exact sequence continues however I am not interested in the further terms. Edit: I now
– A.H
Commented Feb 7 at 10:32

So I think I found a solution to this. The key was in Kashiwara's original thesis theorem 2.4.2 which states that a finitely presented $$D$$-module $$M$$ with a so-called good regular filtration satisfies $$Ext^i_{D_X}(M,O_X)=0$$. Note that any finitely generated $$D$$ module has a good filtration, and by the proof of Lemma 3.3.2 of this reference, we can take $$U\subset X$$ small enough that this filtration becomes free over $$O_U$$. This makes it regular according to definition 2.4.1 . Since any holonomic $$D$$-module $$M$$ satisfies these assumptions, the result follows.
Edit: This makes the study of the solution complex $$Sol(M)=RHom(M,\mathcal{O})$$ highly confusing to mee though...