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 \begin{equation} 0\rightarrow M_1 \rightarrow M_2 \rightarrow M_3\rightarrow 0 \end{equation} 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 \begin{equation} 0\rightarrow Sol(M_3)\rightarrow Sol(M_2) \rightarrow Sol(M_1)\rightarrow 0\;, \end{equation} 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 \begin{equation} 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)\;. \end{equation} 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?

  • $\begingroup$ 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 $.... $\endgroup$
    – MPos
    Commented Jan 29 at 13:29
  • $\begingroup$ @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 $\endgroup$
    – A.H
    Commented Feb 7 at 10:32

1 Answer 1


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...


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