Let $X$ be a smooth algebraic variety over $\mathbb{C}$ (or a field of characteristic zero). We have $D_X$ the sheaf of differential operators on $X$, which is a coherent sheaf of rings, and it carries the canonical increasing filtration $D_X^n$ by the order of differential operators, whose associated graded ring $grD_X$ is identified with the function ring of the cotangent bundle of $X$.For $M$ a coherent $D_X$-module (finite type as $D_X$-module and quasi-coherent as $O_X$-module), we can talk about the notion of good filtration on $M$: it is a increasing filtration $M_n$ by quasi-coherent $O_X$-modules, compatible with $D_X^n$ for the $D_X$-module structure of $M$, such that the associated graded module is of finite type over $grD_X$.

For coherent $D_X$-module $M$, good filtration exists locally over $X$, and in many case this is already sufficiently useful. For example it leads to the notion of characteristic cycles, etc. But my question is when do we do have globally defined good filtration $M_n$ for $M$, such that $D_X^nM_m=M_{n+m}$ for all $m,n\geq 0$?

A natural example one can find is in the D-affine case: $X$ is D-affine if the global section functor establishes an equivalence between coherent $D_X$ modules and coherent $\Gamma(X,D_X)$-modules. In this case a coherent $D_X$-module $M$ with global section $N$ is equipped with a global good filtration $D_X^nN$. Conversely, if every coherent $D_X$-module admits a global good filtration, how far is $X$ from being $D_X$-affine? This seems to be too optimistic.

Another question: if we only consider coherent $D_X$-modules coming from flat connection on vector bundles (coherent as $O_X$-modules), do we always have global good filtration?

By the way, for $X$ a Zariski open subset of $Y$ a smooth variety, does the D-affinity of $Y$ implies the D-affinity of $X$?

Many thanks!

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    $\begingroup$ For your last two questions: 1) a D-module is a vector bundle with connection if and only if the trivial filtration (with a jump in only one degree) is a good filtration, so yes, and 2) $\mathbb{A}^2\setminus\{(0,0)\}$ is not D-affine. $\endgroup$ Nov 12, 2011 at 1:31

1 Answer 1


Every coherent D-module admits a global good filtration. This is theorem 2.1.3 in the book by Hotta, Takuechi, and Tanisaki, D-modules, Perverse Sheaves and Representation Theory. They first prove that there is a $\mathcal O_X$ coherent $\mathcal O_X$-submodule $M_0$ which generates $M$, then define the filtration by

$F_mM = (F_mD_X) M_0$.

In particular $F_nD_X (F_m M) = F_{n+m}M$ for all $m,n \geq 0$. I believe that you other questions have been answered in the comments by Moosbrugger.

  • $\begingroup$ This seems only to work in the case when $M$ is generated by global sections (e.g., the D-affine case). $\endgroup$ Nov 12, 2011 at 4:11
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    $\begingroup$ I am not sure I understand your objection. When I say a submodule $M_0$ which generates $M$, I mean $M_0$ is a subsheaf of $M$ such that the map $D_X \otimes {\mathcal O_X} M_0 \to M$ is a surjective map of sheaves (so that any section of $M$ can be locally written as $Pm_0$ for some local sections $P$ of $D_X$ and $m_0$ of $M$). Similarly, when I write $F_nD_X(M_0)$, I mean everything as sheaves. I never need to talk about global sections of anything. Certainly theorem 2.1.3 applies in general (D-affineness is not needed). What am I missing? Have I misunderstood the question? $\endgroup$ Nov 12, 2011 at 6:19
  • $\begingroup$ It just occurred to me that you might be asking about a filtration on the global sections $\Gamma (M)$ rather than a globally defined filtration on $M$... $\endgroup$ Nov 12, 2011 at 6:44
  • $\begingroup$ Sorry -- my problem was that I was trying to choose the coherent guy $M_0$ in a stupid way. But objection retracted! $\endgroup$ Nov 12, 2011 at 13:34
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    $\begingroup$ @algchen: The associated gradeds can indeed be different, but in both cases the set-theoretic support of the modules is the zero section of the cotangent bundle. (Note that in the VHS case the action of vector fields is nilpotent on the associated graded.) $\endgroup$ Nov 18, 2011 at 2:36

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