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Let $X$ be a smooth variety over a field of characteristic zero. Let $M$ be in the derived category of holonomic $\mathcal{D}_{X}$-modules, $D^{b}_{h}(\mathcal{D}_{X})$.

We know that if we assume that $M$ is regular then the map given by the de Rham functor,

$DR_{X}(M):RHom_{\mathcal{D}_{X}}(M,M)\rightarrow RHom_{\mathbb{C}_{X}}(DR(M),DR(M)),$

is an isomorphism. It's proved in the corollary 3.1.15 at the IHES paper of Mebkhout.

Do we have the converse? Say, if $DR_{X}(M)$ is an isomorphism, then M is regular?

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2 Answers 2

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I don't think so. Here's a possible counter example.

Take $X=\mathbb{A}^1_\mathbb{C}$, $M=(\mathcal{O}_X,\nabla)$ where $\nabla(f)=df-fdz$, $z$ being the co-ordinate on $\mathbb{A}^1_\mathbb{C}$. Then $RHom_{\mathcal{D}_X}(M,M)=End_\nabla(M)=\{g\in \mathcal{O}_X \mid dg=0 \}=\mathbb{C}$, and the locally constant sheaf of horizontal sections of $M$ on $X^\mathrm{an}$ is just isomorphic to the constant sheaf $\mathbb{C}_{X^\mathrm{an}}$, with basis the globally defined exponential map. Hence $RHom_{\mathbb{C}_{X^\mathrm{an}}}(DR(M),DR(M))=\mathbb{C}$. But $M$ is not regular.

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  • $\begingroup$ Why don't you have any Ext^{1}? $\endgroup$
    – prochet
    Commented Sep 30, 2013 at 12:20
  • $\begingroup$ An easy way to see that there is no $\mathrm{Ext}^1$ on the de Rham side is to note that $M$ is rank one, i.e. invertible. We have the dual connection $M^{\vee}=(\mathcal{O}_X,\nabla^\vee)$ where $\nabla^\vee(f)=df+fdz$, and $M\otimes M^\vee$ is just the constant integrable connection $\mathcal{O}_X$. Hence we can pass between $\mathrm{Ext}_\nabla^1(M,M)$ and $\mathrm{Ext}_\nabla^1(\mathcal{O}_X,\mathcal{O}_X)$. $\endgroup$
    – ChrisLazda
    Commented Sep 30, 2013 at 15:43
  • $\begingroup$ your connexion is regular as it's holomorphic. $\endgroup$
    – prochet
    Commented Sep 30, 2013 at 22:25
  • $\begingroup$ Are you sure? If we let $w=z^{-1}$, then the connection becomes $\nabla(f)=df+f\frac{dw}{w^2}$, so it doesn't have logarithmic poles around $w=0$. $\endgroup$
    – ChrisLazda
    Commented Oct 1, 2013 at 8:53
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It seems that the converse is a recent theorem of Jean-Baptiste Teyssier, see http://jbteyssier.com/papers/jbteyssier_caracterisation_modules_reguliers.pdf

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