It is well known that if I have a differentiable manifold (holomorphic manifold) $M$, then I have a functor from the category of vector bundles on $M$ with flat connections to the category of local systems on $M$, given by $$(V,\nabla)\mapsto V^{\nabla}$$ and this functor is an equivalence of categories.

Now if I change the setting a little bit. I assume $X$ is a smooth scheme over a field $k$ of characteristic 0. Do I still have a functor like the above? Or in other words, is it true that for any vector bundle $(V,\nabla)$ on $X$ the sheaf of horizontal sections $V^{\nabla}$ (i.e. the kernel of $\nabla$ as $k$-vector spaces) is still a locally constant sheaf of $k$-vector spaces?

I think it is true, and it should be written somewhere, could you tell me the reference for that?

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    $\begingroup$ Over the complex numbers, the functor $(V,\nabla)\mapsto(V^{an},\nabla)$ is an equivalence of categories between "regular connections" on the base algebraic variety to "integrable connections" on the corresponding complex manifold (see, e.g., Malgrange's article in Algebraic D-Modules, Borel et al.). But, as has been noted, there is an additional problem that $(V^{an})^{\nabla}$ may only be locally constant for the complex topology (not for the etale topology). $\endgroup$
    – mephisto
    Commented May 21, 2011 at 13:09
  • $\begingroup$ Jacob Lurie has just given a nice series of lectures for what happens in a characteristic p oder p-adic setting: hcm.uni-bonn.de/events/eventpages/felix-klein-lectures/… $\endgroup$ Commented Dec 11, 2022 at 19:44

2 Answers 2


Obviously you can still define the functor. But it won't have nice properties because the Zariski open sets are just too big for the concept of locally constant sheaf to apply in an interesting way and the solutions of algebraic differential equations are not algebraic functions in general.

Just take a trivial vector bundle $O_X^2$ on $X = \mathbb{G}_m$ with connection $$ \nabla \begin{pmatrix} f_1 \cr f_2\end{pmatrix} = d\begin{pmatrix} f_1 \cr f_2\end{pmatrix} - \begin{pmatrix} 0 & 0 \cr 1 & 0 \end{pmatrix} \begin{pmatrix} f_1 \cr f_2\end{pmatrix} \frac{dz}{z} $$ Holomorphic horizontal sections are linear combinations $A \begin{pmatrix} 0 \cr 1 \end{pmatrix} + B \begin{pmatrix} 1 \cr log(z) \end{pmatrix}$. You get a rank 2 local system.

But if you look at algebraic horizontal sections you will only get the constant sheaf $A \begin{pmatrix} 0 \cr 1 \end{pmatrix}$.

PS: For the same reason (Zariski open sets are too big), one needs to use the hypercohomology of the algebraic de Rham complex to define a reasonable algebraic de Rham cohomology. Also for the same reason (not enough algebraic solutions) only the holomorphic solution complex or the holomorphic de Rham complex of a D-module are relevant to the Riemann-Hilbert correspondence.

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    $\begingroup$ Even simpler I guess: on the affine line with parameter z, $D(f)=df-fdz$ has no algebraic horizontal sections at all because the holomorphic horizontal section is the exponential. $\endgroup$ Commented May 21, 2011 at 11:15
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    $\begingroup$ That was my first idea but I didn't want to use an irregular equation. $\endgroup$
    – AFK
    Commented May 21, 2011 at 11:25
  • $\begingroup$ I don't think the "bigness" of Zariski open subsets have anything to do with how you need to compute the algebraic de Rham cohomology. After all when the variety is affine you can just take the cohomology of the complex of global sections. In general you need to take a covering by open affines but this is no different from the analytic case where you take an open covering by Stein subsets. The only difference is that in the analytic case you may also use $C^\infty$-forms in which case you can always take just global sections. $\endgroup$ Commented May 21, 2011 at 12:02
  • $\begingroup$ ...I see now that you may have referred to the algebraic de Rham complex of an integrable connection (not just the one for the trivial connection). Then you do get something different if the connection has irregular singularities but not when they are regular. In your example $\log(z)$ is not a global section as it is many-valued. However, in the irregular case it is not clear that the algebraic cohomology is unreasonable just that it is different from the analytic one. $\endgroup$ Commented May 21, 2011 at 12:08

As the previous answer points out you have to consider local systems for a finer topology than the Zariski topology. It is natural to consider the étale topology. The category of étale local systems of finite dimensional $k$-vector spaces form a tannakian category whose group is the étale fundamental group (more precisely it is a pro-(constant finite) algebraic group). You can find this and much more in Saavedra's book on tannakian categories (LNM Volume 265).

The category of bundles with connexion (with a regularity condition at infinity if needed) also form a tannakian category (if $X$ is smooth over an algebraically closed field of characteristic zero say), but the corresponding pro-algebraic group is much larger (Deligne call this the algebraic fundamental group). He computes this group in the example of the affine line in his famous article "Le Groupe Fondamental de la Droite Projective Moins Trois Points" (PDF).

So to an étale local system of finite dimensional $k$-vector space you can associate a bundle with connexion, and this will give you a fully faithful Riemann-Hilbert functor, but in this way you will obtain only those bundles with connexion whose monodromy group is finite and étale (in other words those trivialized by an étale cover).


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