Let $X$ be a complex analytic space. It is a 'well known fact' that the categories of local systems on $X$ (i.e. locally constant sheaves with stalk $C^n$), and of (holomorphic) vector bundles on $X$ with flat connection, are equivalent. I've been looking for a proof of this, but every reference I can find merely says something like 'this is well known' without further argument. Does anyone know of a proof?
Why are local systems on a complex analytic space equivalent to vector bundles with flat connection?
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ It would be better to change your question's title to something involving flat vector bundles and local systems. $\endgroup$– Kevin H. LinCommented Nov 4, 2009 at 22:33
-
$\begingroup$ A reference in which this is argued in some detail (only for complex manifolds and only at the level of objects) is Voisin's Hodge Theory and Complex Algebraic Geometry I, Section 9.2 $\endgroup$– PedroCommented Jul 11, 2023 at 8:12
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
8
The important point of the proof is that either of these objects can be locally trivialized with transition functions on each double overlap given by a constant element of GL(n). So, given a local system, you just build the vector bundle with flat connection that has the same transition functions, and vice versa.
EDIT: Brian Conrad points out below that while this is a fairly complete sketch in the smooth case, it requires more work in the singular case.
answered Nov 4, 2009 at 19:40
-
2$\begingroup$ I'd suspected this, but I wasn't able to make it explicit. Now that you mention it, it ought to have been obvious. Thanks! $\endgroup$ Commented Nov 4, 2009 at 20:16
-
3$\begingroup$ How does this answer address the real difficulty in the question, which is the proof that the kernel of the flat connection is locally constant of the "expected" rank when the base space is an arbitrary (not necessarily smooth) complex-analytic space? Going from the local system to the bundle with flat connection is the easy direction; the other one requires new work when the base is not assumed to be smooth. I don't think this is at all obvious. Deligne's proof in his thin SLN book is very beautiful, and requires a real idea. $\endgroup$– BCnrdCommented Mar 25, 2010 at 3:02
-
$\begingroup$ Brian- Obviously it doesn't. In my interpretation of the question, that was not what the OP was confused about (checking the rank is the difficult part of the question if you understand how the bijection should work). It would great to see an answer which did cover this point. I would certainly vote it up, and I wouldn't blame the OP for unaccepting my answer and accepting a more complete one. $\endgroup$– Ben Webster ♦Commented Mar 25, 2010 at 17:40
-
1$\begingroup$ See 2.23 in Deligne's book for a brilliant inductive proof in the smooth case over an arbitrary analytic base (allowed non-smooth); taking base to be point is smooth case which was the focus of interest in the question. I wrote up the smooth case with base a point in notes on Riemann-Hilbert correspondence on my webpage (see Theorem 2.6, Lemma 1.6 there). I think my memory got confused about Deligne working in relative case over any (possibly non-smooth) base; most likely smoothness of structure map to the base cannot be dropped. Feel free to delete comments about non-smoothness; mea culpa. $\endgroup$– BCnrdCommented Mar 27, 2010 at 4:04
-
1$\begingroup$ @random123 Seemingly they were talking about "Equations différentielles à points singuliers réguliers". $\endgroup$– Z. MCommented Jul 30, 2021 at 22:03
$\begingroup$
$\endgroup$
1
You might also want to read Carlos Simpson's paper "Moduli of representations of the fundamental group of a smooth projective variety", parts I and II. He explains in great detail how to make the set of objects of each of these categories into the points of an algebraic variety, and why these algebraic varieties are analytically but not algebraically isomorphic. He doesn't use the category theoretic perspective much, as I recall, but he is invaluable for understanding how to work with concrete moduli spaces.
answered Nov 4, 2009 at 20:35
-
$\begingroup$ Do you maybe know of a place where the sheaf cohomology of a local system with fiber $G$ is compared with the cohomology of the chain complex $Hom_{\mathbb{Z}[\pi_1]}(C_i\tilde X,G)$, where $\tilde X$ is the universal cover of $X$? $\endgroup$ Commented Jul 28, 2017 at 21:41
$\begingroup$
$\endgroup$
3
More explicitly, given a local system $\mathcal{V}$ you take the vector bundle to be $\mathcal{E}=\mathcal{V} \otimes_{\mathbb{C}} \mathcal{O}_X$, where $\mathcal{O}_X$ is the structure sheaf, and use $d: \mathcal{O}_X \to \Omega_X$ to define the flat connection on $\mathcal{E}$. Conversely, given $D:\mathcal{E} \to \mathcal{E} \otimes\Omega_X$ you take $\mathcal{V}$ to be $\text{ker}(D)$.
And if the analytic space is connected, one can add one more equivalence: once we choose a point $x$ on the space (choosing a "fiber functor"), those two categories are equivalent to complex representations of $\pi_1(X,x)$ (the "Tannaka dual").
-
1$\begingroup$ Is there a more explicit way to show E and O_X \otimes ker(D) isomorphic besides 'they have the same transition functions'? I feel like there ought to be an obvious map, but I can't come up with one that works. $\endgroup$ Commented Nov 4, 2009 at 20:23
-
1$\begingroup$ Depends on your definition of vector bundles. The construction is given, to show that it works, generally you should be able to prove this directly from definitions. $\endgroup$ Commented Nov 4, 2009 at 21:12
-
2$\begingroup$ And don't forget to check the category axioms as well as
\oplusand\otimes! $\endgroup$ Commented Nov 4, 2009 at 21:13