Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
It would be better to change your question's title to something involving flat vector bundles and local systems. –  Kevin H. Lin Nov 4 '09 at 22:33
add comment

3 Answers 3

up vote 6 down vote accepted

The important point of the proof 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.

share|improve this answer
    
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! –  Ketil Tveiten Nov 4 '09 at 20:16
1  
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. –  BCnrd Mar 25 '10 at 3:02
    
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. –  Ben Webster Mar 25 '10 at 17:40
    
I was thinking of the smooth case (Riemann surfaces, actually), but the general case would be interesting too. Which book by Deligne are you talking about? I have 'Equations Differentielles', but there it was 'well known'. –  Ketil Tveiten Mar 25 '10 at 18:25
    
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. –  BCnrd Mar 27 '10 at 4:04
add comment

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.

share|improve this answer
add comment

More explicitly, given a local system V you take the vector bundle to be E=V \otimes_{C} O_X, where O_X is the structure sheaf, and use d: O_X --> \Omega_X to define the flat connection on E. Conversely, given D:E --> E\otimes\Omega_X you take V to be 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").

share|improve this answer
    
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. –  Ketil Tveiten Nov 4 '09 at 20:23
    
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. –  Ilya Nikokoshev Nov 4 '09 at 21:12
    
And don't forget to check the category axioms as well as \oplus and \otimes! –  Ilya Nikokoshev Nov 4 '09 at 21:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.