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I would like to know what in the Donaldson's proof make it work only for Riemann surfaces.

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closed as not a real question by Yemon Choi, Kevin Walker, Chandan Singh Dalawat, Deane Yang, Qiaochu Yuan Jan 18 '12 at 7:17

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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Alexander: the state of being closed is precisely for that. –  Mariano Suárez-Alvarez Jan 18 '12 at 7:27
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Since nick "cedric" seems to be new on MO, I would suggest at least give some clear suggestions to him what should be done. Otherwise it does not seems to be friendly attitude to new-comers. Although I, of course, agree that the formulation of the question is beyond reasonable level of clearness and standards of asking questions on any public site - MO in particular. –  Alexander Chervov Jan 18 '12 at 7:52
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@Cedric I would suggest to clarify the question. Do you really interested in that 1) "why it works in 2d but not for higher dimensional manifolds ?" 2) It would be polite if you give at least some brief comments on what is Narasimhan-Seshadri theorem is about - this is natural requirement of at least trying to make question self-contained written in MO's "FAQ". –  Alexander Chervov Jan 18 '12 at 7:57
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@Cedric "closed" - does not mean "deleted" - it can be opened again. I hope this will happen if you clarify yours question. –  Alexander Chervov Jan 18 '12 at 8:01
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Alexander Chervov: Even though I voted to close the question, I applaud your attitude in the above comments. I would happily vote to reopen the question if it were edited to make it clear just what the question is. I tend to be pretty charitable on this issue, but this question fails to meet even my permissive standards. –  Kevin Walker Jan 18 '12 at 15:37
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1 Answer 1

Briefly speaking the answer is because moment map for the action of gauge group is CURVATURE in 2-d and (hence moment level = 0 you get FLAT connections) and in higher - dimensions the appropriately understood moment map will have form curvature*omega^n , where omega is the kaehler form and so what you get is related to ant-self dual connections in 4d and to something well-known (but I do not remember) in higher dimensions.


Some more details.

Actually from some point of view you are not quite right. Actually in moral sense it works in higher dimensions - but what you get is not flat connections but (anti) - self-dual connections and what you get is Ulenbeck-Yau (??) theorem that moduli of holomorphic bundles is the same as moduli and moduli space of (anti)-self-dual connections.

May be it worth to understand the moral of the proofs which is quite simple and then it clarifies the conclusions.

The key idea for understanding - is the following finite-dimensional fact: Let G - be complex semi-simple group, U - its compact subgroup, M - kaehler manifold.

Then M/G = M//U where "//" is symplectic reduction.

This is a quite non-trivial fact.

All theorems above are applications of the above principle in infinite-dimensional setup.

When you consider M - moduli space of all smooth connections on some manifold N. G - is "gauge group" - group of smooth automorpisms of vector bundle. U - is subgroup of some orthogonal automorphims ( you need to choose metric on your bundle).

Since evrything is infinite-dimensional you cannot apply theorem above directly, but you as a guidence principle it works and people were able to overcome infinit-dim. difficulties and get the desired results.

PS

I am not sure I presented details correctly, but idea I am sure is correct.

Most of this I heard from Misha Verbitsky - may be if You alert him about this question he will answer you here (as you can see now he is not often on MO).

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