I would like to know what in the Donaldson's proof make it work only for Riemann surfaces.
closed as not a real question by Yemon Choi, Kevin Walker, Chandan Singh Dalawat, Deane Yang, Qiaochu Yuan Jan 18 '12 at 7:17
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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.
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).