Theorem of Narasimhan-Seshadri for genus 0 and 1

The theorem of Narasimhan-Seshadri is classically phrased for Riemann surfaces of genus bigger than one.

It then characterises holomorphic bundles which are induced by unitary representations of the fundamental group. Donaldson's theorem on the existence of a Hermitian Einstein metric (which gives a different proof) does not contain the genus condition.

Does anyone know why the "genus >1" condition arise? (or was it just laziness because Grothendieck and Atiyah had aldready covered vector bundles over the Riemann sphere and the torus?)

-
Well I wouldn't exactly call it laziness. In general, there are many good reasons for separating the genus $g=0$, $g=1$ and $g\ge 2$ cases. To be more specific, in this case, it correspond to the trichotomy $\pi_1$ trivial, abelian, or nonabelian. Narasimhan-Seshadri is often phrased as a correspondence between the sets or spaces of irreducible rank $r$ unitary representations and degree $0$ rank $r$ stable bundles. There are no such representations when $g\le 1$ and $r>1$, so there are no stable bundles, which says something, but perhaps it was not very exciting. – Donu Arapura Mar 11 '12 at 20:57
Actually, if you look at their paper (there are a bunch, I mean the annals paper), they seem to consider general $g$ for a lot of the beginning, but it quickly becomes clear that things collapse for $g \leq 1.$ – Igor Rivin Mar 12 '12 at 0:05

I agree with Donu Arapura's comments above: "There are no such representations when g<1, r>1 and , so there are no stable bundles". However the theorem holds true "morally" (if you do no insist on semistability): if g=0 - $\pi_1$ is trivial so we have only 1 holomorphic structure on trivial bundle; of g=1 it is more interesting: $\pi_1$ is abelian so matrices A,B which give you its representation commute, so they can be simultaneously diagonalized - so you get numbers $a_1,...,a_n;b_1...b_n$ and you can make from them DECOMPOSABLE holomorphic bundle $O(D_1)\oplus ... O(D_n)$, – Alexander Chervov Apr 9 '12 at 7:27
continued: where divisors $D_i$ a constructed roughly speaking as z=a+bi/Z^2 - i.e. using NS correspondence for r=1. – Alexander Chervov Apr 9 '12 at 7:28