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Apr 9, 2012 at 7:28 comment added Alexander Chervov continued: where divisors $D_i$ a constructed roughly speaking as z=a+bi/Z^2 - i.e. using NS correspondence for r=1.
Apr 9, 2012 at 7:27 comment added Alexander Chervov 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 $O(D_1)\oplus ... O(D_n)$,
Mar 11, 2012 at 21:40 history answered cliff1753 CC BY-SA 3.0