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It is well-known that the space of S-equivalence classes of rank 2 semistable holomorphic vector bundles with trivial determinant on a genus 2 surface M is $CP^3$ (more concretely $PH^0(Jac(M),L(2\theta)$). Especially, the points corresponding to semistable (and not stable) bundles are smooth points. On the open dense subspace consisting of points corresponding to stable bundles, there is a natural symplectic structure, compatible with the natural Riemannian metric. It defines a Kaehler structure. It should be true that this Kaehler structure extends to the semistable points (and therefore $CP^3$ is equipped with its natural (only) Kaehler structure). Does anyone know a reference, or a proof of this? Thank you.

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The word "Bundles" in the title was changed to "vundles". This seems awfully strange to me, but I'll wait for some corroboration that "vundle" is not the hip new term for "vector bundle" before suggesting a rollback. –  Pete L. Clark Jul 23 '10 at 11:43
What natural Riemannian metric are you referring to? I would say that what this moduli space possesses, very naturally, is a complex structure. The questions are then whether the Goldman form v(x,y) -- the natural symplectic form -- extends over the semistable locus where it is not a priori defined, and whether g(x,y) := v(ix,y) is positive definite. If so, since we know v is closed, we would conclude that g is automatically Kähler (see p. 107 of Griffiths & Harris). However, I don't know offhand whether the answers are positive or negative. –  Michael Thaddeus Jul 23 '10 at 18:21
It depends on ypor point of view. In the differential geometric setup, the metric is quite natural. The moduli space can be identified with the moduli space of flat SU(2) connections, where the stable bundle correspond to the irreducible connections. The tangent space at some connection A can be identified with the space of harmonic $\phi\in\Omega^1(su(V)).$ This space has a natural metric $\int trace(\phi\wedge*\tilde\phi).$ –  Sebastian Jul 26 '10 at 6:20
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