It is wellknown that the space of $S$equivalence classes of rank 2 semistable holomorphic vector bundles with trivial determinant on a genus 2 Riemann 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.

5$\begingroup$ 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. $\endgroup$ – Pete L. Clark Jul 23 '10 at 11:43

1$\begingroup$ 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. $\endgroup$ – Michael Thaddeus Jul 23 '10 at 18:21

1$\begingroup$ It depends on your 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 $\int trace(\phi\wedge*\tilde\phi).$ $\endgroup$ – Sebastian Jul 26 '10 at 6:20
Disclaimer: This answer is rewritten in response to Chris Woodward's insightful comments.
The moduli space of rank 2 semistable holomorphic vector bundles with trivial determinant on a genus 2 surface $\Sigma$ is homeomorphic to the character variety $$\mathfrak{X}_\Sigma(SU(2)):=\mathrm{Hom}(\pi_1(\Sigma),SU(2))/SU(2).$$ This homeomorphism restricts to a diffeomorphism between stable bundles and irreducible representations.
In the case of fixed determinant Higgs bundles on $\Sigma$ there is a similar homeomorphic correspondence with $$\mathfrak{X}_\Sigma(SL(2,\mathbb{C})):=\mathrm{Hom}(\pi_1(\Sigma),SL(2,\mathbb{C}))/\!/SL(2, \mathbb{C}).$$ Carlos Simpson has shown under this latter correspondence all points, even singular points, have isomorphic corresponding étale neighborhoods (Isosingularity Theorem). In short, they are locally isomorphic. However, it is known these moduli spaces are not even biholomorphic let alone biregular, since the complex structure of the Higgs moduli space depends on the complex structure of $\Sigma$ as a Riemann surface whereas the complex structure on the character variety only depends on the group $SL(2,\mathbb{C})$.
The moduli space of holomorphic vector bundles naturally embeds into the moduli space of Higgs bundles as those with trivial Higgs field. Likewise, $\mathfrak{X}_\Sigma(SU(2))$ embeds into $\mathfrak{X}_\Sigma(SL(2,\mathbb{C}))$, and the homeomorphisms above respect these embeddings.
So it is natural to think that the stratified analytic smooth structures on the moduli space of vector bundles and that of the semialgebraic space $\mathfrak{X}_\Sigma(SU(2))$ also correspond; as they do generically and also do on their "complexifications". However, as pointed out by Chris Woodward, Johannes Huebschmann proves in Smooth Structures on Certain Moduli Spaces for Bundles on a Surface that this is not the case. In particular, in Section 8 he explicitly addresses the case of a genus 2 surface where the moduli spaces in question are homeomorphic to $\mathbb{C}P^3$, showing explicitly that the natural smooth structure on $\mathfrak{X}_\Sigma(SU(2))$ differs from that of $\mathbb{C}P^3$.
That is sufficient to answer the original question as: NO, if one interprets the question as "Is $\mathfrak{X}_\Sigma(SU(2))$ Kähler isomorphic to $\mathbb{C}P^3$?"
Remark: With regard to the (singular) symplectic structure on $\mathfrak{X}_\Sigma(SU(2))$. The Goldman Poisson structure, a Lie algebra structure and derivation, is defined globally on the coordiante ring of the character variety and makes sense even in a singular setting. In general, such a Poisson structure gives a foliation of the smooth locus by symplectic submanifolds. In this case there is one leaf (since the surface in question is closed).
Remark: For a direct proof, not using the theory of holomorphic bundles, that the character variety is $\mathbb{C}P^3$ see here. We can see the singularities directly in Choi's construction.

1$\begingroup$ My impression was that the answer was no, because the smooth structures are "different" in the sense of Huebschmann last section of arxiv.org/pdf/dgga/9411008.pdf. $\endgroup$ – Chris Woodward Jun 3 '18 at 12:08

1$\begingroup$ p.s. The Kahler structure on the moduli space depends on the choice of complex structure on the underlying curve; having this structure be the a toric structure would mean that the complex structure would be invariant under the Goldman flow which is not true for a general curve, see the paper www2.math.umd.edu/~raw/papers/local.pdf by DaskalopolousWentworth. $\endgroup$ – Chris Woodward Jun 3 '18 at 12:36

$\begingroup$ Thanks Chris! These are great links. I will edit my post. I see where I got confused, and will try to note it. $\endgroup$ – Sean Lawton Jun 3 '18 at 15:08