Here is a shot in the dark (Disclosure: I really know nothing about this problem).
Let G:=mathsfSU(2) act on G3 by simultaneous conjugation; namely, gcdot(a,b,c)=(gag−1,gbg−1,gcg−1).
The evaluation map shows that the character variety mathfrakX:=mathrmHom(pi1(Sigma),G)/G is homeomorphic to G3/G, where Sigma is an elliptic curve with two punctures.
Fixing generic conjugation classes around the punctures, by results of Mehta and Seshadri (Math. Ann. 248, 1980), gives the moduli space of fixed determinant rank 2 degree 0 parabolic vector bundles over Sigma (where we now think of the punctures are marked points with parabolic structure). In particular, these subspaces are projective varieties.
Letting the boundary data vary over all possibilities gives a foliation of mathfrakXcongG3/GcongS6. Therefore, we have a foliation of S6 where generic leaves are projective varieties; in particular, complex.
Moreover, the leaves are symplectic given by Goldman's 2-form; making them Kähler (generically). The symplectic structures on the leaves globalize to a Poisson structure on all of mathfrakX.
Is it possible that the complex structures on the generic leaves also globalize?
Here are some issues:
- As far as I know, the existence of complex structures on the leaves is generic. It is known to exist exactly when there is a correspondence to a moduli space of parabolic bundles. This happens for most, but perhaps not all, conjugation classes around the punctures (or marked points). So I would first want to show that all the leaves of this foliation do in fact admit a complex structure. Given how explicit this construction is, if it is true, it may be possible to establish it by brute force.
- Assuming item 1., then one needs to show that the structures on the leaves globalize to a complex structure on all of mathfrakX. Given that in this setting, the foliation is given by the fibers of the map: mathfrakXto[−2,2]times[−2,2] by [rho]mapsto(mathrmTr(rho(c1)),mathrmTr(rho(c2))) with respect to a presentation pi1(Sigma)=langlea,b,c1,c2|aba−1b−1c1c2=1rangle, it seems conceivable that the structures on the leaves might be compatible.
- Moreover, mathfrakX is not a smooth manifold. It is singular despite being homeomorphic to S6. So lastly, one would have to argue that everything in play (leaves, total space and complex structure) can by "smoothed out" in a compatible fashion. This to me seems like the hardest part, if 1. and 2. are even true.
Anyway, it is a shot in the dark, probably this is not possible...just the first thing I thought of when I read the question.
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