Here is a shot in the dark (Disclosure: I really know nothing about this problem).
Let $G:=mathsf{SU}(2)$ act on $G^3$ by simultaneous conjugation; namely, $$gcdot(a,b,c)=(gag^{-1},gbg^{-1},gcg^{-1}).$$ Then the quotient space is homeomorphic to $S^6$ (see Bratholdt-Cooper).
The evaluation map shows that the character variety $mathfrak{X}:=mathrm{Hom}(pi_1(Sigma),G)/G$ is homeomorphic to $G^3/G,$ 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 $mathfrak{X}cong G^3/Gcong S^6$. Therefore, we have a foliation of $S^6$ 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 $mathfrak{X}$.
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 $mathfrak{X}$. Given that in this setting, the foliation is given by the fibers of the map: $mathfrak{X}to [-2,2]times [-2,2]$ by $[rho]mapsto (mathrm{Tr}(rho(c_1)),mathrm{Tr}(rho(c_2)))$ with respect to a presentation $pi_1(Sigma)=langle a,b,c_1,c_2 | aba^{-1}b^{-1}c_1c_2=1rangle$, it seems conceivable that the structures on the leaves might be compatible.
- Moreover, $mathfrak{X}$ is not a smooth manifold. It is singular despite being homeomorphic to $S^6$. 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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