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, $$g\cdot(a,b,c)=(gag^{-1},gbg^{-1},gcg^{-1}).$$ 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 $\mathfrak{X}\cong G^3/G\cong S^6$. 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 $\pi_1(\Sigma)=\langle a,b,c_1,c_2\ |\ aba^{-1}b^{-1}c_1c_2=1\rangle$, 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.