Let $\Sigma$ be a closed, oriented, smooth surface. Denote by $\mathcal{M}^{1}(\Sigma)$ the deformation space of unit volume Riemannian metrics on $\Sigma:$ here we consider two metrics equivalent if they differ via an orientation preserving diffeomorphism which is homotopic to the identity. For the time being, let us ignore whether we choose a Sobolev topology on this space, or the more natural, but difficult, Frechet topology. Given a metric $g\in\mathcal{M}^{1}(\Sigma),$ we can identify the tangent space $T_{g}(\mathcal{M}^{1}(\Sigma))$ with the space of $g$-divergence free, symmetric two tensors on $\Sigma.$ Given two such tensors $\alpha, \beta \in T_{g}(\mathcal{M}^{1}(\Sigma)),$ the $L^2$-pairing is defined via the expression (in local coordinates), \begin{align} \langle \alpha, \beta \rangle_{g}:=\int_{\Sigma} \alpha^{ij} \beta_{ij}\ dV_{g}, \end{align} where the indices of $\alpha$ are raised using the metric $g.$ The Hodge star operator associated to $g$ yields an almost-complex structure on $T_{g}(\mathcal{M}^{1}(\Sigma))$ given by viewing a symmetric two tensor as a one form with values in the cotangent bundle $T^{*}(\Sigma),$ and applying the Hodge star operator to the one form part. The question is the following: ignoring the fact of topologies and that we're in infinite dimensions, does the $L^2$-metric with this almost complex structure make $\mathcal{M}^{1}(\Sigma)$ a Kahler manifold. In particular, is this almost-complex structure constant along geodesics for the $L^2$-metric.
Some history is in order, the space of isotopy classes of constant negative curvature metrics $\mathcal{F}(\Sigma)$ on $\Sigma$ is identified (via the uniformization theorem) with the space of istopopy classes of complex structures on $\Sigma,$ which is the classical Teichmuller space. Under this identification, it has been known for a long while (nicely exposed in the book of Tromba "Teichmuller theory in Riemannian Geometry) that the Weil-Petersson pairing of holomorphic quadratic differentials can be identified with the $L^2$-metric. This should not be surprising as the Weil-Petersson pairing is an $L^2$-pairing. It was shown, first by Ahlfors, that the Weil-Petersson metric is Kahler. This is evidence for the question I ask above.
Thanks in advance for any insight into this question!
