Let $\Sigma$ be a compact, connected, oriented surface, and let $\rho\in\Omega^2(\Sigma)$ be a fixed volume form. Then any (almost) complex structure $J\in\Omega^0(M;\operatorname{End}TM)$ compatible with the orientation gives us a Riemannian metric via $$g(-,-) = \rho(-,J-).$$ Now let $\psi:\Sigma\to\Sigma$ be a diffeomorphism. Then $\tilde{J} = \psi^*J$ induces a possibly different metric $\tilde{g}$. We have $$\begin{align} \tilde{g}(-,-) = & \rho(-,\tilde{J}-)\\ = & \rho(-,d\psi^{-1}(J\circ\psi)d\psi-)\\ = & ((\psi^{-1})^*\rho)(d\psi-,(J\circ\psi)d\psi-) \end{align}$$ and as $(\psi^{-1})^*\rho$ at $\psi(m)$ is given by $f(\psi(m))\rho_{\psi(m)}$, where $f:\Sigma\to\mathbb{R}_{>0}$ is some function, we obtain $$\tilde{g} = \psi^*(fg).$$ I need to find $f$ in function of $\psi$ (as this would allow me, for example, to understand how the curvature changes when we change $J$ by the pullback via a diffeomorphism). Are there known results for this? Any help or hint on how to proceed would be appreciated.
Notes: The following things are true in the case I need, but I don't know whether they are useful:
- The genus of $\Sigma$ is at least $2$.
- The diffeomorphism $\psi$ can be taken to be homotopic to the identity.
- As $\rho$ can be thought of as a symplectic form, we are in fact working with Kähler metrics.
My progress:
Write $\phi=\psi^{-1}$, $x=\psi(m)$. Let $v_1,v_2\in T_x\Sigma$, $w_1,w_2\in T_{\phi(x)}\Sigma$ be bases and write $$d\phi_x\cdot(a^iv_i) = d\phi_i^jw_j.$$ Then (using Einstein summation convention): $$\begin{align} (\phi^*\rho)_x(a^iv_i,b^jv_j) = & \rho_{\phi(x)}(d\phi_i^ka^iw_k,d\phi_j^lb^jw_l)\\ = & d\phi_i^ka^id\phi_j^lb^j\rho_{\phi(x)}(w_k,w_l) \end{align}$$ and using the fact that $\rho_{\phi(x)}(w_k,w_l)$ is zero if $k=l$ and writing down explicitly all the summands: $$\begin{align} = & \rho_{\phi(x)}(w_1,w_2)\big(d\phi_i^1a^id\phi_j^2b^j - d\phi_i^2a^id\phi_j^1b^j\big)\\ = & \rho_{\phi(x)}(w_1,w_2)\big(d\phi_i^1a^id\phi_j^2b^j - d\phi_i^2a^id\phi_j^1b^j\big)\\ = & \det\pmatrix{b^1&a^1\\b^2&a^2}\det(d\phi_x)\rho_{\phi(x)}(w_1,w_2). \end{align}$$ At the same time, we have: $$\rho_x(a^iv_i,b^jv_j) = \det\pmatrix{b^1&a^1\\b^2&a^2}\rho_x(w_1,w_2),$$ so that we conclude: $$(\phi^*\rho)_x = \frac{\rho_{\phi(x)}(w_1,w_2)}{\rho_x(v_1,v_2)}\det(d\phi_x)\rho_x.$$ The first term compensates for the changes in the determinant coming from a change of basis, so that this expression is independent of choice.
Is there a nicer way to write down this formula (without having to write terms a priori dependent from the choice of basis)?
Let us use local coordinates $x^1$, $x^2$ around $x_0$, and local coordinates $y^1$, $y^2$ around $y_0 = \phi(x_0)$. Assume that around $x_0$, we have
$\rho = h(x) dx^1 \wedge dx^2$
and that around $y$, we have
$\rho = k(y) dy^1 \wedge dy^2$.
We then have
$\phi^*(\rho) = k(\phi(x)) \det(\frac{\partial y^i}{\partial x^j}) dx^1 \wedge dx^2$
Comparing, we get that $f(x) = \frac{k(\phi(x))}{h(x)} \det(\frac{\partial y^i}{\partial x^j})$
This is equivalent to the OP's formula (with a shorter proof perhaps). There is probably some way of writing this more invariantly, but I don't think it can be simplified much further, to answer the OP's question.