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It is well-known that on closed oriented surfaces $S$, conformal classes of metrics on $S$ correspond bijectively to complex structures on $S$. My understanding is that this correspondance goes as follows: Given a metric $g$ on $S$, there is a unique almost-complex structure $J$ (which is automatically integrable in dimension 2) such that $g(u,Ju) = 0$ and that $u \wedge Ju$ is positively oriented for all tangent vectors $u$. i.e. the metric $g$ gives a way to "rotate 90 degrees counter-clockwise". Clearly, given a conformally equivalent metric $e^ug$, this construction gives the same $J$. Conversely, given an almost-complex structure $J$, we can take any metric $g$ and form the metric $h(u,v) = g(u,v) + g(Ju,Jv)$. Then since it is symmetric, we have $h(u,Ju) = h(Ju,JJu) = -h(u,Ju)$ so $h(u,Ju) = 0$, confirming that we have a bijection.

Now let $S_1$ and $S_2$ be two Riemann surfaces and let $[g_1]$ and $[g_2]$ be the conformal classes of metrics corresponding to their complex structures. If $f : S_1 \to S_2$ is holomorphic, how are the two classes $[g_1]$ and $f^*[g_2]$ on $S_1$ related? In particular, if $f$ is a biholomorphism, does it follow that $[g_1] = f^*[g_2]$?

If $f : (S_1,J_1) \to (S_2,J_2)$ is a holomorphic map, then $df \circ J_1 = J_2 \circ df$ where $df$ is the differential of $f$. So if $f$ is actually a biholomorphism and the underlying spaces $S_1 = S_2 = S$ are the same, then the two complex structures $J_1$ and $J_2$ would be conjugate; $(df) \circ J_1 \circ (df)^{-1} = J_2$. This suggests that my last statement in the second paragraph should not be true as is since these two almost-complex structures don't seem to give the same orthogonal vectors in the construction of the first paragraph. Still, something almost as strong should be true but I can't exactly figure out what.

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First question:

If $f : S_1 \to S_2$ is holomorphic, how are the two classes $[g_1]$ and $f^*[g_2]$ on $S_1$ related? In particular, if $f$ is a biholomorphism, does it follow that $[g_1] = f^*[g_2]$?

Yes, provided that the derivative of $f$ is nowhere zero. At critical points of $f$, the pullback $f^*g_2$ is not a metric.

Second question:

So if $f$ is actually a biholomorphism and the underlying spaces $S_1 = S_2 = S$ are the same, then the two complex structures $J_1$ and $J_2$ would be conjugate; $(df) \circ J_1 \circ (df)^{-1} = J_2$. This suggests that my last statement in the second paragraph should not be true as is since these two almost-complex structures don't seem to give the same orthogonal vectors in the construction of the first paragraph.

In the situation you described, you said $S_1=S_2=S$, but you didn't say that $J_1$ and $J_2$ are the same complex structure on $S$. So yes, if $f$ is a biholomorphism from $(S,J_1)$ to $(S,J_2)$, then the two complex structures $J_1$ and $J_2$ are merely conjugate in the sense you described.

My guess is that what you were really thinking about was a biholomorphism $f$ from a Riemann surface $(S,J)$ to itself; in this case, $J_1=J_2=J$, and the relation between complex structures becomes $(df) \circ J \circ (df)^{-1} = J$. The meaning of this equation might be a little clearer if we indicate the points where everything is being evaluated. Let $p$ be an arbitrary point of $S$, and set $q=f(p)$. Then the equation reads $$ (df_p) \circ J_p \circ (df_p)^{-1} = J_q. $$ What this says is that $df_p$ is a complex-linear map from $T_pS$ to $T_qS$.

Another way to look at it is that if $f\colon S\to S$ is a diffeomorphism, and $J$ is a complex structure on $S$, then we can define a new complex structure $f^*J$ by $$ (f^*J)_q = (df_p) \circ J_p \circ (df_p)^{-1}, $$ where $p=f^{-1}(q)$. Then the statement that $f$ is holomorphic with respect to $J$ is equivalent to $f^*J=J$.

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  • $\begingroup$ Thank you for your answer. I really meant $J_1 \neq J_2$ in my second question. If I understand correctly, if $f : (S,J_1) \to (S,J_2)$ is a biholomorphism with $df$ never zero, then with $[g_1]$ and $[g_2]$ the associated conformal structures, we have $(f^*g_2)(u,J_1u) = g_2((df)u,J_2 (df)u) = 0$ so that $[g_1] = [f^*g_2]$. So if $(S,J_1)$ and $(S,J_2)$ are biholomorphic, the strongest statement is that $(S,g_1)$ is pointwise conformal to some $(S,g)$ which is isometric as a Riemannian manifold to $(S,g_2)$, but we don't necessarily have $[g_1] = [g_2]$. $\endgroup$ – jef808 Oct 19 '14 at 18:03
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    $\begingroup$ Yes, exactly. For example, if $(S,J_2,g_2)$ is the sphere with its standard complex structure and standard metric, and $f\colon S\to S$ is any non-conformal diffeomorphism, we can set $J_1 = f^*J_2$ and $g_1 = f^*g_2$, and then $f\colon (S,J_1)\to (S,J_2)$ is biholomorphic, but $g_1$ is not pointwise conformal to $g_2$. $\endgroup$ – Jack Lee Oct 19 '14 at 18:09

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