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Let $X$ be a complex algebraic curve, or equivalently a compact Riemann surface, of genus $a\ge2$. Denote $\omega_1, \dots , \omega_a$ a basis of holomorphic differentials, and consider the Riemaniann metric $g = \omega_1 \circ \overline{\omega_1}+\dots + \omega_a \circ \overline{\omega_a}$ on $X$. I am interested in comparing $g$ with the hyperbolic metric $h$ (i.e. with constant curvature -1) on $X$ which is conformal to $g$.

I am interested in results comparing $h$ and $g$. In particular, there exist constants $c_1, c_2$ (depending on $a$ and $\omega_1, \dots , \omega_a$) such that $c_1 \ell_h(\gamma) \le \ell_g(\gamma) \le c_2 \ell_h(\gamma)$, where $\gamma$ is a curve in $X$ and $\ell_{h}, \ell_g$ denotes the lengths with respect to the two different Riemaniann structures?

I am not an expert on this topic, and I believe such comparisons should be well-studied in the literature, but I couldn't find an appropriate reference or keyword to help my search.

I am in particular interested in metrics of the form $g = \omega_1 \circ \overline{\omega_1}+\dots + \omega_a \circ \overline{\omega_a}$ because this is the metric on $X$ induced by the embedding of the curve in the Jacobian $X \to J(X) \cong \mathbb{C}^a\big/\Lambda$ given by the Abel-Jacobi map, expressed using the basis $\omega_1, \dots , \omega_a$.

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  • $\begingroup$ I think there is a typo in your definition of $g$. You should also specify what the constants $c_i$ depend on (genus or the complex structure of $X$, or a choice of a basis...). $\endgroup$ Commented Nov 20 at 18:24
  • $\begingroup$ Thank you for the comment: I have added the dependence on the genus and the basis of holomorphic differentials. $\endgroup$
    – LzB
    Commented Nov 20 at 18:44

1 Answer 1

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First of all, the constants $c_i$ will have to depend on the complex structure of $X$ since without prescribing a complex structure one cannot talks about dependence on a basis of the space of holomorphic differentials on $X$. Now, $h$ and $g$ are both Riemannian metrics on a compact manifold and, thus, the identity map $$ (X,g)\to (X,h) $$ is bi-Lipschitz with Lipschitz constants depending, of course, on $g$ and $h$. Suppose that $(X, g), (Y, h)$ are Riemannian manifolds and $f: (X, g)\to (Y, h)$ is an $L$-Lipschitz map. Then for every closed geodesic $c$ in $(X,g)$, the $h$-length $\ell_h(c)$ of $f(c)$ is at most $L\ell_g(c)$. Hence, if $c^*$ is a shortest closed geodesic in the free homotopy class of $f(c)$, we have $$ \ell_h(c^*)\le L\ell_g(c). $$ For $L$-bi-Lipschitz maps one obtains the reverse inequality by considering $f^{-1}$. Thus, for every free homotopy class $[c]$ in $X$ you obtain the inequality $$ L^{-1}\ell_g([c]) \le \ell_h([f(c)])\le L \ell_g([c]), $$ where $\ell_\bullet([\bullet])$ stands for the length of a shortest loop in the given free homotopy class.

Edit. Here is, what I think, the right question in the context of holomorphic differentials and hyperbolic metrics. For a fixed compact hyperbolic Riemann surface $X$ of genus $p$ let $L(X)$ denote the smallest bi-Lipschitz constant of identify maps $$ (X,h)\to (X,g), $$ where $h$ is the hyperbolic metric on $X$, the infimum is taken over all Riemannian metrics $g=\sum_{k=1}^p \omega_k \bar\omega_k$, where $\{\omega_1,...,\omega_p\}$ are bases in $\Omega^1(X)$. (Note that the identity map is conformal with respect to the metrics $h$ and $g$.) Let $\mathcal M_p$ denote the moduli space of compact Riemann surfaces of genus $p$.

Question. (1) Is $$ L_p:=\sup_{X\in {\mathcal M}_p} L(X) $$ finite? (2) If it is finite, what are the asymptotics of $L_p$ as $p\to\infty$?

I do not know how to answer this question and I do not remember seeing any work on this. (Clearly, for getting finiteness it suffices to analyze the behavior of degenerating families of genus $p$ Riemann surfaces. One should probably also analyze the case of Riemann surfaces of finite type so that one can run induction arguments. My guess is that $L_p$ is always infinite.)

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  • $\begingroup$ Thank you for the helpful answer. I was hoping for a specialized analysis for the special g I described in the question, so maybe I will wait for other suggestions before marking this as the accepted answer. $\endgroup$
    – LzB
    Commented Nov 21 at 22:10

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