Got it, thank you.

Thank you for your comment. The section s is a section of a fiber bundle over X whose fibers are isomorphic to the upper half-plane. I would like to know how to obtain a map from X to X using the section s.

I have understood the sentence “The field is contracting the area form". But I don't understand how to get the contradiction. Is it because the above sentence contradicts “the volume of $M$ is finite"? Could you please explain why? Thank you.

[1] R. C. Gunning, \textit{Special coordinate coverings of Riemann surfaces.} [2] R. Mandelbaum, \textit{Branched structures on Riemann surfaces.}

We can construct an affine connection (which can be found in [1] or [2]) on $\Sigma$, then we get a contradiction by a property of affine connection on a compact Riemann surface ([2] Lemma 2). We proved that $h:=\{h_{\alpha}:=\frac{f''}{f'}:U_{\alpha}\rightarrow\overline{\mathbb{C}}\}$ is an affine connection on $\Sigma$, where $\{U_{\alpha},z_{\alpha}\}$ is a complex atlas on $\Sigma$. If $\Sigma$ is $\mathbb{C}$, $\mathbb{C}-\{0\}$ or the punctured tori $\mathbb{T}\setminus\{q\}$, we also can obtain the conclusion. Moreover, we proved special cases for subgroup $B$ or $C$.

Thanks a lot. What do you mean by "this field is contracting the metric"? Let $\Sigma$ be a Riemann surface and $\{p_{i}\}_{i=1}^\infty\subset\Sigma$ is a closed discrete subset. $\mathrm{d} s^2$ is a (singular) hyperbolic metric with cusp or conical singularities at $p_{i}$. $f:\Sigma-\{p_{i}\}_{i=1}^\infty\longrightarrow\mathbb{H}$ is a developing map of $\mathrm{d} s^2$ and the monodromy of $f$ lies in $A$. So your answer proves $\Sigma$ is not compact. We proved it in another way, then we conjectured that $\Sigma$ must be a hyperbolic surface. The following are partial results.

The unit disk is a hyperbolic Riemann surface and the unit disk removes a finite number of points is a hyperbolic Riemann surface. We can construct a negative subharmonic function on it: $\log\mid z\mid$, where $z$ is the complex coordinate on the disk. The complex plane $\mathbb{C}$ and $\mathbb{C}$ removes a finite number of points are parabolic Riemann surfaces. The surface does not need to have finite genus and finite Euler characteristic.

Thank you for your consideration. There are some equivalent definitions for hyperbolic Riemann surface. Let $M$ be an open Riemann surface, then the following are equivalent: (1) There exist a Green's function on $M$ (with singularity at some point $P\in M$). (2) There exist a non-constant negative subharmonic function on $M$ (= there exists a non-constant bounded subharmonic function on $M$). (3) Brownian motion on $M$ is transient. (4) The maximum principle does not hold (for every compact set $K$).