Let $M$ be an open Riemann surface and $f$ a multivalued holomorphic function from $M$ to $\mathbb{H}$, where $\mathbb{H}$ is the upper half plane. Suppose that the monodromy of $f$ lies in the two-dimensional Lie subgroup $A$ of $PSL(2,\mathbb{R})$, i.e. $A=\left\{\begin{pmatrix} a & b \\ 0 & \frac{1}{a}\end{pmatrix}:a>0, b\in\mathbb{R}\right\}.$ I conjecture that $M$ must be a hyperbolic Riemann surface (The surface which is not compact and carries a negative non-constant subharmonic function). But I cannot prove it. This problem has been bothering me for a long time, I will be very grateful for any answers or suggestions.

Special case: If the monodromy of $f$ lies in the one-dimensional subgroup $B=\left\{\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}:b\in\mathbb{R}\right\}\subset A$, then $Im f$ is a positive harmonic function. We know that there exists on nonconstant positive harmonic function on a parabolic Riemann surface (The surface which is not compact and does not carry a negative non-constant subharmonic function). Since $f$ is not constant, then $Im f$ is not constant, $M$ is a hyperbolic Riemann surface. If the monodromy of $f$ lies in the one-dimensional subgroup $C=\left\{\begin{pmatrix} a & 0 \\ 0 & a^{-1}\end{pmatrix}:a>0 \right\}\subset A$, I also can prove it by considering the maximal abelian cover of $M$.

This problem comes from our consideration on monodromy properties of singular hyperbolic metrics on Riemann surfaces.

  • $\begingroup$ Dear Yu, is there some other characterisation of hyperbolic surfaces that doesn't use sub-harmonic functions? It seems to me that according to your definition a surface is hyperbolic if an only if it is not a punctured surface? If I am wrong, could you please give me the simplest example that contradict my statement? Also, do you assume that your surface has finite genus and has finite Euler characteristic? $\endgroup$ – Dmitri Panov Jan 30 '20 at 15:45
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    $\begingroup$ 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 $K$). $\endgroup$ – Yu Feng Jan 30 '20 at 21:22
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    $\begingroup$ 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. $\endgroup$ – Yu Feng Jan 30 '20 at 21:22

I want to say something less trivial, so I would like to propose an approach to this question in the case when the Riemann surface $M$ is a punctured Riemann surface (without boundary) of finite genus. I claim that such an example doesn't exist. This goes in the direction of what the question is asking for.

Proof. The main idea of the proof is that in the situation as I've described, if we take the pull-back $f^*(g)$ of the hyperbolic metric on $\mathbb H$, the metric $f^*(g)$ is a hyperbolic metric with cusps and a finite number of conical points of angles $ 2\pi n$ ($n\in \mathbb Z_+$) on $M$. I will justify this statement later on, but first I want to say how to get a contradiction from it.

Getting contradiction. Indeed, the monodromy of the metric $f^*(g)$ (with respect to the developing map) is just the same as the monodromy of $f$, and I claim that the monodromy of a punctured surface with cusps at punctures and a finite number of conical points of angles $2\pi n$ can not lie in $A$.

Indeed, suppose by contradiction that the monodromy lies in $A$. Let us take the vector field $y\frac{\partial}{\partial y}$ on $\mathbb H$. Let $v=f^*(y\frac{\partial}{\partial y})$ be the pullback. Note that this pull-back field is well defined on $M\setminus \{crit( f)=conical \ points\}$ since $A$ preserves $y\frac{\partial}{\partial y}$ . Furthermore, this field $v$ has norm $1$ in the hyperbolic metric, it is analytic, and it is defined everywhere apart from the branching locus $crit (f)$. So the flow of this field is well defined outside of a subset of $M$ of measure zero. However, this field is contracting the area form corresponding to the metric (this is clear for the field $y\frac{\partial}{\partial y}$ on $\mathbb H$, in time $t$ it contracts the area form by the factor $e^{-t}$. The flow is an isometry on vertical lines and a contraction of horizontal lines, corresponding to horocircles of infinity on $\mathbb H$). At the same time, by Gauss-Bonnet formula, since $M$ has only finite number of cusps and finite number of conical points, the volume of $M$ with respect to the pull-back hyperbolic metric is finite. This is a contradiction.

Let us now prove that $f^*(g)$ is indeed a metric with finite number of cusps (close to the punctures of $M$) and finite number of conical points. Let $p$ be one of punctures of $M$ and let $\dot D\subset M$ be a punctured disk in $M$ whose puncture is at $p$. Then the monodromy around the puncture is an element $\rho\in A$. Clearly, the map $f$ induces to us a map $\tilde f: \dot D\to \mathbb H/\rho$. Now, the quotient $\mathbb H/\rho$ is either a) punctured disk or b) a cylinder, or c) a disk (if $\rho$ is the identity). In cases b) and c) the map $\tilde f$ extends to the whole disk $D$ and so in reality the monodromy around $p$ is trivial and we can extend $f$ to this puncture. If, on the other hand $\mathbb H/\rho$ is a punctured disk, again we can extend $\tilde f$ to the map from the whole $D$ to the one point completion of $\mathbb H/\rho$. In such a case $f^*(g)$ has a cusp at $p$. It is also not hard to see that the number of critical points of $f$ is finite, they can not accumulate towards the cusp. This finishes the proof of the statement.

Old answer. In this answer I was considering the trivial case when the monodromy is $\mathbb Z$.

If the monodromy of $f$ is $A=\left\{\begin{pmatrix} a & b \\ 0 & \frac{1}{a}\end{pmatrix}:a>0, b\in\mathbb{R}\right\},$ let $\mathbb Z$ denote the corresponding group acting on $\mathbb H$ and consider $\mathbb H/\mathbb Z$. This quotient is biholomorphic to either a punctured disk - in case $a=1, b\ne 0$, or a cylinder, if $a\ne 1$, or $\mathbb H^2$ if the monodromy is trivial. Clearly, from the multivalued map $f: M \to \mathbb H^2$ we get a genuine holomorphic map $\tilde f: M\to \mathbb H^2/\mathbb Z$.

Note finally that on a punctured disk, on a cylinder, or on $\mathbb H^2$, we can always construct a negative non-constant harmonic function $\phi$, and we can just take the pull-back $\tilde f^*(\phi)$ on $M$, which will also be negative, non-constant and harmonic on $M$.

  • $\begingroup$ Thanks for your comment. I don't understand why $\widetilde{f}$ is single-valued? For the case $a=1, b\neq0$, the quotient group is the additive integer group. If I consider the universal covering from $\mathbb{H}$ to $\mathbb{D}^{*}$, $z\mapsto e^{iz}$, where $\mathbb{D}^{*}$ is the punctured disk, how can we get $\widetilde{f}$? $\endgroup$ – Yu Feng Jan 28 '20 at 8:10
  • $\begingroup$ Maybe I have misunderstood what you mean by monodromy. But I think that from what you write in your question it follows that at each point $x$ of $M$ the values of $f$ lies in one orbit of the action of $\mathbb Z$ on $\mathbb H$. Correct? What I say is that the space of such orbits is the quotient $\mathbb H/\mathbb Z$. In this quotient space to each point $x\in M$ corresponds only one point - the corresponding orbit. So the map $\tilde f$ is the composition of $f$ with the quotient map $\mathbb H \to \mathbb D^*$. Does this make sense? $\endgroup$ – Dmitri Panov Jan 28 '20 at 8:30
  • $\begingroup$ Let $M$ be a Riemann surface and $x_{0}$ a base point on $M$. Suppose that the monodromy of $f$ lies in the group $A=\left\{\begin{pmatrix} a & b \\ 0 & \frac{1}{a}\end{pmatrix}:a>0, b\in\mathbb{R}\right\},$ i.e. its monodromy representation is a group homomorphism $\mathcal{M}_{f}: \pi_{1}(M, x_{0})\longrightarrow A.$ For each single-valued analytic branch $\mathfrak{f}$ of $f$, $\mathcal{M}_{\gamma}(\mathfrak{f})=\frac{a_{\gamma}\mathfrak{f}+b_{\gamma}}{a_{\gamma}^{-1}}$, where $\begin{pmatrix} a_{\gamma} & b_{\gamma} \\ 0 & a_{\gamma}^{-1}\end{pmatrix}\in A$, $\gamma\in \pi_{1}(M, x_{0})$. $\endgroup$ – Yu Feng Jan 29 '20 at 17:15
  • $\begingroup$ In other words, the monodromy of $f$ is the image of its monodromy representation. So at each point $x$ of $M$, there is a one-to-one correspondence between the values of $f$ at $x$ and its monodromy group. Actually I don't quite understand your proof. I think the function $\widetilde{f}$ you construct is not necessarily single-valued. $\endgroup$ – Yu Feng Jan 29 '20 at 17:15
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    $\begingroup$ I already understand what you mean. Sorry for the misunderstanding. I think I should edit my question again to make it more clear and complete. $\endgroup$ – Yu Feng Jan 30 '20 at 21:15

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