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 $A=\left\{\begin{pmatrix} a & b \\ 0 & \frac{1}{a}\end{pmatrix}:a>0, b\in\mathbb{R}\right\}.$ 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 $B=\left\{\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}:b\in\mathbb{R}\right\}$, then $Im f$ is a positive harmonic function. We know that there exists no 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 a constant, then $Im f$ is not a constant, $M$ is a hyperbolic Riemann surface.

  If the monodromy of $f$ lies in $C=\left\{\begin{pmatrix} a & 0 \\ 0 & a^{-1}\end{pmatrix}:a>0 \right\}$, I also can prove it by considering the maximal abelian cover of $M$.

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.

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 $A=\left\{\begin{pmatrix} a & b \\ 0 & \frac{1}{a}\end{pmatrix}:a>0, b\in\mathbb{R}\right\}.$ 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 $B=\left\{\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}:b\in\mathbb{R}\right\}$, 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 a constant, then $Im f$ is not a constant, $M$ is a hyperbolic Riemann surface.

If the monodromy of $f$ lies in $C=\left\{\begin{pmatrix} a & 0 \\ 0 & a^{-1}\end{pmatrix}:a>0 \right\}$, I also can prove it by considering the maximal abelian cover of $M$.

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 $A=\left\{\begin{pmatrix} a & b \\ 0 & \frac{1}{a}\end{pmatrix}:a>0, b\in\mathbb{R}\right\}.$ 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 $B=\left\{\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}:b\in\mathbb{R}\right\}$, then $Im f$ is a positive harmonic function. We know that there exists no 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 a constant, then $Im f$ is not a constant, $M$ is a hyperbolic Riemann surface.

If the monodromy of $f$ lies in $C=\left\{\begin{pmatrix} a & 0 \\ 0 & a^{-1}\end{pmatrix}:a>0 \right\}$, I also can prove it by considering the maximal abelian cover of $M$.