Riemann surface arising as a quotient of the upper half-plane. Let $H$ be the upper half-plane $\{z \in \mathbb C \mid \Im(z) > 0\}$. For a fixed real $\lambda > 0$, let be the automorphism $$d_\lambda : H \to H, z \mapsto \lambda z .$$
Denote $\Gamma$ the subgroup of $\mathrm{Aut(H)}$ generated by $d_\lambda$. The goal is to determine the Riemann surface $H/\Gamma$.
Let's discard the trivial case $\lambda=1$. [Edit, false : $H/\Gamma$ is open as the open set $\{ z \in H \mid |z|<1 \}$ is a fundamental domain of the holomorphic projection.] One have $\pi_1(H/\Gamma) \simeq \Gamma \simeq \mathbb Z$. Hence, as no compact Riemann surface has $\mathbb Z$ as fundamental group, $H/\Gamma$ must be an open Riemann surface.
As I don't know a lot of open Riemann surfaces, I'm tempted to test $\mathbb C^\ast$ and the punctured unit disk $\mathbb D^\ast$ that matches those properties. But $\mathbb C^\ast$ has universal cover $\mathbb C \not\simeq H$, and so $H/\Gamma \not\simeq \mathbb C^\ast$. This leaves me with $\mathbb D^\ast$. How can I show that $H / \Gamma$ is (or isn't) biholomorphic to $\mathbb D^\ast$ ? If it isn't, does this quotient look like a well-know surface ?
P.S. Does that kind of quotient have a name ? I'm aware of the name of modular curve when $\Gamma$ is a subgroup of $PSL(2,\mathbb Z)$ which is not the case here.
 A: Without loss of generality, assume that $\lambda>1$. Then $\mathbb{H}/\Gamma$ is biholomorphic to the annulus 
$$A(\lambda):=\{z\in\mathbb{C}\mid e^{-\frac{2\pi^2}{\log\lambda}}<|z|<1\}.$$
To see this, let us fix a branch of the logarithm function on $\mathbb{H}$ as follows: 
$$\log(re^{i\theta})=\log r+i\theta,\quad~\forall r>0,~ \theta\in(0,\pi)~.$$ 
Define 
$$f:\mathbb{H}\to \mathbb{C},\quad z\mapsto e^{\frac{2\pi i\log z}{\log \lambda}}.$$ 
By definition, $f$ is holomorphic. It is easy to verify that $f(\mathbb{H})=A(\lambda)$ and  $f:\mathbb{H}\to A(\lambda)$ is a covering map with deck transformation group $\Gamma$. Since $\mathbb{H}$ is simply connected, the conclusion follows.

Remark: On the one hand, the generator $d_\lambda\in Aut(\mathbb{H})$ of $\Gamma$   is hyperbolic. When considered as an element in $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm I\}$, the conjugate class of $d_\lambda$ is uniquely determined by $(\mathrm{tr}~d_\lambda)^2$. On the other hand, up to biholomorphic equivalence, a hyperbolic Riemann surface with fundamental group $\mathbb{Z}$ is uniquely determined by its modulus. For $\mathbb{H}/\langle d_\lambda\rangle\cong A(\lambda)$, there is a $1$-$1$ correspondence between $(\mathrm{tr}~d_\lambda)^2$ and the modulus of  $A(\lambda)$.
A: Partial answer:
Let $f \colon H → ℂ^1 × (0,π), z ↦ (e^{2πi·\log_λ |z|},\operatorname{Arg}(z))$, where $ℂ^1$ is the unit circle.
This map is surjective, open, continuous, and $f(z) = f(z') ⇔ Γ_λ z = Γ_λ z'$, because:


*

*Any $(e^{2πiα},θ)$ in $ℂ^1 × (0,π)$ is hit by $λ^α·e^{2πiθ} ∈ H$.

*It’s a composition and product of open and continuous maps.

*$∃k ∈ ℤ:\, \log_λ |z|  = \log_λ |z'| + k ⇔ |z| = λ^k|z'|$, so:

*$f(z) = f(z') ⇔ \big(\operatorname{Arg}(z) = \operatorname{Arg}(z') ∧ ∃k ∈ ℤ:\, \log_λ |z|  = \log_λ |z| + k\big) ⇔ ∃k∈ℤ:\, z = d_{λ^k}z'$.
This implies that topologically $H/Γ \cong ℂ^1 × (0,π)$.
