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.

  • $\begingroup$ For $λ = 1$ isn’t $d_λ = \operatorname{id}$, so $Γ_λ = \{\operatorname{id}\}$ and $H/Γ_λ = H$? Edit: Oh, I think I misinterpreted your paragraph beginning with “Let’s discard …”. My bad, nevermind. … No, I didn’t? You claim $\{z ∈ H;; |z| < 1\}$ to be a fundamental domain. This isn’t true? $\endgroup$ – k.stm May 11 '13 at 9:46
  • $\begingroup$ I think, you think $〈d_λ〉 = \{d_{kλ};\; k ∈ ℤ\}$, but it really is $\{d_{λ^k};\; k ∈ ℤ\}$. Note that $d_λ ∘ d_λ = d_{λ^2}$. $\endgroup$ – k.stm May 11 '13 at 9:53
  • $\begingroup$ Yes the case $\lambda=1$ is trivial and $H/\Gamma = H$ then. By discarding it, I meant that I now suppose $\lambda \neq 1$. Then every $z_0 \in H$ as a equivalent one in $\{z \in H : |z| < 1\}$ as $(\lambda^{-k}z_0)_{k \geq 0}$ or $(\lambda^{k}z_0)_{k \geq 0}$ converge to $0$ in $\mathbb C$. Right ? (Actually, every neighborhood of $0$ in $\mathbb C$ intersected with $H$ is a fundamental domain.) $\endgroup$ – Pece May 11 '13 at 9:56
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    $\begingroup$ Don’t you require a fundamental domain to contain unique representatives of an orbit? In my view, $\{z ∈ H;\; 1 < |z| < λ\}$ is a fundamental domain for $λ > 1$. $\endgroup$ – k.stm May 11 '13 at 10:03
  • $\begingroup$ Oh sorry, I just meant a set $U$ such that $p(U) = H/\Gamma$ (with $p$ the projection $H \to H/\Gamma$). But I just realize that it does not show that $H/\Gamma$ is not compact. However, it seems to be true because of the fundamental group being isomorphic to $\mathbb Z$. $\endgroup$ – Pece May 11 '13 at 10:12

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 $A(\lambda)$.

  • $\begingroup$ Thanks, it seems to solve the question. Knowing the covering map $H \to \mathbb D^\ast, \, z \mapsto e^{2i\pi z}$ (for the case $\Gamma = \{z \mapsto z+k : k \in \mathbb Z\}$), I was trying to get something similar but invariant under the described $\Gamma$ without success. Your $f$ does the job ! $\endgroup$ – Pece May 11 '13 at 18:29
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    $\begingroup$ @Pece: You are welcome. Please be noted that the generator of $\Gamma$ in your comment is called parabolic, but the one in your question is called heperbolic. These are two different types of Möbius transformations with different geometric features. $\endgroup$ – 23rd May 11 '13 at 18:35
  • $\begingroup$ @Pece: For further reference, I added a remark in my answer. $\endgroup$ – 23rd May 11 '13 at 19:16

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 $H/Γ \cong ℂ^1 × (0,π)$.

  • $\begingroup$ Also, one easily has $ℂ^1 × (0,π) \cong π D^*$, but I don’t know if one can combine these maps to get an isomorphism of Riemann surfaces. $\endgroup$ – k.stm May 11 '13 at 10:58

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