Let \[ A(R)=\{z:1<|z|<R\}. \]
Give an explicit analytic covering map \[ \pi:\mathbb H\to A(R). \]
Compute the hyperbolic metric on \(A(R)\).
Show that \(|z|=\sqrt R\) is a closed geodesic in \(A(R)\), and compute its hyperbolic length.
Solution
Write \(L=\log R\).
1. An explicit universal covering is\[ \boxed{ \pi(\zeta) =\exp\!\left(-\frac{iL}{\pi}\Log\zeta\right), \qquad \zeta\in\mathbb H,} \]where \(0<\arg\zeta<\pi\).
The principal logarithm on \(\mathbb H\) maps it conformally to the horizontal strip \[ 0<\Im u<\pi. \] Multiplication by \(-iL/\pi\) maps this strip conformally to \[ V=\{w:0<\Re w<L\}. \] Finally, \(w\mapsto e^w\) is a covering \(V\to A(R)\), since \[ 1<|e^w|=e^{\Re w}<e^L=R, \] with deck transformations \(w\mapsto w+2\pi i k\). Their composition is the displayed covering map.
2. The hyperbolic metric on the strip \(V=\{0<\Re w<L\}\) is\[ \rho_V(w) =\frac{\pi}{L} \frac{|dw|}{\sin(\pi\Re w/L)}. \]
The map \[ F(w)=\exp\!\left(\frac{i\pi w}{L}\right) \] sends \(V\) conformally onto \(\mathbb H\). If \(w=u+iv\), then \[ |F'(w)|=\frac{\pi}{L}|F(w)|, \qquad \Im F(w)=|F(w)|\sin(\pi u/L). \] Pulling back the metric \(|d\zeta|/\Im\zeta\) on \(\mathbb H\) therefore gives \[ F^*\rho_{\mathbb H} =\frac{\pi}{L} \frac{|dw|}{\sin(\pi u/L)}. \]
3. The hyperbolic metric on \(A(R)\) is\[ \boxed{ \rho_{A(R)}(z) =\frac{\pi}{\log R} \frac{|dz|} {|z|\sin\!\left(\pi\log|z|/\log R\right)}.} \]
Use the covering \(z=e^w\) from \(V\) to \(A(R)\). Then \[ |dw|=\frac{|dz|}{|z|}, \qquad \Re w=\log|z|. \] Substituting these identities into 2 gives exactly the displayed metric.
4. The line \(\Re w=L/2\) is a geodesic in \(V\).
Under the conformal isometry \(F\) from 2, \[ w=\frac L2+iv \quad\longmapsto\quad F(w)=i e^{-\pi v/L}, \] which is the positive imaginary axis. The latter is a hyperbolic geodesic in \(\mathbb H\), so its inverse image is a geodesic in \(V\).
5. This geodesic descends to the closed curve \(|z|=\sqrt R\) in theannulus.
Under \(z=e^w\), the line \(\Re w=L/2\) satisfies \[ |z|=e^{L/2}=\sqrt R. \] The deck translation \(w\mapsto w+2\pi i\) identifies one period of the line, so its image is a closed geodesic in \(A(R)\).
6. Its hyperbolic length is\[ \boxed{\frac{2\pi^2}{\log R}}. \]
On \(\Re w=L/2\), the sine factor in 2 equals \(1\). One period is \(w=L/2+iv\), \(0\le v\le2\pi\). Hence \[ \operatorname{length} =\int_0^{2\pi}\frac{\pi}{L}\,dv =\frac{2\pi^2}{L} =\frac{2\pi^2}{\log R}. \]