Hyperbolic Metric with respect to $\pi: \mathbb{H} \rightarrow {{\Delta}^{*}}$

Question

I have the following problem:

Find the unique metric $\rho=\rho(z)\left|dz\right|$ on the punctured unit disk $\Delta^{*}$ such that $\pi^{*}(\rho)=\left|dz\right|/(\mathrm{Im}(z))$ where $\pi: \mathbb{H} \rightarrow \Delta^{*}$ is the universal covering map. Then, show that the volume

$V(r)=\int_{0<\left|z\right|<r}\rho(z)^{2}\left|dz\right|^{2}$

is finite for each $0<r<1$ and compute its value. Finally, verify that $V(r)=2V(r^{2})$ and give the intuition here.

Here is what I have done thus far: I used the universal covering map $\pi(z)=e^{iz}$ which then induces a metric $\rho(z)=\frac{1}{z\log(z)}dz$. We get this because $\frac{(dx)+(dy)}{y}$ is the metric on the upper half plane that clearly corresponds to $\left|dz\right|/(r\log(r))$ where $\left|z\right|=r$ and $y=\log(r)$. This tells us that the circle has a length of $\frac{2{\pi}}{\log(r)}$, which tends to zero as $r$ goes to zero. I think this is all pretty reasonable so far. Now we check our pullback condition imposed on the universal covering map. That is ${\pi}^{*}(\rho(z))=\rho(\pi(z))\left|dz\right|$, which is clearly equivalent to $\left|dz\right|/(\mathrm{Im}(z))$ given how we have defined the universal covering map and the induced metric on the punctured unit disk.

All of this seems to work, but I am having difficulties computing the volume. I'm not exactly sure what is going wrong, but if someone could show me the finiteness of the integral and the computation I would be grateful. Also, what does it mean exactly to compute this volume with respect to the hyperbolic metric?

Answer 1

The density of a metric should be positive; what you really meant was $$ \rho(z)=\frac{1}{|z|(-\log |z|)} \,|dz| $$ Also, $\frac{dx+dy}{y}$ should be $\frac{\sqrt{dx^2+dy^2}}{y}$ or simply $\frac{|dz|}{\operatorname{Im}z}$.

The integration of $\rho^2$ is an exercise with polar coordinates. It helps that $\rho$ depends only on the polar radius, denoted $t$ below: $$ V(r) = 2\pi \int_0^r \frac{1}{t^2 \log^2 t}\,t\,dt = 2\pi \int_{-\infty}^{\log r} \frac{1}{s^2} \,ds = \frac{2\pi}{-\log r} $$ One interpretation of $V(r^2)=2V(r)$ would be: $z\mapsto z^2$ is locally an isometry of the punctured disk onto itself, and it covers it twice.

Also, what does it mean exactly to compute this volume with respect to the hyperbolic metric?

Volume with respect to metric $\rho(z)\,|dz|$ is the integral of $\rho^2$ over the relevant region.