The punctured unit disc has the complete riemannian metric with constant curvature -1

Question

Find how to construct this metric, find the distance under the metric between $(e^{-2\pi},0)$ and $(-e^{-\pi},0)$ This is a very interesting question, I have an idea ,construct Riemannian covering space from Upper Half plane. Is it right? There are several method to construct ,is the distance invariant regardless any construction? Help to solve this! This question is very important to me! Thanks!

Answer 1

If you are talking about the upper half plane with the Poincaré metric, the distance between those two points is not defined as those two points lie on the circle at infinity of the upper half plane (I am talking about hyperbolic geometry).

Answer 2

Interpreting the question as saying that the punctured unit disk is parametrized by exponentiating $z\rightarrow e^{iz}$ from the upper half-plane $H$, then, yes, the punctured disk inherits the "hyperbolic" structure from $H$. The two points are images of $\pi i$ and $2\pi i$, luckily lying on the imaginary axis. The distance between two points in the disk would be the inf of the distances between their pre-images in the upper half-plane. The hyperbolic metric $ds^2=(dx^2+dy^2)/y^2$ is easy to understand on the imaginary axis, namely, $d(ia, ib)=|\ln(a/b)|$. Thus, $d(\pi i,2\pi i)=\ln 2$.