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

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!

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Henry it usually helps if you separate the actual question from other comments you want to make in asking it, and use proper punctuation and less exclamation :) – Mariano Suárez-Alvarez Jul 3 '11 at 14:40

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, $d(\pi i,2\pi i)=\ln 2$.
Garett You need the requirement that $b > a$. In this case it is fine. – user38268 Jul 3 '11 at 22:24
@paul garrett we can also exponentiating $z\rightarrow e^{i \pi z}$ from the upper half-plane $H$,then the inherit distance will change? – henry Jul 4 '11 at 1:35
In response to comments: the absolute values on the log take care of the $b>a$-or-not issue. The fact that there are many inverse images is taken care of by taking inf. Yes, if a different uniformization map is used, the metric changes by a constant, so there is an issue of normalization, inevitably. That is, the distances themselves are not as canonical as ratios of distances. – paul garrett Jul 4 '11 at 15:31