Upper half plane is complete with the Lobatchevski metric

Question

How do I show that the Upper half plane is complete with the Lobatchevski metric? I tried to use the fact that $M$ is complete iff the lengh of any divegert curve is unbounded,but did not get any results.thanks.

Answer 1

Here's one possible approach:

1. If a Riemannian manifold is homogeneous (meaning for any pair of points there is an isometry moving one to the other), then it is complete.

2. The upper half plane with Lobatchevski metric is homogeneous.

To prove 1, argue as follows: Pick a point $p$. Then for some $\epsilon > 0$, the exponential map is defined on all vectors of length less than $\epsilon$. By homogeneity, this $\epsilon$ works at all points. Intuitively, this means that from any point and in any direction, a geodesic is allowed to flow a least a distance $\epsilon$. This, in turn, easily implies all geodesics are defined for all time.

To prove 2, recall the metric is $ds^2 = \frac{1}{y^2}(dx^2 + dy^2)$. Now, show that if $T_a(x,y) = (a+x,y)$, then $T$ is an isometry. This implies we can move any point to one of the form $(0,y)$. Next, show the map $D_\lambda(x,y) = (\lambda x, \lambda y)$ is an isometry for $\lambda > 0$.

Putting these together shows the hyperbolic plane is homogeneous: To move $(x,y)$ to $(x',y')$, move $(x,y)$ to $(0,y)$ using $T_{-x}$, then use $D_{y'/y}$ to move $(0,y)$ to $(0,y')$, then use $T_{x'}$ to move $(0,y')$ to $(x',y')$.

Answer 2

Try to show the following: if a sequence of points $(x_n,y_n)$ is Cauchy wrt to the Lobatchevskii/hyperbolic/Poincare metric, then $(x_n)$ and $(\log y_n)$ are Cauchy sequences of real numbers. This will imply that $(x_n,y_n)\to (x,y)$ with $y>0$.