Upper half plane is complete with the Lobatchevski metric 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.
 A: Here's one possible approach:


*

*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.

*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 $(x',y')$.
A: 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 $y>0$.
A: I can't add a comment. So I'll post as an answer what is actually a comment. To see that $T_a$ is a isometry, you can proceed as follows. Clearly, $T_a$ is a bijection $\mathbb{H}^2\rightarrow\mathbb{H}^2$ which preserves the Riemannian distance induced from the Lobachevsky metric, since $dT_a$ preserves the length of curves. Indeed, if $\alpha : \left[ a,b\right]\rightarrow\mathbb{H}^2,t\mapsto\alpha(t)=(x(t),y(t))$, is any (sectionally) $C^1$ curve, then $T_a\circ\alpha$ is also a (sectionally) $C^1$ curve, with $(T\circ\alpha)'(t)=\alpha'(t)$ for all $t\in\left[a,b\right]$, so that $L(T_a\circ\alpha)=L(\alpha)$. Moreover, , given $p,q\in\mathbb{H}^2$, there exists a bijection $$\{\textrm{sectionally $C^1$ curves joining}\,\,p\,\,\textrm{to}\,\,q\}\leftrightarrow\{\textrm{sectionally $C^1$ curves joining}\,\,T_a(p)\,\,\textrm{to}\,\,T_a(q)\}.$$ (If $\beta$ is any sectionally $C^1$ curve joining $T_a(p)$ to $T_a(q)$, then $T_{-a}\circ\beta$ is a sectionally $C^1$ curve joining $p$ to $q$. It's clear that $L(\beta)=L(T_{-a}\circ\beta)$.) Thus $d(p,q)=d(T_a(p),T_a(q))$ for all $ p,q\in\mathbb{H}^2 $. Now, we can use a beautiful result due to Myers and Steenrod (vide Petersen, Riemannian Geometry, theorem 18, page 147) which establishes that a bijection $F:(M,g)\rightarrow (N,\overline{g})$ between Riemannian manifolds is a isometry provided that $F$ is distance-preserving. (In the case we're dealing with, we've a diffeomorphism, but this result shows that we only need a distance-preserving bijection.)
