# 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.

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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 $(x',y')$.

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Are you sure that $T_a$ is a isometry? – Jr. Jul 9 '12 at 4:58
@Jr.: Yes because $d(x+a) = dx$, since $a$ is constant. – Jason DeVito Jul 9 '12 at 12:29
I found: $(dT_a)_p(v_1,v_2) = (a+v_1,v_2)$ is it correct? – Jr. Jul 12 '12 at 15:38
I should be $(dT_a)_p(v_1, v_2) = (v_1,v_2)$. To see this, let $\gamma(t) = p + (tv_1,0)$. Then $\gamma'(0) = (v_1,0)$. But then $T_a(\gamma(t)) = a+p +(tv_1, 0)$ so $(T_a(\gamma(t))'|_{t=0} = (v_1,0)$ as well. – Jason DeVito Jul 12 '12 at 18:18
As you said "$\gamma '(0)= (v_1,0)$" but you should take a curve such that $\gamma '(0)= (v_1,v_2)$. Take $\gamma (t)=(x(t),y(t))$ then $(T_a \circ \gamma)(t)=(a+x(t),y(t))$ – Jr. Jul 12 '12 at 22:35

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 $y>0$.

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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.)

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