# How can i prove that the Hyperbolic space is complete by using Divergent Curves?

Let $$H_+^2=\{(x,y)\in\mathbb{R}^2:\ y>0\}$$ and consider the Lobatchevski metric on $H_+^2$: $$g_{11}=g_{22}=\frac{1}{y^2},\ g_{12}=0$$

How can one prove the completeness property of this space by using divergence curves?

Here you can find the definition of Divergent Curve: http://ocw.mit.edu/courses/mathematics/18-994-seminar-in-geometry-fall-2004/lecture-notes/chapter19.pdf

Thanks

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Let $\gamma = (\gamma_1(t), \gamma_2(t))$ be any divergent curve. We need to show its length is infinite. To that end, let $N$ be any positive number. We'll show that the length of $\gamma$ is bigger than $N$.

Note that the length of $\gamma$ is $L = \int_0^\infty \frac{1}{\gamma_2(t)} \sqrt{\gamma'_1(t)^2 + \gamma'_2(t)^2} dt$

To that end, consider the rectangle $[\gamma(0) - L, \gamma(0) + L] \times [\epsilon, M]$ where $L$, $M$, and $\epsilon$ are to be determined. Note that this rectangle is compact since the topology is the usual one on the plane.

Since $\gamma(0)$ is in the rectangle, there is some maximum time $t_0$ such that $\gamma(t)$ is not in the rectangle for $t>t_0$. By considering a linear reparameterization $t' = \frac{t}{t_0}$ (which doesn't change lengths), we may assume wlog that $t_0 = 1$.

Now, there are 4 cases to consider. Either $\gamma(1)$ is on the top, bottom, left, or right edge of the rectangle. (If it hits a corner, say, the top left corner, then it's on top edge and left edge simultaneously).

Suppose $\gamma(1)$ hits the top or bottom. Then we have \begin{align*} \int_0^1 \frac{1}{\gamma_2(t)} \sqrt{\gamma_1'(t)^2 + \gamma_2'(t)^2} dt &\geq \int_0^1 \frac{|\gamma_2'(t)|}{\gamma_2(t)} dt\\ &\geq \left|\int_0^1 \frac{\gamma_2'(t)}{\gamma_2(t)} dt\right| \\ &= \left|\left.\ln(\gamma_2(t))\right]_0^1\right|\\ &= |\ln(\gamma_2(1)) - \ln(\gamma_2(0))|.\end{align*}

Now, if $\gamma(1)$ hits the top, then $\gamma_2(1) = M$, so the last line is $|\ln(M) - \ln(\gamma_1)|.$ Clearly, by picking $M$ large enough, we can make this bigger than $N$. If $\gamma(1)$ hits the bottom, then the last line is $|\ln \epsilon - \ln(\gamma(1))|$ and again, by choosing $\epsilon$ sufficiently close to $0$, we can make this bigger than $N$.

Finally, assume $\gamma(1)$ hits one of the sides of the rectangle. Then we have \begin{align*}\int_0^1 \frac{1}{\gamma_2(t)} \sqrt{\gamma_1'(t)^2 + \gamma_2'(t)^2} dt &\geq \int_0^1 \frac{|\gamma_1'(t)|}{\gamma_2(t)} dt\\ &\geq \int_0^1 \frac{|\gamma_1'(t)|}{M} \\ &\geq \left| \frac{1}{M}\int_0^1 \gamma_1'(t)dt \right| \\ &= \left| \left.\frac{1}{M} \gamma_1(t)\right]_0^1 \right| \\ &= \left| \frac{1}{M}(\gamma(0)\pm L - \gamma(0))\right| \\ &= \frac{L}{M}. \end{align*} So again, it's clear that if $L$ is big enough, then the length of $\gamma$ is at least $N$.

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Very nice answer Jason, i really liked it. Thank you. – Tomás Nov 24 '12 at 15:10
@Tomás: Glad I could answer at least one of your questions succesfully ;-) – Jason DeVito Nov 24 '12 at 15:17
Nah, you answer the other one too and i really appreciate your efforts for this. By the way, i posted a answer to that question (the energy question), please verify for me if it is correct. – Tomás Nov 24 '12 at 15:24

I keep posting this, but...

We can write down all the unit-speed geodesics in this space, very easily. There are only two types. Together, given a fixed point and a direction vector, we can find the geodesic passing through that point in the given direction.

Given real numbers $A$ of any sign, and $B > 0,$ unit speed parametrizations are:

$$A + i \, e^t \, ,$$ $$A + B \, \tanh t + i \, B \, \mbox{sech} \; t \,.$$

That's it, vertical lines that never reach the real axis, and semicircles with centers on the real axis that also never quite reach the real axis.

If you prefer to have your geodesic pass through your point at a specific time, you may replace $t$ by $t-t_0$ without harm.

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