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I am trying to understand the geodesic flow on the following 2-dimensional Riemannian manifold $M$. As a set, $M$ is the interior of the standard 2-simplex,

$$M=\{(x,y)\in\mathbb{R}^2\mid x,y>0~\text{and}~x+y<1\}.$$

The metric on $M$ is given by

$$ds^2=\left(\frac{1}{x}+\frac{1}{1-x-y}\right)\,dx^2+2\left(\frac{1}{1-x-y}\right)\,dxdy+\left(\frac{1}{y}+\frac{1}{1-x-y}\right)\,dy^2.$$

A direct calculation shows that $M$'s Gaussian curvature is given by the positive constant $K=\frac{1}{4}$. Therefore, the Killing-Hopf theorem tells us that $M$ is not geodesically complete; if it were, Killing-Hopf, together with the simple observation that $M$ is simply connected, would imply that $M$ is isometric to the sphere, which is not true.

My question has two parts:

(1) Is the dimension of $M$'s isometry group greater than zero? Note that, even though $M$ is locally isometric to the sphere, with its 3-dimensional isometry group, $M$'s isometry group must have fewer dimensions than 3. If the dimension was $3$, $M$ would be a homogeneous space, and would therefore be geodesically complete (impossible for the reason stated above).

(2) Can you find an example of a geodesic that is not defined on all of $\mathbb{R}$? I'm having trouble visualizing "why" $M$ is not-geodesically complete. Is the problem that certain geodesics can run off the simplex in finite time, or is it something else?

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  • $\begingroup$ Is $f(x,y) = (y,x)$ an isometry? It seems as though, by the symmetry in the coefficients, that this should be an isometry. I don't know if it has any other isometries. $\endgroup$ – Jason DeVito Feb 26 '15 at 2:23
  • $\begingroup$ You are right! What I meant to ask was if the dimension of the isometry group is greater than zero. I'll fix this issue. $\endgroup$ – Josh Burby Feb 26 '15 at 2:34
  • $\begingroup$ Wouldn't it suffice to find a curve (I don't think it has to be a geodesic) joining a point on the interior to a point on the boundary that has finite length? I think if you picked any curve that approaches the boundary and computed its length, you would find that it's finite. $\endgroup$ – Phillip Andreae Feb 26 '15 at 3:47
  • $\begingroup$ @PhillipAndreae I think I understand what you're suggesting. If you had some finite-length curve $\gamma:[0,1)\rightarrow M$ without a limit (in $M$) as $t\rightarrow 1$, then you could use it to construct a non-convergent Cauchy (in the distance metric) sequence. Is that roughly correct? $\endgroup$ – Josh Burby Feb 28 '15 at 5:27
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I will address your second question, in which you asked for a geodesic that isn't defined on all of $\mathbb{R}$.

Consider the curve $\gamma(t) = \left( t, t \right)$ for $t \in \left(0, \frac{1}{2} \right)$. Now, $\gamma$ isn't a geodesic (because its speed isn't constant), but its image is the image of a geodesic; i.e., a reparametrization of $\gamma$ is a geodesic. The reason for this is that the image of $\gamma$ lies in the set $\{y=x \} \subset M$, which is the fixed point set for the map $(x,y) \to (y,x)$, which is an isometry, as mentioned in the comments. Fixed point sets of isometries are totally geodesic submanifolds (see this MSE question), so in particular the set $\{y=x\}$ is the image of a geodesic.

So we could reparametrize $\gamma$ to obtain a geodesic $\tilde{\gamma}$ with the same image such that $\tilde{\gamma}$ is parametrized with respect to arc length. Let's say the maximal domain of $\tilde{\gamma}$ is some interval $I \subset \mathbb{R}$, where the length of $I$ is the length of $\tilde{\gamma}$. If $M$ were geodesically complete, $I$ would have to be all of $\mathbb{R}$, meaning in particular the length of $\tilde{\gamma}$ would be infinite. But length is invariant under reparametrizations, so $\gamma$ and $\tilde{\gamma}$ have the same length, and we can directly compute (or at least estimate) the length of the original curve $\gamma$ and show that it's finite.

I'll briefly outline the computation of the length of $\gamma$. $\gamma'(t) = \frac{\partial}{\partial x} + \frac{\partial}{\partial y}$, so $$||\gamma'(t) || = \sqrt{ \frac{2}{t} + \frac{4}{1 - 2t}}$$ We just need to show that the following improper integral converges, i.e., is finite: $$\int_0^{1/2} \sqrt{ \frac{2}{t} + \frac{4}{1 - 2t}} \, dt$$ I'll leave the details to you! (I wouldn't try to compute it exactly, just estimate it from above.)

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  • $\begingroup$ Thanks for this! The value of the integral is $\pi$!Is it true more generally that if a particular geodesic does not exist on all of $\mathbb{R}$, then that geodesic "runs off to infinity" like this one does? $\endgroup$ – Josh Burby Feb 28 '15 at 5:13

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