1
$\begingroup$

Let $M$ be a compact Riemannian manifold and let $SM$ be its sphere bundle,

$$SM = \{(x,\xi) \in TM : \|\xi\| = 1\}.$$

There is a well-defined function $\ell : SM \rightarrow [0,\infty)$ defined by mapping $(x,\xi)$ to the length of the unique maximally extended geodesic $\gamma$ with $\gamma(0) = x$ and $\gamma'(0) = \xi$. Is $\ell$ continuous?

If $\ell$ is continuous, then compactness immediately implies that the lengths of maximally extended geodesics on $M$ are bounded away from 0. If it turns out that $\ell$ is not continuous, is there a different way to prove this statement?

$\endgroup$
  • $\begingroup$ I guess I would call the length of such a geodesic 2; the length of its image. Is this standard? $\endgroup$ – user15464 Sep 6 '12 at 1:23
5
$\begingroup$

If I understand your definitions, the function $\ell$ isn't continuous. Consider a flat torus, $\mathbb{R}^2 / \mathbb{Z}^2$. The geodesic through $(0, 0)$ can be infinitely long if its angle is an irrational multiple of $\pi$, and will fluctuate wildly on the rational multiples of $\pi$.

$\endgroup$
  • $\begingroup$ Do you know of an alternative proof of the second statement, that lengths of geodesics are bounded away from 0? $\endgroup$ – user15464 Sep 6 '12 at 2:39
  • $\begingroup$ If you had closed geodesics that were arbitrarily small, by compactness of $M$ you could find a sequence of these geodesics that converged to a point $P$ in $M$. But in a small enough neighborhood of any point on a Riemannian manifold there won't be any closed geodesics. $\endgroup$ – Martin M. W. Sep 6 '12 at 3:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.