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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?

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

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up vote 4 down vote accepted

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 $\pi$.

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Do you know of an alternative proof of the second statement, that lengths of geodesics are bounded away from 0? –  user15464 Sep 6 '12 at 2:39
    
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. –  Martin M. W. Sep 6 '12 at 3:04

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