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Let $M$ be a compact Riemmanian manifold. Let $G$ denote the set of all geodesics of $M$. If $\gamma\in G$ let $l(\gamma)$ denote its length. Let $$S=\sup\{l(\gamma): \gamma\in G\}$$

Suppose $S<\infty$. How can we estimate $S$ geometrically?

Edit: I changed some assumptions.

Thanks

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    Well, yes, if you start a line in the square (of the torus) with *irrational* slope.2012-10-08
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    Does "manifold" mean with or without boundary?2012-11-01
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    Do you mean the length of the *image* of $\gamma$? On a compact manifold, all geodesics have infinite length in the sense that $\int_{-\infty}^\infty |\gamma| dt = \infty$.2012-11-02
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    @JasonDeVito, yes it is the length of the image.2012-11-02
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    In this case, all geodesics are closed. Such manifolds are very special - think (but I'm not sure), all known simply connected examples are diffeomorphic (but not necessarily isometric!) to spheres and projective spaces.2012-11-02
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    @JasonDeVito, interesting. Is there anyway to estimate $S$ in this case? If we think this manifold embedded in some $\mathbb{R}^N$, what is the relation between $S$ and the diamenter of the manifold as a set of $\mathbb{R}^N$.2012-11-02
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    @Tomás: I don't really know anything about it. Besse has a book called "Manifolds all of whose geodesics are closed" which might contain the kind of information you're looking for. Here's a single data point: In a circle of intrinsic diameter $\pi r$, all geodesics are closed for stupid reasons and $S = \pi r$. On the otherhand, we can embed the circle isometrically into $\mathbb{R}^3$ as a tightly wound coil, so that the extrinsic diameter can be made as small as we like. This means that the diameter of the image of an embedding tells you nothing about $S$.2012-11-02

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