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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. –  Berci Oct 8 '12 at 13:12
    
Does "manifold" mean with or without boundary? –  Chris Eagle Nov 1 '12 at 23:28
    
Do you mean the length of the image of $\gamma$? On a compact manifold, all geodesics have infinite length in the sense that $\int_{-\infty}^\infty |\gamma| dt = \infty$. –  Jason DeVito Nov 2 '12 at 0:45
    
@JasonDeVito, yes it is the length of the image. –  Tomás Nov 2 '12 at 10:31
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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. –  Jason DeVito Nov 2 '12 at 12:27

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