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Cartan Theorem: Let $M$ be a compact riemannian manifold. Let $\pi_1(M)$ be the set of all the classes of free homotopy of $M.$ Then in each non trival class there is a closed geodesic. (i.e a closed curve which is geodesic in all of its points.)

My question: Why free classes? Why the theorem does not apply if we exchange free classes for classes with a fixed base point?

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Well my immediate guess is that given a basepoint there won't necessarily be a closed geodesic passing through it. – Chris Gerig Jun 12 '12 at 21:04
Dear Eduardo Siva: I think that the compactness of $M$ is a necessary hypothesis in Cartan-Hadamard Theorem. – Giuseppe Jun 12 '12 at 21:08
Chris Gerig: it's a good point. Can you give me an example? – user27456 Jun 12 '12 at 21:18
@ZhenLin: I honestly do not understand the purpose of your comment. – user27456 Jun 13 '12 at 0:36
@EduardoSiva: I believe that Zhen Lin is pointing out that to produce bold text, you used LaTeX instead of using the proper Markdown formatting, i.e. $\bf{text}$ instead of **text**. See this page for info on how to format text on the site using Markdown. – Zev Chonoles Jun 13 '12 at 4:08
up vote 5 down vote accepted

Consider the Klein bottle $K$ with flat metric. I'm thinking of $K$ as a square where the left and right sides are identified in the "right" way (like on the torus), while the top and bottom are identified in the "wrong" way (like on $\mathbb{R}P^2$). In this picture, geodesics are straight lines that wrap around depending on the identifications on the edges.

Take your basepoint $p$ to be the center of the square. Consider the geodesic $\gamma$ emanating from $p$ with slope 1. So, it starts in the middle of the square moving towards the top right corner. After it gets to the top right corner, due to the identifications we're making, it becomes a straight line emanating from the bottom right corner with slope -1 until it eventually hits $p$ again, i.e., it closes up. However, it is not a closed geodesic because it makes a corner at $p$.

Further, I claim no other geodesic emanating from $p$ is in the same homotopy class as $\gamma$. To see this, work in the univeral cover, $\mathbb{R}^2$ (thought of as being tiled by squares with identification arrows as approrpriate, with corners on integer lattice points). Geodesics are still straight lines, but now there is a unique straight line from $(\frac{1}{2},\frac{1}{2})$ to $(\frac{3}{2},\frac{3}{2})$, given by lifting $\gamma$.

This means the Cartan's theorem fails on based loops, at least in this particular case.

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To find the closed geodesic in the same class of a given closed loop, you have to "tighten" the loop: you try to shrink it to make it as short as possible. Either it will shrink to a point (if its free homotopy class is trivial) or eventually it will "catch" around some hole in your manifold, in which case it will be a closed geodesic. (If it wasn't a geodesic, then locally at some point there would be a way to tighten it a bit more to make it shorter.) In this tightening process, you can't expect the loop to remain passing through any particular point.

Added: A typical closed manifold may not have all that many closed geodesics on it, e.g. only countably many. (By the Baire category theorem, most points of the manifold will then not lie on a closed geodesic.) This is the case with hyperbolic surfaces, for example, where one has uniqueness of the closed geodesic representative of a free homotopy class. (And I imagine that this is the typical behaviour.)

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"In this tightening process, you can't expect the loop to remain passing through any particular point." ... the question is why... – Chris Gerig Jun 13 '12 at 2:39
@ChrisGerig: Dear Chris, E.g. because there may not be enough closed geodesics. (See my edit.) Regards, – Matt E Jun 13 '12 at 3:53

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