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CompuChip

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"Let M be a compact Riemannian manifold. [...] Likewise, every homotopy class of closed curves in M contains a curve which is shortest and geodesic."

That's theorem 1.5.1 from this book. Also the proof is given there (and luckily for you, within the previewable pages). Actually I suspect (I didn't look at it nor try it) that it's not very hard so maybe you want to try it yourself first.

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Well, The question I posted is still unsolved.

first, thank CompuChip. But what I am asking is the existence of the geodesic in the "homology" class of a given simple closed curve, rather than the existence of the geodesic in the "homotopy" class of a simple closed curve. Can you help? Thanks a lot for your prospective help.

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can anyone help?

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Not sure of the proof but I would try covering the curve with open balls then deforming it to a geodesic segment in each ball. This would give you a piece wise smooth geodesic in the same homotopy class. It will locally minimize length but maybe not globally. You can then try to show that by smoothing out the kinks you get a geodesic that is even shorter.

This argument should work but as is begs the question because there still may be even a shorter curve that is not a geodesic.

So maybe you start out with a length minimizing curve in the same homotopy class. If it is not a geodesic apply this procedure of geodesic approximation to get even a shorter curve.

Details to be worked out.

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Hurkyl

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Surely knowing something about the homotopy class tells you something about the homology class?

first, thank CompuChip. But what I am asking is the existence of the geodesic in the "homology" class of a given simple closed curve, rather than the existence of the geodesic in the "homotopy" class of a simple closed curve. Can you help? Thanks a lot for your prospective help.

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