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Since in a compact Riemannian manifold $M$ the only totally convex subset is the whole manifold itself, see Closed manifold has no nontrivial totally convex subset?, it should follow that for every point $p\in M$ there exists a geodesic loop starting and ending at $p$ (not necessarily "closing" smoothly). I was looking for a (possibly simple) direct proof of this fact without passing by the general theorem (whose proof is not easy, in my opinion). Thanks.

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(I take it you don't count the constant loop.) Of course, if $M$ has nontrivial $\pi_1$, then you can find a geodesic with fixed endpoints representing any nontrivial homotopy class by minimizing the energy. In the simply-connected case I don't see an immediate proof. – Theo Johnson-Freyd May 20 '13 at 0:32
Suppose $p$ is totally convex, so there is no geodesic loop based at $p$. By the main theorem of the Morse theory the loop space $\Omega_p M$ has a homotopy type of a CW complex with cells correponding to geodesic loops at $p$, so $\Omega_p M$ is contractible, i.e. $\pi_i(M)=0$ for $i>0$. Since $M$ is connected, it is contractible. No closed manifold is contractible by Poincare diality. I do not think there is a proof without Morse theory in disguise. – Igor Belegradek May 20 '13 at 2:59
In fact, one needs very little from the Morse theory because there is no critical points (e.g. the index theorem is not needed); basically, replace the loop space by the finite dimensional manifold of piecewise geodesic loops, and note that the energy functional has no critical points, so the finite dimensional manifold is contractible. The rest is elementary algebraic topology. What could be easier? – Igor Belegradek May 20 '13 at 14:44
@Igor, the manifold of broken geodesics is not complete. So you have to be bit more careful, to make sure that there is a critical point. In other words, read my answer. – Anton Petrunin May 20 '13 at 18:29
Anton: the argument in my first comment surely works. The argument in the second comment does not give complete detail, but I think the point is that the gradient flow of the energy functional gives a deformation retraction of the manifold of broken geodesics loops onto a small neighborhood of a constant loop at $p$, and that small neighborhood is contractible. – Igor Belegradek May 21 '13 at 3:56
up vote 2 down vote accepted

Here is a standard argument, I learned it from "Comparision Theorems in Riemannian Geometry" by Cheeger and Ebin. It avoids use of infinite dimensional spaces.

Choose the smallest $k>0$ such that $\pi_kM\ne 0$. Choose a spheroid which represents a nontricial element of $\pi_kM$. We can assume that the spheroid is swapped by an $\mathbb S^{k-1}$-parameter family of broken geodesics such that the length of each edge is smaller than the injectivity radius of $M$; denote it by $i_M$.

Start a natural curve-shortening process --- velocity vector of every vertex should depend on the directions and lengths of the two edges coming from it. You have to choose one which keep the lengths of the edges below $i_M$ and such that the rate of length (or energy) decay is estimated through rate of change of the broken geodesic.

Note that there is a lower bound for the maximal length of broken geodesics; "maximal" means "maximal in the $\mathbb S^{k-1}$-family after spending arbitrary time in the process". Say, this value can not go below $i_M$.

It follows that after long time in the process, one broken line in the family almost does not change the length. Hence this it almost does not move; the later implies that and all the angles almost $\pi$. Pass to the limit and you get the loop.

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Anton: You need to use minimax argument here since you are looking for an unstable critical point of energy. – Misha May 20 '13 at 2:08
@Misha, Yes, I do use minimax. I made an update, it should be more clear now. – Anton Petrunin May 20 '13 at 3:30

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