# Closed geodesic on a non-simply connected Riemannian manifold

I am looking for a reference or proof of the following property of a Riemannian manifold $M$: if $M$ is not simply connected, then it contains a non-trivial closed curve of minimum length. I am fine with making assumptions about $M$ such as compact, smooth, analytic etc.

I have elementary knowledge of differential geometry. Based on the definition of a simply connected manifold I know that a closed curve cannot be continuously deformed to a point. However, I do not know how to formalize this procedure, or why some sequence of deformations should converge or even result in a reduction of the length of the curve.

Edit: The manifold needs to be closed. I am looking for a proof that on a closed but not simply connected Riemannian manifold there exists at least one closed curve (or closed geodesic) which is a local minimum of the length function in the space of smooth curves.

• Wait, what if $M = \mathbb{R}^2\setminus\{(0,0)\}$? There's no minimum length curve here, right? – mathworker21 Jul 8 '18 at 6:50
• The claim is not correct, as the example of @mathworker21 shows. It is even not correct for complete surface, since you can equip the example from that comment with a complete metric (look at the surface you get by rotating the graph of $\ln(x)$ or $1/x$ aroung a third coordinate axis). What you need is a positive lower bound for the length of all continuous curves in some given homotopy class, and some reason why a minimizing sequence in that homotopy class should converge to something which will then be a minimizer for the length. The latter requirement is fullfilled for complete manifolds. – Thomas Jul 8 '18 at 7:23
• Note that if $M$ is compact, you will also get a positive lower bound for the lengths of curves in a given nontrivial homotopy class. One way to see this is to prove that any curve which is shorter than some constant depending on $M$ will be contained in a coordinate neighbourhood. – Thomas Jul 8 '18 at 7:31
• Sorry, you are right about the punctured plane. Closedness is necessary. It occured to me before, but slipped my mind when I posed the question. – Asdf Jul 8 '18 at 7:55

First, if $M$ is not compact, the result is not true. As said in the comments, the punctured plane with the induced metric is an example.

If $M$ is compact, the result is true. In fact, we have that the following is true:

• If $M$ is not simply connected, then for every conjugate class of $\pi_1(M)$ there exists a loop which minimizes length on such class.

Those loops are necessarily geodesics, by variational reasons.

One way to see why this is true is via a Morse-theory approach, with a full proof being given in Klingenberg - Lectures on Closed Geodesics. The gist of it is to pick a loop close to the infimum on the class and flow it under (minus) the gradient flow of the energy functional in the free loop space (this has an obvious analogy with the finite-dimensional case when we flow via gradient and converge to a critical point).

To see how the result we mentioned implies what you want, just notice that a variation of a geodesic stays in the same class. Thus, if the geodesic minimizes length on the class, it is a local minimizer.

EDIT: The book by Klingenberg gives a reference which does not use the machinery behind the proof I alluded to above. He says:

(...) [O]ne considers the universal Riemannian covering manifold $\widetilde{M}$ of $M$. Any non-trivial element $\tau$ of the fundamental group $\pi_1 M$ operates as a fixed-point free isometry on $M$. One shows that the function $f(\widetilde{p}):=$ distance from $\widetilde{p}$ to $\tau \widetilde{p}$ assumes its infimum on $\widetilde{M}$. Let $\omega$ be this infimum. Then $\omega > O$. Let $\widetilde{p} \in M$ such that the distance $d(\widetilde{p}, \tau \widetilde{p}) = \omega$. Let $\widetilde{c}$ be a geodesic from $\widetilde{p}$ to $\tau \widetilde{p}$ of length $\omega$. Then the projection of $\widetilde{c}$ into $M$ gives a closed geodesic $c$ on $M$ having minimal $E$-value among the closed curves homotopic to $c$.

He attributes the proof (in special cases) to Hadamard and Cartan (although he says that the later has an error), and says that Berger - Lectures on geodesics in Riemannian geometry has a full proof following this idea (I have not seen it).

• That's, of course, correct, but unless my memory deceives me (it's 20 years now, admittedly) way too complicated machinery for this result. The technique you are referring to even allows you to find closed geodesics which are not minimizers, e.g. closed geodesics on homotopically trivial manifolds. Minimizers in non-trivial homoty classes can be found with the direct methods of the Calculus of Variations, just by looking at a minimizing sequence in a suitably chosen space of curves. – Thomas Jul 8 '18 at 8:31
• I also do not understand your first statement. The question does ask for a closed geodesic, and there was not edit by now. – Thomas Jul 8 '18 at 8:41
• Is there a difference between a closed curve of minimum length and a closed geodesic of minimum length under these circumstances (up to parametrization)? I thought the minimality would ensure that the curve is a geodesic? – Asdf Jul 8 '18 at 10:42
• @Thomas, Asdf I got tangled up on the vacuity in the example by mathworker21, due to the fact that there are circles which go to zero in length but they are not geodesics and there are no closed geodesics (I even made a wrong comment before). Essentially, I read the question as intending to mean: "Among the closed geodesics, if they are non-empty, is there one of minimum length?", which is why I ended up making the first comment (and the first example). – Aloizio Macedo Jul 8 '18 at 11:20
• @Thomas Yes, there is a relatively heavy machinery behind the process, but the philosophy is precisely the same of the variational approach. I think that the only difference (and what makes it more complicated) is in building up the geometric-analytical framework (for example, establishing condition (C) of Palais-Smale, the fact that the loop space is a Hilbert manifold if you consider $W^{1,2}$ curves etc) so that we have things like flows and the sort, allowing us to transmit the intuition of the finite dimensional case comfortably. – Aloizio Macedo Jul 8 '18 at 11:25