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Let $(M,g)$ be a compact Riemannian manifold. It is well known that there always exists a nontrivial closed geodesic in $M$, which is the so-called Lusternik-Fet theorem.
But such a geodesic could well self-intersect (in a transversal way).

What is known about the existence of geodesics which do not self-intersect? (I do not know whether 'simple' is the right term for describing this)

Do they always exist? Or are there counterexamples in which all closed geodesics are self-intersecting?

Mizar
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  • pretty sure the shortest curve in a nontrivial free homotopy (i.e. no base point specified) class is a simple closed geodesic. – Will Jagy Sep 16 '15 at 22:54
  • https://books.google.com/books?id=t22O0XBtyJsC&pg=PA205&lpg=PA205&dq=shortest+curve+in+a+free+homotopy+class&source=bl&ots=PB622t0Ami&sig=2G_bAL7VNnsTQ2u426a7Lumo4vo&hl=en&sa=X&ved=0CDAQ6AEwAmoVChMIlPWehNP8xwIVkqSICh0YPgrk#v=onepage&q=shortest%20curve%20in%20a%20free%20homotopy%20class&f=false – Will Jagy Sep 16 '15 at 22:56
  • I see, it really is the Busemann book, one site did not specify: https://books.google.com/books?id=DOdx1FPmGNMC&pg=PA205&lpg=PA205&dq=shortest+curve+in+a+free+homotopy+class&source=bl&ots=Rtpakm8RsW&sig=zAQqr2u7_v3hlRbt2KrAxKXj58A&hl=en&sa=X&ved=0CDIQ6AEwA2oVChMIlPWehNP8xwIVkqSICh0YPgrk#v=onepage&q=shortest%20curve%20in%20a%20free%20homotopy%20class&f=false – Will Jagy Sep 16 '15 at 22:57
  • and compactness is critical. If homotopy is trivial, not sure what to say – Will Jagy Sep 16 '15 at 22:58
  • see http://math.stackexchange.com/questions/148642/why-are-we-interested-in-closed-geodesics – Will Jagy Sep 16 '15 at 23:02
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    Why is it simple? I guess you are letting the homotopy class to vary, right? But then I don't see why a shortest curve exists. – Mizar Sep 16 '15 at 23:05
  • well, it must be a nontrivial free homotopy class as well as a compact manifold. In any case, see the question i found from this site in 2012, question number 148642 – Will Jagy Sep 16 '15 at 23:07
  • Yep, thanks for the link. I didn't understand your last comment: where do you choose the shortest curve and why isn't it self-intersecting? – Mizar Sep 16 '15 at 23:15
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    @WillJagy: If $[\gamma]$ is a homotopy class that is represented by a simple closed geodesic, it is not clear why $2[\gamma]$ can also be represented by a simple closed geodesic. –  Sep 17 '15 at 04:14
  • Perhaps related: Can a polygon with minimal perimeter self-intersect? – Han de Bruijn Oct 19 '15 at 14:26
  • @Mizar : It seems that there is always one for non-simply connected compact manifold. –  Oct 25 '15 at 08:12
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    It also seem that this is an open problem. related –  Oct 26 '15 at 05:48
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    Perfect (even if the paper could be a little outdated..)! Please post it as an answer, including a sketch of the proof for non-simply connected compact manifolds (now this is clear to me but it could benefit the community). – Mizar Oct 26 '15 at 11:32

1 Answers1

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We can easily find a simple closed geodesic if $M$ is not simply connected:

Proof: Let $\tilde M$ be the universal cover of $M$. Then one can pull back the metric to $\tilde M$, as $\pi : \tilde M \to M$ is a local diffeomorphism. Thus $\tilde M$ is also a Riemannian manifold and $\pi$ is an local isometry. As $M$ is compact, it's complete, and one can check that $\tilde M$ is also complete.

Fix $p \in M$ and let $\tilde p \in \pi^{-1}(p)$ be fixed. Define $$ C(p) = \min_{ q\in \pi^{-1}\ (p)\setminus \{\tilde p\} } d(\tilde p, q).$$

Note that $C(p) >0$ and is realized by some $q\in \pi^{-1}(p)$, as $\pi^{-1} (p)$ is a discrete set in $\tilde M$. One can check $C : M \to \mathbb R$ is continuous. Let $p_0 \in M$ be its minimum. Let $\tilde p \in \pi^{-1}(p)$ and $\tilde q \in \pi^{-1}(p) \setminus \{\tilde p\}$ such that $C(p) = d(\tilde p, \tilde q)$. Let $\eta : [0,d] \to \tilde M$ be a shortest geodesic joining $\tilde p$ and $\tilde q$. Such a geodesic exists as $\tilde M$ is complete. $\eta$ is obviously simple as it's length minimizing. Let $\gamma = \pi \circ \eta$. $\gamma$ is a simple geodesic loop ($\gamma(0) = \gamma(d)$), which might not be smooth at $p$.

However, consider $p_1 = \gamma (d/2) \in M$. As $C(p_1) \ge C(p)$, the same $\gamma$ is a curve which is the shortest curve along all curves homotopic to $\gamma$ with the same base point $p_1$. Thus $\gamma$ must be smooth at $p$ and so $\gamma$ is simple closed geodesic in $M$.

Remark When $M$ is simply connected, the general question seems to be an open problem. For $\mathbb S^2$, the answer is affirmative. I am no expert in this question so I think I cannot say more.

Mizar
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