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According to Jairo comment on the first version of this question I revise the question as follows; Let $g$ be a real analytic Riemannan metric on $S^{2}$. Is it true to say that:

There are at most a finite number of disjoint simple closed geodesics on $S^{2}$.

If the answer is yes put $m$= the sup of the number of such disjoint closed geodesics.

What is a geometric interpretation for this geometric invariant $m$?

For a given $n\in \mathbb{N}$, is there a real analytic Riemannian metric on $S^{2} $ for which $m=n$

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Ali Taghavi
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    Your first question seems interesting, although I usually do not like the real-analytic realm in differential geometry - here it makes sense. As for the geometric interpretation you ask, I don't quite see what non-trivial reformulation you are hopping for. Concerning your last question, $n=1$ is obviously feasible, and for larger $n$ a sphere of revolution with $n$ narrow necks should do the trick. – Benoît Kloeckner Dec 21 '13 at 09:51
  • @BenoîtKloeckner perhaps my sensivity to the realm of real analytic objects comes from the book by Ilyashenko and the underlying subject study in that book 'A finitness theorem of limit cycles". This subject is discussed here at my MO question. https://mathoverflow.net/questions/171988/the-error-in-petrovski-and-landis-proof-of-the-16th-hilbert-problem – Ali Taghavi Nov 16 '20 at 11:56

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As it was shown by Igor, there is no univeral bound on number of such geodesics. Let me show that the number can not be infinite.

Assume it is possible to get infinite number of such geodesics, say $\gamma_n$, $n\in\mathbb N$. Note that the geodesics $\gamma_i$ for $i\le n$ cut $\mathbb S^2$ into surfaces with geodesic boundaries. By Gauss--Bonnet formula most of these surfaces are cylinders.

By passing to a subsequence, we can assume that for each $n$, the geodesics $\gamma_i$ for $i\le n$ cut $\mathbb S^2$ into two discs and cylinders between $\gamma_i$ and $\gamma_{i+1}$.

Denote by $\gamma_\infty$ the limit of $\gamma_n$ as $n\to\infty$. Note that this limit is defined and the limit geodesic $\gamma_\infty$ is stable.

Given a point $p$ near $\gamma_\infty$ denote by $\ell(p)$ the length of mimimal geodesic loop based at $p$ which goes sufficiently close to $\gamma_\infty$. Note that $\ell$ is an analytic function and its derivatives vanish on $\gamma_\infty$. It follows that $\ell\equiv 0$; i.e. $\gamma_\infty$ lies in a one parameter family of closed geodesics which sweep a neighborhood of $\gamma_\infty$.

Pass to the analytical extension of this one parameter family, lets denote it by $\xi_\tau$. Note that the geodesics in the family stay simple and disjoint locally. Globally, it only may happen that $\xi_0=\xi_c$ for some parameter $c\ne0$. Moreover, since the surface is compact it actually happens. In this case the surface is a total space of a circle bundle, a contradiction.

Anton Petrunin
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    That's a nice construction, but why would this have an infinite sequence of simple geodesics that are pairwise disjoint? – Robert Bryant Dec 21 '13 at 21:15
  • @RobertBryant Ups, sorry I thought "distinct". – Anton Petrunin Dec 21 '13 at 21:57
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    @RobertBryant, it is rewritten, now the answer is NO. That is the first time I used analyticity in my live; hopefully I did it right. – Anton Petrunin Dec 22 '13 at 03:42
  • Can you expand on the analytic family argument? I'm aware of results of Bohm-Tomi that prove a similar thing for minimal surfaces. But I haven't found the statement you're making in the literature. One way one could prove it is to show that any closed simple geodesic in a surface has a neighborhood foliated by constant curvature curves (these should solve some relative isoperimetric inequality). Intuitively, these curves should be bubbles obtained by blowing air into an annular region near the geodesic. – Ian Agol Dec 22 '13 at 04:14
  • If there were a sequence of geodesics converging, then by the maximum principle, they would have to be leaves of the foliation. But the curvature should be an analytic function of the area, which is therefore constant, so the constant curvature curves would be geodesics. But I haven't found such an argument (giving a CMC foliated neighborhood) in the literature. – Ian Agol Dec 22 '13 at 04:14
  • @IanAgol, I think your argument works perfectly. – Anton Petrunin Dec 22 '13 at 04:23
  • @AntonPetrunin: I'm aware that the geodesic is stable (in the sense of having no negative eigenvalues of the Jacobi operator). One may use a converging sequence of geodesics to form a Jacobi field. And analyticity should imply that this Jacobi field along the geodesic gives rise to a Jacobi vector field in a neighborhood (and therefore globally by analyticity). But I'm not sure how this argument is made precise, which is what I'd like you to expand on (if that's what you had in mind). – Ian Agol Dec 22 '13 at 05:00
  • @IanAgol, are you happy now? – Anton Petrunin Dec 22 '13 at 05:53
  • @AntonPetrunin you wrote "according to Igor construction, there is no a universal upper bound" please look at my first question: We fix a reimannian metric (we do not change it) and we search for a uniform upper bound for the number of disjoint simple closed geodesic. In the Igor construction the geometry changes namely for each n, he give a riemannian metric on $S^{2}$ for which there are at least n disjoint simple closed geodesics. – Ali Taghavi Dec 22 '13 at 06:33
  • I emphasis on "real analytic" because my question is influenced by Ilyashenk Eckel theorem which says every analytic vector filed on $S^{2}$ has only a finit number of isolated closed orbit. In my question, closed geodesics play the role of closed orbit in the later theorem – Ali Taghavi Dec 22 '13 at 06:35
  • Ok, that's a nice argument, much simpler than what I suggested. Note that the surface could be a Klein bottle too. – Ian Agol Dec 22 '13 at 15:40
  • @AliTaghavi, right, I proved what you asked, given a Riemannian metric on $\mathbb S^2$ there is at most finite number of disjoint closed simple geodesics. – Anton Petrunin Dec 22 '13 at 17:36
  • @AntonPetronin Thank you for the answer. I honestly admit that I need to some reference to underestand the detail of your proof. for example analytic extension, stability... could you please give a reference( with minimal necessary background). Ihave also some question: Is not possible that $\gamma_{\infty}$ would be a triangle (not necessarily closed geodesic)? What is the domain of $\ell$? Why $\ell$ can be well defined? Iapologize if my questions are elementary, but I need to a reference to underestand the details of your proof.Thanks again for your help – Ali Taghavi Dec 22 '13 at 21:33
  • @AliTaghavi, these are very basic questions, ask any geometer on your math department, he/she should be able to help, or take any book in Riemannian geometry and read about geodesics. – Anton Petrunin Dec 22 '13 at 23:13
  • @AntonPetrunin Can you explain why $\gamma_{\infty}$ can not have vertex? – Ali Taghavi Dec 23 '13 at 08:08
  • What about the remaining part of my question: for a fixed analytic metric g, is the invariant $m$ defined in my question, finite? For a natural number $n$, is there an analytic metric $g$ for which $m=n$? – Ali Taghavi Dec 23 '13 at 15:34
  • @AliTaghavi, I think the example in Benoît's comment (which is the same as in Igor's answer) should answer your question; it should be easy to show equality if the necks are almost invisible (i.e., the sphere looks almost like cylinder with two spherical caps). – Anton Petrunin Dec 23 '13 at 17:53
  • @AntonPetrunin when we pass from a max to a min, the sign of gaussian curvature, changes. so it seems that we can not count the number of disjoint simple closed geodesics. other subject;was my question on $\gamma_{\infty}$ ,trivial?(two my previous comment) – Ali Taghavi Dec 26 '13 at 10:13
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A simpler example seems to be an "accordion surface" (take a sinusoid $y = \sin x,$ rotate around the line $y= 3$) It will have as many parallel simple geodesics as you like.

Igor Rivin
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The flat torus has infinitely many disjoint simple closed geodesics.

Jairo Bochi
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