# Number of disjoint simple closed geodesics

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 $m=n$

• 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 9:51

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.

• 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
• @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 3:42
• 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 4:14
• 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 4:14

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.

• This surface is not $S^{2}$ and after compactification we possibly loose real analytic structure. am I right? – Ali Taghavi Dec 21 '13 at 20:07
• The surface can easily be made real analytic. – Igor Rivin Dec 21 '13 at 20:11
• after compactification? – Ali Taghavi Dec 21 '13 at 20:15
• How can we control, the structure of riemannian metric at infinity, to be real analytic? – Ali Taghavi Dec 21 '13 at 20:57
• @AntonPetrunin I didn't claim we could, and in fact I am sure we cannot, by compactness... – Igor Rivin Dec 21 '13 at 23:08

The flat torus has infinitely many disjoint simple closed geodesics.

• What about $S^{2}$ with a given real analytic metric? What about the remaining part of my question? – Ali Taghavi Dec 21 '13 at 9:15
• thank you for the answer. So it reasonable That I revised to question with " A compact surface other than torus.." – Ali Taghavi Dec 21 '13 at 9:22