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Let $S_{0,4}$ be the 4-punctured sphere equipped with a hyperbolic metric of finite volume (thus all punctures are cusps). Consider $\gamma$ to be a simple geodesic of $S_{0,4}$ (not necessarily closed). If $L$ is the closure of $\gamma$ in $S_{0,4}$, we know that $L$ is a geodesic lamination of $S_{0,4}$ (union of disjoint simple closed geodesics, including $\gamma$).

My objective is to understand $L$ in the particular case of $S_{0,4}$.

Question: Suppose that $\gamma$ is not a closed geodesic and that it is contained in a compact region of $S_{0,4}$, i.e. it doesn't go to the punctures.

Conjecture: Then $\gamma$ has to accumulate in a closed simple geodesic?

I know it has to accumulate in a simple geodesic, but I don't know if it has to be closed. It seems to me that it has to be closed.

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The answer to your question is "no". There are simple geodesics in $S = S_{0, 4}$ that lie in a compact subsurface of $S$, but whose closures contain no simple closed geodesic.

Here is the usual "first example", which is still, unfortunately, not particularly easy to understand. Suppose that $\sigma_1$ and $\sigma_2$ are the right half-twists interchanging the first puncture with the second and the second with the third, respectively. Let $f = \sigma_1 \circ \sigma_2^{-1}$. Thought of as a braid, the iterates of $f$ are the usual French braid - it also appears as the simplest braid for making Challah. We also note that the isotopy class of $f$ is a pseudo-Anosov" mapping class.

We now build a geodesic lamination. Let $\alpha_0$ be any simple closed geodesic in $S$. Recursively, define $\alpha_{n+1}$ to be the geodesic representative of $f(\alpha_n)$. Let $\alpha_\infty$ be the geometric limit of the sequence of geodesics $(\alpha_n)_{n \in \mathbb{N}}$. Then $\alpha_\infty$ is a geodesic lamination with uncountably many leaves, none of which are closed, and all of which are dense (in $\alpha_\infty$).

I highly recommend Casson and Bleiler's book Automorphisms of surfaces after Nielsen and Thurston as a reference.

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  • $\begingroup$ Thank you for your answer Sam! I have a few questions. I don't see why this sequence of geodesics forms a disjoint family of geodesics. If they are not disjoint, how do we see this geometric limit $a_\infty$ as a geodesic lamination? (very good reference, love this book) $\endgroup$
    – Nelson Schuback
    Commented Feb 11 at 8:16
  • $\begingroup$ I have a different version of this question that maybe you can help me with: Not thinking about geodesic laminations, but instead about limit of simple geodesics. If a simple geodesic accumulate somewhere, I know that they should accumulate in a simple geodesic. But in $S_{0,4}$ it seems to me that they should accumulate in a simple closed geodesic. That what I would like to prove. Do you have any idea on how I could prove that? Thank you so much for your interest in the question! $\endgroup$
    – Nelson Schuback
    Commented Feb 11 at 8:22
  • $\begingroup$ @NelsonSchuback - Regarding your first comment. It is an exercise to show that any pair of simple closed geodesics in $S_{0, 4}$ intersect. (In a bit more detail - either they are identical, or they intersect transversely - they are never disjoint.) In my construction, if $i \neq j$ then $\alpha_i \neq \alpha_j$, and they intersect transversely. The point is that the $\alpha_i$ approach $\alpha_\infty$ "in direction" but not "from the side". In fact, every $\alpha_i$ crosses (all leaves of) $\alpha_\infty$. $\endgroup$
    – Sam Nead
    Commented Feb 11 at 10:08
  • $\begingroup$ Regarding your second comment - I don't see that this is very different from your original question. Think about the lamination $\alpha_\infty$. It has no closed leaf. And every leaf is dense. So fix any leaf $\gamma$ of $\alpha_\infty$. This leaf $\gamma$ answers your question (in the negative). $\endgroup$
    – Sam Nead
    Commented Feb 11 at 10:09

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