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This question may be easy but I could not come up with a proof.

Let $F$ be a hyperbolic surface of finite type (with finitely many boundary and finitely many puncture). Let $\gamma$ be a closed non-simple geodesic. $\gamma$ is not homotopic to a point, a puncture or a boundary. Let $p$ be a self intersection point of $\gamma.$

1) Can there be a simple closed geodesic passing through $p$?

If not then

2) Can there be a closed geodesic passing through $p$ different from $\gamma$?

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    $\begingroup$ I do not get question 2: $\gamma$ is a closed geodesic passing through $p$! $\endgroup$ – Benoît Kloeckner Sep 20 '14 at 11:11
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The answer to (1) is yes.

Take $P$ a hyperbolic surface with one geodesic boundary, called $\delta$, and two punctures. Form $S$, a sphere with four punctures, by doubling $P$ across $\delta$. Note that $S$ has a reflection symmetry $f$ that fixes $\delta$ pointwise. Let $\gamma$ be a figure of eight curve, about two of the punctures of $S$, choosen so that the reflection $f$ fixes $\gamma$ setwise. Deduce that the self-intersection point $p \in \gamma$ lies on $\delta$.

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  • $\begingroup$ Thanks for the example. Can this happen in closed surface? $\endgroup$ – Cusp Sep 20 '14 at 14:19
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    $\begingroup$ Yes. Replace the punctures by equal length geodesic boundary components. Do the above, and then glue to get a surface of genus two. $\endgroup$ – Sam Nead Sep 20 '14 at 17:35

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