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Does there exist a closed geodesic on a closed genus 2 orientable surface (with hyperbolic metric) that self-intersects at only one point thrice?

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2 Answers 2

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Yes, such curves exist on closed hyperbolic surfaces.

As mentioned by Sam Nead, one can think of such a curve as lying on a subsurface which is a 4-holed sphere or 2-holed torus (genus one with two boundary components).

I'll first point out that one of the possible configurations cannot occur inside of a 3-holed sphere.

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The argument uses Gauss-Bonnet, and is the same as given in Example 7 of this paper. Labeling the three angles $\alpha, \beta, \gamma$, then $\alpha+\beta+\gamma=\pi$. But we get a geodesic triangle on the back side which implies $\alpha+\beta+\gamma < \pi$, a contradiction.

There are two possible configurations of immersed curve on an orientable surface with a single triple point. We may immerse a regular neighborhood of such curves in the plane in these two ways:

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The first lives inside a 4-holed sphere, the second in a 2-holed torus. We'll show that the second configuration is realizable. It sits on a genus 2 surface like this:

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The purple and blue curves intersect this immersed curve each in one point.

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Suppose that we can realize both resolutions of this immersed curve as geodesics in a hyperbolic metric.

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Then there must be some surface with the triple point. We can pinch the blue and purple curve to get a noded Riemann surface / hyperbolic surface with 4 cusps, in the boundary of moduli space (say in the Deligne-Mumford compactification). The curve will look like this:

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The picture should be thought of a lying on the Riemann sphere with four punctures. A noded surface is created by identifying the blue and purple points in pairs.

One can see that this is realized as a single triple point intersection in the hyperbolic metric by symmetry. On the other hand, one may also realize both resolutions, by "squeezing" the two blue punctures together, limiting to a configuration without a triple point.

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Hence, both configurations are realized by noded surfaces in the boundary of moduli space. Perturbing to the interior of moduli space, we see that both configurations are realized, and hence a surface with a triple point exists.

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  • $\begingroup$ Very nice. Here is the case you left for us -- let $\gamma$ be the given curve in the four-holed sphere. Pinch down the boundary components to get a four-punctured sphere $X$. Suppose that $X$ is the Riemann sphere with punctures at the vertices of a regular tetrahedron. The symmetry argument implies $\gamma$ has a triple point. Now open the punctures symmetrically and glue. $\endgroup$
    – Sam Nead
    Sep 7, 2018 at 1:59
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    $\begingroup$ @SamNead: that was my first try, but it doesn’t work: one boundary component will be three times as long as the other three by a symmetry argument. And I could only show one resolution on a general 4 punctured sphere by going to the boundary pinching a curve. $\endgroup$
    – Ian Agol
    Sep 7, 2018 at 2:03
  • $\begingroup$ One might be able to do a similar trick to the other case, pinching curves that meet the trefoil, but I didn’t think about it seriously. $\endgroup$
    – Ian Agol
    Sep 7, 2018 at 2:04
  • $\begingroup$ Urk. What I should have said was: "open the three punctures in the monogons symmetrically". The last puncture can be opened any amount without breaking the needed dihedral group. Or am I again missing something? $\endgroup$
    – Sam Nead
    Sep 7, 2018 at 11:23
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    $\begingroup$ Consider the quotient orbifold by rotating by the cyclic group of order 3. This is an annulus with a single cone point of order 3. If the geodesic has a trippple point, then it goes to a loop going through the cone point making equal angles on both sides (π/3). But the angles are equal iff the boundary components of the annulus have the same length. $\endgroup$
    – Ian Agol
    Sep 7, 2018 at 13:51
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The (very short!) paper Double points on hyperbolic surfaces by Jørgensen and Sandler gives a necessary condition for a triple point. They do not assume that the triple point is the only self-intersection. So they do not answer your question. But their techniques are directly relevant.

You should think about your curves as lying in the four-holed sphere or twice-holed torus. Then, hopefully, the extra boundary components will pair up and give the desired examples in genus two.

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