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

1$\begingroup$ Does the answer depend on the metric? I think so. Therefore, could you clarify what metric (or metrics) are you considering? $\endgroup$ – Aknazar Kazhymurat Sep 3 '18 at 4:23
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 4holed sphere or 2holed torus (genus one with two boundary components).
I'll first point out that one of the possible configurations cannot occur inside of a 3holed sphere.
The argument uses GaussBonnet, 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:
The first lives inside a 4holed sphere, the second in a 2holed torus. We'll show that the second configuration is realizable. It sits on a genus 2 surface like this:
The purple and blue curves intersect this immersed curve each in one point.
Suppose that we can realize both resolutions of this immersed curve as geodesics in a hyperbolic metric.
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 DeligneMumford compactification). The curve will look like this:
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.
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.

$\begingroup$ Very nice. Here is the case you left for us  let $\gamma$ be the given curve in the fourholed sphere. Pinch down the boundary components to get a fourpunctured 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 '18 at 1:59

1$\begingroup$ @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 '18 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 '18 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 '18 at 11:23

1$\begingroup$ 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 '18 at 13:51
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 selfintersection. So they do not answer your question. But their techniques are directly relevant.
You should think about your curves as lying in the fourholed sphere or twiceholed torus. Then, hopefully, the extra boundary components will pair up and give the desired examples in genus two.