# How many simple closed geodesics in a given primitive homology class?

It is well-known that an essential closed curve on a hyperbolic surface (possibly with boundary) is homotopic to a unique closed geodesic. Moreover, if the curve under consideration is simple, then so is the geodesic homotopic to it. A reference is "A primer on mapping class groups" of Farb and Margalit (propositions 1.3 and 1.6).

It is proved here that for a torus with 1 puncture $\Sigma_{1, 1}$ (endowed with a complete hyperbolic metric) every primitive homology class $h \in H_1(\Sigma_{1, 1}, \mathbb{Z})\approx \mathbb{Z}^2$ contains a unique simple closed geodesic. This can be surprising for a beginner like me since the preimage of $h$ under abelianization map $\mathrm{ab}:\pi_1(\Sigma_{1, 1})\approx F_2\rightarrow H_1(\Sigma_{1, 1}, \mathbb{Z})$ is infinite. Every homotopy class in this preimage contains a closed geodesic yet only one contains a simple closed geodesic.

My question is: are there examples of hyperbolic surfaces of different topology such that every primitive homology class contains exactly one simple closed geodesic? What are the results/references in this general direction?

The thrice punctured sphere has no simple closed geodesics. The four-times punctured sphere has a unique simple geodesic in each homology class. In general, it is a result of I. Rivin that the number of simple closed geodesics of length bounded above by $L$ grows like $L^{6g - 6 + 2 c},$ where $c$ is the number of punctures. We can restrict to closed surfaces for simplicity, so $c=0.$ Then, the number of homology classes where there is a simple closed geodesic of length $\leq L$ grows no faster than $L^{2g},$ which means that for uniqueness you have to have $g<2,$ which rules out every hyperbolic surface.
• also, could you please explain why 'the number of homology classes where there is a simpel closed geodesic...' grows no faster than $L^{2g}$ for a closed surface? – user74900 Jul 3 '18 at 20:25
• @AknazarKazhymurat No, it does not matter if you have boundary (with the minor difference of whether or not you consider the boundary geodesic). As for the homology classes, it's because the minimal length of a homology class defines a norm on homology (and homology is a $2g$ dimensional vector space (if considered over \$\mathbb{R}),, but the minimal length is sometimes represented by a multi-curve (for a punctured torus, it's always a connected curve). – Igor Rivin Jul 3 '18 at 21:33