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A translation surface is a Riemann surface equipped with a holomorphic 1-form $\omega$ and a Riemannian metric $g=\omega \bar \omega$ with conical singularities. It is well-known that there exists closed regular geodesics, i.e., those not going through singular points, but usually shortest curves in a given homotopy class pass through the singular points.

Question 1: Is the first homology group generated by regular closed geodesics (at least over $\mathbb Q$)? I am mainly interested in the case of genus $5$ surfaces with $\omega$ having four distinct double zeros.

Question 2: If question 1 has a negative answer, I would like to know if there is another way to determine the periods of $\omega$ in terms of the lengths of the regular closed geodesics $\gamma$ and their angles (defined by the (constant) modulus of $\omega(\gamma')$).

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The answer to your Question 1 is no in general. In fact, there are two examples of square-tiled surfaces in genera 3 and 4 (sometimes called "Eierlegende Wollmilchsau" and "Ornithorynque") with the following properties: each of them decomposes completely into two homologous cylinders in all directions and the Veech group is $SL(2,\mathbb{Z})$. These properties imply that the $\mathbb{Q}$-span of the regular closed geodesics (core curves of cylinders in periodic directions) is 2-dimensional (generated by the horizontal and vertical regular closed geodesics). You can find more details about these two examples in this paper here http://w3.impa.br/%7Ecmateus/files/matheus-yoccoz1.pdf (for instance).

On the other hand, it is not known whether "similar" ("Shimura-Teichmuller") examples can exist in genus 5, but Martin Moller showed here http://arxiv.org/pdf/math/0501333v2.pdf that they do not appear for genus $\geq 6$.

In a similar vein, I think these two examples also answer your Question 2: if I'm not mistaken, it is not possible to recover the periods of these examples only from the knowledge of the periods of regular closed geodesics.

Best,

Matheus

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  • $\begingroup$ Dear Matheus, somehow, I have not realized that my question was answered until now. Thanks. $\endgroup$
    – Sebastian
    Commented Apr 30, 2015 at 8:22

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