Length of simple closed curve in half-translation surface

Question

Let $R$ be a Riemann surface of genus $g\ge 2$ and $q$ an holomorphic quadratic differential on $R$. Together they determine a semi-translation structure: an atlas on $X$ such that its changes of charts are of the form $z\mapsto \pm z+c$ and a singular flat metric $|q|$ on $X$.

Suppose that $q$ has at least one singularity of odd order: I can't completely grasp what the $-1$ holonomy implies when considering the lengths of simple closed curves on $(X,q)$. In particular, let $c\subset X$ be a closed simple geodesic for $|q|$ which doesn't meet any singular point of $|q|$. Then, is it true that there is always a cylinder in $(X,q)$ foliated by closed geodesics parallel to $c$ (as in the case of translation surfaces with trivial holonomy)?

If no, then is it still always possible to assume the existence of a concatenation of saddle connections of $q$ parallel to $c$ and with the same length?

Answer 1

Yes, this is true, and as far as I can tell the argument doesn't seem any different whether the surface is a pure translation surface or a semi-translation surface. The point is that the argument is purely local, and from this argument you can deduce from the given conditions on $c$ that the rotational holonomy around $c$ is trivial.

Here's the argument. If $c$ is a simple closed geodesic disjoint from the singular points, then by compactness of $c$ there exists $t>0$ such that the "$t \times t$ square" centered on each point of $c$ is disjoint from the singular points. Depending on the slope of $c$ with respect to the horizontal and vertical foliations, there exists $s \in [t,t\sqrt{2}]$ such that the union of these squares forms the metric neighborhood of $c$ of radius $r$. This metric neighborhood is isometric to a Euclidean product annulus, namely the product of $c$ with the interval $[-s,+s]$. It follows from this description that the $\mathbb{Z}/2\mathbb{Z}$-rotational holonomy around $c$ is trivial, since that's the only possibility for Euclidean product annuli.

Answer 2

Yes, it is, of course, true. Even though foliations defined by $q$ are not orientable, the genus $2$ surface is orientable, so an $\varepsilon$-neighborhood of a closed geodesic $\gamma$ is just the cylinder $\gamma\times [-\varepsilon, \varepsilon]$.

This statement holds more generally for orientable surfaces with flat metric and conical singularities. If you find a simple closed geodesic (smooth one and disjoint from conical points) its $\varepsilon$-neighborhood is a cylinder.