Singular leaf of Strebel differential

Question

Let $R$ be a Riemann surface.Let $\gamma$ be a loop which is non-trivial in $H_{1}(R,\mathbb{Z})$. By the Jenkins–Strebel Theorem we know the following: there exists a holomorphic quadratic differential $\phi$ whose horizontal foliation $\mathcal{F}_{h}(\phi)$ consists of closed leaves foliating one cylinder with core curves in the homotopy class $[\gamma]$. Let $\Gamma$ be the union of leaves both beginning and ending at a critical point of $\phi$ i.e. a zero of $\phi$. This is called the critical graph of $\phi$. For genus $g \geq 2$ one can show that $R - \Gamma$ is a disjoint union of cylinders. $\gamma$ will lie in one of these cylinders. Pick now this cylinder. What is the boundary of this cylinder? Is it made of just one singular leaf (leaf of the foliation which contains zeros of $\phi$)?

Greetings Ben

Answer 1

First of all, if you start with only one loop, the Jenkins-Strebel differential will have only one cylinder (not many cylinders as you seem to be implying).

If the Riemann surface $R$ is closed, then there are two possibilities:

• if $\gamma$ is separating, then the critical graph $\Gamma$ has two connected components;
• if $\gamma$ is non-separating, then $\Gamma$ is connected.

If $R$ has punctures, then it depends on the precise location of the punctures. A puncture may be an endpoint of the critical graph where the quadratic differential has a simple pole, or it may be a regular point or a zero. In the latter cases, removing the puncture may create more connected components for the critical graph.