# Connection between Strebel differentials, ribbon graphs, and Belyi maps

In this paper, a nice story is woven regarding the connection between quadratic differentials on Riemann surfaces, so-called 'ribbon graphs' drawn on those surfaces, and Belyi maps. However, I am running into trouble when I try to apply the results of that paper in a concrete example, and I am wondering whether anyone can help to identify the source of my confusion. Although there are a lot of words below, I hope the question will be reasonably easy to answer!

First, to briefly recap: Consider a meromorphic quadratic differential $q=\phi\left(x\right)\mathrm{d}x^2$ with only second order poles, drawn on a Riemann surface $\mathcal{C}$ of genus $g$ and with $n$ punctures. For special values of the parameters in $\phi\left(x\right)$, $q$ will satisfy the definition of a so-called Strebel differential, as given in Theorem 4.2. When the quadratic differential is Strebel, so-called 'horizontal trajectories' drawn on $\mathcal{C}$ using $q$ can can be used to construct a unique ribbon graph on $\mathcal{C}$ corresponding to that Strebel differential; the ribbon graph has vertices for the zeroes of $q$ and one face for every second-order pole of $q$.

Each ribbon graph can be interpreted as a dessin d'enfant in the sense of Grothendieck, i.e. as a bipartite graph on $\mathcal{C}$. To do this, we colour every vertex of the ribbon graph white, and insert a black node into every edge. In the paper, this is referred to as the "edge refinement of the ribbon graph". To each dessin there corresponds a unique Belyi map, i.e. a holomorphic map to $\mathbb{P}^1$ ramified at only $\left \{ 0,1,\infty \right \}$ (the dessin is the inverse image of the line segment $\left[0,1\right]$ by the Belyi map).

One of the main results of the paper in question is Theorem 6.5. Take a Strebel differential $q$ on $\mathcal{C}$. To $q$ there corresponds a ribbon graph, which we can interpret as a dessin, and to that dessin there corresponds a unique Belyi map $\beta$. It turns out that $q$ is the pullback by $\beta$ of a meromorphic quadratic differential on $\mathbb{P}^1$ with three punctures, so that:

$$q = \beta^* \left(\frac{\left(\mathrm{d}\zeta\right)^2}{4\pi^2\zeta\left(1-\zeta\right)}\right)$$

It then seems to follow (but isn't explicitly stated in the paper) that:

$$q = \left( \frac{\left(\mathrm{d}\beta\right)^2}{4\pi^2\beta\left(1-\beta\right)} \right)$$

My trouble comes when I attempt to apply this theorem in a concrete example. For instance, take $\mathcal{C}$ to be the four-punctured sphere, so that $g=0$ and $n=4$. As drawn in figure 5 here, there are five topologically distinct possible ribbon graphs. Let's just pick one of these, and draw it in dessin form:

Since the ribbon graph has four vertices and four faces, we expect the Strebel differential to have four zeroes (for the vertices) and four second order poles (for the faces). The Belyi map associated to the above dessin is known to be:

$$\beta\left(t\right) = \frac{t^3\left(t+6\right)^3\left(t^2-6t+36\right)^3}{\left(t-3\right)^3\left(t^2+3t+9\right)^3}$$

(The details are given in this paper. In short, the above dessin can be associated to the modular subgroup $\Gamma\left(3\right)$, and the associated Belyi map is given here.)

When we substitute our expression for $\beta\left(t\right)$ into our expression for $q$ in terms of $\beta$ above, we would expect to recover an expression with four zeroes and four second order poles, but we don't! So it looks like something isn't consistent. So, in summary, my question is: why don't I recover the right form of the Strebel differential, for a certain given ribbon graph, when I substitute the associated Belyi map into the expression above for $q$ in terms of $\beta$?

It turns out that my working in the above question went wrong for the following reason: the $\beta\left(t\right)$ I was using was incorrect! This can be checked by explicitly computing its ramification indices. In fact, the correct Belyi map for the tetrahedral clean dessin on $\mathbb{P}^1$ is:
$$\beta\left(z\right) = -\frac{64 z^3 \left(-1+z^3\right)^3}{\left(1+8 z^3\right)^3}$$
Plugging this into the expression for $q$ in terms of $\beta$ gives a quadratic differential what four zeroes and four second order poles, as required.