Given a quadratic differential $q$ on a surface of genus $g$, we say that $q\in \mathcal Q(k_1,\ldots,k_n)$ if $q$ has $n$ distinct zeroes of order $k_1,\ldots,k_n$ respectively. The set $\mathcal Q(k_1,\ldots,k_n)$ is called the stratum of $q$.

A polygon $P$ with angles equal to rational multiples of $\pi$ gives rise to a compact Riemann surface by reflecting $P$ in the complex plane across each side repeatedly, and identifying $z$ with $z+b$ if the map $w\mapsto w+b$ preserves the edges of the polygons. This has an induced abelian differential $dz$, and squaring this gives rise to a quadratic differential $q$.

I'm interested in determining what stratum $q$ lies in from the the angles of $P$. In the case of isosceles or right triangles, the following classification holds:

If $P$ is isosceles with angles $2\pi\frac{a}{b},2\pi r,2\pi r$, $\gcd(a,b)=1$

- and $b=3,4,6$ then $q$ is constant.
- and $b\neq 3,4,6$ and $b\equiv 0\mod 4$ then $q\in\mathcal Q(2a-2,4br-2)$.
- and $b\neq 3,4,6$ and $b\equiv 2\mod 4$ then $q\in\mathcal Q(2a-2,2br-2,2br-2)$.
- and $b\neq 3,4,6$ and $b\equiv 1,3\mod 4$ then $q\in\mathcal Q(2a-2,2a-2,8br-2)$.

If $P$ is right with an angle $\pi\frac{a}{b}\neq \frac{\pi}{2}$, $\gcd(a,b)=1$ then the classification is the same as for $P$ isoceles with angles $2\pi\frac{a}{b},2\pi\left(\frac{1}{4}-\frac{a}{2b}\right),2\pi\left(\frac{1}{4}-\frac{a}{2b}\right)$. It can be verified that choosing the other angle does not change the result.

As pointed out Matheus' answer, each vertex of the polygon corresponding to an angle of $\pi\frac{a}{b}$ in the unfolding is associated with a zero of order $2a-2$. However, identifications of these vertices mean that it is not possible in general to determine the stratum from this alone, as can be seen in the case of isosceles triangles with $b\equiv 0$. Because the number of zeros of $q$ counting multiplicity is $4g-4$ where $g$ is the genus, we know that for a $k$-gon with angles $\pi m_i/n_i$ $$\text{# of zeros of }q=2\gcd(n_1,\ldots,n_k)\left(k-2-\sum_{i=1}^k\frac{1}{n_i}\right)$$ thanks to a result of Masur in the work cited by Matheus. Unfortunately, similar calculations do not seem to give the number of distinct zeros or their orders.

My question is whether there is a generalization of this classification to all triangles, or even all polygons $P$. I would also appreciate a reference to any such classification in literature, so that I can check the correctness of my own work and do not need to reproduce a proof.

**Edits:** The construction of the Riemann surface and quadratic differential from a polygon has been clarified. The classification of strata of quadratic differentials for non-right, non-isosceles triangles has been deleted due to a major error. Discussion relevant to Matheus' answer has been added.