# Strata of quadratic differentials from rational billiards

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.

-
You say $P$ "gives rise to a compact Riemann surface by reflecting P in the complex plane across each side repeatedly and identifying parallel sides", but that is not enough information. The same description could apply to some finite cover of the Riemann surface, in which case the description of the stratum would have different numbers of repetitions of the angles that parameterize the stratum, or even to a finite branched cover in which you would have different angles altogether. –  Lee Mosher Jul 24 '12 at 19:34
Yes, I guess I need to be more precise. I reflect the polygon $P$ in the complex plane and then identify $z\in\mathbb C$ with $z+b$ so long as the map $w\mapsto w+b$ carries edges of $P$ to the same edges on a different copy. –  Alex Becker Jul 24 '12 at 20:59

Dear Alex Becker,

You can find the answer to your question in the excellent survey "Rational billiards and flat structures" of H. Masur and S. Tabachnikov: for a free online version see here (http://math.uchicago.edu/~masur/handbook.dvi).

As you can check in this survey, it is not very hard to deduce the singularity pattern of q from the angles of a rational $k$-gon $P$. Indeed, for each $j=1,\dots,k$, let $v_j$ be the $j$th vertex of $P$, let $\pi m_j/n_j$ be the angle around $v_j$ (where $m_j$ and $n_j$ are coprime), and let $N$ be the least common multiple of $n_j$'s. From the arguments on the survey (see Lemma 1.2 of the .dvi online version), after the unfolding procedure (of reflecting sides), $v_j$ gives rise to $N/n_j$ points with total angle $2\pi m_j$. Thus, the Abelian differential $\omega$ induced by $dz$ on $P$ has a zero of order $m_i-1$ on each of these points, that is, its stratum is $$\mathcal{H}(\underbrace{m_1-1,\dots,m_1-1}_{N/n_1},\dots,\underbrace{m_k-1,\dots,m_k-1}_{N/n_k})$$ In particular, the quadratic differential $q=\omega^2$ belongs to the stratum $$\mathcal{Q}(\underbrace{2(m_1-1),\dots,2(m_1-1)}_{N/n_1},\dots,\underbrace{2(m_k-1),\dots,2(m_k-1)}_{N/n_k})$$ because every time $\omega$ has a zero $x$ of order $a$, it is written locally near $x$ as $\omega=z^a dz$, so that $q:=\omega^2=z^{2a}dz^2$ locally near $x$.

Just for "double check", the genus is given by the formula $$2g-2=\sum \textrm{ orders of zeroes of }\omega$$ or equivalently, $$4g-4=\sum \textrm{ orders of zeroes of }q$$ for $q=\omega^2$. By applying this formula to the unfolding of $P$, we get that the genus satisfies $$2g-2=\sum(m_j-1)N/n_j$$ Since the sum of the inner angles of $P$ is $\pi(k-2)$, we have that $\sum m_j/n_j=k-2$ and thus the previous equation becomes $$2g-2=N(k-2-\sum(1/n_j))$$ that is exactly the same formula for the genus as in the survey. Of course, by multiplying everything in this equation we get the correct corresponding statement for the quadratic differential $q$.

Finally, concerning the triangle associated to the octagon, I think you wish to unfold the triangle with angles $\pi/8, \pi/2, 3\pi/8$ and not $\pi/4,3\pi/8,3\pi/8$: indeed, by unfolding the first you get a nice octagon but by unfolding the second you get two copy of the octagon (that is, a Riemann surface that is not connected) and this explains why your comment doesn't contradict the facts claimed in the survey.

Best regards,

Matheus

-
This doesn't seem to give the whole picture. Consider an isosceles triangle with angles $(\pi/4,3\pi/8,3\pi/8)$. Then the unfolding produces an octagon with opposite sides identified, which has Euler characteristic $8-12+2=-2$, thus has genus $2$. We would then expect the quadratic differential to have $4\cdot 2-4=4$ zeros, yet naïve application of the formula you give would lead me to conclude that there are $2\cdot 2(3-1)=8$. The issue here is that the two vertices with zeros of order $4$ are identified, but I'm not sure how to know which vertices are identified in general. –  Alex Becker Jul 24 '12 at 20:29
Furthermore, in general we have that the genus of a $k$-gon with angles $\pi m_j/n_j$ is $1+\frac{N}{2}(k-2-\sum\frac{1}{n_j})$ where $N$ is the gcd of $n_1,\ldots,n_k$. Thus the sum of the orders of the zeros of the quadratic differential is $2N(k-2-\sum\frac{1}{n_j})$, which is not in general equal to $2(\sum m_j) - k$. Thus the stratum cannot be determined from this alone. –  Alex Becker Jul 24 '12 at 21:52
Thank you very much, this is quite helpful. I see now my problem was that I was not properly keeping track of the orientations of the copies of $P$ in my unfolding. –  Alex Becker Jul 25 '12 at 20:06