Timeline for Strata of quadratic differentials from rational billiards
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 25, 2012 at 20:06 | vote | accept | Alex Becker | ||
| Jul 25, 2012 at 20:06 | comment | added | Alex Becker | 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. | |
| Jul 25, 2012 at 8:55 | history | edited | Matheus | CC BY-SA 3.0 |
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| Jul 25, 2012 at 8:33 | comment | added | Matheus | Please see the edited answer for some clarifications motivated from your comments. | |
| Jul 24, 2012 at 21:52 | comment | added | Alex Becker | 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. | |
| Jul 24, 2012 at 20:29 | comment | added | Alex Becker | 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. | |
| Jul 24, 2012 at 20:20 | history | answered | Matheus | CC BY-SA 3.0 |