# Is there an algorithm to compute a Belyi map for the Riemann surface?

Let $$y^2=x^5-x-1$$ be an affine model of a projective complex curve, is there an algorithm to compute the Belyi map (preferably of small degree), i.e., map to the projective line ramified only at $$\{0,1,\infty\}$$.

In my attempts to do this by hand, I get ramification at 4 points and, subsequently, using Shabat polynomial will skyrocket the degree of the map. Is there a way to avoid increasing the degree to thousands or hundreds? I have $$\beta=h\circ g\ \circ f$$, where $$f$$ is projection on $$x$$, $$g=x^5-x-1$$ and $$h=\frac{( 12500 x + 18750 x^2 + 12500 x^3 + 3125 x^4)}{-2869}$$. $$\beta$$ gives ramification at four points $$\{0,1, \infty, \frac{3125}{2869}\}$$. Also, it is possible to use $$g=x^5-x$$ with different $$h$$, but the Shabat polynomial will still be of very big degree....

After some work, I have obtained maps: $$f$$ projection on $$x$$ ramified at five roots of $$x^5-x-1$$, $$g(z)=z^5-z$$ ramified at $$\pm \frac{4}{5\sqrt[4]{5}}$$, $$h=z^4$$ ramified at $$0$$. As a result $$\beta=h\circ g\circ f$$ is ramified at $$\{0, \frac{256}{3125}, 1, \infty\}$$, which is slightly better. Using Shabat map $$\frac{m}{m+n}=\frac{256}{256+2869}$$ gives $$\alpha(z)=\frac{(m+n)^{m+n}}{m^mn^n}z^m(1-z)^n=\frac{(3125)^{3125}}{256^{256} 2869^{2869} }z^{256}(1-z)^{2869}$$. The resulting map $$\alpha\circ\beta$$ is of huge degree....

• First of all, there are a lot of distinct Belyi maps for a given curve. I guess there is not a method to compute the one of smallest degree, unless you have a "special curve" (modular, large automorphism, etc). Usually the maps have very high degree, since there is a finite number of curves with a given genus $>0$ and degree of the Belyi map. – Xarles Jun 24 at 14:29
• @Xarles If by "method" you mean algorithm, then I just wanted to point out that there is an algorithm to compute a Belyi map of smallest degree on any curve (which is not always very practical though): arxiv.org/pdf/1711.00125.pdf – Ariyan Javanpeykar Jun 24 at 19:16
• Possibly of interest: Vincent Beffara's "Dessins d'enfants for analysts" arxiv.org/abs/1504.00244 featuring lots of pictures – j.c. Jun 24 at 23:43
• @AriyanJavanpeykar Thank you. I have tried to follow the algorithm and can see why it is not practical... However, the bounds give some hope – SamMar Jun 25 at 0:12
• @AriyanJavanpeykar When I was writing that sentence I said to myself that that was probably wrong (so one could make up a very impractical algorithm to compute it, may be just computing all the dessins for a given degree and genus, and so on), but I am happy I wrote that, so you mentioned your nice paper. – Xarles Jun 25 at 8:28