Covering Space for 3-punctured sphere

Question

I have been considering whether we can find a branched covering map $p:Y \rightarrow \mathbb{CP}^1$, having 3 critical values, such that exactly five points are mapped to those critical values. That is,$|p^{-1}(\text{Crit}(p))| = 5$

Answer 1

I'm not an expert, but I hope this helps:

Let $Y$ refer to a compact Riemann Surface of genus $g$, there is a useful form of the Riemann-Hurwitz formula: if $p:S\to S'$ is a map of degree $N$ between surfaces of genus $g,g'$ then $$ r-2+2g = N(b-2+2g') $$ where $b$ is the number of branch points (critical values) and $r$ es the number of ramification points (ie whose image is a branch point). For the data you provide a function can exist only if $$2g + 3 = N$$ so if there is such map it will be of degree $2g+3$.

For $g=0$ we get $N=3$ and it easy to find a funcion with exactly 5 ramification points ie from the sphere into itself we could use $f(z) = -2z^3+3z^2$ which has critical values $\{0,1,\infty\}$ and ramification points $\{0,1,\infty, 2/3, -1/2 \}$. I understand that this function is essentially unique modulo mobius transformations.

For genus 1 you have to find a degree 5 function. I describe below how to obtain one concrete example but I'm not sure if there is such a function for every curve:

If we write the curve $Y$ in Weierstrass form $$ y^2 = x^3 + ax + b $$ Let $p:Y \to \mathbb C$ be the function we are looking for, let's try the function $$ p(x,y) = (x + A)y $$ we know that $x$ has a pole of order 2 and $y$ a pole of order 3 at infinity, so $p$ has a point of multiplicity 5 at infinity. So we need to make sure that there are only two more branch points with exactly two points above each one.

to find the ramification let's compute the differential $$ dp = (x+A) (3x^2+a) \frac{dx}{2y} + y dx $$ so simplifying $$ dp=(5x^3+ 3Ax^2 + (A+2a) x + A^2+2b) \frac{dx}{2y} $$ as $dx/2y$ is nowhere vanishing the zeros of the differential $dp$ are the zeros of the polynomial. So let's take $$ A = a = 5\quad \text{and}\quad b=-10 $$ then the polynomial is $$ 5(x^3 + 3x* 2+3x + 1) = 5(x+1)^3 $$ and then the points $(x,y)=(-1,\pm 4i)$ have multiplicity 4.

So finally in the curve $$ y^2 = x^3 + 5x - 10 $$ the function $$ p(x,y) = (x+5)y $$ has ramification points at infinity of multiplicity 5 and at the points $(-1,\pm 4i)$ with multiplicity 4. And three branch points $\infty$ and $\pm 16i$. As it is of degree five, it wil have two additional ramification point with multiplicity 1, one above each branch point.

For other genus we could use similar ideas but I suppose it becomes harder as the genus increases.