Verify Hurwitz's formula for $\frac{z^3}{1-z^2}$

Question

This is an exercise from Miranda's book "Algebraic curves and Riemann surfaces".

Consider $f(z)=\frac{z^3}{1-z^2}$ as a holomorphic map from the Riemann sphere $\mathbb{C}_\infty$ to itself. Verify Hurwitz's formula.

I found 1 zero and 3 poles.

1. Zeros: $0$, with an order and multiplicity of $3$.
2. Poles: $1, -1$ and $\infty$, with order $-1$ and multiplicity $1$.

These orders add up to $0$, which looks good. The degree of the map is $3$. Plug into the Hurwitz's formula:

$$2g-2=\deg(F)(2g-2)+\sum_{p\in \mathbb{C}_\infty} [\text{mult}_p(F)-1]$$

And get $$-2=3\cdot(-2) + (3-1)$$ That is $-2=-4$. What did I do wrong?

Answer 1

I try to write a complete answer that might be useful.

We need to find the local coordinates in the domain $\mathbb{C}_\infty$ and target $\mathbb{C}_\infty$ such that locally $F$ can be written as $z\mapsto h(z)$ for some holomorphic function $h(z)$. One can then compute the zeros of $h(z)$ to compute the multiplicities appearing in Hurwitz's Formula. There are two charts in the domain and two charts in the target that you need to consider: Let's call the ones in the domain to be $U_1$ and $U_2$ and the ones in the target to be $V_1$ and $V_2$. Then,

• Local coordinates on $U_1$ and $U_2$ are $z$ and $\frac{1}{z}$, respectively;
• local coordinates on $V_1$ and $V_2$ are $f(z)$ and $\frac{1}{f(z)}$, respectively.

We then have

$$h(z)= \begin{cases}\frac{z^3}{1-z^2}, \qquad &F^{-1}(V_1)\cap U_1, \\ \frac{1}{z(z^2-1)}, \qquad &F^{-1}(V_1)\cap U_2, \\ \frac{1-z^2}{z^3}, \qquad &F^{-1}(V_2)\cap U_1, \\ z(z^2-1), \qquad &F^{-1}(V_2)\cap U_2. \end{cases}$$

We can now use Lemma 4.4:

• On $F^{-1}(V_1)\cap U_1$, we have $h'(z)=-\frac{z^2(z-\sqrt{3})(z+\sqrt{3})}{(1-z^2)^3}$. This will give three ramification points:

$$\begin{equation}\begin{aligned} z&=0, &\qquad& \text{mul}_{z=0}(F)=1+2=3, \\ z&=+\sqrt{3}, &\qquad& \text{mul}_{z=+\sqrt{3}}(F)=1+1=2, \\ z&=-\sqrt{3}, &\qquad& \text{mul}_{z=-\sqrt{3}}(F)=1+1=2, \end{aligned}\end{equation}$$

• On $F^{-1}(V_1)\cap U_2$, we have $h'(z)=-\frac{3z^2-1}{z^2(z^2-1)^2}$. The roots are $\pm 1/\sqrt{3}$, which are the same as $\pm\sqrt{3}$ we encountered above (note that in this patch $1/z$ is the local coordinate).

• On $F^{-1}(V_2)\cap U_1$, we have $h'(z)=+\frac{z^2(z-\sqrt{3})(z+\sqrt{3})}{z^6}$. Again, there are no new ramification points.

• On $F^{-1}(V_2)\cap U_2$, we have $h'(z)=3z^2-1$. Again, there are no new ramification points.

The sum of orders of poles or the order of zero of $f(z)$ is $+3$, which is the degree of $F$. We can now check Hurwitz's Formula in Theorem 4.16:

$$\begin{equation} \begin{aligned} \text{LHS}&=2\times 0-2=-2, \\ \text{RHS}&=3(2\times 0-2)+(3-1)+(2-1)+(2-1) \\ &=-6+2+1+1 \\ &=-2. \end{aligned} \end{equation}$$