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

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

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    $\begingroup$ You need to look for critical values: where $f'(z)=0$. These correspond to ramification points. $\endgroup$ – Lord Shark the Unknown Jan 15 at 1:55
  • $\begingroup$ Thanks I got it now. Besides 0, there are other 2 ramification points $\pm\sqrt{3}$ whose orders are 2. Hence they contribute another 2 to the right side and the equation holds. $\endgroup$ – Xipan Xiao Jan 15 at 7:28

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