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I am studying finite group action on Riemann surfaces from the book Algebraic Curves and Riemann Surfaces by Rick Miranda and there is are few statements about the genus that I am not able to understand:

Let $X$ be a Riemann sphere and let $G$ is a finite group acting holomorphically and effectively (or faithfully) on $X$.

  1. If $X = \Bbb{C}_{\infty}$ (the Riemann sphere), then the genus of $X/G$ is zero.

  2. If $X$ is a Riemann surface of genus 1, then $X/G$ has genus at most 1.

How to prove these? I know that the genus is the number of holes in $X$ and for compact Riemann surfaces it is related to the Euler characteristic by the following equation:

$$\chi(X) = 2 - 2g(X).$$

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Roughly, from the Riemann-Hurwitz Frormula you deduce that $$\chi(X)= |G| \cdot \chi(X/G) - \sum_\text{x fix pt.}( n(x) -1)$$ where $n(x)\geq 1$ denotes the cardinality of the quotient of $G$ for the stabiliser of the fixed point. Thus, if we assume that $\chi(X) \geq 0$, we have that
$$ \cdot \chi(X/G) = \frac{\chi(X)+ \sum_\text{x fix pt.}( n(x) -1)}{|G|} \geq \frac{\chi(X)}{|G|} \geq 0$$ with a possibility for the equality only in the case in which $\chi(X)=0$.

Antonio Alfieri
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