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I want to solve this problem

Let $X$ a compact Riemann surface and $f:X\rightarrow \mathbb{P}^1$ a non constant meromorphic function in $X$. Let $h\in\text{Aut }(X) $ and automorphism of finite order $\text{ord } (h)$. If the number of fixed points of $h$ is bigger than $2\,\text {deg }(f)$ then $f=f\circ h$ and $\text {deg } (f)$ is a multiple of $\text {ord }{h}$.

I've been playing around with the Riemann-Hurwitz equations without arriving to something useful.

Any hint?

EQJ
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1 Answers1

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$f$ is holomorphic $X\to\Bbb{P}^1$ and meromorphic $X\to\Bbb{C}$.

Composing $f$ with a Möbius transformation we can assume the given fixed points of $h$ are not poles of $f$.

How many zeros does $f-f\circ h$ have ? How many poles does $f$ thus $f\circ h$ have ?

reuns
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  • I know that $f-f\circ h$ has at least as many zeros as fixed points $h$ has. $f$ has $\text{deg }(f)$ poles and $f\circ h$ has the same number of poles. ... But it is still unclear to me how to continue. – EQJ Oct 08 '19 at 14:33
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    So $f-f \circ h$ has $\ge 2\deg(f) +1$ zeros and $\le 2\deg(f)$ poles. A non-zero meromorphic function on a compact Riemann surface has the same number of zeros & poles. – reuns Oct 09 '19 at 09:12