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I have the following problem:

Let $h : Σ → Σ$ be a conformal self-map, different from the identity, of a compact Riemann surface $Σ$ of genus $p$. Show that $h$ has at most $2p+2$ fixed points.

The problem has a suggestion. Hint: Consider a meromorphic function $f : Σ → S^2$ with a single pole of order $≤ p + 1$ at some $z_0$ which is not a fixed point of $h$, and study $f(z) − f(h(z))$.

So let's go to what I thought ...

By theorem (see for example Jost p. 239), exists a non-constant meromorphic function $f : Σ → S^2$ with only one pole, of order $≤ p + 1$. Then for $z_0 \in Σ $ such that $h(z_0) \neq z_0$, we consider $f$ like mentioned with only one pole of order $≤ p + 1$ at $z_0$. Let $g(z):= f(z)-f(h(z))$.

Now i need study the set zeros of $g$ and conclude that $≤ 2p + 2$. My difficults is exactly here! After that i conclude that $h$ has at most $2p+2$ fixed points.

So that's it ... any help to solve my difficulty is welcome!Thanks!

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    $\begingroup$ If $D$ is a principal divisor then $deg(D) = 0$. Take $D = div(g)$ $\endgroup$ – reuns Feb 26 '17 at 18:57
  • $\begingroup$ I can not continue. It's still not clear to me. $\endgroup$ – Manoel Feb 26 '17 at 20:50
  • $\begingroup$ if $f(z)$ and $f(h(z))$ have $p+1$ poles then $f(z)- f(h(z))$ has at most $2p+2$ poles $\endgroup$ – reuns Feb 26 '17 at 20:53
  • $\begingroup$ So I was not doing well with my idea ... because I conclude that $f$ has only one pole of order $≤p+1$... $\endgroup$ – Manoel Feb 26 '17 at 21:42
  • $\begingroup$ one pole of order $p+1$ is (almost) the same as $p+1$ poles of order $1$ $\endgroup$ – reuns Feb 27 '17 at 17:42

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