I want to prove that the number of fixed points of a non-identity automorphism on a compact Riemann surface $X$ is at most 2g+2.

Following hints given, I have considered the divisor $D = (g+1)P$, where $P \in X$ and $F(P) \neq P$, where $F$ is an automorphism. Then by Riemann-Roch: \begin{align*} \mbox{dim}L((g+1)P) \geq g+1 + 1-g = 2 \end{align*} So there exist meromorphic functions $f \in L((g+1)P)$ with poles of order at most $(g+1)$ at $P$. In the next part, I am given the hint to consider the function $h = f - f \circ F$.

I don't know how to continue the proof though. Can someone please give me a push?

  • 3
    $\begingroup$ Your $f$ is holomorphic away from $P$, which means you know quite well where $h$ has poles, and you have a bound on the degree of poles at $P$ and $F^{-1}(P)$. On the other hand, a fixed point of $F$ gives a zero of $h$, and degree of $Div(h)$ is 0. Can you proceed from here? $\endgroup$ – user27126 Nov 18 '13 at 7:00
  • $\begingroup$ Yes, thankyou! It worked. $\endgroup$ – user82235 Nov 18 '13 at 8:13
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    $\begingroup$ Could you please edit your proof as answer because the hint of user 27126 sketches such a nice application of Riemann-Roch theorem. $\endgroup$ – Jo Wehler Oct 31 '14 at 9:04

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