6
$\begingroup$

Let $X$ be a compact Riemann surface of genus $g>1$, $f\in Aut(X)$, a biholomorphism of $X$ onto itself, $x\in X$ a fixed point of $f$. Since tangent map of a holomorphic map (on the real tangent space) is of the form

\begin{bmatrix} r\cos(\theta) & -r\sin(\theta) \\ r\sin(\theta) & r\cos(\theta) \end{bmatrix}

The local Lefschetz number is always $1$, unless the tangent map is identity, which will impose $f$ to be the identity map. This fact can be easily seen when $f$ is lifted to the universal covering of $X$.

Can we carry on this idea to say something more about the group of automorphism of a compact Riemann surface of genus $g>1$? Recall that on such a Riemann surface, we have exactly $g(g-1)(g+1)$ Weierstrass points, counted with weight, and a biholomorphism must take Weierstrass points to Weierstrass points.

Thank you!

  • 0
    I am sorry. I am a newcomer and I'm not familiar with typing math symbols yet. The row vector should have been a 2 by 2 matrix, I don't know what goes wrong.2011-12-07
  • 1
    The software interprets backslashes in a funny way (it processes Markdown before LaTeX). You need to use three instead of two between lines.2011-12-07
  • 0
    I'm a little confused by your question. Automorphisms of Riemann surfaces need not have fixed points, and they can be any finite group... so what kind of flavour of result are you hoping for? -- why would you hope for an interesting connection here?2011-12-10
  • 0
    I am just beginning to learn Riemann surfaces. I was trying to prove that the automorphisms of a compact Riemann surface is finite. Since there are only finitely many Weierstrass points on a compact Riemann surface and automorphisms take Weierstrass points to Weierstrass points, if the group is large, it must fix some Weierstrass points. I hope this can tell us something. It is just a very vague idea.2011-12-11

2 Answers 2