Existence of orientation preserving, finite order self homeomorphism on a genus 2 surface without fixed point

Let $M$ be a compact 2-manifold of genus 2. Does there exist an orientation preserving homeomorphism $f:M\to M$, so that $f^n=id$ for some integer $n$, and $f$ doesn't have fixed points?

Using Lefschetz formula and the cohomology ring structure, I can prove that the order of such $f$ must be a multiple of 4 or 6. I vaguely remember that I have seen such an $f$ before, but I'm not sure. Is there anyone familiar with this kind of examples?

• You can view the classification of Seifert fibred spaces as a dictionary for enumerating conjugacy classes of finite-order automorphisms of surfaces, with or without fixed points, it's up to you. With that tool it becomes a simple enumeration problem. You just have to determine precisely which Seifert fiber spaces fiber over the circle -- this means they have a horizonal incompressible surface. – Ryan Budney Sep 4 '14 at 17:22

The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction.

Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $M = \partial (T \times [-1, 1])$. Let $f : M \to M$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f$ satisfies the conditions in the question, with $n = 6$.

All finite group actions on a genus 2 surface have been classified; see Classifying finite group actions on surfaces of low genus by A. Broughton, Journal of Pure and Applied Algebra Volume 69, Issue 3, 7 January 1991, Pages 233–270. You can get the article for free from the journal web page. The result is given on page 252 in a table that has fixed point data, from which you can answer your question.

Here is a concrete example: let $C$ be the smooth projective model of the complex plane curve $y^2 = x^6-1$. Then $C$ is genus 2 and has a fixed point free automorphism given by $(x,y)\mapsto (\exp (2\pi i/3)x,-y)$.

Geometrically: let $C$ be the double branched cover of the Riemann sphere, ramified over the 6th roots of unity. The $Z/6Z$ action generated by rotating the Riemann sphere by $2\pi/3$ and simultaneously swapping the sheets of the cover is fixed point free (but the points over 0 and $\infty$ are orbits of size two).

If you are even more topologically minded, you can picture this as follows. Consider the graph with 2 vertices and three edges, each going from one vertex to the other. This graph has an order six automorphism generated by cyclically permuting the three edges and swapping the vertices. Now thicken this graph in three space to be a handle body whose boundary is a genus 2 surface. The previously mentioned automorphism acts on this surface in a fixed point free fashion.

• The projectivization of this curve seems to be singular at $[0,1, 0]$. – Daniele Zuddas Sep 4 '14 at 4:57
• Are you sure of this example? The Euler number of a genus $2$ surface is $-2$, it seems to me that this forbids free cyclic actions on it. Am I missing something? – Francesco Polizzi Sep 4 '14 at 8:38
• The curve is smooth but, as Francesco observes, the automorphism has 2 fixed points at infinity. One should take e.g. $(x,y)\mapsto (\omega x,-y)$ for the same curve, with $\omega$ a nontrivial third root of unity. – abx Sep 4 '14 at 8:52
• @Francesco Polizzi: the Euler number excludes a free action of a finite group, but your automorphism may have no fixed points while some power has fixed points. – abx Sep 4 '14 at 8:54
• @abx: ok, you are right. – Francesco Polizzi Sep 4 '14 at 8:57

The answer is actually yes, and here is an example with $n=6$ (inspired by Broughton's classification quoted by Danny and by Jim's answer).

Take the group $G = \mathbb{Z}/ 6 \mathbb{Z}$ presented as follows:$$G = \langle x, \, y \; | \; x^2=y^3=1, \, [x, y]=1 \rangle.$$ By Riemann Existence Theorem, any Galois cover of $\mathbb{P}^1:=\mathbb{P}^1(\mathbb{C})$ with group $G$ is given by a finite collection of generators $g_1, \ldots, g_k \in G$ whose product is the identity. Now choose $$g_1=x, \quad g_2=x, \quad g_3=y, \quad g_4=y^2.$$

The corresponding Galois cover $\pi \colon M \to \mathbb{P}^1$ has four branch points. The points over two of them have stabilizer of order $2$ (generated by $x$) and the points over the remaining two have stabilizer of order $3$ (generated by $y$).

By Hurwitz formula we have $$2g(M)-2 = |G| \bigg(2g(\mathbb{P}^1)-2+ 2\big(1-\frac{1}{2} \big) + 2 \big(1-\frac{1}{3} \big) \bigg) = 2,$$ that is $M$ is a compact manifold of genus $2$. The automorphism of $M$ corresponding to the generator $g:=xy \in G$ has order $6$ and is fixed point-free (whereas both $g^2=y^2$ and $g^3=x$ have fixed points).

see the paper: A note on the free degrees of homeomorphisms on genus 2 orientable compact surfaces