Let \(\Sigma_3\) be the closed orientable surface of genus \(3\). Suppose \(\mathbb{Z}_3\) acts on \(\Sigma_3\); show that there must be at least two fixed points.
Solution
1. Any action of \(\mathbb Z/3\) on the orientable surface \(\Sigma_3\) is orientation-preserving.
The orientation character of the action is a homomorphism \(\mathbb Z/3\to\mathbb Z/2\), which must be trivial.
2. Let \(r\) be the number of fixed points and let the underlying quotient surface have genus \(h\).
Since the group has prime order, every nonfree orbit has stabilizer equal to the whole group and hence is a fixed point.
3. The Riemann–Hurwitz formula gives\[ 2-2\cdot3=3(2-2h)-2r. \]
Away from the \(r\) fixed points the quotient map has degree \(3\). Each fixed point is a branch point of order \(3\), contributing ramification defect \(3-1=2\).
4. Thus\[ -4=6-6h-2r, \qquad\text{so}\qquad r=5-3h. \]
Rearrange the equation in 3.
5. Since \(r\ge0\) and \(h\ge0\), the only possibilities are\[ (h,r)=(0,5)\quad\text{or}\quad(1,2). \]
If \(h\ge2\), then \(5-3h<0\).
6. Therefore\[ \boxed{r\ge2}. \]
Both possible values of \(r\) in 5 are at least two.