Inequivalent free $\Bbb Z/n\Bbb Z$-actions on orientable compact bordered surface

Question

Let $S_{g,b}$ denote the orientable connected compact surface of genus $g$ with $b$ boundary components. A group homomorphism $\varphi\colon G\to \text{Homeo}^+(S_{g,b})$ is said to be free $G$-action if $\varphi(a)$ has no fixed point for all non-trivial $a\in G$. Two free group actions $\varphi_1,\varphi_2\colon G\to \text{Homeo}^+(S_{g,b})$ are said to be equivalent if there is $\mathscr H\in \text{Homeo}^+(S_{g,b})$ such that $\varphi_2(a)=\mathscr H^{-1}\circ \varphi_1(a)\circ \mathscr H$ for all $a\in G$.

A theorem of Nielsen says that any two free $\Bbb Z/n\Bbb Z$-actions on a closed orientable connected surface are equivalent.

Does there exist a classification theory of inequivalent free $\Bbb Z/n\Bbb Z$-actions on every $S_{g,b}\ (b\neq 0)$?

Any reference/idea will be helpful.

Answer 1

Did you try to use the double $T$ of the surface $S$? Any fixed point-free action of $\mathbf Z/n$ on $S$ induces a fixed point-free action on the closed orientable surface $T$. Moreover, the induced action commutes with the natural orientation-reversing involution on $T$. By Nielsen's Theorem that you mentioned, you've reduced to study orientation-reversing involutions on $T$ that commute with a given action of $\mathbf Z/n$ on $T$ and whose set of fixed points is homeomorphic to a disjoint union of $b$ circles. Such involutions induce orientation-reversing involutions on the quotient surface $U=T/(\mathbf Z/n)$. Now, orientation-reversing involutions on a closed orientable surface $U$ whose set of fixed points is a disjoint union of circles are easily classified. It should not be too difficult to decide which ones lift to orientation-reversing involutions on $T$ having a set of fixed points composed of $b$ circles.

With this approach you already get conditions on the mere existence of fixed point-free actions of $\mathbf Z/n$ on $S$. Indeed, the double $T$ of $S$ has genus $2g+b-1$. So that $n$ has to divide $$ -\chi(T)=2(2g+b-1)-2=4g+2b-4 $$ in order for a fixed point-free action of $\mathbf Z/n$ on $S$ to exist. For example, if $S$ is a pair of pants, i.e., $g=0$ and $b=3$, the natural number $n$ has to divide $2$. Adopting the above strategy, I think you only get $1$ fixed point free action of $\mathbf Z/2$ on the pair of pants $S$: the one where the nontrivial element of $\mathbf Z/2$ acts on $S$ by exchanging $2$ boundary circles, and by acting antipodally on the third. That case probably was clear anyway.