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.

  • 1
    $\begingroup$ I guess a relevant result would be the classification of cyclic group actions on closed surfaces (allowing fixed points). It seems that this has been done in some sense, i.e. classifying the number of fixed points and branching order etc. (Example 3.3 here core.ac.uk/download/pdf/82024929.pdf), however it is not clear (to me) if one can obtain it up to conjugating in the homeomorphism group. $\endgroup$
    – Nick L
    Dec 11, 2021 at 19:49
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    $\begingroup$ Yes, it can be stated in the language of orbifolds since a free action on a surface with boundary extends to an action on a surface without boundary that has at most one fixed point in each added disk. $\endgroup$ Dec 13, 2021 at 14:36

1 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.


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