$\DeclareMathOperator\Mod{Mod}$Let $S$ be a surface and $P=\{a_1,...,a_n\}$ be a pants decomposition of $S$. Denote by $\Mod(S)$ the mapping class group of $S$. Define the stabilizer of $\Mod(S)$ on $P$ to be $$A=\{f\in \Mod(S), f(a_i)=a_i, i =1,...,n\}.$$ What is $A$? I was once told that it is generated by Dehn twists and hyperelliptic involution, but I cannot find a reference for that. Is it true, and why? Could you give a reference for this?

I saw this on Lemma 3.2 in this paper by U. Wolf, which says that $A$ is generated by Dehn twists and half twists But I didn't understand it yet.


1 Answer 1


Suppose that $S$ is closed (without boundary), connected, and oriented.

If $S$ has genus two then the stabiliser is generated by Dehn twists about the $a_i$, the hyperelliptic, and a reflection.

If $S$ has genus greater than two, then there is no hyperelliptic symmetry, but there will still be a reflection symmetry.

This is because any mapping class that stabilises the "cuffs" of the pants (the curves $a_i$) either permutes the pants (and so there are exactly two pants) or preserves each pants setwise (and thus is isotopic to the identity, or a reflection, off of a small neighbourhood of the cuffs).

  • $\begingroup$ I've made a small edit to clarify the meaning. $\endgroup$
    – Sam Nead
    Commented Mar 13, 2022 at 12:30

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