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Let $S$ be an orientable surface of genus $g$ with $b>0$ boundary components, and let $\mathrm{Mod}(S)$ be its mapping class group, that is, the group of isotopy classes of its homeomorphisms modulo isotopy. Homeomorphisms are allowed to reverse orientation or permute boundary components.

  1. What can one say about its torsion elements?
  2. Is it true that every mapping class of order 2 is orientation reversing?

Now assume that $S$ also has marked points on the boundary, and homeomorphisms are also allowed to permute them.

  1. Is it true that every mapping class of order 2 is orientation reversing?

It is well-known that if a mapping class preserves orientation and fixes each point of $\partial S$, then it cannot have finite order. It is not clear to me what happens when we allow orientation reversing or permutations.

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  • $\begingroup$ 2 and 3 are certainly wrong, examples being a plenty (e.g., hyperelliptic involutions). Maybe, not on any surface, though. $\endgroup$ – Alex Degtyarev Feb 22 '15 at 19:02
  • $\begingroup$ hyperelliptic involutions on surfaces with boundary? $\endgroup$ – yanglee Feb 22 '15 at 19:03
  • $\begingroup$ Why not? Say, two boundary components permuted? Or even one boundary component "about" one fixed point? $\endgroup$ – Alex Degtyarev Feb 22 '15 at 19:05
  • $\begingroup$ Sorry, what I mean is that you can take any closed surface and any finite group acting on it, and remove disks about any orbit of this group. So, examples are a plenty. $\endgroup$ – Alex Degtyarev Feb 22 '15 at 19:08
  • $\begingroup$ It seems all your questions have been answered in the comments. It's not so clear what you mean by your question (1) -- torsion elements appear to be able to do pretty much anything. If you need examples to help you think through these types of questions, there's a pencil-and-paper technique to enumerate through the conjugacy classes of finite-order elements of the mapping class group (even allowing orientation-reversing automorphisms). It's the classification of Seifert-fibred 3-manifolds, restricted to the class that fibre over $S^1$. $\endgroup$ – Ryan Budney Feb 22 '15 at 19:30
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Collecting some of the answers above, and editing a bit:

What can one say about its torsion elements?

$\newcommand{\Mod}{\mathrm{Mod}}$As with lattices in Lie groups, and in word-hyperbolic groups, the mapping class group of a surface has finitely many torsion subgroups, up to conjugacy. This means, in principle, if you fix $g$ and $b$ then you can list all finite subgroups of $\Mod(S_{g,b})$. This follows (for example) from "Nielsen realization" stating that for any finite subgroup $G$ in $\Mod(S_{g,b})$ there is a hyperbolic metric on $S$ where $G$ acts via isometries.

Despite this one should note that for any finite group $H$ there is a $g > 0$ so that $H$ is a subgroup of $\Mod(S_g)$.

Also, note that if you require that the surface has boundary, and require all mapping classes and isotopies to fix the boundary pointwise, then the finite order elements go away. For example, the braid group has no torision.

Is it true that every mapping class of order 2 is orientation reversing?

$\newcommand{\ZZ}{\mathbb{Z}}$No. In the closed case the most famous examples are the "hyperelliptic involutions". As somewhat different source of examples, take any surface $S$ and take any double cover $T$. Then the deck group of $T$ over $S$ is a copy of $\ZZ/2\ZZ$ inside of $\Mod(T)$.

Now assume that $S$ also has marked points on the boundary, and homeomorphisms are also allowed to permute them. Is it true that every mapping class of order 2 is orientation reversing?

Again no, as above.

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