Given an orientable compact surface S with genus g b and b boundary components and p punctures, is there a maximun order of finite order elements in the mapping class group of S? If yes is there a formula to determine it?
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1$\begingroup$ Yes: as a group of automorphisms of some variety in char 0, the mapping class group is virtually torsion-free and hence there is a bound on the order of its finite subgroups. The precise bound might depend on some details (e.g., whether you all orientation-reversing elements in MCG). It is also natural to ask about multiplicative bounds (e.g. statements of the form: finite orders of elements should divide some given number, or some number among a small list), and in particular exclude cyclic group of large enough prime order (much below the largest possible order). $\endgroup$– YCorCommented Jul 5, 2017 at 18:50
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3$\begingroup$ That MCGs are virtually torsion-free is certainly proved in Farb--Margalit somewhere. For a crude upper bound, at least in the non-closed case, one can use the fact that the natural map to $Sp(n,\mathbb{Z}/3)$ is known to have torsion-free kernel; so the order of torsion is bounded by the maximal order of an element in $Sp(n,\mathbb{Z}/3)$. (Apparently this fact is proved in a 1968 paper by Baumslag and Taylor.) $\endgroup$– HJRWCommented Jul 5, 2017 at 18:56
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$\begingroup$ On reflection, that they're virtually torsion-free also follows immediately from Nielsen realisation. In particular, any finite subgroup is necessarily of order at most 42(2g-2) in the closed case (since this is the degree of a covering map to the minimum-area 2d orbifold). $\endgroup$– HJRWCommented Jul 5, 2017 at 20:12
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$\begingroup$ @HJRW also it yields a bound in the non-closed case (since one can "close" the surface gluing pairwise the boundary components and closing the possible last one with a 1-punctured torus). (This works at least if the realization problem can be solved with a group of self-homeomorphisms acting as identity on the boundary). On the other hand, I can see why this provides a bound on the order of finite subgroups, but why does this imply virtual torsion-freeness? $\endgroup$– YCorCommented Jul 5, 2017 at 20:45
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$\begingroup$ @YCor, you're right: I guess I was implicitly using a couple of other related facts (only finitely many conjugacy classes of finite subgroups, and residual finiteness). $\endgroup$– HJRWCommented Jul 5, 2017 at 21:11
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1 Answer
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The answers to all the questions can be found in that fond of wisdom Farb-Margalit:
Farb, Benson; Margalit, Dan, A primer on mapping class groups, Princeton Mathematical Series. Princeton, NJ: Princeton University Press (ISBN 978-0-691-14794-9/hbk; 978-1-400-83904-9/ebook). xiv, 492 p. (2011). ZBL1245.57002.
To be precise, in Chapter 7, they note that if $b>0,$ then there the mapping class group is torsion free, while for a closed surface, the maximal order of an element is $4g + 2.$ The latter uses Nielsen realization, not surprisingly.
EDIT As Dan points out in the comments, you lose torsion only if you have boundary components which you wish to preserve pointwise. Punctures keep the number at $4g+2.$
answered Jul 5, 2017 at 21:08
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2$\begingroup$ By the way, if you have a surface of genus g > 1 and p punctures, then the maximum order is still 4g+2 since a finite order homeomorphism of the punctured surface induces a finite order homeomorphism of the closed surface, with the same order. (And thanks!) $\endgroup$ Commented Jul 9, 2017 at 1:47
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$\begingroup$ @DanMargalit Corrected, thanks! The book is amazing, thanks! $\endgroup$ Commented Jul 9, 2017 at 13:59