5
$\begingroup$

Let $S_{g,b}$ be an oriented surface with $b$ boundary components and $S_g^b$ be an oriented surface with $b$ punctures. Let $\mathrm{Mod}(S_{g,b})$ and $\mathrm{Mod}(S_g^b)$ their (orientation preserving) mapping class group fixing boundary/punctures pointwise. There is a short exact sequence:

$$1 \to \mathbb Z^b \to \mathrm{Mod}(S_{g,b}) \to \mathrm{Mod}(S_g^b) \to 1$$

  1. Is $\mathbb Z^b$ undistorted in $\mathrm{Mod}(S_{g,b})$ ?

  2. What is the maximal rank of an undistorted Abelian subgroup in $\mathrm{Mod}(S_g^b)$ ?

Thank you in advance.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

Farb, Lubotzky, and Minsky proved in

B. Farb, A. Lubotzky and Y. Minsky, Rank-1 phenomena for mapping class groups, Duke Math. J. 106 (2001), no. 3, 581–597.

that all abelian subgroups of the mapping class group are undistorted.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Could you say a bit as to why their paper shows that? Skimming the paper they only specifically mention cyclic subgroups are undistorted. $\endgroup$
    – user35370
    Commented Sep 27, 2018 at 16:11
  • $\begingroup$ I guess it is also because abelian subgroups, up to finite index, are generated by elements supported on disjoint subsurfaces. I think this is done in Abelian and solvable subgroups of the mapping class group by Birman, Lubotzky, McCarthy. $\endgroup$
    – user35370
    Commented Sep 27, 2018 at 16:43
  • $\begingroup$ @PaulPlummer: That is correct. $\endgroup$
    – Andy Putman
    Commented Sep 27, 2018 at 19:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .