Geometric subgroups of mapping class groups

Abstract

This paper is a study of the subgroups of mapping class groups of Riemann surfaces, called "geometric" subgroups, corresponding to the inclusion of subsurfaces. Our analysis includes surfaces with boundary and with punctures. The centres of all the mapping class groups are calculated. We determine the kernel of inclusion-induced maps of the mapping class group of a subsurface, and give necessary and sufficient conditions for injectivity. In the injective case, we show that the commensurability class of a geometric subgroup completely determines up to isotopy the defining subsurface, and we characterize centralizers, normalizers, and commensurators of geometric subgroups.

Commensurability of geometric subgroups of mapping class groups

“…Some of our results were previously known in the case of a connected subsurface of an orientable surface [12,13]. The novelty of our work is that we allow the subsurfaces to be disconnected and not necessarily injective (i.e.…”

“…, a n on M define a pantalon decomposition of M if each connected component of M \ n i=1 a i is a pantalon. It is known [13] that an orientable surface has a pantalon decomposition if and only if 2g + r + s > 2 and M = M 3 0 . As observed in Section 5 of [16], one needs to add two more "pieces" in order to decompose nonorientable surfaces, namely a Möbius strip with one puncture and a Möbius strip with an open disk removed.…”

“…, a n be generic, pairwise nonisotopic, pairwise disjoint and two-sided circles on M. Then the homomorphism , then I(a, b) = 0. Although the following proposition is stated in [13] only for orientable surfaces, the proof applies to the nonorientable case verbatim.…”

“…We will use induction on the number m of isotopy classes of circles that have more than one representant in A. For m = 0 the statement follows from Proposition 3.10 of [13] (although the statement in [13] is only for orientable surfaces, the proof applies to the nonorientable case verbatim). Now assume that m ≥ 1 and up to the permutation of elements of A and B we can assume that a 1 ≃ a 2 ≃ .…”

On representations of Artin–Tits and surface braid groups

“…r Proof. It su‰ces to recall that the group M nÀ1 g; p is centerless (see [15]). If x belongs to the center of D n g; p , it follows that x belongs to (the center of ) ker p n , which is a free group; therefore x ¼ 1. r Theorem 4.…”