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# Geometric group theory: Homeomorphism and mapping class groups

Surjective homomorphisms from surface braid groups- Lei Chen (Caltech)

Abstract: In this talk, I will present that any surjective homomorphism from the surface braid group to a torsion-free hyperbolic group factors through a forgetful map. This extends and gives a new proof of an earlier result of the author which works only for free groups and surface groups.

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Inverse limits of covering spaces- Curtis Kent (Brigham Young University)

Abstract: Many techniques that have proved effective in the study of finitely generated groups at first approach seem insufficient for the study of uncountable groups, e.g. big mapping class groups or fundamental groups of complicated spaces. We will discuss how to use inverse limits to study such groups through approximation by finitely generated groups. Specifically we will show that inverse limits of covering spaces, when path-connected, satisfy a Galois type correspondence and how this leads to a splitting of the first homology for many spaces.

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Universal bounds for torsion generating sets of mapping class groups- Justin Lanier (Georgia Tech)

Abstract: I showed with Margalit that the conjugacy class of any periodic element is a generating set for Mod(S_g), as long as g is at least 3 and the element isn’t trivial or a hyperelliptic involution. Since Mod(S_g) is finitely generated, these infinite torsion generating sets can always be whittled down to finite generating sets. Is there a universal upper bound on the number of conjugates required, independent of g and the element chosen? In general, the answer is no. However, for elements with order at least 3 there is a universal upper bound; our proof shows that 60 conjugates always suffice. I will record two short talks: a first talk that describes the history and context of this result and that gives an overview of the proof strategy, and then a second talk that goes into some of the nuances of the proof.

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Analogs of the curve graph for infinite type surfaces- Alexander Rasmussen (Yale)

Abstract: The curve graph of a finite type surface is a crucial tool for understanding the algebra and geometry of the corresponding mapping class group. Many of the applications that arise from this relationship rely on the fact that the curve graph is hyperbolic. We will describe actions of mapping class groups of infinite type surfaces on various graphs analogous to the curve graph. In particular, we will discuss the hyperbolicity of these graphs, some of their quasiconvex subgraphs, properties of the corresponding actions, and applications to bounded cohomology.

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The 'what' and 'why' of framed mapping class groups- Nick Salter (Columbia)

Abstract: The 'what': a framed mapping class group is the stabilizer of an isotopy class of vector field on a surface. Despite being infinite-index subgroups, Aaron Calderon and I have shown that these are generated by simple collections of finitely many Dehn twists. The 'why': Many natural families of Riemann surfaces are also equipped with the extra data of a vector field. This is true of families of translation surfaces (strata), as well as the family of Milnor fibers of an isolated plane curve singularity. Understanding the behavior of these families thus requires an understanding of the framed mapping class group. Work on the framed mapping class group and on strata is joint with Aaron Calderon, and work on plane curve singularities is joint with Pablo Portilla Cuadrado.

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## 23 replies on “Geometric group theory: Homeomorphism and mapping class groups”

[…] Homeomorphism and mapping class groups […]

lvzhou.chensays:

Hi Lei, thanks for your interesting talk.

Just to clarify, in the second remark on page 6 of your slides (around 13:24 of your video), by “abelian finite index subgroup”, do you mean “finite index normal subgroup with abelian quotient”? This sounds more consistent with the concluding comments of your talk, and also I don’t think any finite index subgroup of PB_n(S_g) could be abelian (e.g by considering the image under p_i).

Another question regarding your comments on the Ivanov conjecture: When you realized PB_n(S_g) as a subgroup of some mapping class group, is it obvious that you can make it finite index? If not, how does it relate to the Ivanov conjecture? By taking intersections with finite index subgroups of MCG?

frafoursays:

Hey, thanks again Lei for the talk! I asked this question during the office hours and I’ll write it again here in case someone else has some ideas. What are some classes of groups, other than free groups or surface groups, in which all finitely generated normal subgroups are necessarily trivial or of finite index?

Nick.Saltersays:

Ah I’ve been asking this question for years! I don’t know of any examples beyond free and surface groups, but here’s a bold conjecture: perhaps every hyperbolic group of cohomological dimension at most 2 has this property?

frafoursays:

I was thinking about hyperbolic one-relator groups, so these two guesses line up! 🙂

kevinschrevesays:

If a finite type group G has an infinite, normal, infinite index subgroup, then the first $L^2$-Betti number of G vanishes. So, any group with nontrivial first $L^2$-Betti number has this property. These are generally hard to compute, but sometimes you can do it…

On the other hand, there are plenty of 2-dimensional hyperbolic groups with infinite normal subgroups, for example many free-by-cyclic groups are hyperbolic.

Nick.Saltersays:

Hi Kevin,

Thanks, that’s a very nice observation!

-Nick

lvzhou.chensays:

Not quite related to finite generation, but normal subgroups in higher rank irreducible lattices are either finite or finite index (Margulis normal subgroup theorem).

Stefano_Vidussisays:

Hi, the property you mention is referred to as f.g.n. (finitely generated normal) in work of Karrass and Solitar. In particular, in Proc. Amer. Math. Soc. 37 (1973), they show that if A has the f.g.n. property and and U < A is a slender infinite index subgroup, then the free product A *_U B of A and B amalgamated along U (where U < B) is f.g.n. (No further assumptions on B are needed. The proof builds on their earlier work on amalgamated free products.) With that result it is possible to build new examples out of free or surface groups.

frafoursays:

Thanks for the great answers everyone!

A question for Lei: What property about hyperbolic groups that allows you to prove the result about the non-triviality of the Euler class of a central extension implying that any homomorphism from the extension to a hyperbolic group must factor through the quotient?

Thanks for the excellent talk.

For Justin: Hi Justin!

You mention this result that given a prime p, there is a unique surface with a non-reducible mapping class of order p (madness!). Is there an analogous result if you don’t assume primeness, but maybe the conclusion is that there are only finitely many surfaces or something like that?

Thanks for the great talk as always!

Justin.Laniersays:

Hi Ty. Thanks for the question.

Indeed, there is such an analogue: for any n>1, there are only finitely many g such that Mod(S_g) contains an irreducible periodic mapping class of order n. This follow directly from one of my favorite results, due to Kasahara (Reducibility and orders of periodic automorphisms of surfaces, Theorem 4.1): if f is periodic and irreducible, then $n \geq 2g+1$. So for all g large compared to n—and not even that large!—every element of order n is reducible.

Sweet, reliable as always! Thanks.

For Nick: Hey Nick!

Given a framed surface (with necessarily finitely many punctures), is it always possible to realise it as a translation surface where one of the 2 framing vector fields is simply the horizontal vector field (of constant length or whatever the correct adjective here is to make a theorem true)?

Nick.Saltersays:

Hi Ty,

The answer is “sometimes”. There is one fundamental obstruction for this to be possible, and then a couple of sporadic problems in very low genus.

The fundamental obstruction has to do with the winding numbers assigned to boundary components. If you have a framing inherited from a translation surface, then the winding number of each boundary component (relative to the framing, of course) recovers the order of the zero of the differential. In particular, all of the winding numbers have a definitive sign (using our conventions, they are always negative). So having a framing with any boundary components of non-negative winding number obstructs the existence of a translation surface giving rise to this.

(You could be more general and ask whether these framings come from *meromorphic* differentials on a Riemann surface. I think the answer here is “yes”, but I am under the impression that things are more subtle. You should ask Aaron!)

In the case where the framing is “of holomorphic type”, the answer is typically “yes” – the framing can be realized from a translation surface. Basically, the idea is that whatever your framing is, you can always find a “spanning configuration” of admissible curves (curves of 0 winding number). Then you can run Thurston-Veech on this configuration to turn these admissible curves into cylinders on a translation surface; this translation surface will give rise to the original framing.

-Nick

Aaron_Calderonsays:

Hi Ty,

Adding onto what Nick said: there are in fact framings of meromorphic type which *can’t* come from meromorphic abelian differentials.

For an easy example, note that a meromorphic differential can’t have only a single simple pole (otherwise the residue theorem says it has 0 residue, but then it doesn’t have a pole!). But you absolutely can cook up framings that have winding number 0 around one boundary.

For a more subtle example, fix a holomorphic framing of a surface with one puncture; by the argument Nick gave, you can realize it (well, an isotopic framing) as the horizontal vector field on a surface. Do this so that all of the curves of our generating set are cylinders.

Now in framing-world, you can locally split apart the single zero of order 2g-2 into a “zero” of order 2g and a “pole” of order -2 (equivalently, split the one boundary of winding number 1-2g into one with winding number -2g-1 and one with winding number 1) without disturbing the rest of the framing.

But this can’t correspond to a deformation of differentials! The argument is by proving that certain degenerations don’t exist in the boundary of the corresponding stratum, but the main point is that in flat-world you can never locally split a zero into a zero and a single pole. This follows from the fact that any meromorphic differential on the Riemann sphere with a single zero necessarily has nonzero residue at some pole. The magic words here are the “global residue condition” of Bainbridge, Chen, Gendron, Gruschevsky, and Möller.

Aaron_Calderonsays:

Oh, and to contrast the two examples, the first one fails because the stratum ΩMg(2g-1, -1) is empty, whereas in the second one, the stratum ΩMg(2g, -2) is nonempty, you just can’t realize the framing.

Aaron_Calderonsays:

Just to keep adding onto this: the framing I discuss above is not realizable, but an isotopic one *is* realizable. The issue is that at least one curve which used to be a cylinder (before breaking the zero into a zero and a pole) must not be a cylinder any more on any isotopic realization — the isotopy necessarily destroys cylinders!

Excellent, that’s great, thanks!

For Nick: Hey Nick!

Given a framed surface (with necessarily finitely many punctures), is it always possible to realise it as a translation surface where one of the 2 framing vector fields is simply the horizontal vector field (of constant length or whatever the correct adjective here is to make a theorem true)?

Thanks for the excellent talk.

Such a good talk I had to ask the question twice.

Aaron_Calderonsays:

Hi Lei,
I really enjoyed your talk! I had a question about your cohomology decomposition result: is there any hope a similar theorem about the decomposition of the cohomology might also be true for infinite-index subgroups with abelian quotients (e.g., the commutator subgroup)?