One of low dimensional topology’s most popular groups is the mapping class group of a surface. We care about mapping class groups because of how they relate to Heegaard splittings of 3-manifolds. Number theorists and physicists care about mapping class groups because mapping class groups are orbifold fundamental groups of moduli spaces of Riemann surfaces.
Being an unsophisticated sort of a bloke when it comes to group theory, nothing makes me feel that I understand a group like a good concrete presentation of that group, with generators and relations. In my opinion, a good presentation of a group should have the following properties, which I’ll call MIST for fun.
- Memorable- The presentation should be easy to remember.
- Informative- The generators and the relations should be conceptually meaningful and natural, and should tell me something enlightening about the group.
- Simple- The presentation should be easy to work with, by which I mean mainly that it should be short, with as few generators and relations as possible, and with the relations being as short as possible.
- Typical- It should fit into a bigger picture, by which I mean mainly that it should relate easily to similar presentations of similar groups.
Lots of work has been done to find the best presentation for a mapping class group. But I think there’s still a (relatively short) way to go; and I think that, if the work’s worth doing, then it’s worth doing right. I’ve been thinking a bit about presentations of mapping class groups for a summer course I’m teaching, and I wish I could teach a MIST presentation. But not enough is known, and I don’t know enough about what is known, so in this post I’ll briefly summarize my understanding of the state of the art for mapping class group presentations for genus (higher genus next time), and you can tell me where I’m wrong and where there’s more to be known!
Let’s first decide on a set of generators. I claim that Dehn twists are the best sort of generators. Indeed, the Dehn-Lickorish Theorem states that a mapping class group is generated by a finite number of Dehn twists. Generating mapping classes by Dehn twists is rather like taking handle decompositions of 3-manifolds, so I can understand why it might be a sensible thing to do.
Humphries proved the surprising result that any mapping class is generated by Dehn twists, as given below, and that this number is minimal (if we didn’t require our generators to be Dehn twists, we could do with just two, by a result of Wajnryb. Actually, one of these could be taken to be a Dehn twist by a result of Korkmaz). There’s a fun (and difficult) computer game asking you to decompose Dehn twists into products of Humphries generators. Matsumoto likens it to solving a Rubik’s cube.
Next, let’s find relations. There are two relations between Dehn twists on a surface which are pretty obvious.
- If are disjoint, then Dehn twists around them commute (satisfy the disjointness relation): .
- If are intersect at a single point, then Dehn twists around them satisfy the braid relation: .
Completing work started by Bergau and Mennicke, Birman and Hilden showed that this is essentially a complete set of relations for compact oriented genus 2 surfaces with a single boundary component, so that mapping class group is basically just the braid group on 6 strands. Their paper is my personal favourite on the topic. The idea is to realize the surface as a 2-fold covering space of a punctured disc, where automorphisms of the punctured discs which permute punctures (i.e. braids) lift to Dehn twists on the surface. This is what I consider a MIST presentation- it’s short, informative, and easy to remember.
Genus 3 is already much more difficult. That’s because genus 3 is the first time you really need to get to grips with the strangely protruding Humphries generator . It gives rise to at least one additional relation between the generators, aside from disjointness and the braid relation. For example, it gives rise to the Lantern relation, originally considered by Max Dehn, but rediscovered by Johnson in the 1970′s. The core issue is that the hyperelliptic involution (rotating the once-punctured surface by 180 degrees) no longer commutes with all of the Humphries generators, so the Birman-Hilden approach no longer works, because there are relations “upstairs” in the mapping class group of the surface which don’t come from “downstairs” automorphisms of the punctured disc (the base). But I think that genus 3 might also now have a MIST presentation, in a paper by Looijenga.
I don’t understand Looijenga’s paper very well, because it’s an algebraic geometry paper and I don’t know very much algebraic geometry, but the presentation is beautiful, simple, easy to remember, and looks informative. Recall that a mapping class group is an orbifold fundamental group of a moduli space of Riemann surfaces . Well, Looijenga constructs a variety comprised of the Eilenberg-Maclane space of the centralizer of the hyperelliptic involution in the mapping class group (which is called the hyperelliptic mapping class group, and is what Birman-Hilden looked at), together with a regular neighbourhood of a certain locus of the Deligne-Mumford boundary (whatever that is, and whatever that means). Anyway, the inclusion of this variety in induces an isomorphism on fundamental groups. So what Looijenga has done is to realize the mapping class group of a genus 3 oriented surface as the fundamental group of a space which he understands and which has an obvious natural presentation.
So what do Looijenga’s relations look like? Well, first you draw a diagram with a vertex for each Humphries generator, with one more generator chained into (so you have 8 generators instead of 7- that’s how Looijenga’s presentation works), and a line between them if they intersect at one point.
You get the affine Dynkin diagram for ! Any two disjoint vertices give rise to a disjointness relation, and any two vertices connected by an edge give rise to a braid relation- so an intersection diagram gives rise to an Artin group . The quasi-centre of an Artin group is defined as the set of elements which commute with the generators corresponding to the vertices of . In our case, when is finite and connected, the quasi-centre of is an infinite cyclic group generated by a special element called the fundamental element.
Now, take the affine Dynkin diagram of , and look at all it’s sub-Dynkin-diagrams, which are affine Dynkin diagrams for and for with . Denote the fundamental elements of and of by and respectively. The added relations are now just:
- .
- .
If any reader can explain to me why this is the result that I should expect, I’d love to know! Surely it’s just a matter of understanding Looijenga’s construction. I’d really love an executive summary of Looijenga’s work.
So Looijenga’s presentation is memorable and simple. It’s surely informative (although I don’t understand it yet). It looks as though it surely should be typical.
Question: Starting from an arbitrary Coxeter graph , coming from an intersection diagram of closed essential curves on a surface, maybe with boundary and punctures, what relations do I need to add in order to obtain a presentation for the subgroup of the mapping class group which the curves generate?
The dream would be that, starting from a completely arbitrary finite connected Coxeter graph , there would be a simple informative complete set of relations between the fundamental elements corresponding to affine Dynkin diagrams in . I think that this would be the “job well done” with regard to finding a MIST presentation for the mapping class group, and that to find it (or, more modestly, just to extend Looijenga’s result to arbitrary genus) would be a big step forward for mathematics.
Next time: general genus, Hatcher-Thurston, Harer, Wajnryb, Gervais, Hirose, Benvenuti, Matsumoto, Szepietowski, all that jazz…
Could I ask a very ignorant question? I’m interested in the best presentation of the mapping class group of the -punctured sphere. A brief Google search suggests that it is closely connected to braid groups and also to the groups you are considering above (which makes your post extraordinarily timely — many thanks for it), which suggests to me that a MIST presentation of these groups exists and is well known. Is it easy to say what it is? (I’ve got as far as knowing what Dehn twists are and not much further.) Pointing me towards a reference would be good enough, at least if the reference is friendly to non-experts.
Comment by gowers — May 18, 2011 @ 5:10 am |
This is a preliminary answer- I’ll edit in a few hours when I get to my office. The mapping class group of an n-punctured sphere is the braid group with n strands over the sphere! So it’s very easy to say what it is- you have the disjointness and the braid relations, plus one extra relation which comes from taking a strand “all the way around the back”. So, imagining the punctures all “on top of the sphere”, a curve around the first j punctures can be swept all the way around the sphere to go around the last n-j punctures, flipping its orientation. People looked at this group about 40 years ago, then forgot all about it, and now it’s making a small comeback because of “braid groups over surfaces” (discussed here for example) and because of the amazing Berrick-Cohen-Wong-Wu paper which connects Brunnian braids over a sphere to homotopy groups of ! Their result is one of those mysterious alluring connections which makes mathematics worth doing. John Baez wrote a nice blog post about that story.
Braid groups over are discussed in Murasugi and Kurpita’s book. This is the only book I know which gives them much space. The most interesting non-trivial result I know about these groups is a combing algorithm for braids over - I’ll find the reference.
The genus 2 mapping class group is essentially the 6-strand braid group over , by Birman-Hilden (Bergau-Mennicke).
PS: I should have typed some of this as an answer to your MO question when you asked it- sorry about that!
Comment by Daniel Moskovich — May 18, 2011 @ 7:28 am |
I think it’s not quite true that the spherical braid group on n strings equals the mapping class group of the n-punctured sphere. To get from the latter to the former you need to quotient out by a full twist of all the strings, if I understand correctly a paper you linked to on MO. But the square of that twist does seem to be the identity in the spherical braid group.
Comment by gowers — May 23, 2011 @ 5:55 pm
I looked everything up, and posted it as an answer on MO- see there for the relevant references and for more details.
Comment by dmoskovich — May 18, 2011 @ 3:08 pm |
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