Presentations of mapping class groups in dimension $3$

For any closed oriented surface $$M$$, its mapping class group $$MCG(M)$$ can be generated by Dehn twists along certain curves on $$M$$. A presentation for the group $$MCG(M)$$ was found in [1] and then simplified in [2].

How about in dimension $$3$$? The first question is that, unlike in dimension $$2$$ in which the classification of spaces is easy (genus and number of boundaries), in dimension $$3$$ the classification is complicated. One either uses

1. Thurston's geometrization
2. Lickorish and Kirby's presentation using links

Neither of them is easy, so I'd expect it much harder to get presentation for the mapping class group.

Question: Nevertheless, is there any presentation known?

In the first view-point, the closest answer I have seen is this, in which Allen Hatcher claims that the MCG of three-manifolds are essentially known [3]. There, the natural map from MCG to $$Out(\pi_1(M))$$ is considered. While the kernel of this map is understood in the nonprime case [3. section 2], it is not necessarily onto. Even if it's onto, we still do not have a presentation for MCG.

In the second view-point, I have not heard of any result.

Reference

• [1] A presentation for the mapping class group of a closed orientable surface-[Hatcher and Thurston]

• [2] A simple presentation for the mapping class group of an orientable surface-[Bronislaw Wajnryb]

• [3] Stabilization for mapping class groups of 3-manifolds-[Allen Hatcher and Nathalie wahl]

• It is certainly true that there is an algorithm to compute a presentation of a mapping class group of a given 3-manifold, but I don't think this is a reasonable thing to describe in practice; certainly not in an answer to an MO question, and possibly not ever. As indicated in @SamNead's answer (see my comment below), even if you restrict attention to closed hyperbolic manifolds, every finite group arises. If you really want to understand all irreducible 3-manifolds, you need to learn about the JSJ decomposition, how to compute the mapping class groups of the pieces...
– HJRW
May 24, 2021 at 10:42
• ... and how to "reassemble" them into the mapping class group of the whole thing. If you are also interested in the reducible case (e.g. doubles of handlebodies, as @SamNead mentions) then the answer is more complicated still. So it would help a great deal if you were interested in some more specific class of 3-manifolds. If you really want the answer in full generality, a better question might be to try to write down a roadmap to understanding all the different components.
– HJRW
May 24, 2021 at 10:46
• By the way, the "second viewpoint" isn't really useful for computing the mapping class group -- you have to first translate to the "first viewpoint" by computing the Kneser--Milnor and JSJ decompositions, and then the geometric structures of the pieces. (Algorithms are known for all of these things.) I suspect there are examples of 3-manifolds with very similar link presentations but with radically different JSJ decompositions (and hence mapping class groups).
– HJRW
May 24, 2021 at 10:48
• In general, it's very natural to want to know that a group is finitely presentable. Specific relations are also often useful. But actually writing down an explicit finite presentation for a very complicated group is often not a good way to understand it (and I say this as someone whose PhD thesis was all about finding a presentation for a specific subgroup of the mapping class group of a surface!). May 24, 2021 at 18:00
• Lots of presentations are known. By and large, viewing things through the lens of presentations might be limiting. Often times you have nice models for $Diff(M)$ or $BDiff(M)$. If you are particularly motivated you can derive presentations for the mapping class group from that. May 24, 2021 at 19:04

Suppose that $$M$$ is finite volume hyperbolic. Then the mapping class group is finite and there are algorithms to build its multiplication table. On the other hand, there is definitely no overall pattern to these groups - they are “just” the finite quotients of torsion free lattices in $$\mathrm{SL}(2, \mathbb{C})$$.

And this is only the tip of the iceberg... the mapping class groups of surfaces appear when thinking about Seifert fibered spaces, outer automorphism groups of free groups show up when dealing with the doubles of handlebodies, and if you connect sum the above together you get more craziness...

• This is the first viewpoint - thanks! Do people know about how crazy it is? And this is also why I asked the second viewpoint - conceptually, unlike the first viewpoint which has many kinds of building blocks and even more kinds of combination (thus the craziness), the second viewpoint "only" uses links. Can one start with a link presentation $P$ for a link $L$ (with the $3$-fold obtained from surgery along $L$ denoted $M_L$), and build a group presentation from $P$ for the group $MCG(M_L)$? By the way, any reference/pointer will be appreciated :) May 22, 2021 at 20:12
• And indeed, every finite group arises as the isometry group (i.e. mapping class group) of a closed hyperbolic 3-manifold, by a 1988 theorem of Kojima.
– HJRW
May 24, 2021 at 10:39

For certain 3-manifolds (irreducible), if you are willing to take finite index subgroups, the description is relatively easy, and has to do with dehn-twists too.

Let me add the following paper by McCullough which was useful to me when considering this question: https://projecteuclid.org/journals/journal-of-differential-geometry/volume-33/issue-1/Virtually-geometrically-finite-mapping-class-groups-of-3-manifolds/10.4310/jdg/1214446029.full

I think this goes in line with the comments by HJRW which are more detailed, but I thought this might be useful.