# Are the mapping class groups of manifolds finitely presentable?

The mapping class group of a manifold is the group $\pi_0 Diff(M)$ of components of the diffeomorphism group. There are several variations: oriented manifolds and orientation preserving diffeomorphisms, etc.

I am interested in finding out if we know any sorts of general properties of these groups. Are they finitely presented? finitely generated? ... ?

The mapping class groups of spheres can be identified, via a clutching construction, with the group of exotic spheres one dimension higher. Via Kervaire-Milnor, which relates these these groups to the stable homotopy groups of spheres, and the fact that the stable homotopy groups of spheres are finite, we know that these mapping class groups are finite. However this uses a lot of heavy machinery from surgery. The general case seems far from obvious (though maybe there is an obvious counter example out there?).

I would also be interested in statements about the higher homotopy groups as well.

These things can be pretty wild. For instance, for $$n \geq 5$$ the mapping class group of the $$n$$-torus is not even finitely generated : it is a split extension of $$\text{GL}_n(\mathbb{Z})$$ by an infinite rank abelian group. See Theorem 4.1 of

A. E. Hatcher, Concordance spaces, higher simple-homotopy theory, and applications. Proc. Sympos. Pure Math., XXXII, Part 1, pp. 3-21, Amer. Math. Soc., Providence, R.I., 1978.

and Example 4 of

Hsiang, W. C.; Sharpe, R. W. Parametrized surgery and isotopy. Pacific J. Math. 67 (1976), no. 2, 401–459.

I also recommend Hatcher's survey about the homotopy type of the diffeomorphism groups of manifolds here; it discusses a lot of things that are known about $$\pi_0$$.

It should be remarked that the above trouble happens because we are considering manifolds with infinite fundamental groups, which often causes terrible things to happen in high-dimensional topology. If $$M$$ is a compact simply-connected manifold of dimension at least $$5$$, then Sullivan proved that the mapping class group of $$M$$ is an arithmetic group, and hence has every possible finiteness property. This was extended to manifolds with finite fundamental groups by Triantafillou. For all of this, I recommend the survey

Triantafillou, Georgia The arithmeticity of groups of automorphisms of spaces. Tel Aviv Topology Conference: Rothenberg Festschrift (1998), 283–306, Contemp. Math., 231, Amer. Math. Soc., Providence, RI, 1999.

• Nothing is known in dimension $4$. In dimension $3$, they are always finitely presentable. This is very explicit for prime $3$-manifolds. For instance, the mapping class groups of hyperbolic $3$-manifolds are finite by Mostow rigidity, and one can show that the mapping class groups of Seifert fibered spaces are closely connected to the mapping class groups of their base surfaces. For non-prime $3$-manifolds, things are a little more complicated. Hatcher's survey mentioned in my answer has a good description of all of this. – Andy Putman Mar 29 '14 at 4:24
• Some things are known for smooth 4-manifolds. I proved (A polynomial invariant of diffeomorphisms of 4-manifolds. Geom. Topol. Monogr. 2, 473–488, 1998) that for many simply-connected manifolds, the kernel of the natural map $\pi_0(Diff(M)) \to \pi_0(Homeo(M))$ is infinitely generated. In fact, for simply-connected manifolds, $\pi_0(Homeo(M))$ equals $Aut(H_2(M),\cdot)$, so this says that the `Torelli group' can be infinitely generated. But many finitely generated groups have infinitely generated subgroups, so I don't know if the actual mapping class group is infinitely generated. – Danny Ruberman Apr 2 '14 at 15:28
• @Andy Putman I realized that I do in fact know that the MCG is infinitely generated, and am going to amend my comment to make it an answer. – Danny Ruberman Apr 2 '14 at 17:31
• @Andy Putman After looking at my paper some more, I realized that it doesn't immediately imply the infinite generation of $\pi_0(Diff(M))$. I think this is true, but it requires a little thought. Thanks to Chris for raising an interesting question! – Danny Ruberman Apr 2 '14 at 17:57
• Sullivan proved important results about MCGs of simply-connected high-dimensional manifolds, hut he did not quite prove that they are arithmetic, as was recently observed by M. Krannich and O. Randal-Williams, see dpmms.cam.ac.uk/~or257/nonarithmeticmcg.pdf The reason Sullivan's result was misunderstood by many who knew about it was probably his unusual usage of the term "commensurability". – Jens Reinhold Mar 9 '20 at 10:01

The proof of this amounts to arguing that $$Diff(S^1 \times D^3)$$ acts transitively on the reducing discs, i.e. the properly-embedded $$3$$-discs in $$S^1 \times D^3$$ that are non-separating. The isotopy classes of reducing discs form a group, and that is not finitely-generated.

We show this by arguing that any component of the space of embeddings $$S^1 \to S^1 \times S^3$$ has a not-finitely-generated fundamental group. This part of the argument is fairly classical: Haefliger-style double-point considerations. This step appears to have been known to Dax, as well as Arone and Szymik.

Then we look at the fibre bundle $$Diff(S^1 \times S^3) \to Emb(S^1, S^1 \times S^3)$$. The fiber over the "degree 1 component" of the embedding space is the diffeomorphisms of $$S^1 \times S^3$$ that fix the knot pointwise, which is closely related to $$Diff(S^1 \times D^3)$$. The key argument is showing that none of our "interesting" elements of $$\pi_1 Emb(S^1, S^1 \times S^3)$$ are in the image of the map from $$\pi_1 Diff(S^1 \times S^3)$$. By an isotopy-extension argument they are turned into non-trivial isotopy classes of diffeomorphisms of $$S^1 \times D^3$$.