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Let $N$ be a compact smooth manifold. By "mapping class group" I will mean

$$\pi_0 \operatorname{Diff}(N)$$

i.e. the isotopy-classes of diffeomorphisms of $N$.

My presumption is that this mapping class group is likely to be finitely generated if and only if either $n = \dim N \leq 3$ or when $n \geq 4$ the group $\pi_1 N$ has finitely-many conjugacy classes, i.e. the path-components of the free loop space on $N$ are finite.

i.e. other than in low dimensions, finite-generation of the mapping class group corresponds to the fundamental group being finite.

Are there any obvious counter-examples to this statement I might be missing? I'm wondering if it's a reasonable conjecture or not. I would imagine this conjecture would remain the same in the PL and topological categories. At least, all the non-finite-generation examples I know of are category-independent, and smoothing theory I believe would say this question is category-independent in high dimensions.

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    $\begingroup$ I think what you say is known in dimensions $\le 3$ (at least of for orientable manifolds). For prime 3-manifolds the mapping class group is known, and I think in the non-prime case the finite generation is also known (Hatcher says so in his 50 year diffeo groups survey). So the conjecture is really about dimensions $\ge 4$, and in this case it seems rather bold. $\endgroup$
    – Igor Belegradek
    Commented Jun 28, 2023 at 23:25
  • $\begingroup$ Anything could go wrong in dim 4 in my opinion. $\endgroup$
    – Anubhav Mukherjee
    Commented Jun 28, 2023 at 23:27
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    $\begingroup$ While Hatcher (among others) proved that the MCG of $T^6$ is not finitely generated, he cites Hatcher-Lawson that $T^6\#(S^3\times S^3)$ has pseudo-isotopy implies isotopy. Does this imply finitely generated MCG? $\endgroup$
    – Ben Wieland
    Commented Jun 29, 2023 at 1:55
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    $\begingroup$ @HJRW: yes, the hyperbolic case and the argument behind it (barbell diffeos and things like it) is one of the key motivators of this question. $\endgroup$
    – Ryan Budney
    Commented Jul 4, 2023 at 16:00
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    $\begingroup$ As for the infinite fundamental group case, this is completely out of reach. It is an open question whether there is an infinite finitely presented group with finitely many conjugacy classes, which would be a promising counterexample. What if we restrict to fundamental group $\mathbb Z$? If a manifold has higher homotopy groups that are not finitely generated, is its group of homotopy equivalences not finitely generated? Probably. Is its subgroup realizable by diffeomorphisms not finitely generated? Perhaps is it the kernel of a surjection to an arithmetic subgroup, which would show not fg. $\endgroup$
    – Ben Wieland
    Commented Jul 24, 2023 at 20:00

1 Answer 1

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For closed connected smooth manifolds $N$ of dimension $d \geq 6$ with $\pi_1(N)$ finite, it is known that $\pi_0(\mathrm{Diff}(N))$ has a classifying space which is a finite CW-complex. This is due to Sullivan if $\pi_1(N)$ is trivial (this even works in dimension $d=5$) and to Triantafillou if $\pi_1(N)$ is finite. There were several oversights in her proof, which were corrected in a paper by Mauricio Bustamante, Manuel Krannich, and myself.

Conversely, if $d \geq 6$ and $\pi_1(N)$ has infinitely many conjugacy classes not equal to their inverse then Hsiang and Sharpe proved under mild conditions that $\pi_0(\mathrm{Diff}(N))$ is not finitely generated. Some more groups can be handled by techniques from the Farell-Jones conjectures.

Ass I know the converse direction is not known in general, and in dimension $d=5$ both directions remain open. As you mention, the PL and topological cases imply the smooth case by smoothing theory if $d \geq 5$.

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  • $\begingroup$ Your examples make me realize the conjecture probably should be that $\pi_1 N$ has finitely many conjugacy classes. These barbell diffeo implantations are invariants of the path components of the free loop space on $N$. $\endgroup$
    – Ryan Budney
    Commented Sep 6, 2023 at 3:07

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