Let $M$ be a smooth orientable compact connected (with boundary) manifold of dimension $4$. In addition $M$ is assumed to be aspherical and acyclic.

Question: is there a "classification" of such manifolds? Or can they be classified in any effective way?


2 Answers 2


There are plenty of such manifolds, but as Danny indicates in his answer, there is not a known classification.

Take any acyclic group $G$ with a finite aspherical 2-complex $C$ with $\pi_1(C)=G$. Then one can create an aspherical 4-manifold with boundary having $G$ as fundamental group. We may assume that the 1-skeleton $C^{(1)}$ of $C$ is a wedge of $k$ circles. Then take a 4-dimensional handlebody $H$ with $k$ 1-handles, with a spine of $C^{(1)}$. There are $k$ disks attached to the 1-skeleton in $C$. Attach $2$-handles to $H$ in such a way that the core of the attaching map is homotopic to the attaching map in the 2-skeleton $C^{(1)}$ to get a manifold $W$ with handle structure so that $C$ is a deformation retract of $W$, and hence $\pi_1(W)\cong G$. By the Poincaré-Lefschetz theorem, $\partial W$ is a homology 3-sphere. But in general we may get many different boundaries depending on the choice of isotopy class and framing of the boundary of the cores of the 2-handles.

To get such groups $G$, one can choose a small-cancellation $C'(\frac16)$ balanced presentation with $H_1(G)=0$. Then a presentation complex $C$ will be aspherical and acyclic. Added: See an explicit example due to Rylee Lyman in the comments. A simpler presentation of the Higman group is given (which perfect and has aspherical presentation complex).

The difficulty here is that one has no idea what 3-manifold the boundary of such a manifold will be. Moreover, it's not clear what the homeomorphism classification of such manifolds is, even if they have the same aspherical 2-skeleton spine and boundary.

Presumably there are also examples which do not have a 2-dimensional spine. The only obvious restriction I see is that the fundamental group must have cohomological dimension three.

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    $\begingroup$ Isn't it not even known whether PD(3) groups are fundamental groups of 3-manifolds? I thought this was only known for PD(1) groups (Stallings) and PD(2) groups (Linnell, Muller, and Eckmann). $\endgroup$ Oct 8, 2020 at 4:11
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    $\begingroup$ That's correct, I was adding a reference when you added your comment. $\endgroup$
    – Ian Agol
    Oct 8, 2020 at 4:12
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    $\begingroup$ Perhaps it's worth commenting that decision problems about acyclic, aspherical 2-complexes are wide open. For instance, it's a longstanding open problem if there's an algorithm to decide whether or not such a 2-complex is simply connected. Since any meaningful "classification" should make it possible to decide if the manifold is simply connected, your answer shows that any such classification should decide this algorothmic problem too. $\endgroup$
    – HJRW
    Oct 8, 2020 at 8:15
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    $\begingroup$ @DannyRuberman, I think the canonical easiest example of an acyclic group is Higman's group. See, for example, the answers to this MO question. mathoverflow.net/questions/352015/… (Like Ian, I tend to reach for small-cancellation first!) $\endgroup$
    – HJRW
    Oct 9, 2020 at 12:03
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    $\begingroup$ Maybe I should also point out that, in the comments on an answer to the question I linked to above, Rylee Lyman gives the following example of a 2-generator, 2-relator acyclic group: $\langle a,b\mid a(aba^{-1}b^{-1})^5, b^6a^2b^3a^{-1}b^{-8}a^{-1} \rangle$. $\endgroup$
    – HJRW
    Oct 9, 2020 at 12:06

I would say no. If M is simply connected, then it is contractible and hence determined topologically by its boundary. But there's no current smooth classification; the case when the boundary is $S^3$ would be the unsolved 4-dimensional Poincaré conjecture.

For more general examples, you would also have to specify the boundary, even in the topological setting. (The smooth setting would continue to be hopeless at present.) An additional issue you'd have to face for extending the topological classification when $\pi_1 \neq 0$ is that the fundamental group might not be a `good' group, ie one for which surgery theory works in the topological category.

I'm also a bit confused as to whether there are any non-simply connected manifolds of the sort you describe.

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    $\begingroup$ How come a contractible manifold is determined up to homeomorphism by its boundary? I found the existence in Theorem 1.4' of Freedman's paper (that any homology sphere bounds a contractible 4-manifold), but I couldn't find the uniqueness. $\endgroup$
    – Ian Agol
    Oct 8, 2020 at 5:09
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    $\begingroup$ It's the same argument as in the closed case; you find a relative h-cobordism by surgery theory. Steve Boyer worked out the general theorem for simply connected manifolds in the 1980s. (Simply-connected 4-manifolds with a given boundary. Trans. Amer. Math. Soc. 298 (1986), no. 1, 331–357.) When the boundary is a homology sphere it's basically the same as the closed case, but it's a fair amount more complicated in general. $\endgroup$ Oct 8, 2020 at 12:18

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