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I am interested in knowing if there is a closed, (smooth) aspherical manifold $M$ (hyperbolic would be best) with superperfect fundamental group (that is to say, with $H_1(\pi_1(M);\mathbb{Z}) = H_2(\pi_1(M);\mathbb{Z}) = 0$; note $H_1(\pi_1(M);\mathbb{Z}) = 0$ is equivalent to perfect) and non-trivial center ($\mathbb{Z}_2$ would be best, but any f.g. Abelian group will do).

Also, assuming there is a manifold that fits the criteria, I would likely need a handlebody decomposition for the manifold, assuming the "standard handlebody procedure" for producing a closed, smooth manifold from a prescribed finite presentation of a/the fundamental group does not yield the smooth manifold in question (e.g., the "standard manifold" is not aspherical).

I found many hyperbolic 3-manifolds with superperfect fundamental group using SnapPy, but SnapPy evidently doesn't have a center "method" for the fundamental group method/class attached to 3-manifolds. Sage/GAP/MAGMA also appear not to be able to compute the center for an infinite finitely-presented fundamental group.

Thanks much in advance. I realize this is kind-of "shooting for the moon/stars".

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    $\begingroup$ In every dimension, the fundamental group of a hyperbolic manifold has trivial centre. There are no examples of what you want in dimension 2, but in dimension 3, there are the Brieskorn homology spheres $\Sigma(p,q,r)$ for $1/p+1/q+1/r <1$. These are aspherical homology spheres which are Seifert fibred, so their fundamental groups have centre $\mathbb{Z}$. $\endgroup$
    – HJRW
    Commented Dec 20, 2019 at 15:32
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    $\begingroup$ Glossary: $G$ superperfect means $H_1(G,\mathbf{Z})=H_2(G,\mathbf{Z})=0$. $\endgroup$
    – YCor
    Commented Dec 20, 2019 at 15:32
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    $\begingroup$ The fundamental groups of aspherical manifolds are always torsion free for homological reasons, so a centre of $\mathbb{Z}$ is the best you can do. $\endgroup$
    – HJRW
    Commented Dec 20, 2019 at 15:38
  • $\begingroup$ @HJRW That's right, and the hyperbolic ones, at least (I think?), can't have a $\mathbb{Z} \times \mathbb{Z}$ or higher rank free Abelian group as a subgroup; I had looked up both those facts last night but forgot them. It's truly $\mathbb{Z}$ or nothing, thanks. $\endgroup$
    – Jeffrey Rolland
    Commented Dec 20, 2019 at 15:44
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    $\begingroup$ One more point--there are well-known genus 2 Heegaard splittings (derived from well-known surgery diagrams) for the Brieskorn homology spheres with 3 exceptional fibers. See eg Boileau-Zieschang [link] (numdam.org/article/AST_1988__163-164__247_0.pdf) $\endgroup$
    – Danny Ruberman
    Commented Dec 20, 2019 at 16:02

1 Answer 1

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Per Lee's request, I've turned the discussion into an answer.

Suppose that $M$ is a closed connected oriented hyperbolic $n$ manifold, for $n$ at least two. Then the fundamental group contains no torsion elements; also its center is trivial. Both claims are exercises from the classification of isometries of hyperbolic $n$-space.

Thus there are no hyperbolic homology three-spheres with the property you desire.

However, there are many three-manifolds that are (integral) homology spheres (and so $\pi_1$ is super-perfect) and where $\pi_1$ has non-trivial abelian centre. These are found among the Seifert fibered spaces. The most famous of these is the Poincare homology sphere, but this example is ruled out by your requirement that $M$ be aspherical.

More specifically, you should consider the "aspherical Brieskorn homology spheres" $\Sigma(p, q, r)$. These are described at the Wikipedia page linked to immediately above, which also briefly sketches a presentation of their fundamental group. If you want Snappy to compute things for you, then consult Figure 1 of this paper (say) for a surgery description of $\Sigma(p, q, r)$. This paper also gives a short but useful discussion of the fundamental group.

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  • $\begingroup$ I was and am planning on collecting the above in an answer, but was waiting for the book smile.amazon.com/Invariants-Homology-3-Spheres-Nikolai-Saveliev/… to come in at the library to make sure I have all I need for the presentation of the fundamental group and handlebody decomposition of the minimal Brieskorn homology sphere to meet my criteria; sorry if I dragged my feet posting this $\endgroup$
    – Jeffrey Rolland
    Commented Dec 30, 2019 at 23:45
  • $\begingroup$ I also need to know a product of generators representing a generator of the center; I can't edit the above $\endgroup$
    – Jeffrey Rolland
    Commented Dec 31, 2019 at 0:06
  • $\begingroup$ @JeffreyRolland - the presentation given by Boden, Herald, and Kirk (linked to in my answer) includes the generator $h$ of the center as one of the generators of the fundamental group. They also clearly indicate $h$ in their Figure 1. $\endgroup$
    – Sam Nead
    Commented Jan 1, 2020 at 14:35
  • $\begingroup$ I got the book today, and it answered all my questions and more; I'll post an "answer" and my exciting MATLAB .m script to compute the Bézout coefficients corresponding to the relatively prime integers in the Brieskorn homology sphere tomorrow $\endgroup$
    – Jeffrey Rolland
    Commented Jan 3, 2020 at 1:30
  • $\begingroup$ Indeed, the book by Saveliev also explicitly includes the generator ℎ of the center as one of the generators of the fundamental group; thanks so much for the response $\endgroup$
    – Jeffrey Rolland
    Commented Jan 3, 2020 at 1:32

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