12
$\begingroup$

In answer to the question "Is there a flat manifold with trivial first homology?" I proposed choosing a finite perfect group $P$ and a surjection $\phi:F\to P$ where $F$ is a free group of finite rank. If $K$ is the kernel of $\phi$ then the group $G:=[F,F]/[K,K]$ is perfect, torsion-free, finitely generated and abelian-by-finite.

This construction is not novel. I believe it is a standard strategy for obtaining torsion-free perfect groups.

The Hirsch length of $G$ is the dimension of a closed flat manifold with fundamental group $G$. One might hope to achieve the smallest possible Hirsch length by taking $P$ to be the alternating group $A_5$ and taking $F$ to be free on two generators. In that case the kernel $K$ is free on 61 generators by the Nielsen-Schreier Theorem, and $F/[K,K]$ has Hirsch length 61. Since $F/[F,F]$ has Hirsch length 2 we find that $G$, in this example, is a perfect f.g. abelian-by-finite group of Hirsch length 59. Is 59 the least dimension for which this phenomenon occurs? Maybe it is possible to factor out some of the rank 59 abelian normal subgroup in $G$ without sacrificing torsion-freeness. $G$, like any crystallographic group, has a unique largest abelian subgroup of finite index $A$ and it would be useful to know the constituents of the $\mathbb QP$-module $V:=A\otimes\mathbb Q$ as a $\mathbb QP$-module.

From this discussion, and the known fact that upto dimension 3 there are no perfect flat manifolds, the answer to the question lies in the range $4 \le n \le 59$. Can this be pinned down?

Addendum 16 January 2023: the comments below strongly support n=15 for the minimum possible.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ In terms of groups, I think you're asking for the smallest $d\ge 1$ such that there exists a perfect, torsion-free, virtually $\mathbf{Z}^d$ group. $\endgroup$
    – YCor
    Commented Dec 22, 2022 at 15:31
  • 3
    $\begingroup$ There is apparently a complete classification of closed flat 4-manifold, see arxiv.org/pdf/1306.6613.pdf, and according to their tables 4 and 5 none of these groups are perfect. $\endgroup$
    – Igor Belegradek
    Commented Jan 2, 2023 at 20:07
  • 1
    $\begingroup$ @IgorBelegradek: It looks like the relevant table is Table 3, which handles the orientable case. Tables 4 and 5 handle the non-orientable case, which will always surject $C_2$. $\endgroup$
    – HJRW
    Commented Jan 3, 2023 at 8:25
  • $\begingroup$ @HJRW: thanks for the correction! $\endgroup$
    – Igor Belegradek
    Commented Jan 3, 2023 at 15:55
  • 1
    $\begingroup$ The minimal dimension is indeed 15. Lutowski and Szczepański just showed in arxiv.org/abs/2302.11368 that 15 is the minimal dimension in which there exists a non-solvable Bieberbach group. Since nontrivial solvable groups aren't perfect, this does the job. $\endgroup$
    – Igor Belegradek
    Commented Feb 24, 2023 at 4:45

1 Answer 1

Reset to default
7
$\begingroup$

For $G=A_5=\text{PSL}_2(\mathbb{F}_5)$, the minimal dimension of a flat manifold with holonomy group $G$ is 12 15 according to Theorem (V.1) in W. Plesken:``Minimal dimensions for flat manifolds with prescribed holonomy’’, Math. Ann., 284,477–486, 1989.

Improve this answer
$\endgroup$
11
  • $\begingroup$ DOI link to Plesken's paper (behind expensive Springer paywall) $\endgroup$
    – YCor
    Commented Jan 2, 2023 at 20:22
  • $\begingroup$ @KasperAnderson This is definitely a strong hint at 12. It seems implausible that allowing other perfect holonomy groups in place of $A_5$ would enable a reduction to 11 or less. I am just holding back from ticking this answer while I think about this last ingredient. I accept that at a common sense level, Plesken has addressed the question head on and has answers. The Plesken paper is quite technical and in some ways I feel it would be attractive to have a clean simple route into this question. $\endgroup$
    – Peter Kropholler
    Commented Jan 6, 2023 at 20:33
  • $\begingroup$ @KasperAnderson J. A. Hillman emailed me to make some comments one of which was that the rank should be 15 not 12. I have gone over Plesken's formula applying Theorem V.1 when p=5 and I agree with Hillman: the minimal dimension with A_5 as holonomy seems to come out as 15 if we accept Plesken's formula. (I have not gone through the detail of Plesken's proof, I am just using the displayed formula on page 483 of his paper. $\endgroup$
    – Peter Kropholler
    Commented Jan 7, 2023 at 15:06
  • $\begingroup$ @PeterKropholler Thanks, I computed this too quickly.... I also found the result 15 for $A_5$ in Hiss' presentation math.rwth-aachen.de/~Gerhard.Hiss/Presentations/… (page 14). However Hiss claims that for $\text{PSL}_2(7)$ the answer is 23, I get 15 in this case as well... $\endgroup$
    – Kasper Andersen
    Commented Jan 9, 2023 at 14:26
  • $\begingroup$ Can you expand on why minimality of dimension implies the manifold has trivial first homology? $\endgroup$
    – Jason DeVito - on hiatus
    Commented Feb 5, 2023 at 18:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .