Timeline for What is the least $n\ge1$ for which there is an $n$-dimensional closed flat manifold with perfect fundamental group?
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| Feb 27, 2023 at 15:20 | comment | added | Jason DeVito - on hiatus | @PeterKropholler: Well, I suspect I am missing the key point because this answer has 7 upvotes. Your question at the top is about the lowest dimension supporting a flat manifold with perfect fundamental group. This answer addresses the lowest dimension supporting a flat manifold with perfect holonomy group. These two groups are related (perfect fundamental group obviously implies perfect holonomy group), so this answer (and Igor's comment above) provides a lower bound of $15$ for your question. But as far as I can tell, it doesn't imply the answer to your question is precisely $15$. | |
| Feb 27, 2023 at 5:58 | comment | added | Peter Kropholler | @JasonDeVito I think it likely that for a given perfect (finite) holonomy group $\phi$ there be more than one flat manifold with perfect fundamental group and with that holonomy, maybe of different dimensions. That maybe suggests that you would not expect the inequality $\dim N\le \dim M$. Or am I still missing some key point? | |
| Feb 25, 2023 at 18:34 | comment | added | Jason DeVito - on hiatus | @PeterKropholler: I agree with everything you just wrote. The way I read Igor's answer is that we start with a flat manifold $M$ with holonomy $\phi$ with $\phi$ chosen to be perfect. Via your answer, or the Holt-Pleskin reference, we obtain a new flat manifold $N$ with holonomy $\phi$, where $\pi_1(N)$ is perfect; equivalently, $H_1(N) = 0$. One way to rephrase my question is: is it true that $\dim N \leq \dim M$? If not, I still don't see why isolating the minimal dimension in which $\phi$ can be perfect answers the question (though it does provide a lower bound). | |
| Feb 25, 2023 at 16:57 | comment | added | Peter Kropholler | @JasonDeVito You are right that the holonomy group determines neither the fundamental group nor the first homology of the flat manifold. But it is always true that the abelianisation of the fundamental group is isomorphic to the first homology of the manifold. | |
| Feb 24, 2023 at 15:12 | comment | added | Jason DeVito - on hiatus | @Peter: I am still confused, because the holonomy group doesn't determine the fundamental group, nor does it determine $H_1$. As Igor mentions in his answer in the post you linked to, if $\pi$ is the fundamental group of flat manifold $M^n$ with holonomy $\phi$, then $\pi\oplus \mathbb{Z}$ is also the fundamental group of a flat manifold $N^{n+1}$ with holonomy $\phi$, and clearly $H_1(N)\cong H_1(M)\oplus \mathbb{Z}$ in this case. Of course, minimality of the dimension in which you get holonomy $\phi$ rules out this particular construction, but why couldn't there be other constructions? | |
| Feb 24, 2023 at 13:09 | comment | added | Peter Kropholler | @JasonDeVito Flat manifolds are aspherical which means that the homology of the manifold and its fundamental group coincide. On the group theory side, the fist homology of a group with trivial integer coefficients is isomorphic to the largest abelian quotient of the group. So perfect fundamental group amounts to the same condition as trivial first homology for these manifolds. In particular, it's not really a feature that is influenced by minimality. | |
| Feb 5, 2023 at 18:35 | comment | added | Jason DeVito - on hiatus | Can you expand on why minimality of dimension implies the manifold has trivial first homology? | |
| Jan 9, 2023 at 14:26 | comment | added | Kasper Andersen | @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... | |
| Jan 9, 2023 at 14:22 | history | edited | Kasper Andersen | CC BY-SA 4.0 |
added 10 characters in body
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| Jan 7, 2023 at 15:06 | comment | added | Peter Kropholler | @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. | |
| Jan 6, 2023 at 20:33 | comment | added | Peter Kropholler | @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. | |
| Jan 2, 2023 at 20:22 | comment | added | YCor | DOI link to Plesken's paper (behind expensive Springer paywall) | |
| Jan 2, 2023 at 19:14 | history | answered | Kasper Andersen | CC BY-SA 4.0 |