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S Aug 28, 2021 at 22:08 vote accept piper1967
S Aug 28, 2021 at 22:08 vote accept piper1967
S Aug 28, 2021 at 22:08
Aug 28, 2021 at 22:08 vote accept piper1967
S Aug 28, 2021 at 22:08
Aug 4, 2021 at 21:45 history became hot network question
Aug 4, 2021 at 16:08 answer added mme timeline score: 7
Aug 4, 2021 at 15:01 answer added Jens Reinhold timeline score: 16
Aug 4, 2021 at 14:32 comment added Tim Campion Note that if $G = \pi_1(X)$ is perfect, you are asking for a finite acyclic space $X$ with fundamental group $G$. There's a lot of literature about acyclic spaces which could be relevant.
Aug 4, 2021 at 14:15 comment added mme In fact there is a 2-dimensional simplicial complex whose fundamental group is $2I$, the "binary icosahedral group", which is nonabelian and perfect and order 120, and whose homology groups $H_i$ are all zero for all $i \geq 1$. This can be constructed by identifying appropriate twists of antipodal faces on a dodecahedron (and triangulating each pentagon), but to the modern eye the fastest description is probably "This is the punctured Poincare homology sphere".
Aug 4, 2021 at 14:15 review Close votes
Aug 6, 2021 at 9:55
Aug 4, 2021 at 14:02 comment added piper1967 I edited the question. I meant finite non-abelian group in $\pi_1.$
Aug 4, 2021 at 14:01 history edited piper1967 CC BY-SA 4.0
edited title
Aug 4, 2021 at 14:00 answer added Johannes Hahn timeline score: 5
Aug 4, 2021 at 13:54 comment added Tim Campion Is there anything wrong with a bouquet of circles?
Aug 4, 2021 at 13:45 review First posts
Aug 4, 2021 at 14:15
Aug 4, 2021 at 13:44 history asked piper1967 CC BY-SA 4.0