Timeline for For what topological groups $G$ can we take $EG \rightarrow BG$ to be of the form $S^{\infty} \rightarrow BG$?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 10, 2021 at 14:26 | vote | accept | Noel Lundström | ||
| Feb 9, 2021 at 15:51 | comment | added | Tyrone | $S^\infty$ in the CW topology is not metrisable, so certainly not homeomorphic to either a Hilbert space or a Hilbert cube. As a (weakly) contractible $\mathbb{R}^\infty$-manifold, $S^\infty$ is homeomorphic to $\mathbb{R}^\infty$ (everything in CW topology). | |
| Feb 9, 2021 at 15:39 | answer | added | Chris Schommer-Pries | timeline score: 20 | |
| Feb 9, 2021 at 12:27 | comment | added | Jim Belk | If $G$ is a countable, discrete, torsion-free group, shouldn't the left regular representation of $G$ induce a free, properly discontinuous action of $G$ on $S^\infty$? | |
| Feb 9, 2021 at 12:06 | comment | added | Neil Strickland | @FernandoMuro I presume that here $S^\infty=\bigcup_nS^n$ rather than the whole unit sphere in a Hilbert space $H$, so I don't think that $S^\infty$ is homeomorphic to $H$. However, this $S^\infty$ is homeomorphic to $\mathbb{R}^\infty=\bigcup_n\mathbb{R}^n$ and to many other spaces such as such as the Stiefel manifold $V_n\mathbb{R}^\infty$, the complement $\mathbb{R}^\infty\setminus(\text{a finite set})$, the ordered configuration space $F_n\mathbb{R}^\infty$ etc. I am not sure where to find these facts in the literature, though. | |
| Feb 9, 2021 at 8:59 | comment | added | Maxime Ramzi | @TheoJohnson-Freyd : yes it does, in fact there's almost a converse (in "Periodic resolutions for finite groups", Swan proves that a group with periodic cohomology will act freely on a simplicial complex homotopy equivalent to a sphere). Not sure about SU(2) though | |
| Feb 9, 2021 at 1:45 | history | edited | Noel Lundström |
edited tags
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| Feb 9, 2021 at 0:33 | comment | added | Fernando Muro | If I'm not mistaken, $S^\infty$ is homeomorphic to a whole Hilbert space, whose unitary group is probably huge, although contractible. Maybe if you edit the tags you can obtain better answers from people in representation theory, analysis, operator algebras, etc. | |
| Feb 9, 2021 at 0:11 | comment | added | Theo Johnson-Freyd | @TomGoodwillie If a group acts freely on a sphere, doesn't it imply that its cohomology is periodic? And I thought that the only groups of periodic cohomology were subgroups of SU(2). Herm, maybe that's just for finite groups. | |
| Feb 8, 2021 at 23:06 | comment | added | Noel Lundström | I know $S^7$ is not a group, but it might contain some group which is not a subgroup of $S^3$. I don't know enough about $S^7$ to exclude such a possibility. @TomGoodwillie | |
| Feb 8, 2021 at 23:00 | comment | added | Tom Goodwillie | (1) $S^7$ is not a group. (2) Any group that can act freely on $S^n$ can also act freely on $S^\infty$. (3) I wonder about the converse. In particular, can a non-cyclic group of order act freely on $S^\infty$. (4) There are papers about which groups can act freely on (some finite-dimensional) spheres, but I don't know the state of the art. | |
| Feb 8, 2021 at 22:53 | comment | added | Noel Lundström | I'm not quite sure I follow, what would a subgroup of $GL(n)$ be an example of? Something with a free action on $S^{\infty}$ which is not a subgroup of $S^0,...,S^7$ for big enough $n$? @PhilTosteson | |
| Feb 8, 2021 at 22:44 | comment | added | Phil Tosteson | Often for infinite dimensional spaces homotopy equivalence often implies homeomorphism. See the discussion here: mathoverflow.net/questions/293382/… I don't understand this phenomenon well enough to confirm this, but it seems likely that many contractible infinite dimensional manifolds are homeomorphic to $S^\infty$. This would give quite a few examples: e.g. any subgroup of $GL_n$. | |
| Feb 8, 2021 at 22:32 | history | edited | Noel Lundström | CC BY-SA 4.0 |
added 16 characters in body
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| Feb 8, 2021 at 22:23 | history | asked | Noel Lundström | CC BY-SA 4.0 |