Timeline for More four-dimensional counterexamples
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 6, 2011 at 13:30 | vote | accept | Igor Rivin | ||
| Jul 4, 2011 at 21:15 | answer | added | Igor Belegradek | timeline score: 9 | |
| Jul 4, 2011 at 20:49 | comment | added | Igor Rivin | I believe the canonical reference is Freedman-Quinn (for polycyclic groups) (in "Topology of 4-manifolds") | |
| Jul 4, 2011 at 19:57 | comment | added | Igor Belegradek | @Anton: thanks, I wrote this in haste. In fact, I do not even know the 4-dimensional Borel conjecture for all subexponential growth groups, but all virtually nilpotent groups are okay. I will give references later today. | |
| Jul 4, 2011 at 8:01 | comment | added | Anton Petrunin | @Igor: "if both the fiber and the base have zero Euler characteristic, then I think the fundamental group" --- what about $T^2$-bundle over $S^1$? | |
| Jul 4, 2011 at 5:30 | comment | added | Igor Rivin | Cool! I would assume that "Borel's conjecture" would be based on some set of cases where Borel actually knew this to be true... Also, what's the reference for the "subexponential growth" case? | |
| Jul 4, 2011 at 4:17 | comment | added | Igor Belegradek | Incidentally, if both the fiber and the base have zero Euler characteristic, then I think the fundamental group of the total space has subexponetial growth in which this case the Borel's conjecture is true. | |
| Jul 4, 2011 at 3:58 | comment | added | Igor Belegradek | Borel's conjecture predicts that a homotopy equivalence of closed aspherical manifolds is homotopic to a homeomorphism. I would be very surprized if the conjecture were not true for surface bundles over surfaces. | |
| Jul 4, 2011 at 0:36 | history | asked | Igor Rivin | CC BY-SA 3.0 |