Timeline for Which manifolds are homeomorphic to simplicial complexes?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 23, 2022 at 12:43 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
http -> https (the question was bumped anyway)
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| Mar 12, 2013 at 1:43 | history | edited | Andy Putman | CC BY-SA 3.0 |
added 254 characters in body
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| Oct 29, 2010 at 1:10 | comment | added | Charles Rezk | Ah, I misread it. | |
| Oct 29, 2010 at 1:03 | comment | added | Paul | @IB the relationship is explained in Akbulut-McCarthy's book in detail but I've forgotten the argument. There's an outline here math.niu.edu/~rusin/known-math/96/Triangulations which asserts that if E8 were triangulable it could be smoothed in the complement of its vertices. Removing a nbd of each vertex yields a smooth manifold with boundary a union of homotopy (?) 3-spheres whose total Rohlin invariant is 1 (since $\sigma(E8)=1$). But Casson's invariant is zero on homotopy spheres and is a lift of Rohlin. | |
| Oct 28, 2010 at 23:19 | comment | added | Igor Belegradek | It seems that the fact that E8 is not triangulable should follow from basic properties of Casson invariant (of which I know next to nothing). I am curious to see how the argument goes. | |
| Oct 28, 2010 at 22:49 | comment | added | Igor Belegradek | The slides say (on page 5) that NOT every 4-manifold is triangulable. | |
| Oct 28, 2010 at 22:38 | comment | added | Charles Rezk | Thanks! Those slides say that the problem is solved in dimension 4 (all manifolds are triangulable), attributed to Casson. | |
| Oct 28, 2010 at 22:03 | history | answered | Andy Putman | CC BY-SA 2.5 |