Timeline for Exotic differentiable structures on R^4?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 10, 2018 at 9:45 | comment | added | Aru Ray | Just a note for future visitors - as Andy correctly bet, simply connected at infinity is sufficient. That is, Corollary 1.2 of Freedman's The Topology of Four-Dimensional Manifolds says that any topological 4-manifold which has trivial $\pi_1$ and $H_2$, and is simply connected at infinity, is homeomorphic to $\mathbb{R}^4$. | |
| Dec 12, 2015 at 1:26 | comment | added | pro | .....MIND=BLOWN. | |
| Aug 12, 2010 at 15:30 | comment | added | Andy Putman | @Outis : That's correct. | |
| Aug 5, 2010 at 5:56 | comment | added | Outis | Andy, just to clarify, in "Cartesian products of contractible open manifolds", I guess you are referring to Theorem 2? This concerns a product that is "topologically" $\mathbb{R}^4$, implying just homeomorphism, but you are saying that the manner of the proof implies something stronger -- piecewise linear isomorphism. Correct? | |
| May 20, 2010 at 12:50 | comment | added | Andy Putman | @Peter : It's a smooth embedding! For a nice introductory discussion of 4d topology (including stuff like this), I recommend Scorpan's book "The Wild World of 4-Manifolds". | |
| May 20, 2010 at 6:42 | comment | added | Peter Samuelson | The third paragraph is definitely surprising (to me, at least). This might be a dumb question, but is the embedding U -> R^4 a smooth embedding or just a topological one? | |
| May 18, 2010 at 12:59 | comment | added | Igor Belegradek | Thanks, Andy! I missed that McMillan works in PL category. | |
| May 17, 2010 at 23:19 | comment | added | Andy Putman | Yes (the following might not be a geodesic to the proof, but it works <grin>). Let U be a smooth contractible 3-manifold. In "Cartesian products of contractible open manifolds", McMillan proved (assuming the Poincare conjecture, which I suppose we now all believe) that UxR is PL isomorphic to R^4. You can now quote Corollary 6.6 of "Obstructions to the smoothing of piecewise-differentiable homeomorphisms" by Munkres, which says that any smooth manifold which is PL isomorphic to R^n is diffeomorphic to R^n. | |
| May 17, 2010 at 13:18 | comment | added | Igor Belegradek | Speaking of stabilization, I think it is true that the product of any contractible 3-manifold with $\mathbb R$ is homeomorphic to $\mathbb R^4$; is it diffeomorphic to $\mathbb R^4$? | |
| May 17, 2010 at 12:44 | history | answered | Andy Putman | CC BY-SA 2.5 |