Timeline for Parallelizability of the Milnor's exotic spheres in dimension 7
Current License: CC BY-SA 4.0
8 events
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| Mar 3, 2019 at 1:09 | comment | added | John Klein | @IanAgol right. But not the converse. | |
| Mar 3, 2019 at 1:05 | history | edited | John Klein | CC BY-SA 4.0 |
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| Mar 2, 2019 at 6:05 | comment | added | Ian Agol | Lemma 2.17 in Allen Hatcher's book "vector bundles and K-theory" gives a proof that if $S^n$ is parallelizable, then it is an H-space. pi.math.cornell.edu/~hatcher/VBKT/VBpage.html | |
| Mar 5, 2013 at 1:20 | comment | added | Dylan Wilson | A direct construction of the H-space map is also given as thm 10.5.7 in Aguilar-Gitler-Prieto (algebraic topology from a homotopical viewpoint) | |
| Mar 11, 2011 at 21:14 | history | edited | John Klein | CC BY-SA 2.5 |
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| Mar 11, 2011 at 20:18 | comment | added | John Klein | Let $P \to M$ be the frame bundle of a Riemannian manifold $M$: at $x\in M$ its fiber is given by linear isometries $\phi: \Bbb R^n \to M$. Let $Q \to M$ be the fibration whose fiber at $x\in M$ is given by the space of homotopy equivalences $f: S^{n-1} \to S_x$, where $S-x$ is the image of a small sphere under the exponential map. Then when $M$ is an exotic sphere we get $Q \simeq G_{n+1}$. The map $P \to Q$ is given by restricting the isometry $\phi$ to a small sphere at the origin, and then applying the exponential map. How does that sound to you? | |
| Mar 11, 2011 at 17:10 | comment | added | Johannes Ebert | It seems to be valid, but not sufficient. According to Adams, Dold showed that if $S^n$ is parallelizable with respect to some smooth structure, then it is an H-space. You would have to create a map from the frame bundle of the exotic structure to $G_{n+1}$ and I do not see this map. | |
| Mar 11, 2011 at 16:59 | history | answered | John Klein | CC BY-SA 2.5 |