Timeline for If a manifold suspends to a sphere...
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18 events
| when toggle format | what | by | license | comment | |
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| Aug 7, 2023 at 3:30 | comment | added | Michael Albanese | A proof of the fact that Igor Belegradek mentioned in the comments from 2011 (if two closed manifolds become homeomorphic after multiplying by $\mathbb{R}$, then they are h-cobordant), can be found in this answer, written following Igor's suggestions. | |
| Mar 30, 2022 at 21:30 | comment | added | Igor Belegradek | @MichaelAlbanese Actually my "metastable range" statement contradicts Proposition 8.2 mentioned above, so it is wrong. If $2k\ge n+3$ the obvious homotopy equivalence $\Sigma\to S^n\times\mathbb R^k$ is homotopic to a smooth embedding, and its tubular neighborhood is diffeomorphic to $S^n\times\mathbb R^k$ when $k\ge 3$. The tubular neighborhood is the total space of a vector bundle over $\Sigma$, but the bundle may not be trivial. It only is stably trivial. | |
| Mar 30, 2022 at 2:36 | comment | added | Igor Belegradek | @MichaelAlbanese: if $\Sigma$ bounds a parallelizable manifold, then $k=3$, see e.g. remark 5.8 in arxiv.org/abs/0912.4869. In general, the answers are more complicated, see Proposition 8.2 in arxiv.org/abs/1104.4136. Certainly, stable range is an overkill; metastable range suffices: if $2k\ge n+3$, then $\Sigma\times\mathbb R^k$ and $S^n\times\mathbb R^k$ are diffeomorphic. | |
| Mar 29, 2022 at 23:08 | comment | added | Michael Albanese | @IgorBelegradek: Right, I forgot about the potential existence of exotic 4-spheres. If I'm not mistaken, it follows from the fact that the tangent bundles of $\Sigma$ and $S^n$ are isomorphic that $\Sigma\times\mathbb{R}^{n+1}$ and $S^n\times\mathbb{R}^{n+1}$ are diffeomorphic. I wonder what the smallest value of $k$ is such that $\Sigma\times\mathbb{R}^k$ and $S^n\times\mathbb{R}^k$ are diffeomorphic. By your comment, $k > 2$. | |
| Mar 29, 2022 at 13:12 | comment | added | Igor Belegradek | @MichaelAlbanese: what you say is correct, except that the "in particular" part needs $n\ge 5$ so that the h-cobordism theorem applies (the argument does not apply to exotic 4-spheres if they exist). In fact, a similar argument applied twice shows that $\Sigma\times\mathbb R^2$ and $S^n\times\mathbb R^2$ are not diffeomorphic: if there were then $\Sigma\times S^1$ and $S^n\times S^1$ would be h-cobordant, hence diffeomorphic, and passing to the universal cover gives a diffeomorphism of $\Sigma\times\mathbb R$ and $S^n\times\mathbb R$. | |
| Mar 29, 2022 at 9:51 | comment | added | Michael Albanese | @IgorBelegradek: If I'm not mistaken, your argument also works smoothly: if $M$ and $N$ are closed smooth manifolds such that $M\times\mathbb{R}$ and $N\times\mathbb{R}$ are diffeomorphic, then $M$ and $N$ are smoothly $h$-cobordant. In particular, if $\Sigma$ is an exotic $n$-sphere, then $\Sigma\times\mathbb{R}$ and $S^n\times\mathbb{R}$ are not diffeomorphic. Is that correct or have I oversimplified? | |
| May 6, 2011 at 0:31 | comment | added | Joel Fine | @Igor again, I'm sorry, having reread all of the comments more carefully I see that this line of reasoning was clearly already apparent to you! | |
| May 6, 2011 at 0:30 | comment | added | Joel Fine | @Igor, no me neither! I guess it's the case that $M$ must be a homotopy sphere and so it follows from Poincaré. But since we are assuming a homeomorphism it feels like we're giving ourselves strictly more information than in the Poincaré conjecture. Although I could easily be wrong on that. | |
| May 5, 2011 at 23:54 | comment | added | Igor Belegradek | @Joel Fine: I am not sure how to prove your "only if" assertion. | |
| May 5, 2011 at 22:43 | comment | added | Joel Fine | I'm no expert but perhaps it's not necessary to invoke all of these deep theorems to prove this. I'm fairly sure the following is true. Given an $n$-manifold $M$, consider the cone $CM$ on $M$. Then the vertex of $CM$ has a neighbourhood homeomorphic to an open set in $\mathbb{R}^{n+1}$ if and only if $M$ is the $n$-sphere. If this really is true then the only manifold whose suspension is again a manifold is $S^n$. Am I right here? | |
| May 5, 2011 at 21:25 | comment | added | Pete L. Clark | @Willie: but then what do we call it? It's not any one person's theorem... | |
| May 5, 2011 at 20:58 | comment | added | Willie Wong | I think we can stop calling it a "conjecture" now. :-) | |
| May 5, 2011 at 20:11 | comment | added | Igor Belegradek | In fact, one need not involve h-cobordisms at all: just note that $M$ and $S^n$ are homotopy equivalent and use Poincare's conjecture. I guess, I just like to advertize that fact that if two closed manifolds become homeomorphic after multiplying by $\mathbb R$, then they are $h$-cobordant. :) | |
| May 5, 2011 at 19:38 | comment | added | Igor Belegradek | To see that $M$ and $S^n$ are h-cobordant consider a homeomorphism $h$ of their products with $\mathbb R$, and use excision in homology to show that the submanifolds $S^n\times 0$ and $h(M\times t)$ bound an h-cobordism, where $t$ need to be sufficiently large to ensure that the submanifolds are disjoint. | |
| May 5, 2011 at 19:29 | vote | accept | James Cranch | ||
| May 5, 2011 at 19:15 | comment | added | Igor Belegradek | Topological h-cobordism theorem for simply-connected manifolds holds in all dimensions (due to Freedman in dimension 4, to Perelman in dimension 3, and to Newman in dimensions >4). | |
| May 5, 2011 at 19:10 | comment | added | Benoît Kloeckner | You need $n>4$, though, don't you? | |
| May 5, 2011 at 19:06 | history | answered | Igor Belegradek | CC BY-SA 3.0 |