Timeline for Decide a manifold via its boundary
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 21, 2016 at 7:32 | comment | added | Greg Friedman | In the answer, "bounds a sphere" = "bounded by a sphere"? | |
| Jun 17, 2014 at 1:03 | vote | accept | Chih-Wei Chen | ||
| Nov 10, 2011 at 12:54 | comment | added | algori | Bruno -- re "there are no exotic balls as far as I remember": indeed, by the h-cobordism theorem. | |
| Nov 9, 2011 at 21:02 | comment | added | Bruno Martelli | A compact contractible smooth manifold bounded by a sphere $S^{n-1}$ is actually also diffeomorphic to a disc $D^n$ provided that $n>4$: there are exotic spheres, but there are no exotic balls as far as I remember. In fact, every known exotic sphere is the union of two standard smooth balls glued via a nonstandard smooth map along their boundaries. So if you remove a ball from an exotic sphere you get a standard ball. The dimension-4 case is of course still open and of completely different flavour. | |
| Nov 1, 2011 at 4:37 | comment | added | algori | Dear Chih-Wei -- a torus $T^k$ inside some manifold $M$ does not necessarily split $M$: take $M=T^k\times T^1$. And yes, no contractible manifold has $T^k$ as boundary unless $k=1$. | |
| Nov 1, 2011 at 4:35 | comment | added | Chih-Wei Chen | In the comment above, $k$ is less than the dimension $n$ of $M$. And I forget to put the power $n-k$ on the interval $[-1,1]$, sorry. In case that $k<n-1$, it seems quite different and hard to our discussion before. | |
| Nov 1, 2011 at 4:28 | comment | added | Chih-Wei Chen | Dear algori and Ryan, thanks for your answers. I'd like to learn more about this. Suppose we have a complete non-compact contractible manifold $M$ with a $T^k\times [-1,1]$ neck inside, can we say that the $T^k$ must divide $M$ into two parts and one of them is compact and contractible (so that $T^k$ shrinks to a point inside)? (or such neck cannot exist in a contractible manifold, by Ryan's second comment?) Thank you a lot. | |
| Nov 1, 2011 at 4:12 | comment | added | Ryan Budney | But if an $n$-manifold bounds a contractible manifold, by Poincare Duality it must be a homology sphere -- meaning having the same homology has $S^n$. | |
| Nov 1, 2011 at 4:08 | comment | added | Ryan Budney | ... and in dimension $n \geq 3$ there are $n$-manifolds which are not spheres that bound contractible manifolds. Google "Mazur manifold" for details. | |
| Nov 1, 2011 at 4:03 | history | answered | algori | CC BY-SA 3.0 |