Timeline for When is the union of embedded smooth manifolds a smooth manifold?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 20, 2023 at 14:22 | comment | added | Moishe Kohan | Indeed, the "claim" is false as Umberto noted. | |
| Oct 7, 2016 at 20:21 | comment | added | Umberto Lupo | Then $M_1$ and $M_2$ are (I think) both embedded one-dimensional submanifolds, their intersection is the interval $(0,1)$ on the $x$-axis which is again a one-dimensional embedded submanifold, and drawing some figures one sees that the conditions on the closures as written in this answer are satisfied. But the union $M_1 \cup M_2$ is a "lasso" figure, which is an immersed but not an embedded submanifold of the plane. Have I gone wrong somewhere? | |
| Oct 7, 2016 at 20:19 | comment | added | Umberto Lupo | @RyanBudney I am guessing that the submanifolds in your answer are intended to be embedded submanifolds -- so that the inclusion is a homeomorphism onto its image with the subspace topology. But then can we not construct a counterexample to your claim as follows? Let the ambient manifold be $M=\mathbb{R}^2$. Let $M_1$ be the interval $(-1,1)$ on the $x$-axis, and $M_2$ be the one-dimensional submanifold traced by going along the $x$-axis from $0$ (not included) to $1$, traveling up a little and then back again to the origin, approaching the latter in a direction transverse to the $x$-axis. | |
| Oct 23, 2011 at 22:31 | comment | added | Mirco | Maybe its a manifold, iff the coproduct exists as a manifold. This is just an idea I had today and I have to go deeper into it. But maybe someone else knows more... | |
| Oct 22, 2011 at 7:30 | comment | added | Mark Grant | Its worth noting that every closed manifold arises as such a union: just take a finite atlas. | |
| Oct 21, 2011 at 21:09 | comment | added | Ryan Budney | What do you mean? Certainly it can be true, unless you add some extra hypothesis that nullifies this possibility. For example, in $\mathbb R^2$ you can take the set $\{(x,y): x > -1, y=0 \}$ together with $\{(x,y): x < 1, y=0 \}$. | |
| Oct 21, 2011 at 18:44 | comment | added | Mirco | Unfortunately the requirement $dim(M_i \cap M_j)=m$ for $dim(M_i)=m$ is not true in the simplicial setting | |
| Oct 21, 2011 at 18:01 | history | undeleted | Ryan Budney | ||
| Oct 21, 2011 at 18:01 | history | deleted | Ryan Budney | ||
| Oct 21, 2011 at 18:00 | history | answered | Ryan Budney | CC BY-SA 3.0 |
