Timeline for When is the closure of a manifold a manifold with boundary?
Current License: CC BY-SA 4.0
11 events
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| Feb 18 at 23:56 | comment | added | Brian Seguin | @PietroMajer This is my intuition as well, but it is unclear to me exactly how to use the local inverse theorem here since $\bar{\cal S}$ is closed and this theorem deals with open sets. If you know how the details would go, I'd be grateful if you could writeup an answer. If you do, it might be better to do so under this other post as it asks specifically this question: mathoverflow.net/questions/464387/… | |
| Feb 18 at 18:55 | comment | added | Pietro Majer | I think that adding the condition that the boundary is a $C^1$ curve, then it is true. The condition that the normal extends to a continuous $\nu: \overline S\to\mathbb R^3$ should imply by the local inverse theorem that $\overline S$ is locally a graph of a $C^1$ function defined on a $C^1$ manifold with boundary in $\mathbb R^2$ | |
| Feb 18 at 12:26 | comment | added | Martin Väth | I realized that my example is not correct without additional work: The boundary is normally not $C^1$ at $x=0$. It might be possible to save the example by rounding the corners in a very special way, but it needs more consideration. | |
| Feb 18 at 11:08 | comment | added | Daniele Tampieri | @MartinVäth Why don't you write down your comment as an answer? I'll upvote it. | |
| Feb 18 at 11:07 | comment | added | Pietro Majer | @MartinVäth Could you give more details? it is nor clear to me... thx | |
| Feb 18 at 11:00 | comment | added | Martin Väth | Take a $C^1$ function on $(0,1]$ which has a continuous but not differentiable extension to $[0,1]$. Now let the manifold be the open “rounded” square (the corners rounded smoothly), and move its $z$ coordinate by that function w.r.t. the $x$-coordinate: The normal has a continuous extension to the boundary, the boundary is a $C^1$-curve, but the manifold is not smoother than continuous at the boundary | |
| Feb 18 at 10:51 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing
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| Feb 18 at 0:42 | comment | added | Brian Seguin | Ah, good one. Thanks for the counterexample! Going back to the larger question of what conditions are necessary to ensure the closure is a $C^1$ surface, what if we also assume that the boundary is a $C^1$ curve. Is that enough to ensure the closure is a $C^1$ manifold? | |
| Feb 17 at 8:36 | comment | added | Pietro Majer | But for instance, what about a plane domain in {z=0}, interior to a continuous nondifferentiable curve? | |
| S Feb 17 at 6:53 | review | First questions | |||
| Feb 17 at 9:27 | |||||
| S Feb 17 at 6:53 | history | asked | Brian Seguin | CC BY-SA 4.0 |