Timeline for isotopy inverse embeddings vs. diffeomorphisms
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jun 27, 2012 at 3:34 | vote | accept | Ricardo Andrade | ||
| Jun 27, 2012 at 2:41 | comment | added | Ricardo Andrade | @Sergey: Thanks for the clarification. | |
| Jun 27, 2012 at 2:16 | comment | added | Sergey Melikhov | Ricardo: by the isotopy extension theorem (see Agol's answer). A non-local knot in $W$ is smoothly isotopic to a knot in $h(W)\subset B\subset W$; by the isotopy extension theorem this isotopy is covered by a smooth isotopy of $W$. Less trivially, Smythe proved that a knot in a $3$-manifold, PL isotopic to the unknot (by a possibly non-locally-flat PL isotopy), is contained in a ball (see Lemma 2.2 in arxiv.org/abs/math.GT/0103114 ). By the way, I of course meant to say that Zack's remark suffices to answer the case $M=\Bbb R^3$ of your question, not the full 3D case. | |
| Jun 27, 2012 at 0:18 | comment | added | Ian Agol | @Zack: I think the issue with your clever suggestion is that if you have an exotic $\mathbb{R}^4$ E, and embeddings $E\subset \mathbb{R}^4\subset E$, then how do you know that the induced embedding $E\subset E$ is smoothly isotopic to the identity? (I say clever because when I first read it I upvoted it!). In any case, I think my answer shows that it cannot be. | |
| Jun 26, 2012 at 23:55 | comment | added | Ricardo Andrade | @Sergey and @Zack: I have a perhaps silly question. How do you guarantee that non-local knots are preserved under isotopies (through embeddings, not necessarily diffeomorphisms) of the identity? | |
| Jun 26, 2012 at 23:47 | comment | added | Sergey Melikhov | ("nontrivial" should be "non-split") | |
| Jun 26, 2012 at 22:19 | comment | added | Sergey Melikhov | A knot in a $3$-manifold is called nonlocal if it is not contained in any closed ball. An open $3$-manifold containing no nonlocal knots is homeomorphic to $\Bbb R^3$, ams.org/journals/proc/1971-028-01/S0002-9939-1971-0271919-1 so Zack's remark suffices to answer the 3-dimensional case of the question. To see an explicit nonlocal knot in the Whitehead manifold $W$ it suffices to know that the $n$th Whitehead link is nontrivial (and this is how Whitehead originally proved that $W\not\cong\Bbb R^3$). | |
| Jun 26, 2012 at 6:13 | answer | added | Ian Agol | timeline score: 26 | |
| Jun 24, 2012 at 23:22 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
clarification; edited body
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| Jun 24, 2012 at 23:13 | comment | added | Ricardo Andrade | @Ben: Thank you for pointing out that my edit introduced an obvious counter-example. I partially reversed the edit to avoid your counter-examples. | |
| Jun 24, 2012 at 23:05 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
remove easy counter-example
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| Jun 24, 2012 at 22:40 | comment | added | Zack | @Lee: The image of $Whitehead\subset Ball\subset Whitehead$ doesn't contain any nonlocal knots. | |
| Jun 24, 2012 at 18:05 | comment | added | Ben Wieland | Does your edit mean you invite examples with boundary? A closed interval and an open interval. For a compact example, take manifolds that are $h$-cobordant but not diffeomorphic. Crossing with an open interval gives diffeomorphic manifolds and crossing with a closed interval gives an example of your property. | |
| Jun 24, 2012 at 13:51 | comment | added | Lee Mosher | @Misha: I don't see the contradiction. If you are thinking about the case $Whitehead \subset Ball \subset Whitehead$, one can imagine a highly nonproper isotopy, whose first half pierces some of the sphere but not the rest, and ends with a self embedding of Whitehead that does not have compact closure but also does not contain the sphere; then the 2nd half of the isotopy continues to the identity map. This avoids the sequence of 2-spheres you are concerned about. A similar imaginary isotopy avoids this contradiction in the case $Ball \subset Whitehead \subset Ball$. | |
| Jun 24, 2012 at 3:42 | comment | added | Misha | @Ricardo: Embedding a Whitehead manifold W to an open ball in $R^3$ would not work. (Since an isotopy would yield a sequence of 2-spheres which exit the end of W, which is impossible.) | |
| Jun 24, 2012 at 3:25 | comment | added | Ricardo Andrade | @Zack: I was actually thinking of simply using the Whitehead manifold together with its embedding into ${\mathbb R}^3$, but I know next to nothing about the space of self-embeddings of the Whitehead manifold. | |
| Jun 24, 2012 at 2:34 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
deleted 17 characters in body; edited body
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| Jun 24, 2012 at 2:33 | comment | added | Zack | Would small exotic $\mathbb R^4$s work? | |
| Jun 24, 2012 at 2:13 | history | asked | Ricardo Andrade | CC BY-SA 3.0 |