Timeline for Recognizing sections up to isotopy
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29 at 15:34 | vote | accept | Matthew Kvalheim | ||
| Feb 21, 2023 at 21:21 | answer | added | Gael Meigniez | timeline score: 3 | |
| Nov 10, 2022 at 17:17 | comment | added | Connor Malin | An embedded isotopy is an embedding $i:M \times [0,1] \rightarrow N$, notably it gives an isotopy between $i_0$ and $i_1$. This is stronger than the usual notion of isotopy. | |
| Nov 10, 2022 at 16:22 | comment | added | Matthew Kvalheim | @MarkGrant thank you for your help. That is indeed a condition I have been thinking about, but I believe it is implied by the existence of a free homotopy from $h$ to a section of $\pi$. I was hoping that there might be some "computable" way of ruling out existence of an isotopy even when such a free homotopy exists. I was actually wondering whether this could be achieved using some fancier intersection number (e.g. of submanifolds of some jet bundle) than the one you described, but maybe that isn't possible? The $s$-cobordism theorem also seems relevant - see my reply to Ryan Budney. | |
| Nov 10, 2022 at 16:17 | comment | added | Matthew Kvalheim | @RyanBudney thank you for your help. I did indeed mean free homotopies like you said. In my situation of interest I know that a free homotopy exists, but I was hoping that there still might be some "computable" way of ruling out existence of an isotopy. On the other hand, for the special case $E = B\times \mathbb{R}$ mentioned in my question, it seems that for $\dim(B)\geq 5$ the $s$-cobordism theorem provides necessary and sufficient conditions for the existence of an isotopy, but I don't know how "computable" those conditions are. | |
| Nov 10, 2022 at 16:10 | comment | added | Matthew Kvalheim | @ConnorMalin interesting - what is an "embedded isotopy"? | |
| Nov 10, 2022 at 7:29 | comment | added | Mark Grant | Your embedding $h$ needs to have (homological) intersection number $1$ with every fiber. | |
| Nov 10, 2022 at 3:53 | comment | added | Ryan Budney | You get isotopy for free if your homotopy lives in the space of sections. But presumably you are interested in free homotopies and not homotopies in the space of sections. But the question of having a homotopy in the space of sections would be a good partial answer, which you could also approach with obstruction theory. | |
| Nov 10, 2022 at 3:38 | comment | added | Ryan Budney | Obstruction 1: The bundle needs to have a section. Obstruction theory gives you a tool to find these, or determine if one does not exist. For example, the Hopf fibration $S^3 \to S^2$ does not have a section. | |
| Nov 10, 2022 at 3:20 | comment | added | Connor Malin | An example where such an embedded isotopy doesn't exist can be constructed from inertial h-cobordisms, i.e. nontrivial h-cobordisms which start and end at $M$. Suppose $M \xrightarrow{i_1} H \xleftarrow{i_2} M$ is an inertial h-cobordism. Embed $H$ into $M \times [0,1]$ via $j$ so that $j \circ i_1$ is the inclusion $M \times \{0\} \rightarrow M \times [0,1]$. Then $j \circ i_2$ is homotopic to a section, but there cannot be an embedded isotopy which terminates in a section because that would imply that the h-cobrdism was trivial. I'm not sure if this example also contradicts plain isotopy. | |
| Nov 10, 2022 at 3:03 | history | edited | Matthew Kvalheim | CC BY-SA 4.0 |
added 25 characters in body
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| Nov 10, 2022 at 2:58 | history | asked | Matthew Kvalheim | CC BY-SA 4.0 |