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Let $E$, $B$ be smooth manifolds, $\pi\colon E\to B$ be a smooth fiber bundle, and $h:B\to E$ be a smooth embedding. I would like to learn what is known about the following

Question. When does there exist an isotopy from $h$ to a section of $\pi$?

A necessary condition for the existence of such an isotopy is that $h$ be homotopic to a section of $\pi$. Are there invariants that can be used to show that an isotopy does not exist even when a homotopy exists?

I am mainly interested in the special case that $B$ is a closed manifold, $E = B\times \mathbb{R}$, and $\pi:(b,t)\mapsto b$ is the trivial bundle. I would be grateful to learn about this special case or the general case above.

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    $\begingroup$ 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. $\endgroup$
    – Connor Malin
    Commented Nov 10, 2022 at 3:20
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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Nov 10, 2022 at 3:38
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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Nov 10, 2022 at 3:53
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    $\begingroup$ Your embedding $h$ needs to have (homological) intersection number $1$ with every fiber. $\endgroup$
    – Mark Grant
    Commented Nov 10, 2022 at 7:29
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    $\begingroup$ @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. $\endgroup$
    – Matthew Kvalheim
    Commented Nov 10, 2022 at 16:17

1 Answer 1

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I guess that by "isotopy" you mean isotopy of embeddings.

If $B$ is the $n$-sphere and $E=B\times R$ then your question is equivalent to the $(n+1)$-dimensional Schönfliess theorem: known for $n\neq 3$, and a big open question for $n=3$.

To answer your complementary question: if there is a self-diffeomorphism $f$ of $R^{n+1}$ sending the unit sphere $S^n\subset R^{n+1}$ to your embedded sphere $h(S^n)$, then you can always arrange that $f$ is isotopic to the identity (with compact support). Indeed, you can first arrange that $f$ is orientation-preserving; hence $f$ restricted to the unit ball $B^{n+1}$ is an orientation-preserving embedding of the ball; and such an embedding is necessarily isotopic to the inclusion, by the "Alexander trick" (shrinking the ball to its center).

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  • $\begingroup$ Many thanks (belatedly) for your answer. One question: do you mean the Palais disk theorem rather than the Alexander trick? $\endgroup$
    – Matthew Kvalheim
    Commented May 30 at 18:23
  • $\begingroup$ Yes, the Palais disk theorem if you like better, it is almost the same. $\endgroup$
    – Gael Meigniez
    Commented Jun 13 at 6:55

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