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Suppose $X, Y$ are compact sets in $\mathbb{R}^2$ and $F$ is an ambient isotopy carrying $X$ onto $Y$.

Is there an ambient isotopy $F'$ agreeing with $F$ on $X$ and which is constant in a neighborhood of $\infty$?

Basically, to move one compact set to another, you don't need to radically deform the whole space. I was trying to prove it using the Annulus Theorem, and reduced it to this question:

If $J$ is a Jordan curve and $D$ is an open disc containing the trace of $J$ under some isotopy $G$ on $J$, then $G$ can be extended to $\overline{D}$ such that it's fixed on $\partial D$.

In other words, you'll have a Jordan curve sliding around inside a fixed, larger circle and want to extend it to the larger disc in a well-behaved way.

Does anyone know a reference or proof of this? If you can extend it in the annular region, then that's sufficient; you can concoct the extension inside $J$ from that. I've been stuck on it for a few days, now; I know how to prove it when $F$ is smooth, but I don't know how to use that to get the topological version.

Thanks!

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    $\begingroup$ Is your ambient isotopy a smooth family $[0,1] \times \mathbb{R}^n \to \mathbb{R}^n$? If so, the answer is yes. Look for example at Hirsch's proof of the isotopy extension theorem in his textbook. He gives strong control over the support of the isotopy. $\endgroup$
    – Ryan Budney
    Commented Mar 12, 2021 at 2:52
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    $\begingroup$ Mine won't be smooth, it will just be some isotopy of a Jordan curve in the plane, that I want to extend to an annulus such that it's fixed on the outer boundary component. If it happens in a disc instead of annulus, that's also sufficient; I can just replace what happens inside the Jordan Curve with the original $F$. I think I'll restate the question like that, since it seems more natural. $\endgroup$
    – John Samples
    Commented Mar 12, 2021 at 3:05
  • $\begingroup$ There is also a fiber bundle version: We have the closed plane annulus $A$ and a product $A \times I \subset \mathbb{R}^3$ with slices parallel to the $xy$-plane, where the outer cylindrical shell is just the unit circle times $I$ but the inner cylindrical shell just has a continuous family of Jordan curves. Is there an isotopy to a 'standard' annular prism that's slice? $\endgroup$
    – John Samples
    Commented Mar 16, 2021 at 3:33
  • $\begingroup$ For the two-dimensional version you can use complex analysis: surely a Riemann mapping for the region bounded by a Jordan curve depends on the curve continuously in whatever sense you need. $\endgroup$
    – Tom Goodwillie
    Commented Mar 17, 2021 at 23:06

1 Answer 1

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While this might have been known earlier, one way to derive this result is to apply Corollary 1.2 in

Edwards, R. D.; Kirby, R. C., Deformations of spaces of imbeddings, Ann. Math. (2) 93, 63-88 (1971). ZBL0214.50303.

enter image description here

To apply this result (I will do it in $R^n$), start with an isotopy $F(\cdot ,t), t\in [0,1]$ carrying a compact $C=X\subset R^n$ to $Y$ (the second space will play no role); in their notation, $h_t(x)=F(x,t)$, in particular, $h_0=id_X$. Let $K$ be the image $F(X\times [0,1])$ and let $M$ be a closed $n$-ball $B(0,R)$ whose interior contains $K$. According to Corollary 1.2 above, $F$ extends to an isotopy $H: M\times [0,1]\to M$. It is then easy to extend $H$ to an isotopy of $R^n$ which is the identity outside the ball $B(0,2R)$.

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  • $\begingroup$ I'll have to look at this paper, but seems promising! Is $M$ allowed to have boundary? Does their 'isotopy' mean $H$ is onto? If it's constantly a homeo on $M$ then yes we'll have a nice, circular boundary and can explicitly slide the points outside $M$ along the action of $H$ on $\partial(M)$ radially, and then use your gradient trick (on the MSE thread) to get it eventually constant. I've now also been told that it's in researchgate.net/publication/… by Oversteegen but I looked in here already; will have to look again x_x $\endgroup$
    – John Samples
    Commented Mar 18, 2021 at 1:14
  • $\begingroup$ Ok, I don't understand the proof yet but this Edwards-Kirby paper contains several theorems which imply the above result. I will post another answer in this thread in a while (weeks?) extracting/translating a direct proof of these results in the plane, which I assume is much easier. I already planned to look at the Oversteegen papers soon, so I will hopefully have some interesting things to say. $\endgroup$
    – John Samples
    Commented Mar 18, 2021 at 4:25
  • $\begingroup$ Alright, I've gone through it and everything checks out! Especially, $M$ can have a boundary and they're using "isotopy" in the usual sense for that result. $\endgroup$
    – John Samples
    Commented Apr 14, 2021 at 20:40
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    $\begingroup$ Oh yeah, I should add for future readers that you'll actually want $B(0,R)$ to contain the trace of some neighborhood $U$ of $C$. It always exists, but to directly apply the result it should be mentioned. $\endgroup$
    – John Samples
    Commented Apr 14, 2021 at 21:17

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