# Stability Question for Isotopies Between Compact Sets

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!

• 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. – Ryan Budney Mar 12 at 2:52
• 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. – John Samples Mar 12 at 3:05
• 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? – John Samples Mar 16 at 3:33
• 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. – Tom Goodwillie Mar 17 at 23:06

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 $$B(0,2R)$$.
• 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 – John Samples Mar 18 at 1:14
• 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. – John Samples Apr 14 at 20:40
• 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. – John Samples Apr 14 at 21:17