Are diffeomorphic sets smoothly deformable into each other?

Question

Given connected, bounded and open sets $U, V\subset \mathbb{R}^n$ and an orientation preserving diffeomorphism $F:U\to V$, is there always an isotopy $H:[0,1]\times \mathbb{R}^n\to\mathbb{R}^n$, s.t. $H(0,\cdot)=id_{\mathbb{R}^n}$ and $H(1,\cdot)_{|U}=F$ (what if we change the domain of $H$ to $[0,1]\times U$ ?)

Answer 1

First, a 2-dimensional example: Let $U\subset {\mathbb C}$ be the open annulus $$ \{z: 2^{-1}<|z|<2\} $$ and $f(z)=z^{-1}, f: U\to U$ is an orientaion-preserving diffeomorphism. It sends the inner circle to the outer circle (turning the annulus "inside-out"). It is now not hard to see that the map $f: U\to U\subset {\mathbb C}=R^2$ is not isotopic to the identity.

Here is an answer to a more interesting version of your question: Given two diffeomorphic connected subsets $U, V$ of $R^n$, is there a always a diffeomorphism $f: U\to V$ which is isotopic to the identity map of $U$?

The answer to this is positive in dimensions 1 (easy) and 2 (tricky) and negative in dimensions $>2$. Namely, take $U$ and $V$ to be open tubular neighborhoods of the trivial and a nontrivial knots in $R^n$ respectively. (Recall that a knot in $R^n$ is a smoothly embedded codimension 2 sphere.) For instance, in the case $n=3$ you can take $U$ to be a tube around a round circle and $V$ to be a tube around the trefoil knot.

On the positive side, a smooth isotopy exists whenever $U$ is diffeomorphic to $R^n$:

Theorem. Suppose that $U, V\subset R^n$ are open subsets diffeomorphic to $R^n$. Then every orientation-preserving diffeomorphism $f: U\to V$ is smoothly isotopic (as a map $U\to R^n$) to the identity map $U\to U$.

Proof. I will need two basic ingredients for the proof:

Lemma 1 (concatenation lemma). Smooth isotopy is an equivalence relation: If $f\sim g, g\sim h$ then $f\sim h$, where $f, g, h$ are diffeomorphic embeddings $M\to N$ between two fixed smooth manifolds and $\sim$ is smooth isotopy.

Using this lemma one reduces the proof of the theorem to the case $U=R^n$. I also need another lemma which is a pleasant exercise in linear algebra:

Lemma 2. The group $Aff_+(R^n)$ of orientation-preserving affine (invertible) maps $R^n\to R^n$ is connected.

Now, we can prove the theorem. Since $f$ is orientation-preserving, we get (by the Taylor formula at $0$) $$ f(x)= A(x) + O(|x|^2), $$
(as $x\to 0$) where $A\in Aff_+(R^n)$. By postcomposing $f$ with $A^{-1}$ and applying the Lemma 2 (and using, again, the concatenation property for smooth isotopies), one reduces the problem to the case $f(0)=0$ and $Df(0)=I$, (the derivative at $0$ is the identity matrix). Now, consider the smooth map $$ F(x,t)= t^{-1} f(tx), x\in R^n, t\in (0,\infty). $$ Clearly, $F(x,1)=f(x)$ and (by the definition of derivative at the origin!) $$ \lim_{t\to 0+} F(x,t)=Df(0)(x)=x. $$ Then one checks that the formula $F(x,0)=x$ defines a smooth extension of the original map $F$ to a map $F: R^n\times [0,\infty)\to R^n$, which is the required isotopy between $f$ and the identity map. qed