Let ${\Omega _1}$,${\Omega _2}$ be two open sets in $\mathbb R^n$ and $f$ is a diffeomorphism between them. For every $x$ in ${\Omega _1}$, is there an open set $\Omega_{x} \subset \Omega_1$ and a diffeomorphism $g$ of $\mathbb R^n$ such that $g=f$ when restricted to $\Omega_{x}$?
The extension of diffeomorphism
1 Answers
Assume $x\in \Omega_1$ such that $Df(x) = A$. Wolg $x=0$ and $f(0) =0$, otherwise the following reasoning needs to be 'translated'. Now let $\phi (x)= \psi(||x||)$ be a smooth radially symmetric cutoff function which is identically $0$ outside of $B_t(0)$ and $=1$ inside, say, $B_{t/2}(0)$, s.t. $\psi'(t) \le O(1/t)$ . Assume $B_t(0)\subset \Omega_1.$
Now look at
$ F(x):= (1-\phi(x)) Ax + \phi(x) f(x) $
which is defined for every $x\in\mathbb{R}^n$ (it's easy to see that this is possible).
For $||x|| \ge t$ this is just $Ax$.
For $||x||< t$ denote $r=||x||$ and calculate for $v$ such that $||v||= 1$ $DF(x) v = -\psi'(r)<\frac{x}{r},v> Ax +(1-\phi(x))Av +\psi'(r)<\frac{x}{r}, v> f(x) + \phi(x) Df(x)v\\ = Av + \phi(x)(Df(x)v-Av) +\psi'(r) <\frac{x}{r}, v>(f(x)-Ax)$
The first term on the RHS is nonzero for nonzero $v$, since $A$ is invertible. Note that it does not depend on $t$, and since the sphere is compact it's norm is bounded from below by a constant.
The second term is arbitrarily small, since $ f$ is continuously differentiable (I do assume continuity of $Df$) and $\phi$ is bounded, if only $t$ is small enough.
Since $f(x)-Ax = o(x)$ the third term also becomes arbitrarily small given the growth rate of $\psi$.
In response to a comment from Leonid Kovalev: from $F(x)-F(y) =Ax-Ay - \phi(x)(Ax-f(x)) +\phi(y)(Ay - f(y))$ it is easy to see that $F$ is 1-1, using the $f(x)-Ax=o(||x||)$ property and the fact that $A$ invertible. The fact that $F$ is onto follows, e.g., by elementary degree theory from the fact that $F=A$ on the boundary of $B_t(0)$. It is probably possible to see this with a more elementary reasoning, for which I don't have the time right now.
An alternative reasoning which you may find more satisfying: according to Thm. 1.7 in Hirsch's Differential Topology the set of diffeomorphisms is open in $C^1(\mathbb{R}^n, \mathbb{R}^n)$. The $F$ I have constructed is arbitrarily close to $x\mapsto Ax$ in $C^1$.
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0Nice proof. BTW, that $F$ is a diffeomorphism is easy if we notice that $F$ is proper. – 2012-05-24