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Let $a,b \in R^n$.

Show that there is a diffeomorphism of $R^n$ carrying $a$ to $b$ which is the identity outside of an open set.

I think I have a proof of this using integral curves but I would like to see if there's another proof; and more specifically if such a diffeomorphism can be described explicitly (using bump functions or otherwise).

Any help appreciated.

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2 Answers

up vote 2 down vote accepted

Let $\Psi$ be a bump function such as the one described here: http://en.wikipedia.org/wiki/Bump_function#Examples

Let $f(x)=ub+(1-u)x$, where $u=\Psi(|x-a|^2)/\Psi(0)$.

[EDIT] Bernard has a good point, which is that it's not obvious that $f^{-1}$ is smooth. In fact, I think the real issue is whether or not $f$ is invertible. So here's a patched-up version of the above map.

$u=\Psi(k|x-a|^2)/\Psi(0)$

$k=(b-a)^{-2}/1000$

$f(x)=(x+b-a)u+x(1-u)$

The specific bump function $\Psi(z)$ given in the WP article is clearly smooth, and has a smooth inverse, everywhere except possibly at $z=0$ and $z\ge 1$. The factor of $1/1000$ ensures that $f$ is always invertible, since $\Psi$ never has a slope as large as 1000. I don't think we have an issue at $z=0$, since the constancy of $\Psi$ there just means that $f$ looks like an identity map at $a$. Ditto at $z\ge1$.

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OK, clearly f is smooth, but how would you show that $f^{-1}$ is smooth? –  Bernard Oct 12 '12 at 3:38
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Find an open ball containing both points, then since the line joining $a$ and $b$ is closed, we can use Urysohns lemma to produce a smooth function $\rho$ equaling 1 on the line and having compact support in our big ball. Now let $X$ be the constant vector field on $R^n$ which points in the same direction as the line joining the two points. so that $\rho(x) X$ is a smooth, compactly supported vector field. The flow of $\rho(x)X$ will give the diffeomorphism you want.

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Yes, that's the solution I meant above when I mentioned integral curves. I am asking for other solutions (not using flows). –  Bernard Oct 12 '12 at 3:39
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