A differentiable transformation of R^n at each point has an invertible derivative. Does it imply that the transformation is a global diffeomorphism?
3
-
4$\begingroup$ Look on the exponential map on the real line... $\endgroup$– PetyaCommented Dec 4, 2011 at 17:37
-
$\begingroup$ I think you mean the exponential function on the complex numbers (considered as ${\mathbb R}^2$). The statement would be true for $C^1$ maps of $\mathbb R$ to itself. $\endgroup$– Robert IsraelCommented Dec 4, 2011 at 18:21
-
2$\begingroup$ Robert -- I think Petya meant that the exponential is not a diffeomorphism $\mathbb{R}\to\mathbb{R}$. $\endgroup$– algoriCommented Dec 4, 2011 at 18:36
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Let $f:\mathbb{R}^n\rightarrow M\subseteq\mathbb{R}^n$ be the differentiable transformation, with $M=f(\mathbb{R}^n)$. If $M\neq\mathbb{R}^n$, then obviously $f$ isn't an global diffeomorphism of $\mathbb{R}^n$. But it is global diffeomorphism between $\mathbb{R}^n$ and $M$.
*This was intended to be a comment not an answer, but I cannot comment, sorry. *
-
1$\begingroup$ stephanos -- re "it is a global diffeomorphism ...": it isn't in general: take $f$ to be the complex exponential, then $M=\mathbb{R}^2$ minus the origin, which is not diffeomorphic to the plane. Even if the image of $f$ is the whole $\mathbb{R}^n$, then $f$ is not necessarily a diffeomorphism. To see this set $g$ to be a degree 3 complex 1-variable polynomial whose derivative has no double roots. The universal cover of $U=\mathbb{C}\setminus$ the roots of $f'$ is $\mathbb{R}^2$; set $h$ to be the universal covering map $\mathbb{R}^2\to U$. Set $f=g\circ h$.... $\endgroup$– algoriCommented Dec 4, 2011 at 21:08
-
$\begingroup$ ... this is a local diffeomorphism with image $\mathbb{C}=\mathbb{R}^2$ but not a global diffeomorphism as it takes each value infinitely many times. $\endgroup$– algoriCommented Dec 4, 2011 at 21:10