3
$\begingroup$

Assume that we have a $1$ dimensional foliation of $\mathbb{R}^2$. Is there a global diffeomorphism of the plane which maps all leaves of the foliation to curves with non zero curvature? One can consider the same question for a $1$ dimensional foliation of $\mathbb{R}^n$ requiring that all leaves to be transform ed to Frenet curves.

A motivation for this question:

When I was thinking to the following question, I was thinking to the orthonormal frame $\{\gamma',\gamma'' \}$ as a possible resolution to find a metric compatible to our vector field:

Limit cycles as closed geodesics(in negatively or positively curved space)

$\endgroup$

1 Answer 1

1
$\begingroup$

Restrict the standard foliation of $\mathbb R^2$ by horizontal lines to the open subset $$U:=\{(x,y)\in\mathbb R^2\mid y^2\le 1+x^2-x^4\}.$$ Clearly, $U$ is diffeomorphic to $\mathbb R^2$. But I don't think that there exists a diffeomorphism which maps all the leaves to curves without inflection points.

$\endgroup$
4
  • $\begingroup$ Thanks for your answer. Can a bioholomorphic maps in the plane carry $y=x^2$ to $y=x^3$? I mean that: is not the Riemann mapping theorem an obstruction for your statement? $\endgroup$ Jan 31, 2018 at 9:57
  • $\begingroup$ The only biholomorphic maps $\mathbb C\to\mathbb C$ are of the form $z\mapsto az+b$, but there exist many diffeomorphisms $\mathbb R^2\to \mathbb R^2$. $\endgroup$ Jan 31, 2018 at 15:51
  • $\begingroup$ I mean locally (not necessarily global bioholomorphism). $\endgroup$ Jan 31, 2018 at 15:53
  • 1
    $\begingroup$ Yes, locally this is possible: both a real analytic curves, and any real analytic diffeomorphism between these curves extends to a holomorphic diffeomorphism between neighbourhoods of these curves. That's called analytic continuation. $\endgroup$ Jan 31, 2018 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.