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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)

Ali Taghavi
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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.

André Henriques
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  • 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? – Ali Taghavi Jan 31 '18 at 09:57
  • The only biholomorphic maps $\mathbb C\to\mathbb C$ are of the form $z\mapsto az+b$, but there exist many diffeomorphisms $\mathbb R^2\to \mathbb R^2$. – André Henriques Jan 31 '18 at 15:51
  • I mean locally (not necessarily global bioholomorphism). – Ali Taghavi Jan 31 '18 at 15:53
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    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. – André Henriques Jan 31 '18 at 15:56