3
$\begingroup$

Let $f(x)=\frac{1}{\sin(\pi x)}$ for $x\in (0, 1)$ and let $\Gamma=\left\{(x,f(x)): x\in (0, 1)\subset \mathbb{R}^2\right\}$ be its graph.

For any set $X\subset \mathbb{R}^2$ and $\lambda>0$ and $\mathbf{v}\in \mathbb{R}^2$, let $\lambda X+\mathbf{v}=\left\{\lambda \mathbf{x}+\mathbf{v}: \mathbf{x}\in X\right\}$.

Let $\mathcal{F}_0$ be the (smooth) foliation of $(1,2)\times \mathbb{R}$ whose leaves are $\Gamma+t \mathbf{e}_2+\mathbf{e}_1$ for $t\in \mathbb{R}$. For any integer $n$ let $\mathcal{F}_n=2^n \mathcal{F}_0$ be the corresponding (smooth) foliation of $(2^n, 2^{n+1})\times\mathbb R$.

We construct a "foliation," $\mathcal{F}$, of $\mathbb{R}^2$ by taking the leaves of $\mathcal{F}$ to be the leaves of $\bigcup_{n=-\infty}^\infty \mathcal{F}_n$ together with the vertical lines at $x=2^n$ for all integers $n$ and the vertical lines $x=t$ for all $t\leq 0$.

My question is whether this is actually a $C^0$ foliation. I'm having a lot of trouble seeing how it could be true for any point on the line $x=0$, but I also don't really know how to show that it isn't (it's obviously not a $C^1$ foliation).

Added

By $C^0$ foliation, I mean that for each point $p\in \mathbb{R}^2$, there is a neighborhood $U_p$ and a homeomorphism $\psi:U_p\to \mathbb{R}_x\times \mathbb{R}_y$ so that $\psi^{-1}(\mathbb{R}\times \{ s\})$ is a connected component of $U_p\cap \sigma$ for $\sigma$ a leaf of the foliation.

$\endgroup$
5
$\begingroup$

No, your partition of $\mathbb R^2$ into curves is not a foliation in the sense you provide. To see this, you only have to notice that any neighborhood $U$ of $(0 , y_0)$ disconnects some curves and does not others. Being given a rectifying chart $(\psi , U)$ and a strict subdomain $U'\subset U$ with smooth boundary, there exists a limiting point $p\notin \{x=0\}$ where two components belonging to the same global curve meet (it is a tangency point between $\partial U'$ and the curve). The image $\psi(U')$ must therefore disconnect some lines $\mathbb R\times\{s\}$ arbitrary close to $\psi(p)$. As a conclusion it is not homeomorphic to another $\mathbb R^2$ through an homeomorphism leaving the vertical lines globally invariant (which is what you want to achieve).

That being said, there does exist a non-constant continuous first integral $F : \mathbb R^2\to\mathbb R$. (This fact underlines the genuine difference with $C^1$ foliations, for which the local inverse theorem applies)

Define $$g : (x,y)\mapsto y-\frac{1}{\sin(\pi x)}.$$ Notice that the function $$F_0 : (x,y)\longmapsto\frac{1}{1+|g(x,y)|}$$ is a continuous first integral of $\mathcal F_0$ on the closed rectangle $[1,2]\times \mathbb R$. Define in the same fashion $F_n$ on $[2^n,2^{n+1}]\times \mathbb R$ by $$F_n(x,y):=2^{-|n|}F_0(2^{-n}x , 2^{-n}y)$$ which is a first integral of $\mathcal F_n$ on the closed rectangle.

These functions patch together, for they vanish along the vertical lines $\{x=2^n\}$, to give a function $F$ continuous on $]0,+\infty[\times \mathbb R$. Simply define $F(x,y):=x$ if $x\leq 0$. This extension is continuous because each $F_n$ is bounded by $2^{-|n|}$.

$\endgroup$
  • $\begingroup$ Sorry I should have specified. By $C^0$ foliation, I mean that for each point $p\in \mathbb{R}^2$, there is a neighborhood $U_p$ and a homeomorphism $\psi:U_p\to \mathbb{R}_x\times \mathbb{R}_y$ so that $\psi^{-1}(\mathbb{R}\times \{ s\})$ is a connected component of $U_p\cap \sigma$ for $\sigma$ a leaf of the foliation. This seems to be a stronger condition than what you have shown to exist, but I'm not sure... $\endgroup$ – foliations Feb 18 '15 at 13:08
  • $\begingroup$ @foliations Yes indeed, your requirement is stronger. I will amend my answer. $\endgroup$ – Loïc Teyssier Feb 18 '15 at 15:17
  • $\begingroup$ Thanks for the update. I'm having a bit of trouble following the logic of the first paragraph, do you mind expanding on it? $\endgroup$ – foliations Feb 18 '15 at 23:00
  • $\begingroup$ After sleeping on it, I now understand. Thanks so much for your assistance! $\endgroup$ – foliations Feb 19 '15 at 23:18
13
$\begingroup$

Wanted to add a picture of the foliation but could not place the image into a comment (can it be done?) so put it in an answer

a sketch of the foliation

(just a sketch of it)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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