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Consider the linear functional $\omega \in (\mathbb{R}^3)^{*}$, gived by $ω(x,y,z)=dy+y⋅dx$ (The base ${dx,dy,dz}$ for $(\mathbb{R}^3)^{*}$ is dual in the canonical base of $\mathbb{R}^3$ $\{ e_1,e_2,e_3 \})$.The $2$-distribution $D_2:\mathbb{R}^3→T\mathbb{R}^3;(x,y,z)↦ker(ω(x,y,z))$, is $D_2$ integrable?

First, i'm trying to find the kernel, so:

$\begin{equation} \begin{aligned} \omega(x,y,z) &= dy(x,y,z) - z\cdot dx(x,y,z) \\ & = y - zx \\ \end{aligned} \end{equation}$

Then, the kernel is the surface $y-zx=0$, but, i don't know how to express that surface as a span of vectors in $\mathbb{R}^3$.

sango
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    You're still making the same mistake I warned you about earlier. $\omega$ is actually NOT in $(\Bbb R^3)^*$, as it varies from point to point. But at the point $(x,y,z)$, you want to apply to a tangent vector $(u,v,w)$, and $(u,v,w)\in\ker \omega(x,y,z) \iff v+yu=0$. – Ted Shifrin Sep 28 '19 at 16:45
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    @TedShifrin, you right. But i still without understand how to get the vector fields that span the kernel. – sango Sep 28 '19 at 17:35
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    Well, note that $(0,0,w)$ always satisfies this equation. And so does $(u,-yu,0)$. In standard notation, $\partial/\partial z$ and $\partial/\partial x - y,\partial/\partial y$ give a basis. – Ted Shifrin Sep 28 '19 at 17:38
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    @TedShifrin the second basis not is $\frac{\partial}{\partial x} - x \frac{\partial}{\partial y}$ ? – sango Sep 28 '19 at 17:45
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    No. Note that $(dy+y,dx)(\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}) = -y + y = 0$. – Ted Shifrin Sep 28 '19 at 17:49
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    @TedShifrin I've commeted a mistake in my question. $\omega(x,y,z)$ is $dy-z⋅dx$. But, with your help i've comenced to understand, with that $\omega$ i've founded that the vectors $(0,0,v),(u,wu,w)$ satisfies the ecuation then the basis is $\partial / \partial z, \partial / \partial x + z \partial / \partial y+\partial / \partial z$ ? – sango Sep 28 '19 at 18:19
  • No, this is wrong. Please be careful. And this $\omega$ will not be integrable. – Ted Shifrin Sep 28 '19 at 18:39
  • Let us continue this discussion in chat. – sango Sep 28 '19 at 21:12

1 Answers1

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This distribution is integrable. Before I explain why lets prove a small lemma.

Lemma: Let $\omega$ be an one form in $\mathbb R^3$ which is non zero in every point. The distribution $D:=\ker \omega$ is integrable if, and only if, $\omega \wedge d\omega =0$.

Proof: We can write $\omega = \langle N,\cdot \rangle$ for a vector field $N$. If $X$ and $Y$ are vector fields such that $X(p),Y(p) \in D_p$ for all $p$, then we have $$\omega \wedge d\omega(N,X,Y)=\omega(N)d\omega(X,Y)+\omega(Y)d\omega(N,X)+\omega(X)d\omega(Y,N).$$

Now, the exterior derivative can be written as $$d\omega(X,Y) = X \omega(Y)- Y \omega(X)-\omega([X,Y]).$$

By our choice of $X$ and $Y$ we have $\omega(X)=\omega(Y)=0$.

Therefore, $$\omega \wedge d \omega(N,X,Y) = -\langle N,N \rangle \omega([X,Y]).$$ The identity above proves the lemma.

Now, lets analyze your problem. We have $$\omega = dy +ydx.$$ Notice $$\omega \wedge d\omega = (dy+ydx)\wedge(dy\wedge dx) = 0.$$

Therefore, your distribution is integrable.

Hugo
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