I've been given the following vector field on the manifold $M = \mathbb{R}^3 \setminus \{\text{$z$-axis}\}$ $$X = (x-2y)\,\frac{\partial}{\partial x} + (2x+y)\, \frac{\partial}{\partial y} + z\, \frac{\partial}{\partial z},\quad Y = -y\,\frac{\partial}{\partial x} + x\, \frac{\partial}{\partial y}$$
It was straightforward to show that they're linearly independent and commute, therefore they define an involutive rank $2$ distribution, say $D$. The question is find and describe its integral submanifolds. I tried the following methods:
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I was able to figure out that the form $\omega = – xz\,dx – yz\,dy + (x^2 + y^2)\,dz$ annihilates $X$ and $Y$ and $d\omega = 3x\,dx\,dz + 3y\,dy\,dz$. Since our distribution is involutive, naturally $d\omega\wedge \omega = 0$. But I've been unable to figure out an $\alpha$ such that $d\omega = \omega\wedge\alpha$. This could then help me figure out the integral submanifolds
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Alternatively, we can the flows of $X$ and $Y$, which I did after a painstaking computations but I'm not sure how to proceed after that.
Any help would be appreciated.
Best Answer
Note that there is an easy integrating factor to guess for your $\omega$: Multiply through by $f=\dfrac1{(x^2+y^2)z}$ and the resulting form is exact. Indeed, $f\omega = d(-\log r + \log z)$, where $r=\sqrt{x^2+y^2}$. Integral manifolds are then $z=cr$ for various constants $c$.
Following Sophus Lie, there is a good way to look for such integrating factors. If you see an obvious one-parameter group leaving the distribution invariant (or sending $\omega$ to a functional multiple of $\omega$), let $V$ be its infinitesimal generator and let $f=1/\omega(V)$. Then you can check that $f\omega$ will be closed. In your case, the obvious one-parameter group is given by radial stretches, and $V=x\dfrac{\partial}{\partial x}+y\dfrac{\partial}{\partial y}+z\dfrac{\partial}{\partial z}$. Unfortunately, the numbers work out to $\omega(V)=0$ — in other words, $\mathscr L_V\omega = 0$ rather than just a general functional multiple of $\omega$. But, still, we will see the appearance of $(x^2+y^2)z$ when we do this computation.
EDIT: A natural way to avoid the issue with $z=0$ about which you're worried is to introduce a homogenizing variable $u$ with $z=ur$, $r=\sqrt{x^2+y^2}$. Then $$\omega = -zr\,dr + r^2\,dz = -ur^2\,dr + r^2(u\,dr + r\,du) = r^3\,du,$$ and so, naturally, $\dfrac 1{r^3}\omega = du$ is exact. (So, here, in the new coordinate system, $1/r^3$ does become the integrating factor!) Oh, and finally, the integral manifolds are patently obvious: They're given by $u=c$ for all constants $c$.