Calculation of integral manifolds through 1-forms

Question

I will focus to the example that i have and is not at all understandable. Suppose we have the following vector fields on $\mathbb{R^3}$: $X=(1,0,2xz/(1+x^2+y^2))$ and $Y=(0,1,2yz/(1+x^2+y^2))$. We want to find the integral manifolds of the distribution $D$ spanned by those vector fields. In this example want to find an 1-form $ω$ such that $ω(X)=0$ and $ω(Y)=0$. Without any explanation it says that one 1-form that works is $ω=(-1-x^2-y^2)\,d(z/(1+x^2+y^2))$. Then by solving $ω=0$ we get the equations of the integral manifold. I cannot see how one can think that the certain function will do the work. I guess there is an algorithmic method but in any case could some one explain how to work in such problems?? I have almost none knowledge of differential forms theory.

Answer 1

So here's how you should approach it. Let's take $\omega = a\,dx+b\,dy+c\,dz$ In order for $\omega(X)=0$ to hold, we see that if $c=1$, then $a=-2xz/(1+x^2+y^2)$, and in order for $\omega(Y)=0$, we see that if $c=1$, then $b=-2yz/(1+x^2+y^2)$. So we write down $$\omega = -\frac{2xz}{1+x^2+y^2}dx - \frac{2yz}{1+x^2+y^2}dy + dz.$$ Of course, we could take $f\omega$ for any nonvanishing function $f$. Regardless, for this choice of $\omega$ we do have $$\omega = (1+x^2+y^2)d\left(\frac z{1+x^2+y^2}\right),$$ as you can verify immediately with the product rule. I'm not sure why they put in the negative sign.