# Explicit form of the foliation associated to a differential one-form

I'm writing because of a problem in constructiong the foliation associated to a differential one-form.

Explicitely I have the following differential one-form $\theta$ on Minkowski spacetime $\mathcal{M}$:

$$\theta=\frac{1}{\sqrt{(x^{1})^{2}-(x^{0})^{2}}}\left(x^{1}dx^{0}-x^{0}dx^{1}\right)$$ and I want to find the foliation associated to it.

The differential one-form $\theta$ is integrable, in the sense of Frobenius theorem, i.e. $\theta\wedge d\theta=0$.

I'm sorry but I can not provide any more insights because I'm not very confident with foliations.

In the concrete I want to find the explicit form of the leaves of the foliation.

Thank You.

One way is to look at the vector fields annihilate by $\theta$. It is easy to see that these are spanned by $x^0\partial_0 + x^1\partial_1$. This is the radial vector field so the foliation is by lines through the origin.
Note: If you were talking about four-dimensional space (the foliation only cares about the underyling manifold so this is really just a question of $\mathbb R^4$ and not Minkowski space), then a leaf would just be the cartesian product of $\mathbb R^2$ with a line through the origin in (another copy of) $\mathbb R^2$. Also, for $\theta$ to be well-defined you of course want to restrict to the open subset where $x^0 \ne \pm x^1$.
• Thank You Eric O. Korman. For physical reasons I have to restrict myself to $(x^{1})^{2}\geq(x^{0})^{2}$. Why do you say the foliations is by lines through the origin? I guess You refer to the integral curves of $x^{0}\partial_{x^{0}} + x^{1}\partial_{x^{1}}$. Am I wrong? – Ittiolo Apr 9 '14 at 14:41
• Correct, the leaves are the integral curves of $x^0\partial_0 + x^1\partial_1$. So explicitly the leaves are parameterized by $x^0 = v^0 t, y^0 = v^1 t$, where $v^0,v^1$ are constants. – Eric O. Korman Apr 9 '14 at 14:54
• I think that the integral curves of the vector field are determined by the equations $\frac{dx^{0}(t)}{dt}=x^{0}(t)$ and $\frac{dx^{1}(t)}{dt}=x^{1}(t)$, thus I think that we have $x^{0}(t)=x^{0}_{0}\exp(t)$ and $x^{1}(t)=x^{1}_{0}\exp(t)$ where $x^{0}_{0}$ and $x^{1}_{0}$ are constants. – Ittiolo Apr 9 '14 at 15:03