I'm preparing for some exams and this problem is from an older exam:
"On $\mathbb{R}^4$, with standard coordinates $(x, y, z, t)$, consider the 1-form $\theta = x \, dy − y \, dx + z \, dt − t \, dz$. Is there a smooth immersion $j : \mathbb{R}^3 \to \mathbb{R}^4$ such that $j^∗ \theta = 0$ everywhere?"
I'm not sure of where to start besides of course supposing there is such an immersion and seeing if it leads to a contradiction. I've also differentiated the 1-form just to see if any patterns come of it but there are none as far as I can tell. However, the 1-form does look very symmetric and I'm sure it was chosen with care.
I was told to look at some integrability conditions such as the kernel of some map and apply Frobenius' Theorem. However, I only know of the version of Frobenius' theorem in the context of taking the Lie bracket of vector fields. Any help or hints are helpful, thank you!
Recall that for any smooth fields $X$ and $Y$, one has: $$\mathrm{d}\theta(X,Y)=X(\theta(Y))-Y(\theta(X))-\theta([X,Y]).$$ Let $\xi:=\ker(\theta)$, then according to Frobenius' theorem $\xi$ is integrable if, and only, if $\theta\wedge\mathrm{d}\theta=0$.
In this case, one has $\theta\wedge\mathrm{d}\theta\neq 0$, so that there is no immersed hypersurface of $\mathbb{R}^3$ with tangent bundle $\xi$. Therefore, such an immersion cannot exist, since $j^*\theta=0$ exactly means that $Tj(\mathbb{R}^3)\subset\xi$.