The following question comes up in the study of metrics with the same unparameterized geodesics in Riemannian and Finsler geometry:
Question. Let $M$ be a closed manifold of dimension $2n+1$ and let $\omega_1$ and $\omega_2$ be two maximally non-degenerate closed $2$-forms on $M$ (i.e. their kernels are one-dimensional). Assume that for every integer $k$ from $0$ to $n-1$, the differential $2n$-form $\omega_1^k \wedge \omega_2^{n-k}$ is a constant multiple of $\omega_1^n$. Does this imply that $\omega_2$ is a constant multiple of $\omega_1$?
Added after Sebastian's answer: In abstracting the situation from the Riemannian and Finsler motivation I threw the baby out with the bath water. For future reference (to my next question) I'll add a few remarks:
If $M$, $\omega_1$, and $\omega_2$ are as in the question, the characteristic foliations of $\omega_1$ an $\omega_2$ coincide if and only if the $2n$-form $\omega_2^n$ equals $\omega_1^n$ times a nowhere-zero function. This function, let's call it $\nu$, is necessarily constant on the leaves of the foliation. Indeed, if $X$ is a vector field tangent to the foliation $$ 0 = \mathcal{L}_X \omega_2^n = \mathcal{L}_X \nu \omega_1^n = X(\nu)\omega_1^n + \nu \mathcal{L}_X \omega_1^n = X(\nu)\omega_1^n. $$ So if every smooth function constant along the characteristics is a constant, then the form $\omega_2^n$ is a constant multiple of $\omega_1^n$. The same happens for the forms $\omega_1^k \wedge \omega_2^{n-k}$ $(0 \leq k < n)$: they are multiples of $\omega_1^n$ and the multiples are "integrals of motion" so that if all integrals of motions are constant, then these forms are all constant multiples of $\omega_1^n$. Sebastian's answer shows that the converse is not true: these $2n$ forms may all be constant multiples of each other and the characteristic foliation may nevertheless admit lots of non-constant integrals of motion.
