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. 



A question on differential forms and integral invariants
No,
take a 5 dimensional real torus, and $\omega\_1=dx\_1\wedge dx\_2+dx\_3\wedge dx\_4$ and $\omega\_2= dx\_1\wedge dx\_2+dx\_3\wedge dx\_4+dx\_1\wedge dx\_3.$


