Every $2$-form $\omega\in \Omega^2(\mathbb{R}^{2n+1})$ induces a skew-symmetric map
$$
\omega(-,-)\colon\Gamma(T\mathbb{R}^{2n+1})\otimes \Gamma(T\mathbb{R}^{2n+1}) \to C^\infty(\mathbb{R}^{2n+1})
$$
where $\Gamma(T\mathbb{R}^{2n+1})$ denotes the space of vector-fields on $\mathbb{R}^{2n+1}$. It is clear that at every point there exists a non-zero tangent vector $v$ such that $w(p)(v,-)=0$.


Does there always exists a non-zero vector field $\varphi$ such that $\omega(\varphi,-)$ is the constant zero function? Does there always exist such a vector field if $\omega$ is assumed to be closed? In other words is the above pairing over $C^\infty(\mathbb{R}^{2n+1})$ non-degenerate?


The problem obviously lies in the fact that the null spaces of the forms do not have to form a bundle and I don't see a reason why there can not be a two form, where the union of the null spaces does not contain a linebundle.



$2$-Form inducing a non-degenerate form on $\Gamma(T\mathbb{R}^{2n+1})$
The example in $\mathbb{R}^3$ given by
$$
\omega = x\,\mathrm{d}y\wedge\mathrm{d}z + y\,\mathrm{d}z\wedge\mathrm{d}x + z\,\mathrm{d}x\wedge\mathrm{d}y,
$$
shows that the kernel of $\omega$ need not be a line bundle. Away from the origin, the kernel is spanned by the radial vector field
$$
R = x\,\frac{\partial\ }{\partial x} + y\,\frac{\partial\ }{\partial y} + z\,\frac{\partial\ }{\partial z},
$$
and there cannot be a continuous nonvanishing vector field that is a multiple of $R$ everywhere except at the origin.


A similar argument works in the case of the closed $2$-form
$$
\omega = x\,\mathrm{d}y\wedge\mathrm{d}z + y\,\mathrm{d}z\wedge\mathrm{d}x - 2z\,\mathrm{d}x\wedge\mathrm{d}y.
$$


