Let $M$ be a smooth oriented $d$-dimensional manifold, and let $\omega \in \Omega^k(M)$, $k<d$. Does there always exist a closed $(d-k)-$degree form $\eta \in \Omega^{d-k}(M)$ such that
$\omega_p \neq 0 \Rightarrow \omega_p \wedge \eta_p \neq 0$?
I think that if we relax the requirement $\eta$ should be closed, then I have a solution:
For the local case, we can write $\omega=f_Idx^I$, where $dx^I=dx^{i_1} \wedge \dots \wedge dx^{i_k}$ for a multi-index $I=(i_1,\dots,i_k)$.
Then we can choose $\eta=f_Idx^{I^c}$, where $I^{c}$ is the complement of $I$ in $(1,\dots,n)$. We arrange the order of the indices in $I^c$ to make sure $dx^I \wedge dx^{I^c}=dx^1 \wedge \dots \wedge dx^n$. Then $$ \omega \wedge \eta=(f_Idx^I)\wedge(f_Jdx^{J^c})=f_If_J dx^I \wedge dx^{J^c}=f_I^2 dx^I \wedge dx^{I^c}=(\sum_I f_I^2)dx^1 \wedge \dots \wedge dx^n$$
is non zero, whenever $\omega$ is non-zero.
Now, we can use a partition of unity argument* to obtain a global solution. (Note our local solution can always be chosen in such a way that $ \omega \wedge \eta|_{\text{loc}}$ would be a positive top-form; convex combination of such positive forms is also positive, of course).
I am interested to know if $\eta$ can be chosen to be closed, even in the local problem? (Does it help if I know $\omega$ is closed?)
*I don't think the formula $\eta=f_Idx^{I^c}$ is coordinate-independent, so using a partition seems necessary here. (Perhaps there is an immediate way to see this formula is indeed not well-defined when we change coordinates?).
This is not possible in general, even if $\omega$ is closed.
To see why, let $M^d$ be any closed oriented $d$-dimensional manifold with $d \geq 2$ and let $\Omega$ be a volume form on $M$.
Choose a Morse function $f \colon M \rightarrow \mathbb{R}$ and set $Z = \{ p \in M \, | \, df|_{p} = 0 \}$. Since $f$ is Morse, $Z$ is a finite set of isolated points. Let $\omega = df \in \Omega^1(M)$ and assume that we can find a closed $\eta \in \Omega^{d-1}(M)$ such that
$$ p \notin Z \implies \omega_{p} \wedge \eta_{p} \neq 0. $$
We can always write $\omega \wedge \eta = g\Omega$ for a smooth $g \colon M \rightarrow \mathbb{R}$ and the condition above implies that if $p \notin Z$ then $g(p) \neq 0$. Since $d \geq 2$, the space $M \setminus Z$ is connected and so we have either $g(p) > 0$ for all $p \in M \setminus Z$ or $g(p) < 0$ for all $p \in M \setminus Z$. By replacing $\eta$ with $-\eta$ we can assume that $g(p) > 0$ for all $p \in M \setminus Z$ (and $g(p) \geq 0$ for all $p \in M$). But then
$$ 0 = \int_{\partial M} f\eta = \int_M d(f \eta) = \int_M df \wedge \eta = \int_M \omega \wedge \eta = \int_M g \Omega > 0, $$
a contradiction.