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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?).

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    $\begingroup$ BTW, you can write your solution without a partition of unity argument by choosing a Riemannian metric and letting $\eta = \star \omega$. Then $\omega \wedge \eta = \omega \wedge (\star \omega) = \left<\omega, \omega \right> \Omega$ so if $\omega_p \neq 0$ then $(\omega \wedge \eta)|_{p} \neq 0$. In a local orthonormal frame, your formula for $\eta$ is precisely the formula for $\star \omega$. $\endgroup$ – levap Mar 16 '18 at 10:40
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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.

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