Constructing a 2 form that does not vanish on a space from a one form that does

Question

This may be a silly question...

Given a nonzero one-form $\omega$ that vanishes on a subspace $W$ (with dimension larger than $2$), is it possible to find another one form $\phi$ such that $\omega\wedge\phi$ does not vanish on $W$?

Answer 1

It's not possible. I assume you're just thinking about multilinear algebra since you say "subspace," but the reason I will give below translates easily to differential forms on manifolds.

By the definition of the wedge product, we have that for $1$-forms $\omega$ and $\phi$ on a vector space $V$, $$(\omega \wedge \phi)(u, v) = \frac{1}{2}(\omega(u)\phi(v) - \omega(v)\phi(u))$$ for any $u, v \in V$. Now if $W$ is a subspace of $V$ such that $\omega(u) = 0$ for all $u \in W$, the above formula implies that \begin{align} (\omega \wedge \phi)(u, v) & = \frac{1}{2}(\omega(u)\phi(v) - \omega(v)\phi(u)) \\ & = \frac{1}{2}(0 \cdot \phi(v) - 0 \cdot \phi(u)) \\ & = 0 \end{align} for all $u, v \in W$. Hence if $\omega$ vanishes on $W$, $\omega \wedge \phi$ vanishes on $W$ for any $k$-form $\phi$ (the argument extends to $k > 1$ by just writing the general definition of the wedge product).

Answer 2

A different way to see that it's not possible is to read what it means for a form to vanish on a submanfold.

When you have a submanifold $W\subseteq M$ and an $\omega\in \Omega^k(M)$, then $\omega$ vanishes on $W$ when the pullback by the embedding map $\iota\colon W\to M$ is zero: $\Omega^k(W)\ni\iota^*\omega=0$.

But pullback commutes with wedge product, so for any $\eta$ we have $$\iota^*(\omega\wedge \eta)=\iota^*\omega\wedge\iota^*\eta=0\wedge \iota^*\eta=0$$