TITLE: How "far" a differential form is from an exterior product
QUESTION [6 upvotes]: Consider two differential manifolds $X$ and $Y$.
Consider now a differential form (of any order) $\omega$ on $X\times Y$. The easiest example is taking $\omega=\xi\wedge\eta$, where $\xi$ is a differential form on $X$ and $\eta$ is a differential form on $Y$. 
In general, any form $\omega$ is not such an exterior product. Is there a way to "measure" how much $\omega$ differs from being such a decomposition?
I'm looking for something in the "spirit" of: 

The rank of a matrix measures how much the matrix differs from being a tensor product of two vectors. 

Is there an analogous quantity for differential forms?
Thanks.

REPLY [4 votes]: Let $V$ denote an $n$-dim vector space (over $\mathbb{R}$, say).  Recall that a $p$-form $\alpha \in \Lambda^p(V^*)$ is decomposable (or simple) iff it can be written as a wedge product of $1$-forms -- i.e., $\alpha = \omega_1 \wedge \cdots \wedge \omega_p$, with each $\omega_i \in \Lambda^1(V^*) = V^*$.
Definition: Let $\alpha \in \Lambda^p(V^*)$ be a $p$-form.  The space of linear divisors of $\alpha$ is
$$L_\alpha := \{\omega \in V^* \colon \omega \wedge \alpha = 0\} \subset V^*.$$

Exercise: Let $\alpha \in \Lambda^p(V^*)$ be a $p$-form.  Then:
(a) $\dim(L_\alpha) \leq p$.
(b) $\dim(L_\alpha) = p$ $\iff$ $\alpha$ is decomposable.

The reason for the name "linear divisors" has to do with the following proposition and corollary:

Prop: Let $\alpha \in \Lambda^p(V^*)$.  Let $\{\omega^1, \ldots, \omega^q\}$ be a basis of $L_\alpha$.  Then we can write
$$\alpha = \omega^1 \wedge \cdots \wedge \omega^q \wedge \pi$$
for some $\pi \in \Lambda^{p-q}(V^*)$.

That is, we can "divide" $\alpha$ by one-forms $q$ times.  Said another way, $q = \dim(L_\alpha)$ tells us how many times we can divide $\alpha$ by one-forms.

Cor: Let $\alpha \in \Lambda^p(V^*)$.  Let $\omega \in V^*$, $\omega \neq 0$.  Then $\alpha = \omega \wedge \psi$ for some $\psi \in \Lambda^{p-1}(V^*)$ if and only if $\alpha \wedge \omega = 0$.

That is, $\omega$ "divides" $\alpha$ if and only if $\omega \in L_\alpha$.  This explains the name.

References
Agricola, Friedrich, "Global Analysis: Differential Forms in Analysis, Geometry, and Physics."
Bryant, Chern, Gardner, Goldschmidt, Griffiths, "Exterior Differential Systems."