Decomposable k-form

A differential k-form ω in the exterior algebra ΛV for certain vector space V is said to be decomposable if there exist k 1-forms αi such that

ω=α1αk.

Forms of degree 0, 1, dimV1 and dimV are always decomposable.

Example of non decomposable: in R4, the 2-form e1e2+e3e4.

Important property:

Lemma
Suppose V is a n-dimensional vector space and η is a k-form in the exterior algebra ΛV. Consider the linear map

η^:VΛk1V

defined by η^(v)=ivη (interior product). The η is a decomposable k-form if and only if Ker(η^) has codimension k in V.

Proof
Found here.
If η is decomposable, η=e1ek, with ej linearly independent. We can complete them to a basis for V, e1,,en, with dual basis e1,,en. For an arbitrary v=vieiV we have

η^(v)=i=1k(1)kvie1ei^ek

where the hat indicates omission. This is zero if and only if v1=v2==vk=0, so Ker(η^) has codimension k.

Conversely, if Ker(η^) has dimension nk, we can look for a basis (e1,,en)of V such that Ker(η^)=span({ek+1,,en}). Suppose

(1)η=JηJej1ejk

for increasing multiindices J=(j1,,jk). If J is such that jk>k then ejkKer(η^) and so

ηJ=η(ej1,,ejk)=(1)k1η^(ejk)(ej1,,ejk1)=0

so the only term that survive in (1) is the associated to J=(1,,k), and therefore

η=ce1ek.

Lemma. Let ωΛk(U) and X a vector field such that Xω0. Then Xω is decomposable if and only if ω is decomposable.
Proof. It is in @barcoThesis Lemma 3.2.2.

As a consequence we have:

Corollary. Let D=S({Y1,,Ym}) a distribution of vector fields on an open subset of Rn, and mn1. Then

Y1Ym(dx1dxn)

is a decomposable (nm)-form. In particular is the wedge product of nm independent generators of the Pfaffian system D (see dual description of the distribution).

More info about decomposable k-forms: here.

I am not sure but: