A differential -form in the exterior algebra for certain vector space is said to be decomposable if there exist 1-forms such that
Forms of degree , , and are always decomposable.
Example of non decomposable: in , the 2-form .
Important property:
Lemma
Suppose is a -dimensional vector space and is a -form in the exterior algebra. Consider the linear map
defined by (interior product). The is a decomposable k-form if and only if has codimension in .
Proof
Found here.
If is decomposable, , with linearly independent. We can complete them to a basis for , , with dual basis . For an arbitrary we have
where the hat indicates omission. This is zero if and only if , so has codimension .
Conversely, if has dimension , we can look for a basis of such that . Suppose
for increasing multiindices . If is such that then and so
so the only term that survive in is the associated to , and therefore
Lemma. Let and a vector field such that . Then is decomposable if and only if is decomposable. Proof. It is in @barcoThesis Lemma 3.2.2.
As a consequence we have:
Corollary. Let a distribution of vector fields on an open subset of , and . Then
is a decomposable -form. In particular is the wedge product of independent generators of the Pfaffian system (see dual description of the distribution).