Decomposable k-form

A differential $k$ -form $ω$ in the exterior algebra $Λ V^{*}$ for certain vector space $V$ is said to be decomposable if there exist $k$ 1-forms $α_{i}$ such that

ω = α_{1} \land \dots \land α_{k} .

Forms of degree $0$ , $1$ , $d i m V - 1$ and $d i m V$ are always decomposable.

Example of non decomposable: in $R^{4}$ , the 2-form $e^{1} \land e^{2} + e^{3} \land e^{4}$ .

Important property:

Lemma
Suppose $V$ is a $n$ -dimensional vector space and $η$ is a $k$ -form in the exterior algebra $Λ V^{*}$ . Consider the linear map

\hat{η} : V \to Λ^{k - 1} V^{*}

defined by $\hat{η} (v) = i_{v} η$ (interior product). The $η$ is a decomposable k-form if and only if $Ker (\hat{η})$ has codimension $k$ in $V$ . $◼$

Proof
Found here.
If $η$ is decomposable, $η = e^{1} \land \dots \land e^{k}$ , with $e^{j}$ linearly independent. We can complete them to a basis for $V^{*}$ , $⟨ e^{1}, \dots, e^{n} ⟩$ , with dual basis $⟨ e_{1}, \dots, e_{n} ⟩$ . For an arbitrary $v = \sum v^{i} e_{i} \in V$ we have

\hat{η} (v) = \sum_{i = 1}^{k} (- 1)^{k} v^{i} e^{1} \land \dots \land \hat{e^{i}} \land \dots \land e^{k}

where the hat indicates omission. This is zero if and only if $v_{1} = v_{2} = \dots = v_{k} = 0$ , so $Ker (\hat{η})$ has codimension $k$ .

Conversely, if $Ker (\hat{η})$ has dimension $n - k$ , we can look for a basis $(e_{1}, \dots, e_{n})$ of $V$ such that $Ker (\hat{η}) = span ({e_{k + 1}, \dots, e_{n}})$ . Suppose

\begin{matrix} (1) & η = \sum_{J} η_{J} e^{j_{1}} \land \dots \land e^{j k} \end{matrix}

for increasing multiindices $J = (j_{1}, \dots, j_{k})$ . If $J$ is such that $j_{k} > k$ then $e_{j_{k}} \in K e r (\hat{η})$ and so

η_{J} = η (e_{j_{1}}, \dots, e_{j_{k}}) = (- 1)^{k - 1} \hat{η} (e_{j_{k}}) (e_{j_{1}}, \dots, e_{j_{k - 1}}) = 0

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

η = c e^{1} \land \dots \land e^{k} .

$◼$

Lemma. Let $ω \in Λ^{k} (U)$ and $X$ a vector field such that $X ⌟ ω \neq 0$ . Then $X ⌟ ω$ 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 $D = S ({Y_{1}, \dots, Y_{m}})$ a distribution of vector fields on an open subset of $R^{n}$ , and $m \leq n - 1$ . Then

Y_{1} ⌟ \dots ⌟ Y_{m} (d x_{1} \land \dots \land d x_{n})

is a decomposable $(n - m)$ -form. In particular is the wedge product of $n - m$ independent generators of the Pfaffian system $D^{\circ}$ (see dual description of the distribution). $◼$

More info about decomposable k-forms: here.

I am not sure but:

a $k$ -form decomposable define a distribution involutive if $d ω = η \land ω$
$Y$ is a symmetry if $L_{Y} ω = β ω$ bla bla bla...