A differential $k$-form $\omega$ in the exterior algebra $\Lambda V^*$ for certain vector space $V$ is said to be decomposable if there exist $k$ 1-forms $\alpha_i$ such that
$$ \omega=\alpha_1\wedge \ldots \wedge \alpha_k. $$Forms of degree $0$, $1$, $dim V-1$ and $dim V$ are always decomposable.
Example of non decomposable: in $\mathbb{R}^4$, the 2-form $e^1\wedge e^2+e^3\wedge e^4$.
Important property:
Lemma
Suppose $V$ is a $n$-dimensional vector space and $\eta$ is a $k$-form in the exterior algebra $\Lambda V^*$. Consider the linear map
$$ \hat{\eta}:V\to \Lambda^{k-1}V^* $$defined by $\hat{\eta}(v)=i_v\eta$ (interior product). The $\eta$ is a decomposable k-form if and only if $\mbox{Ker}(\hat{\eta})$ has codimension $k$ in $V$. $\blacksquare$
Proof
Found here.
If $\eta$ is decomposable, $\eta=e^1\wedge \ldots \wedge e^k$, with $e^j$ linearly independent. We can complete them to a basis for $V^*$, $\langle e^1,\ldots,e^n\rangle$, with dual basis $\langle e_1,\ldots,e_n\rangle$. For an arbitrary $v=\sum v^i e_i \in V$ we have
$$ \hat{\eta}(v)=\sum_{i=1}^{k}(-1)^{k} v^{i} e^{1} \wedge \cdots \wedge \widehat{e^{i}} \wedge \cdots \wedge e^{k} $$where the hat indicates omission. This is zero if and only if $v_1=v_2=\ldots=v_k=0$, so $\mbox{Ker}(\hat{\eta})$ has codimension $k$.
Conversely, if $\mbox{Ker}(\hat{\eta})$ has dimension $n-k$, we can look for a basis $(e_1,\ldots,e_{n})$of $V$ such that $\mbox{Ker}(\hat{\eta})=\mbox{span}(\{e_{k+1},\ldots,e_n\})$. Suppose
$$ \eta=\sum_{J} \eta_{J} e^{j_{1}} \wedge \cdots \wedge e^{j k} \tag{1} $$for increasing multiindices $J=(j_1,\ldots,j_k)$. If $J$ is such that $j_k>k$ then $e_{j_k}\in Ker(\hat{\eta})$ and so
$$ \eta_{J}=\eta\left(e_{j_{1}}, \ldots, e_{j_{k}}\right)=(-1)^{k-1} \hat{\eta}\left(e_{j_{k}}\right)\left(e_{j_{1}}, \ldots, e_{j_{k-1}}\right)=0 $$so the only term that survive in $(1)$ is the associated to $J=(1,\ldots,k)$, and therefore
$$ \eta=c e^{1} \wedge \cdots \wedge e^{k}. $$$\blacksquare$Lemma. Let $\omega \in \Lambda^k(U)$ and $X$ a vector field such that $X \lrcorner \omega \neq 0$. Then $X\lrcorner \omega$ is decomposable if and only if $\omega$ is decomposable.$\blacksquare$
Proof. It is in @barcoThesis Lemma 3.2.2.$\blacksquare$
As a consequence we have:
Corollary. Let $D=\mathcal{S}(\{Y_1,\ldots,Y_m\})$ a distribution of vector fields on an open subset of $\mathbb R^n$, and $m\leq n-1$. Then
$$ Y_1\lrcorner\ldots \lrcorner Y_m (dx_1\wedge \cdots \wedge dx_n) $$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). $\blacksquare$
More info about decomposable k-forms: here.
I am not sure but:
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Author of the notes: Antonio J. Pan-Collantes
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