# Is a pointwise decomposable differential form smoothly decomposable?

Let $\omega$ be a smooth differential form on a smooth manifold $M$. Suppose $\omega$ is pointwise decomposable, that is for every $p \in M$, $\omega_p=e^1_p \wedge e^2_p \wedge \dots \wedge e^k_p$ for some $e^i_p \in T_p^*M$.

Is it true that $\omega$ is smoothly decomposable? at least locally?

That is, does there exist (locally) smooth one-forms $\omega_i$ such that $\omega=\omega_1 \wedge \omega_2 \dots \wedge \omega_k$?

• Suppose $\omega \in \Lambda^k V$ for a vector space $V^n$. Then the linear map $f_\omega\colon V\rightarrow \Lambda^{k+1}V ,\theta \mapsto \omega \wedge \theta$ has rank $\ge n- k$ if and only if $\omega$ is decomposable. Maybe you can use this. – Jan Bohr Jun 28 '18 at 12:09
• Is there an easy way to determine whether an algebraic form becomes decomposable after suitable change of basis? The only related condition I am aware of is that $\omega\wedge \omega = 0$ is necessary for $\omega$ to be decomposable (but I doubt that is a sufficient condition!) – Jason DeVito Jun 28 '18 at 12:16

Lemma. Suppose $V$ is a finite-dimensional vector space and $\eta$ is a nonzero alternating $k$-tensor on $V$. Define a linear map $\widehat\eta\colon V\to \Lambda^{k-1}V^*$ by $$\widehat\eta (v) = i_v \eta$$ (where $i_v\eta$ denotes interior multiplication). Then $\eta$ is decomposable if and only if $\operatorname{Ker} \widehat\eta$ has codimension $k$ in $V$.
Proof. Let $n$ denote the dimension of $V$. First suppose $\eta$ is decomposable, and write $\eta = e^1\wedge\dots\wedge e^k$ for $1$-tensors $e^1,\dots,e^k$. Since the $e^i$'s are linearly independent (otherwise their wedge product would be zero), we can complete them to a basis $(e^1,\dots,e^n)$ for $V^*$. Let $(e_1,\dots,e_n)$ be the dual basis for $V$. If we write an arbitrary vector $v\in V$ as $v = \sum_{i=1}^n v^i e_i$, then $$\widehat\eta(v) = \sum_{i=1}^k (-1)^k v^i \, e^1 \wedge \dots \wedge \widehat{e^i} \wedge\dots \wedge e^n,$$ where the hat indicates that $e^i$ is omitted. This is zero if and only if $v^1 = \dots = v^k = 0$, which is to say that $v$ lies in the $(n-k)$-dimensional subspace spanned by $e_{k+1},\dots,e_n$.
Conversely, suppose $\operatorname{dim}\operatorname{Ker} \widehat\eta = n-k$. Let $(e_1,\dots,e_n)$ be a basis for $V$ such that $\operatorname{Ker} \widehat\eta$ is spanned by $e_{k+1},\dots,e_n$. In terms of this basis, we can write $$\eta = \underset{J}{\sum\nolimits'} \eta_J \,e^{j_1}\wedge \dots \wedge e^{j_k},$$ where the primed summation means summation over increasing multi-indices $J = (j_1,\dots,j_k)$. For any $J$ such that $j_k>k$, we have $$\eta_J = \eta(e_{j_1},\dots,e_{j_k}) = (-1)^{k-1} \widehat\eta(e_{j_k})(e_{j_1},\dots,e_{j_{k-1}}) = 0,$$ since $e_{j_k}\in \operatorname{Ker}\widehat\eta$. Thus the only increasing $J$ for which $\eta_J\ne 0$ is $J = (1,\dots,k)$, showing that $\eta = c\, e^1\wedge\dots\wedge e^k$ for some real number $c$, so it is decomposable. $\square$
Theorem. Let $M$ be a smooth manifold. If $\omega$ is a smooth $k$-form that is pointwise decomposable, then in a neighborhood of each point there are smooth $1$-forms $\omega^1,\dots,\omega^k$ such that $\omega = \omega^1\wedge\dots\wedge\omega^k$.
Proof. Define a smooth bundle map $\widehat \omega\colon TM \to \Lambda^{k-1}T^*M$ by $\widehat\omega(v) = i_v\omega$. The lemma implies that $\widehat\omega$ has constant rank. The kernel of a smooth bundle map with constant rank is a smooth subbundle (see, for example, Theorem 10.34 in my Introduction to Smooth Manifolds, 2nd ed.). Thus in a neighborhood of each point, there are smooth vector fields $V_{k+1},\dots,V_n$ that span $\operatorname{Ker}\widehat\omega$ at each point. In a possibly smaller neighborhood, we can choose smooth vector fields $V_1,\dots,V_k$ such that $(V_1,\dots,V_n)$ is a smooth local frame for $M$. Let $(\varepsilon^1,\dots, \varepsilon^n)$ denote the dual coframe. Then a computation just like the one in the proof of the lemma shows that $\omega$ can be written locally as a function multiplied by $\varepsilon^1\wedge\dots\wedge\varepsilon^k$, and that function is smooth because $\omega$ is smooth. $\square$