Can we always span a decomposable form via constant coefficients?

Question

Let $M$ be a smooth manifold of dimension $d$. Let $1 < k <d$ be an integer. Let $\omega^i$ be a local frame of the exterior power bundle $\Lambda_{k}(T^*M)$.

Does there exists numbers $a_i \in \mathbb{R}$ , such that $a_i \omega^i$ is decomposable?

(I am using the summation convention here).

Of course, there always exist functions $a_i$ satisfying the above, since by definition, if $\omega^i$ is a frame, we can span every smooth element with it.

My guess is that the answer might be negative, since being decomposable element is $\Lambda_k(V)$ a closed condition (rather than an open one, see Plucker relations). Here $V$ is a fixed real vector space .

I am OK with shrinking the local neighbourhood on where the $\omega_i$ are defined.

Answer 1

Of course $a_1 = \cdots = a_{{}_dC_k} = 0$ is always a solution. Excluding this case, however, the answer in general is no.

Here's how we can construct a counterexample: The lowest values $d, k$ for which there are indecomposable (local) $k$-forms on a $d$-manifold are $d = 4$, $k = 2$, and for $k = 2$ we have a lemma useful here:

A $2$-form $\alpha$ is locally decomposable iff $\alpha \wedge \alpha = 0$.

For any local coframe $(\theta^i)$ of $TM$, define $\theta^{ab} := \theta^a \wedge \theta^b$, and for a coordinate function $\tau$, set \begin{alignat*}{3} \omega^1 &:= \cos \phantom{2} \tau \,\theta^{12} - \sin \phantom{2} \tau \,\theta^{34},& \qquad \omega^2 &:= \sin \phantom{2} \tau \,\theta^{12} + \cos \phantom{2} \tau \,\theta^{34} \\ \omega^3 &:= \cos 2 \tau \,\theta^{13} - \sin 2 \tau \,\theta^{24},& \qquad \omega^4 &:= \sin 2 \tau \,\theta^{13} + \cos 2 \tau \,\theta^{24} \\ \omega^5 &:= \cos 3 \tau \,\theta^{14} - \sin 3 \tau \,\theta^{23},& \qquad \omega^6 &:= \sin 3 \tau \,\theta^{14} + \cos 3 \tau \,\theta^{23} . \\ \end{alignat*} It's straightforward to check that this is a frame of $\bigwedge^2 T^*M$.

By the lemma, if $\Omega := \sum a_i \omega^i$ (with constant $a_i$) is decomposable then $\Omega \wedge \Omega = 0$; so, to show that this frame gives a counterexample, it's enough to show that this equation forces $a_1 = \cdots = a_6 = 0$.

Substituting gives $$f(\theta) \, \theta^1 \wedge \theta^2 \wedge \theta^3 \wedge \theta^4 = 0,$$ and applying angle sum identities to expand $f$ in the linearly independent functions $\cos j \tau, \sin j \tau$, $j \in \{0, 1, 2, \ldots\}$ gives $$ f(\tau) = 2 a_5 a_6 \cos(6 \tau) + (a_6^2 - a_5^2) \sin (6 \tau) + \cdots = 0 $$ where $\cdots$ denotes terms with $j \neq 6$. By linear independence these coefficients are both zero, so $a_5 = a_6 = 0$. Then successively considering $j = 4$ and $j = 2$ gives that the remaining $a_i$ are zero, too, and we're done.