In a previous question, I asked an utterly trivial question, which Deane Yang correctly pointed out was utterly trivial. I will now ask a similar question, which is the one I meant to ask last time; I hope it's not similarly trivial.

I am working on $\mathbb R^n$, although in fact any manifold with volume form is good enough. And my question is local: I have an open neighborhood $U \ni 0$, and you are allowed to make it smaller if you want.

Recall that a **constant-rank distribution** $D$ on $U$ is a vector subbundle of the tangent bundle ${\rm T}U$. Let's fix the rank to be $k\leq n$, and suppose that everything is smooth: $\Gamma(D)$ is a $C^\infty$-submodule of $\Gamma({\rm T}U)$. The distribution is **involutive** if $\Gamma(D)$ is a Lie subalgebra of $\Gamma({\rm T}U)$ (over $\mathbb R$ — the Lie bracket on $\Gamma({\rm T}U)$ is not $C^\infty$-linear). The distribution $D$ is **smooth** if $\Gamma(D)$ is a free rank-$k$ module over $C^\infty$, i.e. if $\Gamma(D)$ has a **basis** $\{v_1,\dots,v_k\}$ so that $\Gamma(D) = \operatorname{Span}_{C^\infty}\{v_1,\dots,v_k\}$.

So, suppose that on $U \subseteq \mathbb R^n$ I have a smooth involutive rank-$k$ distribution. Given a basis $v_1,\dots,v_k \in \Gamma(D)$ (and I will use coordinates $x^1,\dots,x^n$ on $\mathbb R^n$, so I will write $v_a = \sum_i v_a^i(x) \frac{\partial}{\partial x^i}$), then I can define the **structure coefficients** $f_{a,b}^c(x)$, $a,b,c = 1,\dots,k$, via $[v_a,v_b] = \sum_c f_{a,b}^c v_c$, or, in coordinates:
$$ \sum_i \left( v_a^i(x)\,\frac{\partial v_b^j}{\partial x^i} - v_b^i(x)\,\frac{\partial v_a^j}{\partial x^i}\right) = \sum_c f_{a,b}^c(x)\,v_c^j(x) $$

Question:Does there exist a basis $\{v_a\}$ for a given smooth involutive constant-rank distribution so that for each $a=1,\dots,k$ and each $x\in U$, we have $\displaystyle \sum_b f^b_{a,b}(x) = \sum_i \frac{\partial v^i_a(x)}{\partial x^i}$?

For example, by Frobenius theorem (my utterly trivial question), I can find a basis so that the LHS vanishes for each $a$. Or, by another of my trivial questions, I could make the basis entirely divergence-free. But I don't think I can simultaneously make the basis consist of divergence-free vector fields.

Question:If so, how many choices of such a basis do I have? Clearly ${\rm GL}(k,\mathbb R)$ acts on the set of choices; are there more?