My question is local and coordinate-full: I have an open neighborhood $0 \in U \subseteq \mathbb R^n$, and I'm allowed to make it smaller around $0$. On this neighborhood, I have a constant-rank-$k$ smooth integrable distribution. "Smoothness" means that I have $k\leq n$ many vector fields $v\_1,\dots,v\_k$ — I will work in coordinates $x^1,\dots,x^n$, so $v\_a = \sum\_{i=1}^n v\_a^i(x) \frac{\partial}{\partial x^i}$ — and the distribution at $x$ is the span in ${\rm T}\_xU$ of $\{v\_1(x),\dots,v\_k(x)\}$. "Constant-rank" means that at every $x\in U$, the vectors $\{v\_a(x)\}$ are linearly independent. And "integrable" means that the commutator of any two vector fields on the distribution is in the distribution, i.e. there is a tensor $f^c\_{a,b}(x)$ such that 
$$\sum\_{i=1}^k \left( v\_a^i(x)\, \frac{\partial v\_b(x)^j}{\partial x^i} - v\_b^i(x)\, \frac{\partial v\_a(x)^j}{\partial x^i} \right) = \sum\_{c=1}^k f^c\_{a,b}(x)\,v^j\_c(x) $$


So $\{v\_1,\dots,v\_k\}$ is a basis for the distribution. A different basis is given by $\{w\_1,\dots,w\_k\}$ if $w\_a^i(x) = \sum\_b A\_a^b(x)\,v\_b^i(x)$ for some pointwise-invertible matrix-valued function $A$.



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> My question is whether there is a basis, which in abuse of notation I will call $\{v\}$, such that the structure constants $f^c\_{a,b}$ satisfy $\sum\_a f^a\_{a,b}(x) = 0$ for every $x\in U$?
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Does every smooth integrable constant-rank distribution have a basis in which the structure constants are traceless?
Am I missing something? By the Frobenius theorem, there exist co-ordinates $y^1, \dots, y^n$ such that $\partial/\partial y^1, \dots, \partial/\partial y^k$ span the same distribution. So these $k$ co-ordinate vector fields form a basis where the structure constants vanish identically.


