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 $A$.

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 $x\in U$?


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.

  • $\begingroup$ No, it was me who was missing something. Thanks. $\endgroup$ – Theo Johnson-Freyd May 24 '10 at 4:52
  • $\begingroup$ Do you happen to know what would happen if I drop "constant rank"? $\endgroup$ – Theo Johnson-Freyd May 24 '10 at 4:53
  • $\begingroup$ I don't know what happens if the constant rank assumption is dropped. But I believe this has been studied in the smooth category, maybe by Sussman and Sontag? There are probably better results in the real or complex analytic categories. $\endgroup$ – Deane Yang May 24 '10 at 11:55
  • $\begingroup$ Ah, I see what happened. I was typing up the question late at night, and misremembered what was the result that I wanted. $\endgroup$ – Theo Johnson-Freyd May 24 '10 at 23:27

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