Suppose $u$ and $v$ are two vector fields in the plane, with $u$ and $v$ linearly independent and Lie bracket $[u,v]$ zero everywhere. Does it follow that $u$ and $v$ are both curl-free? Intuitively, I would think that commutativity of the flows of $u$ and $v$ is stronger than integrability of each individual field, but is there an algebraic way to recover $\nabla \times u=0$ from $[u,v]=0$?
If this fact is true, the linear independence of $u$ and $v$ must play some essential role, since of course $[u,u]=0$ for any $u$.
We shouldn't expect this to be true. Consider any noncurlfree, nonvanishing vector field $U$. In (local) flowbox coordinates $(s, t)$ we have $U = \partial_s$, and $[\partial_s, \partial_t] = 0$, so not only is the local version of the statement false, the condition $[U, V] = 0$ doesn't impose any local restriction on $U$ near any point where $U$ is nonzero.
Example The pair $$U := \exp x \,\partial_y, \qquad V := \partial_x + y \partial_y$$ is (pointwise) linearly independent and satisfies $[U, V] = 0$ but $\nabla \times U = \exp x$. (In this example, we actually have $\nabla \times V = 0$, but this is not necessary either, as we can replace $V$ with $\tilde V = x V$, which satisfies the same conditions but $\nabla \times \tilde V = y$.)