I found this question on an old qual and I have a start on a solution. However, I got stuck. Thanks for any suggestions in order to proceed.

Let $X_1, X_2, X_3$ be smooth vector fields on $\mathbb{R}^3$ such that $\{X_1(0), X_2(0), X_3(0)\}$ forms a basis for $\mathbb{R}^3$. For each $i = 1, 2, 3$, let $\phi_s^i$ ($s\in \mathbb{R}$) denote the flow generated by $X_i$. Prove that there is an open ball $B$ centered at the origin such that for each $x\in B$, there is $(t_1, t_2, t_3)\in \mathbb{R}^3$ such that $$x = \phi_{t_3}^3 \circ \phi_{t_2}^2 \circ \phi_{t_1}^1(0).$$

My attempt at a solution: Since we have vector fields, we can generate local flows around the origin. The $B$ be the intersection of the open sets where the flows are defined. Then, for $x\in B$, we can write $x$ as $(a_1X_1(0), a_2X_2(0), a_3X_3(0))$ since we are given a basis for $\mathbb{R}^3$. I had hoped that $a_i$ could correspond to the $t_i$ in the problem, but I do not know what $\phi_{t_1}^1(0)$ looks like. Can I tell what is happening to the flow?

  • $\begingroup$ I've been working on this problem with some other students, are we still stumped! Can anyone expand the hint or give a solution? We are studying for a qualifying exam. $\endgroup$ – phenomenalwoman4 Aug 15 '13 at 21:10

Hint: Consider $F\colon U\to\mathbb R^3$ where $U$ is a neighborhood of $0\in\mathbb R^3$, $F(t_1,t_2,t_3)=\phi_{t_3}^3 \circ \phi_{t_2}^2 \circ \phi_{t_1}^1(0)$. Compute $DF(0)$.


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