Question on the image of the parameter space under two non-commuting flows.

Question

Let $\phi^t_1$ and $\phi^s_2$ be two smooth flows on $\mathbb{R}^2$ defined for $t\in[-1,1]$ and $s\in [-1,1]$. Assume

(1) $\phi^0_1\big((0,0)\big)=\phi_2^0\big((0,0)\big)=0$

(2) $(\phi^t_1)'|_0$ and $(\phi^s_2)'|_0$ span the tangent space at $(0,0)$

Question: Does there exists an $\varepsilon>0$ such that $$B_\varepsilon(0)\subset\{\phi_1^t\circ\phi_2^s\big((0,0)\big): t\in [-1,1], s\in[-1,1]\}$$

Of course this is trivially true if they commute since the corresponding vector field can be simultaneously straightened. So I am curious what happens when they don't commute. Of course integrability conditions are satisfied in a neighborhood of $(0,0)$ since we are in 2 dimensional situation and have two vector fields. But what I am asking is slightly different from Frobenius Theorem.

Any help appreciated

Answer 1

Yes, there exists such an $\varepsilon$:

Consider the map $F(t,s)=(\phi_1^t\circ\phi_2^s)(0)$. We have $F(0,0)=0$ and $D_{(0,0)}F$ has columns $(\phi^t_1)'|_0$ and $(\phi^s_2)'|_0$, and so is invertible. It follows from the inverse function theorem that $F$ is a local diffeomorphism, which implies the desired property.