Chow–Rashevskii theorem

Remarks

The set of points reachable from $p$ is the set of endpoints of finite compositions of flows of the fields $\pm X_i$. Since the system is symmetric under $u \mapsto -u$, reachability is an equivalence relation, and the equivalence classes partition $M$. Flowing back and forth along two fields moves a point, to second order, in the direction of their bracket. With the bracket condition, this shows that for suitable $Y_j$ among the $\pm X_i$, the map $(t_1, \ldots, t_N) \mapsto e^{t_N Y_N} \circ \cdots \circ e^{t_1 Y_1}(p)$ has an image with nonempty interior. So each class is open, and connectedness of $M$ leaves only one class. Example: on $\R^3$, the fields $X_1 = \partial_x - \frac{y}{2} \partial_z$ and $X_2 = \partial_y + \frac{x}{2} \partial_z$ satisfy $[X_1, X_2] = \partial_z$, so $X_1, X_2, [X_1, X_2]$ span $\R^3$ at each point and every point is reachable, although only two directions are controlled. The bracket condition cannot be dropped: for $X_1 = \partial_x$ and $X_2 = \partial_y$ on $\R^3$ all brackets vanish, and the trajectories from $p$ stay in the plane through $p$ where $z$ is constant.

Going deeper

For a student

The theorem is the basic controllability result of geometric control theory for systems without drift. In sub-Riemannian geometry it shows that the distance defined by curves tangent to a bracket-generating distribution is finite, and it is the starting point of that theory. For systems with drift $x' = X_0(x) + \sum u_i X_i(x)$ the bracket condition only gives that the reachable sets have nonempty interior, not controllability.