Two vector fields commute if and only if their local flows commute

Assume $[X,Y]=0$. By [L1] and [L2], the field $Y$ is invariant under the $X$-flow. Therefore, for each admissible $t$, the diffeomorphism ${\mathrm{\Phi }}_t^X$ sends $Y$-integral curves to $Y$-integral curves with the same parameter.

Conversely, assume the local flows commute whenever both sides are defined. Differentiate the identity ${\mathrm{\Phi }}_t^X({\mathrm{\Phi }}_s^Y(p))={\mathrm{\Phi }}_s^Y({\mathrm{\Phi }}_t^X(p))$ with respect to $s$ at $s=0$. This yields $({\mathrm{\Phi }}_t^X{)}_{\ast }Y=Y$ on the common domain. By [L2], ${\mathcal{L}}_{X}Y=0$, and then [L1] gives $[X,Y]=0$.

Fix $p$ and admissible $s,t$. The curves $r\mapsto {\mathrm{\Phi }}_t^X({\mathrm{\Phi }}_r^Y(p))\text{and}r\mapsto {\mathrm{\Phi }}_r^Y({\mathrm{\Phi }}_t^X(p))$ are both $Y$-integral curves through the point ${\mathrm{\Phi }}_t^X(p)$ at $r=0$. By uniqueness of integral curves, they agree for all common $r$, and evaluating at $r=s$ gives ${\mathrm{\Phi }}_t^X({\mathrm{\Phi }}_s^Y(p))={\mathrm{\Phi }}_s^Y({\mathrm{\Phi }}_t^X(p))\mathrm{.}$

Therefore two smooth vector fields commute if and only if their local flows commute on their common domains.