Flow of sum of commuting vector fields

Question

I'm trying to understand why the flow of sum of commuting vector fields is the composition of their flows.

This is apparently supposed to be obvious but I don't see how.

Answer 1

I hope someone answers in stricter math, but for intuition consider an affine input nonlinear dynamical system: $ \dot{x} = f(x) + g(x) u $ with state $x$, input $u$ and $f,g$ vector fields. Now you can ask what happens, if I just let the syste do its thing, thus $u=0$. The trajectories of the system are now the flow of $f$. If then you decide to influence the system using your input $u$, you can just "add" that to the autonomous dynamics.

It goes somewhere along the lines of the solution of an inhomogenious diff. eq.: first you look for the general case and then a particular solutionn for the inhomogenious term - and add them.

Answer 2

This is not obvious. Let $X, Y$ commute, and consider a point where $X\not =0$ (if $X(p)=Y(p)=0$ the flows at $p$ commute). Using the "flow box theorem", we can find a coordinate system $x,y_1,...,y_n$ such that $X= \partial _x$, and $Y= a\partial _x+ b_1\partial _{y_1}++++b_n\partial {y_n}$. The commutation $[X,Y]$ means that $\partial _x a= \partial _x b_i=0$. The flow $\Phi _t$ of $X$ is just $(x,y_1,...y_n)\to (x+t, y_1,...y_n)$ The flow $\Psi _t$ of $Y$ can be computed $(x,y_1,...y_n)\to (x+\int _0^t a(\psi(u,y))du, \psi(t,y))$, where $\psi(u,y)$ is the flow of the "vertical" vector field $b_1\partial _{y_1}++++b_n\partial {y_n}$ (which is independant of $x$). Then $\Phi _t\circ \Psi _t (x,y)= (x+t+\int _0^t a(\psi(u,y))du, \psi(t,y)).$ Now deriving this formula for $t=0$ one obtains $ (1+a, b_1,...b_n)= X+Y(0)$ . The same computation proves also that the two flows commute.