Flow of sum of non-commuting vector fields

Let $V,W\in\Gamma(M)$ be any two vector fields. Is there any “nice” expression for the flow of $V+W$ in terms of the flow of $V$ and the flow of $W$? It would be sufficient for me to have some sort of expansion in $\epsilon$ for the flow of $V+\epsilon W$, at least the first few terms (or even just at the order of $\epsilon$).

In the case where $[V,W]=0$, it is pretty easy to show that
$$\varphi_{V+W}^t = \varphi_V^t\circ\varphi_W^t,$$
but the non commuting case is of greater interest to me.

Solutions Collecting From Web of "Flow of sum of non-commuting vector fields"

In the general non-commuting case, the flow $\phi^t_{V+W}$ equals to first order both $\phi^t_V \circ \phi^t_W$ and $\phi^t_W \circ \phi^t_V$. Morally, the second order approximation should be ‘halfway between’ the two aforementioned flows. Since $\phi^{t}_V \circ \phi^{t}_W \circ \phi^{-t}_V \circ \phi^{-t}_W$ is approximated by $\phi^{t^2}_{[V,W]}$, we expect to have

$$ \phi^t_{V+W}(x)= \left(\phi^{t^2}_{\frac{1}{2}[V,W]} \circ \phi^t_W \circ \phi^t_V\right)(x) \, .$$

It happens to be the first few terms of the Zassenhaus formula (in reverse order) for the exponential map ; Notice that we can interpret a vector field on a manifold $M$ as an element of the Lie algebra of the infinite dimensional Lie group $\mathrm{Diff}(M)$, so that taking the exponential map $\mathfrak{diff}(M) \to \mathrm{Diff}(M)$ corresponds to integrating vector fields.

This is going to be a partial answer based on the contents of the paper

A Lie Group Structure on the Space of Time-Dependent Vector Fields”
(A. Poscilicano)

In that paper the author deals with time dependent vector fields. A usual vector field is a particular case of that so that will work in your case.

If $X$ and $Y$ are time-dependent vector fields on a manifold $M$ the following identity holds $$\Phi^{X+(\Phi_{t, 0}^X)_*(Y_t)}_{t, s}=\Phi^X_{t, 0}\circ \Phi^Y_{t, s}\circ \Phi^X_{0, s}.$$ You’re invited to check this. Then, instead of using the vector field $Y$ we are going to apply this to the time dependent vector field $(\Phi_{0, t}^X)_*(Y_t)$. Using that $$(\Phi_{t, 0}^X)_*(\Phi_{0, t}^X)_*(Y_t))= (\Phi_{t, 0}^X\circ \Phi_{0, t}^X)_*(Y_t)=(\Phi^X_{t, t})_*(Y_t)=Y_t,$$ we are lead to $$\Phi^{X+Y}_{t, s}=\Phi^X_{t, 0}\circ \Phi^{(\Phi_{0, t}^X)_*(Y_t)}_{t, s}\circ \Phi^X_{0, s}.$$ Generally, if $X$ is a vector field then we can see it as a time-dependent vector field and the following relation holds $$\Phi^X_{t, s}=\Phi^X_{t-s},$$ where the right hand side is the usual flow of $X$. Use this to apply to your case.