Commuting time dependent vector fields and pullback invariance

Question

Let $X_t, Y_t \in C^\infty(\mathbb{R}; \mathfrak{X}^\infty(M))$ be (smooth, or something else if it's necessary) time dependent vector fields.

Is there some analogue of the following fact in finite dimensions (with no time dependence, identifying Lie group elements with exponentials of algebra elements could also be false when you involve time):

$[A,B] = 0 \implies \operatorname{Ad}_{e^A}B = e^A B e^{-A} = B$

Where now I want to say something along the lines of $(\phi^X_t)_* Y_t = Y_t$

I know infinite dimensional Lie groups are very easy to get misled by intuition from the finite case, so I'm looking for a reference that takes care of the details.

Answer 1

Let $\phi^t$, $\psi^t$ be the respective flows of two vector fields $A$, $B$ at time $t$ on a manifold $M$ (assume to fix ideas that these flows do exist for every time). If $A, B$ do not depend on time, then of course one has an equivalence between the 4 properties

i) $[A,B]=0$;

ii) $A$ is invariant by every $\psi^t$;

iii) $B$ is invariant by every $\phi^t$;

iv) $\phi^s\circ\psi^t=\psi^t\circ\phi^s$ for every two times $s$, $t$.

(see e.g. Spivak, "A Comprehensive Introduction To Differential Geometry" volume 1).

For time-dependant vector fields, nothing of the like can hold. For example, on $M=R$ the real line, consider $A(t)=f(t)d/dx$ and $B(t)=g(t)xd/dx$ where $f:R\to R$ and $g:R\to R$ are two smooth functions such that $f(t)=0$ except for $t\in(0,1/2)$ and $g(t)=0$ except for $t\in(1/2,1)$ and $\int_Rf=\int_Rg=1$. Certainly, at every time $t$ one has $[A(t),B(t)]=0$ since at least one of the two vector fields is identically $0$ at this time! But their respective flows at time $1$ are $\phi^1:x\mapsto x+1$ and $\psi^1:x\mapsto ex$, which don't commute. Also, if say $g(3/4)\neq 0$, then clearly the translation $\phi^{3/4}=\phi^1$ does not preserve the linear vector field $B(3/4)$.