# Why do two vector fields commute iff their flows do?

I have the following definition, that is two vector fields $$X$$ and $$Y$$ commute if $$[X,Y]=0$$. There is result that states that two vector fields commute iff their flows do. This proposition is proved by using the following fact: given $$F$$ and $$G$$ the flows of $$X$$ and $$Y$$ respectively, we have that:

$$F_t \circ G_s (p)- G_s \circ F_t(p)=st[X,Y](p) + o (s^2 + t^2)$$

where $$F_t$$ and $$G_s$$ obviously mean the flow at times $$t$$ and $$s$$. This tells us that the bracket tells us "how much the two vector fields don't commute".

The question is: if the flows commute, it is obvious that $$[X,Y]=0$$. But for the other implication, why can I ignore the terms in $$o(s^2 + t^2)$$?

• math.stackexchange.com/questions/128195/… seems to mostly answer your question, as it implies $\partial_t\phi_t^*Y=0$, where $\phi_t$ is the flow of $X$, so $\phi_t^*Y=Y$ for all $t$, which implies the commutation of the flows. – Mindlack Sep 12 '19 at 16:40
• Seems right. That should be the Lie derivative in the particular case when you are deriving a vevctor field. I was wondering, though, what is inside that $o(s^2 +t^2)$? – tommy1996q Sep 12 '19 at 16:45
• I don't think this formula is going to do it to prove the reverse direction. – Ted Shifrin Sep 12 '19 at 20:09
• @TedShifrin So you say that the right approach would be the one of the questin suggested above? Alos, only because I am curious, do you know what would be the terms contained on $o(s^2 +t^2)$? – tommy1996q Sep 12 '19 at 22:16
• I've never seen this formulation, but if you want to get a feel for it, write out the Taylor series for $e^{tX}e^{sY}-e^{sY}e^{tX}$ for square matrices $X$ and $Y$. – Ted Shifrin Sep 12 '19 at 22:38