Exponentiation of linear combination of commuting Vector fields proof other direction

Question

I have already asked this question, but I forgot to include the if and only if. So the question has already been answered in one direction, but I have to prove the converse as well. So If $$e^{a\partial/ \partial\lambda +b \partial / \partial\mu}=e^{a\partial/ \partial\lambda}e^{b\partial/ \partial\mu}$$ then $$[\partial/ \partial\lambda,\partial/ \partial\mu]=0.$$ I thought of doing it by Taylor expanding all the three exponentials in the first equation and then bringing everything to one side. Then clearly the zeroth and first order terms cancel and the second order term yields the commutator. And so if one expands it up to second order one indeed gets that the commutator has to vanish. I guess that higher order terms will yield nested commutators. So for the first equation implies that the commutator has to vanish. But how can I show this in an elegant way for higher order terms as well?

Answer 1

$\newcommand{\ex}[1]{\mathrm{e}^{#1}}$A quick way to see it is to add an auxiliary parameter, $s$ in the exponentials so that $$ \ex{s(a\, \partial_\lambda + b\,\partial_\mu)} = \ex{s\,a\, \partial_\lambda}\; \ex{s\, b\,\partial_\mu}.\tag{1}$$ Then you can Taylor expand both sides and take the second derivative with respect to $s$ evaluated at $s=0$, namely, since (1) holds, $$ \frac{\partial^2}{\partial s^2}\;\ex{s(a\, \partial_\lambda + b\,\partial_\mu)}\Bigg\vert_{s=0} = \frac{\partial^2}{\partial s^2}\;\ex{s\,a\, \partial_\lambda}\; \ex{s\, b\,\partial_\mu}\Bigg\vert_{s=0},$$ also holds. This keeps only the quadratic terms in the Taylor expansion, so you can prove $[\partial_\lambda, \partial_\mu]=0$.