Product of exponential of matrices

Question

Suppose that $e^Xe^Y=e^Ye^X$.

It is known that, in general, this equality does not imply $XY=YX$. However, an exercise in Stillwell's Naive Lie Theory Page 154, Exercise 7.6.3 asks to show that $XY=YX$ (My understanding from section 7.6 in the book is that, this exercise requires the implicit assumption that the norms of all matrices involved are small). I would appreciate possibly different hints/solutions for doing this problem.

Edit: Fortunately, I could add the the link to the exercise through googlebooks. Please follow the link for the precise statement of the exercise and the context in which it appears.

Answer 1

"Small norms" may mean that the sought equality holds only approximately. $$ e^{X} e^{Y} = (I + X + \mbox{(negligible)}) (I + Y + \mbox{(negligible)}) = I + X + Y + XY + \mbox{(negligible)} $$ Similarly, $$ e^{Y} e^{X} = I + X + Y + YX + \mbox{(negligible)} $$ Since you asked only for hints, I'll stop here.

Answer 2

In terms of the commutators, you've got $$ 0=[e^X,e^Y] \\ =[1+X+\frac{1}{2}X^2+O(X^3),1+Y+\frac{1}{2}Y^2+O(Y^3)] \\ =[X,Y]+\frac{1}{2}[X^2,Y]+\frac{1}{2}[X,Y^2]+O(X^3,Y^3), $$ which implies that either $X$ and $Y$ commute, or else the commutator of $X$ and $Y$ is $O(X^2,Y^2)$, since the remaining terms are all at least that small and need to cancel it out. If you want to be precise about what you're proving, I believe you can show that if $[e^{cX},e^{cY}]=0$ whenever $|c| < \varepsilon$ (for some positive $\varepsilon$), then $[X,Y]=0$. (In other words, $e^{cX}$ and $e^{cY}$ can "accidentally" commute for some values of $c$, but if they do so for arbitrarily small values of $c$, then $X$ and $Y$ must themselves commute.)