Prove that $e^{t(X+Y)}=e^{tX} e^{tY}$ implies $[X,Y]=0$

Solution 1:

Take $ A=\begin{pmatrix} 0&0 \\ 0&2i\pi \\ \end{pmatrix}$, $\quad B=\begin{pmatrix} 0&1\\ 0 & 2i\pi \end{pmatrix}.$ Then you can show that $\exp(A)=\exp(B)=\exp(A+B)=I_2$, and $AB \neq BA$.

For the proof of $e^{Xt+Yt}=e^{Xt} e^{Yt} \Rightarrow [X,Y]=0$ : note that $$\exp(\frac{Xt}{n}).\exp(\frac{Yt}{n}).\exp(-\frac{Xt+Yt}{n}) = {\rm Id} + \frac{[Xt,Yt]}{2n^2} + o(n^{-2}),$$ hence $$\lim_{n \to +\infty} \left( \exp(\frac{Xt}{n}).\exp(\frac{Yt}{n}).\exp(-\frac{Xt+Yt}{n}) \right)^{2n^2} = \exp([Xt,Yt]).$$ since the sequence is constant to $1$, we have $\exp([Xt,Yt])=1$. Take the derivative at $t=0$ to conclude that $[X,Y]=0$.

Solution 2:

The formula of Baker-Campbell-Hausdorff computes

$$log (\ exp(X) \ exp(Y))$$

as power series in X, Y and their higher commutators, e.g., Wikipedia. If you replace $X$ by $tX$ and $Y$ by $tY$ you obtain by equating powers of t:

$$exp(tX)exp(tY)=exp(tX+tY+(t^2/2)[X,Y]), t\in \mathbb R \implies [X,[X,Y]] + [Y,[Y,X]] = 0.$$

Of course the formula also gives the previous result

$$exp(tX)exp(tY)=exp(tX+tY), t\in \mathbb R \implies [X,Y] = 0.$$