$e^a e^b e^{-a} e^{-b} = e^{[a, b] + o(\|a\|^2 + \|b\|^2)}$, matrix exponentials. Fix a norm $\|\cdot\|$ on $\text{M}_n(\mathbb{R})$. For $a, b \in \text{M}_n(\mathbb{R})$ write $[a, b] := ab - ba$. How do I see the following formula?$$e^a e^b e^{-a} e^{-b} = e^{[a, b] + o(\|a\|^2 + \|b\|^2)}$$In the above, $o(\|a\|^2 + \|b\|^2)$ stands for a function $f: \text{M}_n(\mathbb{R}) \times \text{M}_n(\mathbb{R}) \to \text{M}_n(\mathbb{R})$ such that$$\lim_{(a, b) \to (0, 0)} {{\|f(a, b)\|}\over{\|a\|^2 + \|b\|^2}} = 0.$$

Let $t$ be a small real and $X,Y$ be complex matrices. According to BCH formula,  $e^{tX}e^{tY}=e^{t(X+Y)+t^2/2[X,Y]+O(t^3)}=I+t(X+Y)+t^2/2[X,Y]+t^2/2(X+Y)^2+O(t^3)$
In the same way,$e^{-tX}e^{-tY}=e^{-t(X+Y)+t^2/2[X,Y]+O(t^3)}=I-t(X+Y)+t^2/2[X,Y]+t^2/2(X+Y)^2+O(t^3)$ and the product is $e^{tX}e^{tY}e^{-tX}e^{-tY}=I+t^2[X,Y]+O(t^3)=e^{t^2[X,Y]+O(t^3)}$. 
