$$\mathrm{exp}(X)\mathrm{exp}(Y)=\mathrm{exp}\left(X+Y+\frac{1}{2}[X,Y]+\mathrm{\cdots}\right)$$ where the dots indicate higher order terms that can all be written in terms of iterated brackets of $X$ and $Y$ .
The BakerCampbellHausdorff formula
The BakerCampbellHausdorff formula
In the last video, we found a local inverse for the matrix exponential $\mathrm{exp}:\U0001d524\U0001d529(n,\mathbf{R})\to GL(n,\mathbf{R})$ . In other words, we found a neighbourhood $U$ of $0\in \U0001d524\U0001d529(n,\mathbf{R})$ and a neighbourhood $V$ of $I\in GL(n,\mathbf{R})$ such that ${\mathrm{exp}}_{U}:U\to V$ is a bijection whose inverse we called $\mathrm{log}:V\to U$ .
We also wrote down a power series expansion: $$\mathrm{log}(I+M)=M\frac{1}{2}{M}^{2}+\frac{1}{3}{M}^{3}\mathrm{\cdots}$$ valid for small $M$ (as $V$ is a neighbourhood of the identity, anything in $V$ can be written as $I+M$ for some small $M$ ).
If $XY=YX$ then we saw that $\mathrm{exp}(X)\mathrm{exp}(Y)=\mathrm{exp}(X+Y)$ . What happens if $XY\ne YX$ ? We can figure out a formula for $\mathrm{log}\mathrm{exp}(X)\mathrm{exp}(Y)$ using our power series for $\mathrm{log}$ .
Let's compute. $$\mathrm{log}(\mathrm{exp}(X)\mathrm{exp}(Y))=\mathrm{log}\left(\left(I+X+\frac{{X}^{2}}{2}+\mathrm{\cdots}\right)\left(I+Y+\frac{{Y}^{2}}{2}+\mathrm{\cdots}\right)\right).$$ Multiplying the power series together, we get $$\mathrm{log}\left(I+X+Y+XY+\frac{{X}^{2}}{2}+\frac{{Y}^{2}}{2}+\frac{X{Y}^{2}}{2}+\frac{{X}^{2}Y}{2}+\mathrm{\cdots}\right).$$ Taking $M=X+Y+XY+\frac{{X}^{2}}{2}+\frac{{Y}^{2}}{2}+\mathrm{\cdots}$ and using our power series for $\mathrm{log}(I+M)$ we get $$M\frac{1}{2}{M}^{2}+\frac{1}{3}{M}^{3}+\mathrm{\cdots}=X+Y+XY+\frac{1}{2}{X}^{2}+\frac{1}{2}{Y}^{2}+\mathrm{\cdots}\frac{1}{2}({X}^{2}+XY+YX+{Y}^{2}+\mathrm{\cdots})+\mathrm{\cdots}$$ where the dots stand for terms which are at least cubic in $X$ and $Y$ . For now we'll ignore all cubic and higher order terms.
This simplifies to give $$\mathrm{log}(\mathrm{exp}X\mathrm{exp}Y)=X+Y+\frac{1}{2}(XYYX)+\mathrm{\cdots}$$
This quantity $XYYX$
is going to turn out to be very important in this course. It's called the
We're now saying that $\mathrm{log}(\mathrm{exp}(X)\mathrm{exp}(Y))$ is:

to first order just $X+Y$ ,

has a second order correction term $[X,Y]$ , which vanishes if $X$ and $Y$ commute,
and remarkably, all the cubic and higher order correction terms can be expressed as iterated commutators of $X$ and $Y$ . For example, the cubic term is $$\frac{1}{12}[X,[X,Y]]\frac{1}{12}[Y,[X,Y]].$$ This fact that the higher order terms can be written in this way is genuinely miraculous and not at all obvious: proving it is one of the possible projects you could do.
You can write a general formula (due to Dynkin) for the $n$ th order term, but this is not very illuminating. We will state the existence of this formula as a theorem:
The main point of the BakerCampbellHausdorff formula is that the group multiplication of the invertible matrices $\mathrm{exp}(X)$ and $\mathrm{exp}(Y)$ in $GL(n,\mathbf{R})$ is determined by the bracket operation (on $\U0001d524\U0001d529(n,\mathbf{R})$ ).
By keeping track of the cubic terms, check that the cubic term in the BakerCampbellHausdorff formula is $\frac{1}{12}[X,[X,Y]]\frac{1}{12}[Y,[X,Y]].$
Preclass exercise
Verify the BakerCampbellHausdorff formula for $\mathrm{exp}\left(\begin{array}{ccc}\hfill 0\hfill & \hfill {a}_{1}\hfill & \hfill {c}_{1}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {b}_{1}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\mathrm{exp}\left(\begin{array}{ccc}\hfill 0\hfill & \hfill {a}_{2}\hfill & \hfill {c}_{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {b}_{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$ . [Hint: In this case, the formula is $\mathrm{exp}\left(X+Y+\frac{1}{2}[X,Y]\right)$ with no further correction terms. Can you see why?]