The Baker-Campbell-Hausdorff formula

0.00 In the last video, we found a local inverse for the matrix exponential $\exp\colon\mathfrak{gl}(n,\mathbf{R})\to GL(n,\mathbf{R})$ . In other words, we found a neighbourhood $U$ of $0\in\mathfrak{gl}(n,\mathbf{R})$ and a neighbourhood $V$ of $I\in GL(n,\mathbf{R})$ such that $\exp|_{U}\colon U\to V$ is a bijection whose inverse we called $\log\colon V\to U$ .

0.45 We also wrote down a power series expansion:

$\log(I+M)=M-\frac{1}{2}M^{2}+\frac{1}{3}M^{3}-\cdots$ valid for small

$M$ (as

$V$ is a neighbourhood of the identity, anything in

$V$ can be written as

$I+M$ for some small

$M$ ).

1.10 If $XY=YX$ then we saw that $\exp(X)\exp(Y)=\exp(X+Y)$ . What happens if $XY\neq YX$ ? We can figure out a formula for $\log\exp(X)\exp(Y)$ using our power series for $\log$ .

2.25 Let's compute.

$\log(\exp(X)\exp(Y))=\log\left(\left(I+X+\frac{X^{2}}{2}+\cdots\right)\left(I+% Y+\frac{Y^{2}}{2}+\cdots\right)\right).$ Multiplying the power series together, we get

$\log\left(I+X+Y+XY+\frac{X^{2}}{2}+\frac{Y^{2}}{2}+\frac{XY^{2}}{2}+\frac{X^{2% }Y}{2}+\cdots\right).$ 3.25 Taking

$M=X+Y+XY+\frac{X^{2}}{2}+\frac{Y^{2}}{2}+\cdots$ and using our power series for

$\log(I+M)$ we get

$M-\frac{1}{2}M^{2}+\frac{1}{3}M^{3}+\cdots=X+Y+XY+\frac{1}{2}X^{2}+\frac{1}{2}% Y^{2}+\cdots-\frac{1}{2}(X^{2}+XY+YX+Y^{2}+\cdots)+\cdots$ 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.

5.20 This simplifies to give

$\log(\exp X\exp Y)=X+Y+\frac{1}{2}(XY-YX)+\cdots$

6.25 This quantity $XY-YX$ is going to turn out to be very important in this course. It's called the commutator bracket or Lie bracket of $X$ and $Y$ and is written $[X,Y]$ . This vanishes if $X$ and $Y$ commute.

7.00 We're now saying that $\log(\exp(X)\exp(Y))$ 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.

8.30 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:

Theorem (Baker-Campbell-Hausdorff formula):

$\exp(X)\exp(Y)=\exp\left(X+Y+\frac{1}{2}[X,Y]+\cdots\right)$ where the dots indicate higher order terms that can all be written in terms of iterated brackets of

$X$ and

$Y$ .

10.25 The main point of the Baker-Campbell-Hausdorff formula is that the group multiplication of the invertible matrices $\exp(X)$ and $\exp(Y)$ in $GL(n,\mathbf{R})$ is determined by the bracket operation (on $\mathfrak{gl}(n,\mathbf{R})$ ).

Exercise:

By keeping track of the cubic terms, check that the cubic term in the Baker-Campbell-Hausdorff formula is $\frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]].$

Pre-class exercise

Exercise:

Verify the Baker-Campbell-Hausdorff formula for $\exp\begin{pmatrix}0&a_{1}&c_{1}\\ 0&0&b_{1}\\ 0&0&0\end{pmatrix}\exp\begin{pmatrix}0&a_{2}&c_{2}\\ 0&0&b_{2}\\ 0&0&0\end{pmatrix}$ . [Hint: In this case, the formula is $\exp\left(X+Y+\frac{1}{2}[X,Y]\right)$ with no further correction terms. Can you see why?]