TITLE: On the relationship between the commutators of a Lie group and its Lie algebra
QUESTION [13 upvotes]: I was trying to teach myself some basic Lie theory, and I came across this statement on Mathworld, relating the commutator of a group, $\alpha\beta\alpha^{-1}\beta^{-1}$, to the commutator of its Lie algebra, $[A,B] = AB-BA$:

For instance, let $A$ and $B$ be square matrices, and let $\alpha(s)$ and $\beta(t)$ be paths in the Lie group of nonsingular matrices which satisfy
  $$\begin{align}
  \alpha(0)=\beta(0)    &=    I    \\
  \left.\frac{\partial\alpha}{\partial s}\right|_{s=0}    &=    A    \\
  \left.\frac{\partial\beta}{\partial s}\right|_{s=0}    &=    B,
  \end{align}$$
  then
  $$\left.\frac{\partial}{\partial s}\frac{\partial}{\partial t}\alpha(s)\beta(t)\alpha^{-1}(s)\beta^{-1}(t)\right|_{(s=0,t=0)}=2[A,B].$$

When I tried to derive this for myself, using the fact that
$$\left.\frac{\partial\alpha^{-1}}{\partial s}\right|_{s=0} = \left.-\alpha^{-1}\frac{\partial\alpha}{\partial s}\alpha^{-1}\right|_{s=0} = -A,$$
I expanded the expression to get
$$\left.\frac{\partial}{\partial s}\frac{\partial}{\partial t}\alpha\beta\alpha^{-1}\beta^{-1}\right|_{(s=0,t=0)}$$
$$=\left.\left( \frac{\partial\alpha}{\partial s}\frac{\partial\beta}{\partial t}\alpha^{-1}\beta^{-1} + \alpha\frac{\partial\beta}{\partial t}\frac{\partial\alpha^{-1}}{\partial s}\beta^{-1} + \frac{\partial\alpha}{\partial s}\beta\alpha^{-1}\frac{\partial\beta^{-1}}{\partial t} + \alpha\beta\frac{\partial\alpha^{-1}}{\partial s}\frac{\partial\beta^{-1}}{\partial t} \right)\right|_{(s=0,t=0)}$$
$$=AB - BA - AB + AB$$
$$=[A,B].$$
The difference is that the factor of 2 is missing. This seems to agree with the lecture notes I found on MIT OCW, which state (in Ch. 2, PDF 1) that if $X, Y \in \mathfrak{g}$,

$\exp(-tX)\exp(-tY)\exp(tX)\exp(tY) = \exp\{t^2[X,Y]+O(t^3)\}$.

Since this is not my area of expertise, I wanted to make sure I got things right before I contacted MathWorld about a typo. Have I done something wrong somewhere, or is the MathWorld statement actually an error?

REPLY [6 votes]: Ignoring terms of order $3$ and higher, and inverting $1+X$ using the power series,
$$\begin{align}&(1+A)(1+B)(1-A+A^2)(1-B+B^2)\\
=\;& (1 + (A+B) + AB) (1 - (A+B) + (A^2 + AB + B^2))\\
=\;& 1 - (A+B)^2 + AB + A^2 + AB + B^2 \\
=\;& 1 - (A^2 + AB + BA + B^2) + 2AB + A^2 + B^2\\
=\;& 1 + AB - BA   \\
=\;& 1 + [A,B]\end{align}$$