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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?

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Don't believe everything that Wolfram's minions tell you :-) – Robin Chapman Oct 26 '10 at 18:33
Ignoring terms of order 3 and higher, and inverting 1+X using the power series, (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]. So no factor of 2. – T.. Oct 26 '10 at 18:36
@Robin, @T..: That's good to know! Do either of you want to post an answer so I can accept it? – Rahul Oct 26 '10 at 21:50
P.S. I have sent MathWorld a message about this. – Rahul Oct 27 '10 at 1:56
@Rahul: I saw your recent meta thread and, looking at your question postings, realized that this discussion might have been the intended example. I undeleted the answer which, after posting, seemed to fit better as a comment, being a short and basically single-bit YES or NO answer to a [verification-request]. – T.. Nov 14 '10 at 20:04
up vote 2 down vote accepted

Ignoring terms of order $3$ and higher, and inverting $1+X$ using the power series,

$$\begin{align}&(1+A)(1+B)(1-A+A^2)(1-B+B^2)\\ =\;& (1 + (A+B) + AB) (1 - (A+B) + (A^2 + AB + B^2))\\ =\;& 1 - (A+B)^2 + AB + A^2 + AB + B^2 \\ =\;& 1 - (A^2 + AB + BA + B^2) + 2AB + A^2 + B^2\\ =\;& 1 + AB - BA \\ =\;& 1 + [A,B]\end{align}$$

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