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If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?

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I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.

Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $\exp(\theta G_x)$ is a rotation through angle $\theta$ about the $x$-axis, and $\exp(\phi G_y)$ is a rotation through angle $\phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.

When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $\theta$ and $\phi$ both have infinitesimal magnitude $\epsilon\ll 1$, then the rotations $\exp(\theta G_x)$ and $\exp(\phi G_y)$ commute with each other to first order in $\epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $\exp(\theta G_x)\exp(\phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $\epsilon^2\ll\epsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $\epsilon$.

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  1. Illustrative example: It is straightforward to prove that the Lie group $$ SO(3)~:=~\{ M\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid M^tM=\mathbb{1}_{3\times 3}, ~\det(M)=1\} $$ of 3D rotations is generated by the corresponding Lie algebra $$ so(3)~:=~\{ m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^t=-m\} $$ of real $3\times 3$ antisymmetric matrices, which clearly do not all commute.

  2. Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!

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While \begin{align} e^{\epsilon A} e^{\epsilon B}&=(1+\epsilon A+\textstyle\frac{1}{2}\epsilon^2 A^2+\ldots)(1+\epsilon B+\frac{1}{2}\epsilon^2 B^2+\ldots)\tag{1}\, ,\\ &= 1+ \epsilon (A+B)+ \frac{1}{2} \epsilon^2 (A^2+AB +B^2)+\ldots \end{align} we have \begin{align} e^{\epsilon B} e^{\epsilon A}&=(1+\epsilon B+\frac{1}{2}\epsilon^2 B^2+\ldots)(1+\epsilon A+\textstyle\frac{1}{2}\epsilon^2 A^2+\ldots)\, ,\tag{2}\\ &= 1+ \epsilon (B+A)+ \frac{1}{2} \epsilon^2 (A^2+BA +B^2)+\ldots \end{align} so:

  1. To order $\epsilon$, the transformations are the same,
  2. To order $\epsilon^2$ they are different.

Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $\epsilon^2$ in (1) and (2) to see that they differ by a commutator.

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