Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 302 / 3

The Lorentz group consists of linear maps $x\mapsto \Lambda x$ satisfying ${\Lambda }^T\eta \Lambda =\eta$. The Poincare group consists of affine isometries $x\mapsto \Lambda x+a$ and has multiplicationThus it is the semidirect product ${\mathbb{R}}^{1,3}⋊O(1,3)$, with the Lorentz group as the subgroup fixing the spacetime origin.

With cyclic spatial indices,Substitution in the given Poincare algebra brackets yields

Because the translations commute, $P^{\sigma }{P}^{\mu }$ is symmetric in $(\sigma ,\mu )$, whereas the Levi-Civita tensor in the Pauli-Lubanski pseudovector is antisymmetric, so $W_{\mu }{P}^{\mu }=0$. Moreover,because the two terms cancel after relabeling and each remaining momentum product is symmetric.

In the rest frame $P^{\mu }=(m,0,0,0)$, antisymmetry gives $W_0=0$. Taking ${ϵ}_{0123}=+1$ and $J_3=M^{12}$ givesHenceThe sign of the second eigenvalue reverses if the opposite Levi-Civita convention is chosen.