Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 302 / 4

Use the conventionfor the group commutator.

Invariance for every $x$ requiresThus $G=O(1,3)$, the Lorentz group; its identity component is $S{O}^{+}(1,3)$.

At first order, ${\Lambda }^T\eta \Lambda =\eta$ gives
${\omega }^T\eta +\eta \omega =0$. Therefore

Using ${\Lambda }_a^{-1}=I-{\omega }_a+{\omega }_a^2+O({\omega }_a^3)$,Thus $({\omega }_c{)}^{\mu }{}_{\nu }=({\omega }_2{)}^{\mu }{}_{\rho }({\omega }_1{)}^{\rho }{}_{\nu }-({\omega }_1{)}^{\mu }{}_{\rho }({\omega }_2{)}^{\rho }{}_{\nu }$.

Comparison of $U[{\Lambda }_c]$ with the operator group commutator gives the Lorentz algebraThere is no factor of $i$ because the generators are anti-Hermitian.

Substituting $J_m={ϵ}_{mij}{M}^{ij}/2$ and $K_i=M^{0i}$ yields

For $J_i^{\pm }=(J_i\pm i{K}_i)/2$,

The two commuting triples each generate a copy of the complexified SU(2) Lie algebra. Hence the chiral decomposition of the complex Lorentz algebra isIn physicists' compact notation this is
$\mathfrak{su}(2{)}_{+}\oplus \mathfrak{su}(2{)}_{-}$; the real Lorentz algebra relates the two factors by complex conjugation.