 Solution

Use the Clifford algebra to write ${\gamma }^{\nu }{\gamma }^{\mu }=2{\eta }^{\mu \nu }{I}_4-{\gamma }^{\mu }{\gamma }^{\nu }$. The Lorentz generator from gamma-matrix commutators is thereforeThis also checks the diagonal case $\mu =\nu$, where $S^{\mu \mu }=0$ because $({\gamma }^{\mu }{)}^2={\eta }^{\mu \mu }{I}_4$. The metric term in the printed expression is understood to multiply the identity matrix. The generators are antisymmetric in their two spacetime indices.