I am asked to show that the matrix

$S=1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}$

represents the infinitesimal Lorentz transformation

${\mathrm{\Lambda}}^{\mu}{}_{\nu}={\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu},$

in the sense that

${S}^{-1}{\gamma}^{\mu}S={\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

I have already proven that ${S}^{-1}={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}$, so I can begin with the left-hand side of this third equation:

$\begin{array}{rl}{S}^{-1}{\gamma}^{\mu}S& ={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}{\gamma}^{\mu}S\\ & ={\gamma}^{0}(1+\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}){\gamma}^{0}{\gamma}^{\mu}(1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu})\\ & ={\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}+{\gamma}^{0}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}\frac{i}{4}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}{\gamma}^{0}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}\left({\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\right){\gamma}^{\mu}.\end{array}$

Remember, we want this to be equal to

${\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}=({\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu}){\gamma}^{\nu}={\gamma}^{\mu}+{\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

The first term is there already, but I have no idea how that second term is going to work out to ${\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}$. Can anyone give me a hint, or tell me what I’ve done wrong?

$S=1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}$

represents the infinitesimal Lorentz transformation

${\mathrm{\Lambda}}^{\mu}{}_{\nu}={\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu},$

in the sense that

${S}^{-1}{\gamma}^{\mu}S={\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

I have already proven that ${S}^{-1}={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}$, so I can begin with the left-hand side of this third equation:

$\begin{array}{rl}{S}^{-1}{\gamma}^{\mu}S& ={\gamma}^{0}{S}^{\u2020}{\gamma}^{0}{\gamma}^{\mu}S\\ & ={\gamma}^{0}(1+\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}){\gamma}^{0}{\gamma}^{\mu}(1-\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu})\\ & ={\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}+{\gamma}^{0}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}{\gamma}^{0}{\gamma}^{\mu}-{\gamma}^{0}\frac{i}{4}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}\frac{i}{4}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}{\gamma}^{0}{\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\gamma}^{0}{\gamma}^{\mu}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\\ & ={\gamma}^{\mu}+\frac{1}{16}\left({\sigma}_{\rho \tau}{\epsilon}^{\rho \tau}{\sigma}_{\mu \nu}{\epsilon}^{\mu \nu}\right){\gamma}^{\mu}.\end{array}$

Remember, we want this to be equal to

${\mathrm{\Lambda}}^{\mu}{}_{\nu}{\gamma}^{\nu}=({\delta}^{\mu}{}_{\nu}+{\epsilon}^{\mu}{}_{\nu}){\gamma}^{\nu}={\gamma}^{\mu}+{\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}.$

The first term is there already, but I have no idea how that second term is going to work out to ${\epsilon}^{\mu}{}_{\nu}{\gamma}^{\nu}$. Can anyone give me a hint, or tell me what I’ve done wrong?