Algebra of the rotation generators of the Lorentz Group I’m studying the generators of Lorentz group in QFT.
How can I prove the algebra of the rotation generators $J_i$
$$\left [  J_i,J_j\right ] =i\epsilon_{ijk}J_k $$
from the algebra of the Lorentz group generators $M^{\mu\nu}$?
 A: Write $J_i$ in terms of $M_{\mu \nu}$, namely, $J_i=\tfrac 1 2\varepsilon_{ijk}M_{jk}$ and compute the commutator.
Let's use only the spatial part
\begin{eqnarray}
[M_{ij},M_{kl}]=i \left ( M_{ik}\delta_{jl}-M_{jk}\delta_{il}-M_{il}\delta_{jk}+M_{jl}\delta_{ik}\right),
\end{eqnarray}
then
\begin{eqnarray}
[J_p,J_q]& = &\frac{1}{4}\varepsilon_{ijp}\varepsilon_{klq}[M_{ij},M_{kl}]=
\frac{i}{4}\varepsilon_{ijp}\varepsilon_{klq} \left ( M_{ik}\delta_{jl}-M_{jk}\delta_{il}-M_{il}\delta_{jk}+M_{jl}\delta_{ik}\right)\\
& = &
\frac{i}{4}\left ( M_{ik}\varepsilon_{ijp}\varepsilon_{kjq}-M_{jk}\varepsilon_{ijp}\varepsilon_{kiq}-M_{il}\varepsilon_{ijp}\varepsilon_{jlq}+M_{jl}\varepsilon_{ijp}\varepsilon_{ilq}\right)=i M_{ik}\varepsilon_{jip}\varepsilon_{jkq} \\
& = &
i M_{ik}(\delta_{ik}\delta_{pq}-\delta_{iq}\delta_{pk})=iM_{pq}=i\varepsilon_{pq k} J_k.
\end{eqnarray}
The way it is supposed to be. Do not forget that there is no $\frac{1}{2}$ for the inverse relation between $J$ and $M$, namely, that $M_{jk}=\varepsilon_{jki}J_i$. To see that multiply $J_i=\tfrac 1 2\varepsilon_{ijk}M_{jk}$ by $\varepsilon_{ipq}$, that will lead to
\begin{equation}
\varepsilon_{pqi} J_i=\frac{1}{2}\varepsilon_{ipq}\varepsilon_{ijk}M_{jk}=\frac{1}{2}(\delta_{pj}\delta_{qk}-\delta_{pk}\delta_{qj})M_{jk}=\frac{1}{2}(M_{pq}-M_{qp})=M_{pq}.
\end{equation}
