First of all, I am a pedestrian in group theory. I have a general question and two particular ones.
General question: I am trying to show that $so(3,1)\simeq sl(2,\mathbb{C})$ by redefining its generators, namely, in the spirit of the following analog procedure for its Lorentzian counterpart, the isomorphism $$so(2,2)\simeq so^+(2,1)\oplus so^-(2,1)\simeq sl^+(2,\mathbb{R})\oplus sl^-\left(2,\mathbb{R}\right).$$ This question has been asked in a lot of PSE questions, e.g., Refs. 1, 2, 3, but it's not clear to me.
The $so(2,2)$ algebra is \begin{align} \left[J_a,J_b\right]=\eta^{cd}\epsilon_{abc}J_d\,,\quad \left[J_a,P_b\right]=\eta^{cd}\epsilon_{abc}P_d\,,\quad \left[P_a,P_b\right]=-\Lambda\eta^{cd}\epsilon_{abc}J_d\,, \end{align} where $\Lambda=-1/\ell^2<0$ the cosmological constant, $\ell$ the AdS$_3$ radius and $a,b,c,d=0,1,2$. If we define $Y_a^\pm=\frac{1}{2}\left(J_a\pm\ell P_a\right)$, then \begin{align} \left[Y_a^\pm,Y_b^\pm\right]=\eta^{cd}\epsilon_{abc}Y_d^\pm\,. \end{align} Thus, we proved the first isomorphism. The final is addressed through \begin{align} L_{-1}^\pm=-2Y_0^\pm\,,\quad L_0^\pm=Y_2^\pm\,,\quad L_1^\pm=Y_1^\pm\,, \end{align} from where we find \begin{align} \left[L_n^\pm,L_m^\pm\right]=(n-m)L_{n+m}^\pm\,. \end{align} On the other hand, the $so(3,1)$ algebra is \begin{align} \left[\tilde{J}_a,\tilde{J}_b\right]=\delta^{cd}\epsilon_{abc}\tilde{J}_d\,,\quad \left[\tilde{J}_a,\tilde{P}_b\right]=\delta^{cd}\epsilon_{abc}\tilde{P}_d\,,\quad \left[\tilde{P}_a,\tilde{P}_b\right]=-\frac{1}{\ell^2}\delta^{cd}\epsilon_{abc}\tilde{J}_d \end{align} In this regard, I ask the following particular questions.
- It is possible to do the same, or an analog procedure, to reveal the isomorphism $so(3,1)\simeq sl(2,\mathbb{C})$? If not, how one can do it by considering redefinition of the $so(3,1)$ generators?
- Why for reals one obtains two copies of $sl(2,\mathbb{R})$ and for complex-valued generators not?
