# Is $\mathfrak sl(2,\Bbb C)_\Bbb R \cong \mathfrak so(1,3)_\Bbb R$?

That is: is $$\mathfrak sl(2,\Bbb C)$$ isomorphic to $$\mathfrak so(1,3)$$ when both are considered as real algebra?

I am using the following six generators for $$\mathfrak sl(2,\Bbb C)_\Bbb R$$, as by definition $$X \in \mathfrak sl(2,\Bbb C)$$ if $$tr(X)=0$$

$$X_1=\left( \begin{matrix} 1 & 0\\ 0 & -1\\ \end{matrix} \right )\,, X_2=\left( \begin{matrix} i & 0\\ 0 & -i\\ \end{matrix} \right )\,, X_3=\left( \begin{matrix} 0 & 1\\ 0 & 0\\ \end{matrix} \right )\,, X_4=\left( \begin{matrix} 0 & 0\\ 1 & 0\\ \end{matrix} \right )\,, X_5=\left( \begin{matrix} 0 & i\\ 0 & 0\\ \end{matrix} \right )\,, X_6=\left( \begin{matrix} 0 & 0\\ i & 0\\ \end{matrix} \right )\,,$$

However, when I calculate the brackets for $$\mathfrak sl(2,\Bbb C)_\Bbb R$$ what I get is different from the usual $$\mathfrak so(3,1)_\Bbb R$$ brackets. As an example:

$$[X_1,X_2]=0\,, [X_1,X_3]=2X_3\,, [X_1,X_4]=-2X_4\,, [X_1,X_5]=2X_5\,,[X_1,X_6]=-2X_6$$

I would expect to get brackets like $$[X_1, X_2]=2iX_3$$

• Yes, the title equation is correct. This is explained in my Phys.SE answer here. Related post by OP: physics.stackexchange.com/q/669776/2451 Oct 5, 2021 at 16:34
• Show some calculations of basis generators. Oct 5, 2021 at 18:06
• I have added all the backets calculation for the first generator Oct 5, 2021 at 19:23
• I have just realized that if I factor out the $i$ I get $[X_1,X_2]=0\,, [X_1,X_3]=-2iX_5\,, [X_1,X_4]=2iX_6\,, [X_1,X_5]=2iX_3\,,[X_1,X_6]=-2iX_4$ Oct 5, 2021 at 19:58

Take $$σ_x,σ_y,σ_z$$ (or perhaps their negatives) as the generators of boosts and $$iσ_x,iσ_y,iσ_z$$ (or perhaps their negatives) as the generators of spatial rotations.
Your $$X_3$$ through $$X_6$$ generate parabolic transformations.
Just as a matrix of $$R \in \mathrm{SO}(3)$$ expands to first order as $$R = I + i \delta \omega^i J_i$$ where the $$\delta \omega^i$$ are real and the (Hermitian, without the convention of using an $$i$$ here they would be anti-Hermitian) $$J_i$$ satisfy $$[J_i,J_j] = i \varepsilon_{ijk} J_k$$ a matrix $$M \in \mathrm{SU}(2)$$ can be expanded to first order as $$M = I + i \delta \omega^i \sigma_i$$ where the $$\delta \omega^i$$ are real and the $$\sigma_i$$ satisfy $$[\sigma_i,\sigma_j] = i \varepsilon_{ijk} \sigma_k,$$
Just as a matrix of $$\mathrm{SO}(1,3)^{+}$$ expands to first order as $$R = I + \delta \omega^i J_i + \delta \omega'^i K_i$$ where the $$\delta \omega^i, \delta \omega'^i$$ are real and $$J_i,K_i$$ satisfy $$[J_i,J_j] = + i \varepsilon_{ijk} J_k$$ $$[J_i,K_j] = + i \varepsilon_{ijk} K_k$$ $$[K_i,K_j] = - i \varepsilon_{ijk} J_k$$ a matrix of $$\mathrm{SL}(2,\mathbb{C})$$ can be expanded to first order as $$M = I + i \delta \omega^i \sigma_i + i \delta \omega'^i (+i\sigma_i)$$ where the $$\delta \omega^i, \delta \omega'^i$$ are real and the $$J_i = \frac{1}{2}\sigma_i, K_i = \frac{1}{2}(+i \sigma_i)$$ satisfy $$[J_i,J_j] = + i \varepsilon_{ijk} J_k$$ $$[J_i,K_j] = + i \varepsilon_{ijk} K_k$$ $$[K_i,K_j] = - i \varepsilon_{ijk} J_k$$ We can also consider $$M = I + i \delta \omega^i \sigma_i + i \delta \omega''^i (- i\sigma_i)$$ where the $$\delta \omega^i, \delta \omega''^i$$ are real and the $$J_i = \frac{1}{2}\sigma_i, K_i' = \frac{1}{2}(-i \sigma_i)$$ satisfy $$[J_i,J_j] = + i \varepsilon_{ijk} J_k$$ $$[J_i,K_j'] = + i \varepsilon_{ijk} K_k'$$ $$[K_i',K_j'] = - i \varepsilon_{ijk} J_k$$ Thus, while the abstract commutation relations are the same for $$\mathrm{SO}(1,3)^+$$ and $$\mathrm{SL}(2,\mathbb{C})$$, showing they are isomorphic, we found two different (inequivalent) ways of associating the generators of $$\mathrm{SO}(1,3)^{+}$$ to those of $$\mathrm{SL}(2,\mathbb{C})$$, i.e. two different (inequivalent) representations of the Lie algebra of $$\mathrm{SO}(1,3)^{+}$$, and they can't be equivalent because any intertwiner $$U$$ has to send the $$J_i$$'s of one rep into the $$J_i$$'s of the other, so if $$U$$ is such that $$J_i \to J_i$$ via $$\sigma^i \to U \sigma ^i U^{-1} = \sigma^i$$ then it simply can't change the sign of the $$\sigma^i$$ in the $$K_i$$'s.
Your choice of combinations $$X_1 = 2 J_3 \ , \ X_2 = 2 K_3 \ ,$$ $$X_3 = J_1 + K_2 \ , \ X_4 = J_1 - K_2 \ , \$$ $$X_5 = K_1 - J_2 \ , \ X_6 = K_1 + J_2 \ ,$$ thus naturally satisfy commutation relations like $$[X_1,X_3] = [2J_3,J_1+K_2] = 2 i J_2 - 2 i K_1 = 2 K_2 + 2 J_1 = 2(J_1 + K_2) = 2 X_3$$ but the $$i$$ should be kept outside since we've assumed the generators $$J_i,K_i$$ are Hermitian hence the $$X_i$$ combinations are Hermitian and so the commutator should be anti-Hermitian on both sides, so we should really write $$[X_1,X_3] = - 2 i X_5.$$