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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 $[X_1, X_2]=2iX_3$

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    $\begingroup$ 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 $\endgroup$
    – Qmechanic
    Commented Oct 5, 2021 at 16:34
  • $\begingroup$ Show some calculations of basis generators. $\endgroup$
    – DanielC
    Commented Oct 5, 2021 at 18:06
  • $\begingroup$ I have added all the backets calculation for the first generator $\endgroup$
    – Andrea
    Commented Oct 5, 2021 at 19:23
  • $\begingroup$ I have just realized that if I factor out the $i$ I get $[X_1,X_2]=0\,, [X_1,X_3]=-2iX_5\,, [X_1,X_4]=2iX_6\,, [X_1,X_5]=2iX_3\,,[X_1,X_6]=-2iX_4$ $\endgroup$
    – Andrea
    Commented Oct 5, 2021 at 19:58

2 Answers 2

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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.

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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 $$[X_1,X_3] = - 2 i X_5.$$

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