The Lie algebra of $SL(2,C)$ is given by $[X_1,X_2]=2X_2$, $[X_1,X_3]=-2X_3$ , and $[X_2,X_3]=X_1$, while the Lie algebra of the Lorentz group is given by: $\left[J_{i}, J_{j}\right]=\epsilon_{i j k} J_{k}, \quad\left[J_{i}, K_{j}\right]=\epsilon_{i j k} K_{k}, \quad\left[K_{i}, K_{j}\right]=-\epsilon_{i j k} J_{k}$ . Why are these two Lie algebras isomorphic?

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    $\begingroup$ Does this answer your question? What's the relationship between $SL(2,\mathbb{C})$, $SU(2)\times SU(2)$ and $SO(1,3)$? $\endgroup$ – ohneVal Mar 23 at 23:08
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    $\begingroup$ The exact question posed has an answer in the Qmechanic's answer here: physics.stackexchange.com/questions/28505/…. More precisely, if one shows that the Lie groups $\text{SL}(2,\mathbb C)$ and $\frak{Lor}$$(1,3)$ are homeomorphic, there is a theorem in differential geometry which states that their Lie algebras are homeomorphic. If, moreover, the two groups are locally isormorphic, then their Lie algebras are isomorphic. $\endgroup$ – DanielC Mar 24 at 0:00
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    $\begingroup$ By the way, you offer a $\mathbb C$-basis for the vector space of $\frak{sl}$ $(2,\mathbb C)$ which has only three generators. You need an $\mathbb R$-basis which has six. Then you have a premise: the two Lie algebras are isomorphic as real Lie algebras. $\endgroup$ – DanielC Mar 24 at 0:05