# Conjugate Representations of Lie Algebra of Lorentz Group

I'm trying to understand the Lie algebra of the Lorentz group and am almost there, but am stuck at the final hurdle! It's easy to prove that

$$\frak so(1,3)^\uparrow_{\mathbb{C}}=sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$$

by considering generators. Indeed $\frak so(1,3)^\uparrow$ has generators $J_i$ for rotations and $K_i$ for boosts. The complexification has basis

$$L_i^{\pm}=J_i\pm iK_i$$

and it's not hard to show [$L_i^{\pm}, L_j^{\pm}]=\epsilon_{ijk}L^\pm_k$ and $[L_i^+,L_j^-]=0$ yielding two commuting copies of the complexification of $\frak su(2)$ which is $\frak sl(2,\mathbb{C})$. Is this correct?

Now my notes say that a generic representation of $\frak so(1,3)^\uparrow_{\mathbb{C}}$ is the tensor product of the spin-$j_1$ representation of $\frak sl(2,\mathbb{C})$ and the spin-$j_2$ conjugate representation of $\frak sl(2,\mathbb{C})$. Where does this conjugate business come from? I can't make head or tail of it!

Note: I know that this makes physical sense, since then the $(0,\frac 12)$ representation yields right handed spinors and the $(\frac 12,0)$ representation gives left handed spinors. But where does it come from mathematically?!

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The generic representation of the Lorentz algebra is the tensor product of two spin representations of $\mathfrak{sl}(2,\mathbb{C})$, labelled $(j_1,j_2)$. Now we can see that the $(j_1,j_2)$ representation is conjugate to the $(j_2,j_1)$ representation, by plugging in the definitions of $J,K$ in terms of $L$ and seeing what happens.
This means that one can regard the $(0,j)$ representation as the conjugate of the $(j,0)$ representation. Now identifying the $(j,0)$ representation with the spin-$j$ representation of $\mathfrak{sl}(2,\mathbb{C})$ as a complex Lie algebra, the nomenclature makes sense.