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?!
Many thanks in advance for your help!
For anyone who is interested - I've worked out the solution myself. It turns out it is just sloppy wording.
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.
It's a pretty circular way of looking at things though, and I certainly won't be using this terminology in any of my own work!