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The Lorentz group can be divided into two separate $SU(2)$ algebras and thus we label such representations with two spins $(j_{1},j_{2})$.

The first and second spins correspond to generators $$J^{\pm}_{i}=\frac{1}{2}(J_{i}\pm iK_{i})$$ where $J_{i}$ and $K_{i}$ are the usual rotation and boost generators, respectively.

Take the $(\frac{1}{2},\frac{1}{2})$ representation. Here we have two separate $SU(2)$ algebras with spin-$\frac{1}{2}$ and thus generators that are half the Pauli matrices. How do I reconcile these two separate algebras to obtain back the usual Lorentz group generators?

For example using $J^{\pm}_{i}=\frac{1}{2}\sigma_{i}$ and solving for the $J_{i}$ and $K_{i}$ does not help since the result is $2\times 2$ matrices, whereas we need $4\times 4$ matrices. I believe a tensor product is needed but I don't really understand how to make this work, so would anybody be able to help?

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The Lorentz group cannot be divided into "two separate $\mathrm{SU}(2)$ algebras". This is impossible because a group and an algebra are two different things and because the actual relation is that the complexification of the Lorentz algebra, $\mathfrak{so}(1,3)_\mathbb{C}$ is isomorphic to two copies of the complexification of $\mathfrak{su}(2)$, i.e. $$ \mathfrak{so}(1,3)_\mathbb{C}\cong \mathfrak{su}(2)_\mathbb{C}\oplus\mathfrak{su}(2)_\mathbb{C}.\tag{1}$$ The labels $(j_1,j_2)\in \frac{1}{2}\mathbb{Z}\times\frac{1}{2}\mathbb{Z}$ refer to the $j_1$-spin representation of the first $\mathfrak{su}(2)$ and the $j_2$-spin representation of the second one. This corresponds to your description by noting that $$ J^\pm_i \mapsto \frac{1}{2}(J_i \pm \mathrm{i}K_i)\tag{1'}$$ provides the isomorphism from the r.h.s. to the l.h.s, labelling the generators of the first $\mathfrak{su}(2)$ as $J^+$ and that of the second as $J^-$.

In general, for two Lie algebras $\mathfrak{g},\mathfrak{h}$, we obtain irreducible representations of $\mathfrak{g}\oplus\mathfrak{h}$ as tensor products of irreducible representations of the individual algebras, so indeed the $(j_1,j_2)$-labelled representation is supposed to be the tensor product of the $j_1$-spin and the $j_2$-spin representation.

If you want to obtain the matrix form of the $J^\pm$ (and therefore of the $J_i$ and $K_i$ by inverting the isomorphism) in a certain representation, you need to use the Kronecker product of matrices to obtain the group elements $\mathrm{e}^{\mathrm{i}J_i^\pm}$ from the known expression for the pauli matrices $\mathrm{e}^{\mathrm{i}\sigma_i}$. On the level of the algebra, you get that the representation works as $$ \pi_{(j_1,j_2)}(g,h) = \pi_{j_1}(g)\otimes\mathbf{1}_2 + \mathbf{1}_2\otimes \pi_{j_2}(h)\tag{2}$$ where $\otimes$ is again the tensor/Kronecker product, $(g,h)\in\mathfrak{su}(2)_\mathbb{C}\oplus\mathfrak{su}(2)_\mathbb{C}$ and $\pi_{j_i}$ are the standard spin-$j_i$ representations. So compute the forms of $J^\pm_i$ by this formula - the matrix of e.g $J^+_i$ is $\pi_{j_1,j_2}(J^+_i,0)$ - and then obtain the forms of the $J_i,K_i$ by inverting the map (1').

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