Relationship of $\mathfrak{so}(1,3)$ to $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ I'm sure I'm going to miss some mathematical detail in this, but I want to make sure that I have the gist of the $\mathfrak{so}(1,3) \to \mathfrak{su}(2)\oplus\mathfrak{su}(2)$ relationship correct.
$\mathfrak{so}(1,3)$ is formed of the six generators of proper orthochronous Lorentz group, $\mathrm{SO}(1,3)$, three $J_{i}$ and three $K_{i}$, with the Lie brackets:
$$
[J_i,J_j] = i\varepsilon_{ijk}J_k \\
[K_i,K_j] = -i\varepsilon_{ijk}J_k \\
[J_i,K_j] = i\varepsilon_{ijk}K_k
$$
If you redefine these as $A_p = \frac{1}{2}(J_p + iK_p)$ and $B_p = \frac{1}{2}(J_p - iK_p)$, these new generators span a subset of complexification of $\mathfrak{so}(1,3)$, $\mathfrak{so}(1,3)_{\mathbb{C}}$, that is in fact isomorphic to $\mathfrak{so}(1,3)$.
The new Lie brackets are then given by:
$$
[A_p,A_q] = i\varepsilon_{pqr}A_r \\
[B_p,B_q] = i\varepsilon_{pqr}B_r \\
[A_p,B_q] = 0
$$
so in fact the $\{A_p\}$ are isomorphic to $\mathfrak{su}(2)$, as are the $\{B_p\}$ independently. Hence $\mathfrak{so}(1,3)$ is isomorphic to a restriction of the full $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ Lie algebra to just generators of the form:
$$
A_p = \sigma_p \otimes I \\
B_p = I \otimes \sigma_p
$$
Then, by rearranging $A_p = \frac{1}{2}(J_p + iK_p)$ and $B_p = \frac{1}{2}(J_p - iK_p)$ we get:
$$
J_p = A_p + B_p \\
K_p = -i(A_p - B_p) \\
$$
from which it follows that:
$$
J_p = \sigma_p \otimes I + I \otimes \sigma_p \\
K_p = -i(\sigma_p \otimes I - I \otimes \sigma_p)
$$
which is the standard result that's on the Wikipedia page (barring the extra minus sign I seem to have in $K_p$...).
Hence, you can find the representations of $J_p$ and $K_p$ in terms of the representations of $\mathfrak{su}(2)$, which are well-known. What I don't understand is why you have the freedom to choose different dimensions for the representations of the two copies of $\mathfrak{su}(2)$. Is it just because the tensor product preserves the dimension of the product of square matrices under commutation?
 A: I am not sure about your conventions, but it is not true that $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ are isomorphic or that $\mathfrak{so}(1,3)$ is a subalgebra of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. The both have dimension $6$ and they both have the same complexification, namely $\mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C)$. However, $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ is a compact real form, while $\mathfrak{so}(1,3)$ is non-compact (which can be seen from that fact that the Killing form is definite respectively indefinite). 
Nonetheless, the real Lie algebras $\mathfrak{so}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ have the same complex representations. This is because any complex representation of a real Lie algebra uniquely extends to the complexification, so in both cases complex representations are equivalent to complex representations of $\mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C)$. In the language of $\mathfrak{so}(1,3)$ the fact that there are "two components" of a representation is best seen when building up all representations from spinors. 
