I want to find the irreducible representations of the Lie algebra $\mathrm{so}(3,1)$. I know the standard procedure goes on like
- Complexify the Lie algebra, obtaining the complex Lie algebra $A_1\oplus A_1$;
- Obtain all the irreducible representations of $A_1$
- Build a representation of $A_1\oplus A_1$ from the product of two irreducible representations of $A_1$.
Now I should go back to the real algebra $\mathrm{so}(3,1)$, but I don't know how. I know that if we have a (complex-linear) representation of a complex Lie algebra $L$ we can use its Weyl canonical basis (constructed using the Cartan subalgebra and the root system) to create a real-linear representation of the compact real section of $L$. In my case, however, the compact real section is $\mathrm{so}(4)$ (or $\mathrm{su}(2)\oplus\mathrm{su}(2)$ if you like), not $\mathrm{so}(3,1)$.
In its article on the representation theory of the Lorentz group, Wikipedia says that
[...] all irreducible representations of $\mathrm{so}(3,1)_{\mathbb{C} }$, and, by restriction, those of $\mathrm{so}(3,1)$ are obtained.
What "restriction" is it about? Would someone explain how can we construct a representation of a real Lie algebra from the representations of its complexification?
Let $\mathfrak{g}$ be a Lie algebra, then $\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ is a complex Lie algebra where we extend the Lie bracket by complex linearity. The map $i : \mathfrak{g} \to \mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ given by $X \mapsto X\otimes 1$ is a Lie algebra homomorphism.
Recall that a representation of a Lie algebra is just a Lie algebra homomorphism to $\mathfrak{gl}(V)$. Therefore, if $\rho : \mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C} \to \mathfrak{gl}(V)$ is a representation of $\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$, then $\rho\circ i : \mathfrak{g} \to \mathfrak{gl}(V)$ is a representation of $\mathfrak{g}$ (because the composition of Lie algebra homomorphisms is a Lie algebra homomorphism).