By showing that the complexified Lie algebra of the proper Lorentz group ${\rm SO(3,1)}$ is equivalent to two the direct sum of two complexified ${\rm SU(2)}$ Lie algebras $$\mathfrak{so}(3,1)_\mathbb{C}=\mathfrak{su}(2)_\mathbb{C}\oplus \mathfrak{su}(2)_\mathbb{C}.$$ With this at our disposal, it is possible to find all the irreducible representations of that algebra.
With this technology, I do not need to know or mention anything about the group ${\rm SL(2, C)}$, its Lie algebra or its representations. The knowledge of ${\rm SO(3,1)}$ and ${\rm SU(2)}$ seems to be sufficient, I think.
- So why learn about ${\rm SL(2, C)}$ in the first place? It seems to me to be unnecessary. Please tell me if I am missing something on a conceptual level.
Best Answer
To put it simply, to know that the Lie group $SL(2,\mathbb{C})$ of $2\times 2$ complex matrices with unit determinant is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$ gives a simple and direct way to understand how the restricted Lorentz group can act on a Weyl spinor $\psi\in\mathbb{C}^2$. In contrast, the group action (on a Weyl spinor) is somewhat mysterious/less intuitive from the point of $4\times 4$ spacetime Lorentz transformations $\Lambda\in SO^+(1,3;\mathbb{R})$.
Similarly, the subgroup $SU(2)\subseteq SL(2,\mathbb{C})$ is (the double cover of) the 3D rotation group $SO(3)\subseteq SO(1,3;\mathbb{R})$.
There is a corresponding double-copy version for the complexified proper Lorentz group $SO(1,3;\mathbb{C})$. For more details, see this & this related Phys.SE posts.