Isomorphisms of the Lorentz group and algebra

Question

I'm trying to read a few books on QFT and some seem to say the Lorentz algebra obeys $\mathfrak{so}(1,3)\otimes \mathbb{C} \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2)$ while others say $\mathfrak{so}(1,3)\otimes \mathbb{C} \cong \mathfrak{su}(2)\otimes \mathfrak{su}(2)$. However, I didn't think one could take tensor products of Lie algebras and get another Lie algebra. Does anyone know how to decompose $\mathfrak{so}(1,3)$ into combinations of $\mathfrak{su}(2)$. Also, how does this work for the corresponding Lie groups; i.e., is $SO(1,3)\otimes \mathbb{C} \cong SU(2) \oplus SU(2)$, or $SO(1,3) \cong SU(2)\otimes SU(2)$, etc. ?

Answer 1

The (vector space) tensor product of two Lie algebras isn't naturally a Lie algebra; the obvious choice should fail to satisfy the Jacobi identity. The sources you've been reading probably mean the direct product $\mathfrak{su}(2) \times \mathfrak{su}(2)$, which is just the direct sum.

(Also, the statement is slightly incorrect. $\mathfrak{su}(2)$ should be replaced with its complexified form $\mathfrak{su}(2) \otimes \mathbb{C} \cong \mathfrak{sl}_2(\mathbb{C})$.)

Lie groups also don't have a notion of tensor product. The correct construction in the Lie group case is the direct product again, although the correct statement is complicated; there isn't obviously a notion of the complexification of a Lie group in the same way that there is a notion of the complexification of a Lie algebra.