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. ?

  • $\begingroup$ I am not sure, but can you draw the Dingkin diagram for both? If they coincide they should be isomorphic. However, I admit I do not know how to draw the Dingkin diagram for the direct sum of two Lie algebras... $\endgroup$ – Bombyx mori Apr 16 '13 at 6:56

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.

  • $\begingroup$ still up at this point...T_T $\endgroup$ – Bombyx mori Apr 16 '13 at 6:53
  • $\begingroup$ So, are you saying it should be $\mathfrak{so}(1,3)\otimes \mathbb{C} \cong \mathfrak{su(2)} \oplus \mathfrak{su}(2)$ and $SO(1,3)\otimes \mathbb{C} \cong SU(2) \oplus SU(2)$? $\endgroup$ – user71210 Apr 16 '13 at 6:59
  • $\begingroup$ No, I'm saying that the first thing should be $\mathfrak{so}(1, 3) \otimes \mathbb{C} \cong \mathfrak{sl}_2(\mathbb{C}) \times \mathfrak{sl}_2(\mathbb{C})$ (I don't want to call this construction the direct sum) and that the second thing is not obviously meaningful to do (and the direct product of Lie groups definitely shouldn't be called the direct sum). $\endgroup$ – Qiaochu Yuan Apr 16 '13 at 7:05
  • $\begingroup$ Ok, thanks. But, why don't you want to call the direct product the direct sum in the algebra case? Is it because the Lie bracket is not closed under direct sums? Also, is there a combination of $SU(2)$'s that is related to $SO(1,3)$? $\endgroup$ – user71210 Apr 16 '13 at 7:15
  • $\begingroup$ It's because direct sum should mean biproduct (en.wikipedia.org/wiki/Biproduct), and the direct product of Lie algebras is not a biproduct (biproducts include the direct sum of abelian groups and vector spaces). And no, not directly, because $\text{SU}(2)$ is compact and $\text{SO}(1, 3)$ isn't, but they are related through their Lie algebras. $\endgroup$ – Qiaochu Yuan Apr 16 '13 at 7:21

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