# Show the Lie algebra is the same for $SU(2) \times SU(2)$ and Lorentz group [duplicate]

So I know: $$[\sigma_{I},\sigma{j}] = 2i \epsilon_{ijk} \sigma_{k}$$ So two products of this should give us the Lorentz group: $SO(4) = SU(2) \times SU(2)$

Where $SO(4)$ has 3 Lie algebra which can be expressed as one by denoting: $$J_{I}^{\pm} = \frac{1}{2} (J_{I} \pm ik_{I})$$

This gives two Lie algebra: $$[J_{I}^{\pm}, J_{j}^{\pm}] = i\epsilon_{ijk}J_{k}^{\pm}$$

But how is this related to our original product of SU(2)? I was under the impression if you have two Lie groups, $(g_{1},h_{1})$ and $(g_{2}, h_{2})$, then the product would be $(g_{1}g_{2},h_{1}h_{2})$. However, I tried that with the Pauli matrices and it doesn't work to be the same as $SO(4)$ Lie algebra. How should this be done?

• – Qmechanic Apr 20 '14 at 20:54
• A nitpick: QFT and its notations are not too wonted or known to me, but as an outsider I get confused when you call $SO(4)$ the Lorentz group. Do you mean to say $SO(3,\,1)$ (i.e. nontrivial signature)? – Selene Routley Apr 22 '14 at 0:05
• The conclusion of your first statement is NOT correct. As Lie groups $$SU(2) \times SU(2) \neq SO(4)$$ rather $$(SU(2) \times SU(2)) / \mathbb{Z}_2 \simeq SO(4)$$ As Lie algebras $$\mathfrak{su}(2) \oplus \mathfrak{su}(2) \simeq \mathfrak{so}(4)$$ These are inequivalent statements, albeit they are closely related. – Flint72 May 21 '14 at 14:02