# How do I construct the $\operatorname{SU}(2)$ representation of the Lorentz Group using that $\text{SU}(2)\times\text{SU}(2)\cong \text{SO}(3,1)$?

This question is so mathematical that I think I'll have more luck asking it in the mathematics section, than I would in the physics section.

This is problem II.3.1 in Anthony Zee's book Quantum Field Theory in a Nutshell (I'm reading this for fun- it isn't a homework problem.)

Show, by explicit calculation, that $(1/2,1/2)$ is the Lorentz Vector.

In other words, How do I construct the $\operatorname{SU}(2)$ representation of the Lorent Group using the fact that $\operatorname{SU}(2)\times\operatorname{SU}(2)\cong \operatorname{SO}(3,1)$?

Here is some background information:

Zee has shown that the algebra of the Lorentz Group is formed from two separate $\operatorname{SU}(2)$ algebras [$\operatorname{SO}(3,1)$ is isomorphic to $\operatorname{SU}(2)\times \operatorname{SU}(2)$] because the Lorentz Algebra satisfies:

$[J_{+i},J_{+j}]= i*e_{ijk} [J_{k+}]$

$[J_{-i},J_{-j}]= i*e_{ijk} [J_{k-}]$

$[J_{+i},J{-j}]=0$

The representations of $\operatorname{SU}(2)$ are labelled by $j=0,\frac{1}{2},1,\ldots$ so the $\operatorname{SU}(2)\times \operatorname{SU}(2)$ rep is labelled by $(j_+,j_-)$ with the $(1/2,1/2)$ being the Lorentz 4-vector because and each representation contains $(2j+1)$ elements so $(1/2,1/2)$ contains 4 elements.

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I now see that the generators of SU(2) are the Pauli Matrices and the generators of SO(3,1) is a matrix composed of two Pauli Matrices along the diagonal. Is it always the case that the Direct Product of two groups is formed from the generators like this? – Barry May 17 '12 at 17:32