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

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.

• 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? May 17, 2012 at 17:32
• Crossposted from physics.stackexchange.com/q/28505/2451 May 15, 2017 at 6:43
• I’m voting to close this question because it has been crossposted and answered on Physics Stack Exchange. Aug 22, 2021 at 12:27
• I’m voting to close this question because it has been crossposted and answered on Physics Stack Exchange. Oct 13, 2021 at 22:16