# $3+3$ representation of $SO(4)$

In Zee's Group Theory in a Nutshell book, he says that the antisymmetric tensor $$A^{ij}$$ furnishes a 6 dimensional representation of $$SO(4)$$. He further argues that this 6 dimensional representation can be broken up into two 3 dimensional representations by listing the following grouping ($$A^{14}$$, $$A^{24}$$, $$A^{34}$$) and ($$A^{12}$$, $$A^{23}$$, $$A^{31}$$). Why does this make sense? Why do the elements in each group only transform amongst themselves?

•  Which page? – Qmechanic Jun 24 at 4:35
• It's on page 197. – LaserTotoro Jun 24 at 10:45
• @LaserTotoro We are talking about complexified Lie algebras here, right? – MadMax Jun 25 at 21:23

1. OP is correct: The Lie algebra isomorphism $$so(4)~\cong~ so(3)_+~\oplus~ so(3)_-\tag{1}$$ uses (anti)self-dual real antisymmetric $$4\times 4$$ matrices, respectively, cf. e.g. this Math.SE post.
2. However Ref. 1 is trying to make another point. Instead Ref. 1 is discussing branching rules for the subgroup $$H:=SO(3)~\ni R ~\mapsto~ \begin{pmatrix} R & \vec{0}\cr \vec{0}^t & 1 \end{pmatrix}_{4\times 4}~\in ~ G:=SO(4).\tag{2}$$ In this specific embedding (2) of the subgroup, the explicit splitting of the $$G$$-representation into irreducible $$H$$-representations $${\bf 6}_G\cong {\bf 3}_H\oplus {\bf 3}_H\tag{3}$$ is as Zee writes: $$A^{\prime 4i}~=~R^i{}_j A^{4j}, \qquad A^{\prime k\ell} ~=~ R^k{}_iA^{ij} R^{\ell}{}_j, \qquad i,j,k,\ell~\in\{1,2,3\}.\tag{4}$$
• I agree. I think the way to fix it is to consider $\frac{A^{ij}+B^{ij}}{2}$ and $\frac{A^{ij}-B^{ij}}{2}$ where $B$ is the dual of $A$. – LaserTotoro Jun 24 at 10:42
•  Yes, exactly. – Qmechanic Jun 24 at 10:48
The Lie algebras of SO(4) and SU(2)$$\times$$SU(2) are isomorphic, so the you can get representations of SO(4) former by taking the tensor product of two representations of SU(2).