$3+3$ 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?
 A: *

*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.


*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:
$$\begin{align} A^{\prime 4i}~=~&R^i{}_j A^{4j}, \cr
A^{\prime k\ell} ~=~& R^k{}_iA^{ij} R^{\ell}{}_j, \cr i,j,k,\ell~\in~&\{1,2,3\}.\end{align}\tag{4} $$
References:

*

*A. Zee, Group Theory in a Nutshell for Physicists, 2016; p. 197.

A: 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).
