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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?

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  • $\begingroup$ $ $ Which page? $\endgroup$ – Qmechanic Jun 24 at 4:35
  • $\begingroup$ It's on page 197. $\endgroup$ – LaserTotoro Jun 24 at 10:45
  • $\begingroup$ @LaserTotoro We are talking about complexified Lie algebras here, right? $\endgroup$ – MadMax Jun 25 at 21:23
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  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: $$ A^{\prime 4i}~=~R^i{}_j A^{4j}, \qquad A^{\prime k\ell} ~=~ R^k{}_iA^{ij} R^{\ell}{}_j, \qquad i,j,k,\ell~\in\{1,2,3\}.\tag{4} $$

References:

  1. A. Zee, Group Theory in a Nutshell for Physicists, 2016; p. 197.
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  • $\begingroup$ 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 $A$. $\endgroup$ – LaserTotoro Jun 24 at 10:42
  • $\begingroup$ $ $ Yes, exactly. $\endgroup$ – Qmechanic Jun 24 at 10:48
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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).

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