# Representation of $SO(3)$ acting on 3-index tensors in Zee's Group Theory Book

In page 193, the book Group Theory in a Nutshell for Physicists started to explain inductively how we need to consider only the traceless symmetric (in all indices) tensors to construct $$3^j$$-dimensional representation of $$SO(3)$$ acting on $$j$$-index tensors\$.

When he proved this fact for $$j=3$$, he considered antisymmetric combination,$$T^{[ij]k}= T^{ijk}-T^{kij}.$$

Since we have the the following relation, $$B^{lk}=\varepsilon^{ijl}T^{[ij]k}$$ we can say that $$T^{[ij]k}$$ transforms like 2-index tensors.

When we construct 3^3-dimensional representation of $$SO(3)$$, is it true that the part of the matrix that is furnished by antisymmetric combinations has dimensions of $$9 = 5 \bigoplus 3 \bigoplus 1$$?

In addition, we know that the dimension of representation furnished by totally symmetric traceless tensors is $$7 = 3*2 + 1$$. Dimension of representations furnished by three different traces is 3.

We obtained $$9 + 7 + 3 = 19$$ dimensions. So, we still need 8 dimensional representation to reach $$3^3$$. Does this 8-dimension correspond to representation associated to $$T^{\{ij\}k} = T^{ijk}+T^{jik}$$?

The book that I referred to: https://press.princeton.edu/titles/10773.html

• You might, or might not, be assisted in your reduction of your matrix by recalling that combining three spin one objects gives you a spin 3 one, two spin 2 s, and three spin 1s, and a singlet. So 27=7⨁5⨁5⨁3⨁3⨁3⨁1. So there are seven separate irreducible pieces in your 3x3x3 array. – Cosmas Zachos Apr 19 '19 at 15:03

It is true that a two-index tensor has 9 components, and reduces to $$3\otimes 3= 5\oplus 3\oplus 1 ,$$ familiar from adding two spin ones to get a spin 2, a spin one, and a spin 0 singlet. This is Tony's lowest rung. The 3 is antisymmetric, whereas the 5 and singlet are symmetric.

A 3-index tensor has 27 components, $$3\otimes 3\otimes 3= 7\oplus 5\oplus 5\oplus 3 \oplus 3 \oplus 3 \oplus 1 ,$$ so three spin ones yield a spin 3, two spin 2 s, three spin 1s, and a singlet. The "why" follows.

The 9 components of $$T^{[ij]k}$$ thus amount to the lowest rung reduction mentioned, while the remaining 18 in $$3 T^{\{ ij\} k} = ( T^{\{ ij\} k}+ T^{\{ jk \}i} + T^{\{ki \} j}~) +( T^{\{ ij\} k}- T^{\{ jk \}i} ~) + ( T^{\{ ij\} k} - T^{\{ki \} j}~)$$ consist of a significant fully symmetric "new" piece, $$S^{ijk}$$, the first parenthesis, with 10 components, so $$7\oplus 3 ,$$ (recall, as Tony emphasizes and you may check by inspection, only one trace is independent!); and the other two parentheses' terms drop "down the inductive ladder" to the lower rung as being effectively rank two.

The author does not remind you what you may also check, namely that these two terms may be rewritten as linear combinations of perms of the rank-two type $$T^{[ij]k}$$ downladder terms, so they are not independent!

They, in fact, introduce only 8 independent components together, so only introduce a novel $$5\oplus 3$$.

Grouping all seven irreducible blocks together, one has the full reduction of the 27 components displayed above.

• Sir, thank you for the answer. The addition-of-angular-momentum approach to the problem makes it easier to find irreducible representations. – Here is the thing Apr 20 '19 at 8:06