Origin of antisymmetric $\ell=2$ irrep in direct product of two symmetric second-rank tensors

Question

In the excerpt below from Chapter 18 Section 6 of the textbook Group Theory -- Application to the Physics of Condensed Matter by Dresselhaus, Dresselhaus, and Jorio, the irreducible representations of the fourth rank elasticity tensor are derived from a tensor product of two symmetric second rank tensors (with irreps $ 0 \oplus 2$). Because the elasticity tensor is itself symmetric, our degrees of freedom only stem from symmetric irreps.

My question is, why is one of the copies of $\Gamma_{\ell=2}$ antisymmetric?

Answer 1

1. The tensor $$C~=~\sum_{i,j,k,l=1}^3 C_{ij,kl} (e^i\odot e^j)\otimes (e^k\odot e^l)$$ with symmetry $$C_{ji,kl}~=~C_{ij,kl}~=~C_{ij,lk}$$ is split into symmetric and antisymmetric tensor product $${\bf 36}~\cong~{\bf 6}\otimes{\bf 6} ~\cong~{\bf 6}\otimes_{(s)}{\bf 6}\oplus{\bf 6}\otimes_{(a)}{\bf 6} ~\cong~{\bf 6}\odot{\bf 6}\oplus{\bf 6}\wedge{\bf 6} ~\cong~{\bf 21}_{(s)}\oplus{\bf 15}_{(a)}.$$

2. Recall that symmetric $3\times3$ matrices decompose into traceless and traceful irreps $${\bf 6}~\cong~{\bf 5}\oplus{\bf 1}$$ under the 3D rotation group. Next use the distributive law to rewrite in terms of irreps $$ {\bf 6}\otimes{\bf 6} ~\cong~({\bf 5}\oplus{\bf 1})\otimes ({\bf 5}\oplus {\bf 1}) ~\cong~{\bf 5}\otimes{\bf 5}\oplus({\bf 5}\otimes{\bf 1}\oplus{\bf 1}\otimes{\bf 5})\oplus {\bf 1}\otimes{\bf 1},$$ where $$ {\bf 5}\otimes{\bf 5}~\cong~({\bf 9}\oplus{\bf 5}\oplus{\bf 1})_{(s)}\oplus({\bf 7}\oplus{\bf 3})_{(a)} ,$$ and $$ {\bf 1}\otimes{\bf 1}~\cong~{\bf 1}_{(s)} .$$

3. Returning to OP's question, the antisymmetric copy ${\bf 5}_{(a)}$ comes from the mixture of traceless and traceful parts: $${\bf 5}\otimes{\bf 1}\oplus{\bf 1}\otimes{\bf 5} ~\cong~{\bf 5}\otimes_{(s)}{\bf 1}\oplus{\bf 5}\otimes_{(a)}{\bf 1} ~\cong~{\bf 5}\odot{\bf 1}\oplus{\bf 5}\wedge{\bf 1} ~\cong~{\bf 5}_{(s)}\oplus{\bf 5}_{(a)}.$$