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?

  • 1
    $\begingroup$ It is confusingly worded! Presumably, by "antisymmetric" they mean antisymmetric in the exchange of the two tensors, not antisymmetric in the exchange of pairs of indices on either one of the tensors. That is, for $\Gamma_{ij} \Gamma_{kl}$ they mean $(i, j) \leftrightarrow (k, l)$ and not $(i, j) \leftrightarrow (j, i)$ or $(k, l) \leftrightarrow (l, k)$. $\endgroup$
    – knzhou
    Dec 24, 2019 at 19:48

1 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: $${\bf 5}\otimes{\bf 1}\oplus{\bf 1}\otimes{\bf 5} ~\cong~{\bf 5}\otimes_{(s)}{\bf 1}\oplus{\bf 5}\otimes_{(a)}{\bf 1} ~\cong~{\bf 5}\odot{\bf 1}\oplus{\bf 5}\wedge{\bf 1} ~\cong~{\bf 5}_{(s)}\oplus{\bf 5}_{(a)}.$$

  • $\begingroup$ Thanks Qmechanic. I'm unfamiliar with the $\bigodot$ and $\bigwedge$ notation and their use for that specific expansion. Could you please recommend a reference? $\endgroup$
    – Tess
    Dec 24, 2019 at 22:12
  • $\begingroup$ I added some links. $\endgroup$
    – Qmechanic
    Dec 25, 2019 at 0:22

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