I know the decomposition of general rank two tensors $X_{ab}$

\begin{align} X_{ab} &= X_{[ab]} + \frac{1}{n}\delta_{ab}\delta^{cd}X_{cd} + \left(X_{(ab)}-\frac{1}{n} \delta_{ab}\delta^{cd}X_{cd}\right) \\ & = X_{[ab]} + X_{(ab)} \end{align}

How about the general rank three tensors?


This answer has been directly copied from my answer to Finding the irreducible components of a rank 3 tensor on the Mathematics Stack Exchange site.

Using the bold-number notation to denote the representation of dimension $2\ell+1$ (with total angular momentum number $\ell$), the decomposition you give of rank-2 tensors is equivalent to the decomposition $$ \mathbf{3}\otimes\mathbf{3} = \mathbf{1} \oplus \mathbf{3} \oplus \mathbf{5} $$ for the $\rm SO(3)$ rotation group, with the $\mathbf{1}$, $\mathbf{3}$ and $\mathbf{5}$ corresponding, respectively, to the scalar, vector, and symmetric traceless tensors. (In terms of dimensions, you can write $\dim(\mathbf{3}\otimes\mathbf{3} )=3\times 3=9$ and $\dim(\mathbf{1} \oplus \mathbf{3} \oplus \mathbf{5})= 1 + 3 + 5=9$.)

For rank 3, you basically do the same: \begin{align} \mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3} & = (\mathbf{3}\otimes\mathbf{3})\otimes\mathbf{3} \\ & = (\mathbf{1} \oplus \mathbf{3} \oplus \mathbf{5})\otimes\mathbf{3} \\ & = (\mathbf{1}\otimes\mathbf{3}) \oplus (\mathbf{3}\otimes\mathbf{3}) \oplus (\mathbf{5}\otimes\mathbf{3}) \\ & = \mathbf{1} \oplus (\mathbf{1} \oplus \mathbf{3} \oplus \mathbf{5}) \oplus (\mathbf{3} \oplus \mathbf{5} \oplus \mathbf{5}) \\ & = \mathbf{1} \oplus \mathbf{3} \oplus \mathbf{3} \oplus \mathbf{3} \oplus \mathbf{5} \oplus \mathbf{5} \oplus \mathbf{7}. \end{align} (In terms of dimensionality, $$\dim(\mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3})=3^3=27$$ whereas \begin{align} \dim(\mathbf{1} \oplus \mathbf{3} \oplus \mathbf{3} \oplus \mathbf{3} \oplus \mathbf{5} \oplus \mathbf{5} \oplus \mathbf{7}) & = 1+3+3+3+5+5+7 = 27, \end{align} which works out fine.)

This means that the decomposition into irreducible representations must contain

  • one scalar representation,
  • three vector representations,
  • two quadrupole representations (isomorphic to symmetric traceless tensors of rank 2), and
  • one octupole representation.

It is important to note that this is only true for a general rank-3 tensor, with no symmetry relations between the various indices. If there are symmetry constraints, the picture will obviously change.

So, the next question is, of course, what are these tensor representations?

The structure of these representations is encased within the symmetry of the tensor, so it's important to break it down as such. Thus, we separate $T$ into $$T_{ijk} = S_{ijk} + A_{ijk} + N_{ijk},$$ where $S$ is totally symmetric under exchange of indices, $A$ is totally antisymmetric, and $N$ is a remainder which is neither symmetric nor antisymetric.

With that in hand, let's see what representations are in each symmetry sector.

  • The symmetric part $S$ contains the first vector representation, by contracting it with a Kronecker delta, $\alpha_k = \delta_{ij}S_{ijk}$ (its "trace part"), or $$\alpha_k = \frac13\left( \delta_{ij}T_{ijk} + \delta_{ij}T_{ikj} + \delta_{ij}T_{kij}\right)$$ in terms of our initial tensor. This can be lifted back to a full rank-3 tensor which contributes directly to $T$ in the form $$K_{ijk} = \frac15 \left( \alpha_i \delta_{jk} + \alpha_j \delta_{ik} + \alpha_k \delta_{ij} \right).$$

  • The remaining part of $S$, defined as $R$ in $S_{ijk} = K_{ijk}+ R_{ijk}$, is the octupole, $\ell=\mathbf 3$, representation. (In other words, the totally symmetric part $S$ of $T$ is already "almost" an invariant subspace under $\rm SO(3)$, and all we needed to do was remove a vector component.)

  • Turning now to the antisymmetric part, this is already an irreducible representation as it stands, because it is actually just a one-dimensional subspace. This is the scalar representation, which can be seen because there is only a single rank-3 totally antisymmetric isotropic tensor, the Levi-Civita tensor, so we must have $A_{ijk} = A\, \epsilon_{ijk}$, with $A=\frac16 \epsilon_{ijk}A_{ijk}=\frac16 \epsilon_{ijk}T_{ijk}$ the relevant (pseudo)scalar.

The rest of the game (two vector and two quadrupole representations) must therefore be encoded in the mixed-symmetry remainder, $N_{ijk}$. This mixed-symmetry part looks quite intimidating (starting with, what does mixed-symmetry even actually imply?), but here is the simple trick to make sense of it: just do a partial contraction with the Levi-Civita symbol, i.e. something of the form $$ A_{ij} = \epsilon_{ikl}T^{jkl}, \quad B_{ij} = \epsilon_{ikl}T^{kjl}, \quad C_{ij} = \epsilon_{ikl}T^{klj}, $$ which gives you three separate rank-2 tensors directly from $T$. However, it's important to note that these are not all independent, since one can show that their sum, $$ A_{ij} + B_{ij} + C_{ij} = 3\epsilon_{ikl}A_{jkl}, $$ reduces to the antisymmetric part. (On the other hand, the symmetric part $S$ does not contribute at all to the $B$ and $C$ tensors.) The upshot from that is that we can forget about $A_{ij}$ (which is then just a re-encoding of the scalar $A_{ijk} = A\, \epsilon_{ijk}$) and use it to get two independent rank-2 tensors, $$ \tilde B_{ij} = B_{ij} - 2A\delta_{ij} \quad\text{and}\quad \tilde C_{ij} = C_{ij} - 2A\delta_{ij}, $$ (whose definition ensures they are all traceless), which are given directly in terms of the mixed-symmetry part $N$ as $$ \tilde B_{ij} = \epsilon_{ikl}N^{kjl} \quad\text{and}\quad \tilde C_{ij} = \epsilon_{ikl}N^{klj}. $$ These two rank-2 tensors are linearly independent (as are any two linearly independent combinations of $\tilde B_{ij}$ and $\tilde C_{ij}$ with $\tilde A_{ij} = \epsilon_{ikl}N^{jkl}$). As traceless rank-2 tensors they each have dimension 8, which (times 2) accounts for the full dimensionality (16) of the mixed-symmetry part $N$.

And, as traceless rank-2 tensors, each of $\tilde B_{ij}$ and $\tilde C_{ij}$ contains

  • one quadrupole representation (symmetric traceless rank-2 tensor), and
  • one vector representation (antisymmetric rank-2 tensor).

This concludes the count.


My account of this decomposition is strongly based on

Decomposition of third-order constitutive tensors. Yakov Itin and Shulamit Reches. arXiv:2009.10752 (2020).

... which is honestly a much better description (including better notation, and a complete account of the decomposition in terms of the $\rm GL(3)$, $\rm O(3)$ and $\rm SL(3)$ groups as part of the build-up to $\rm SO(3)$), though it is longer and more technical.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.