# Irreducible decomposition of a Dyadic operator

Recently, I've been studying $$SO(3)$$ Group Theory for physics application and for a couple of days I'm struggling to understand how to get a 3D cartesian tensor, like: $$T_{ij}=a_i\,b_j$$ to be decomposed as $$\frac{\vec{a}\cdot\vec{b}}{3}\delta_{ij}+\frac{a_i\,b_j-a_j\,b_i}{2}+\left(\frac{a_i\,b_j+a_i\,b_j}{2}-\frac{\vec{a}\cdot\vec{b}}{3}\delta_{ij}\right),$$ i.e., this irreducible representation is invariant under rotation and it's commonly said that $$3\otimes3=1+3+5$$. I already know that a second order tensor can be decomposed as antisymmetric and symmetric parts, but I don't get it why in decomposition above there is a trace part (you may say that is because is a traceless tensor, but why it has to be?).

• The traceful part is proportional to I, and so invariant under rotations, $R^T I R=I$, whereas the symmetric traceless part S does transform nontrivially like a quintet, $R^T S R= S'$. You need the two decoupled irreps then to be trace-orthogonal, which fixes the normalizations. Dec 3, 2020 at 15:34

The so-called natural form of a rank $$N$$ tensor is symmetric in all its indices and is traceless. It has $$2N+1$$ degrees of freedom that transform like $$2N+1$$ order-$$N$$ spherical harmonics: $$Y_{l=N}^{m}(\theta, \phi)$$.

So for rank-2, the natural form is:

$$N_{ij} = \frac 1 2 (T_{ij}+T_{ji}) - \frac 1 3 \delta_{ij} T_{kk}$$

$$N_{ij} = S_{ij} - \frac 1 3 \delta_{ij} T_{kk}$$

where $$S_{ij}$$ refers to the symmetric part (but not trace free).

The relation between spherical($$T^{l,m}$$) form and cartesian is:

$$T^{2,\pm 2} = \frac 1 2 [S_{xx}-S_{yy}\pm 2iS_{xy}]$$

$$T^{2,\pm 1} = \frac 1 2 [S_{zx}\pm iS_{yz}]$$

$$T^{2,0} = \sqrt{\frac 2 3} S_{zz}$$

You will find that the $$T^{2,m}$$ are rotated just as the $$Y_2^m(\theta,\phi)$$ are.

At higher ranks, it gets involved. The symmetrized rank-3 tensor is:

$$S_{ijk} = \frac 1 6 [T_{ijk}+T_{kij}+T_{jki}+T_{kji}+T_{jik}+T_{ikj}]$$

and the trace free version is:

$$N_{ijk} = S_{ijk} - \frac 1 {30}[(\delta_{ij}+\delta_{ji})(T_{llk}+T_{lkl}+T_{kll})+ (\delta_{ik}+\delta_{ki})(T_{llj}+T_{ljl}+T_{jll})+(\delta_{kj}+\delta_{jk})(T_{lli}+T_{lil}+T_{ill})]$$

where

and the $$(2\cdot 3+1)=7$$ spherical tensors that transform like $$Y_3^m$$ are (according to my notes):

$$T^{3,\pm3} = \frac 1 {\sqrt 8}[(-S_{xxx}+3S_{xyy}) \mp i(S_{yyy}-S_{xxy})]$$

$$T^{3,\pm2} = \frac 1 2[-S_{xxz}-S_{yyz} \mp 2iS_{xyz}]$$

$$T^{3,\pm 1} = \frac {\sqrt{15}} 3\big(\frac 1 {\sqrt 2}[\mp S_{zzz}-iS_{zzz}] + \frac 1 {\sqrt 8}[\mp(S_{xxx}-S_{xyy})+i(S_{yyy}\pm S_{xxy}]\big)$$

$$T^{3,0} =\frac {\sqrt{10}} 3[\frac 1 {\sqrt 2}(S_{xzz}+iS_{yzz}) + S_{zzz}]$$

Rank-3 breaks down according to:

$${\bf 3} \otimes {\bf 3} \otimes {\bf 3} = {\bf 10} \oplus {\bf 8} \oplus {\bf 8} \oplus {\bf 1}$$

(where the $${\bf 1}$$ is the familiar $$\epsilon_{ijk})$$. So what happened in creating the natural form tensor $$S_{ijk}$$ is that we subtracted off a vector trace: $$S_{ijj}$$ from the original 10 degrees-of-freedom:

$${\bf 10} \rightarrow {\bf 7} + {\bf 3}$$

Likewise, the octet is split into a rank-2 like object and a vector trace:

$${\bf 8} \rightarrow {\bf 5} + {\bf 3}$$

See Physical Review A, "Irreducible fourt-rank Cartesian tensors", Andrews and Ghoul, Volume 25, Number 5, Page 2647, [1992], for Rank-4, though it has typos in the indices, and looks something like this: