Shape of 3-order tensors in $O_h, O, T_d$ and $D_3$ point gruops How does one calculate the shape of higher order $(Dimension>2)$ tensors in respect of point group symmetry?
I understand that you have to use transformation matrix corresponding to a symmetry operation of a group and then equate the obtained coefficients to the old ones (since after a symmetry operation the system does not change). 
I need to find shapes of 3 order tensors in $O_h, O,  T_d$ and  $D_3$ point groups.
An example of $C_3(x_3)$ on a $T_{ijk} $ tensor would be much appreciated. How do I do the matrix tensor multiplication?
Thank You in advance!
 A: In real space, given a vector of coordinates $(x_1, x_2, x_3)$, the tensor computes a quantity
$$s=T_{ijk}x_ix_jx_k,$$
where there is an implicit summation from 1 to 3 on repeated indices. Applying a point group transformation,
$$x'_i = R_{ij}x_j,$$
and we want the new coordinates of the tensor to satisfy
$$s=T'_{ijk}x'_ix'_jx'_k.$$
Thus
$$T_{ijk}x_ix_jx_k=s=T'_{ijk}R_{il}R_{jm}R_{kn}x_lx_mx_n$$
has to be valid for any $x_1, x_2, x_3$ and therefore
$$T_{lmn} = T'_{ijk}R_{il}R_{jm}R_{kn},\text{ for all } l,m,n=1,2,3.$$
Requiring the tensor to be invariant is requiring that $T_{ijk}=T'_{ijk}$ for all $i,j,k=1,2,3$ and here are your equations:
$$T_{lmn} = T_{ijk}R_{il}R_{jm}R_{kn},\text{ for all } l,m,n=1,2,3.\tag{1}$$
Let's work out the example you wanted. Working with the $R$ setting (not the same $R$ as above, just in case you wonder!), the group has the following single generator
$$R=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix},$$
which simply circularly permutes the basis vector. I chose that setting because this makes $R$ as sparse as possible, thus simplifying the writing of equations (1). Actually, we can even explicitly characterise the non-zero elements of the rotation matrix: $R_{ij}$ is non-zero iff $\newcommand{\modulo}[2]{#1\,(\textrm{mod}\,#2)}\modulo{i=j+1}{3}$. Thus equations (1) read
$$T_{lmn} = T_{\modulo{l+1}{3},\modulo{m+1}{3},\modulo{n+1}{3}},\ \forall l,m,n=1,2,3.$$
Of course, in general, there is not such a neat formula because the rotation matrix sparse pattern is not that neat. You can't avoid the tedium, or you can use a CAS, or sling a small script in your preferred language.
