How can the isotropic and deviatoric parts of fourth- and fifth-rank tensors be obtained? For a second-order rank, we know that the isotropic and deviatoric parts is given by
$$
\frac{1}{2} \left( A_{ij} + A_{ji} \right) - \frac{1}{3} A_{ss} \delta_{ij} \, , 
$$
whereas for a third-rank tensor, it is given by
$$
\frac{1}{6} \left( A_{ijk} + A_{ikj} + A_{jik} + A_{jki} + A_{kij} + A_{kji} \right)
- \frac{1}{15} 
\left[
 \left( A_{ssi} + A_{sis} + A_{iss} \right) \delta_{kj} 
+  \left( A_{ssj} + A_{sjs} + A_{jss} \right) \delta_{ik} 
+ \left( A_{ssk} + A_{sks} + A_{kss} \right) \delta_{ij} 
\right] \, .
$$
Then, how can this be generalized for fourth- and fifth-rank tensors?
Thank you!
 A: $
\def\tss{\mathop{\bullet}\limits}
\def\dss{\mathop{\odot}\limits}
\def\bbR#1{{\mathbb R}^{#1}}
\def\a{\alpha}\def\b{\beta}\def\d{\delta}\def\t{\theta}
\def\A{{\cal A}}\def\E{{\cal I}}\def\B{{\cal B}}
\def\L{\left}\def\R{\right}
\def\LR#1{\L(#1\R)}
\def\BR#1{\Big(#1\Big)}
\def\iso#1{\operatorname{iso}\LR{#1}}
\def\dev#1{\operatorname{dev}\LR{#1}}
\def\Diag#1{\operatorname{Diag}\LR{#1}}
\def\sym#1{\operatorname{sym}\LR{#1}}
\def\skew#1{\operatorname{skew}\LR{#1}}
\def\trace#1{\operatorname{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\c#1{\color{red}{#1}}
$Let $\A$ denote an arbitrary $k^{th}$ order tensor, $\E$ the totally symmetric isotropic tensor of the same order, and $(\tss)$ the $k^{th}$ order contraction product, i.e.
$$\A\tss\B
 = \sum_{i=1}^n\sum_{j=1}^n\ldots\sum_{\ell=1}^n
\;{\A_{ij\ldots\ell}\;\B_{ij\ldots\ell}}$$
Then the isotropic part of $\A$ is defined as
$$\eqalign{
\iso{\A} = \LR{\frac{\A\tss\E}{\E\tss\E}} \E \\
}$$
and the deviatoric part of the tensor is what's left over
$$\eqalign{
\dev{\A} &= \A - \iso{\A}\qquad \\
}$$
Note that $\LR{\A\tss\E}$ and $\LR{\E\tss\E}\,$ are fully contracted scalar quantities.
Update
It occurs to me that you might be studying irreducible tensor decompositions, in which case you should probably study this post.
