Rotation invariant tensors, isotropic tensors, follow up question

Question

A previous question established that the only rank 2 and 3 isotropic tensors are the $\lambda \delta_{ij}$ and the $\lambda \epsilon_{ijk}$ tensors respectively. How do we know ... "All isotropic tensors of higher rank are obtained by combining $\delta_{ij}$ and $\epsilon_{ijk}$ tensor products, contractions, and linear combinations." IA Vector Calculus That is, how do we know that, for example, every rank 4 isotropic tensor has the form $\alpha \delta_{ij} \delta_{kl} + \beta \delta_{ik} \delta_{jl}+ \gamma \delta_{il} \delta_{jk}$

Note: I'm pretty sure we can show, using rotations as in the link, that the only non-zero terms in an isotropic rank 4 tensor have the form $T{iijj},T{ijij},T{ijji}$ and that $T{ijij} = T{ikik}$ etc, so the terms in T are the ones appearing in form above, so ....

Answer 1

There is a quick easy answer. Let V = R2 to make it simple.

By rotating the axes 90 deg., f1 = e2, and f2 = -e1 it's easy to show the non-zero components of the isotropic rank 4 isotropic tensor T are [t1111 0 0 t1122; 0 t1212 t1221 0; 0 t1221 t1212 0; t1122 0 0 t111].

Why? t1111 corresponds to e1xe1xe1xe1 in e basis coordinates and f1xf1xf1xf1 (= e2xe2xe2xe2) in f basis coordinates, so t1111 = t2222, and similarly for components with 2 1s and 2 2s. If a component has 1 2 and 3 1s, for example t2111, it corresponds to e2xe1xe1xe2 in e basis coordinates and f2xf1xf1xf1 (= (-e1)xe2xe2xe2) in f basis coordinates and is hence 0. Similarly for a component with 1 1 and 3 2s.