I was doing a question on Tensors and hit a roadblock
The Question:
Suppose that $C_{ij}$ and $D_{ij}$ satisfy the quadratic relationship $C_{ij} = T_{ijklmn} D_{kl}D_{mn}$,
where $T_{ijklmn}$ is an isotropic tensor of rank 6. If $C_{ij}$ is symmetric and $D_{ij}$ is antisymmetric, find the most general non-zero form of $T_{ijklmn} D_{kl} D_{mn}$ and prove that there are only two independent terms.
My Attempt
It is immediate that $T_{ijklmn}$ is symmetric in indices $(i,j)$ and antisymmetric in indices $(k,l)$ and $(m,n)$. Also, by using the general form of a rank 2 antisymmetric tensor $\epsilon_{ijk}\omega_k$ it is easy to conclude that $C_{ij}=2T_{ijklkl}\omega_p\omega_p - 4T_{ijklkn}\omega_n\omega_k$. However, I don't know what to do from here, or if what I have done is remotely correct.
Edit: Oh also, idk if it is useful, but I suspect $T_{ijklmn}$ might be equal to $\epsilon_{ikl}\epsilon_{jmn}$
One of the hints given is that we don't need to use the general form of a rank 6 isotropic tensor.
Right, I solved this, turns out I was being really dumb haha.
Since $D_{ij}$ rank 2 antisymmetric means that $D_{ij} = \epsilon_{ijk}v_{k}$ for some $v_k$
But then it is easy to conclude the most general non-zero form of $T_{ijklmn}D_{kl}D_{mn}$.
As for the last part, this is easy as now $$C_{ij}=T_{ijklmn} \epsilon_{klp} \epsilon_{mnq}v_{p}v_{q}$$ Easy to see that if you contract on $(k,m)$ and $(l,n)$ then the right hand side is equivalent to $${T^{'}_{ijpq}}v_pv_q$$ where $T^{'}_{ijpq}$ is a rank 4 isotropic tensor. Using the general form of a rank 4 isotropc tensor, which implies the second result easily.