Confusing question on tensors

Question

A tensor of rank $4$ satisfies $T_{ijkl}=T_{jilk}=-T_{jikl}$ and $T_{ijij}=0$. I need to show that: $$T_{ijkl}=-\varepsilon_{ijp}\varepsilon_{klq}T_{rqrp}$$ Could someone offer a hint? I have tried expanding the Levi-Civita product into Kronecker deltas but it is a mess and gives nothing.

Answer 1

first note that the tensor is antisymmetric under $i \leftrightarrow j$ and $k \leftrightarrow l$. Therefore you could "dualise" along the $(ij)$ and $(kl)$ induces by defining a tensor $Q_{pq}$:

(1) $Q_{pq}=\epsilon_{pij}\epsilon_{qkl}T_{ijkl}$,

which should have the same number of degrees of freedom as $T_{ijkl}$. Now note that tensor $Q_{pq}$ is traceless:

(2) $Q_{pp}=\epsilon_{pij}\epsilon_{pkl}T_{ijkl}=(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})T_{ijkl}=T_{ijij}-T_{ijji}=2T_{ijij}=0$

Now multiply equation (1) by $\epsilon_{i'j'p}\epsilon_{k'l'q}$ you will get (using the contraction of the epsilon symbols):

(3) $T_{i'j'k'l'}=\frac{1}{4}\epsilon_{i'j'p}\epsilon_{k'l'q}Q_{pq}$

The only thing that is left is to obtain $Q_{pq}$ in terms of $T_{ijkl}$. To this end just contract the indices $i'$ and $k'$ in eq. (3), Using again the contraction of the epsilon symbols and the tracelessness of $Q_{pq}$ one easily gets that:

(4) $T_{i'j'i'l'}=-\frac{1}{4}Q_{l'j'}$, substituting this in (3) you arrive at:

(5) $T_{ijkl}=-\epsilon_{ijp}\epsilon_{klq}T_{rqrp}$, where I skipped the primes and renamed some indices to make it look exactly as the expression that you need.

I hope this helps