# Three index entities summing to product of Kronecker deltas

While looking at this problem I ended up having to find entities $a_{ijp}$ and $b_{klp},$ with $i,j,k,l,p = 1, \ldots, n,$ such that $$\sum_{p=1}^{n} a_{ijp} \, b_{klp} = \delta_{ik} \, \delta_{jl},$$ where $\delta_{ij}$ is the Kronecker delta.

Probably $a$ and $b$ can be taken equal so that we want $$\sum_{p=1}^{n} a_{ijp} \, a_{klp} = \delta_{ik} \, \delta_{jl}.$$

Does anyone have know if this is even possible, and if so, how to choose $a_{ijp}$?

In the actual case, $n=3,$ but a generic solution is preferred.

How did this equation come up?

I asserted that $\sigma^{ij} \, \partial_i u_j \, dV$ could be rewritten as $\omega \wedge \pi$ where $\omega = \alpha_{kl}^{ij} \sigma_{ij} \, dx^k \wedge dx^l$ and $\pi = \beta_p^{mn} \partial_m u_n \, dx^p.$ This led to $$\alpha_{kl}^{ij} \, \beta_p^{mn} \, \sigma_{ij} \, \partial_m u_n \epsilon^{klp} \, dV = \sigma^{mn} \, \partial_m u_n \, dV.$$ Setting $\gamma^{ijp} = \epsilon^{klp} \, \alpha_{kl}^{ij}$ then gives $$\beta_p^{mn} \gamma^{ijp} = \delta^{mi} \, \delta^{nj}.$$ For the presentation in the post the names were just changed and the summation over $p$ was made explicit.

• You probably should give us some background. What is $a_{ijk}$ and $b_{ijk}$ actually. – Sou Jul 27 '18 at 19:47
• Not quite the Levi-Civita Symbol en.wikipedia.org/wiki/Levi-Civita_symbol – Donald Splutterwit Jul 27 '18 at 20:23
• @Sou. I've added some background. – md2perpe Jul 27 '18 at 22:38
• @DonaldSplutterwit. That's close. Perhaps some modification of $\epsilon$ can be used. – md2perpe Jul 27 '18 at 22:39