Proving that $\sum_{k=1}^3 \Gamma_{i,j,k}\Gamma_{m,n,k} = \delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$

Question

For a variable $\Gamma_{ijk}$ I know that $\Gamma_{123}=\Gamma_{231}=\Gamma_{312}=1$ and $\Gamma_{132}=\Gamma_{321}=\Gamma_{213}=-1$, otherwise $\Gamma_{ijk}=0$, whereas $i,j,k$ run between $1$ and $3$. I would now like to show:

$$\sum_{k=1}^3 \Gamma_{ijk}\Gamma_{mnk} = \delta_{im}\delta_{jn}-\delta_{in}\delta_{jm}$$

With $\delta_{ij}$ denoting the Kronecker delta.

I am not quite sure what's the best way to prove it... I was thinking about case analysis, but I would not know where to start. For example if I say in my first case that my RHS is zero, then I can assume that $(i\ne m\vee j\ne n) \land (i\neq n \vee j\ne m)$. What would I do from here? I do not think that this gives me enough information to say anything about the LHS of the sum above.

Any help and hints are greatly appreciated!

Answer 1

The only choices of $i,\,j,\,k,\,m,\,n$ that contribute are those for which $i,\,j,\,k$ and $m,\,n,\,k$ both permute $1,\,2,\,3$, as otherwise matching indices imply a $\Gamma$ cancels. So $i,j$ must permute $m,\,n$. The first term on the right-hand side is the case $i=m$, and the second is $i=n$.