Product of Levi-Civita Symbols I was reviewing Levi-Civita symbols and came across this identity:
$$ \epsilon_{ijk} \epsilon_{ijn} = 2 \delta_{kn}$$
My first thought was the identity that involves a determinant:
$$\epsilon_{ijk}\epsilon_{lmn}=\det\left|
   \begin{array}{cccc}
      \delta_{il}   & \delta_{im}  & \delta_{in}   \\
      \delta_{jl}   & \delta_{jm}  & \delta_{jn}   \\
      \delta_{kl}   & \delta_{km}  & \delta_{kn}   
   \end{array}
\right|
$$
which is frequently used to prove other identities, such as
$$\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
If I were to employ this approach, I would take the determinant and replace $l$ with $i$ and $m$ with $j$, which is obviously easy to do. However, I seem to recall there being a significantly more elegant approach that doesn't resort to using this determinant-based definition or the identity that follows - does anyone remember what it is?
 A: The expression $\epsilon_{ijk} \epsilon_{ijn}$ is only nonzero when $i \neq j \neq k$ and $i \neq j \neq n$, so it is only nonzero if $k = n$, so it is proportional to $\delta_{kn}$. To figure out the constant of proportionality, set $k = n = 3$. We want to evaluate
$$\epsilon_{ij3} \epsilon_{ij3}.$$
The only nonzero terms are when $(i, j) = (1, 2)$ and $(i, j) = (2, 1)$, giving contributions of $1^2$ and $(-1)^2$ respectively. Then the constant is $2$. 
A: Recalling that $\epsilon_{ijk}$ is an invariant tensor for $so(3)$, the result of the various contractions must clearly be proportional to another 2-indexed invariant tensor, the only one being $\delta_{kn}$.  The proportionality factor can be found by inspection by setting $k=n=3$, $\epsilon_{123}\epsilon_{123}+\epsilon_{213}\epsilon_{213}=2$. One can easily generalize it to the analog identity for $\epsilon_{\mu_1\ldots\mu_{n-1}\mu_n}\epsilon_{\mu_1\ldots\mu_{n-1}\nu_n}$ and $\delta_{\mu_n \nu_n}$, using the same argument for the invariant tensor of $so(n)$, as well as to minkowskian metrics for e.g. $so(n,1)$.
