Product of Levi-Civita Symbols

Question

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?

Answer 1

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 $2$.

Answer 2

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. $so(n,1)$.