In order to prove the following identity:
$$\sum_{k}\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$
Instead of checking this by brute force, Landau writes thr product of Levi-Civita symbols as:
$$\epsilon_{ijk}\epsilon_{lmn}=\det\left|
\begin{array}{cccc}
\delta_{il} & \delta_{im} & \delta_{in} \\
\delta_{jl} & \delta_{jm} & \delta_{jn} \\
\delta_{kl} & \delta_{km} & \delta_{kn}
\end{array}
\right|
$$
The proof that the equalty holds is quite straightforward if you consider what values the indices can take. But I've been told that there's a much more profound and elegant demonstration based on the representation of the symmetric group.
Does anybody know this approach based on group theory?
Best Answer
I know this from physics courses but only when you sum from 1 to 3.
And the argument is quite simple:
For $\epsilon_{ijk}$ to be nonzero $ijk$ has to be pairwise different. So $(ijk)$ can be understood as permutation, than $\epsilon_{ijk} = \text{sign}(ijk)$. For $\epsilon_{ijk}\epsilon_{lmk}$ to be nonzero you have only two situations:
1) i=l and j=m, so $\text{sign}(ijk)=\text{sign}(lmk)$
2)i=m and j=l, so $\text{sign}(ijk)=-\text{sign}(lmk)$
Therefore $ \sum_{k}\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$
note that these aruguments can only be made when your indices can range from 1 to 3