Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory 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?
 A: I made the same answer in the physic forum (here https://physics.stackexchange.com/questions/134156/how-to-prove-the-levi-civita-contraction/134163#134163), but i leave it here too for completion.
One way to see this is to consider the fact that the vector space of rank (3,3) completely antisymmetric tensors ($ \Lambda_3^3(R^3) $) has dimension one (it's just a linear algebra exercise). Then define the tensor:
$$
M_{ijk}^{lmn} = \delta_i^{[l} \, \delta_j^m \, \delta_k^{n]} = \frac{1}{3!} \sum_{\sigma \in S_3} sgn(\sigma) \, \delta_i^{\sigma(l)} \, \delta_j^{\sigma(m)} \, \delta_k^{\sigma(n)} 
$$
where we are summing over all the permutations of three numbers $\sigma$, and $sgn(\sigma)$ denotes the sign of the permutation. It's worthy to note that
$$
M_{ijk}^{lmn}=\frac{1}{3!}\begin{vmatrix} \delta_i^l & \delta_i^m & \delta_i^n\\ 
\delta_j^l & \delta_j^m & \delta_j^n \\ \delta_k^l & \delta_k^m & \delta_k^n \end{vmatrix}
$$
by the Leibniz formula of the determinant (http://en.wikipedia.org/wiki/Leibniz_formula_for_determinants). So we have $M \in \Lambda_3^3(R^3) $. Since $M \neq 0$, $M$ is a basis for the space $ \Lambda_3^3(R^3)  $. Now consider the tensor
$$
\epsilon_{ijk} \, \epsilon^{lmn} = B_{ijk}^{lmn}
$$
Since $B \in \Lambda_3^3(R^3) $ and $M$ is a basis, there exists a constant $k$ such that
$$
B_{ijk}^{lmn} = k \, M_{ijk}^{lmn} \implies \epsilon_{ijk} \, \epsilon^{lmn} = k \, \delta_i^{[l} \, \delta_j^m \, \delta_k^{n]} 
$$
Now to determine $k$ contract $\epsilon_{lmn}$ on both sides and use the fact that $\epsilon_{lmn} \, \epsilon^{lmn} =3!$ (since you sum $3!$ terms equal to one)
$$
\epsilon_{ijk} \, \epsilon^{lmn} \, \epsilon_{lmn} = 3! \, \epsilon_{ijk} = k \, \delta_i^{[l} \, \delta_j^m \, \delta_k^{n]} \epsilon_{lmn} = k \, \delta_i^{l} \, \delta_j^m \, \delta_k^{n} \, \epsilon_{[lmn]} =  k \, \delta_i^{l} \, \delta_j^m \, \delta_k^{n} \, \epsilon_{lmn} = k \, \epsilon_{ijk} 
$$
So we finally get $k=3!$ and
$$
\epsilon_{ijk} \, \epsilon^{lmn}=\begin{vmatrix} \delta_i^l & \delta_i^m & \delta_i^n\\ 
\delta_j^l & \delta_j^m & \delta_j^n \\ \delta_k^l & \delta_k^m & \delta_k^n \end{vmatrix}
$$
And you can get the identity you want by contracting. The same argument could be used in any dimension to show that
$$
\epsilon_{i_1,\dots,i_n} \, \epsilon^{j_1,\dots,j_n} = \begin{vmatrix} \delta_{i_1}^{j_1} & \dots & \delta_{i_1}^{j_n}\\ 
\vdots &  & \vdots \\ \delta_{i_n}^{j_1} & \dots & \delta_{i_n}^{j_n} \end{vmatrix}
$$
A: 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
