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?


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

  • $\begingroup$ hehe I checked wiki and levi-civita symbol with three indices is only defined in 3-dimensions $\endgroup$ – tom Apr 22 '13 at 20:20

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 $$ \epsilon_{i_1,\dots,i_n} \, \epsilon^{j_1,\dots,j_n} = \begin{vmatrix} \delta_{i_1}^{j_1} & \dots & \delta_{i_1}^{j_n}\\ \vdots & & \vdots \\ \delta_{i_n}^{j_1} & \dots & \delta_{i_n}^{j_n} \end{vmatrix} $$


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