# How to prove the Levi-Civita contraction?

I want to prove the following relation

\begin{align} \epsilon_{ijk}\epsilon^{pqk} = \delta_{i}^{p}\delta_{j}^{q}-\delta_{i}^{q}\delta_{j}^{p} \end{align}

I tried expanding the sum \begin{align} \epsilon_{ijk}\epsilon^{pqk} &=& \epsilon_{ij1}\epsilon^{pq1} + \epsilon_{ij2}\epsilon^{pq2} + \epsilon_{ij3}\epsilon^{pq3} \end{align} I made the assumption that $\epsilon_{ij1} = \delta_{2i}\delta_{3j} - \delta_{3i}\delta_{2j}$, then I tried to argue the following using cyclical permutations \begin{align} \epsilon_{ijk}\epsilon^{pqk} &=& (\delta_{2i}\delta_{3j}-\delta_{3i}\delta_{2j})(\delta^{2p}\delta^{3q}-\delta^{3p}\delta^{2q}) \\&+& (\delta_{3i}\delta_{1j}-\delta_{1i}\delta_{3j})(\delta^{1p}\delta^{3q}-\delta^{1p}\delta^{3q}) \\&+& (\delta_{1i}\delta_{2j}-\delta_{2i}\delta_{1j})(\delta^{1p}\delta^{2q}-\delta^{2p}\delta^{1q}) \end{align} and then I realized that this was getting long and messy and I lost my way.

How does one prove the Levi-Civita contraction?

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 $$\epsilon_{i_1,\dots,i_n} \, \epsilon^{j_1,\dots,j_n} = \begin{vmatrix} \delta_{i_1}^{j_1} & \dots & \delta_{i_1}^{j_n}\\ \vdots & & \vdots \\ \delta_{i_n}^{j_1} & \dots & \delta_{i_n}^{j_n} \end{vmatrix}$$
You need to use the fact that the only isotropic (invariant under rotations) tensors are $\epsilon_{ijk}$ and $\delta_{ij}$. Any other isotropic tensor must be expressible in terms of these two tensors. Clearly, $T_{ijlm}=\epsilon_{kij}\epsilon_{klm}$ is isotropic. It is a tensor that is (i) antisymmetric under the exchanges: $i\leftrightarrow j$, $l\leftrightarrow m$ and (ii) symmetric under the simultaneous exchange $(i,j)\leftrightarrow (l,m)$. This implies that $$T_{ijlm}= A\ (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\ ,$$ where $A$ is a constant. A simple computation, say $T_{1212}=1$, shows that $A=1$.
The product $\epsilon_{ijk}\epsilon^{pqr}$ has certain symmetry properties. They are the same properties as the determinant
$$\begin{vmatrix} \delta_i^p & \delta_i^q & \delta_i^r\\ \delta_j^p & \delta_j^q & \delta_j^r \\ \delta_k^p & \delta_k^q & \delta_k^r \end{vmatrix}$$
It's a rank 6 tensor, it changes sign under exchange within $ijk$ and $pqr$ (swapping a row or column). It is equal to 1, -1, 0 under the same conditions that the Levi Cevita product is e.g. repeated index within $ijk$ or $pqr$ makes two rows/columns equal so the determinant is 0. They are the same thing.
From here, just contract via $\delta_r^k$ and expand the determinant.