§ About the Levi-Civita symbol

One can convince themself that this definition suffices to define the totally anti-symmetric symbol. Indeed, for the case \(n=3\) we may simply write: $$ \begin{align} \det\{A\} &= \det \begin{pmatrix} A^1{}_1 & A^1{}_2 & A^1{}_3 \\ A^2{}_1 & A^2{}_2 & A^2{}_3 \\ A^3{}_1 & A^3{}_2 & A^3{}_3 \end{pmatrix} =\\ &= \hphantom{+}A^1{}_1A^2{}_2A^3{}_3 + A^3{}_1A^1{}_2A^2{}_3 + A^2{}_1A^3{}_2A^1{}_3 +\\ &\hphantom{=}- A^3{}_1A^2{}_2A^1{}_3 - A^1{}_1A^3{}_2A^2{}_3 - A^2{}_1A^1{}_2A^3{}_3 \end{align} $$ which clearly sets \(\epsilon_{123}=\epsilon_{312}=\epsilon_{231}=1\), \(\epsilon_{321}=\epsilon_{132}=\epsilon_{213}=-1\) being all the components with repeated indices zero.

Let us now take the canonical basis \(\{\mathbf{e}_i\}_{i=1}^n\) of the \(n-\)dimensional \(\mathbb{R}-\)vector space \(\mathbb{R}^n\), where each vector has \(a-\)th component given by \((\mathbf{e}_i)^a=\delta^a_i\). Similarly, we introduce the dual canonical basis \(\{\mathbf{e}^i\}_{i=1}^n\) satisfying \((\mathbf{e}^i)_a=\delta^i_a\). There is the natural identification of \(\{\mathbf{e}_i\}_{i=1}^n\) with column-vectors and of \(\{\mathbf{e}^i\}_{i=1}^n\) with row vectors. This way of viewing these vectors allows us to define the following matrices: $$ E = \begin{pmatrix} \mathbf{e}_{\sigma(1)} & \mathbf{e}_{\sigma(2)} & \cdots & \mathbf{e}_{\sigma(n)}\end{pmatrix} \in \text{Mat}_{n\times n}(\mathbb{K}) $$ where \(\sigma \in S_n\), being \(S_n\) the symmetric group on \(n\) elements. The matrix \(E\) clearly verifies that its transpose is given by $$ E^T = \begin{pmatrix} \mathbf{e}^{\sigma(1)} \\ \mathbf{e}^{\sigma(2)} \\ \vdots \\ \mathbf{e}^{\sigma(n)}\end{pmatrix} \in \text{Mat}_{n\times n}(\mathbb{K}) $$ In index notation, \(E^a{}_b=(\mathbf{e}_b)^a=\delta^a_b\), whereas \((E^T)^a{}_b=(\mathbf{e}^a)_b=\delta^a_b\). This implies that $$ \det\{E\}= \epsilon_{i_1i_2\ldots i_n} \delta^{i_1}_{\sigma(1)}\delta^{i_2}_{\sigma(2)}\cdots \delta^{i_n}_{\sigma(n)} = \epsilon_{\sigma(1)\sigma(2)\ldots \sigma(n)} $$ and also, using this time the Levi-Civita with the upper indices for computing the determinant of the transpose $$ \det\{E^T\}= \epsilon^{i_1i_2\ldots i_n} \delta_{i_1}^{\sigma(1)}\delta_{i_2}^{\sigma(2)}\cdots \delta_{i_n}^{\sigma(n)} = \epsilon^{\sigma(1)\sigma(2)\ldots \sigma(n)} $$ and since \(\det\{E\}=\det\{E^T\}\) we have: $$ \epsilon_{\sigma(1)\sigma(2)\ldots \sigma(n)} = \epsilon^{\sigma(1)\sigma(2)\ldots \sigma(n)} $$ in other words, we do not really care about whether our indices are upstairs or downstairs. Let us now try to compute the following product: $$ \begin{align} \epsilon^{\sigma(1)\sigma(2)\ldots \sigma(n)}\epsilon_{\tau(1)\tau(2)\ldots \tau(n)} &= \det\{E^T\}\det\{E\}=\det\{E^TE\}=\\ &= \det\left\{\begin{pmatrix} \mathbf{e}^{\sigma(1)} \\ \mathbf{e}^{\sigma(2)} \\ \vdots \\ \mathbf{e}^{\sigma(n)}\end{pmatrix} \begin{pmatrix} \mathbf{e}_{\tau(1)} & \mathbf{e}_{\tau(2)} & \cdots & \mathbf{e}_{\tau(n)}\end{pmatrix}\right\}=\\ &= \det \begin{pmatrix} \mathbf{e}^{\sigma(1)}\cdot \mathbf{e}_{\tau(1)} &\mathbf{e}^{\sigma(1)}\cdot \mathbf{e}_{\tau(2)} & \cdots & \mathbf{e}^{\sigma(1)}\cdot \mathbf{e}_{\tau(n)} \\ \mathbf{e}^{\sigma(2)}\cdot \mathbf{e}_{\tau(1)} &\mathbf{e}^{\sigma(2)}\cdot \mathbf{e}_{\tau(2)} & \cdots &\mathbf{e}^{\sigma(2)}\cdot \mathbf{e}_{\tau(n)} \\ \vdots & \vdots & \ddots & \vdots\\ \mathbf{e}^{\sigma(n)}\cdot \mathbf{e}_{\tau(1)} &\mathbf{e}^{\sigma(n)}\cdot \mathbf{e}_{\tau(2)} & \cdots & \mathbf{e}^{\sigma(n)}\cdot \mathbf{e}_{\tau(n)} \\ \end{pmatrix}=\\ &= \det\begin{pmatrix} \delta^{\sigma(1)}_{\tau(1)} & \delta^{\sigma(1)}_{\tau(2)} & \cdots & \delta^{\sigma(1)}_{\tau(n)}\\ \delta^{\sigma(2)}_{\tau(1)} & \delta^{\sigma(2)}_{\tau(2)} & \cdots & \delta^{\sigma(2)}_{\tau(n)}\\ \vdots & \vdots & \ddots & \vdots\\ \delta^{\sigma(n)}_{\tau(1)} & \delta^{\sigma(n)}_{\tau(2)} & \cdots & \delta^{\sigma(n)}_{\tau(n)}\\ \end{pmatrix} \end{align} $$ As before, let us consider the three dimensional case and denote \(\sigma(1)=i, \sigma(2)=j, \sigma(3)=k\) and \(\tau(1)=l, \tau(2)=m, \tau(3)=n\) $$ \epsilon^{ijk}\epsilon_{lmn}= \det \begin{pmatrix} \delta^i_l & \delta^i_m &\delta^i_n \\ \delta^j_l & \delta^j_m &\delta^j_n \\ \delta^k_l & \delta^k_m &\delta^k_n \\ \end{pmatrix} $$ Should \(l=i\), then for this product to be non zero clearly \(j\neq i\neq k\), and \(m\neq l \neq n\) and thus, $$ \epsilon^{ijk}\epsilon_{imn}= \det \begin{pmatrix} 1 & 0 & 0 \\ 0 & \delta^j_m &\delta^j_n \\ 0 & \delta^k_m &\delta^k_n \\ \end{pmatrix} = \delta^j_m\delta^k_n-\delta^j_n\delta^k_m $$