The Levi-Civita permutation symbol is a special case of the generalized Kronecker delta symbol. Using this fact one can write the Levi-Civita permutation symbol as the determinant of an $n\times n$ matrix consisting of traditional delta symbols. See the entry on the generalized Kronecker symbol for details.

When using the Levi-Civita permutation symbol and the generalized Kronecker delta symbol, the Einstein summation convention is usually employed. In the below, we shall also use this convention.

Properties

• When $n=2$ , we have for all $i,j,m,n$ in $\{1,2\}$ , \begin{eqnarray} \label{eq0} \varepsilon_{ij} \varepsilon^{mn} &=& \delta_i^m \delta_j^n - \delta_i^n \delta_j^m, \\ \label{eq1} \varepsilon_{ij} \varepsilon^{in} &=& \delta_j^n,\\ \label{eq2} \varepsilon_{ij} \varepsilon^{ij} &=& 2. \end{eqnarray}
• When $n=3$ , we have for all $i,j,k,m,n$ in $\{1,2,3\}$ , \begin{eqnarray} \label{eq3} \varepsilon_{jmn} \varepsilon^{imn} &=& 2\delta^i_j, \\ \label{eq4} \varepsilon_{ijk} \varepsilon^{ijk} &=& 6. \end{eqnarray}

Let us prove these properties. The proofs are instructional since they demonstrate typical argumentation methods for manipulating the permutation symbols.

Proof. For equation , let us first note that both sides are antisymmetric with respect of $ij$ and $mn$ . We therefore only need to consider the case $i\neq j$ and $m\neq n$ . By substitution, we see that the equation holds for $\varepsilon_{12} \varepsilon^{12}$ , i.e., for $i=m=1$ and $j=n=2$ . (Both sides are then one). Since the equation is anti-symmetric in $ij$ and $mn$ , any set of values for these can be reduced the above case (which holds). The equation thus holds for all values of $ij$ and $mn$ . Using equation , we have for equation \begin{eqnarray*} \varepsilon_{ij}\varepsilon^{in} &=& \delta_i^i \delta_j^n - \delta^n_i \delta^i_j\\ &=& 2 \delta_j^n - \delta^n_j\\ &=& \delta_j^n. \end{eqnarray*}Here we used the Einstein summation convention with $i$ going from $1$ to $2$ . Equation follows similarly from equation . To establish equation , let us first observe that both sides vanish when $i\neq j$ . Indeed, if $i\neq j$ , then one can not choose $m$ and $n$ such that both permutation symbols on the left are nonzero. Then, with $i=j$ fixed, there are only two ways to choose $m$ and $n$ from the remaining two indices. For any such indices, we have $\varepsilon_{jmn} \varepsilon^{imn} = (\varepsilon^{imn})^2 = 1$ (no summation), and the result follows. The last property follows since $3!=6$ and for any distinct indices $i,j,k$ in $\{1,2,3\}$ , we have $\varepsilon_{ijk} \varepsilon^{ijk}=1$ (no summation).

Examples and Applications.

• The determinant of an $n\times n$ matrix $A=(a_{ij})$ can be written as $$ \det A = \varepsilon_{i_1\cdots i_n} a_{1i_1} \cdots a_{ni_n},$$ where each $i_l$ should be summed over $1,\ldots, n$ .
• If $A=(A^1, A^2, A^3)$ and $B=(B^1, B^2, B^3)$ are vectors in (represented in some right hand oriented orthonormal basis), then the $i$ th component of their cross product equals $$ (A\times B)^i = \varepsilon^{ijk} A^j B^k.$$ For instance, the first component of $A\times B$ is $A^2 B^3-A^3 B^2$ . From the above expression for the cross product, it is clear that $A\times B = -B\times A$ . Further, if $C=(C^1, C^2, C^3)$ is a vector like $A$ and $B$ , then the triple scalar product equals $$ A\cdot(B\times C) = \varepsilon^{ijk} A^i B^j C^k.$$ From this expression, it can be seen that the triple scalar product is antisymmetric when exchanging any adjacent arguments. For example, $A\cdot(B\times C)= -B\cdot(A\times C)$ .
• Suppose $F=(F^1, F^2, F^3)$ is a vector field defined on some open set of with Cartesian coordinates $x=(x^1, x^2, x^3)$ . Then the $i$ th component of the curl of $F$ equals $$ (\nabla \times F)^i(x) = \varepsilon^{ijk}\frac{\partial}{\partial x^j} F^k(x). $$