Levi-Civita permutation symbol

Definition 1.

Let $k_{i}\\in\\{1,\\cdots,n\\}$ for all $i=1,\\cdots,n$ .The Levi-Civita permutation symbols $\\varepsilon_{k_{1}\\cdots k_{n}}$ and $\\varepsilon^{k_{1}\\cdots k_{n}}$ aredefined as

\\varepsilon_{k_{1}\\cdots k_{m}}=\\varepsilon^{k_{1}\\cdots k_{m}}=\\left\\{\\begin{%array}[]{ll}+1&\\mbox{when}\\,\\{l\\mapsto k_{l}\\}\\mbox{ is an even permutation (%of $\\{1,\\cdots,n\\}$),}\\\\-1&\\mbox{when}\\,\\{l\\mapsto k_{l}\\}\\mbox{ is an odd permutation,}\\\\0&\\mbox{otherwise, i.e., when\\,}k_{i}=k_{j},\\ \\mbox{for some }\\ i\eq j.\\\\\\end{array}\\right.

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

When using the Levi-Civita permutation symbol and the generalized Kronecker deltasymbol, 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 $\\{1,2\\}$ ,
$\\displaystyle\\varepsilon_{ij}\\varepsilon^{mn}$ $\\displaystyle=$ $\\displaystyle\\delta_{i}^{m}\\delta_{j}^{n}-\\delta_{i}^{n}\\delta_{j}^{m},$ (1)
$\\displaystyle\\varepsilon_{ij}\\varepsilon^{in}$ $\\displaystyle=$ $\\displaystyle\\delta_{j}^{n},$ (2)
$\\displaystyle\\varepsilon_{ij}\\varepsilon^{ij}$ $\\displaystyle=$ $\\displaystyle 2.$ (3)
•
When $n=3$ , we have for all $i, j, k, m, n$ in $\\{1,2,3\\}$ ,
$\\displaystyle\\varepsilon_{jmn}\\varepsilon^{imn}$ $\\displaystyle=$ $\\displaystyle 2\\delta^{i}_{j},$ (4)
$\\displaystyle\\varepsilon_{ijk}\\varepsilon^{ijk}$ $\\displaystyle=$ $\\displaystyle 6.$ (5)

Let us prove these properties. The proofs are instructional since theydemonstrate typical argumentation methods for manipulating thepermutation symbols.

Proof. For equation 1, let us first note that both sidesare antisymmetric with respect of $i j$ and $m n$ . We therefore only needto consider the case $i\eq j$ and $m\eq n$ . By substitution, we see thatthe 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 isanti-symmetric in $i j$ and $m n$ , any set of values for these can bereduced the above case (which holds). The equationthus holds for all values of $i j$ and $m n$ .Using equation 1, we have for equation 2

$\\displaystyle\\varepsilon_{ij}\\varepsilon^{in}$	$\\displaystyle=$	$\\displaystyle\\delta_{i}^{i}\\delta_{j}^{n}-\\delta^{n}_{i}\\delta^{i}_{j}$
	$\\displaystyle=$	$\\displaystyle 2\\delta_{j}^{n}-\\delta^{n}_{j}$
	$\\displaystyle=$	$\\displaystyle\\delta_{j}^{n}.$

Here we used the Einstein summation convention with $i$ going from $1$ to $2$ .Equation 3 follows similarly from equation 2.To establish equation 4, let us first observe that both sidesvanish when $i\eq j$ . Indeed, if $i\eq j$ , 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 remainingtwo 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 anydistinct indices $i, j, k$ in $\\{1,2,3\\}$ , we have $\\varepsilon_{ijk}\\varepsilon^{ijk}=1$ (no summation). $\\Box$

Examples and Applications.

•
The determinant of an $n\\times n$ matrix $A=(a_{ij})$ can be writtenas
$\\det A=\\varepsilon_{i_{1}\\cdots i_{n}}a_{1i_{1}}\\cdots a_{ni_{n}},$
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 $\\mathbbmss{R}^{3}$ (represented in some right hand oriented orthonormal basis), thenthe $i$ 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 $B$ , thenthe 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 isantisymmetric 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 someopen set of $\\mathbbmss{R}^{3}$ with Cartesian coordinates $x=(x^{1},x^{2},x^{3})$ . Thenthe $i$ th component of the curl of $F$ equals
$(\abla\\times F)^{i}(x)=\\varepsilon^{ijk}\\frac{\\partial}{\\partial x^{j}}F^{k}(%x).$