Levi-Civita Symbol
Definition
The $\epsilon_{ijk}$ defined as follows is called the Levi-Civita symbol.
$$ \epsilon_{ijk} = \begin{cases} +1 & \text{if} \ \epsilon_{123}, \epsilon_{231}, \epsilon_{312} \\ -1 & \text{if} \ \epsilon_{132}, \epsilon_{213}, \epsilon_{321} \\ 0 & \text{if} \ i=j \ \text{or} \ j=k \ \text{or} \ k=i \end{cases} $$
Description
While the Kronecker delta only considers whether the indices are equal, the Levi-Civita symbol, as shown in its definition, is also affected by the order of the indices. $\epsilon_{ijk}$ results in $+1$ if $i$, $j$, $k$ are in ascending order without duplicates$(1\to 2\to 3\to 1)$, $-1$ if in descending order without duplicates$(3\to 2\to 1\to 3)$, and $0$ if there’s any duplication. Simply put, that’s the summary, and if you list all values, there are a total of $3\times 3\times 3=27$, out of which $6$ are not $0$.
$$ \begin{array}{|c|c|c|c|}\hline i=1 & k=1 & k=2 & k=3 \\ \hline j=1 & \epsilon_{111}=0 & \epsilon_{112}=0 & \epsilon_{113}=0 \\ j=2 & \epsilon_{121}=0 & \epsilon_{122}=0 & \epsilon_{123}=1 \\ j=3 & \epsilon_{131}=0 & \epsilon_{132}=-1 & \epsilon_{133}=0 \\ \hline \end{array}\quad \begin{array}{|c|c|c|c|}\hline i=2 & k=1 & k=2 & k=3 \\ \hline j=1 & \epsilon_{211}=0 & \epsilon_{212}=0 & \epsilon_{213}=-1 \\ j=2 & \epsilon_{221}=0 & \epsilon_{222}=0 & \epsilon_{223}=0 \\ j=3 & \epsilon_{231}=1 & \epsilon_{232}=0 & \epsilon_{233}=0 \\ \hline \end{array} \\ {} \\ \begin{array}{|c|c|c|c|}\hline i=3 & k=1 & k=2 & k=3 \\ \hline j=1 & \epsilon_{311}=0 & \epsilon_{312}=1 & \epsilon_{313}=0 \\ j=2 & \epsilon_{321}=-1 & \epsilon_{322}=0 & \epsilon_{323}=0 \\ j=3 & \epsilon_{331}=0 & \epsilon_{332}=0 & \epsilon_{333}=0 \\ \hline \end{array} $$
It is an example of a $3$th order tensor.
Examples
Cross Product
Using the Levi-Civita symbol, the cross product of two vectors can be expressed very simply. In a 3-dimensional orthogonal coordinate system, the cross product of two vectors is as follows.
$$ \begin{align*} \mathbf{A} \times \mathbf{B} =&\ \hat{\mathbf{e}}_{x} (A_{y}B_{z}-A_{z}B_{y}) + \hat{\mathbf{e}}_{y} (A_{z}B_{x}-A_{x}B_{z}) + \hat{\mathbf{e}}_{z} (A_{x}B_{y}-A_{y}B_{x}) \\ =&\ \begin{vmatrix} \hat{\mathbf{e}}_{x} \quad \hat{\mathbf{e}}_{y} \quad \hat{\mathbf{e}}_{z} \\ A_{x} \quad A_{y} \quad A_{z} \\ B_{x} \quad B_{y} \quad B_{z} \end{vmatrix} \end{align*} $$
Here, if we set $x=1$, $y=2$, $z=3$, then the cross product of two vectors can be expressed using the Levi-Civita symbol as below.
$$ \sum\limits_{i=1}^{3} \sum\limits_{j=1}^{3} \sum\limits_{k=1}^{3} {\epsilon_{ijk} \hat{\mathbf{e}}_{i} A_{j}B_{k}} $$
If we expand only the terms that are not $0$, it can be written as follows.
$$ \begin{align*} & \sum\limits_{i=1}^{3} \sum\limits_{j=1}^{3} \sum\limits_{k=1}^{3} {\epsilon_{ijk} \hat{\mathbf{e}}_{i} A_{j}B_{k}} \\ =&\ \epsilon_{123} \hat{\mathbf{e}}_{1}A_{2}B_{3} + \epsilon_{132} \hat{\mathbf{e}}_{1}A_{3}B_{2} + \epsilon_{ 231 }\hat{\mathbf{e}}_{2}A_{3}B_{1} + \epsilon_{213}\hat{\mathbf{e}}_{2}A_{1}B_{3} + \epsilon_{312}\hat{\mathbf{e}}_{3}A_{1}B_{2} + \epsilon_{321}\hat{\mathbf{e}}_{3}A_{2}B_{1} \end{align*} $$
By substituting $\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1$, $\epsilon_{132}=\epsilon_{213}=\epsilon_{321}=-1$ and rearranging, we obtain the following.
$$ \begin{align*} & \sum\limits_{i=1}^{3} \sum\limits_{j=1}^{3} \sum\limits_{k=1}^{3} {\epsilon_{ijk} A_{i}B_{j}}\hat{\mathbf{e}}_{k} \\ =&\ \hat{\mathbf{e}}_{1}\left( A_{2}B_{3} - A_{3}B_{2} \right) + \hat{\mathbf{e}}_{2}\left( A_{3}B_{1} - A_{1}B_{3} \right) + \hat{\mathbf{e}}_{3}\left( A_{1}B_{2} - A_{2}B_{1} \right) \end{align*} $$
Finally, by substituting $1$, $2$, $3$ with $x$, $y$, $z$ respectively, we get the following.
$$ \hat{\mathbf{e}}_{x}\left( A_{y}B_{z} - A_{z}B_{y} \right) + \hat{\mathbf{e}}_{y}\left( A_{z}B_{x} - A_{x}B_{z} \right) + \hat{\mathbf{e}}_{z}\left( A_{x}B_{y} - A_{y}B_{x} \right) $$
Therefore, we obtain the below result.
$$ \mathbf{A} \times \mathbf{ B } = \sum \limits_{i=1}^{3} \sum \limits_{j=1}^{3} \sum \limits_{k=1}^{3} \epsilon_{ijk}\hat{\mathbf{e}}_{i}A_{j}B_{k} $$
Using Einstein notation, it can be expressed as follows.
$$ \mathbf{A} \times \mathbf{ B } = \epsilon_{ijk}\hat{\mathbf{e}}_{i}A_{j}B_{k} $$
Each component of the cross product can be easily expressed as shown in the above equation, where the $i$ component of $(\mathbf{A} \times \mathbf{B})$ is as follows.
$$ (\mathbf{A} \times \mathbf {B})_{i}=\epsilon_{ijk}A_{j}B_{k} $$
Determinant
The determinant of matrix $3 \times 3$, $A = [a_{ij}]$ is expressed as follows.
$$ \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{vmatrix} = \sum\limits_{i,j,k=1}^{3} \epsilon_{ijk}a_{1i}a_{2j}a_{3k} $$
The demonstration is simple. By expanding the determinant and carefully observing the second index of each matrix component, one can see that the index’s Levi-Civita symbol corresponds to the sign of each term.
$$ \begin{align*} & \det A \\ &= a_{11}(a_{22}a_{33} - a_{23}a_{32}) + a_{12}(a_{23}a_{31} - a_{21}a_{33}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \\ &= a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} + a_{12}a_{23}a_{31} - a_{12}a_{21}a_{33} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} \\ &= a_{1\textcolor{red}{1}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{3}} - a_{1\textcolor{red}{1}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{2}} + a_{1\textcolor{red}{2}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{1}} - a_{1\textcolor{red}{2}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{3}} + a_{1\textcolor{red}{3}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{2}} - a_{1\textcolor{red}{3}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{1}} \\ &= \epsilon_{\textcolor{red}{123}}a_{1\textcolor{red}{1}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{3}} + \epsilon_{\textcolor{red}{132}}a_{1\textcolor{red}{1}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{2}} + \epsilon_{\textcolor{red}{231}}a_{1\textcolor{red}{2}}a_{2\textcolor{red}{3}}a_{3\textcolor{red}{1}} + \epsilon_{\textcolor{red}{213}}a_{1\textcolor{red}{2}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{3}} + \epsilon_{\textcolor{red}{312}}a_{1\textcolor{red}{3}}a_{2\textcolor{red}{1}}a_{3\textcolor{red}{2}} + \epsilon_{\textcolor{red}{321}}a_{1\textcolor{red}{3}}a_{2\textcolor{red}{2}}a_{3\textcolor{red}{1}} \\ &= \sum\limits_{i,j,k=1}^{3} \epsilon_{ijk}a_{1i}a_{2j}a_{3k} \end{align*} $$
Formula
(a) When one index is the same: $\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$
(b) When two indices are the same: $\epsilon_{ijk}\epsilon_{ljk}=2\delta_{il}$
(c) When three indices are the same: $\epsilon_{ijk}\epsilon_{ijk}=6$