Product of Two Levi-Civita Symbols
Theorem
The $\epsilon_{ijk}$, defined as follows, is referred to as the Levi-Civita symbol.
$$ \epsilon_{ijk} = \begin{cases} +1 & \text{if} \ \epsilon_{123}, \epsilon_{231}, \epsilon_{312} \\ -1 & \text{if} \ \epsilon_{132}, \epsilon_{213}, \epsilon_{321} \\ 0 & \text{if} \ i=j \ \text{or} \ j=k \ \text{or} \ k=i \end{cases} $$
The $\delta_{ij}$, defined as follows, is referred to as the Kronecker delta.
$$ \delta_{ij} := \begin{cases} 1,&i=j \\ 0, & i\ne j \end{cases} $$
Between the product of two Levi-Civita symbols and the Kronecker delta, the following relationships hold:
(a) When one index is the same: $\epsilon_{ijk}\epsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}$
(b) When two indices are the same: $\epsilon_{ijk}\epsilon_{ijm}=2\delta_{km}$
(c) When all three indices are the same: $\epsilon_{ijk}\epsilon_{ijk}=6$
Explanation
Note that the summation symbol $\sum$ is omitted throughout this text, adhering to the Einstein notation. This applies to the formulas above as well. Memorizing (a) can be very useful as it’s frequently used. A simple way to remember it is as follows.
Proof
(a)
Let $\mathbf{e}_{i}$ $(i=1,2,3)$ be the standard unit vectors in 3-dimensional space.
$$ \mathbf{e}_{1} = (1, 0, 0),\quad \mathbf{e}_{2} = (0, 1, 0),\quad \mathbf{e}_{3} = (0, 0, 1) $$
Let $P_{ijk}$ be a $3 \times 3$ matrix whose 1st, 2nd, and 3rd rows are $\mathbf{e}_{i}$, $\mathbf{e}_{j}$, and $\mathbf{e}_{k}$, respectively.
$$ P_{ijk} = \begin{bmatrix} \text{— } \mathbf{e}_{i} \text{ —} \\ \text{— } \mathbf{e}_{j} \text{ —} \\ \text{— } \mathbf{e}_{k} \text{ —} \end{bmatrix} $$
Then, by the properties of determinants, it’s easy to see that $\det P_{ijk} = \epsilon_{ijk}$. Initially, $P_{123}$ is the identity matrix, hence its determinant is $1$. Moreover, the value of the determinant remains unchanged when swapping different rows an even number of times, hence,
$$ \det P_{123} = \det P_{231} = \det P_{312} = 1 $$
When different rows are swapped an odd number of times, the sign of the determinant changes, hence,
$$ \det P_{132} = \det P_{213} = \det P_{321} = -1 $$
The determinant of a matrix with two or more identical rows is $0$, hence the rest of the cases are all $0$. Therefore, $\det P_{ijk} = \epsilon_{ijk}$ holds true. The product of two Levi-Civita symbols with one identical index can be expressed as follows, using the properties of determinants.
$$ \begin{align*} \epsilon_{ijk}\epsilon_{ilm} &= \det \begin{bmatrix} \text{— } \mathbf{e}_{i} \text{ —} \\ \text{— } \mathbf{e}_{j} \text{ —} \\ \text{— } \mathbf{e}_{k} \text{ —} \end{bmatrix} \det \begin{bmatrix} \text{— } \mathbf{e}_{i} \text{ —} \\ \text{— } \mathbf{e}_{l} \text{ —} \\ \text{— } \mathbf{e}_{m} \text{ —} \end{bmatrix} \\ &= \det \begin{bmatrix} \text{— } \mathbf{e}_{i} \text{ —} \\ \text{— } \mathbf{e}_{j} \text{ —} \\ \text{— } \mathbf{e}_{k} \text{ —} \end{bmatrix} \det \begin{bmatrix} \vert & \vert & \vert \\ \mathbf{e}_{i} & \mathbf{e}_{l} & \mathbf{e}_{m} \\ \vert & \vert & \vert \end{bmatrix} & (\because \det A = \det A^{T}) \\ &= \det \left( \begin{bmatrix} \text{— } \mathbf{e}_{i} \text{ —} \\ \text{— } \mathbf{e}_{j} \text{ —} \\ \text{— } \mathbf{e}_{k} \text{ —} \end{bmatrix} \begin{bmatrix} \vert & \vert & \vert \\ \mathbf{e}_{i} & \mathbf{e}_{l} & \mathbf{e}_{m} \\ \vert & \vert & \vert \end{bmatrix} \right) & \Big(\because (\det A) (\det B) = \det (AB) \Big) \\ &= \det \begin{bmatrix} \mathbf{e}_{i} \cdot \mathbf{e}_{i} & \mathbf{e}_{i} \cdot \mathbf{e}_{l} & \mathbf{e}_{i} \cdot \mathbf{e}_{m} \\ \mathbf{e}_{j} \cdot \mathbf{e}_{i} & \mathbf{e}_{j} \cdot \mathbf{e}_{l} & \mathbf{e}_{j} \cdot \mathbf{e}_{m} \\ \mathbf{e}_{k} \cdot \mathbf{e}_{i} & \mathbf{e}_{k} \cdot \mathbf{e}_{l} & \mathbf{e}_{k} \cdot \mathbf{e}_{m} \end{bmatrix} \end{align*} $$
Since $\mathbf{e}_{i}$ are standard unit vectors, $\mathbf{e}_{i} \cdot \mathbf{e}_{j} = \delta_{ij}$ holds true.
$$ \epsilon_{ijk}\epsilon_{ilm} = \det \begin{bmatrix} \delta_{ii} & \delta_{il} & \delta_{im} \\ \delta_{ji} & \delta_{jl} & \delta_{jm} \\ \delta_{ki} & \delta_{kl} & \delta_{km} \end{bmatrix} $$
Note that we are only considering cases where $i$ is different from $j, k, l, m$. This is because if $j, k, l, m$ includes $i$, then $\epsilon_{ijk}\epsilon_{ilm} = 0$, rendering the result meaningless. Therefore, the result is
$$ \epsilon_{ijk}\epsilon_{ilm} = \det \begin{bmatrix} 1 & 0 & 0 \\ 0 & \delta_{jl} & \delta_{jm} \\ 0 & \delta_{kl} & \delta_{km} \end{bmatrix} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl} $$
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(b)
This is the case where $l=j$ in (a). Thus, it can be expressed as follows.
$$ \epsilon_{ijk}\epsilon_{ijm} = \delta_{jj}\delta_{km} - \delta_{jm}\delta_{kj} $$
Here, $\delta_{jj}=3$ and $\delta_{jm}\delta_{kj}=\delta_{mk}$ hold, leading to the following.
$$ \epsilon_{ijk}\epsilon_{ijm} = \delta_{jj}\delta_{km} - \delta_{jm}\delta_{kj} = 3\delta_{km} - \delta_{mk} = 2\delta_{km} $$
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(c)
This is the case where $m=k$ in (b), hence,
$$ \epsilon_{ijk}\epsilon_{ijk} = \sum_{k=1}^{3}2\delta_{kk} = 2\delta_{11} + 2\delta_{22} + 2\delta_{33} = 2 + 2 + 2 = 6 $$
Alternatively, by explicitly writing out all non-zero terms, the following can be obtained.
$$ \begin{align*} \epsilon_{ijk}\epsilon_{ijk} &=\sum \limits _{i=1} ^{3}\sum \limits _{j=1} ^{3}\sum \limits _{k=1} ^{1} \epsilon_{ijk}\epsilon_{ijk} \\ &=\epsilon_{123}\epsilon_{123}+\epsilon_{231}\epsilon_{231}+\epsilon_{312}\epsilon_{312}+\epsilon_{132}\epsilon_{132}+\epsilon_{213}\epsilon_{213}+\epsilon_{321}\epsilon_{321} \\ &=6 \end{align*} $$
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