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  3. The best answers are voted up and rise to the top spouted off and threw out with no motivation $$\epsilon_{ijk} = \frac{1}{2}(i - j)(j - k)(k - i) \, \forall \, \, k \in \{1, 2, 3\}$$ where $ \epsilon_{i_1i_2i_3...i_n} = \left\{ \begin{array}{rcl} +1 & \mbox{if } (i_1, i_2, i_3, ..., i_n) \text{ is an even permutation of } (1, 2, 3, ..., n) & \\ -1 & \mbox{if } (i_1, i_2, i_3, ..., i_n)\text{ is an odd permutation of } (1, 2, 3, ..., n) \\ 0 & \mbox{if } (i_1, i_2, i_3, ..., i_n) \text{ is NOT a permutation of } (1, 2, 3, ..., n) \end{array}\right. $

What is the intuition or derivation? I tried looking online but found nothing.

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The key properties of the Levi-Civita symbol are its antisymmetry and normalization, $\epsilon_{1\cdots n}=1$. We wish to capture, with the integers $\{i_1,\ldots,i_n\}$, $i_j\in\{1,\ldots,n\}$, these properties of $\epsilon$. It is natural to consider products of the form $\prod_{j,k}(i_j-i_k)$, since the product vanishes if any of the $i$s are the same. In addition, we can't have an even number of factors of the form $(i_j-i_k)$, since we'll have no hope of capturing the antisymmety property of $\epsilon$. A simple ansatz is $$\begin{eqnarray*} e(i_1,\ldots,i_n) &=& c_n\prod_{1\le j<k\le n}(i_k-i_j) \\ &=& c_n \prod_{k=2}^n\prod_{j=1}^{k-1}(i_k-i_j) \\ &=& c_n [(i_2-i_{1})] \\ && \times [(i_3-i_1)(i_3-i_2)] \\ && \cdots \\ && \times [(i_{n-1}-i_{1})\cdots(i_{n-1}-i_{n-2})] \\ && \times [(i_n-i_{1})\cdots(i_n-i_{n-1})] \end{eqnarray*}$$ where $c_n$ is some constant that we'll determine shortly.

From the form of the product we can see that permuting adjacent $i$s in the product will simply introduce a factor of $-1$. (For example, $e(i_2,i_1,\ldots,i_n) = -e(i_1,i_2,\ldots,i_n)$ since we'll pick up one factor of $-1$ from the factor $(i_2-i_1)$.) This is enough to show that the product has the antisymmetry property of $\epsilon$.

If any of the $i$s are not distinct the product is zero. All other products can be obtained by permutations of the product $e(1,\ldots,n)$. All that remains is to determine $c_n$. We have $$\begin{eqnarray*} e(1,\ldots,n) &=& c_n \prod_{k=2}^n\prod_{j=1}^{k-1}(k-j) \\ &=& c_n \prod_{k=2}^n\prod_{l=1}^{k-1}l \hspace{5ex} (\textrm{let }m=k-j) \\ &=& c_n \prod_{k=2}^n (k-1)! \\ %&=& c_n \prod_{k=1}^{n-1}k \\ &=& c_n \, \mathrm{sf}(n-1), \end{eqnarray*}$$ where $\mathrm{sf}(n)=\prod_{k=1}^{n}k!$ is the superfactorial. (Starting from $n=0$, the sequence of superfactorials is $1,1,2,12,288,\ldots$.) Therefore, $$\begin{eqnarray*} \epsilon_{i_1\cdots i_n} &=& e(i_1,\ldots,i_n) \\ &=& \frac{1}{\mathrm{sf}(n-1)} \prod_{1\le j<k\le n}(i_k-i_j). \hspace{10ex} \textrm{(1)} \end{eqnarray*}$$ Some results for small $n$ follow.

$n=2$ $$\epsilon_{ij} = j-i, \quad i,j\in\{1,2\}$$

$n=3$ $$\begin{eqnarray*} \epsilon_{ijk} &=& \frac{1}{2}(j-i)(k-i)(k-j), \quad i,j,k\in\{1,2,3\} \\ &=& \frac{1}{2}(i-j)(j-k)(k-i) \end{eqnarray*}$$

$n=4$ $$\begin{eqnarray*} \epsilon_{ijkl} &=& \frac{1}{12}(j-i)(k-i)(k-j)(l-i)(l-j)(l-k), \quad i,j,k,l\in\{1,2,3,4\} \end{eqnarray*}$$


Since $\epsilon_{i_1\cdots i_n}^{2m+1} = \epsilon_{i_1\cdots i_n}$ for $m\in\mathbb{N}$, we immediately find an infinity of other possible representations for $\epsilon$, that is, the product $$\begin{eqnarray*} \frac{1}{\mathrm{sf}(n-1)^{2m+1}} \prod_{1\le j<k\le n}(i_k-i_j)^{2m+1} \end{eqnarray*}$$ is also a perfectly good representation of $\epsilon$. The principle of parsimony dictates that representation (1) is preferable.

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