Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.