I am not sure I understand what the short-hand anti-symmetrization means. I.e. I know that $$\delta_{cd}^{[ab]} ~=~ \frac{1}{2}(\delta_{c}^{a}\delta_{d}^{b} - \delta_{c}^{b}\delta_{d}^{a})$$ but how do we expand "bigger" analogues? I.e. $$\delta_{b_1 b_2 b_3 b_4}^{[a_1 a_2 a_3 a_4]} ~=~ ?$$

What is the order of permutation in this case?


1 Answer 1


$$ \delta^{[\mu_1\mu_2\ldots \mu_n]}_{\nu_1\nu_2\ldots \nu_n}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} \prod_{i=1}^n \delta^{\mu_{\pi(i)}}_{\nu_i}. $$

More generally,

$$ T^{[\mu_1\mu_2\ldots \mu_n]}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} T^{\mu_{\pi(1)}\mu_{\pi(2)}\ldots \mu_{\pi(n)}}. $$

Here $S_n$ is the symmetric group of permutations, and $\pi\in S_n$ is a permutation with signature ${\rm sgn(\pi)}$. The order of $S_n$ is $$|S_n|~=~n!~.$$

See also this and this related Phys.SE posts.

  • $\begingroup$ Ehm, could you give me an example to understand this better? $\endgroup$
    – Marion
    Commented Dec 16, 2014 at 18:37
  • $\begingroup$ Updated the answer. $\endgroup$
    – Qmechanic
    Commented Dec 16, 2014 at 20:22
  • 2
    $\begingroup$ @Qmechanic - I suggest you write out an explicit example maybe for $n=3$ or $n=4$. $\endgroup$
    – Prahar
    Commented Dec 16, 2014 at 20:31
  • 1
    $\begingroup$ I got it. It is rather long to write for $n=4$ or even for $n=3$. I have done it already and thanks for the answers. $\endgroup$
    – Marion
    Commented Dec 16, 2014 at 23:21
  • 1
    $\begingroup$ @Marion: Perfect, that saves me the trouble of typing it out. For $n=3$, see also Prahar's answer. $\endgroup$
    – Qmechanic
    Commented Dec 16, 2014 at 23:23

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