# Confusion about square bracket notation in the the Kronecker delta symbol

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?

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

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 $$|S_n|~=~n!~.$$

• @Qmechanic - I suggest you write out an explicit example maybe for $n=3$ or $n=4$. – Prahar Dec 16 '14 at 20:31
• 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. – Marion Dec 16 '14 at 23:21
• @Marion: Perfect, that saves me the trouble of typing it out. For $n=3$, see also Prahar's answer. – Qmechanic Dec 16 '14 at 23:23