I know that $A_{[a} B_{b]} = \frac{1}{2!}(A_{a}B_{b} - A_{b}B_{a})$

But how can write $E_{[a} F_{bc]}$ like the above?

Can you provide a reference where this notational matter is discussed?


I use following method:


Set indices into determinant

$\left| \begin{matrix} i & j & k \\i & j & k \\i & j & k \end{matrix} \right| = ijk + jki + kij - ikj - jik - kji.$

Next, apply 6 indices and sign into $T$, so we get

$T_{[ijk]} = \dfrac{1}{3!} (T_{ijk}+T_{jki}+T_{kij}-T_{ikj}-T_{jik}-T_{kji}).$

  • $\begingroup$ So how does this translate to $E_{[a}F_{bc]}$? $\endgroup$ – Kyle Kanos Feb 13 '16 at 16:22
  • $\begingroup$ It can be done by same method; apply $[abc] \rightarrow abc + bca + cab - acb - bac - cba$ to $E_{[a} F_{bc]}$ and we get $E_{[a} F_{bc]} = \frac{1}{3!} (E_a F_{bc} + E_b F_{ca} + E_c F_{ab} - E_a F_{cb} - E_b F_{ac} - E_c F_{ba})$. $\endgroup$ – Um6r41 Mar 16 '16 at 13:23
  • $\begingroup$ The point I was trying to make was that your answer would be better if it made that connection more explicit (though in reviewing this, I couldn't see the other answers that neglect this step as well) $\endgroup$ – Kyle Kanos Mar 16 '16 at 13:29

The antisymmetric part is defined as $$ A_{[a_1 \cdots a_n]} = \frac{1}{n!} \sum\limits_{\sigma \in P(n)} \text{sgn}(\sigma)A_{a_{\sigma(1)} \cdots a_{\sigma(n)}} $$ where $P(n)$ is the set of all permutations of the set $\{1,\cdots,n\}$. $\text{sgn}(\sigma)$ is called the sign of the permutation and is positive of $\sigma$ is obtained from the identity $\sigma_0 = \{1,\cdots,n\} \in P(n)$ using an even number of exchanges and negative otherwise.

Applying this to $n=3$, we find $$ P(3) = \{\{1,2,3\},\{1,3,2\},\{2,1,3\},\{3,2,1\},\{2,3,1\},\{3,1,2\}\} $$ With signs $$ \text{sgn}(\sigma) = \{1,-1,-1,-1,1,1\} $$ We then find $$ A_{[a_1a_2a_3]} = \frac{1}{3!} \left( A_{a_1a_2a_3} + A_{a_3a_1a_2} + A_{a_2a_3a_1} - A_{a_1a_3a_2} - A_{a_2a_1a_3} - A_{a_3a_2a_1} \right) $$

  • $\begingroup$ +1: As a notational note, I personally think it's a bit less confusing (and more standard) to denote the sign of a permutation $\sigma$ as $\mathrm{sgn}(\sigma)$ which then equals $(-1)^m$ where $m$ is the number of inversions in the decomposition of $\sigma$ as a product of transpositions. See, for example, en.wikipedia.org/wiki/Parity_of_a_permutation $\endgroup$ – joshphysics Sep 30 '13 at 21:49

This is paraphrasing Wald - General Relativity, section 2.4. Antisymmetrizing $n$ indices means summing over all permutations of the indices, times the sign of each permutation. Since there are $n!$ permutations, it's a sane convention to divide by $n!$ (not all authors do this).

For your example, there are $3! = 6$ permutations of $(abc)$. The even ones are simply $(abc)$, $(bca)$ and $(cab)$, whereas the odd ones are $(acb)$, $(cba)$ and $(bac)$. So

$$E_{[a}F_{bc]} = \frac{1}{6} \left[ E_aF_{bc} + E_bF_{ca} + E_cF_{ab} - E_aF_{cb} - E_bF_{ac} - E_cF_{ba} \right].$$

If you've never heard of the distinction between even and odd permutations, you need to open up a textbook about undergraduate group theory -- it's easy to understand but takes quite some space to do the explicit proof that thing is a well-defined thing and you do need to understand the structure of the permutation group $S_N$.


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