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In Hobson's General Relativity: An Introduction for Physicists book, pg. 95, the totally antisymmetric part of a rank 3 tensor $t_{abc}$ is defined as

$$t_{[abc]} = \frac{1}{6}(t_{abc}-t_{acb}+t_{cab}-t_{cba}+t_{bca}-t_{bac}).$$

Why did the author include a factor of $1\over 6$ in the expression? Is he trying to anticipate that the tensor $t_{abc}$ can be made up in some way using the totally antisymmetric tensor $t_{[abc]}$?

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Adding the $\frac{1}{n!}$ factor in the antisymmetrisation is just a convention (and some authors omit it) - but this convention ensures that for totally antisymmetric tensors, $T_{\mu\nu...\sigma} = T_{[\mu\nu...\sigma]}$.

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