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According to the book I read, a general $p$-form can be written as: $$\omega=\omega_{a_1\ldots a_p} dx^{a_1}\wedge\ldots\wedge dx^{a_p},\hspace{0.5cm} a_1>a_2>\ldots>a_p$$

where I have used a local coordinate basis $\{dx^a \}$ for 1-forms and each $a$ runs from $1$ to $n$.

I can't understand why it's $a_1>a_2>\ldots>a_p$. Ok for $a_i=a_j,i\neq j$ it's $dx^{a_i}\wedge dx^{a_j}=0$, so we don't have to include them. But what about the other components like $\omega_{1,2\ldots n} dx^{1}\wedge dx^{2}\wedge\ldots \wedge dx^{n}$?

For example, a $2$-form, in the two dimensional case ($n=2$) should be $$\omega=\omega_{12}dx^{1}\wedge dx^{2}+\omega_{21}dx^{2}\wedge dx^{1}=(\omega_{21}-\omega_{12})dx^{2}\wedge dx^{1}$$

and not $\omega=\omega_{21}dx^{2}\wedge dx^{1}$ as in the definition. Is $\omega_{12}=-\omega_{21}$ so that $\omega=2\omega_{21}dx^{2}\wedge dx^{1}$? But still we get the $2$ in front.

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No it's just notation: your $\omega_{12}-\omega_{21}$ is precisely their definition of $\omega_{12}$. Just absorb constants into one constant. – Chris Gerig Mar 27 '13 at 9:46
Is $\omega_{12}=-\omega_{21}$? If so, they should have put $1/p!$ in the definition. – Andyk Mar 27 '13 at 9:56
It is going to be unique only if we have $\omega_{b_1,b_2,..,b_n}=0$ for all nondecreasing sequences $(b_1,b_2,..,b_n)$. – Berci Mar 27 '13 at 10:56
up vote 1 down vote accepted

As you know, $u\land v=-v\land u$, so we can group the coefficients of all terms $dx^{a_{\sigma(1)}}\land dx^{a_{\sigma(2)}}\land \dots\land dx^{a_{\sigma(k)}}$ for $\sigma\in Sym(\{1,2,..,k\})$ together into one coefficient of $dx^{a_1}\land dx^{a_2}\land\dots\land dx^{a_k}$ for any fixed order $(a_1,a_2,a_3,..,a_k)$. E.g. we have $$ f\cdot (dx\land dz\land dy) +g\cdot (dz\land dx\land dy) = (f-g)\cdot (dz\land dy\land dx)\,,$$ So here $\omega_{3,2,1}=f-g$ and we don't have such separate things as $\omega_{1,2,3}$ or $\omega_{3,1,2}$ -- or, if you insist on having them, these are all defined to be $0$ -- as these are all presented in $\omega_{3,2,1}$.

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