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I think I have a little confusion with index notation (concerning p-form). For example, do the coordinates on $\mathbb{R}^n$ $\{x^1,x^2, \dots , x^k\}$ and $\{x^{i_1},x^{i_2}, \dots , x^{i_k}\}$ mean the same? Can they be used equivalently (if no, why not?)?

To illustrate this more claerly: Suppose we have a p-form $\omega$ on $\mathbb{R}^n$. Then I find just expression like $$\omega = \omega_{{i_1}, \dots,{i_p}}dx^{i_1} \wedge \dots \wedge dx^{i_p}.$$ Can I also write instead: $$\omega = \omega_{{1}, \dots,{p}}dx^{1} \wedge \dots \wedge dx^{p} ?$$ Is there a difference, and why? I am confused because if you perfom the exterior derivative of $\omega$ (here I use the first representation) according to this formula $$ d\omega = \sum_{i=1}^n \frac{\partial \omega_{{i_1}, \dots,{i_p}} }{\partial x^i} dx^i \wedge dx^{i_1} \wedge \dots \wedge dx^{i_p}, $$ I don't understand how this can be the same as (now I use the second representation) $$d\omega = \sum_{i=1}^n \frac{\partial \omega_{{1}, \dots,{p}} }{\partial x^i} dx^i \wedge dx^{1} \wedge \dots \wedge dx^{p}.$$

Performing the derivative for a 1-form on $\mathbb{R}^2, \omega = \omega_{{i_1}}dx^{i_1} $ I get: $$ d\omega = \frac{\partial \omega_{{i_1}}}{\partial x^1} dx^1 \wedge dx^{i_1} + \frac{\partial \omega_{{i_1}}}{\partial x^2} dx^2 \wedge dx^{i_1} . $$ If I use the second representation I get: $$ d\omega = \frac{\partial \omega_{{1}}}{\partial x^1} dx^1 \wedge dx^{1} + \frac{\partial \omega_{{1}}}{\partial x^2} dx^2 \wedge dx^{1},$$ where $\frac{\partial \omega_{{1}}}{\partial x^1} dx^1 \wedge dx^{1}$ cancels, and this derivative is indeed right.

Then, how do I get to this result using the first derivative? Why is $\frac{\partial \omega_{{i_1}}}{\partial x^1} dx^1 \wedge dx^{i_1} $ equal to 0 and $ dx^2 \wedge dx^{i_1}$ equal to $ dx^2 \wedge dx^{1} $?

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    $\begingroup$ The difference is that if you have $i_1,\dotsc,i_k$, you sum over repeated indices, which you do not do with explicit numbers (you should have met this by now if you're studying differential forms: it's called summation convention, and is very common in both vector calculus and differential geometry. $\endgroup$ – Chappers Oct 20 '15 at 19:59
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The tuple ${x^1,...,x^k}$ has clearly determined order and elements. The tuple ${}{x^{i_1},...,x^{i_p}}$ means that every element of the tuple can be any coordinate, the indices are numbered however to distinguish them, so that each indexed element can take on values independent of other indexed elements.

If $\omega$ is an $n$-form on $\mathbb{R}^n$ (eg. it is a MAXIMAL order form), then the notations $$ \omega_{1,2,...,n}dx^1\wedge...\wedge dx^n $$ and $$ \omega_{i_1...i_n}dx^{i_1}\wedge...\wedge dx^{i_n} $$ are essentially same, except for an $1/n!$ factor in the second expression. It is because the $n$th exterior power of an $n$ dimensional vector space is one-dimensional, so all $n$-forms have one independent component only and that is the one in the first expression.

What the second expression contains, is basically that one independent component as many times as many permutations of ${1,...,n}$ is, and that is $n!$, so you have to divide it by $n!$ to get the same thing as the first one.

However, if you take a form that is NOT of maximal order, eg. a $p$ form when $p<n$, then you have to take a summation, because those forms will generally have more than one independent component.

However, it might be helpful to restrict the summation to lexicographical order to avoid having terms with multiplicity. The correct form for a $p$-form should be $$ \omega=\sum_{i_1<i_2<...<i_p}\omega_{i_1...i_p}dx^{i_1}\wedge...\wedge dx^{i_p}=\frac{1}{p!}\omega_{i_1...i_p}dx^{i_1}\wedge...\wedge dx^{i_p}, $$ where in the second case, the summation wasn't restricted.

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Example. In $\mathbb R^3$, the form $dx\wedge dy$ is clearly different from $dy\wedge dz$ and from $dy\wedge dx$ (which is the first one multiplied by $-1$). Indeed, the integral of the first form over the unit disk in the $x,y$ plane defined by $$\{(x,y,z)\in\Bbb R^3 \mid x^2+y^2\le 1, z=0\}$$ is the area of the disk; by contrast, the integral of the second form over the same disk is $0$. The opposite happens when integrating over the unit disk in the $y,z$ plane, and both integrals vanish over the disk in the $x,z$ plane.

To sum up, the form $dx^{i_1}\wedge\dots\wedge dx^{i_k}$ will depend on which numbers your $\{i_j\}_{j=1}^k$ stand for and in which order they appear. In the first example above, we had $i_1=1, i_2=2$, whereas in the second example we had $i_1=2$ and $i_2=3$ (under the convention that $x=x^{1}, y=x^{2}, z=x^{3}$).

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