# Compute the $n$th exterior power of a differential $2$-form

Suppose you have a differential form $\omega$ written in local coordinates as $$\omega=\sum_{i=1}^ndx_i\wedge dy_i.$$ Can anyone help me showing the following equality: $$\omega^n=n!(dx_1\wedge dy_1\wedge\ldots \wedge dx_n\wedge dy_n).$$ Only as a motivation: the above equality is useful for showing every symplectic manifold is orientable.

• By $\omega^n$ you mean $\omega$ wedged with itself $n$ - times? – user38268 Jun 12 '13 at 11:49
• Try with $n=1$ (it is ok, right?). The case $n=2$ helps you with manipulating the indices. Then use induction and remember that $d x_i \wedge d y_j=- d y_j\wedge d x_i$ – Avitus Jun 12 '13 at 11:53
• @O.L., I tried using induction but I lose myself in the calculations.. – PtF Jun 12 '13 at 12:25
• @BenjaLim yes, $\omega^n=\underbrace{\omega\times \ldots\times \omega}_{n-times}$. – PtF Jun 12 '13 at 12:25
• @Avitus, I'll try.. – PtF Jun 12 '13 at 12:26

A motivating example with $n=2$. Let $\omega=dx_1\wedge dy_1 + dx_2\wedge dy_2$ be our 2-form. Then

$\omega\wedge\omega=(dx_1\wedge dy_1 + dx_2\wedge dy_2)\wedge (dx_1\wedge dy_1 + dx_2\wedge dy_2)=dx_1\wedge dy_1\wedge dx_1\wedge dy_1+ dx_2\wedge dy_2\wedge dx_2\wedge dy_2+ dx_1\wedge dy_1\wedge dx_2\wedge dy_2+dx_2\wedge dy_2\wedge dx_1\wedge dy_1$,

where all brackets are removed due to associativity of $\wedge$. Now

$dx_1\wedge dy_1\wedge dx_1\wedge dy_1=-dx_1\wedge dx_1\wedge dy_1\wedge dy_1$

and

$dx_2\wedge dy_2\wedge dx_2\wedge dy_2=-dx _2\wedge dx_2\wedge dy_2\wedge dy_2$

by antisymmetry of $\wedge$. As $dx_i\wedge dx_i=dy_i\wedge dy_i=0$ (again by antisymmetry!) the two above expressions are euqal to $0$. It remains that

$\omega\wedge\omega= dx_1\wedge dy_1\wedge dx_2\wedge dy_2+dx_2\wedge dy_2\wedge dx_1\wedge dy_1$. But

$dx_2\wedge dy_2\wedge dx_1\wedge dy_1=$(moving twice $dy_2$ to the right) $dx_2\wedge dx_1\wedge dy_1\wedge dy_2=$(moving twice $dx_2$ to the right) $dx_1\wedge dy_1\wedge dx_2\wedge dy_2$.

In summary

$\omega\wedge\omega= 2(dx_1\wedge dy_1\wedge dx_2\wedge dy_2)=2!(dx_1\wedge dy_1\wedge dx_2\wedge dy_2)$,

as wished.

• thanks I'll try to generalizate it =D – PtF Jun 12 '13 at 12:44
• if you find the answer useful, you can upvote it and flag it :-) All you need is to remember that "whenever I exchange two symbols $dx$ and $dy$ in the wedge product I get a minus" :-) – Avitus Jun 12 '13 at 12:47
• @Avitus +1 for patient explanation – Start wearing purple Jun 12 '13 at 12:49

I'd begin by checking (if you haven't already seen this fact) that, even though $1$-forms anticommute, $2$-forms commute. In particular, all the summands $dx_i\land dy_i$ in your $\omega$ commute with each other. So you can multiply out $\omega^n$ without worrying about signs as long as you leave each pair $dx_i$ and $dy_i$ adjacent to each other and in the original order. (In other words, don't change $dx_i\land dy_i$ to $dy_i\land dx_i$, and don't insert any other factors between $dx_i$ and $dy_i$.) When you multiply out $\omega^n$ in this way, a lot of terms have repeated factors; they vanish because (now using anticommutativity) you can move two matching factors next to each other, where they cancel because $dx_i\land dx_i=0$). The surviving terms in $\omega^n$ are those where each of the summands $dx_i\land dy_i$ occurs exactly once as a factor. Each of those terms equals $dx_1\land dy_1\land dx_2\land dy_2\land\dots\land dx_n\land dy_n$ because these summands commute. And there are $n!$ such terms.

• Thanks, I'll try that @Andreas =D – PtF Jun 12 '13 at 16:54