# Show that $\omega^{2}\wedge\cdots\wedge\omega^{2}$ n times is equal to

Consider $\mathbb{R}^{2n}$ with coordenates $x^{1},\cdots,x^{2n}$ and the following differential form of grade two $$\omega^{2}=dx^{1}\wedge dx^{n+1}+dx^{2}\wedge dx^{n+2}+\cdots+dx^{n}\wedge dx^{2n}$$ Show that $\omega^{2}\wedge\cdots\wedge\omega^{2}$ n times, is equal to $(-1)^{\frac{n(n-1)}{2}}n!dx^{1}\wedge\cdots\wedge dx^{2n}$

• A hint: you know that $\omega^2\wedge^{(n)} \omega^2$ is a differential form of grade $2n$; there's (fundamentally) only one such form. You just need to figure out what terms in your wedge product can produce a factor of $dx^1\wedge\cdots dx^{2n}$, what the appropriate sign is, and how many there are. – Steven Stadnicki Dec 1 '15 at 3:30

Sketch: Write $y^i = dx^i \wedge dx^{n+i}$, where $i=1, 2, \cdots, n$. Note that $y^i \wedge y^j = y^j \wedge y^i$ and $y^i \wedge y^i = 0$. So symbolically, you are expanding $$(y^1 +\cdots + y^n)^n$$ and counting the number of terms corresponding to $y^1 \wedge y^2 \wedge \cdots \wedge y^n$. It is easy to see that (as everything commutes) $$(y^1 + y^2 + \cdots + y^n)^n = n!\ y^1\wedge y^2 \wedge \cdots \wedge y^n.$$ That extra sign come from reordering $y^1 \wedge y^2 \cdots \wedge y^n$ back to $dx^1 \wedge \cdots \wedge dx^{2n}$.
• Do you mean $\omega^2$? @asdasdasd – user99914 Dec 2 '15 at 4:35
• Yes, I know that a k-fom is called closed, if $d\omega^{k}=0$, but I am confused – asdasd asd Dec 2 '15 at 4:37
• Do you know how to calculate $d( dx^1 \wedge dx^{n+1})$? @asdasdasd – user99914 Dec 2 '15 at 5:56
• And $d(dx^1) =0$ as $d^2 =0$. @asdasdasd – user99914 Dec 3 '15 at 5:54