The $n$-fold wedge product of a $2$ form

Question

For the $2$-form $\omega$ on $\Bbb R^{2n}$, $$\omega = \sum_{i = 1}^{2n-1} dx_i \wedge dx_{i + 1}$$

why is $\bigwedge_{i = 1}^n \omega \neq 0$?

I thought that if $n = 1$, (testing just $3$ terms) $\omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4 $.

So $\omega \wedge (\omega \wedge \omega) = \omega \wedge (2 \, dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4) = 0$

But apparently this isn't supposed to be $0$?

I got the result from here, but I don't understand why my reasoning is wrong.

Answer 1

You have an off-by-one error. The example you show is a 2-form on $\mathbb R^4$, so $n = 2$. And the claim is that the 2-fold wedge product of $\omega$ with itself (i.e. $\omega \wedge \omega$) is nonzero, which you've already shown: it's $2dx_1 \wedge ... \wedge dx_4$. The 3-fold wedge product would be a 6-form on $\mathbb R^4$, which is necessarily zero.

(You also, in writing out $\omega$, left out $dx_2 \wedge dx_3$, by the way.)