Let $ \omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n} \in \mathbb{R}^{2n}$. Find $\omega^n$ (in respect to $\wedge$)

Question

Let $ \omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n} \in \mathbb{R}^{2n}$. Find $\omega^{n}$ (in respect to $\wedge$)

When I say "$\omega^{n}$ (in respect to $\wedge$)" I mean the exterior product of $n$ $\omega$'s.

I tried writing

$$(dx_1 \wedge dx_2) (v_1,v_2) = \det \begin{pmatrix} d{x_1}(v_1) & d{x_1}(v_2)\\ d{x_2}(v_1) & d{x_2}(v_2) \end{pmatrix} = d{x_1}(v_1)d{x_2}(v_2) - d{x_1}(v_2)d{x_2}(v_1) $$

$$ (dx_3 \wedge dx_4) (v_1,v_2) = \det \begin{pmatrix} d{x_3}(v_1) & d{x_3}(v_2)\\ d{x_4}(v_1) & d{x_4}(v_2) \end{pmatrix} = d{x_3}(v_1)d{x_4}(v_2) - d{x_3}(v_2)d{x_4}(v_1) $$

but that wouldn't lead me nowhere.

I know the property $ \omega \wedge (\phi_1 + \phi_2) = \omega \wedge \phi_1 + \omega \wedge \phi_2 $

Then:

$$ \begin{align} \omega \wedge \omega = {} & (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \\ & {} \wedge (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \\[8pt] = {} & (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_1 \wedge dx_2) \\ & {} + (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_3 \wedge dx_4) + \cdots \\ & {} + (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_{2n-1} \wedge dx_{2n}) \end{align} $$ but I'm not sure that's what I'm supposed to do...

Thanks.

Answer 1

Write $\omega = \alpha_1 + \dots + \alpha_n$ where $\alpha_i = dx_{2i - 1} \wedge dx_{2i}$. The important observation is that if you expand

$$ \omega^n = (\alpha_1 + \dots + \alpha_n) \wedge \dots \wedge (\alpha_1 + \dots + \alpha_n) = \sum_{i_1, \dots, i_n} \alpha_{i_1} \wedge \dots \wedge \alpha_{i_n} = \sum_{I} \alpha_I$$

then if $I$ contains a repeated index the $2n$-form $\alpha_I$ will be a wedge of $2n$ one-forms in which (at least) two factors will be the same and so $\alpha_I = 0$. Thus,

$$ \omega^n = \sum_{\sigma \in S_n} \alpha_{\sigma(1)} \wedge \dots \wedge \alpha_{\sigma(n)}. $$

Since $\alpha_i$ are two-forms, we have $\alpha_{\sigma(1)} \wedge \dots \wedge \alpha_{\sigma(n)} = \alpha_1 \wedge \dots \wedge \alpha_n$ and so

$$ \omega^n = n! ( \alpha_1 \wedge \dots \wedge \alpha_n) = n! (dx_1 \wedge dx_2 \wedge \dots \wedge dx_{2n-1} \wedge dx_{2n}). $$

To demonstrate the argument for $n = 2$, we have

$$ \omega^2 = (dx_1 \wedge dx_2 + dx_3 \wedge dx_4) \wedge (dx_1 \wedge dx_2 + dx_3 \wedge dx_4) = \\ (dx_1 \wedge dx_2) \wedge (dx_1 \wedge dx_2) + (dx_3 \wedge dx_4) \wedge (dx_3 \wedge dx_4) + \\ (dx_1 \wedge dx_2) \wedge (dx_3 \wedge dx_4) + (dx_3 \wedge dx_4) \wedge (dx_1 \wedge dx_2) = \\ 2(dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4).$$