# 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 $\wedge$) [duplicate]

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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.

## marked as duplicate by Steven Stadnicki, C. Falcon, Shailesh, Daniel W. Farlow, M. VinayJun 24 '16 at 4:07

• (This is actually a multifold duplicate; I'm pretty sure I've seen this question at least 5 times. But that one actually came up in the 'related' tab.) – Steven Stadnicki Jun 23 '16 at 21:20
• Sorry. I'm still learning how to search with latex... – user286485 Jun 23 '16 at 21:29

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

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