# Wedge product of 1-Forms

I'm trying to write down the wedge product of 2 1-forms on an n-dimensional Manifold.

$\alpha = \alpha_1 dx^1 + \alpha_2 dx^2 + \cdots + \alpha_n dx^n$

and

$\beta = \beta_1 dx^1 + \beta_2 dx^2 + \cdots + \beta_n dx^n$

I know how to do this for the 2 and 3 dimensional case. But I'm having a problem with the n-dimensional case. More specifically, what sign the individual 2-forms get?

What I mean by this is, let's concider a few terms of $\alpha \wedge \beta$:

$\cdots \alpha_1 \beta_2 dx^1\wedge dx^2 + \alpha_2 \beta_1 dx^2 \wedge dx^1 \cdots$

Now the problem I'm having is, I think $dx^2\wedge dx^1 = -dx^1\wedge dx^2$. Which allows me to combine the above two terms. For 2 and 3 dimensions, this seems easy, as I can just look at the permutations, but in n-dimensions,

(1) how does this permutation look like? Or is it always minus if I switch the $dx$?

(2) Does this mean that $\alpha\wedge\alpha = 0$ always for 1 forms?

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If you switch two $dx$'s, yes, you get a $-1$. And yes, $\alpha\wedge\alpha=0$ for all $1$-forms. – Mariano Suárez-Alvarez Jun 8 '12 at 15:44
If you write the result as $\sum_{i < j} \; c_{ij} \; dx^i \wedge dx^j,$ you get $c_{ij} = \alpha_i \beta_j - \alpha_j \beta_i.$ – Will Jagy Jun 8 '12 at 15:51

You get $$\alpha\wedge\beta = \sum_{1\le i<j \le n}(\alpha_i\beta_j-\beta_i\alpha_j)dx^i\wedge dx^j$$