# Computing wedge products

Compute $\omega = (e_1^* + e_2^* + \cdots+ e_n^*) \wedge (e_1^* + e_2^*) \wedge (e_1^* + e_3^*) \wedge \cdots \wedge(e_1^* + e_n^*)$ in the standard form.

I first thuoght I'd pick a value from the first bracket and wedge it with a value in the next bracket (one I don't already have) and keep doing this and when I do this, I end up getting:

$$(e_1^* \wedge e_2^* \wedge \cdots \wedge e_n^*) + (e_2^* \wedge e_1^* \wedge e_3^* \wedge \cdots \wedge e_n^*) + (e_3^* \wedge e_2^* \wedge e_1^* \wedge e_4^* \wedge \cdots \wedge e_n^*) + \cdots + (e_n^* \wedge e_{n-1}^* \wedge e_{n-2}^* \wedge \cdots \wedge e_1^*),$$

but I don't know how to simplify this. But then I realised, I've been wedging the same value more than once. So for example, if I pick $e_1^*$ in the first bracket, the I haved wedged it in the second bracket with $e_2^*$, but then when I pick $e_3^*$, I have wedged that with $e_2^*$ again. So how would I compute this wedge product?

• You should use the fact that if $\sigma \in S_{n}$ is a permutation then $x_{\sigma(1)} \wedge x_{\sigma(2)} \wedge \cdots \wedge x_{\sigma(n)} = \text{sign}(\sigma )(x_{1} \wedge x_{2} \wedge \cdots \wedge x_{n})$. – tharris Apr 17 '13 at 21:48
• It is an $n$ form so all you have to do is find the right coefficient. – guaraqe Apr 17 '13 at 21:49

I agree with your calculation except for the last term, which I would have said is $$e_n^*\wedge e^*_2\wedge\cdots\wedge e^*_{n-1}\wedge e_1^*.$$ Note that each term except the first one in your calculation differs from $e_1^*\wedge\cdots\wedge e_n^*$ by a single transposition, so you get $$e_1^*\wedge\cdots\wedge e_n^*-e_1^*\wedge\cdots\wedge e_n^*-\cdots-e_1^*\wedge\cdots\wedge e_n^*=??$$
Make use of the following obvious identity $$e_1^*+e_2^*+\cdots+e_n^*=(e_1^*+e_2^*)+(e_1^*+e_3^*)+\dots+(e_1^*+e_n^*)-(n-2)e_1^*.$$
If you substitute the right-hand side into $\omega$, almost all terms in the first bracket will cancel. The only one left is the last one, $-(n-2)e_1^*$. So,
$$\omega=-(n-2)e_1^*\wedge(e_1^*+e_2^*)\wedge(e_1^*+e_3^*)\wedge\cdots\wedge(e_1^*+e_n^*)$$ and $$\omega=-(n-2)e_1^*\wedge e_2^*\wedge e_3^*\wedge\cdots\wedge e_n^*.$$