How to compute the Wedge product in $\bigwedge ^3$ as well as $\bigwedge ^2$

Question

I want to know on how to compute the Wedge Product, since this is the first time I am doing it. I wanted to compute:

$$e_1 \wedge(e_2 - \frac{1}{2}e_4) \wedge (-2 e_1 +e_4) $$

Though currently I have no idea how to do it, so I tried my calculation onto the inner term:

$$(e_2 - \frac{1}{2}e_4) \wedge (-2 e_1 +e4)$$

from which I got:

$$-2e_2 \wedge e_1 + e_2\wedge e_4 + e_4\wedge e_1 $$

and the term $ e_4\wedge e_4$ disappears due to the wedge with itself is zero due to the alternating property.

I just want a runthrough on how to do this really, on how to shorten terms as well for continuous computation.

Answer 1

I'd reorder the wedges to evaluate, moving the $e_1$ wedge to the end (this flips the sign twice):

$$\begin{aligned}e_1 \wedge(e_2 - \frac{1}{2}e_4) \wedge (-2 e_1 +e_4) &=(-1)^2 (e_2 - \frac{1}{2}e_4) \wedge (-2 e_1 +e_4) \wedge e_1 \\ &= (e_2 - \frac{1}{2}e_4) \wedge \left( { (-2 e_1 +e_4) \wedge e_1 } \right) \\ &= (e_2 - \frac{1}{2}e_4) \wedge \left( { e_4 \wedge e_1 } \right) \\ &= e_2 \wedge e_4 \wedge e_1.\end{aligned}$$

The term with $e_4$ wedged with itself can be dropped because $e_4 \wedge e_4 \wedge e_1 = (e_4 \wedge e_4) \wedge e_1 = 0 \wedge e_1 = 0.$