With p-forms, is $2\alpha \wedge \alpha = 0$?

Question

I am having troubles computing the following problem.

For, the 1-form and 2-form

$$\eta = x^1 \mathrm{d}x^2-x^2\mathrm{d}x^1+x^3\mathrm{d}x^4-x^4\mathrm{d}x^3$$ $$\alpha = \mathrm{d}x^1 \wedge\mathrm{d}x^2+ \mathrm{d}x^3 \wedge \mathrm{d}x^4$$

I need to compute $\mathrm{d}\eta \wedge \alpha$

$\mathrm{d} \eta = \mathrm{d} (x^1 \mathrm{d}x^2-x^2\mathrm{d}x^1+x^3\mathrm{d}x^4-x^4\mathrm{d}x^3)$ $ = \mathrm{d} x^1 \wedge \mathrm{d}x^2-\mathrm{d}x^2\wedge \mathrm{d}x^1+\mathrm{d}x^3\wedge \mathrm{d}x^4-\mathrm{d}x^4\wedge \mathrm{d}x^3$ $= 2(\mathrm{d}x^1\wedge \mathrm{d}x^2 + \mathrm{d}x^3 \wedge \mathrm{d}x^4)$

Therefore,

$\mathrm{d}\eta \wedge \alpha = 2(\mathrm{d}x^1\wedge \mathrm{d}x^2 + \mathrm{d}x^3 \wedge \mathrm{d}x^4) \wedge (\mathrm{d}x^1 \wedge\mathrm{d}x^2+ \mathrm{d}x^3 \wedge \mathrm{d}x^4)$

Now, my understanding is that this should be $0$ since that is the same as $2\alpha \wedge \alpha$ but if I expand it I get

$= 2(\mathrm{d}x^1\wedge\mathrm{d}x^2)\wedge(\mathrm{d}x^1\wedge\mathrm{d}x^2) +2(\mathrm{d}x^1 \wedge \mathrm{d}x^2)\wedge (\mathrm{d}x^3 \wedge \mathrm{d}x^4) + 2(\mathrm{d}x^1 \wedge \mathrm{d}x^2) \wedge (\mathrm{d}x^1 \wedge \mathrm{d}x^2) + 2(\mathrm{d}x^3 \wedge \mathrm{d}x^4) \wedge (\mathrm{d}x^3 \wedge \mathrm{d}x^4)$

$= 0 + 2(\mathrm{d}x^1 \wedge \mathrm{d}x^2)\wedge (\mathrm{d}x^3 \wedge \mathrm{d}x^4) + 0 + 2(\mathrm{d}x^3 \wedge \mathrm{d}x^4) \wedge (\mathrm{d}x^1 \wedge \mathrm{d}x^2)$

$= 2(\mathrm{d}x^1 \wedge \mathrm{d}x^2\wedge \mathrm{d}x^3 \wedge \mathrm{d}x^4) + 2(\mathrm{d}x^3 \wedge \mathrm{d}x^4 \wedge \mathrm{d}x^1 \wedge \mathrm{d}x^2)$

$= 2(\mathrm{d}x^1 \wedge \mathrm{d}x^2\wedge \mathrm{d}x^3 \wedge \mathrm{d}x^4) + (-1)^22(\mathrm{d}x^1 \wedge \mathrm{d}x^2 \wedge \mathrm{d}x^3 \wedge \mathrm{d}x^4)$ (Since I switched places twice, pulling out 2 negative signs)

$= 4(\mathrm{d}x^1 \wedge \mathrm{d}x^2\wedge \mathrm{d}x^3 \wedge \mathrm{d}x^4) = 4\alpha \neq 0$

Answer 1

In the algebra of real differential forms we have the commutation rule $$\beta\wedge\alpha=(-1)^{mn}\alpha\wedge\beta$$ when $\alpha$ is an $m$-form and $\beta$ an $n$-form. Therefore $$\alpha\wedge\alpha=(-1)^{m^2}\alpha\wedge\alpha.$$ When $m$ is odd, this says $\alpha\wedge\alpha=-{\alpha\wedge\alpha}$, that is $\alpha\wedge\alpha=0$. But when $m$ is even, it says $\alpha\wedge\alpha=\alpha\wedge\alpha$ which is not terribly exciting.

We find for $\alpha=dx_1\wedge dx_2+dx_3\wedge dx_4$ that $$\alpha\wedge\alpha=2dx_1\wedge dx_2\wedge dx_3\wedge dx_4\ne0.$$ In general $\alpha\wedge\alpha$ can be nonzero for $\alpha$ of even degree.