A question on wedge product of differential forms

Question

Let $\omega$ be a $k$-form, is it true that $$\omega\wedge\omega=0?$$ is it true that $$d\omega\wedge d\omega=0?$$

Answer 1

Either $\omega\wedge \omega=0$ or $d\omega\wedge d\omega=0$, because firstly, the exterior product is graded anticommutative: $\alpha\wedge\beta=(-1)^{|\alpha||\beta|}\beta\wedge\alpha$, where $|\alpha|$ denotes the degree of $\alpha$. It is 0 for odd-degree forms. And secondly, the exterior derivative increments the degree.

Answer 2

No. For example, over $\mathbb{R}^4$, the $1$-form $\omega = x\; dy + z\; dw$ has $\omega\wedge \omega = 0$ (as any $1$-form has by (graded)-commutativity), but $d\omega = dx\wedge dy + dz \wedge dw$ has $$d\omega \wedge d\omega = 2\; dx\wedge dy\wedge dz\wedge dw\not =0.$$