Is there a $1$-form $\lambda$ on $\mathbb R^5$ such that $(\lambda\wedge d\lambda\wedge d\lambda)_p\neq 0$ everywhere?

Question

Question:

Is there a $1$-form $\lambda$ on $\mathbb R^5$ such that $(\lambda\wedge d\lambda\wedge d\lambda)_p\neq 0$ for all $p\in \mathbb R^5$?

I tried to take $\lambda=\sum_{i=1}^5 \lambda_idx_i$, $d\lambda=\sum_{i=1}^5 (d\lambda_i\wedge dx_i)$, but after writing out the expression for $\lambda\wedge d\lambda\wedge d\lambda$, I don't have clue about how to solve $\lambda_1,\cdots,\lambda_5$. So I am wondering whether there is some intuitive example to answer the question or another way to address the problem. Any help is appreciated!

Thank you very much!

Answer 1

Try $\lambda = dx_5+x_1\,dx_2+x_3\,dx_4$.

The main example to keep in mind of a $2$-form $\omega$ on $\Bbb R^{2n}$ with $\underbrace{\omega\wedge\dots\wedge\omega}_{n\text{ times}}\ne 0$ is $\omega = dx_1\wedge dx_2+\dots+dx_{2n-1}\wedge dx_{2n}$. This should lead you to an obvious $\lambda$ on $\Bbb R^{2n-1}$.