1-form on $S^n$ with non-degenerate differential.

Question

How it can be solved? Whether there is a $1$-differential form $w$ on $S^n$, such that $dw$ is non-degenerate on $T_aS^n$ for each $a \in S^n$?

Answer 1

The answer is no, there is no such $\omega$ for any $n$.

To see this, assume for a contradiction that such an $\omega$ exists. Then, since $d^2 \omega= 0$, $d\omega$ is closed, and it is non-degenerate by assumption. This implies $(S^n, d\omega)$ is a symplectic manifold.

Much is known about symplectic manifolds. For example, it immediately follows that $n$ is even. Further, the $n$-form $(d\omega)^\frac{n}{2}$ is known be a volume form. In particular, $[d\omega]^\frac{n}{2} \neq 0\in H_{\text{de Rham}}^n(S^n)$.

Of course, this is absurd, because, by definition of de Rham cohomology, $[d\omega] = 0 \in H_{\text{de Rham}}^2(S^n)$. This contradiction shows that no such $\omega$ exists.