nowhere zero forms on Spheres I came across this question about the existence of a no-where zero $1$-form on $S^2$. I argued as follows. 
I know $S^2$ is a symplectic manifold with symplectic $2$-form $\omega$ (say). So we have a bijection from the space of vector fields to $1$-forms given by the interior product
$\mathfrak X(S^2)\longrightarrow \Omega^1(S^2) \\ 
X\longmapsto\iota_X\omega$
But the Hairy Ball Theorem says $\exists p\in S^2$ such that
$X_p=0 \\ 
\implies (\iota_X\omega)_p(v)=w_p(X_p,v)=0 \ \forall \ v\in T_pS^2$
and hence $\iota_X\omega$ is NOT nowhere zero. 
My question is what if we have a similar question for higher dimensional spheres. We know $S^{2n}$ is not symplectic if $n>1$ as $H^2_{dR}(S^{2n})=0$ so my approach as above clearly fails. In general what can we say about the existence of lower dimensional nowhere zero forms because a nowhere zero top form exists iff the manifold is orientable. 

If $g$ is a Riemannian metric on a smooth manifold $M$, there is an isomorphism $\Phi_g : \mathfrak{X}(M) \to \Omega^1(M)$ given by $X \mapsto g(X, \cdot)$. In particular, for any one-form $\alpha$, there is a vector field $X_{\alpha}$ such that $\alpha = g(X_{\alpha}, \cdot)$. It follows that $\alpha_p = 0$ if and only if $(X_{\alpha})_p = 0$, so $M$ admits a nowhere-zero one-form if and only if $M$ admits a nowhere-zero vector field. By the Poincaré-Hopf Theorem, if a closed connected orientable manifold $M$ admits a nowhere-zero vector field, then $\chi(M) = 0$; the converse is also true, see here. Therefore $S^n$ admit a nowhere-zero one-form if and only if $n$ is odd. Viewing $S^{2m-1}$ as the unit sphere in $\mathbb{C}^m$ with coordinates $(x^1, y^1, \dots, x^m, y^m)$, an example of such a form is the restriction of $-y^1dx^1 + x^1dy^1 + \dots - y^mdx^m + x^mdy^m$ to $S^{2m-1}$. 
For other forms, note that if $\alpha$ is nowhere-zero one-form on a closed smooth oriented $n$-dimensional manifold $M$, equipped with a Riemannian metric $g$, then $\ast\alpha$ is a nowhere-zero $(n-1)$-form. For intermediate forms, note that $\operatorname{rank}(\bigwedge^kT^*M) = \binom{n}{k} > n = \dim M$, so it follows from obstruction theory that $\bigwedge^kT^*M$ admits a nowhere-zero section, and hence $M$ admits a nowhere-zero $k$-form. In particular, every sphere $S^n$ admits nowhere-zero $k$-forms for all $k = 2, \dots, n - 2$, as well as $k = 0$ and $n$.
