Decomposition of the volume form on the sphere For which odd values of $n\in \mathbb{N}$, $n \neq 7$, does the volume form $\alpha$ on $S^n$ admit a wedge product decomposition $\alpha = \beta \wedge  \gamma$ such that neither $\beta$ nor $\gamma$ is a $1$-form?
For which even values of $n\in \mathbb{N}$ does the volume form $\alpha$ on $S^n$ admit a wedge product decomposition $\alpha = \beta \wedge \gamma$ such that neither $\beta$ nor $\gamma$ is a $n$-form?
 A: For all odd $n=2k{+}1>3$ one can write the volume form $\alpha$ on $S^n$ as a wedge product $\alpha=\beta\wedge\gamma$ with $\mathrm{deg}(\beta)$ and $\mathrm{deg}(\gamma)$ both greater than $1$.  Just note that there is a $1$-form $\theta$ such that $\alpha = \theta\wedge(\mathrm{d}\theta)^k$ where $n=2k{+}1$, and set $\beta =\theta\wedge(\mathrm{d}\theta)^{k-1}$ and $\gamma=\mathrm{d}\theta$.
Another construction in the odd case:  Any time one can write $TS^n=B\oplus C$, where $B$ and $C$ are vector bundles of ranks $p$ and $q$, respectively, one can write $\alpha =\beta\wedge\gamma$ where $\mathrm{deg}(\beta) = p$ and $\mathrm{deg}(\gamma)=q$. 
Certainly, for $n=6$, one can write $\alpha=\omega^3$ where $\omega$ is a $2$-form. Then $\beta = \omega$ and $\gamma=\omega^2$ works.  One can also write $\alpha=\Upsilon\wedge\Psi$, for certain $3$-forms $\Upsilon$ and $\Psi$.
Added comment:  In fact, if $p$ satisfies $1<p<n{-}1$, then there always exists a nonvanishing $p$-form $\beta$ on $S^n$ (because the rank of the $p$-form bundle on an $n$-manifold is greater than $n$).  Then $\beta \wedge \ast\beta$ is a positive $n$-form (where $\ast \beta$ is the Hodge dual of $\beta$).  By scaling $\beta$ by a positive function, we can assume that $\beta \wedge \ast\beta=\alpha$.  Thus, setting $\gamma=\ast\beta$, one sees that it's always possible for $n\ge4$, whether $n$ is even or odd.
