I met such an exercise in my smooth-manifolds course, which made me very much puzzled. Let $U=\mathbb{R}^{n} \backslash\{0\}$ and $\omega=\sum_{i=1}^{n}(-1)^{i+1} f_{i} \mathrm{d} x^{1} \wedge \ldots \wedge \widehat{\mathrm{d} x^{i}} \wedge \ldots \wedge \mathrm{d} x^{n}$ be a given $(n-1)$-form on $U$, where $f_{i}: \mathbb{R}^{n} \rightarrow \mathbb{R} \mapsto \frac{x^{i}}{\|x\|^{n}}, i=1, \ldots, n.$ Now the exercise is to prove that this $\omega$ isn't an exact form on $U$.
I calculated directly: suppose $\eta=\sum_{1 \leq j<k \leq n} g_{j k} d x^{1} \wedge \ldots \wedge \widehat{d x^{j}} \wedge \ldots \wedge \widehat{d x^{k}} \wedge \ldots \wedge d x^{n}$ and $d\eta =\omega$, then the system of equations are
$\sum_{j=1}^{i-1}(-1)^{j+1} \frac{\partial g_{j, i}}{\partial x^{j}}+\sum_{k=i+1}^{n}(-1)^{k} \frac{\partial g_{i, k}}{\partial x^{k}}=(-1)^{i+1} f_{i}, i=1, \ldots, n.$
of course I can't solve this equation. I tried to find a contradiction, but I have no idea how to do this. I guess there might be some background or some trick. can anyone give me some hint? thx!
If $\omega$ was exact, so would be $u=\omega_{|S^{n-1}}$.
But one easily sees that $u$ is a volume form that maps any orthonormal basis $B$ of the tangent space of a given point $x$ to $\pm 1$, and $1$ if the basis $(x,B)$ is direct in $\mathbb{R}^n$.
It follows (using oriented charts) that every point $x \in S^{n-1}$ has a neighborhood $U$ such that if $f$ is compactly supported on $U$ and nonnegative and nonzero, $fu$ has positive integral. Using partitions of unity, it follows that $u$ has positive integral. Thus by Stokes, $u$ isn’t exact.