I'd like to construct a closed but not exact $n-1$-form $\omega$ on $\mathbb{R}^n\setminus\{0\}$ in analogy to the winding form: $$\frac{x~dy-y~dx}{x^2+y^2}$$
I think something like $$\omega=\frac{\sum_{i=1}^n x_i(\star dx_i)}{(x_1^2+\dots+x_n^2)^{n/2}}$$ should probably work. $\star$ is the hodge star operator, for example $\star dx_1=dx_2\wedge\dots\wedge dx_n$ and $\star dx_2=-dx_1\wedge dx_3\wedge\dots\wedge dx_n$. $\omega$ is closed since $$d\omega=\sum_{i=1}^nd\left(\frac{ x_i}{(x_1^2+\dots+x_n^2)^{n/2}}\right)\wedge\star dx_i=\sum_{i=1}^n \frac{(\sum x_j^2)^{n/2}-n x_i^2 (\sum x_j^2)^{n/2-1}}{(\sum x_j^2)^{n}} dx_1\wedge\dots\wedge dx_n=0$$
Now it remains to show that $\omega$ is not exact. I guess a direct computation is not the way to do it. I thought about assuming that there exists $d\eta=\omega$ and then using Stoke's theorem to get the contradiction $0\not=\int_{S^{n-1}}\omega=\int_\emptyset \eta=0$. Two questions about this:
- Is the general argument correct?
- How to prove $0\not=\int_{S^{n-1}}\omega$? I guess a direct computation using generalized sphere coordinates is rather tedious.
Best Answer
Yes, this is a reasonable way to go about things.
Let $\psi=\sum_{i=1}^n x_i (\star dx_i)$. Then $\left.\psi\right|_{S^{n-1}}=\left.\omega\right|_{S^{n-1}}$, and so $\int_{S^{n-1}} \omega = \int_{S^{n-1}} \psi$. Moreover, $d\psi$ is a constant multiple of the volume form on $\Bbb{R}^n$. Now, apply Stokes' theorem to $B^n$ and $S^{n-1}$...