Let $\omega$ be the $(n-1)-$form on $\Bbb{R}^n \setminus \{0\}$ defined by
$$\omega = \vert \vert x \vert \vert^{-n} \sum_{j=1}^n(-1)^{j-1}x^j \, dx^1 \wedge … \wedge \, \hat{dx^j} \wedge … \wedge \, dx^n.$$
Where $\hat{}$ denotes omitting the component. Prove $\omega$ is closed but not exact. So for closed I would just need to show $d \omega = 0$, right? And how would I even GO about finding the exterior derivative of $\omega$? And for not being exact, do I suppose
$$\omega = d \eta$$
for some $(n-2)-$form $\eta$ and deduce a contradiction? Thanks for any tips or hints. Been staring at this one for quite some time.
Attempt
To show $\omega$ is not exact, suppose there exists a differential form $\eta$ such that
$$\omega = d \eta.$$
Then by Stokes' Theorem, on one hand one has
$$\int_{S^{n-1}} d \eta = \int_{\partial S^{n-1}} \eta = \int_{\emptyset} \eta = 0.$$
I fail to see why
$$\int_{S^{n-1}} \omega \neq 0.$$
I found a solution here but dont really understand it. Also, I fail to see why $d \omega = 0$. I am having trouble computing the exterior derivative of this sum with the norm on the outside. Also, the solution posted currently doesnt have the $(-1)^{j-1}$ term.
Best Answer
Part I:
Let's show $d\omega = 0$. Rewrite: $$\omega = \sum_{j=1}^n(-1)^{j-1}\bigg(\frac{x^j}{||x||^n}\bigg) dx^1 \wedge ... \wedge \hat{dx^j} \wedge ... \wedge dx^n.$$
Then since $d_k:\Omega^{k}(M)\to \Omega^{k+1}(M)$ is linear, we get: $$d\omega = \sum\limits_{j=1}^n(-1)^{j-1}d\bigg[\bigg(\frac{x^j}{||x||^n}\bigg) dx^1 \wedge ... \wedge \hat{dx^j} \wedge ... \wedge dx^n\bigg]$$ $$:= \sum\limits_{j=1}^n(-1)^{j-1}d\bigg(\frac{x^j}{||x||^n}\bigg)\wedge \big[dx^1 \wedge ... \wedge \hat{dx^j} \wedge ... \wedge dx^n\big] \text{ }\text{ }\text{ }(\star)$$
The 2-norm, $||x|| := \sqrt{(x^1)^2+...+(x^n)^2}$, which implies $\partial_i||x|| = \large \frac{x^i}{||x||}$.
As well, $df:= \sum\limits_{i=1}^n\partial_ifdx^i$ together with the product rule, power rule, and chain rule gives: $$d\bigg(\frac{x^j}{||x||^n}\bigg) = \sum\limits_{i=1}^n\bigg[\frac{\delta^j_i}{||x||^n} +x^j\cdot\bigg(\frac{-n}{||x||^{n+1}}\cdot \frac{x^i}{||x||}\bigg)\bigg]dx^i$$ $$=\bigg[\frac{||x||^2-n(x^j)^2}{||x||^{n+2}}\bigg]dx^j+\sum\limits_{i\neq j}^n\bigg[\frac{-nx^ix^j}{||x||^{n+2}}\bigg]dx^i \text{ }\text{ }\text{ }(\star\star)$$
Noting that $dx^i\wedge dx^i = 0$ for any $i$, substituting $(\star\star)$ into $(\star)$ results in: $$d\omega = \sum\limits_{j=1}^n(-1)^{j-1}\bigg[\frac{||x||^2-n(x^j)^2}{||x||^{n+2}}\bigg]dx^j\wedge \big[dx^1 \wedge ... \wedge\hat{dx^j} \wedge ... \wedge dx^n\big]$$ $$= \sum\limits_{j=1}^n\bigg[\frac{||x||^2-n(x^j)^2}{||x||^{n+2}}\bigg]dx^1\wedge ... \wedge dx^n$$ $$=\frac{1}{||x||^{n+2}}\bigg[n||x||^2-n\sum\limits_{j=1}^n(x^j)^2\bigg]dx^1\wedge ... \wedge dx^n = 0.$$ This shows $\omega$ is closed, as desired. $\square$
Part II:
We want to show $\omega$ is not exact. Applying Stokes' Theorem over the filled hypercube (instead of $\mathbb{S}^{n-1}$) gives the contradiction. We don't need the empty boundary, we just need nonzero integral.
Suppose $\omega$ is exact, then $\exists \eta \in \Omega^{n-2}\big(\mathbb{R}^n\backslash \{0\}\big)$ such that $d\eta = \omega$. Naively: $$0 = \int_Cd(d\eta) = \int_Cd\omega = \int_{\partial C}\omega.$$
Define the face parameterizations in our punctured space as: $$F_{\pm x^i}:(-1,1)^{n-1}\to \mathbb{R}^n\backslash \{0\}$$ $$(s^1,...,s^{n-1})\mapsto (s^1,...,\underbrace{\pm 1}_{i^{th}\text{-coord}},...,s^{n-1}).$$ Then since $d(x^i\circ F_{\pm x^i}) = 0$, the only surviving term in the pullbacks are the ones with $j=i$. So: $$\color{darkblue}{F^*_{\pm x^i}(\omega) = (-1)^{i-1}\bigg(\frac{\pm 1}{(||s||^2+1)^{n/2}}\bigg)ds^1\wedge ... \wedge \widehat{0} \wedge ...\wedge ds^{n-1}}.$$
Lastly, we have: $$\int_{\partial C}\omega = \sum\limits_{\pm x^i}\int_{(-1,1)^{n-1}} F^*_{\pm x^i}(\omega) = 2*\sum\limits_{+x^i}\int_{(-1,1)^{n-1}} F^*_{+x^i}(\omega)$$ because the negative faces reverse orientation (choose outward normal on $\partial C$ and standard on the domain, then look at the differential at a point). Looking at the dark blue expression, we see that the RHS is nonzero, giving us the contradiction. So $\omega$ is not exact. $[\text{ }]$