Is $\omega_n$ exact in $\mathbb R^n -\{0 \}$?

Question

For $n \ge 2$ consider the differential form $\omega_n=r^{-n} \sum_{i=1}^n(-1)^{i-1}x_idx_1 \wedge \ldots \wedge dx_{i-1} \wedge dx_{i+1} \wedge \ldots \wedge dx_n$, defined on $\mathbb R^n \setminus \{ \bf0\}$. (where, $r=\sqrt{x_1^2+\ldots + x_n^2}$). Is it exact there?

I think the answer is no, and I've verified my answer for $n=2,3$ by integrating $\omega_n$ over the unit $(n-1)$-sphere $S^{n-1}$ to a nonzero result. I guess it's true for all values of $n$, but the calculations become very cumbersome (many $(n-1) \times (n-1)$ Jacobians to find...). I also need to prove that the boundaries are empty.

Is there a more sophisticated way of proving that $\omega_n$ is not exact? If not, can anyone help my with the calculations?

Answer 1

Hint: It's enough to show that $i^\ast \omega_n$ is never zero, where $i:S^{n-1}\rightarrow \mathbb{R}^n$ is the inclusion.

The inclusion is nothing but the function which takes a point in $S^{n-1}$ and thinks about it in $\mathbb{R}^n$. The $i^\ast$ thing is called the pullback, but it's basically the restriction map: $\omega_n|_{S^n}$. (In particular, it can only be applied to vectors which are tangent to $S^{n-1}$, but other than that, works exactly as $\omega_n$ does.)

The key fact is the following.

Proposition: If $\omega$ is an $n$-form on $S^n$ and $\omega(p)\neq 0$ for all $p\in S^n$, then $\omega$ cannot be exact.

Believing this proposition, note that if $\omega_n$ is exact, then $\omega_n = d\alpha$, so $i^\ast \omega_n = i^\ast d\alpha = d(i^\ast \alpha)$, so $i^\ast \omega_n$ is also exact, contradiciting the proposition.

Here's how I'd prove that $\omega$ restricted to $S^{n-1}$ is nonzero. Big picture: Show that $\omega$ is nonzero at a particular point and show that it's invariant under a bunch of transformations which, taken as a whole, can move any point to the particular point.

Our particular point will be $p = (1,0,\ldots,0)$. Then if $v_i$ is the tangent vector in the direction of $x_i$, clearly each $v_i$ for $i=2,...,n$ is tangent to the sphere at $p$ and further, $\omega(v_2,\ldots, v_n) = 1\neq 0$.

We now have a lemma. I'll use the notation $dx_{I_i}$ to mean $dx_1 \wedge \ldots \wedge dx_{i-1} \wedge dx_{i+1} \wedge \ldots \wedge dx_n$.

Lemma: For any $\theta$, and any $i\neq 1,2$, \begin{align*} x_i dx_{I_i} = &x_i d(\cos\theta x_1 + \sin \theta x_2)\wedge d(-\sin \theta x_1 + \cos\theta x_2)\wedge dx_3 \wedge \ldots \\ &\wedge dx_{i-1}\wedge dx_{i+1} \wedge \ldots \wedge dx_n \end{align*} and \begin{align*}x_1 dx_{I_1} -x_2 dx_{I_2} = & (\cos\theta x_1 + \sin \theta x_2) d(-\sin \theta x_1 + \cos \theta x_2) \wedge dx_3 \wedge \ldots \wedge dx_n \\ &-(-\sin \theta x_1 + \cos\theta x_2) d(\cos \theta x_1 + \sin \theta x_2) \wedge dx_3 \wedge \ldots \wedge dx_n\end{align*}

Proof: Compute.

Those transformations are nothing but rotations in the $x_1 x_2$ plane. More generally, one can see that any rotation in adjacent coordinates preserves $\omega_n$. Since $\omega_n = \sum x_i dx_{I_i}$, this implies that $\omega_n$ is invariant under these rotations.

Now, given any point $q\in S^{n-1}$ one can use a rotation in the $x_{n-1}x_n$ plane to transform $q$ into one of the form $(x_1,...,x_{n-1}, 0)$. Continuing with a rotation in the $x_{n-1}x_{n-2}$ plane, we may assume $q = (x_1,..., x_{n-2}, 0, 0)$. Continuing in this fashion, we may assume $q = (1,0,0,\ldots ,0)$, where our direct computation above shows $\omega$ is nonzero.

Answer 2

This exact (no pun intended) question was answered by Micah here. His observation was that on the unit sphere $\omega_n $ is the same as $r^n \omega_n$, and since the latter does not have a singularity at the origin, Stokes's theorem can be used: $$ \int_{S} \omega_n=\int_{S}r^n \omega_n = \int_B d(r^n\omega_n) = n \int_B dx_1\wedge\dots \wedge dx_n $$