I am currently studying exterior calculus on manifolds and am not sure if I understand things correctly. In my textbook (R.W.R. Darling, Differential Forms and Connections) there is an example of a $2$-form on $A:=\mathbb{R}^3\setminus\{(0,0,0)\}$, namely
$$\omega = \frac{x(dy\wedge dz) – y(dx\wedge dz)+z(dx\wedge dy)}{(x^2+y^2+z^2)^{3/2}}.$$
Now restricting $\omega$ to the unit sphere $S^2$ gives a $2$-form $\eta\in\Omega^2S^2$ and some calculations show that $d\eta = 0$.
The example ends here, and I am wondering: Does this imply that also $d\omega = 0$? Otherwise I do not see why the author would start with a $2$-form on $A$ (also, I have a similar homework assignment which deals with a similar 2-dimensional case and where $d\omega = 0$ is to be verified on the whole set $\mathbb{R}^2\setminus\{(0,0)\}$). My ideas are the following:
The map $g \colon A \to S^2, g((x,y,z)) := \frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$ is smooth, and (according to the textbook) the pullback of $g$ is a linear map $g^* \colon \Omega^qS^2 \to \Omega^qA$ satisfying $g^*d\alpha = d(g^*\alpha)$ for all differential forms $\alpha$ (I believe this textbook notation of its domain and range is a bit sloppy; $g^*$ is meant to be a map assigning any differential form a differential form of one higher degree). Now if I understand the meaning of the pullback correctly, it holds $g^*\eta = \omega$, hence the above equation applied on the $2$-form $\eta$ implies that $0 = g^*0 = g^*d\eta = d(g^*\eta) = d\omega$.
Is this argumentation correct? Is there an easier way to prove this?
Best Answer
No, you've got it backwards. :)
Let $i : S^2 \hookrightarrow \Bbb R^3$ be the usual inclusion. If $\omega$ is a form on $\Bbb R^3$, then its restriction to $S^2$ is defined as $\eta = i^* \omega$ (the pull-back of $\omega$). Therefore, the fact that $\eta$ is closed does not mean anything relevant for $\omega$ because $0 = \Bbb d \eta = \Bbb d (i^* \omega) = i^* (\Bbb d \omega)$ and...? You can't get $\Bbb d \omega = 0$ from this.
What you are doing with your map $g$ is that instead of pulling $\omega$ back (as I have done above), you are pushing it forward. The problem is that the push-forward $g_*$ is defined only when $g$ is an injective immersion - clearly not the case here. So not only are you doing a push-forward (instead of a pull-back), but you are also doing it wrong! :)
What you denote $g^* \eta$ is not the restriction of $\omega$ to $S^2$, it's just some form on $\Bbb R^3$. In particular, it is not true that $g^* \eta = \omega$.