# Integration of differential form inclusion of $S^2$

Take this example. We have the natural inclusion $$i : S^2 \rightarrow R^3$$ and the differential form: $$\omega = x dy \wedge dx + y dz \wedge dx + z dx \wedge dy$$. How can we say that that the pullback of the inclusion $$i^*\omega = cos(\theta)d\phi \wedge d\theta$$?

That is an expression on local coordinates!

I guess I am in general confused about when we are "allowed" to use coordinate expressions for a coordinate independent object.

We defined the Integral of a differential form $$\omega: TM \rightarrow R$$ over a measurable subset $$A \subseteq M$$ with charts $$(U_\alpha,x_{\alpha})$$

via partition of Unity $$\{ \phi_\alpha \} _{\alpha \in I}$$ s.t. $$\omega= \sum_{\alpha \in I} \phi_{\alpha} \omega$$ in the following way: $$\int_A \omega = \sum_{\alpha \in I} \int_{x_{\alpha}(U_{\alpha} \cap A)} f_{\alpha} dm$$ where $$f_\alpha$$ is the unique function s.t. $$\phi_{\alpha} \omega = x_\alpha ^*(f_\alpha dx^1\wedge ... \wedge dx^n)$$

I think that in the above example $$\int_{S^2} i^* \omega = \int_{x(S^2)}cos(\theta)d\phi \wedge d\theta = \int_{(0,2\pi) \times (-\frac{\pi}{2}, \frac{\pi}{2})}cos(\theta)d\phi d\theta = 4 \pi$$

But how can I rigorously say that? Or just do it?...

EDIT I bascially want to know how to show the first equality in the last line? (I know how to compute it)

• The last equality you wrote is true since $(\phi,\theta)$ is a parametrization of the sphere on a subset of full measure. The remainder is of measure zero and does not contribute to the integral. Commented Dec 2, 2021 at 9:36
• But how do I get the $cos(\theta) d\phi \wedge d\theta$ in the integral? I know how to compute it, but I think I can't just say that $i^* \omega = cos(\theta )d\phi \wedge d\theta$. But then how do I show the first equality in the last line? Commented Dec 2, 2021 at 9:39
• Basically the same way as in the link you provided: the change of coordinates on $\Bbb R^3$ given by the polar coordinates $(r,\phi,\theta)$ allows on to write $\omega$ in the new coordinates system as $r^3 \cos \theta d\phi\wedge d\theta$. Now, pull-back this on the unit sphere $r=1$. Commented Dec 2, 2021 at 9:42
• But that's exactly my confusion! This is a coordinate expression and thus not actually $i^* \omega$, but maybe $x_*(i^* \omega)$ ? But then again, just claim that's true or can I how to argue that this is true? Commented Dec 2, 2021 at 9:45
• I think I don't get what you don't get. You don't have to take any chart, just take the one given by your favourite coordinates, here the spherical coordinates. Commented Dec 2, 2021 at 10:47

Consider $$p\colon (0,2\pi) \times (-\frac{\pi}{2},\frac{\pi}{2}) \to \Bbb S^2$$ the parametrization given by $$p(\theta,\phi) = (\cos\theta\cos\phi,\sin\theta\cos\phi,\sin\phi)$$. First, note that $$p^*(i^*\omega)= (i\circ p)^* \omega$$, and therefore, \begin{align} p^*(i^*\omega) &= p^*i^*(x dy\wedge dy + y dz\wedge dx + z dx \wedge dy)\\ &= (x(i\circ p)) (i\circ p)^* (dy\wedge dz) + (y(i\circ p)) (i\circ p)^*(dz \wedge dx) + (z(i\circ p)) (i\circ p)^* (dw\wedge dy) \end{align} Now, use the fact that the pullback and the wedge product commute; so that \begin{align} (i\circ p)^* (dy\wedge dz) &= ((i\circ p)^*dy)\wedge((i\circ p)^* dz)\\ (i\circ p)^* (dz\wedge dx) &= ((i\circ p)^*dz)\wedge((i\circ p)^* dx)\\ (i\circ p)^* (dx\wedge dz) &= ((i\circ p)^*dx)\wedge((i\circ p)^* dy) \end{align} Note that \begin{align} (x(i\circ p)) &= x\circ i \circ p = \cos\theta\cos\phi\\ (y(i\circ p)) &= y\circ i \circ p = \sin\theta\cos\phi\\ (z(i\circ p)) &= z\circ i \circ p = \sin\phi \end{align} and now, use the chain rule in order to show that \begin{align} (i\circ p)^*dx &= dx \circ d(i\circ p) = d (x\circ i \circ p) = d(\cos\theta\cos\phi)\\ (i\circ p)^*dy &= dy \circ d(i\circ p) = d (y\circ i \circ p) = d(\sin\theta\cos \phi)\\ (i\circ p)^*dz &= dz \circ d(i\circ p) = d (z\circ i \circ p) = d(\sin\phi) \end{align} Expand these equalities, e.g $$d(\cos\theta \cos \phi) = -\sin\theta \cos \phi d\theta - \cos \theta \sin \phi d\phi$$. Gluing these equalities all together and using the fact that $$\cos^2 + \sin^2 = 1$$, you should find $$p^*(i^*\omega) = \cos\phi d\theta \wedge d\phi$$ Since the complementary of $$Im(p)$$ in $$\Bbb S^2$$ has measure zero, and since $$p$$ is a diffeomorphism onto its image, it follows that \begin{align} \int_{\Bbb S^2} i^* \omega &= \int_{Im(p)} i^*\omega \\ &= \int_{(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})}p^*(i^* \omega)\\ &= \int_{(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})} \cos \phi d\theta\wedge d\phi\\ &:= \int_0^{2\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \phi d\theta d\phi \end{align} The last equality being true by definition of the integral of the top form $$\cos\phi d\theta \wedge d\phi$$ in the oriented manifold $$(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})$$.