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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)

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    $\begingroup$ 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. $\endgroup$
    – Didier
    Commented Dec 2, 2021 at 9:36
  • $\begingroup$ 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? $\endgroup$
    – Thomas
    Commented Dec 2, 2021 at 9:39
  • $\begingroup$ 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 $r=1$. $\endgroup$
    – Didier
    Commented Dec 2, 2021 at 9:42
  • $\begingroup$ 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? $\endgroup$
    – Thomas
    Commented Dec 2, 2021 at 9:45
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    $\begingroup$ 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. $\endgroup$
    – Didier
    Commented Dec 2, 2021 at 10:47

1 Answer 1

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Here is a proof.

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 $(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})$.

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  • $\begingroup$ Thank you so much! That made it clear! $\endgroup$
    – Thomas
    Commented Dec 10, 2021 at 18:52

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