Let $\omega=x\,dy\wedge dz +y\,dz\wedge dx+z\,dx\wedge dy$ or in spherical coordinates (unless I had made some mistake) $\omega=r^3\cos \theta\, d\phi\wedge d\theta$. Now I want to find $i^*\omega$ where $i:S\to\mathbb{R}^3$ is inclusion of unit sphere, using $\phi$ and $\theta$. It seems quite easy but I'm not sure how to interprete $i$ and how to use it in $i^* \omega$. Any suggestions?
Differential Geometry – Inclusion and Pullback of Differential Form
differential-formsdifferential-geometryspherical coordinates
Related Solutions
Indeed, $d\theta$ does not shrink near the poles. In fact, it blows up in such a say that $\sin\phi d\phi\wedge d\theta$ is finite and non-zero at the poles.
One way to see this is to convert $d\theta$ into a linear combination of $dx,dy$, and $dz$. Since $\theta = \arctan(y/x)$, it is easy to compute that $$d\theta =\frac{1}{x^2 + y^2}\left( -y dx + x dy\right).$$
Considering $\{dx, dy, dz\}$ to be an orthornomal basis for the cotangent space at each point of $\mathbb{R}^3$, it follows that $|d\theta|\rightarrow \infty$ as $(x,y)\rightarrow (0,0)$.
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 $(0,2\pi)\times (-\frac{\pi}{2},\frac{\pi}{2})$.
Best Answer
As $S$ is the unit sphere in $\mathbb{R}^3$, $S$ is given by the equation $r = 1$ in spherical coordinates. Therefore,
$$i^*\omega = i^*(r^3\cos\theta d\phi\wedge d\theta) = 1^3\cos\theta d\phi\wedge d\theta = \cos\theta d\phi\wedge d\theta.$$