# Computing the restriction of a differential form

Define $\omega$ on $\mathbb{R}^3$ by $\omega = x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy$.

Thus far I have computed $\omega$ in spherical coordinates $(\rho,\phi,\theta)$, as well as computed $d\omega$ in both Cartesian and spherical coordinates. I found $\omega=\rho^3\sin\phi\,d\phi\wedge d\theta$.

But now I'm asked to compute the restriction $\omega|_{S^2} = \iota^*\omega$, where $\iota:S^2\to\mathbb{R}^3$ is the inclusion map, using coordinates $(\phi,\theta)$ on the open subset where they are defined.

So far, this is all I have:

Fix $p\in S^2$ and consider the basis $\left(\frac{\partial}{\partial\phi},\frac{\partial}{\partial\theta}\right)$ on $T_pS^2$. Then

$$(\iota^*\omega)_{(\phi,\theta)}\left(\frac{\partial}{\partial\phi},\frac{\partial}{\partial\theta}\right)=\omega_{\iota(\phi,\theta)}\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)=\sin\phi\,d\phi\wedge d\theta\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)\,.$$ And I suppose I also know that $-\frac{\pi}{2}<\phi<\frac{\pi}{2}$ and $0<\theta<2\pi$.

I've done so few examples, though, that it's unclear to me where to go from here.

Any help is appreciated

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## 1 Answer

You need to write down the inclusion map $\iota: S^2\rightarrow\mathbb{R}^3$ which is given by $$\iota(\phi,\theta)=(\cos\phi\cos\theta,\cos\phi\sin\theta,\sin\phi).$$ Therefore, the differential of $\iota$ is given by $$d\iota=\left[ \begin{array}{cc} -\sin\phi\cos\theta & -\cos\phi\sin\theta \\ -\sin\phi\sin\theta & \cos\phi\cos\theta \\ \cos\phi & 0 \\ \end{array} \right],$$ which implies that $$\iota_*\left(\frac{\partial}{\partial\phi}\right)=-\sin\phi\cos\theta\frac{\partial}{\partial x}-\sin\phi\sin\theta\frac{\partial}{\partial y}+\cos\phi\frac{\partial}{\partial z},$$ $$\iota_*\left(\frac{\partial}{\partial\theta}\right)=-\cos\phi\sin\theta\frac{\partial}{\partial x}+\cos\phi\cos\theta\frac{\partial}{\partial y}.$$ Therefore, $$dy\wedge dz\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)=-\cos^2\phi\cos\theta$$ because $$dy\wedge dz\left(X,Y\right)=\left[ \begin{array}{cc} dy(X) & dz(X) \\ dy(Y) & dz(Y) \\ \end{array} \right].$$ Similarly, we have $$dz\wedge dx\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)=-\cos^2\phi\sin\theta, dx\wedge dy\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)=-\sin\phi\cos\phi.$$ Hence, $$(\iota^*\omega)_{(\phi,\theta)}\left(\frac{\partial}{\partial\phi},\frac{\partial}{\partial\theta}\right)=\omega_{\iota(\phi,\theta)}\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)$$ $$=\cos\phi\cos\theta dz\wedge dy\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)+\cos\phi\sin\theta dz\wedge dx\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)$$ $$+\sin\phi dx\wedge dy\left(\iota_*\left(\frac{\partial}{\partial\phi}\right),\iota_*\left(\frac{\partial}{\partial\theta}\right)\right)$$ $$=-\cos^3\phi\cos^2\theta-\cos^3\phi\sin^2\theta-\sin^2\phi\cos\phi=-\cos\phi.$$

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This is wonderful; thank you for your help! –  Bey Dec 2 '11 at 19:00
Sorry for the necropost; I just saw this question--nicely done, BTW: what would happen if we used a different choice of parametrization for S^2? Does the "form of the form w" depend on the choice of parametrization of S^2? –  user99680 Nov 15 '13 at 3:27