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We define a 2-form $\Omega$ on $\mathbb{R^3}$ by $\Omega=x\;dy \wedge dz+y\;dz\wedge dx+z\;dx \wedge dy$. How can I show that $\Omega|_\mathbb{S^2}$ is nowhere zero? Before proving that how can I compute the restriction $\Omega|_\mathbb{S^2}$?

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  • $\begingroup$ Do you want to show that the restriction is nowhere zero, or that the pullback to $S^2$ is nowhere zero? The first is easy (see Alexander's answer below). The second is slightly more involved. $\endgroup$
    – Aaron
    Commented Dec 30, 2011 at 13:45
  • $\begingroup$ @Aaron I want to show that the restriction is nowhere zero. Thanks. $\endgroup$
    – user20353
    Commented Dec 30, 2011 at 13:56
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    $\begingroup$ @Aaron: restricting a form to $S^2$ is the same thing as pulling it back. $\endgroup$ Commented Dec 30, 2011 at 13:56
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    $\begingroup$ @Eric: It depends on your definitions: differential forms are sections into the exterior algebra of the cotangent-bundle. The restriction sends such a section $\omega: \mathbb R^3 \to \wedge^2 T^\ast\mathbb R^3$ to $\omega\vert_{S^2}: S^2 \to \wedge^2 T^\ast\mathbb R^3$, whereas the pullback along the inclusion $i: S^2 \to \mathbb R^3$ sends it to $i^*\omega: S^2 \to \wedge^2 T^\ast S^2$. Some people also call the latter the restriction. $\endgroup$ Commented Dec 30, 2011 at 14:06
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    $\begingroup$ @Eric: I'm not entirely sure. If $M\subset N$, we have $TM\subset TN|_M\subset TN$. Given a previous answer (now deleted), I think one can reasonably interpret "restriction" to mean restriction to the fibers over $M$ in the tangent bundle of $N$. Pulling back is definitely restricting to the tangent bundle of $M$, but the term "restricting" the form is ambiguous. In particular, if one took the 3-form $dx\wedge dy \wedge dz$, it's restriction to $S^2$ is nowhere zero but the pullback is identically zero. $\endgroup$
    – Aaron
    Commented Dec 30, 2011 at 14:07

3 Answers 3

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Observe that wedging $\Omega$ with $\omega := x dx + y dy + z dz$ gives $(x^2 + y^2 + z^2) dx \wedge dy \wedge dz$, a positive multiple of the volume form. Now $TS^2 = \ker \omega\vert_{S^2}$. Now let $v, w \in T_uS^2$ and think of $u$ as the radial vector on $\mathbb R^3$. Then these three vectors are linearly independent so that $\Omega \wedge \omega (u,v,w) \ne 0$. But $\Omega \wedge \omega(u,v,w)$ is proportional to $\Omega(v,w)$ since $v,w \in \ker \omega$.

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  • $\begingroup$ Small quibble. If $p\in S^2$, then $(p,v)\in T\mathbb R^3$ is in $TS^2$ iff $(p,v)\in \ker \omega,$ as the condition is equivalent to $p\perp v$ where $p$ is viewed as a vector instead of just a point. However, $\ker \omega$ contains vectors with every possible base point, and not just those on the sphere. $\endgroup$
    – Aaron
    Commented Dec 30, 2011 at 14:32
  • $\begingroup$ @Aaron: good point, I've edited it accordingly. $\endgroup$ Commented Dec 30, 2011 at 14:55
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Use the spherical coordinates: $$(x,y,z)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),0<\theta<2\pi,0<\phi<\pi.$$ Restricted to $\mathbb{S}^2$, we have $r=1$, i.e. $$(x,y,z)=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$ which implies that $$dx|_{\mathbb{S}^2}=-\sin\phi\sin\theta\,d\theta+\cos\phi\cos\theta\,d\phi,$$ $$dy|_{\mathbb{S}^2}=\sin\phi\cos\theta\,d\theta+\cos\phi\sin\theta\,d\phi,$$ $$dz|_{\mathbb{S}^2}=-\sin\phi\,d\phi.$$ This gives $$dx\wedge dy\big|_{\mathbb{S}^2}=-\sin\phi\cos\phi\,d\theta\wedge d\phi,$$ $$dy\wedge dz\big|_{\mathbb{S}^2}=-\sin^2\phi\cos\theta\,d\theta\wedge d\phi,$$ $$dz\wedge dx\big|_{\mathbb{S}^2}=-\sin^2\phi\sin\theta\,d\theta\wedge d\phi.$$

Combining all these, we have $$\Omega\big|_{\mathbb{S}^2}=(xdy \wedge dz+ydz\wedge dx+zdx \wedge dy)\big|_{\mathbb{S}^2}$$ $$=-\sin^3\phi\cos^2\theta\,d\theta\wedge d\phi-\sin^3\phi\sin^2\theta\,d\theta\wedge d\phi-\sin\phi\cos^2\phi\,d\theta\wedge d\phi$$ $$=-\sin\phi\,d\theta\wedge d\phi.$$ Since $0<\phi<\pi$, we have $\Omega\big|_{\mathbb{S}^2}\neq 0$.

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if $p=(x,y,z)$ is on the sphere, then any tangent to the sphere at that point is normal to the direction vector of $p$ -- eg the vector $(y, -x,0)^T$ (you can identify the tangent vectors to the sphere with their image under the tangent map, formally you need to pull back the form to the sphere).

If you evaluate your form on pairs of vectors of that type (not equal to each other, because of antisymmetry) you can check whether it's nonzero by simple calculation (I did not).

(The restriction to the sphere is obtained kind automatically if you only insert vectors tangent to the Sphere).

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