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Question

Let $i:S^2 \to \mathbb{R}^3$ be inclusion. Let $\omega=dz$ be a 1-form on $\mathbb{R}^3$. I want to compute $i^* \omega$ and it vainishes exactly two points.

Is it right that $i^* \omega=i^*dz=d(i^*z)=d(z \circ i)$? Why does it vanishes at two points?

Hint: $dz$ is the differential of projection to the $z$-axis. For which two points of $\mathbb{S}^2$ does the projection to the $z$ axis not have full rank? — Neal, Dec 12 '12 at 15:34

Manoel · Answer 1 · 2012-12-12T17:18:03.787

1

$(i^{*}w)(p)v=w(i(p))(di_{p}v)=dz(di_{p}v)=dz(i(v))=dz(v)=0 \Leftrightarrow v=(v_{1}, v_{2},0) $, where $p\in \mathbb{S} ^{2}$,and$ v\in \mathbb{R} ^{3}$.

edited Dec 12 '12 at 17:18

answered Dec 12 '12 at 15:47

Manoel

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I'm a little confused, $(i^{*}w)(p)v=0$, $\forall$ $v=(v_{1}, v_{2}, v_{3}) \in \mathbb{R}^{3}$ such that $v_{3}=0$. I do not see how it vanishes exactly two points. – Manoel Dec 20 '12 at 02:00
@Gobi, My answer did not help? You do not agree? – Manoel Feb 10 '13 at 01:17
wtf???????????? – jk001 Apr 21 '21 at 18:56

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