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Could any one give me hint how to show that the zero section of any smooth vector bundle is smooth?

Zero section is a map $\xi:M\rightarrow E$ defined by $$\xi(p)=0\qquad\forall p\in M.$$

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2 Answers 2

up vote 7 down vote accepted

Smoothness can be checked locally on $M$ and locally $E$ is trivial.

Can you use these two facts to conclude what you want?

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No........................... –  miosaki Aug 7 '12 at 23:48
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@KutukKatuk If $M$ and $N$ are smooth manifolds and $p\in N$, then the inclusion $M\to M\times N$ given by the rule $m\to (m,p)$ is smooth. If you understand this and combine it with Mariano's answer, then you will have found an answer to your original question. –  Amitesh Datta Aug 8 '12 at 0:32
    
I am a little confused; do we explicitly use the mapping to zero in the proof? It seems like local triviality would apply no matter what constant vector we mapped to. –  Tarnation Oct 9 '12 at 2:26
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@Tarnation: but mapping to another «constant vector» would in general not give an actual section of the bundle: most vector bundles do not have non-zero sections! Zero is different: we always have the zero section. –  Mariano Suárez-Alvarez Oct 9 '12 at 4:17

Say $E\to M$ is a smooth vector bundle, and let $p\in M$. Then there is a neighborhood $U\subseteq M$ of $p$ and a smooth local trivialization $\Phi:\pi^{-1}(U)\to U\times\mathbb R^k$ of $E$ over $U$. What can you say about $(\Phi\circ\zeta)(q)$ where $q\in U$? Can you conclude that $\Phi\circ\zeta$ is smooth? Now remember that $\Phi$ is a diffeomorphism.

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