Am I right that a vector bundle $(E,M)$ of rank 0 means that sections of $E$ are functions $f:M \to M$?
Vector bundle $(E,M)$ of rank $0$: sections of $E$?
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differential-geometry
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3Why do you think you are right? Have you tried to prove it? – 2012-12-02
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0@JasonDeVito Well, rank 0 means $E$ and $M$ are diffeomorphic. So a section of $E$ is a map from $M$ to $E \equiv M$. – 2012-12-02
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1A vector bundle comes equipped with a map from $E(\cong M)$ to $M$. Sections have to respect this map somehow. How does that enter in? – 2012-12-02
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2A section of $E$ is a map from $M$ to $E$ that ...... – 2012-12-02
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0Sorry I don't get how I can use that $\Pi \circ S = id_M $ for a section $S$. All I know is that $S\circ d:M \to M$ where $d$ is the said diffeomorphism. – 2012-12-02
1 Answers
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If $E$ is a rank $0$ vector bundle over $M$, identify $E \cong M\times\{0\}$. The projection map $\pi:E\to M$ is projection onto the first factor of the product. Every section of $E$ is a map $s:M\to E$ such that $\pi\circ s = id_M$.
With these three facts, you can completely characterize every single section of $E$ by asking what sort of function $M\to M\times\{0\}$ composed with projection to $M$ gives the identity on $M$.
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0Why did you identify $E$ with $M \times \{0\}$. What's the use of the tag 0? – 2012-12-04
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0The empty basis is a global trivialization of $E$, so we can identify $E$ with the product bundle $M\times\{0\}$. – 2012-12-04