Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Am I right that a vector bundle $(E,M)$ of rank 0 means that sections of $E$ are functions $f:M \to M$?

share|cite|improve this question
Why do you think you are right? Have you tried to prove it? – Jason DeVito Dec 2 '12 at 18:45
@JasonDeVito Well, rank 0 means $E$ and $M$ are diffeomorphic. So a section of $E$ is a map from $M$ to $E \equiv M$. – maximumtag Dec 2 '12 at 18:47
A 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? – Jason DeVito Dec 2 '12 at 18:48
A section of $E$ is a map from $M$ to $E$ that ...... – Neal Dec 2 '12 at 18:49
Sorry 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. – maximumtag Dec 2 '12 at 19:10
up vote 3 down vote accepted

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 $M$.

share|cite|improve this answer
Why did you identify $E$ with $M \times \{0\}$. What's the use of the tag 0? – maximumtag Dec 4 '12 at 13:41
The empty basis is a global trivialization of $E$, so we can identify $E$ with the product bundle $M\times\{0\}$. – Neal Dec 4 '12 at 14:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.