Take the 2-minute tour ×
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.

Could anyone please help me with this question?

(1) Let (E, p, B) be a vector bundle where E is the total space, B is the base, and p is the structure map, that is, p:E->B. Now suppose E' is a subspace of E, B' is the subspace of B, and p' is the restriction of p to E'. If the image of p' is contained in B', then show (E', p', B') is a vector bundle. (2) Prove or disprove: Let s be a section of (E, p, B). Then restriction of s to B' is a section of (E', p', B') if and only if s(b) is in E' for each b in B'.

THANK YOU SO MUCH IN ADVANCE.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

There is a mistake in your formulation of the question, unless you are using the word "subspace" in two different ways. Take $B$ to be a single point, $E = \mathbb{R} \times B$, $B' = B$, and $E' = [0,1] \times B'$.

More likely you meant to assume that $p': E' \to B'$ is such that $p'$ is the restriction of $p$ to $E'$ and additionally $p'^{-1}(b) = p^{-1}(b)$ for every $b \in B'$. If this is the case then (1) is just an exercise in playing around with the subspace topology. Given any $b \in B'$, we need to show that there is a neighborhood $U \subseteq B'$ which trivializes $p': E' \to B'$ in the sense that $p'^{-1}(U)$ is homeomorphic to $U \times \mathbb{R}^n$ in such a way that $p': p'^{-1}(U) \to U$ corresponds to the projection map $U \times \mathbb{R}^n \to U$. But since $p: E \to B$ is a vector bundle there is a trivializing neighborhood $V \subseteq B$ of $b$, and with the assumptions on $p'$ given above one has that $U = V \cap B'$ does the job.

As for (2), you need to check that the restriction $s'$ of $s$ to $B'$ is continuous (obvious) and satisfies $p'(s'(b)) = b$ for every $b \in B'$. But $p'(s'(b)) = p(s(b)) = b$ since $p'$ is the restriction of $p$ and since $s$ is a section, so we're done.

share|improve this answer
    
Thank you so much. But something is not very clear to me. Why did you take E=R×B as a trivial bundle? Again, thanks. –  Feri Nov 20 '11 at 5:50
    
The main point of my example was that even if you take a subspace of $E$ which fibers over $B'$ it is not automatic that the fibers are vector spaces unless you add some additional assumptions, just as $[0,1]$ is a (topological) subspace of $\mathbb{R}$ which is not a vector space. –  Paul Siegel Nov 21 '11 at 15:12

Your Answer

 
discard

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.