Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recall the usual definition of a $k$-dimensional vector bundle (everything is assumed to be continuous/smooth/etc depending on the category):

A $k$-dimensional vector bundle is a triple $(E,B,\pi)$, where $\pi\colon E \to B$, satisfying the following:

    a) The map $\pi$ is onto (I don't know if everyone requires this, but I will).

For each $p\in B$:

    b) The fiber $E_p=\pi^{-1}(p)$ is a (real) vector space.

    c) There's a neighborhood $U\ni p$ and a diffeomorphism $\phi\colon \pi^{-1}(U) \to U \times \mathbb{R}^k$ such that $P_1\circ \phi = \pi$, where $P_1$ is projection onto the first factor.

    d) for each $q\in U$, the restriction $\phi\colon E_q \to {q}\times \mathbb{R}^k$ (where $\phi$ is the diffeomorphism from c)) is a linear isomorphism.

My Question

Can d) be replaced with

    d') the restriction $\phi\colon E_p \to {p}\times \mathbb{R}^k$ is a linear isomorphism.

In the original version, the restriction of $\phi$ to every fiber over $U$ must be a linear isomorphism. In the alternate version, this is only required for the single fiber $E_p$.

(Note: This was posted at Math.SE a few weeks ago but received no answers.)

share|improve this question
    
You mean homeomorphism, not diffeomorphism, and the answer is trivially no if you don't require $B$ to be connected, but it is also surely no in general. You might have a topological bundle with linearity by accident at some point in a coordinate neighborhood. –  Peter May Feb 15 '13 at 3:49
    
a) follows from b). Therefore it is usually not mentioned in the definition, but rather as a property. –  Martin Brandenburg Feb 15 '13 at 4:05
    
For me the usual definition of vector bundle is: you have diffeomorphisms $\phi:\pi^{-1}(U_{\alpha}) \to U_{\alpha} \times \mathbb{R}^k$ over $U_{\alpha}$, where { $U_{\alpha}$ } is a covering of $B$, and the transition functions $\phi_{\beta} \circ \phi_{\alpha}^{-1} : U_{\alpha\beta} \times \mathbb{R}^k\to U_{\alpha\beta}\times\mathbb{R}^k$ over $U_{\alpha \beta}=U_{\alpha}\cap U_{\beta}$ are fiberwise linear maps. –  Qfwfq Feb 15 '13 at 14:44
    
@Qfwfw: Good luck at defining morphisms etc. –  Martin Brandenburg Feb 15 '13 at 16:38
    
@Martin: a morphism is a map $f: E\to F$ over $B$ such that for every $p\in B$, for every $\alpha$ of a covering that trivializes both bundles, the map $\phi_{\alpha}^F \circ f|_{E_p} \circ (\phi_{\alpha}^E)^{-1} : \mathbb{R}^k \to \mathbb{R}^k$, with the obvious notations, is linear. Are there problems with this definition? –  Qfwfq Feb 17 '13 at 0:14
add comment

1 Answer

up vote 3 down vote accepted

No. Take $E=[0,1] \times \mathbb{R} \to [0,1]=B$ with the usual vector space structure at all fibers except for the fiber of $1/2$ where the vector space structure is twisted by any non-linear diffeomorphism of $\mathbb{R}$, for example $x \mapsto x^3+x$.

share|improve this answer
    
@Martin I've posted an elaboration of your answer to the original question on Math.SE. –  Avi Steiner Feb 16 '13 at 0:25
add comment

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.