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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.)

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  • $\begingroup$ 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. $\endgroup$
    – Peter May
    Commented Feb 15, 2013 at 3:49
  • $\begingroup$ a) follows from b). Therefore it is usually not mentioned in the definition, but rather as a property. $\endgroup$
    – Martin Brandenburg
    Commented Feb 15, 2013 at 4:05
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    $\begingroup$ 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. $\endgroup$
    – Qfwfq
    Commented Feb 15, 2013 at 14:44
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    $\begingroup$ @Qfwfw: Good luck at defining morphisms etc. $\endgroup$
    – Martin Brandenburg
    Commented Feb 15, 2013 at 16:38
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    $\begingroup$ @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? $\endgroup$
    – Qfwfq
    Commented Feb 17, 2013 at 0:14

1 Answer 1

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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 $x \mapsto x^3+x$.

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    $\begingroup$ @Martin I've posted an elaboration of your answer to the original question on Math.SE. $\endgroup$
    – Avi Steiner
    Commented Feb 16, 2013 at 0:25

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