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.

Edit: 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 $E_p$.

  • 6
    $\begingroup$ I don't think this might work. The point of a vector bundle is that the fibers should be parameterized in a smooth (or continuous) way. If you weaken d) like you do above you are basically throwing this parameterization away. But those are only my 2 cents, I cannot enforce this opinion of mine with actual counterexamples. $\endgroup$ Jan 28 '13 at 18:54
  • $\begingroup$ Can you be more specific about what it is you're changing between (d) and (d')? To me it looks like the only thing you've dropped is the phrase "for each $q \in U$", but as (I assume) $\phi$ is defined to be the map $\pi^{-1}(U) \to U \times \mathbb R^k$ from (c), and hence depends on $U$, I'm not sure what you mean by dropping this condition. $\endgroup$
    – Jim
    Jan 29 '13 at 7:16

(The following is an elaboration of Martin Brandenburg's answer to the MO cross-post of this question.)

The answer to my question, as the counterexample below shows, is no.

Consider the canonical projection $\pi$ from $E:=[0,1]\times \mathbb{R}$ to $[0,1]$. For $t\neq1/2$. Equip the fiber $E_{1/2}=\{1/2\}\times\mathbb{R}$ with the "twisted" vector space structure induced by any nonlinear homeomorphism $f\colon \mathbb{R}\to\mathbb{R}$ (such as $x\mapsto x^3+x$)---in other words, define $x+'y:=f^{-1}(f(x)+f(y))$ and $r\cdot'x:=f^{-1}(rf(x))$. Equip the each of the other fibers $E_t=\{t\}\times \mathbb{R}$ with the standard vector space structure inherited from $\mathbb{R}$.

This clearly satisfies the alternate definition of a vector bundle---when $t\neq 1/2$, take $\varphi$ to be identity, and when $t=1/2$, take $\varphi$ to be $\operatorname{id}\times f$. However, this does not satisfy the actual definition of a vector bundle. To see this, suppose we have a local trivialization (the original type, not my alternate type) $(\varphi,U)$ near $1/2$. Let $\varphi_2$ be the second component of $\varphi$.

For $t\neq 1/2$, the slice $\varphi_2(t,-)$ is a continuous linear automorphism of the (standard) reals and therefore must be of the form $\varphi_2(t,x)=h(t)x$ for some nonzero $h(t)\in \mathbb{R}$. The continuity of $\varphi_2$ then requires that $h$ be continuous where it's defined.

When $t=1/2$, the slice $\varphi_2(1/2,-)$ is still a continuous linear isomorphism, but this time it's from the "twisted" real numbers $E_{1/2}$ to the standard reals. But by construction, $f$ is also such an isomorphism. So, composing $\varphi_2(1/2,-)$ with $f^{-1}$ gives a continuous linear automorphism of the reals and hence is of the form $\varphi_2(1/2,f^{-1}(u))=cu$ for some nonzero $c\in\mathbb{R}$. Thus, $\varphi_2(1/2,x)=\varphi_2[1/2,f^{-1}(f(x))]=cf(x)$.

Combining the above two results, we have $$\varphi_2(t,x)= \begin{cases} cf(x), &\text{if }t=1/2,\\ h(t)x, &\text{if }t\neq 1/2. \end{cases}$$ So, since $\varphi_2$ is continuous, $$f(x)=\frac{1}{c}\varphi_2(1/2,x)=\frac{1}{c}\lim_{t\to1/2}\varphi_2(t,x)=\frac{1}{c}\lim_{t\to1/2}h(t)x = \left(\frac{1}{c}\lim_{t\to1/2}h(t)\right)x.$$ In other words, $f$ is linear, contradicting the fact that it isn't.


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