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A $\Bbb K$ vector bundle over a space $B$ is a space $E$ and a continuous map $p:E\to B$ so that $p^{-1}(b)$ is a topological $\Bbb K$ vector space for any $b\in B$.

One always includes local triviality in the definition, that is one demands that for any $b\in B$ there is a neighbourhood $U$ of $B$ together with homeomorphism $f:U\times \Bbb R^n \to p^{-1}(U)$ so that $p\circ f$ is the same as the projection onto $U$.

If one does not include this in the definition, can one construct simple examples of vector bundles that are not locally trivial?

s.harp
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    It would really defeat the point of the construction, that locally $B\times E$ behaves like $B \times \mathbb{R}^n$ – Birch Bryant Jun 03 '17 at 15:47
  • I think it's a good question. Consider for example when you have a family of compact complex manifolds, those are not locally trivial (if we ask that the trivialization have compact fibers). It's not obvious to me the sane can't happen here. – Gunnar Þór Magnússon Jun 03 '17 at 16:08
  • @s.harp: Does your proposed idea differ in any way from "a family of (finite-dimensional) vector spaces parametrized by $B$"? If not, there are plenty of examples, even natural ones (e.g., images or kernels of mappings of non-constant rank between locally-trivial vector bundles over $B$). – Andrew D. Hwang Jun 03 '17 at 19:38
  • @Andrew As I read the question, I assumed all the spaces were of the same dimension. Otherwise we of course have the examples you mentioned. – Gunnar Þór Magnússon Jun 03 '17 at 21:55
  • @Gunnar: I see, thank you. (Even with fibres of constant dimension, it appears one can, for example, construct a "subbundle" of $\mathbf{R} \times \mathbf{R}^{2}$ by "picking lines through the origin discontinuously", e.g., the $x$-axis over rationals and the $y$-axis over irrationals.) – Andrew D. Hwang Jun 03 '17 at 22:08
  • @AndrewD.Hwang Oh, I think that might work. Very nice. – Gunnar Þór Magnússon Jun 04 '17 at 08:16
  • Much more frequent in mathematics is the concept of affine bundles. Notice that a map like yours, if a vector bundle (that is, locally trivial) always has the zero section. It is easy to construct examples like yours with no section and thus can not be a vector bundle, but every fiber is an affine space, thus a vector space. If you are unfamiliar, google affine bundles. These vary continuously, dimension does not jump, but yet not a vector bundle. – Mohan Jun 17 '17 at 18:36

1 Answers1

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$\newcommand{\Reals}{\mathbf{R}}$To summarize (and slightly amplify) the comments: Define a vector party (thinking of a disorganized collection, with or without local triviality) over a space $B$ to be a space $E$ together with a continuous map $p:E \to B$ such that $p^{-1}(b)$ is a vector space (over some fixed field) for each $b$ in $B$.

A vector party is far from a vector bundle, merely a collection of vector spaces parametrized by points of $B$. Fix a topological vector space $V$ of dimension at least $2$ (but not necessarily finite).

A vector party can fail to be a vector bundle for multiple reasons, including:

  1. The total space $E$ has "finer topology than it should".

    Let $E = B \times V$ be the Cartesian product with the discrete topology on $B$, and $p:E \to B$ projection to the first factor. If $O$ is an open set in $V$, then for each $b$ in $B$, the set $\{b\} \times O$ is open in $E$; if the base space $B$ is not discrete, $p:E \to B$ is not locally trivial. (In general, $E$ has no continuous, non-constant local section.)

  2. The fibres "vary discontinuously".

    Let $k$ be an integer, $0 < k < \dim V$, let $\sigma:B \to G_{k}(V)$ be an arbitrary mapping from $B$ to the Grassmannian of $k$-planes in $V$ (i.e., "a choice of $k$-dimensional subspace of $V$ for each $b$ in $B$"), let $E$ be the collection $$ E = \bigcup_{b \in B} \{b\} \times \sigma(b) \subset B \times V $$ equipped with the subspace topology, and let $p:E \to B$ be (the restriction of) projection to the first factor.

    If $\sigma$ is discontinuous (e.g., map $\Reals$ to the space of lines through the origin in $\Reals^{2}$ by taking $\sigma(b)$ to be the $x$-axis if $b$ is rational, the $y$-axis if $b$ is irrational, as in the comments), then the subspace topology on $E$ differs from the "local product" topology of a vector bundle. (In general, $E$ has no continuous, non-trivial local section.)

  3. The fibres "vary continuously", but "the dimension jumps".

    If $E$ and $E'$ are (smooth, say) vector bundles over $B$, and if $f:E \to E'$ is a (smooth) bundle morphism, the "bundle of kernels" (whose fibre at $b$ is the kernel of $f_{b}:E_{b} \to E'_{b}$) and the "bundle of images" are vector parties, but generally not vector bundles.

    Think, for example, of the helicoid $E \subset \Reals \times \Reals^{2}$ parametrized by $$ (b, v) \mapsto \bigl(b, (v\cos b, v\sin b)\bigr), $$ the product $E' = \Reals \times (\{0\} \times \Reals)$, a.k.a., the $(x, z)$-plane, and the bundle morphism $f:E \to E'$ given by projection to the $(x, z)$-plane, i.e., $$ \bigl(b, (v\cos b, v\sin b)\bigr) \mapsto \bigl(b, (0, v\sin b)\bigr). $$ The "party of kernels" is the bundle analog of the skyscraper sheaf supported at integer multiples of $\pi$, while the "party of images" results from collapsing these fibres of $E'$ to points.

Andrew D. Hwang
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    Thank you, examples like these was exactly what I was looking for. – s.harp Jun 04 '17 at 12:52
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    Isn't this vector party slightly more than a collection of vector spaces parameterized by points of $B$? It is parameterized by a function $p$ which is continuous, which surely must mean something. Where in your definition of a vector party does it differ from a vector bundle? –  Sep 06 '19 at 14:26