I'm trying to get some intuition for vector bundles. Does anyone have good examples of constructions which are not vector bundles for some nontrivial reason. Ideally I want to test myself by seeing some difficult/pathological spaces where my naive intuition fails me!
Apologies if this isn't a particularly well-defined question - hopefully it's clear enough to solicit some useful responses!
On
Fix $ B = (-1,1) $ to be the base space, and to each point $ b $ of $ B $, attach the vector-space fiber $ \mathcal{F}_{b} \stackrel{\text{def}}{=} \{ b \} \times \mathbb{R} $. We thus obtain a trivial $ 1 $-dimensional vector bundle over $ B $, namely $ B \times \mathbb{R} $. Next, define a fiber-preserving vector-bundle map $ \phi: B \times \mathbb{R} \rightarrow B \times \mathbb{R} $ as follows: $$ \forall (b,r) \in B \times \mathbb{R}: \quad \phi(b,r) \stackrel{\text{def}}{=} (b,br). $$ We now consider the kernel $ \ker(\phi) $ of $ \phi $. For each $ b \in B $, let $ \phi_{b}: \mathcal{F}_{b} \rightarrow \mathcal{F}_{b} $ denote the restriction of $ \phi $ to the fiber $ \mathcal{F}_{b} $. Then $ \ker(\phi_{b}) $ is $ 0 $-dimensional for all $ b \in (-1,1) \setminus \{ 0 \} $ but is $ 1 $-dimensional for $ b = 0 $. Hence, $ \ker(\phi) $ does not have a local trivialization at $ b = 0 $, which means that it is not a vector bundle.
In general, if $ f: \xi \rightarrow \eta $ is a map between vector bundles $ \xi $ and $ \eta $, then $ \ker(f) $ is a sub-bundle of $ \xi $ if and only if the dimensions of the fibers of $ \ker(f) $ are locally constant. It is also true that $ \text{im}(f) $ is a sub-bundle of $ \eta $ if and only if the dimensions of the fibers of $ \text{im}(f) $ are locally constant.
The moral of the story is that although something may look like a vector bundle by virtue of having a vector space attached to each point of the base space, it may fail to be a vector bundle in the end because the local trivialization property is not satisfied at some point. You want the dimensions of the fibers to stay locally constant; you do not want them to jump.
Richard G. Swan has a beautiful paper entitled Vector Bundles and Projective Modules (Transactions of the A.M.S., Vol. 105, No. 2, Nov. 1962) that contains results that might be of interest to you.
Here are two ways one might break the definition a vector bundle.
If one is tricky, one might define a fiber bundle with fiber $\Bbb{R}^n$ that's not a vector bundle, if the structure group isn't linear. For instance, you could bundle $\Bbb{R}$ over the circle but define charts on a two-set open cover such that the transition function would send $(s,r)\in S^1\times\Bbb{R}$ to $(s,r^3)$-generally, bring in any nonlinear homeomorphism of the fiber to itself. This particular example might not qualify as non-trivial, but I don't know any very legitimate cases of this.
Something perhaps a bit more interesting: the condition that the fiber of a (fiber or) vector bundle be constant over the whole base space is pretty strong. On a manifold with boundary, one can define a degenerate tangent "bundle" which is only a half-space on the boundary, which could be quite useful but doesn't qualify as a vector bundle.
Similarly if your almost-manifold has degenerate dimension somewhere for some other reason, as e.g. $z=|x^3|$ embedded in $\Bbb{R}^3,$ which is the union of a surface of two connected components with a $1$-manifold, specifically the line $x=z=0$. You could construct something close to a bundle as the union of the tangent bundle on the $2$-D part and the lines perpendicular tot he $1$-D part, and it wouldn't be a vector bundle.