# Almost A Vector Bundle

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!

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The naïve (i.e. fibrewise) kernel of a bundle homomorphism can fail to be a vector bundle: just a take a bundle homomorphism whose rank jumps. –  Zhen Lin Oct 30 '12 at 22:22

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.

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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.

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