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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

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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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