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Suppose there is a fiber bundle $E$. The base space is $M$ so that $\pi:E\rightarrow M$ is the projection. By the definition, the bundle has a typical fiber $F$ such that the local trivialization over one open subset $U_i$ of $M$ is defined in this way $\phi_i:U_i\times F\rightarrow \pi^{-1}(U_i)$. So we can find out that no matter which open set $U_i$ we choose, we always choose the same typical fiber $F$. I guess this is because we want to have a smooth structure. If at different points of $M$, we choose different fibers, for example, at point $p\in M$ we attach a 3 dimensional vector space, while at point $q\in M$ we attach a 2 dimensional vector space and so on, then we are not going to have a smooth structure.

So my question is: can we attach different fibers to different points of a base manifold? If we insist to do so, is this structure still smooth? Is it still called a fiber bundle?

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    $\begingroup$ Sure. The base could be disconnected. (If the base is connected, then no: it's a nice exercise to show that a fiber bundle over a connected base has homeomorphic fibers, and similarly a smooth fiber bundle over a connected base has diffeomorphic fibers.) $\endgroup$ Commented Nov 27, 2014 at 21:39

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See the notion of fibration:

A fibration is like a fiber bundle, except that the fibers need not be the same space, rather they are just homotopy equivalent.

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The projection $\mathbb R^3\setminus \{0\}\to \mathbb R: (x,y,z)\mapsto x$ is a smooth submersion between connected smooth manifolds and all its fibers are connected.
However the fibers are not diffeomorphic: the fiber of $0\in \mathbb R$ is difffeomorphic to $\mathbb R^2\setminus \{0\}$ whereas all other fibers are diffeomorphic to $\mathbb R^2$.

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    $\begingroup$ Right, but this isn't a fiber bundle. $\endgroup$ Commented Nov 28, 2014 at 9:24
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    $\begingroup$ @Qiaochu: of course it is not a locally fiber bundle, but I was answering the question "can we attach different fibers to different points of the manifold" and I wanted to give an example of a more general situation (which is also a counterexample to Ehresmann's theorem for non proper maps). $\endgroup$ Commented Nov 28, 2014 at 10:03

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