What is a Fibre Bundle? A 5 Minute Introduction

[Greg Bernhardt]

Definition/Summary
Intuitively speaking, a fibre bundle is space E which ‘locally looks like’ a product space B×F, but globally may have a different topological structure.
Extended explanation
Definition:
A fibre bundle is the data group $(E, B,\pi, F)$, where $E, B$, and $F$ are topological spaces called the total space, the base space, and the fibre space, respectively and $\pi : E \rightarrow B$ is a continuous surjection, called the projection, or submersion of the bundle, satisfying the local triviality condition.
(We assume the base space B to be connected.)
The local triviality condition states the following:
we require that for any $x \in E$ that there exist an open neighborhood, $U$ of $\pi (x)$ such that $\pi^-1$$(x)$ is homeomorphic to the product space $U×F$ in such a manner as to have $\pi$ carry over to the first factor space of the product...

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[fresh_42]

See also: https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/ (esp. part 3)

 

[eltodesukane]

"see the first figure"
Where is it? I see no figure.

 

[Greg Bernhardt]

"see the first figure"
Where is it? I see no figure.

I have removed the language for now

 

[ChinleShale]

I think the Examples section is too brief to be informative.

You might describe how the Mobius band and the Klein bottle are fiber bundles and perhaps compare them to the cylinder and the torus. More generally it would seem important say what a trivial bundle is.

Perhaps also mention that one of the most studied classes of fiber bundles are vector bundles where the fiber has the structure of a vector space. While one can derive such a structure from the structure group usually this is defined the other way around where one starts with a vector space structure and deduces the structure group.

 

[eltodesukane]

"In the case of the Möbius strip, the fibre bundle ‘locally looks like’ the flat euclidean space R^2, however the overall topology is markedly different."
-- I don't understand, each point of the space has a fiber. What is different with the Möbius strip? Does each point has 2 fibers?

 

[fresh_42]

If you look at a small neighborhood of the Möbius strip, you will find a flat neighborhood with a one dimensional fiber at each point. This is the same as in the Euclidean plane with perpendicular one dimensional vector spaces attached at each point. However, if you consider the entire total space, then walking along a closed curve on the Möbius strip changes the direction (sign) of a vector in the fiber, whereas it does not on the Euclidean plane.

 

[ChinleShale]

"In the case of the Möbius strip, the fibre bundle ‘locally looks like’ the flat euclidean space R^2, however the overall topology is markedly different."
-- I don't understand, each point of the space has a fiber. What is different with the Möbius strip? Does each point has 2 fibers?

The Mobius strip is a line bundle over the circle. You can see this by taking the base circle to be its equator. Perpendicular to each point on the circle is a line segment. This is the fiber of the bundle.

As a bundle, the structure group of the Mobius strip is not zero but rather is Z/2. The transition function one gets when making the half twist is multiplication by -1.

An easy way to see that the Mobius strip is not a trivial bundle is to notice that its boundary is single circle. It is were trivial its boundary would be two circles.

Also as @fresh_42 points out in post #7 if a person walks along the equator standing perpendicular to it he will return home upside down.

While a Mobius band made out of a piece of paper is geometrically flat, as a fiber bundle it has no canonical geometry. One can have highly warped and wobbly Mobius strips and they are still topologically identical.

Here is something to think about: Suppose instead of a half twist you do a full twist. This is another line bundle over the circle. What bundle is it?

 

[fresh_42]

A nice experiment is to clue a strip from paper and cut it along the ring in the middle.