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How can I explain the concept of a fiber bundle to someone with no mathematical background?

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2 Answers 2

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Draw a picture! For example you could use:

  • the cylinder for a (trivial) $S^1$-bundle over $\mathbb{R}$
  • the Möbius strip for a (non-trivial) $(0,1)$-bundle over $S^1$
  • the volume between two spheres of different radius for a (trivial) $(0,1)$-bundle over $S^2$
  • the torus for a (trivial) $S^1$-bundle over $S^1$

I think that those can already give some kind of intuition on what a fibre bundle is, in particular the Möbius strip is an easy non-trivial example. You can explain how it is different from the others by noticing that you cannot deform the Möbius strip into a cyilinder.

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    $\begingroup$ Also, it would be helpful to talk about a Cartesian product. Some examples might be best made physical -- like all the configurations of an hour hand and a minute hand on a clock as a "physical" $S^1 \times S^1$. $\endgroup$ Aug 13, 2013 at 17:27
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I would say draw an example of a space which can be exhibited as a fiber bundle, draw in the base space as a subspace (if this is possible) and then show that for a sufficiently small neighbourhood around a point in the base space, the fiber of the neighbourhood just looks like the product of the neighbourhood and the fiber. Some easy starting examples would be a compact product space (why not pick a square, cylinder and a torus - all easy to draw) to show that there are trivial bundles, and then something a bit more exotic like a Möbius strip or a covering space (the connected double cover of a circle seems like a good start).

The point is that examples are definitely the best way to get the point across.

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  • $\begingroup$ I like the idea of the connected double cover of the circle, I never thought of it as a fibre bundle. $\endgroup$ Aug 13, 2013 at 17:19
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    $\begingroup$ Covering spaces are really just fiber bundles with discrete fibers. $\endgroup$
    – Dan Rust
    Aug 13, 2013 at 17:20

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