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How does a fibre bundle differ from base space X typical fibre?

  1. Nov 21, 2008 #1
    Let's take a simple example.

    Both a cylinder and Möbius strip consist of a circle with a line segment associated with each point of the circle. The cylinder is considered truly equivalent to the direct product S1 X line segment while the Möbius strip is only locally like S1 X line segment. Ok, what does that mean? Does the direct product have any properties which aren't local?

    I mean, I thought S1 X line segment L is merely [tex]\{(p,x)|p \in S^1, x \in L\}[/tex], period. Both the cylinder and Möbius strip would be instances of this general thing, differing from each other only in additional properties, e.g. the transition functions.
    Last edited: Nov 21, 2008
  2. jcsd
  3. Nov 21, 2008 #2


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    Homework Helper

    While locally the same, there are global, ie, topological differences. Specifically, the two spaces are not homeomorphic. For example, one is orientable and one is not. Also, if you cut along a circle around the strips, the cylinder gets cut into 2 disjoint pieces while the Mobius strip does not.
  4. Nov 22, 2008 #3
    Thanks, StatusX. I think I fully understand the global difference between the cylinder and the Mobius strip.

    The problem is in how they relate to the direct product S1 X line segment.

    For instance from Wikipedia http://en.wikipedia.org/wiki/Fiber_bundle#Trivial_bundle

    Trivial bundle

    Let E = B × F and let π : E → B be the projection onto the first factor. Then E is a fiber bundle (of F) over B. Here E is not just locally a product but globally one. Any such fiber bundle is called a trivial bundle

    (End quote)

    So what I am asking is, "what does 'globally' like a product mean?" As far as I can see, direct product is strictly a local relation.

    Apparently I am missing something since both the cylinder (the "trivial bundle") and the Mobius strip seem to me to be instances of S1 X line segment L, albeit instances which differ from each other by additional (global) properties.

    Yet the literature indicates that the cylinder is identical to S1 X L while the Mobius strip is only "locally" like S1 X L. I don't see the distinction--with regard to S1 X L. (I do see the distinction between the cylinder and the Mobius strip.)
    Last edited: Nov 22, 2008
  5. Nov 22, 2008 #4
    Clarification: in fibre bundle terminology here, the base space is the circle S1 and the (typical) fibre is the line segment L. The total space of the fibre bundle (also called the bundle space) is S1 X L ... for both the cylinder and the Mobius strip?
  6. Nov 22, 2008 #5


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    The difference between cylinder and Möbius strip is in the topology of their total space (E). A fiber bundle consists of two topological spaces (the total space, and the base space), and a projection from the total space to the base space. The topology of the total space determines, if the bundle can be covered continuously with a single product of the base space and the typical fiber, or it is impossible (like in the case of the Möbius strip). The emphasis is on the continuity.
  7. Nov 22, 2008 #6


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    No. Only the cylinder's total space is S1 x L, the total space of the Möbius strip isn't.
  8. Nov 22, 2008 #7
    I think I have it now.

    I was thinking that since both the cylinder and the mobius strip consist of a circle with a line segment associated with each point of the circle and so both are equivalent to S1 X L. However, another way to view S1 X L is as a line segment with a circle associated with each point. When you picture it that way it is easy to see that S1 X L is the cylinder and not the mobius strip.

    Thanks for playing, mma and StatusX.
  9. Nov 23, 2008 #8


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    Good understanding! It is often said that fiber bundle is a Cartesian product who has lost one of his projection. Now you have shown the projection of the Cartesian product cylinder which is lost when we consider it only a fiber bundle with bas space S1 and fiber L. However in the case of the cylinder, this projection can be retrieved as you showed, that's why it is called trivial. You have now shown that if we start from the Cartesian product cylinder, we can create another fiber bundle from it by forgetting it's another projection. That is, we can regard the cylinder not only as a fiber bundle with base space S1 and fiber L, but also as a fiber bundle with base space L and fiber S1. The Möbius strip of course not, because of its topology.
  10. Nov 23, 2008 #9
    I was just trying to wrap my head around this very concept. Thanks for tying it to my question.
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