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When is a foliated manifold a fibre bundle?

  1. May 17, 2014 #1
    I k-foliation of a ##n##-manifold ##M## is a collection of disjoint, non-empty, submanifolds who's union is ##M##, such that we can find a chart ##(U,x^1, \ldots mx^k, y^{k+1}, \ldots, y^n)=(\phi, (x^\mu, y^\nu))## about any point with the property that setting the ##n-k## last coordinates equal to a constant determines a particular submanifold, and varying ##x^\mu## lets us move around on the surface.

    Now I wonder, what conditions would one additionally have to impose for ##M## to become a fibre bundle? Is it's enough that the leaves (submanifolds) of the foliation are diffeomorphic to eachother?
     
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  3. May 17, 2014 #2

    Matterwave

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    I'm not sure what this construction has to do with fiber bundles. What is your base space? What is your fiber? What is your projection, and trivialization and structure group?
     
  4. May 17, 2014 #3
    If one has a foliation, then there excists a chart ##(U,\phi)## with the property that ##\phi(U) = V \times W##, where the points in ##W## determines which leaf we are on. Now since these leaves are submanifold the chart defined by ##(U\cap S, \tilde \phi)=( U\cap S, x^\mu)## for a particular leaf S is a chart that takes ##U\cap S## to ##\phi(U\cap S) = V## where ##U\cap S## is an open subset of the leaf S. Hence, locally ##M \sim ((\text{subset of} \ S) \times W##. Now if we impose conditions so that this is instead the product ##S \times W##, and all the leafs is diffeomorphic to each other (in particular to ##S##), we have a local trivialization with fibre S. If we then construct the quotient ##B=M/S## (the manifold of fibres), we have a base-space. We need no structure group for it to be a fibre bundle..

    So the question is really what conditions are necessary to impose in order for us to have a product ##S \times W## instead of ##\text{subset of} \ S) \times W## and for the leaves to be diffeomorphic to each other. Does the one imply the other, or do we have to impose other coniditions?
     
  5. May 18, 2014 #4

    Matterwave

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    Ah, ok I get where you're going. Sadly, I do not know the answer to this question...sorry.
     
  6. Sep 27, 2014 #5

    lavinia

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    The leaves can be diffeomorphic without the foliation being a fiber bundle.

    Example:

    The Mobius band is foliated by circles but is not a circle bundle over a closed interval. This foliation extends to the Klein bottle where it is not a circle bundle over the circle.
     
  7. Sep 30, 2014 #6
    See, simple foliation and Ehresmann's lemma
     
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