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What is the/a Quotient Bundle.?

  1. May 12, 2010 #1
    Hi, everybody:

    I have read different sources for quotient bundle: Milnor and Stasheff, Wikipedia
    Wolfram, and I still cannot figure out how it's defined. All I know is that it involves
    a space X, a subspace Y, and a restriction.

    Let Tx be a bundle p:M-->X for X, and Tx|y be the restriction of Tx to y, i.e., the
    bundle with top space p<sup>-1</sup> (y):I see a mention of a(n informal) exact
    sequence (since there is no actual algebraic map to speak of a kernel). Tx/y is defined
    as the "completion of the exact sequence"

    0-Ty-Tx|y-Tx/y-0

    where it would seem the first two maps are inclusions.

    Any hints, ideas, please.?

    Thanks.

    Since this is not an exact sequence in the algebraic sense, I don't know how to
    f
     
  2. jcsd
  3. May 12, 2010 #2
    It is an exact sequence in the algebraic sense. Think of the restriction of the tangent bundle to a submanifold. It splits (look at it in a chart) into the tangent bundle of the submanifold and the normal bundle. Once you understand how the normal bundle works, the generalization is straightforward.
     
  4. May 13, 2010 #3

    lavinia

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    you just need to check that the space obtained by modding out the sub-bundle in each fiber is still locally trivial
     
  5. May 17, 2010 #4
    Thanks , both.

    I think I realized the issue (please correct me o.wise): we start with a bundle:

    p:E
    |
    \/
    x

    And y is a subspace of x.

    The sequence is defined pointwise: we have y<x ( y a subspace of x) , so that
    we consider T_py ,the tangent space of p , for any p in x , as a subspace of the
    tangent space of p in x:

    0-T_py-T_px|y-T_px/y-0

    then we consider the quotient vector space T_px|y / T_py

    at every point in y, right.?. But, how do we define the projection map in T_px/y .?.

    I would imagine we compose the original p with a projection onto the first

    component of the quotient , otherwise, it becomes a mess to define p on

    the cosets.

    Is this right.?

    Thanks.
     
  6. May 18, 2010 #5

    lavinia

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    the space X is fixed and you have a sub-bundle of the bundle, E, over X. This sub-bundle is also a bundle over X with fiber a subspace of the fiber of E over X. The exact sequence is an exact sequence of fibers. The quotient bundle is another bundle over X.
     
  7. May 18, 2010 #6
    Thanks, Lavinia, I wanted to know how we defined the projection in the quotient bundle,
    tho, since we are dealing with cosets, and p in E-->X is defined on elelments of E. Do we
    apply p in Tx/y , to the E component of the quotient.?
     
  8. May 18, 2010 #7
    Fibrewise (and bundle-wise over a compact manifold) the exact sequence splits, so you can define the projection map to complete the exact triangle.
     
  9. May 19, 2010 #8
    Thanks again, Zhentil; I forgot that every exact sequence of vector spaces splits,
    and if the sequence doesn't split, then you must acquit:smile: (remember J.Cochran
    on O.J's trial)
     
  10. May 19, 2010 #9

    lavinia

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    yes. though I don't think that cosets are required. the only thing you need is that the quotient fibers are preserved by the coordinate transformations of E. this automatically gives you a bundle.

    Not all bundles are vector bundles yet they still have quotient bundles and these quotients need not be obtained by modding out by a sub-bundle either. For instance, take the unit sphere bundle of a vector bundle with a Riemannian metric. The antipodal map on each fiber commutes with the coordinate transformation so the quotient fibers - which are projective spaces - form a bundle.
     
    Last edited: May 19, 2010
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