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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"


    where it would seem the first two maps are inclusions.

    Any hints, ideas, please.?


    Since this is not an exact sequence in the algebraic sense, I don't know how to
  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


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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:


    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:


    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.?

  6. May 18, 2010 #5


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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


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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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