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Orientation Forms in Different Codimension.

  1. Oct 15, 2012 #1

    WWGD

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    Hi, Everyone:

    A way of defining an orientation form when given a codimension-1 ,

    orientable n-manifold N embedded in R^{n+1} , in which

    the gradient ( of the parametrized image ) is non-zero (I think n(x)

    being nonzero is equivalent to N being orientable), is to

    consider the nowhere-zero normal vector n(x), and to define the

    form w(v)_x : = | n(x) v1 , v2 ,...,v_n-1| (**)

    Where {vi}_i=1,..,n-1 is an orthogonal basis for T_x N , written

    as column vectors, and n(x) is the vector normal to N at x , so that we write:

    | n_1(x) v_11 v_21..... v_n1|
    | n_2(x) v_12 v_22......v_n2|
    ................................

    ................................
    |n_n(x) v_1n v_2n.......v_nn|

    For vi= (vi1, vi2,....,vin )

    Then the vectors in (##) are pairwise orthogonal, and so are

    Linearly-independent.

    *QUESTION* : How do we define a form for a curve of codimension-1,

    and, in general, for orientable manifolds of codimension larger-

    than 1 ? I have seen the expression t(x).v , meaning <t(x),v> ,for the curve. But the

    tangent space of a curve is 1-dimensional, so, how is this a dot product?

    Also, for codimension larger than one: do we use some sort of tensor contraction?

    Thanks.

    Thanks.
     
  2. jcsd
  3. Nov 30, 2012 #2

    lavinia

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    If the normal bundle to the submanifold is trivial then you can get an orientation form by contracting the orientation form of the ambient manifold by a smooth set of lineally independent normal vector fields.

    In Euclidean space you contract the standard volume element.

    If the normal bundle is not trivial i am not sure off of the top of my head but let's see if we can figure it out. It shouldn't be hard.

    The normal bundle of a hypersurface (closed without boundary, codimension 1) of Euclidean space is always trivial but I do not know a direct proof and again would like to work on it with you. I do know a nasty indirect proof if you would like to see it.
     
    Last edited: Nov 30, 2012
  4. Nov 30, 2012 #3

    WWGD

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    Sure, I'll work on it with you. Do you have any suggestions/format in mind?
     
  5. Dec 1, 2012 #4

    lavinia

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    Just looking at ideas plus some reading. My first thought is to see if there is something that can be dome with the Thom class of the normal bundle to the submanifold.
     
  6. Dec 2, 2012 #5

    lavinia

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    I think you can Just contract the volume element of the ambient manifold by a local orthonormal basis for the normal bundle over each coordinate chart. Over each chart you get a local volume form for the submanifold and since the coordinate transformations are in SO(n) they should piece together to give you a global orientation for on the submanifold. Is this right?
     
  7. Dec 2, 2012 #6

    lavinia

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    BTW: it follows that if the normal bundle is orientable then the submanifold must be orientable as well.
     
    Last edited: Dec 2, 2012
  8. Dec 2, 2012 #7

    quasar987

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    I think this is correct. Well done lavinia!
     
  9. Dec 2, 2012 #8

    quasar987

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    More generally, I think if a vector bundle E splits as a direct sum of two subbundle F and G, then 2 of them are orientable iff the third one is.
     
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