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More on Diff. Forms and Distributions as Kernels

  1. Nov 12, 2012 #1

    WWGD

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    Hi, Again:

    I'm trying to show that, given a 3-manifold M, and a plane field ρ (i.e., a distribution on
    TM) on M, there exists an open set U in M, so that ρ can be represented as the kernel of a
    differential form w , for W defined on U.

    The idea is that the kernel of a linear map from R3 --TxM
    is either the whole space, or a two-dimensional space, by , e.g., rank-nullity.


    My idea is to start by choosing the assigned plane ∏m at any point m in M. Then we use the fact that any subspace can be expressed as the kernel of a linear map.

    Specifically, we choose a basis {v1,v2} for ∏m Subset TmM
    and define a form w so that:

    w(v1)=w(v2)=0 .

    Then we extend the basis {v1,v2} into a basis {v1,v2,v3} for TmM , and
    then we declare w(v3)=1 (any non-zero number will do ), so that
    the kernel of w is precisely ∏m, by some linear algebra.

    Now, I guess we need to extend this assignment w at the point m, at least into
    a neighborhood Um of M . I guess all the planes in a subbundle have
    a common orientation, so maybe we can use a manifold chart Wm for
    m, which is orientable ( being locally-Euclidean), and then use the fact that there is
    an orientation-preserving isomorphism between the tangent planes at any two points
    p,q in TpU . Does this allow me to define a form in U whose kernel is ρ ?

    Edit: I think this should work: please critique: for each q in U , the hyperplane ∏
    q can be described as the kernel of a map , as for the case of m. Again,
    we find a basis {w1,w2} for ∏q , and then define an orientation-preserving
    map (which exists because U is orientable) between TmM and
    TqM, sending basis elements to basis elements.



    Thanks.
     
    Last edited: Nov 12, 2012
  2. jcsd
  3. Nov 12, 2012 #2

    quasar987

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    I did not really try to understand what you're saying with orientations and all because you seem to be overcomplicating things: The usual way this is done is simply to take around p a local frame (v1,v,3) for TM such that (v1,v2) is a local frame for your distribution D (This means a set of 3 vector fields defined on a nbhd of p that are linearly independant at each point.) Then as you said, simly define w by w(v1)=w(v2)=0 and w(v3)=1. Done.

    I other words, what you did at p, you can do locally around p for the same cost!
     
  4. Nov 12, 2012 #3

    lavinia

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    It is not clear to me why you can assume that there is a basis field to the field of two planes. This needs to be shown.

    Globally it is not generally true so the proof boils down to showing that in a small enough region the field of two planes can be oriented. While I believe this, I don't see why it is obviously true. The converse is clear. That is you start with a smooth non-zero vector field and construct a 1 form from it. Its kernel will be a smooth field of orientable 2 planes.

    Here is an example of why this can fail globally. Look at the orientable manifold that is the quotient of Euclidean 3 space by the standard lattice (all vectors with integer coordinates) together with the isometry (x.y.z) -> (x+1/2,-y,-z). The planes that are parallel to the (x,y,0) plane project to parallel surfaces in the three manifold. Take the field of 2 planes to be the tangent spaces to these surfaces.

    The (x,y,0) plane projects to a Klein bottle so there is no global basis for the field of 2 planes along it(because the Klein bottle is not orientable). Likewise there is no global normal vector field. So there is no 1 form whose kernel is the tangent spaces to this Klein bottle. Any oriented frame must rotate off of the tangent space. Any vector that starts out normal must eventually become tangent at some point.

    .
     
    Last edited: Nov 12, 2012
  5. Nov 12, 2012 #4
    Hello there.

    Isn't local existence of V1 V2 the definition of a smooth distribution?
     
  6. Nov 12, 2012 #5

    lavinia

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    A field of two planes to me has no given basis. Two independent vector fields - which is what I think of as a distribution - is not the same. Showing that a field of two planes is orientable is the same as finding a distribution - in a three manifold - I think. to me the problem is to show that a field of two planes on the 3 manifold can be locally oriented - or what is the same is locally defined by a distribution.
     
  7. Nov 12, 2012 #6
    In that case, how would you define a smooth field of 2-planes?

    Two independent vector fields determine a distribution, but they are not identified with the distribution. The distribution itself is their span as a bundle over M (i.e. a field of k-planes). The local existence of an independent smooth basis is just the smoothness condition. Alternatively, a k-distribution could be defined (locally) as the kernel of a nonzero (n-k) form, or as the intersection of the kernels of (n-k) 1-forms.
     
  8. Nov 12, 2012 #7

    lavinia

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    you are right - the distribution is just the entire plane. I always think of it in terms of independent vector fields since what you want to know often is whether it is involutive. But this a semantic point. But what you are saying is that smoothness by definition means there is a local smooth basis. Hmm.. Well in that case there is nothing to prove. So what do you say about the induced bundle from a map of the manifold in a Grassmann manifold of k planes in k+n space? that does not give you a smooth distribution? It still seems to me that there is something to prove.

    So I guess what you are saying is that a local basis is in fact the definition of smoothness.I would have thought something like a smooth map into the Grassmann bundle of tangent k planes would be the definition. Sorry for the mistakes.

    BTW: It you can show that a section of the Grassmann bundle is actually a vector bundle then the existence of a local basis would seem to follow from local triviality not from smoothness since it would work even if the section was continuous.
     
    Last edited: Nov 12, 2012
  9. Nov 12, 2012 #8
    I see what you're saying. You could define a distribution as a smooth section of the fiber bundle of k-planes in TM. Somehow that seems cleaner conceptually. It makes sense and presumably it is equivalent, but most textbooks define it in the lower tech way without involving fiber bundles.

    I was working with the definition from Wikipedia: http://en.wikipedia.org/wiki/Distribution_(differential_geometry [Broken])
    That one matches the definition from Frank Warner's book, (and Lee's as well, if I remember correctly). Basically, they circumvent the problem of having to describe the smooth structure on the fiber bundle.

    Also, just to clarify, I did not mean to suggest that the distribution is smooth if there exists a local basis. Rather that the distribution is smooth if there exists a smooth local basis of vector fields.
     
    Last edited by a moderator: May 6, 2017
  10. Nov 13, 2012 #9

    WWGD

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    Thank you all for your replies.

    Yes, I think there is an orientability issue in order for the distribution to be a subbundle
    of the tangent bundle. I thought the fact that charts are oriented --because of the
    chart homeomorphism with R^n -- that the chart U would be orientable, and the
    planes in the distribution would inherit this orientation.

    Actually, I'm curious as to what the obstructions are for defining such a form
    globally. I think orientability is one of them, and we can orient locally using a chart
    and pulling back the orientation from R^n. Maybe the only other obstruction is
    just that of the distribution being a subbundle of the tangent bundle.
     
    Last edited: Nov 13, 2012
  11. Nov 14, 2012 #10

    quasar987

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    I thought that by definitio (Lee), a distribution is a subbundle. What is a distribution to you??
     
  12. Nov 16, 2012 #11

    WWGD

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    ,

    Ah, sorry, I don't know enough about bundles to know if any assignment of
    subspaces is necessarily a subbundle of TM. Maybe there is, e.g., a smoothness
    condition.
     
  13. Nov 17, 2012 #12

    quasar987

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    Yes, there is. We need to be able to find a smooth local frame around every point.
     
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