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Integrable vs. Completely Integrable Distributions

  1. May 20, 2014 #1
    I am a bit puzzled about the distinction between integrable, and completely integrable distributions. Before I pose my question, let me first define the terms:

    A distribution ##D## of dimension ##k## on ##M## is a smooth collection of ##k##-dimensional subspaces ##D_p \subset T_pM## with ##D = \cup_{p\in M}D_p##. Consider ##D##.

    ##D## is said to be integrable if one can find an (immersed) submanifold ##S## for every point ##p \in M## such that ##D_p = T_pS##.

    ##D## is said to be completely integrable if one can find a neighbourhood ##U## about every point, and a corresponding coordinate chart ##(U,\phi)## with the property that the coordinate basis vectors ##\partial_1, \ldots, \partial_k## spans ##D_p##.

    Question: Why is this distinction necessary? Suppose we have an integrable ##k##-distribution ##D## in an ##n##-manifold. It seems to be that if one can find a submanifold about every point #p#, then per definition of a(n embedded) submanifold ##S##, one has a chart ##(U,\phi)## about every point ##p## such that ##U\cap S## is defined by ##x^{k+1}=\ldots= x^n=0##. The first ##k## coordinate basis vectors are tangent to ##S## for ##p \in S\cap U##, so does not this imply that ##D## is also integrable.

    However, there is the word immersed: is it not possible to find a chart ##(U,\phi)## about every point ##p## for an such that ##U\cap S## is defined by ##x^{k+1}=\ldots= x^n=0## for an immersed submanifold ##S##?

    If not, why? And is this the point of the distinction? Or are there more points?
     
  2. jcsd
  3. May 20, 2014 #2

    quasar987

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    Note that in the definition of completely integrable, the coordinate vectors span D at every point of U, whereas for the chart corresponding to an immersed submanifold, the coordinate vectors only span D along the submanifold.


    No, that is precisely the difference between and immersed and embedded submanifold. You should read the chapter on submanifolds in the book by John Lee.

    As it turns out, integrability and complete integrability are equivalent (Frobenius' theorem), so the distinction is only useful up to the point where Frobenius' thm is proved...
     
  4. May 20, 2014 #3

    WWGD

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    One of the issues that may be at play here, I think, is that there may or may not be a global basis for the subspaces in a distribution. I only know that for the case n=2 , i.e., for plane distributions ( actually, this is for the case of 3-manifolds ) there is a global basis of vector fields iff the Euler class is zero, so that, e.g., all plane distributions in ## \mathbb R^n ## have global bases of vector fields. I cannot think at this point of some manifold whose Euler class is not trivial, nor , of course of an example of a 2-distribution that does not have a global basis.
     
  5. May 20, 2014 #4
    Does he proves that this is the case? If so at what page in the second (or first) edition?
     
  6. May 20, 2014 #5

    quasar987

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    He proves that immersed (resp. embedded) submanifolds are precisely the images of immersions (resp. embeddings). He also proves that an immersion is locally an embedding.

    So what is the obstruction for an immersion to be an embedding? It is that globally, an immersed manifold may have "intersections" in the sense of points that do not verify the defining property of an embedded submanifold in terms of charts (ex: the figure 8 in the plane).

    That is what I meant.
     
  7. May 21, 2014 #6

    Come to think of it.. How is the distribution well defined at a point ##p## in ##M## when foliated with immersed submanifolds? Generally a point ##p## is associated with more than one tangent space (ex:center point of the figure 8 in the plane).

    (Never mind. I found out; I was just confused about the cross which seem to indicate two slopes. The whole point about immersions is that the tangent spaces are in bijectively related to each other.)
     
    Last edited: May 21, 2014
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