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Parallelizability vs. hairy ball theorem

  1. Apr 21, 2008 #1

    mma

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    Usually the hairy ball theorem is cited for proving that [tex]S^2[/tex] is not parallelizable. However, hairy ball theorem is too strong for this. This theorem states that there isn't a nowhere vanishing continuous vector field on [tex]S^2[/tex]. Unparalellizable property means only that there aren't two linearly independent vector fields on [tex]S^2[/tex]. Could somebody tell an example of a nonparallelizable n-dimensional manifold on which hairy ball theorem is false, i.e. on which there is continuous nowhere vanishing vector field (but because of the nonparallelizability, n linearly independent aren't)?
     
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  3. Apr 22, 2008 #2

    mathwonk

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    isn't that trivial? (take products.)
     
  4. Apr 22, 2008 #3

    mma

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    Yes it's trivial. Sorry for the stupid question and thanks.
     
  5. Apr 24, 2008 #4

    mathwonk

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    sorry for the smartass answer. i have been enjoying slinging around the word "trivial" for a few weeks.

    and is it really trivial? is it obvious the product of a torus and a sphere is not parallelizable?
     
  6. Apr 25, 2008 #5

    mma

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    Yes, it seems to be obvious. If it were parallelizable then there would be 4 continuous vector fields having 4 linearly independent vectors above each (t,s) point of the product. These vectors would span the tangent space of the product space above (t,s). But this tangent space is a pair of tangent spaces: [tex](T_t,T_s)[/tex], [tex]T_t[/tex] is of the torus above t and [tex]T_s[/tex] of the sphere above s. So the 4 vectors would span both [tex]T_t[/tex] and [tex]T_s[/tex]. But this would contradict to the unparallelizable property of the sphere.
     
    Last edited: Apr 25, 2008
  7. Apr 25, 2008 #6

    mathwonk

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    dont you need just two of them to span the tangent space of the sphere at every point, to get that?
     
  8. Apr 26, 2008 #7

    mma

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    I'm afraid that you are kidding. Yes, 2 from [tex]T_s[/tex] is enough to span [tex]T_s[/tex]. Two from [tex]T_p[/tex] don't span [tex]T_s[/tex]. And one from [tex]T_p[/tex] and one from [tex]T_s[/tex] also don't. But 2 from [tex]T_s[/tex] we don't have at each point because of the Hedgehog Theorem.
     
  9. Apr 26, 2008 #8

    mma

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    Sorry, I missed the indexes. p means t (a point of the torus)
     
  10. Apr 26, 2008 #9

    gel

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    Well, the Klein bottle and Mobius strip are not parallelizable, because they aren't orientable. However it is not too hard to find a nowhere vanishing vector field.
     
  11. Apr 26, 2008 #10

    gel

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    Also, Sn has a nowhere vanishing vector field whenever n is odd.
    To construct one, consider Sn as the elements of Rn+1 with norm 1. Let A be an antisymmeric (n+1)x(n+1) non-singular matrix. As n+1 is even, this can be done by building it out of (n+1)/2 copies of the matrix
    [tex]
    \left(
    \begin{array}{rr}
    0 & 1 \\
    -1 & 0
    \end{array}
    \right).
    [/tex]
    Then, the vector field is v = Ax at each point x of Sn.

    However, except for the cases n=1,3,7, Sn is not parallelizable for any odd n (mentioned here, although I expect that it is very difficult to prove).
     
  12. Apr 26, 2008 #11

    gel

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    It is parallelizable.
    torus x sphere = circle x circle x sphere.
    And, the product of a circle and sphere is parallelizable, which you can see by looking at it as a quotient space of R3-{0}
     
  13. Apr 26, 2008 #12

    gel

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    the what?
     
  14. Apr 26, 2008 #13

    mma

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    Hedgehog Theorem = Hairy Ball Theorem.

    By the way, what error do you see in my proof?
     
  15. Apr 29, 2008 #14

    mma

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    Then, finally, is the product of a sphere and a torus parallelizable or not?
     
  16. Apr 29, 2008 #15

    gel

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    Yes. That's what I was saying in my previous post.
     
  17. Apr 30, 2008 #16

    mma

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    Yes, but on the one hand I didn't understand that (I don't know what quotient space do you mean), and on the other hand you didn't show the error in my proof.

    Perhaps I am wrong and the continuous vector fields of a product space aren't the Cartesian products of the continuous vector fields of the component spaces?
     
  18. Apr 30, 2008 #17

    gel

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    That's right, but I didn't follow how that was supposed to imply that the component spaces are parallelizable.
    If you project linearly independent vectors onto a subspace, they no longer have to be linearly independent or even non-zero.
     
  19. May 1, 2008 #18

    mma

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    Maybe it's trivial but I don't know what qoutient space dou you mean. Could you explain this or give a reference for the explanation?
     
  20. May 1, 2008 #19

    gel

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    The circle can be expressed as the real line R quotiented out by the group of translations x |-> x+n (for integer n). That is, you identify x,y in R if x-y is an integer. This expresses R as the covering space of the circle, S1.

    You can map R x S2 to R3-{0} by (x,y) |-> exp(x) y, which takes the group of translations of R to the group x |-> exp(n) x on R3-{0}, for integer n.

    So, S1 x S2 is diffeomorphic to R3-{0} with points x, exp(n)x identified (integer n).

    If ei is a basis for R3 as a vector space, then you have the basis vi(x) = |x| ei for the tangent space of R3-{0} at each point x. This is invariant under the action of the group x |-> exp(n)x, so it maps to a basis for the quotient space S1 x S2.
     
  21. May 2, 2008 #20

    mathwonk

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    the point is that there are a lot of vectors tangent to the product which are not tangent to either factor.


    i.e. consider a family of vectors tangent to the product R^3, and project those vectors onto the sphere. notice that some of these vectors project to zero on the sphere.
     
    Last edited: May 2, 2008
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