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Number of zeros of (tangent) vector field on sphere

  1. Mar 28, 2008 #1
    Is it possible to have a tangent vector field on the unit 2-sphere x^2+y^2+z^2 =1 in
    3D which vanishes at exactly one point? By the Poincare-Hopf index theorem
    the index of such vector field at the point where it vanishes must be 2. Is that possible? If yes, can one write an explicit formula for such vector field.
     
  2. jcsd
  3. Mar 28, 2008 #2

    Hurkyl

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    Yes. Observe that your question is equivalent to "Does there exist a tangent vector field on R^2 that vanishes at infinity?" Does that give you any ideas?
     
  4. Mar 31, 2008 #3
    Yes, thank you for the hint. I see it now. I was able to compute the following explicit formula for the vector field:

    (x^2/(z-1) + 1, x*y/(z-1), x)

    for any (x,y,z) with x^2+y^2+z^2=1 except (0,0,1).
     
  5. Mar 31, 2008 #4

    mathwonk

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    stick one finger in your eye and a thumb in the other and bring thumb and finger together.

    that gives a vector field with a single zero in the middle of your nose, of index 2.
     
  6. Apr 1, 2008 #5
    All that gave me was a headache...
     
  7. Apr 5, 2008 #6

    mathwonk

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    a headache and a vector field of index 2.
     
  8. Apr 7, 2008 #7
    zero only at infinity

    a constant vector field, say parallel to the x axis, must be slowed down near infinity to a single zero.

    I also found it useful to think about the complex form dz on the Riemann sphere.
     
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