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Tangent bundle on the sphere

  1. Jan 14, 2006 #1
    Hi everybody,

    How can one show that the tangent bundle TS² of the 2-sphere is not trivial ? I know we can use the tools of algebraic topology, but I'm looking for a way to show it only with elementary tools of differential geometry.

    More precisely, I constructed an atlas for TS² by using the stereographic projection on S². I also computed the expression of the change of coordinates for the tangent vectors in TS², following the equator; it is the application defined by the matrix :

    [tex]\left(\begin{array}{cc}-1 & 0 \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc}\cos 2\theta & \sin2\theta \\ & \\ -\sin2\theta & \cos2\theta\end{array}\right)[/tex]​

    where [tex]\theta[/tex] is the polar angle coordinate of the point on the equator.

    From that point on, how can I show that TS² is non trivial ? (--> How can I show that there is no non-vanishing section of TS² ?)
    Last edited: Jan 14, 2006
  2. jcsd
  3. Jan 14, 2006 #2


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    If you could prove every vector field had a zero, would that mean the tangent bundle was nontrivial?
  4. Jan 15, 2006 #3
    I think it would.

    A section of TS² is a vector field on S². If TS² is trivial, then there exists at least one vector field without a zero. Thus, if we can prove that such vector field does not exist on S², it would mean that TS² is non trivial.

    But I don't know how to proceed, by using the change of coordinates above ...
  5. Jan 15, 2006 #4
    Yes, because this would imply that there is a global basis. Let V be the nowhere zero vector field on S^2. At each point p of S^2, V_p and p are orthogonal, so V_p x p (cross product) is a nowhere zero vector on S^2, orthogonal to V_p. Thus, we would have a global basis for TS^2, which then is trivial.
  6. Jan 15, 2006 #5
    But such vector field doesn't exist on S². It's the statement of the hairy ball theorem : there is no non-vanishing vector field on [tex]S^{\ n}[/tex] if n is even.

    Actually, I'm trying to prove that theorem for n=2, but only with the tools I presented in the first message (and I know it's possible).
  7. Jan 15, 2006 #6
    Oh, I'm not saying that such a v.f. actually exists. Hurkyl's question was: does the existence of a nowhere zero v.f. imply that TS^2 is trivial? That was what I was answering.

    However, in this case, what is important is the fact that if TS^2 is trivial, then there must be a nowhere zero v.f. Thus, if you prove that every v.f. on TS^2 is zero somewhere (i.e. the "You can't comb the hair of a bowling ball" Theorem), then that implies that TS^2 is nontrivial.

    The only proof that I could think to solve this using no topology is to look at the two coordinate systems formed by stereographic projection by the North pole and the South pole. Assume that V is a v.f. nowhere zero with respect to the one of the coordinate systems (say, the NP system) and which is globally defined. Rewrite V in the SP coordinates and show somehow that V needs to be 0 at the SP.

    This shouldn't be hard since both systems can be written in complex coordinates and the transition map from one coordinate system to the other is something like z->1/z (Edit: oops, got my orientation wrong, this should be: z-->1/bar(z)).
    Last edited: Jan 15, 2006
  8. Jan 15, 2006 #7


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    quick answer: if the tangent bundle on the sphere were trivial, then so would the cotangent bundle be trivial.

    but there is a function on the sphere x^2 + y^2 + z^2 = 1 with exactly two critical points, namely the height function z, one a mqx and one a min, and hence a covector field with 2 zeroes, both of "index" 1 (sorry for not defining the index). hence there is a covector field with total index 2.

    since all vector fields have the same index, no vector filed can have index 0, so no vector filed can have no zeroes, so the bundle cannot be trivial.
  9. Jan 15, 2006 #8


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    well i glanced again at your original post, and saw the words "differential geometry" so lets think in terms of the gauss map, i.e. the map that takes a point of a surface in 3 space to its normal vector. if there were an everywhere non zero tangent vector field, then it seems to me one could rotate it to the everywhere non zero normal vector field and thus give a map from the sphere to its anti-pode. sorry i am halucinating, so i will slow down.

    ok. suppose v is a non zero tangent (unit) vector field on the 2 sphere S, i.e. for each x on S, v(x) is a vector perpendicular to x. then at each point x, and for each number t between 0 and pi, consider the map f(t) that takes x to the point K(x) = cos(t)x + sin(t)v(x), also on the unit sphere.

    then for t = 0 this point K(x) = x, and for t = 1 this point K(x) = -x.

    hence we have a smooth homotopy taking the identity map into the antipodal map of the 2 sphere.

    but if we integrate the solid angle form over the sphere we get a non zero number, and that number is invariant under smooth homotopy (by stokes theorem), which violates the fact that the integral of the solid angle form over the antipodal map is minus what it was over the identity map.


    this fails for the 3 sphere since the integral of the angle form changes by (-1)^n+1 by the antipodal map on the n sphere. so it is only on the 2n spheres that it is not homotoipic to the identity map.

    sorry for the fuzzy exploanation, but this was a fun take home exam in my sophomore several variables calculus class back in 1972 at central washington state college, and now you know why all the students except one thought it was too hard.
  10. Mar 3, 2009 #9
    The replies to your question are all correct but they do not follow your attempt to use the
    coordinate transformations to prove directly that the tangent bundle to the 2-sphere is non

    I think your approach is interesting and have tried thinking about it. I think your proof is
    not differential geometric. It is topological and not elementary. I apologize that my
    exposition isn't going to be easy to follow but here goes the idea of it anyway.

    If you look at the sphere as two disks glued together along their bounding circles, then the
    tangent circle bundle, the tangent vectors of length 1,is two solid tori glued together
    along their bounding tori. The way to see this is to slice the sphere along the equator
    and realize that the tangent circle bundle over the two hemispheres is just a disk crossed
    (Cartesian product) with a circle. This can be written down explicitly using your
    stereographic coordinates.

    But a disk crossed with a circle i.e. a circle of disks, is just a solid torus.

    Along the equator you have a circle of circles i.e. a regular torus and the two tori from
    the hemispheres are glued together to make the tangent circle bundle. So you have two solid tori glued together along their boundaries.

    More than one manifold can be created by such gluing of two solid tori. The manifold you get depends on the gluing rule.

    If the tangent bundle were trivial then the circle bundle would be the Cartesian product, S2
    X S1,the 2-sphere crossed with a circle.

    If it were not trivial, then it would be one of the other possible manifolds that can be
    made by gluing two sold tori together such as the sphere in 4 dimensions or the three
    dimensional real projective space.

    So the problem as you have stated it is to show that the coordinate transformations glue the two solid tori together so that you don't get the Cartesian product but rather one of these other manifolds.

    If you visualize a solid torus as a circle of disks, you see that any circle that is the
    edge of one of these disks can be shrunk to a point just by collapsing it onto the center of
    its disk. But the equatorial circle at right angles to these can not be shrunk to a point
    nor can the curves that wind around this circle twice or any other number of times.

    If the tangent circle bundle were the Cartesian product this would still be true after the
    gluing. But if it were the sphere in four dimensions then every circle would be shrinkable
    to a point. If it were the projective 3 -space then the twice the equatorial circle would be
    shrinkable to a point. So figuring out the shrinkability of the equatorial circle after
    gluing is the key.

    So what is your gluing map? After a reflection (this is you reflection matrix), the tangent
    circle is glued to its counterpart by the Jacobian of the coordinate transformation, a
    rotation by twice the angle, theta, as you wrote.

    Under this mapping, a shrinkable disk circle, for instance,the circle formed by the radial
    vector pointing away from the center of the disk,is glued to a curve that winds around the corresponding shrinkable disk circle on the other torus while it simultaneously rotates twice around the other torus's equatorial circle. So after gluing this mixed curve will be shrinkable to a point(since it is glued to a shrinkable curve).

    I'm not sure how to explain this, but this also means that twice the equatorial circle by itself is in fact shrinkable to a point. So you get the real projective 3-space and the tangent
    bundle is not trivial.

    I will try to write this out more explicitly and more clearly. Technically, I think it
    involves Van Kampen's Theorem.

    I don't know how many different manifolds can be created by gluing two solid tori together.
    It would be interesting to think about it.

    By the way, another proof that is pretty mean and also doesn't follow your idea is to notice
    that the group of rotations of 3 dimensional space acts on the tangent circle bundle of the
    2-sphere transitively and without fixed points and so is homeomorphic to it. It is well
    known and fairly easy to prove that the rotation group is homeomorphic to real projective 3
    -space. So the tangent bundle of the 2-sphere is not trivial.
    Last edited: Mar 3, 2009
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