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P: 707
 Quote by Sephi 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 : $$\left(\begin{array}{cc}-1 & 0 \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc}\cos 2\theta & \sin2\theta \\ & \\ -\sin2\theta & \cos2\theta\end{array}\right)$$ where $$\theta$$ 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 ?)
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
-trivial.

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.