Thread: Tangent bundle on the sphere View Single Post
 Sci Advisor HW Helper P: 9,453 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. done. 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.