it is clear (geometrically) that is parallelizable because we can consider a vector field of unit tangent vectors at each point (say, in anti-clockwise direction).
I am having trouble understanding how to write down the above argument mathematically ....we need to come up with a smooth vector field , so that .
what is the choice for a smooth function to have the vector field as above??
can we consider the following (coordinate) vector field ??
Thanks a lot.
....I think we have to be carefull here, we need to be smooth in any chart. will it be the case here?
For example, if we consider then I don't think the field will do the same trick...even though it's coordinate functions are constant=1 in some chart....but it is known that any smoth (or even continuous) field on a 2-sphere has at least one point where the tangent is zero.
The key fact is that can be defined on the whole circle in a canonical way, while I don't know what you would mean by on the sphere. Locally, sure it can be defined, but not globally. That's why the argument doesn't translate to the 2-d sphere.
For an explicit proof, you can say that the unit vector field defined by is continuous.
yes, I meant locally.
now, why defines a smooth vector field on the whole circle?
Also, regarding your last example of a vector field , ...I think I have a problem understanding how a vector field on the ambient space (in this case ) translates to a vector field of a submanifold.
I am thinking of tangent vectors as derivations (or germs, in case of complex/analytic manifolds).
Now, can I argue as follows to show that the above vector field defines a smooth vector field on :
Consider the following (smooth) vector field on defined as .
I want to show that this defines a smooth vector field on . In other words, I need to show that this defines a derivation on .
So, let be a smooth function on some open set , so that .
Therefore (by the chain rule) .
So that is a tangent vector to .
Now, why can I just, instead of the above formal argument, say that this defines a tangent vector to simply because is orthogonal to ?
Also, is there a general result that will guarantee that this restricted smooth vector field (restricted to a submanifold) will be smooth again?
Sorry about too many questions, but I feel that there is a gap in my understanding of the subject...
Thanks a lot!
Hi,
I won't have time for a lengthy answer; I hope the following will be of some interest nevertheless:
- the notation is to be taken with care: the definition of at depends on the choice of a chart at ; for instance refers implicitly to a parametrization of by (or with cos/sin) on the open subset or on . These two charts cover the circle, and the map is the "same" in both charts (we need two charts nevertheless because of the lack of injectivity), so that the derivation , defined locally (on each chart), is in fact global (coherent definition for both charts).
- there is a (nontrivial) equivalence between vector fields (i.e. indefinitely differentiable functions ) and derivations.
- in the case of submanifolds of , the tangent subspace can be thought in various equivalent ways: either it is defined intrinsically, like for a general manifold (no need of the surrounding manifold), or it can be viewed as a subspace of tangent to the submanifold (no wonder it is called "tangent" space). For the circle, as a submanifold of , the tangent space at a point can be identified with the line that is tangent to the circle at this point.
- if is a submanifold of , and if is a vector field of such that, for , , then (restricted to ) is a vector field on .