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Math Help - Circle is parallelizable.

  1. #1
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    Circle is parallelizable.

    it is clear (geometrically) that {\mathbb S}^1 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 X: {\mathbb S}^1 -> T{\mathbb S}^1, so that X(p) = X^1(x) d/dx.

    what is the choice for a smooth function X^1(x) to have the vector field as above??

    can we consider the following (coordinate) vector field X(p) = d/dx??
    Last edited by Different; March 20th 2009 at 05:33 PM.
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  2. #2
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    Quote Originally Posted by Different View Post
    it is clear (geometrically) that ${\mathbb S}^1$ 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 $X: {\mathbb S}^1 -> T{\mathbb S}^1$, so that $X(p) = X^1(x) d/dx$.

    what is the choice for a smooth function $X^1(x)$ to have the vector field as above??

    can we consider the following (coordinate) vector field $X(p) = d/dx$??
    I'd say yes... X^1(x)=1 is non-zero and continuous.

    (note: you should write [ math ]...[ /math ] (without spaces) instead of $$, so that your math can be displayed properly)
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  3. #3
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    Quote Originally Posted by Laurent View Post
    I'd say yes... X^1(x)=1 is non-zero and continuous.

    (note: you should write [ math ]...[ /math ] (without spaces) instead of $$, so that your math can be displayed properly)
    Thanks a lot.

    ....I think we have to be carefull here, we need X^1(x)=1 to be smooth in any chart. will it be the case here?

    For example, if we consider {\mathbb S}^2 then I don't think the field X(p) = d/dx + d/dy 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.
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  4. #4
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    Quote Originally Posted by Different View Post
    Thanks a lot.

    ....I think we have to be carefull here, we need X^1(x)=1 to be smooth in any chart. will it be the case here?

    For example, if we consider {\mathbb S}^2 then I don't think the field X(p) = d/dx + d/dy 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 \frac{d}{dx} can be defined on the whole circle in a canonical way, while I don't know what you would mean by X(p) = d/dx + d/dy 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 X({\cos\theta\choose\sin\theta})={-\sin\theta\choose\cos\theta} is continuous.
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  5. #5
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    Quote Originally Posted by Laurent View Post
    The key fact is that \frac{d}{dx} can be defined on the whole circle in a canonical way, while I don't know what you would mean by X(p) = d/dx + d/dy 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 X({\cos\theta\choose\sin\theta})={-\sin\theta\choose\cos\theta} is continuous.
    yes, I meant locally.

    now, why \frac{d}{dx} defines a smooth vector field on the whole circle?

    Also, regarding your last example of a vector field X({\cos\theta\choose\sin\theta})={-\sin\theta\choose\cos\theta}, ...I think I have a problem understanding how a vector field on the ambient space (in this case  {\mathbb R}^2 ) 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  {\mathbb S}^1 :

    Consider the following (smooth) vector field on  {\mathbb R}^2 defined as  X = -y \frac{d}{dx} + x \frac{d}{dy}.
    I want to show that this defines a smooth vector field on  {\mathbb S}^1 . In other words, I need to show that this defines a derivation on  {\mathbb S}^1 .
    So, let  f: U -> {\mathbb R} be a smooth function on some open set  U \subset {\mathbb S}^1 , so that  f(x , y) = f(\cos\theta , \sin\theta) .

    Therefore  X(f) = -\sin\theta \frac{df}{dx} + \cos\theta \frac{df}{dy} = (by the chain rule)  = \frac{df}{d\theta} .

    So that  X is a tangent vector to  {\mathbb S}^1 .

    Now, why can I just, instead of the above formal argument, say that this defines a tangent vector to  {\mathbb S}^1 simply because  (\cos\theta, \sin\theta) is orthogonal to  (-\sin\theta, \cos\theta)?

    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!
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  6. #6
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    Hi,

    I won't have time for a lengthy answer; I hope the following will be of some interest nevertheless:

    - the notation \tfrac{d}{dx} is to be taken with care: the definition of \tfrac{d}{dx} at x\in M depends on the choice of a chart at x; \tfrac{d}{d\theta} for instance refers implicitly to a parametrization of \mathbb{S}^1(\subset\mathbb{C}) by \theta\mapsto e^{i\theta} (or with cos/sin) on the open subset \mathbb{S}^1\setminus\{1\} or on \mathbb{S}^1\setminus\{-1\}. 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 \tfrac{d}{d\theta}, 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 X:x\in M\mapsto X(x)\in T_xM\subset TM) and derivations.

    - in the case of submanifolds of \mathbb{R}^n, 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 \mathbb{R}^n tangent to the submanifold (no wonder it is called "tangent" space). For the circle, as a submanifold of \mathbb{R}^2, the tangent space at a point can be identified with the line that is tangent to the circle at this point.

    - if N is a submanifold of M, and if X is a vector field of M such that, for x\in N, X(x)\in T_xN, then X (restricted to N) is a vector field on N.
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  7. #7
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    fantastic! everything makes more sense now. thanks a lot.
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