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Thread: Circle is parallelizable.

  1. #1
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    Circle is parallelizable.

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

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

    can we consider the following (coordinate) vector field $\displaystyle X(p) = d/dx$??
    Last edited by Different; Mar 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... $\displaystyle X^1(x)=1$ 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... $\displaystyle X^1(x)=1$ 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 $\displaystyle X^1(x)=1$ to be smooth in any chart. will it be the case here?

    For example, if we consider $\displaystyle {\mathbb S}^2$ then I don't think the field $\displaystyle X(p) = d/dx + d/dy$ 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 $\displaystyle X^1(x)=1$ to be smooth in any chart. will it be the case here?

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

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

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

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

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

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

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

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

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

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