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Writing a general curve on a manifold given a metric

  1. Nov 2, 2015 #1
    I have what I think is a basic question. Say I have a manifold and a metric. How do I write down the most general curve for some arbitrary parameter?

    For example in [itex]\mathbb{R}^2[/itex] with the Euclidean metric, I think I should write [itex]\gamma(\lambda) = x(\lambda)\hat{x} + y(\lambda)\hat{y}[/itex]

    But what about the case of the [itex]S^{2}[/itex] parameterized by [itex](\theta,\phi)[/itex] given the metric [itex]\mathrm{diag}(1,\sin^{2}{\theta})[/itex]?

    Is it [itex]\gamma(\lambda) = \theta(\lambda)\hat{\theta} + \phi(\lambda)\hat{\phi}[/itex]
    or something like [itex]\gamma(\lambda) = \theta(\lambda)\hat{\theta} + \sin(\theta)\phi(\lambda)\hat{\phi}[/itex]?
     
  2. jcsd
  3. Nov 2, 2015 #2

    andrewkirk

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    I think you are mixing together a couple of different concepts.
    A curve on a manifold ##M## is a function ##\gamma:U\to M## where ##U## is an interval in ##\mathbb{R}##. For ##\lambda\in\mathbb{R}##, ##\gamma(\lambda)## is not a vector, which is what is implied by the above.

    What is a vector is the velocity ##\dot{\gamma}(\lambda)## of the curve which, given a coordinate system, may be written
    $$\dot{\gamma}(\lambda)=\sum_{\alpha=1}^n \frac{d\gamma^\alpha}{d\lambda}\vec{e}_\alpha$$

    where ##\vec{e}_\alpha## is the coordinate vector for coordinate line ##\alpha##. That coordinate vector will not necessarily be a unit vector. For instance, in polar coordinates, ##\vec{e}_\theta## has modulus ##r##, so we would write the velocity vector as

    $$\dot{\gamma}(\lambda)=\frac{d\gamma^r}{d\lambda}\hat{e}_r
    + \frac{d\gamma^\alpha}{d\lambda}r\hat{e}_\theta$$

    On ##S^2## we would write it as:

    $$\dot{\gamma}(\lambda)=\frac{d\gamma^\theta}{d\lambda}\hat{e}_\theta
    + \frac{d\gamma^\phi}{d\lambda}(\sin\theta)\hat{e}_\phi$$

    So what you wrote second is closer, except that it is for the velocity of the curve, not the curve itself.

    If the coordinate system is orthogonal, the metric tensor will be diagonal in the corresponding basis and the equation for the velocity vector can be written as:

    $$\dot{\gamma}(\lambda)=\sum_{\alpha=1}^n \frac{d\gamma^\alpha}{d\lambda}\hat{e}_\alpha\sqrt{g^{\alpha\alpha}}$$
     
  4. Nov 3, 2015 #3

    WWGD

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    Another thing I think is worth thinking about is whether the velocity vectors associated to the vector live in the tangent space of the manifold. This has to see with covariant differentiation.
     
  5. Nov 3, 2015 #4
    Safe, cheers lads!

    I think my confusion is all cleared up. I'm still in that period of trying to forget what I know about calculus and relearn it in terms of full power differential geometry.
     
  6. Nov 4, 2015 #5
    Sorry, one more clarification actually. If [itex]\gamma(\lambda)[/itex] is not a vector, then what are the [itex]\gamma^{\alpha}[/itex]? Are these just the coordinates? In that case wouldn't we say [itex]\gamma^\alpha = (x,y,z)^\alpha[/itex] (in [itex]\mathbb{R}^3[/itex] for example) This makes it look like [itex]\gamma[/itex] is a vector.
     
  7. Nov 5, 2015 #6

    andrewkirk

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    Yes they are just the coordinates. They are coordinates of a location on a manifold. Only a very few special manifolds, notably the Euclidean spaces, are also vector spaces. So the fact that a point on a manifold has a n-tuple of coordinates does not make that point a vector.

    Of course one can always say that the n-tuple is a vector in ##\mathbb{R}^n##, but that doesn't lead anywhere because the manifold and vector space ##\mathbb{R}^n## has nothing to do with the manifold under discussion.
     
  8. Nov 13, 2015 #7

    lavinia

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    The velocity vector of a curve is a curve in the tangent bundle. Its velocity is therefore in the tangent bundle of the tangent bundle.
     
  9. Nov 13, 2015 #8

    WWGD

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    But can't you also see it as a vector field? Not just by duality (using a choice of Riemannian metric), but assigning to each point its velocity vector? Then it is not necessary to use the tangent bundle of the tangent bundle, but just the tangent bundle of first order. But then again, I think there is a natural identification between T(TM) and TM, isn't there? EDIT: I think this setup between the vector field and the curve may, by itself give the natural/canonical identification between T(TM) and TM.
     
    Last edited: Nov 13, 2015
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