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Tangent to reparameterized curve

  1. Sep 5, 2010 #1
    Given is a curve [tex]\gamma[/tex] from [tex]\mathbb{R} \rightarrow M[/tex] for some manifold M. The tangent to [tex]\gamma[/tex] at [tex]c[/tex] is defined as

    [tex](\gamma_*c)g = \frac{dg \circ {\gamma}}{du}(c)[/tex]

    Now, the curve is to be reparameterized so that [tex]\tau = \gamma \circ f[/tex], with f defining the reparametrization. (f' > 0 everywhere)

    The book I'm reading claims that [tex]\tau_* = f' \cdot \gamma_* \circ f[/tex], however I do not quite see how this result is derived.

    Using the chain rule, I get

    (\tau_*c)g = \frac{dg \circ \gamma \circ f}{du}(c) =\frac{dg \circ \gamma \circ f}{df} \cdot \frac{df}{du}(c)

    The second part of the rhs is obviously f', but how is the first part equal to [tex]\gamma_* \circ f[/tex]?
  2. jcsd
  3. Sep 5, 2010 #2


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    Your notation is really bad. Try this:

    The tangent to [tex]\gamma[/tex] at [tex]\gamma(u_0)[/tex] is the tangent vector [tex]\gamma_{*,u_0}[/tex] defined by

    (\gamma_{*,u_0})g := \frac{d(g \circ {\gamma})}{du}(u_0)

    So, for [tex]\tau:= \gamma \circ f[/tex] a reparametrization, the chain rule yields

    (\tau_{*,t_0})g = \frac{d(g \circ \gamma \circ f)}{dt}(t_0) =\frac{d(g \circ \gamma)}{du}(f(t_0)) \frac{df}{dt}(t_0)=f'(t_0)(\gamma_{*,f(t_0)})g

    That is to say,

    [tex]\tau_{*,t_0}=f'(t_0)\cdot \gamma_{*,f(t_0)}[/tex]

    for all t_0.

    Or, even more compactly,

    [tex]\tau_*=f'\cdot \gamma_{*}\circ f[/tex]

    I highly recommend the book Introduction to Smooth Manifolds by John Lee.
    Last edited: Sep 5, 2010
  4. Sep 5, 2010 #3
    Thank you very much, I think I can work it out now!

    When you say my notation is bad, are you referring to my application of the chain rule (which I believe is flawed), or to the notation in which the problem was posed (which was taken from "Tensor analysis on manifolds", Bishop & Goldberg)?
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