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Weird Issue with the Chain Rule

  1. Apr 17, 2008 #1
    1. The problem statement, all variables and given/known data

    I'm working on a quick problem regarding a presentation that I'm giving, but I've come across an issue that I can't seem to resolve. Namely

    [tex] \displaystyle \left. \frac{d}{dt} \right|_{t=0} f(\phi^p (t+t_0) ) = \left( \phi^p \right) ^\prime (t_0) f [/tex]

    Does anybody see how this is true?

    3. The attempt at a solution

    [tex] \displaystyle \left. \frac{d}{dt} \right|_{t=0} f(\phi^p (t+t_0) ) = f^\prime(\phi^p(t_0)) \left(\phi^p \right)^\prime (t_0) [/tex]

    All we know about f is that it is a smooth function and [itex] t_0 [/itex] was arbitrarily chosen, so I'm not seeing where we make the jump. (Note: [itex] \phi^p(t) [/itex] is a dynamical system flow on a smooth manifold, but I don't see how that should help)
  2. jcsd
  3. Apr 17, 2008 #2


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    Is f some form of exponential, so f'(x)=f(x)?
  4. Apr 17, 2008 #3
    All we know is that f is a smooth function on the manifold.
  5. Apr 17, 2008 #4


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    That first equation looks incomplete to me. There is no argument to f.
  6. Apr 17, 2008 #5
    f is defined as a function on the manifold so the argument is generally omitted; otherwise, we would need to introduce an atlas on the manifold and define a local coordinate system using a coordinate chart. I don't think it's really necessary to the argument. Furthermore, [itex] (\phi^p)^\prime [/itex] is a locally defined vector field, and [itex] \phi^p [/itex] being a flow would imply that vector field is smooth. Thus it acts as an operator on f, in which case the argument for f is again not necessary.
  7. Apr 18, 2008 #6
    I think the answer was staring me in the face the whole time.

    I probably should have defined flows, it would've made this a bit more obvious.

    The flow maps the Cartesian product of a monoid and a manifold to a manifold. But [itex]\frac{d}{dt} [/itex] is a derivative that holds only with respect to the monoid. The way that that is written isn't implying that [itex] \phi^ [/itex] is in the argument of f, it's implying that this is indeed the product of two smooth functions, one of which is completely independent of t!

    I think that must be the answer, but I really blame horrible and ambiguous notation...
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