Weird Issue with the Chain Rule

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Kreizhn
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Homework Statement



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?

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)
 
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All we know is that f is a smooth function on the manifold.
 
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.
 
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...