Tangent to reparameterized curve

[monea83]

Given is a curve \gamma from \mathbb{R} \rightarrow M for some manifold M. The tangent to \gamma at c is defined as

(\gamma_*c)g = \frac{dg \circ {\gamma}}{du}(c)

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

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

Using the chain rule, I get

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

The second part of the rhs is obviously f', but how is the first part equal to \gamma_* \circ f?

 

[quasar987]

Your notation is really bad. Try this:

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

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

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

<br /> (\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&#039;(t_0)(\gamma_{*,f(t_0)})g<br />

That is to say,

\tau_{*,t_0}=f&#039;(t_0)\cdot \gamma_{*,f(t_0)}

for all t_0.

Or, even more compactly,

\tau_*=f&#039;\cdot \gamma_{*}\circ f

I highly recommend the book Introduction to Smooth Manifolds by John Lee.

 

[monea83]

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)?