- #1
monea83
- 20
- 0
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
[tex]
(\tau_*c)g = \frac{dg \circ \gamma \circ f}{du}(c) =\frac{dg \circ \gamma \circ f}{df} \cdot \frac{df}{du}(c)
[/tex]
The second part of the rhs is obviously f', but how is the first part equal to [tex]\gamma_* \circ f[/tex]?
[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
[tex]
(\tau_*c)g = \frac{dg \circ \gamma \circ f}{du}(c) =\frac{dg \circ \gamma \circ f}{df} \cdot \frac{df}{du}(c)
[/tex]
The second part of the rhs is obviously f', but how is the first part equal to [tex]\gamma_* \circ f[/tex]?