How to Derive Tensor Equations in Physics Notes?

[latentcorpse]

On page 14 in the notes attached in this thread:
https://www.physicsforums.com/showthread.php?t=457609

(i)In the definition given just above eqn 22, it says "In other words, $\phi_\alpha \cdot \lambda$ is a smooth map from $I$ to $\mathbb{R}^n$ for all charts $\phi_\alpha$. Where do they pull this from?

(ii) I cannot see how they derive equation 24.

From equation 23, we have

$X_p(f) = \{ \frac{d}{dt} [ f ( \lambda(t))] \}_{t=0}$
and we can rewrite $f \cdot \lambda = f \cdot \phi^{-1} \cdot \phi \cdot \lambda$ and then I am sure it is just an application of the chain rule but I just cannot see it! In particular why evaluate the 1st bracket at $\phi(p)$ and not $t=0$ and also the $f \cdot \phi^{-1}$ go together to give the $F(x)$ in the 1st brack adn teh $\lambda$ is clearly still in the 2nd bracket, but where has the $\phi$ gone?

Thanks

 

[NateD]

!Answer (i): The definition of $\phi_\alpha \cdot \lambda$ as a smooth map from I to $\mathbb{R}^n$ is pulled from the fact that $\lambda$ is assumed to be a smooth curve in $M$ and that $\phi_\alpha$ is a diffeomorphism from $M$ to $\mathbb{R}^n$. Since $\phi_\alpha$ is a diffeomorphism, it is a bijective, smooth function. As a result, the composition of two smooth functions is also smooth. Answer (ii): To derive equation 24, we start by using the definition of $X_p(f)$ given in equation 23, which states that $X_p(f) = \{ \frac{d}{dt} [ f ( \lambda(t))] \}_{t=0}$,where $\lambda : I \rightarrow M$ is a smooth curve such that $\lambda(0)=p$. Then, since $f \cdot \lambda$ is a composition of two smooth functions, we can apply the chain rule to obtain $X_p(f) = \{ \frac{d}{dt} [ f ( \lambda(t))] \}_{t=0} = \{\frac{d}{dt}[f \cdot \phi^{-1} \cdot \phi \cdot \lambda(t)]\}_{t=0} = \{Df_{\phi(p)} \cdot D(\phi^{-1})_{\phi(p)} \cdot \lambda'(0)\}$,where $F(x) = f \cdot \phi^{-1}$ is the function obtained by composing $f$ and $\phi^{-1}$. Finally, we use the fact that, since $\phi$ is a diffeomorphism, we have $D(\phi^{-1})_{\phi(p)} \cdot \phi'(p) = I$, which yields equation 24.