High School Reconciling basis vector operators with partial derivative operators

[Shirish]

Ref. 'Core Principles of Special and General Relativity' by Luscombe. Apologies in advance for the super-long question, but it's necessary to show my thought process.

Let $\gamma:I\to M$ be a smooth curve from an open interval $I\subset\mathbb{R}$ to a manifold $M$, and let $f:M\to\mathbb{R}$ be smooth. We denote the generic argument of $\gamma$ by $u$, so $\gamma(u)\in M$ for any $u\in I$. The set of all smooth functions from $M\to\mathbb{R}$ at $p$ is denoted by $\mathcal{F}_p$ (so that $f\in\mathcal{F}_p$). The definition of tangent space elements is given as:

The tangent space at $p\in M$, the set of all directional derivatives at $p$, is denoted by $T_p(M)$. For fixed $\gamma$, the directional derivative $\mathbf{t}_{\gamma}(f)\equiv d(f\circ\gamma)/du\ |_p$ is an operator that maps functions $f\in\mathcal{F}_p$ onto numbers, $\mathbf{t}_{\gamma}:\mathcal{F}_p\to\mathbb{R}$.

The directional derivative on $M$ was defined indirectly via a curve $\gamma$ as follows:
[/QUOTE]Let $p\in M$ be such that $\gamma(c)=p$ for $c\in I\subset\mathbb{R}$. For a smooth function $f$ on $M$, the composite map $f\circ\gamma:\mathbb{R}\to\mathbb{R}$ is a function on $\mathbb{R}$. The directional derivative on $M$,$$\frac{d(f\circ\gamma)}{du}\bigg|_c\equiv\lim_{s\to 0}\frac{1}{s}[(f\circ\gamma)(c+s)-(f\circ\gamma)(c)]$$
[/QUOTE]
Now to define the basis of $T_p(M)$, we consider coordinate curves $\gamma^i$ passing through $p\in M$. Let $p$, with chart $(U,\phi)$, have the coordinates $(x^1(p),\ldots,x^n(p))$. The coordinate curves are specified by $\phi^{-1}:\mathbb{R}\to M$ as follows:$$\gamma^i(u)\equiv \phi^{-1}(x^1(p),\ldots,x^{i-1}(p),x^i(p)+u,x^{i+1}(p),\ldots,x^n(p))$$
Then it's written:

The directional derivative associated with $\gamma^i$ at $p$ defines the partial derivative of $f$ with respect to $x^i$, $$\frac{d(f\circ\gamma^i)}{du}\ \bigg|_{u=0}=\frac{\partial f}{\partial x^i}\ \bigg|_p$$

And this last part is pretty confusing. I evaluated the LHS as follows:$$\lim_{s\to 0}\frac{1}{s}[(f\circ\phi^{-1})(x^1(p),\ldots,x^i(p)+s,\ldots,x^n(p))-(f\circ\phi^{-1})(x^1(p),\ldots,x^i(p),\ldots,x^n(p))]$$ $$=\frac{\partial (f\circ\phi^{-1})}{\partial x^i}\ \bigg|_{\phi(p)}$$
which is not the same as the RHS in the quoted equation, so I don't know where that came from.

Another doubt is how do I reconcile the basis element of $T_p(M)$ with the differential operator $\partial/\partial x^i$? As per the definition $\mathbf{t}_{\gamma}(f)\equiv d(f\circ\gamma)/du\ |_p$, which means $$\mathbf{t}_{\gamma^i}(f)= d(f\circ\gamma^i)/du\ |_p = d(f\circ\gamma^i)/du\ |_{u=0}$$
I showed above that $d(f\circ\gamma^i)/du\ |_{u=0}$ is equal to $\partial (f\circ\phi^{-1})/\partial x^i\ |_{\phi(p)}$. So finally $$\mathbf{t}_{\gamma^i}(f)=\partial (f\circ\phi^{-1})/\partial x^i\ |_{\phi(p)}$$
I'm pretty sure (at least intuitively) that $\{\mathbf{t}_{\gamma^i}\}$ (the subscript reads $\gamma^i$) forms the basis of $T_p(M)$. I can only claim that $\mathbf{t}_{\gamma^i}=\partial/\partial x^i$ when $\mathbf{t}_{\gamma^i}(f)=\partial f/\partial x^i$ for all $f\in\mathcal{F}_p$. From the above equation, I have no way to isolate $f$ on the RHS, so with what sorcery can I show that $\mathbf{t}_{\gamma^i}(f)=\partial f/\partial x^i$ for all $f\in\mathcal{F}_p$?

 

[Infrared]

$$=\frac{\partial (f\circ\phi^{-1})}{\partial x^i}\ \bigg|_{\phi(p)}$$
which is not the same as the RHS in the quoted equation, so I don't know where that came from.

What is your definition of $\frac{\partial f}{\partial x^i}$ if not the above?

 

[Shirish]

What is your definition of ∂f∂xi if not the above?

Sorry, but I'm not sure that I follow - I'm completely new to the subject. Do you mean there's some notation convention that I'm not catching on to? Would be grateful if you could explain in detail because what's obvious to you (being well-versed in the subject) may not be obvious to me at all.

 

[Infrared]

In order for the statement you quoted to be true or false, you must have defined the expression $\frac{\partial f}{\partial x^i}.$ What is the definition of $\frac{\partial f}{\partial x^i}$ that you're using?

I've usually seen $\frac{\partial f}{\partial x^i}$ $\bf{defined}$ to be $\frac{\partial (f\circ \phi^{-1})}{\partial x^i}$ (and the RHS makes sense because $f\circ\phi^{-1}$ is a function defined on $\mathbb{R}^n$, where we already know how to take partial derivatives), in which case there's nothing left to prove. If you have a different definition of what $\frac{\partial f}{\partial x^i}$ means for a smooth function $f:M\to\mathbb{R}$, then you should let us know what it is.

Edit: Okay, it looks like your source is using the LHS as the definition of the partial derivative of a function on a manifold. Can you explain why you think this is inconsistent with $\frac{\partial f}{\partial x^i}=\frac{\partial (f\circ \phi^{-1})}{\partial x^i}$?

 

[Shirish]

Edit: Okay, it looks like your source is using the LHS as the definition of the partial derivative of a function on a manifold. Can you explain why you think this is inconsistent with $\frac{\partial f}{\partial x^i}=\frac{\partial (f\circ \phi^{-1})}{\partial x^i}$?

I'm not sure about how it'd be inconsistent. But what I did understand from your answer is that since we can't directly calculate the derivative of $f$ (because of its domain), we use some sort of a "proxy function". For example, if we want to differentiate $f$ w.r.t. some parameter $u$ of some curve, we can't directly do it but have to instead calculate it via the derivative of $f\circ\gamma$ w.r.t. $u$, since the domain of $f\circ\gamma$ is an open set in $\mathbb{R}$.

Similarly $\partial f/\partial x^i$ can't be directly calculated because $x^i$'s lie in $\mathbb{R}^n$ while $f$ is defined on $M$. So we again calculate it as the derivative of $f\circ\phi^{-1}$ w.r.t. $x^i$, since the domain of $f\circ\phi^{-1}$ is an open set in $\mathbb{R}^n$.

So it's natural to define derivatives of $f$ implicitly via their "proxy functions". Hope most of my intuition is correct.