the differential of a function $f\in C^{\infty}(M)$: two definitions

Question

Let $M$ be a smooth $n$-dimensional manifold (on $\mathbb R$). If $p\in M$, we have that $\Big\{\frac{\partial}{\partial x^1}\Big|_p,\ldots,\frac{\partial}{\partial x^n}\Big|_p\Big\}$ is a basis for $T_pM$ while $\big\{ dx_1|_p,\ldots ,dx_n|_p \big\}$ is its dual basis for $T^\ast_pM$. Now if $f\in C^{\infty}(M)$ we can recover two definition for the differential of $f$:

1. ${df}_p: T_pM\longrightarrow T_{f(p)}\mathbb R\cong\mathbb R$ is a linear map such that, locally, if the representation of $f$ as function from $\mathbb R^n$ in $\mathbb R$ is $\widehat f$, then the matrix of $df_p$ is $\left(\frac{\partial{\widehat f}(\widehat p)}{\partial x^1},\ldots,\frac{\partial{\widehat f}(\widehat p)}{\partial x^n}\right)$. Clearly $\widehat p$ is the image of $p$ in $\mathbb R^n$ using the charts.

2. $df$ is a differential $1$-form $df_p:T_pM\longrightarrow\mathbb R$ such that, calculating in local coordinates as below, we have that $df_p=\frac{\partial{\widehat f}(\widehat p)}{\partial x^i} dx_i|_p$ (there is a summation over the index $i$).

Clearly in both cases $df_p$ is an element of $T^\ast_pM$, but in which way are conceptually connected the two definitions? Formally we start from two radically different approaches but we arrive at the same concept!

Thanks in advance.

Answer 1

I find both definitions equally bad!
The differential of the function $f\in C^\infty (M)$ at $p$ definitely should not depend on the choice of a system of coordinates $(x^1,\cdots,x^n)$ around $p$.
The correct definition is intrinsic, but of course depends on the choice of the definition of "tangent vector".

A) If one decides that a tangent vector $v\in T_pM$ is a derivation $v:C^\infty (M)\to \mathbb R$, then one must define $df_p(v)=v(f)$.
Similarly, if one decides (and it is a better decision) that $v:C^\infty_{M,p} \to \mathbb R$ is a derivation on the germs of smooth functions, then $df_p(v)=v(\text {germ}_pf)$.

B) If one decides that a tangent vector $v\in T_pM$ is given by the equivalence class of a smooth curve $\gamma:I\to M$ with $I$ some real interval around zero and $\gamma(0)=p$ then, if $v$ is the equivalence class of $\gamma$, one defines $df_p(v)=(f\circ \gamma)'(0)$

C) Similarly, every definition of $v\in T_pM$ will lead to a canonical definition of $df_p(v)$.
Coordinates should not be used in the definition of $df_p$, but only for practical calculations.

Edit: application to original question
1. Once you have chosen a basis $v_1,\cdots ,v_n$ of a vector space $V$ the matrix of a linear form $\phi:V\to \mathbb R$ is $(\phi(v_1),\cdots, \phi(v_n))=(\cdots,\phi(v_j),\cdots)$
So in your context the $1\times n$ matrix representing $df_p$ is $(\cdots,df_p(\frac{\partial}{\partial x^j}),\cdots)=(\cdots,\frac{\partial{\widehat f}(\widehat p)}{\partial x^j},\cdots)$

2.If $(v_i^*)$ is the dual basis in $V^*$ to $(v_i)$, we have the formula for an arbitrary linear form $\phi$: $$\phi=\sum \phi(v_i)v_i^*$$ This explains your formula $df_p=\frac{\partial{\widehat f}(\widehat p)}{\partial x^i} dx^i|_p$

The key point in all this is the formula $$ df_p(\frac{\partial}{\partial x^j})= \frac{\partial{\widehat f}(\widehat p)}{\partial x^j}$$ following from A) .

Answer 2

Think about what $dx_i$ does; if $V = V^j \frac{\partial}{\partial x^j}$ (i.e. $V = (V^1,V^2,\dots,V^n)^T$ in local coordinates), then $$ dx_i(V) = V^i $$ right?

So, how would you write $dx_i$ as a row vector in these coordinates? It would be $$ dx_i = (0,\dots,0,1,0,\dots,0) $$ where the $1$ is in the $i$th position, so $$ (0,\dots,0,1,0,\dots,0)^T\left(\begin{array}{c}V^1 \\ V^2 \\ \vdots \\ V^n \end{array}\right) = V^i. $$

From this point of view, it is straightforward to see that

$$ \frac{\partial \widehat f(\widehat p)}{\partial x^i}dx_i $$ and $$ \left(\frac{\partial \widehat f(\widehat p)}{\partial x^1},\dots,\frac{\partial \widehat f(\widehat p)}{\partial x^n}\right) $$ are the same thing.