General Relativity: Exchanging a field with its infinitesimal components on metric tensor

Question

On a youtube video about Einstein's field equations, the author writes the following equation (https://youtu.be/foRPKAKZWx8?t=1078):

$$d\phi=\sum_{n} \frac{\partial \phi}{\partial x^n} dx^n\tag{1}$$

Where $d\phi$ is the gradient of a tensor field. After some explanations regarding metrics, he changes equation (1) to (https://youtu.be/foRPKAKZWx8?t=2844):

$$dx^m=\frac{\partial x^m}{\partial y^r}dy^r\tag{2}$$

where the summation over $r$ is implied.

Why this can be done? I don't understand the change from $d\phi$ to $dx^m$.

Answer 1

A exterior derivative (the measure of how much an $k$-form varies along an $n$-dimensional manifold) for an $k$-form $\phi=g dx^A = gdx^i \wedge ...\wedge dx^p$ (where $g$ is a 0-form i.e., a smooth function) with $A$ as an index set defined by $A=\{i_{n}\}_{n=1}^{k}$ with $i_{k}=p$, is written in the following form:

$d\phi=\frac{\partial g}{\partial x_i}dx^i \wedge dx^A$ (note that is a $(k+1)$-form)

e.g. if $\phi$ is an 0-form defined by $\phi=g(x)$ the exterior derivative is

$d\phi=\frac{\partial g(x)}{\partial x_{i}}dx^i$

Applies Einstein's summation convention and you will have the desired result.

Answer 2

In general, they are not related. The total differential of a function of $n$ variables $f(x_1, \ldots, x_n)$ is defined to be

$$df = \sum_{i=1}^n \frac{\partial f}{\partial x_i}dx_i. $$

So in your second case $x(y)$ and so you can apply the same formalism.