What's the precise definition of coordinates in Euclidean space?

Question

I have a loose understanding of what coordinates are, but not something rigorous or concrete. For example take the statement of this result below

When the author says let $x = (x^1, \dots x^n)$ denote the "standard coordinates" on $U$ and $y = (y^1, \dots, y^m)$ those on $\widetilde{U}$ what precisely does the author mean?

For example there is a very clear and rigorous definition of what a basis for a vector space is, but I can't seem to find that for coordinates. The closest thing I know to a definition for coordinates is the following:

If I have a smooth manifold say $M$ of dimension $n$ and a smooth chart $(U, \phi)$, then for any $p \in U$ we have $\phi(p) = (x^1(p), \dots, x^n(p))$ where the $x^i : U \to \mathbb{R}$ are the component functions of the homeomorphism $\phi$ and the collection of component functions $(x^1, \dots, x^n)$ are called local coordinates on $M$.

However this notion doesn't seem to apply to the corollary above, because it seems "coordinates" take on a different meaning above, because for example $$\frac{\partial G^i}{\partial y^k}$$ (in the statement of the result above) doesn't make sense since $y^k$ interpreted this way would actually be a function and it's meaningless to take the partial derivative of a function with respect to another function.

Furthermore the only definition of the partial derivative of a function that I'm familiar with is the following

Definition: Let $U \subseteq \mathbb{R}^m$ be an open set and let $f : U \to \mathbb{R}$. The $j$-th partial derivative of $f$ at $a$ is defined to be the directional derivative of $f$ at $a$ with respect to the basis vector $e_j = (0, \dots, 1, \dots, 0)$ (where $1$ is in the $j$-th position) provided the derivative exists $$\frac{\partial f}{\partial x^j} = \lim_{t \to 0} \frac{f(a +te_j) - f(a)}{t}$$

and in this definition above I just think of $x^j$ as a reminder that I'm taking the directional derivative with respect to the $e_j$ basis vector on $\mathbb{R}^m$.

How do I reconcile the definition above with what the authors mean by standard coordinates?

Basically the question I'm asking is, what does

"let $x = (x^1, \dots, x^n)$ denote the standard coordinates on $U$"

mean? Or more generally if $V$ is an open subset of $\mathbb{R}^k$ what would

"let $z = (z^1, \dots, z^k)$ denote the coordinates on $V$"

mean?

Answer 1

About convention/notation:

What does the expression $$\text{Let }(z^1,z^2,\dots,z^n)\text{ denote the coordinates on $V$}$$ mean?

The formalized version of this expression is

If $(V,\varphi)$ is a chart on some $n$-dimensional manifold then let $$(z^1,z^2,\dots,z^n) = (\pi_1\circ\varphi,\pi_2\circ\varphi,\dots,\pi_n\circ\varphi)$$ where $\pi_i$ is the $i^{th}$ projection on $\mathbb{R}^n$. That is, $z^i : V \rightarrow \mathbb{R}$ is defined by $$z^i(p) = \pi_i(\varphi(p))$$ for all $p \in V$.

Note that the expression references, tacitly, some ambient manifold and a chart (which is usually clear from context).

Of course, an equivalent version of this is defined in your post, but I'm stating it here in order to fully address your concern about the Corollary. Also, the above expression is often used where the manifold is smooth (and so is the chart).

About Partial Derivatives:

Firstly,

Partial derivatives are well-defined on any smooth chart $(V,\varphi)$ belonging to a smooth manifold.

Here is how it is defined:

Let $M$ denote an $n$-dimensional smooth manifold, and $(V,\varphi)$ a smooth chart on $M$. Let $(z^1,z^2,\dots,z^n)$ denote the coordinates on $V$. If $f \in C^\infty(V)$, then the $i^{th}$ partial derivative operator $\frac{\partial f}{\partial z^i} : V \rightarrow \mathbb{R}$ defined at $f$ is $$\frac{\partial f}{\partial z^i}(p) = \frac{\partial (f\circ \varphi^{-1})}{\partial x^i}(\varphi(p))$$ for $p \in V$, where the second expression uses the regular $i^{th}$ partial derivative operator in $\mathbb{R}^n$.

I left out a lot of details, as well as the path toward motivation for its definition, for brevity (and focus) of this post. To see these details and why the above is well-defined, see Chapter 3, "Tangent Vectors", of the text (see the section "Computations in Coordinates" in particular).

Note that in the above, you're not "differentiating with respect to" $z^i$ (which is senseless in this context as you've mentioned), you're just using $z^i$ in the notation to reference the chart you're defining the operator with.

This is helpful for your question because in the case of the Corollary, the expression

Let $(x^1,x^2,\dots,x^n)$ denote the standard coordinates on $U \subseteq \mathbb{R}^n$.

means

Let $(x^1,x^2,\dots,x^n)$ denote the coordinates on $U$ (see the section About convention/notation above) where the ambient manifold is $\mathbb{R}^n$ with its standard smooth structure and the smooth chart under consideration is $(U,id_U)$.

so that if $f \in C^\infty(U)$ then using the above definition of partial derivatives we have that for $x \in U$,

$$\frac{\partial f}{\partial x^i}(x) = \frac{\partial (f \circ id_U^{-1})}{\partial x^i}(id_U(x)) = \frac{\partial f}{\partial x^i}(x)$$

That is, in the case where $U$ is given standard coordinates, the operation is the same as the regular partial derivative operator. The $x^i$ on the left corresponds to coordinates on $U$, and the $x^i$ on the right corresponds to the notation used for the definition of the regular partial derivative operator in $\mathbb{R}^n$. For further clarity,

If $(z^1,z^2,\dots,z^n)$ denotes the standard coordinates on $U \subseteq \mathbb{R}^n$, then $$ \frac{\partial f}{\partial z^i}(x) = \frac{\partial (f \circ id_U^{-1})}{\partial x^i}(id_U(x)) = \frac{\partial f}{\partial x^i}(x)$$

Thus, for the case of the notation within the Corollary,

$$\frac{\partial G^i}{\partial y^k}(F(x)) = \frac{\partial (G_i \circ id_\widetilde{U}^{-1})}{\partial x^k}(id_\widetilde{U}(F(x))) = \frac{\partial G_i}{\partial x^k}(F(x))$$

where the $x^k$ on the right corresponds to the notation used for the regular $k^{th}$ partial derivative in $\mathbb{R}^m$.

Caveat:

This is a book about smooth manifolds, so the above partial derivative operator is only defined for cases where $f \in C^\infty(U)$ rather than $f \in C^1(U)$ as required within the Corollary. Of course we know the Corollary to be true from classical analysis, and it can be proved without any direct mention of ideas from the theory of smooth manifolds. The main reason I believe why the author mentioned the notation/convention is to differentiate between the regular partial derivative operator within $\mathbb{R}^n$ and $\mathbb{R}^m$ (since the book only uses the $x^i$ notation to define the regular partial derivative, a scattering of only $x^i$'s would've made the Corollary unpleasant to read) as well as to begin to bring in some of the convention/notation that will be used within the text.

Answer 2

You mention bases in linear algebra - this is very, very similar. When you write $\mathbb{R}^n$ here, you mean an $n$-dimensional $\mathbb{R}$-vector space $V$ with a chosen ordered basis (you could be picky and just think about the set of $n$-tuples of real numbers with the usual addition and scalar multiplication and the ordered basis of tuples like $(1,0,\dots,0)$ that you'd likely choose); let's say $b_1,\dots,b_n \in V$ is this ordered basis. There is a unique ordered basis (the dual basis) in the dual of our vector space $V^*$, which we'll denote by $x^1,\dots,x^n \in V^*$, so that $$x^i(b_j) = \left\{ \begin{array}{ll} 1 & i=j \\ 0 & i\ne j \end{array} \right.$$ for any $1\le i,j\le n$. By definition, these $x^i$ are (linear) functions from our vector space $V$ (though we abuse notation and just write $\mathbb{R}^n$ instead of $V$) to the underlying field $\mathbb{R}$. This is all cooked in when you write $\mathbb{R}^n$. These functions $x^1,\dots,x^n$ are the standard coordinates on $\mathbb{R}^n$.

Now if you take any open subset $U\subset \mathbb{R}^n$, you can just restrict these functions $x^1,\dots,x^n$ to $U$ and call them the standard coordinates on $U$ (even though $U$ is likely no longer a vector space).