Coordinate system vs ordered basis I have an issue with the definition of coordinate system in differential geometry vs the definition of coordinate system in linear algebra. The post is a bit long, but it's necessary so that I get my point across.
Let $V$ be an $n$-dimensional normed space over the reals and equip $V$ with the topology that its norm induces. $V$ can be given a natural smooth structure, making it into a smooth $n$-manifold. Now, let $\{v_1, \dots, v_n\}$ be an ordered basis for $V$. Any $p \in V$ can be written as $p = c^i v_i$, where we are using Einstein notation. This means that $p$ has the coordinate representation $(c^1, \dots, c^n)$, relative to the given basis. This seems to define a coordinate system - not in the usual differential geometric sense, but if $V = \mathbb{R}^n$ then a basis gives us coordinate axes.
However, we also have the usual definition of a coordinate system about $p$: The ordered pair $(U, \varphi)$ is a coordinate system about $p$ if $U \ni p$ and $U$ is open and $\varphi$ is a diffeomorphism onto some open subset of $\mathbb{R}^n$. This allows us to naturally identify $p$ with $(x^1(p), \dots, x^n(p))$ where the $x^i$ are the local coordinates of $\varphi$.
So it seems that the two definitions of coordinate systems above give us the same thing: a way to uniquely identify $p$ with a point of $\mathbb{R}^n$, which is precisely what we want. However, the two definitions are not equivalent. Let me demonstrate:
Given that $V$ is a finite-dimensional vector space, I will make the usual identification of $V$ with $T_p V$ and just write $V$ instead. Similarly, even though $p \in V$, it will also be used interchangeably as both an element of $V$ as well as $T_p V$. Given a coordinate system $(U, \varphi)$, it induces a coordinate basis at $p$, and this is like a coordinate system in the first linear algebraic sense that I described. That's fine. In a differential geometric sense, $p$ is identified with $(x^1(p), \dots, x^n(p))$. In a linear algebraic sense, we can write $p = p^i \frac{\partial}{\partial x^i}$ where $p^i = p(x^i)$. The coordinate representation of $p$, in the linear algebraic sense is then $(p^1, \dots, p^n)$ which is naturally identified with $(x^1(p), \dots, x^n(p))$. So whether we are using the differential geometric or linear algebraic definition of coordinate system, we get the same identification $p \leftrightarrow (x^1(p), \dots, x^n(p))$.
However, the two definitions gave the same identification only because we used a coordinate basis. From what I have previously read (I don't remember the source, but I am sure that you more knowledgeable posters will be aware of this), not every basis for $V$ is a coordinate basis. That is, there could be an ordered basis $\{w_1, \dots, w_n\}$ such that no coordinate chart induces it. This bothers me, because by giving an ordered basis $\{w_1, \dots, w_n\}$, we indeed do have a coordinate system - every element of $V$ has a coordinate representation relative to the basis, BUT, this basis may not necessarily give rise to a coordinate chart. So now we have a coordinate system in one sense (the linear algebraic) but we do not have an equivalent coordinate system in the differential geometric sense. This bothers me a lot!
The differential geometric definition of coordinate system was conceived of for when there is no natural  or useful linear algebraic definition of coordinate system: That is, for when we cannot identify a manifold with its tangent space. But in the case when the manifold is a finite dimensional normed space, we can identify the manifold with its tangent space (for example, $\mathbb{R}^n \leftrightarrow T_p \mathbb{R}^n$), and so in this case, both definitions should be equivalent, i.e. give the same coordinate system, but they do not, as I just demonstrated. How do I reconcile this?
 A: (1) Given any finite-dimensional vector space $\Bbb V$, a choice $(E_a)$ of basis determines a linear isomorphism $\Phi : \Bbb R^n \to \Bbb V$, $n := \dim \Bbb V$, by $$\Phi(x^1, \ldots, x^n) := x^a E_a .$$
The inverse map, $\phi := \Phi^{-1} : \Bbb V \to \Bbb R^n$ defines a preferred global chart on $\Bbb V$ and so realizes $\Bbb V$ as a smooth $n$-manifold. (Pace the claim in the question, this procedure does define coordinates on $\Bbb V$ in the differential-geometric sense.) We might call these charts $\phi$ linear coordinate charts on $\Bbb V$.
In turn, this choice determines a global frame of the tangent bundle, $T\Bbb V$, namely, $(\partial_{x^a})$, which in turn restricts at each point $v \in \Bbb V$ to a basis $(\partial_{x^a}\vert_v)$ of $T_v \Bbb V$. What's special to the case of a vector space is that for each $v \in \Bbb V$ there a canonical identification $\Psi_v : T_v \Bbb V \to \Bbb V$, namely,
$$\Psi_v : v^a \partial_{x^a}\vert_v \mapsto v^a E_a,$$
or just $$\Psi_v : V \mapsto dx^a(V) E_a .$$
Here canonical means that this identification doesn't depend on our choice of basis, it's a natural, built-in feature of $\Bbb V$. Note, by that way, that when $n > 0$, not all coordinate charts on $\Bbb V$ arise from a basis---there is only an $n^2$-parameter family of bases, but uncountably many choices of coordinate charts.
(2) Now, we can see that for any $v \in \Bbb V$, any basis $(F_a)$ of $T_p \Bbb V$ is the restriction of a global frame induced by a choice of basis of $\Bbb V$, namely, $(\Psi_v(F_a))$. More generally, given any smooth manifold $M$, point $p \in M$, and basis $(F_a)$ of $T_p M$, one can construct a smooth chart $(U, \varphi)$ on $M$, $U \ni p$, such that $F_a$ is the restriction of the coordinate frame, that is, such that $T_0 \varphi \cdot F_a = \partial_{x^a}\vert_0$.
On the other hand, not every local frame of $\Bbb V$ is induced by a choice of basis of $\Bbb V$---and indeed, not every local frame on a smooth manifold $M$ is not a coordinate frame! Thus, for our vector space $\Bbb V$ (when $n > 1$) there are proper inclusions
$$
\begin{align}
&\{\textrm{coordinate frames of linear coordinate charts on $\Bbb V$}\} \\
&\qquad\qquad \subsetneq
\{\textrm{(local) coordinate frames on $\Bbb V$}\} \\
&\qquad\qquad\qquad\qquad \subsetneq
\{\textrm{(local) frames on $\Bbb V$}\} \\
\end{align}
$$
One way to see that the latter inclusion holds (for all manifolds of dimension $n > 1$) is to define the Lie bracket operation $[\,\cdot\, , \,\cdot\,] : \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)$. In coordinates it is given by
$$[X, Y] := (X^b \partial_{x^b} Y^a - Y^b \partial_{x^b} X^a) \partial_{x^a} ,$$
but computing the transformation of $[X, Y]$ under a general change of coordinates shows that it does not depend on a choice of coordinates. For any coordinate chart $(U, \phi)$, the above formula gives that the coordinate frame $(\partial_{x^a})$ satisfies
$[\partial_{x^a}, \partial_{x^b}] = 0$ for all $a, b$. On the other hand, most local frames $(E_a)$ do not satisfy $[E_a, E_b] = 0$ and so cannot be the coordinate frames of any coordinate chart.
A: Every choice of a basis $v_1,\ldots,v_n$ to the vector space $V$ induces a linear isomorphism $V\to\mathbb{R}^n$, which is in particular a coordinate chart. On the other hand, not every coordinate chart is a linear isomorphism. In other words, we have an inclusion $$\mathcal{L}\subset\mathcal{C},$$ where $\mathcal{L}$ denotes the space of linear coordinate charts on $V$, and $\mathcal{C}$ denotes the space of not-necessarily-linear coordinate charts on $V$.
Using the language of the post, a coordinate chart in $\mathcal{L}$ preserves the identification $V\cong T_pV$, whereas a general coordinate chart in $\mathcal{C}$ does not.
