# I Coordinate system vs ordered basis

1. May 27, 2017

### ConfusedMonkey

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?

2. May 27, 2017

### Staff: Mentor

To be honest, I don't get your point. You have $\mathbb{R}^n$ as manifold, which you call $V$, $\mathbb{R}^n$ as tangent space $T_p(V)$ and $\mathbb{R}^n$ as Euclidean space where the charts are from. So all of your basis are those of $\mathbb{R}^n$ and you get different representations: $(c^1,\ldots , c^n)\, , \,(x^1(p),\ldots ,x^n(p))\, , \,(p^1,\ldots ,p^n)$ and whatever the coordinates according to $\{w_1,\ldots ,w_n\}$ are. The basis $\{\frac{\partial}{\partial x^1},\ldots ,\frac{\partial}{\partial x^n}\}$ is a orthogonal basis, the others any. Fine until here. Then you write
What should this mean? What is a coordinate basis? And "I have previously read" is a miserable source.
What does "induces it" mean? If you have a basis of a vector space, then the coordinates are the components according to this basis. If you have a manifold, then basis means either one of the tangent space or one of the chart map. The rest are basis transformations, as you always have $\mathbb{R}^n$ everywhere in your example.

The only interesting point is, that the basis vectors of the tangent space above are the directional derivatives of the standard basis in $\mathbb{R}^n$ which makes the notation a bit different than usual.
A basis gives rise to a chart map? A chart maps a part of the manifold on a part of a Euclidean space which has a basis. What did you mean by the first basis you mentioned? If you have a different basis in the tangent space than for your charts, you also have a basis transformation. So what? If you have a basis in your manifold, which only makes sense in an example as the one above, where the manifold itself is a Euclidean space, then you can have again a different basis than the one for the charts. However, since all vector spaces here: $V,T_p(V),\mathbb{R}^n$ are the same, I would chose only one basis for all of them: $(x^i) \textrm{ in }V \, , \,(\frac{\partial}{\partial x^i}) \textrm{ in }T_p(V)\, , \,(\mathfrak{e}^i)\textrm{ in }\mathbb{R}^n$ are noted differently but are basically all the same.

3. May 27, 2017

### ConfusedMonkey

A coordinate basis is a basis for the tangent space which is induced by the coordinate chart. For example, if $(U, \varphi)$ is a coordinate chart for a smooth manifold $M$ containing a point $p$, where $\varphi = (x^1, \dots, x^n)$, then the coordinate basis for $T_p M$ induced by this chart is $\{\frac{\partial}{\partial x^i}, \dots, \frac{\partial}{\partial x^n}\}$.

Here is a source. See post #6: https://www.physicsforums.com/threads/classifications-of-basis-vectors.785676/

Also, I asked the same question here: https://math.stackexchange.com/questions/2299345/coordinate-system-vs-ordered-basis
and was told: "At a single, fixed point, any basis comes from a coordinate chart. But, yes, if you look at a (smooth) choice of basis throughout an open set, it'll only come from a coordinate chart if the pairwise Lie brackets $[w_i,w_j]$ all vanish."

Anyway, my initial post may have been too long winded, obscuring my point. Let me try again.

If we are in a finite dimensional vector space, then any ordered basis $\{v_1, \dots, v_n\}$ is a coordinate system, because for any $p \in V$ we can write $p = c^i v_i$ and so the coordinate representation of $p$ is $(c^1, \dots, c^n)$.

Now suppose we are in a general topological space. In general, it isn't a vector space and so we cannot just impose a coordinate system on the manifold by using a basis. However, what we can do is equip the topological space with local coordinate charts, making it into a manifold.

So we have two definitions of coordinate systems. One that uses basis vectors, and one that uses coordinate charts. But, there are cases, such as $\mathbb{R}^n$ where we could use both definitions of "coordinate systems". We could use basis vectors to define a coordinate system or we could use smooth coordinate charts. However, the two definitions are not equivalent because while every coordinate chart gives a basis for $\mathbb{R}^n$ (we are still using the identification $\mathbb{R}^n \leftrightarrow T_p \mathbb{R}^n$), not every basis for $\mathbb{R}^n$ has a corresponding coordinate chart. So if I wanted a coordinate chart on $\mathbb{R}^n$, which definition do I use? That is what I am confused about - the fact that there are slightly differing notions for "coordinate system" in the case of $\mathbb{R}^n$, or in the slightly more general case where the manifold is a finite-dimensional normed space.

Last edited: May 27, 2017
4. May 28, 2017

### Orodruin

Staff Emeritus
You seem to have several confusions. Let us try to clear them up.
An ordered basis is not a coordinate system. However, it defines a coordinate system on the vector space where the coordinates are taken to be the components of a vector with respect to the ordered basis. Note that this will give you an affine coordinate system, which is only a subset of the possible coordinate systems. There is no reason to expect an arbitrary coordinate system to be affine.

You need a topological space that is locally homeomorphic to $\mathbb R^n$. Just any topological space will not do.

As I said earlier, your first definition defines an affine coordinate system, it is not the most general type of coordinate system.

You are contradicting yourself here. Every vector basis in a vector space does give a corresponding affine coordinate system. Note that the set of basis vectors for a (non-affine) coordinate system changes from point to point. For example, in spherical coordinates the position vector is $r\vec e_r$ everywhere because $\vec e_r$ depends on the position.

There is only one definition of a coordinate chart, it needs to locally uniquely specify a point in the manifold. An affine coordinate system is a type of coordinate system (which includes the subset of Cartesian coordinates when the basis is orthonormal), not all coordinate systems are affine.

5. May 28, 2017

### ConfusedMonkey

I guess I will just have to live with the fact that not every ordered basis for Euclidean space will be a coordinate basis, and whether I use a linear algebraic or differential geometric coordinate system will depend on the case at hand.

EDIT: I posted this at the same time you replied, Orodruin. I will read your post then reply to it some time tomorrow as it is very late where I am.

6. May 28, 2017

### davidge

I'm not sure whether it's appropriate entering into this discussion, but I'm following this thread since the opener started it. I have a question.
What you mean here? e.g. in $\mathbb{R}^3$, take a point $p$ with coordinates $(x,y,z)$; if a vector has components $(x,y,z)$, then the ordered basis $(1,0,0), (0,1,0),(0,0,1)$ defines an affine coordinate system at $p$?

7. May 28, 2017

### Orodruin

Staff Emeritus
Given an ordered basis (it does not need to be orthonormal) $\vec w_i$ in a vector space, you can use the $n$-tuple of numbers $\zeta^i$ as coordinates such that an element in the vector space is written $\vec v = \zeta^i w_i$. The basis itself is a basis, not a coordinate system.

8. May 28, 2017

### ConfusedMonkey

@Orodruin

Yes, a basis is a basis and only defines a coordinate system, I was speaking loosely, which is not a good thing in mathematics. I have realized where I was going wrong. A basis $\{v_i\}$ for $V$ determines a smooth global chart, $(V, \varphi)$ on $V$ and this chart determines a global frame of coordinate vector fields on $TV$. However, the key is that not every frame will be determined by a coordinate chart.