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I recently realized that I was confused by the coordinates of a vector and the coordinates on a manifold. It probably happened because I think of manifolds as subsets of $\mathbb R^n$ and most of the time for convenience $n=3$. Now that I realized that the same word is used to mean two different things: one is the coefficients with respect to a basis and one is a continuous function. I am starting to wonder whether there is any relation between the two. Is there? Or should I start calling the manifold case local coordinates or chart to distinguish the two?

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up vote 4 down vote accepted

The two notions use the same word on purpose, because a manifold looks locally like Euclidean space, and therefore points on a manifold can be described with coordinates of a vector.

The idea is that for any point on a manifold, there is some open set containing the point that is diffeomorphic to an open set of some Euclidean space. We can then translate the coordinates of the open set in Euclidean space to the neighborhood on the manifold via this diffeomorphism.

An example should be helpful. Consider $S^2$, and let $p$ be the point $(1,0,0)$. Now, consider the open set $U\subset S^2$ given by the equation $x>0$. We can define a diffeomorphism as follows:

$$\begin{array}{rcl}(0,\pi)\times(-\pi/2,\pi/2)&\longrightarrow &U\\(\theta,\phi)&\mapsto &(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)\end{array}$$

This diffeomorphism in a sense has taken the 'coordinate patch' $(0,\pi)\times(-\pi/2,\pi/2)$, and glued it onto the sphere, giving the sphere gridlines just like the patch has in $\mathbb{R}^2$. The vertical and horizontal lines on the patch in $\mathbb{R}^2$ correspond to lines of longitude and latitude on $U$. Under this diffeomorphism (also called a 'chart'), our point $p$ has the coordinates $(\pi/2,0)$.

In general, think of the local coordinates of a manifold as a choice of diffeomorphism of neighborhoods, so that the manifold inherits the coordinate axes from Euclidean space.

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Can one think of it like this: in linear spaces the coordinates are numbers expressing a vector with respect to a basis, on manifolds the coordinates are numbers expressing a vector locally with respect to a chart? – student Jul 29 '13 at 9:26
@user84127: I would say "coordinates are numbers expressing a point locally with respect to a chart." Other than that, it seems like you have the right idea. – Jared Jul 29 '13 at 17:21
Thank you! And thank you for this great answer! – student Jul 30 '13 at 5:59
I mean, I'm not sure my previous comment was very clear, just in case it wasn't: I thought that if $U$ is some open set in an $n$-manifold and $\varphi$ is the corresponding chart map then $U$ would be a subset of $\mathbb R^n$ and $\varphi $ would also be a map into $\mathbb R^n$: $$\varphi : U \subseteq \mathbb R^n \to \mathbb R^n$$ – self-learner May 17 at 6:07
@self-learner the 2-sphere $S^2$ is identified with or even defined as an embedded submanifold like $\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\}$; that (global) coordinate map given by Jared might have its arrow written the 'wrong' way to you but recall that the map is expected to be invertible since a diffeomorphism is an isomorphism. the coordinates that we're endowing $S^2$ with indeed comprise a pair like you expected--that's what the $(\theta,\phi)$ are – oldrinb Jul 7 at 19:12

Two different sets of coordinates are defined, speaking about manifolds.

Given a $d$-dimensional manifold $M$, to every point $p\in M$ is attached a tangent space $T_pM$, whose dimension is again $d$. A (tangent) vector on the manifold is just an element of one of these tangent spaces.

Now, local charts and coordinates parametrize the manifold and allow you to label the points in some way of your choice. Let $x^1,\dots,x^d$ be local coordinates on an open set $U\subset M$ which comprises $p$: the local coordinate expression of $p$ will be $\varphi(p)=(x^1_{(p)},\dots,x^d_{(p)})$, where $\varphi\colon M\to\mathbb R^d$ is the diffeomorphism of the local chart.
These local coordinates induce a basis of tangent vectors $\partial_1|_p,\dots,\partial_d|_p$ in the point $p$. Every tangent vector in $p$ may be expressed through its coordinates w.r.t. this basis: $$v\in T_pM \longleftrightarrow v=v^i\partial_i|_p$$ Obviously you may decide to use a different basis of tangent vectors, obtainable from the first via a multiplication by an invertible matrix. Then, also the coordinates of the tangent vector will change, but in the opposite way.

When you speak about $M=\mathbb R^3$, confusion may arise, because the tangent space to a vector space is isomorphic to the vector space: $T_p\mathbb R^3=\mathbb R^3$.
When you say - for example - that the electric field at a point in space has a certain coordinate expression, you are really speaking about its coordinates as a vector of the tangent space; these coordinates may then depend on $x,y,z$, or $r,\theta,\phi$, which are coordinates of the local charts used to label points of the manifold.

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Basically, coordinates are ultimately just a set of numbers which identify a certain element of a set. If our set has additional structure, we of course want to select those numbers in a way that you can see this number directly on those numbers.

For example, in a vector space you've got linearity (indeed, linearity is what defines vector spaces; that's why they are also sometimes called linear spaces). Therefore we want to use coordinates which reflect that linearity. That is, if you do linear combinations of vectors (linearity basically means that you can do linear combinations), we want those to map to linear combinations of the numbers in our coordinates. Now from this it follows that your coordinate space is itself a vector space, namely the vector space $\mathbb R^n$. And the standard basis of that basis maps to some basis of your vector space.

In a Riemannian manifold, you care about the differentiable structure, therefore you want your coordinates to represent that. That is, whenever you've got smooth functions on you manifold, you also want to get smooth functions of your coordinates. That is, you want to have a smooth map from the manifold $\mathbb R^n$ (repectively an open subset of that manifold) to the manifold in question. Now there's the complication that unlike in the case of vector spaces you generally cannot do a single such map, therefore you've got to use several such maps.

Of course if you've got a manifold, you've got tangential spaces which themselves are vector spaces. Therefore you need the manifold coordinates (which identify the point, and therefore also the corresponding tangential space) and in addition the coordinates of the vectors in that space. Now it turns out that from the coordinates of the manifold you get a natural set of basis vectors in your tangential space, which in turn leads to a natural choice of vector coordinates.

BTW, I know the term "components" for the vector coordinates (or more generally for the coordinates of a tensor).

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