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This is a very basic question that I feel that I should know the answer to but haven't been able to think through clearly.

In linear algebra, we learn that the basis of a finite dimensional vector space can be thought of as 'co-ordinates' of that space. And, we model what we intuitively understand as the euclidean plane, using the vector space $\mathbb{R^2}$ equipped with the standard inner product and metric etc. The underlying space is taken to be independent of the choice of basis, that is we understand that properties that are inherent to the space are those that will be invariant under change of basis.

Now, $\mathbb{R^2}$ comes with a canonical basis: this can be understood as saying that given any arbitrary two dimensional vector space and any basis, the vectors $v_i$ of the basis under the co-ordinate map maps to $e_i$. Since, we have also introduced inner products and thus a notion of parallel, the intuitive picture we now have of the co-odrinate grid is a criss-cross of lines. and the 'co-ordinates' are called, imaginatively, 'rectangular co-ordinates'.

In school, we also learn about polar 'co-ordinates' of the plane. The associated picture is of concentric circles and rays fanning out of the origin. However, these 'co-ordinates' do not fit within the 'Basic Linear Algebra' framework (since among other things, $0$ has no unique representation and the functions that change the variables are not linear). The one way of seeing the transformation of the 'rulings' of the plane is to consider $\mathbb{R^2}$ as $\mathbb{C}$ and the change as the map $\exp: \mathbb C \to \mathbb C$.

What is the framework in which the notion of 'co-ordinates' subsumes both these pictures (likewise for cylindrical, spherical, etc in dimension(?) three). My second and connected question: is there a linear algebraic connection for using two numbers to represent points in polar co-ordinate, i.e. is it because the vector space dimension of $\mathbb{E^2}$ is two. My third question is: am I confusing different concepts of 'dimension' here?

Added: Thanks for all the replies. They are all great but given my continuing dissatisfaction, either I have not really understood the answers or haven't been able to communicate the question properly (or perhaps and quite likely, I don't know what I want to ask).

Now, when we think of a (topological) manifold, we think of some object that is locally euclidean, i.e., we already have a 'handle' on $\mathbb{E^n}$ and want to use our knowledge of being able to do things, such as calculus, onto the new object . In my question, I am looking at ways we can 'handle' $\mathbb{E^2}$ itself, so invoking manifolds, seems a bit like putting the cart before the horse. I want to say something like this: The point of polar co-ordinates is to represent the plane by looking at it as $S^1\times \mathbb{R}$, where $S^1$ is a basic object like $\mathbb R$.

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Where is the question in your edit? – Qiaochu Yuan Dec 25 '10 at 11:11
The edit was a clarification of what I have been trying to ask. – Dactyl Dec 25 '10 at 22:55

Great observation; it is quite annoying that nobody bothers to explicitly tell you this. The relevant concept here is the dimension of a manifold, not the dimension of a vector space (although one is defined in terms of the other). What "coordinate" means here is that you are specifying a certain isomorphism (say, differentiable) between $\mathbb{R}^n$ (or some nice submanifold thereof, for example restricting the values of some of the coordinates) and your manifold $M$, and you are using the points of $\mathbb{R}^n$ to parameterize points of $M$. $M$ itself may be of the form $\mathbb{R}^m$, but the isomorphism need not be linear; it is an isomorphism of (say, smooth) manifolds.

The value of requiring the isomorphism to be differentiable is that a differentiable isomorphism $\phi : M \to N$ between smooth manifolds induces an isomorphism in the ordinary vector space sense between their tangent spaces at any point.

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Hi. So by 'co-ordinates' we mean atlases? – Dactyl Dec 24 '10 at 15:47
Atlases, I think, are generally called local coordinates. Here I am talking about global coordinates. – Qiaochu Yuan Dec 24 '10 at 16:17
Ok. And by your answer, we have the two numbers in polar co-ordinate representation because of invariance of domain? – Dactyl Dec 24 '10 at 16:24
We don't have to appeal to invariance of domain here because the map is differentiable, but yes, you can think about it that way. – Qiaochu Yuan Dec 24 '10 at 16:31

Another way of looking at coordinate systems is they are a way of naming points. There are many different systems which have been created because they make it easy to work on certain kinds of problem. Polar coordinates make it easy to express things that happen around a center (compare $r=c$ to $\sqrt{x^2+y^2}=c$), but hard to express things in straight lines unless they pass through the origin and addition of vectors. Wikipedia has a discussion of various definitions of dimension.

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I think the key point is understanding the notion of "howto parametrize points" in a manifold, and in particular the big question is "How can I pass from delightful, amenable coordinates in $\mathbb{R}^n$ to those fancy 'things' with a lot of cosines, logarithms and so on?". Well, the answer (IMO) is "Because I know how to do analysis nowhere but in open balls of $\mathbb {R}^n$". Studying the beginning of Mathematical Physics teached me that anytime I'm considering the speed of a motion over a (hopefully smooth) surface I'm considering a vector in the tangent space (plane) of that surface in a point: in this fashion "local" description of a motion is the description of a "straight" motion of a point $p$, with rules established, instant per instant, by the velocity vector $\vec v$ of the point $p$.

I would like to be more precise. Fix a surface $\mathcal M$ in the euclidean space (if you don't know the precise definition of a 2-dimensional manifold don't worry and go on thinking to $\mathcal M$ as a sheet on which you can smoothly skate)

You want to mathematically capture the idea of $p$ to be bonded to move on $\mathcal M$. So you define its motion to be not the entire three dimensional space $\mathbb R^3$ but only a precise subset of it, which is parametrized by some diffeomorphism (if you don't know what a diffeomorphism is, then learn it on wikipedia, or visualize it as a "highly differentiable homeomorphism" ;) ) $f\colon U\to W\subseteq \mathcal M$, $U$ being an oper set of $\mathbb{R}^2$ and $W$ being an open set (with the subspace topology induced by $\mathbb R^3$) of $\mathcal M$, containing the point $p$. It means in a few words that you're taking a point $q=(q_1,q_2)$ in the affine space of $\mathbb{R}^2$ and you're smoothly sending it in a point $f(q_1,q_2)=(f_1(q),f_2(q),f_3(q))$.

In this neighborhood of $p$, every happens "as if you were on $\mathbb R^2$, upto go back with the inverse of $f$ on your starting $U$. A motion on $\mathcal M$ is totally captured by a motion on $U$, because you can think about a motion on $\mathcal M$ as a "suitably deformed" motion on $U$, where you know how to define the position of a point, using coordinates, and other useful things like (piecewise) smoothness of a curve, tangent lines, et cetera.

The idea to be 'bonden on $\mathcal M$' is finally captured by asking that, given the curve $\gamma\colon ]-\epsilon,\epsilon[\to \mathcal M\subseteq\mathbb{R}^3$ describing the motion of $p$ (i.e., for example $\gamma(0)=p$), the velocity vector $\gamma'(t)$ lies in $\mathbb{R}^3$ but also in the tangent plane $T_p(\mathcal M)$, which can in fact be defined as the collection of all possible tangent vectors to curves lying on $\mathcal M$.

Now notice that the number '2' of independent coordinate you are using to uniquely describe the motion of $p$ on $\mathcal M$ is not at all a random number. It is the dimension of the tangent plane in $p$.

So, let's find a basis of $T_p(\mathcal M)$. Nothing easier: do you remember the function $f\colon U\to W$? Brutally derive it, taking its differential $\text{d}f_q$, where I'm writing $q=f^{-1}(p)$, and you will get a very nice invertible matrix whose columns form a basis of $T_p(\mathcal M)$. Don't you like the canonical basis of $\mathbb{R}^2$, do you? Well, change it and see what's happen (the answer is: upto a linear automorphism of $T_p(\mathcal M)$ I'm doing nothing).

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Another question implicit in what you ask is why use other co-ordinate systems. One reason is that the form of many equations is dramatically simplified in other co-ordinate systems. For example, the rectangular representation of the curve defined by the polar coordinate equation r = 1 + cos(2(theata)) would make it much harder to understand than in its polar form. r= 3 is a circle of radius 3 in polar coordinates. For modeling of certain phenomena polar coordinates makes things much easier, including work involving orbit calculations.

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