# Generic coordinate system representations [closed]

Please excuse the verboseness which follows, as the question is rather basic, so I would like to state it carefully so that it will not be accidentally neglected as automatically trivial. If, after my discussion, it still appears trivial, then I am sorry.

First I begin with some motivation. Say we want to define a vector field (in the sense of advanced calculus, rather than say, abstract manifolds) in R^3. Then we need a way to consistently define (a.) points -- that is, the unique positions for each point in the space, and (b.) a directional basis at each point. (When saying "consistently define", I essentially mean that this information can be prescribed with a single set of parametrizations, rather than say, an uncountable set of {point,basis}.)

In the natural Cartesian coordinates, these requirements are satisfied by providing three particular coordinate functions. These coordinate functions are orthonormal, unit speed curves, placed at a designated origin. The tangent space at each point is also typically given in terms of the same basis.

When we start with a Euclidean space, we have no preferred origin and no preferred directions. The example of Cartesian coordinates seems to suggest that we can axiomatize or automate the generation of an arbitrary coordinate system by specifying three non-coplanar, or possibly everywhere non-coplanar curves.

On the other hand, consider another common choice of coordinates like spherical coordinates. In spherical coordinates (and curvilinear coordinates in general) the basis for a tangent space depends on the point. Furthermore, there does not seem to be a simple prescription of three curves which parametrize the space, since the coordinate curves themselves are re-defined for different points in space.

To go even further, one could choose an entirely arbitrary, non-orthogonal coordinate system. For a fully arbitrary system, we do not even care about its parametrization by the Cartesian coordinates. One begins to wonder, what property of this arbitrary system is actually grounding the satisfaction of the intial requirements, (a.) and (b.)? Specifically,

What are the requirements necessary for an arbitrary coordinate system such that it satisfies (a.) and (b.)?

For example, I might conjecture one answer: the coordinate system needs three functions defined in R^3 such that at all points their gradients are non-co-planar. Unfortunately, I am not even sure if this makes sense, since this answer still requires the specification of the Cartesian coordinates (in order to take the gradients, or convert the gradients into the new coordinates), which we were attempting to circumvent in the first place!

I would especially appreciate any attempt at making this discussion more rigorous or precise. Also I would appreciate any references which give a thorough discussion of these ideas.

-

## closed as off-topic by Ricardo Andrade, abx, Ryan Budney, Stefan Kohl, Chris GodsilAug 28 '14 at 1:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Ricardo Andrade, abx, Ryan Budney, Stefan Kohl, Chris Godsil
If this question can be reworded to fit the rules in the help center, please edit the question.

It seems that what you're trying to pin down is the tangent bundle of a smooth manifold - as a side point, though you specify vector fields "in the sense of advanced calculus rather than say, abstract manifolds", the rest of your discussion on trying to understand changes of coordinates is precisely the motivation to study manifolds in general. John M. Lee's book "An Introduction to Smooth Manifolds" is a book I liked which ought to shed light on what you've asked. – j.c. Aug 12 '10 at 12:00

The first condition implies that the coordinate transformation from standard Euclidean coordinates (or any other coordinate system) is a bijection, while the second implies that it is a $C^1$ diffeomorphism, that is, derivative matrix is non-singular at each point.
In coordinate free language you could say like this: you need (as you say) three globally definded smooth functions $f_1,f_2,f_3$ on $\mathbb{R}^3$ such that a): the values of these functions on any two different points of $\mathbb{R}^3$ are not all the same, and b) the differentials $df_1,df_2,df_3$ are linearly independent at every point of $\mathbb{R}^3$. Differentials are the coordinate-free (or metric-free) version of the gradient, you can find out about them in any good differential geometry book. – Michael Bächtold Aug 12 '10 at 18:20
You need the standard Euclidean coordinates in some way: these are intrinsically used to define $\mathbb{R}^3$. You could view $\mathbb{R}^3$ as an abstract manifold, but then you'd need to specify what the coordinate charts are... So to specify what a generic coordinate system is, you need Euclidean coordinates, but once you have a coordinate system you can define any other coordinate system by a coordinate transformation, i.e. the group of $C^1$ diffeomorphisms acts freely, transitively on coordinate systems. – Jaap Eldering Aug 13 '10 at 12:23