12
$\begingroup$

Apologies if this is a basic question, but I'd really like to clarify the exact meaning of what a Cartesian coordinate system is. Heuristically, is it correct to say that a Cartesian coordinate system is "Euclidean geometry with coordinates" and so by definition the geometry that we are studying has to be Euclidean (i.e. Euclidean metric, parallel postulate, etc. must hold) in order to use this coordinate system. Also, more mathematically, is it correct to say that a Cartesian coordinate system is a special kind of mapping between points of Euclidean space $\mathbb{E}^{n}$ and real space $\mathbb{R}^{n}$. Such a coordinate map makes the intrinsic spatial distance between two points in $\mathbb{E}^{n}$ be directly reflected by the ‘numerical distance’ between their numerical coordinates in $\mathbb{R}^{n}$.

If this is correct then I can understand why Cartesian coordinate maps cannot generally be used for patches on manifolds as the geometry will generally be non-Euclidean. As such the homeomorphisms (coordinate maps) from the patch on the manifold to $\mathbb{R}^{n}$ will necessarily deform the patch, consequently intrinsic distances between points, angles, etc. will not preserved and so Cartesian coordinates cannot be used.

$\endgroup$
9
  • 4
    $\begingroup$ This is hardly a basic question IMHO. +1) $\endgroup$
    – David H
    Jun 21, 2015 at 17:22
  • $\begingroup$ @DavidH I thought that too, but I was worried other people might feel differently. $\endgroup$
    – Will
    Jun 21, 2015 at 17:49
  • $\begingroup$ I'll note that "coordinates" alone do not require Euclidean geometry. In the hyperbolic plane, eg, one can designate a pair of orthogonal axes. From each point $P$ in the plane, one can drop perpendiculars to these axes; the lengths of these perpendiculars, along with appropriate signs for a given quadrant, provide perfectly-useful $(x,y)$-coordinates for $P$. The significant distinguishing characteristic between these coords and coords on the Euclidean plane is the associated distance formula. Clearly, $\sqrt{(x_0-x_1)^2+(y_0-y_1)^2}$ doesn't apply to hyperbolic coordinates. (What does?) $\endgroup$
    – Blue
    Jun 21, 2015 at 17:53
  • $\begingroup$ @Blue That was kind of my thinking, that Cartesian coordinates require the notion of the usual Pythagorean distance formula, which doesn't hold for non-Euclidean geometries and hence why Cartesian coordinate maps are not possible for such geometries?! $\endgroup$
    – Will
    Jun 21, 2015 at 17:56
  • 1
    $\begingroup$ @Will: I up-voted your question, because I think your frustration is valid. I'm just saying that I, personally, consider "Cartesian coordinates" to be independent of "Pythagorean distance formula". These texts you've read seem have me in the minority; I'm okay with that. :) Come to think of it, if I were to discuss hyperbolic coordinates with someone, I might well drop "Cartesian" as a descriptor, just to avoid the kind of confusion you're facing. Even so, I'd expect authors to be more explicit with their assumptions. I, like you, await an answer with better insights on the matter. $\endgroup$
    – Blue
    Jun 21, 2015 at 18:50

1 Answer 1

5
$\begingroup$

In the mathematical literature, the term "Cartesian coordinates" is used most frequently to refer simply to the standard coordinate functions on $\mathbb R^n$, namely the functions $x^1,\dots,x^n\colon \mathbb R^n\to \mathbb R$ defined by $x^i(a^1,\dots,a^n) = a^i$. Somewhat less frequently, I've also seen the term used to refer to any coordinate system on $\mathbb R^n$ obtained by composing the standard coordinates with a rigid motion, which can also be characterized as those coordinates for which the standard coordinate vectors $\partial/\partial x^1,\dots,\partial/\partial x^n$ are orthonormal.

The point is that it only makes sense to talk about "Cartesian coordinates" on $\mathbb R^n$ itself, or on an open subset of $\mathbb R^n$. On an arbitrary smooth manifold, the term has no meaning. Of course, on any smooth manifold $M$, each point has a neighborhood $U$ on which we can find a smooth coordinate chart, and such a chart allows us to identify each point $p\in U$ with its coordinate values $(x^1(p),\dots,x^n(p))\in\mathbb R^n$, and thus to temporarily identify $U$ with an open subset of $\mathbb R^n$; but we would not call these coordinates "Cartesian coordinates on $M$."

If your manifold $M$ is endowed with a Riemannian metric $g$, then there is more that can be said. For example, one could ask whether it's possible to find a coordinate chart in which the given Riemannian metric has the same coordinate expression as the Euclidean metric: $g= (dx^1)^2 + \dots + (dx^n)^2$. If this is the case, then geodesics and distances within this coordinate neighborhood are given by the same formulas as they are in Euclidean space; but that might not hold true elsewhere on the manifold. I think this might be the question you're getting at in your last paragraph, although I would not call these "Cartesian coordinates" because they don't have an open subset of $\mathbb R^n$ as their domain. Off the top of my head, I don't know of any standard nomenclature for such coordinates, but it wouldn't be inconsistent to call them "Euclidean coordinates" or "flat coordinates."

It's a basic theorem of Riemannian geometry that it is impossible to find such coordinates unless the curvature tensor of the Riemannian metric is identically zero on the open subset $U$. You'll find a proof of this fact in virtually any book on Riemannian geometry, such as my Riemannian Manifolds: An Introduction to Curvature (Theorem 7.3). If you want a treatment that doesn't use so much of the machinery of Riemannian manifolds, my Introduction to Smooth Manifolds has a proof that it's impossible to find Euclidean coordinates for the ordinary $2$-sphere in $\mathbb R^3$ (Proposition 13.19 and Corollary 13.20).

$\endgroup$
15
  • $\begingroup$ Thanks for your detailed answer, I think it's cleared up a few things for me. Is it correct to say that when one constructs a homeomorphism from a patch on a manifold to $\mathbb{R}^{n}$ this defines a coordinate system (or coordinate map) that labels points in the patch by coordinates in $\mathbb{R}^{n}$... $\endgroup$
    – Will
    Jun 21, 2015 at 22:10
  • $\begingroup$ ...If the geometry on the manifold is non-Euclidean then when one talks of it being locally homeomorphic to $\mathbb{R}^{n}$ (Euclidean space), is this the statement that the topology of the manifold and $\mathbb{R}^{n}$ are locally identical and not a statement about the geometry being locally Euclidean (assuming that a metric has been defined)? $\endgroup$
    – Will
    Jun 21, 2015 at 22:10
  • $\begingroup$ @Will: Yes, as Peter Franek pointed out in his answer to this question, a homeomorphism is by definition simply a map that is bijective, continuous, and with continuous inverse. This definition says absolutely nothing about geometry, Euclidean or otherwise. $\endgroup$
    – Jack Lee
    Jun 21, 2015 at 22:15
  • $\begingroup$ So does the geometry determine the coordinate maps that one can construct? Would it be in some sense correct to say that Cartesian coordinates are only possible if the manifold is $\mathbb{R}^{n}$ as they are global coordinates? (For Cartesian coordinates to be applicable is it in any sense correct to say that the space has to be locally flat, i.e. have a Euclidean metric?) $\endgroup$
    – Will
    Jun 21, 2015 at 22:44
  • 1
    $\begingroup$ No, the geometry has no bearing on the coordinate maps you can construct, because coordinates do not have to have any direct relationship to a metric. (The definition of a "coordinate chart" makes no reference to a metric.) You can use Cartesian coordinates if and only if your manifold is an open subset of $\mathbb R^n$, regardless of whether it's endowed with the Euclidean metric, some other metric, or no metric. What is true is that unless the metric you're interested in on that open set is the Euclidean one, then Cartesian coordinates will not have any relationship with the metric. $\endgroup$
    – Jack Lee
    Jun 21, 2015 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.