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Euclidean space, euclidean topology and coordinate transformation

  1. Aug 27, 2010 #1

    I have some doubts about the precise meaning of Euclidean space. I understand Euclidean space as the metric space [tex](\mathbb{R}^n,d)[/tex] where [tex]d[/tex] is the usual distance [tex]d(x,y)=\sqrt{\sum_i(x_i-y_i)^2}[/tex].

    Now let's supose that we have our euclidean space (in 3D for simplicity) [tex](\mathbb{R}^3,d)[/tex] and then we consider the transformation to spherical coordinates. The "spherical coordinate space" [tex]\mathbb{R}^3[/tex] of 3-tuples [tex](r, \theta, \phi)[/tex], is itself an euclidean space?

    As I see we have the choice to use also the distance [tex]d[/tex] on this latter space even if it has no physical meaning, but on the other hand it seems more useful to use the "phyisical" distance such that the distance between two points in the "spherical coordinate space" correspond to the physical distance between the points they represent. So, what distance to use? for each choice, is the space [tex]\mathbb{R}^3[/tex] of 3-tuples [tex](r, \theta, \phi)[/tex] also the euclidean space?
  2. jcsd
  3. Aug 27, 2010 #2
    From my understanding, Euclidean geometry deals only with flat surfaces, not curves.

    In other words Euclidean space is any space where two parrallel lines will stay parallel to infinity, on a curved surface two parallel geodesics will eventually meet, or separate.

    Feel free to correct me...
    Last edited: Aug 27, 2010
  4. Aug 27, 2010 #3
    How does that answer my question??
  5. Aug 27, 2010 #4
    The coordinate system and the metric tensor are different entities. The expression you denote for d is the Euclidean metric in Cartesian coordinates. The Euclidean metric expressed in terms of a different coordinate system will be different. The coordinate system does not change the geometry of the underlying metric space; it is just a way of numerically labelling points.
  6. Aug 27, 2010 #5
    Euclidean space can be defined as an affine space, [itex]\mathfrak{E}^n=(\mathbb{E}^n,\mathfrak{D}^n,+)[/itex], where [itex]\mathbb{E}^n[/itex] is the set of points, [itex]\mathfrak{D}^n[/itex] the associated "translation space", that is, the vector space (over the field of real numbers with their usual addition and multiplication) of displacement vectors from one point to another, having the underlying set of vectors [itex]\mathbb{D}[/itex], and [itex]+[/itex] is a function, [itex]+:(\mathbb{E},\mathbb{D}) \rightarrow \mathbb{E}[/itex], relating points and vectors such that, according to the affine space axioms, [itex]\forall p \in \mathbb{E}[/itex] and [itex]\forall \textbf{u}, \textbf{v} \in \mathbb{D}[/itex]

    [tex](1) \enspace (p + \textbf{u}) + \textbf{v} = p + (\textbf{u}+\textbf{v})[/tex]

    and [itex]\forall q \in \mathbb{E}, \exists[/itex] a unique vector [itex]\textbf{v} \in \mathbb{D}[/itex] such that

    [tex](2) \enspace q = p + \textbf{v}[/tex]

    This defines an affine space. In this case, we let [itex]\mathfrak{D}^n[/itex] be an inner product space, with inner product [itex]g : \mathbb{D}^n \times \mathbb{D}^n \rightarrow \mathbb{R}[/itex], and take the square root of [itex]g(\textbf{v},\textbf{v})[/itex] as our distance function, which makes [itex]\mathfrak{E}^n[/itex] also a metric space. Then Euclid's axioms, or some ultra-rigorous modern version of them, define this particular affine space as Euclidean space, its dimension being that of [itex]\mathfrak{D}^n[/itex]. I haven't mentioned the set of n-tuples [itex]\mathbb{R}^n[/itex] yet, but then, as far as I know, Euclid didn't either.

    It's useful, though, to set up some coordinates. Every n-dimensional vector space over a given field, [itex]\mathbb{K}[/itex], is isomorphic to the vector space [itex]\mathfrak{K}^n[/itex] over the same field, whose underlying set is the set [itex]\mathbb{K}^n[/itex] of n-tuples of elements of the underlying set of that field, with addition defined componentwise. In our case, specifically, the Euclidean translation space [itex]\mathfrak{D}^n[/itex] is isomorphic to [itex]\mathfrak{R}^n[/itex], whose vectors are n-tuples of real numbers.

    This allows us to use n-tuples of real numbers as coordinates for the points of [itex]\mathbb{E}^n[/itex] by specifying n one-to-one functions, called coordinate functions, from [itex]\mathbb{E}^n[/itex] to [itex]\mathbb{R}[/itex]. We can do this in many ways. Simplest is to pick one point, arbitrarily, to map to (0,0,...,0), and n orthogonal (but otherwise arbitrary) directions as the axes of this coordinate system. Already these arbitrary choices give us infinite possibilities, but, of course, there are many more. I think of each coordinate system as a different copy of [itex]\mathbb{R}^n[/itex], having the same structure but a different role to play: some serve as Cartesian coordinates, some as spherical coordiantes, and so on. Some can be defined on the whole of our Euclidean space, some only on a part. (If we identify a collection of coordinate systems, and their domains, which meet certain requirements about continuity and differentiability, we can describe Euclidean space as a Riemannian manifold.)

    Just as a point's position is given a different label in different coordinate systems, so certain functions associated with Euclidean space may take on a different guise. In Cartesian coordinates, the usual inner product on [itex]\mathfrak{R}^n[/itex] does the job of the inner product of the Euclidean translation space [itex]\mathfrak{D}^n[/itex]. But in other coordinate systems, this won't generally be the case. In some coordinate systems, the componentwise inner product on [itex]\mathfrak{R}^n[/itex] gives a result that differs from the Euclidean inner product itself associated with [itex]\mathfrak{D}^n[/itex], and so in such coordinate systems, the distance function has to be represented in a different way. We still want to compute the Euclidean distance, but if we want to do so in a particular coordinate system, we have to ask the question in the language of that coordinate system if we want the right answer.
  7. Aug 27, 2010 #6


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    In general, named mathematical structures do not specify a particular, unique thing. Instead, they pick out an equivalence class of things.

    e.g. the set of points on a Euclidean line endowed with an origin, unit, and arithmetic in the usual way? It's the real numbers! The set of Cauchy sequences of rational numbers modulo a certain equivalence relation? It's the real numbers! The set of Dedekind cuts of the rational numbers? It's the real numbers!

    Each of these sets are actually different things -- but they all yield isomorphic results, and for most purposes, it simply isn't all that useful to distinguish them.

    They are lots of ways to realize Euclidean space....

    Wait, do you mean the topological space? The affine space? The metric space? The manifold? The smooth manifold? The real algebraic variety? Let's talk about Euclidean space the metric space, since that's what you brought up in your original post.

    Therefore, "Euclidean space" is an isomorphism class of metric spaces. The set R3 with the metric [itex]d(x,y,z)^2 = x^2 + y^2 + z^2[/itex] is one realization of this space.

    The set of points [itex]\{ (r, \theta, \phi) | r > 0 \wedge 0 \leq \theta < 2 \pi, -\pi/2 < \phi < \pi/2 \}[/itex] together with the additional points O, N, and S (origin, north pole, south pole) together with the appropriately chosen metric is isometric to the space mentioned above. Therefore, this is also "Euclidean space".

    Incidentally, the set of all triples [itex](r, \theta, \phi)[/itex] with the appropriately chosen (pseudo)metric is not a "Euclidean space". The map to Euclidean space is not an isometry, because, e.g., (1,0,0) and [itex](1, 2\pi, 0)[/itex] both have the same image. Also, the same pair of points proves this isn't even a metric space, because the distance between them is zero.

    However, if you impose the equivalence relation that two points are equivalent iff the distance between them is zero, then the result is, in fact, a Euclidean space.
  8. Aug 27, 2010 #7
    You are right, I wrongly referred to the set of all triples [itex](r, \theta, \phi)[/itex] when I actually meant [itex]\{ (r, \theta, \phi) | r > 0, 0 \leq \theta < 2 \pi, -\pi/2 < \phi < \pi/2 \}[/itex].

    Good point. I further explain the motivation behind my question and my mental process. I came across this question when I was thinking about the usual definition of a manifold which is a local homeomorphism to the Euclidean space. All right up to this point... at first glance. But then I moved on to differential forms integration which has to do with the transformation of volumes (wedge product of differentials). So I thought "ok, what's the meaning of the covector [itex]dr[/itex] in spherical coordinates?. What's the meaning of the differential of volume [itex]dV=dr \wedge d\theta \wedge d\phi[/itex]?" In cartesian coordinates is clear that the basic covectors [itex]dx_i[/itex] return the i-th component of a tangent vector, so [itex]dr[/itex] should return the component along the [itex]r[/itex] axis of a vector in the [itex](r, \theta, \phi)[/itex] space, so exactly as in cartesian coordinates we think of the [itex]r[/itex] axis and the [itex]\theta[/itex] axis and the [itex]\phi[/itex] axis as the basis for vectors in that space, so the spherical coordinate space (or any other coordinate system space) is totally indistinguible from the cartesian coordinate and so, how can we theoretically distinguish between these "copies" of 3-tuples in an unambiguous rigorous way? Then the answer came up: the different distances defined on each space is what allows to rigurously distinguish between but I found no reference in literature to the metric (or topology) defined in the space of other coordinate system, all text just talk about the euclidean distance (root square of sum of squares) globally in [itex]\mathbb{R}^n[/itex].

    Therefore, your answer that Euclidean space is a equivalence class of different metric spaces (different coordinate system or "point labeling") with the appropiate metric defined in each one so that they are isometric under coordinate transformation pretty much makes sense.

  9. Aug 28, 2010 #8
    Hurkyl, Rasalhague :
    Would it be accurate to say that (a) Euclidean space is any space that models
    or satisfies the axioms you stated, and that R^n is just a particular model of/for
    these axioms.?
  10. Aug 28, 2010 #9


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    I'm pretty sure things could be phrased that way.
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