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Distances, compactification & Möbius transformations

  1. Dec 31, 2009 #1
    I have two points on a one-dimensional Euclidean submanifold, say the x-axis.
    I want to assume that this subspace is kind of "cyclic". This is often accomplished with the compactification [tex]R\cup \{ \infty \}[/tex]

    The question is: How can I compute distances (up to some constant factor) between two points taking into account this sort of "cyclicness" ?

    My idea was to use the complex plane, translate the x-axis vertically so that it passes through the point [tex](0,i)[/tex] and apply a Möbius transformation [tex]1/z[/tex]. Now all the points [tex]z=x+i[/tex] where [tex]x\in R[/tex] are mapped onto a circle, and I could use the shortest arc between the two corresponding points.
    - Is this actually correct?
    - Is the "shortest arc" length the correct metric to use?
  2. jcsd
  3. Jan 4, 2010 #2
    It's hard to follow what you're trying to do without some background.
    Firstly if you are taking a non-compact one dimensional submanifold of the Euclidean plane and then compactifying it, you may as well be taking one dimensional submanifolds of the sphere.

    There are a whole lot of different notions of distance on the sphere; the standard notion of distance being inherited from its standard embedding in Euclidean 3 space. Then your one dimensional submanifolds (circles) inherit a metric from whatever one you have on the sphere.

    Your suggestion gives a distance function (but so do a boat-load of others), but it is of course different to the distance function in the Euclidean plane (since all distances between points on a circle are finite, but distances along the x-axis are arbitrarily big).

    So to summarise: there is no "correct" metric to use. It depends on the context. I'm pretty sure your idea is a metric, but it is by no means the only one.
  4. Jan 4, 2010 #3
    You are right, I have to formalize my question better. I'll try now.

    I want to consider a one dimensional manifold (the x-axis) in which the segment-of-arc length is given by some metric; for the sake of simplicity let's assume it is an Euclidean metric. So for any pair of points [tex](a,b) [/tex] on the x-axis, we have their distance is [tex]d_{a,b}=\sqrt{(a-b)^2} [/tex]

    Now let´s consider the compactification [tex]R\cup\{\infty\}[/tex].
    I want to define a mapping [tex]f : R\cup\{\infty\} \rightarrow \mathcal{S}^1[/tex] or, why not, more generally as you suggested [tex]f : R\cup\{\infty\} \rightarrow \mathcal{S}^2[/tex]. In both cases we would be mapping into circles.

    The Möbius transformation I used, is indeed a (conformal) mapping of that kind, however it distorts the space.
    I want to compare the transformed points by using a distance [tex]d'[/tex] which is equivalent to the Euclidean one.

    Two distances [tex]d[/tex] and [tex]d'[/tex] are said to be equivalent when for any two pairs of points (a,b) and (c,d):

    [tex]d_{a,b}=d_{c,d} \Leftrightarrow d'_{a,b}=d'_{c,d}[/tex]

    In this sense the shortest-arc that I proposed, is not suitable!
    BTW, if any conformal mapping (Möbius transformation) is a curvilinear coordinates system, shouldn't it be always possible to define arc-length, area elements, geodesics, etc...?
    Last edited: Jan 4, 2010
  5. Jan 5, 2010 #4
    That all sounds perfectly reasonable.

    In order for a map to be defined as conformal both the range and domain must be Riemannian manifolds; so they certainly have geodesics, but they do not in general have a unique notion of distance, and hence arc-length or area. (You can always have a topological metric on any manifold, and hence a notion of arc-length, etc. - depending on your definition of manifold.)
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