# Distances, compactification & Möbius transformations

1. Dec 31, 2009

### mnb96

Hi,
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 $$R\cup \{ \infty \}$$

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 $$(0,i)$$ and apply a Möbius transformation $$1/z$$. Now all the points $$z=x+i$$ where $$x\in R$$ 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. Jan 4, 2010

### fantispug

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.

3. Jan 4, 2010

### mnb96

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 $$(a,b)$$ on the x-axis, we have their distance is $$d_{a,b}=\sqrt{(a-b)^2}$$

Now let´s consider the compactification $$R\cup\{\infty\}$$.
I want to define a mapping $$f : R\cup\{\infty\} \rightarrow \mathcal{S}^1$$ or, why not, more generally as you suggested $$f : R\cup\{\infty\} \rightarrow \mathcal{S}^2$$. 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 $$d'$$ which is equivalent to the Euclidean one.

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

$$d_{a,b}=d_{c,d} \Leftrightarrow d'_{a,b}=d'_{c,d}$$

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
4. Jan 5, 2010

### fantispug

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.)