Talk:Beltrami–Klein model

Moved from the article
I have moved from the article the following text which appears to be an essay with unclear purpose and some errors. Arcfrk (talk) 02:07, 28 March 2009 (UTC)

Angles and orthogonality
Given two intersecting lines in the Cayley–Klein model, which are intersecting chords in the unit disk, we can find the angle between the lines by mapping the chords, expressed as parametric equations for a line, to parametric functions in the Poincaré disk model, finding unit tangent vectors, and using this to determine the angle.

We may also compute the angle between the chord whose ideal point endpoints are $$u$$ and $$v$$, and the chord whose endpoints are $$s$$ and $$t$$, by means of a formula. Since the ideal points are the same in the Cayley–Klein model and the Poincaré disk model, the formulas are identical for each model.

If both chords are diameters, so that $$v=-u$$ and $$t=-s$$, then we are merely finding the angle between two unit vectors, and the formula for the angle $$\theta$$ is
 * $$\cos(\theta) = u \cdot s.$$

If $$v=-u$$ but not $$t=-s$$, the formula becomes, in terms of the wedge product,
 * $$\cos^2(\theta) = \frac{P^2}{QR},$$

where
 * $$P = u \cdot (s-t),$$
 * $$Q = u \cdot u,$$
 * $$R = (s-t) \cdot (s-t) - (s \wedge t) \cdot (s \wedge t)$$

If both chords are not diameters, the general formula obtains
 * $$\cos^2(\theta) = \frac{P^2}{QR},$$

where
 * $$P = (u-v) \cdot (s-t) - (u \wedge v) \cdot (s \wedge t),$$
 * $$Q = (u-v) \cdot (u-v) - (u \wedge v) \cdot (u \wedge v),$$
 * $$R = (s-t) \cdot (s-t) - (s \wedge t) \cdot (s \wedge t).$$

Using the Binet–Cauchy identity and the fact that these are unit vectors we may rewrite the above expressions purely in terms of the dot product, as
 * $$P = (u-v) \cdot (s-t) + (u \cdot t)(v \cdot s) - (u \cdot s)(v \cdot t),$$
 * $$Q = (1 - u \cdot v)^2,$$
 * $$R = (1 - s \cdot t)^2.$$

Determining angles is greatly simplified when the question is to determine or construct right angles in the hyperbolic plane. A line in the Poincaré disk model corresponds to a circle orthogonal to the unit disk boundary, with the corresponding Cayley–Klein model line being the chord between the two points where this intersects the boundary. The tangents to the intersection at the two endpoints intersect in a point called the pole of the chord. Any line drawn through the pole, which is the center of the Poincaré model circle, will intersect the Poincaré model circle orthogonally, and hence the line segments intersect the chord in the Cayley–Klein model, which corresponds to the circle, as perpendicular lines.

Restating this, a chord $$B$$ intersecting a given chord $$A$$ of the Cayley–Klein model, which when extended to a line passes through the pole of the chord $$A$$, is perpendicular to $$A$$. This fact can be used to give an easy proof of the ultraparallel theorem.

simplification rewrite
I am trying to rewrite this to a bit simpler text:

- more about the two dimensional case - and more of that ilk

please help WillemienH (talk) 07:09, 25 April 2015 (UTC)


 * I'm unconvinced that bringing the comparison with the Poincaré disk and the detail of the distance formula into the lead is a simplification. What is it about the prior version that you wanted to simplify? I can see that a section on the two-dimensional case may be beneficial. —Quondum 21:20, 25 April 2015 (UTC)

My idea was especially add more about the 2 dimensional case maybe a bit like what is in the Poincare half-plane model, some constructions, how to calculate angles ed. I did a minor rewrite on the lead, made a history section , but more is to do WillemienH (talk) 04:40, 26 April 2015 (UTC)


 * We need to be clear about whether this article is about the general model of the hyperbolic geometry of any number of dimensions, or whether it is primarily about the plane. I see it as being the former, with the "Beltrami–Klein disk model" being the special case in two dimensions. This would argue for keeping mention of two dimensions to a section (which could be quite extensive and could even become its own page), and for keeping it from becoming prominent in the lead. —Quondum 14:27, 26 April 2015 (UTC)

Thanks for the idea about clarity but my idea is that it more should be primarily about the two dimensional Disk model than about the higher dimensional n-Ball model, and I would like to "banish " all higher dimensional parts to a late seperate section.

My reasons are the following:


 * the two dimensional case is more basic, easier to understand and therefore more usefull to the reader and better as introduction.
 * all illustrations here are 2 dimensional (a wikipedia page is just a 2 dimensional sheet )

I first thought that the page had Disk in its title and therefore this discussion could be considered closed but then realised it hasn't. (but neither there is ball) I would like it a bit conforming to what I did on the hyperbolic geometry page where most higer dimensional parts are moved to hyperbolic space.WillemienH (talk) 20:54, 26 April 2015 (UTC)


 * I tweaked your edits a little, not changing much. I did remove the mention of the line at infinity; searching about, I noticed that its name is the absolute. It is incorrect to call it a line: it is really a conic. Hyperbolic geometry is a projective geometry, and so cohesiveness of terminology is important.  I suspect that some authors may have adopted some terms in an incompatible way, but not many seem to refer to the absolute as "the line at infinity" in this context. —Quondum 02:42, 22 May 2015 (UTC)

Thanks for your edit, we all work together to make it better, most times I first make big rewrite and then later make some smaller tweaks. I do think that some parts of my layout were a bit better (I like lists more than dense text) so I will tweak your edit again (and again and again ...) You are right about the absolute (but we need to add something about that it is at infinity. Also the text on perpendiculary needs improving. (see editing never ends) I was thinking about adding some constructions ( how to reflect a point, how to find the midpoint of a chord) also some questions remain: - What do circles become?(are they ellipses or other strange forms?) -the same for hypercycles. Another point: any ideas about the metric tensor? It looks a completely different beast than the metric tensor at the poincare disk model (see also http://math.stackexchange.com/q/1292707/88985 ). Btw the poincare disk model is my next target. also thanks for the rewrites at hyperbolic geometry.(also a target of me) Some points where I am struggeling with: -How do we call making lines from chords, in this one article, we have extending and lengthening? - I need to improve my english (I know) - do we need more drawings? see lots to improve :) WillemienH (talk) 06:46, 22 May 2015 (UTC)


 * Simply constructing a list as a series of paragraphs is a bit awkward. It comes across as a staccato series of unrelated statements.  What you may want to try is to tie points in a list together by introducing the list followed by bullets. Yes, the ideal points can be stated as being "at infinite distance", rather than "at infinity". Horocycles and hypercycles are circles in the Poincaré disk model (AFAICT), but are funny shapes in the Klein disk model.  The ultraideal points are naturally mapped to points outside the Klein disk model (these points must be expanded on), but the points outside the Poincaré disk model naturally map to points inside the disk (it is just an inversive map of the interior).  Also, in the Poincaré disk model, horocycles would be circles tangent to the border, and hypercycles would be circles intersecting the border at two distinct points (excluding those that are lines).  The metric tensor should not be difficult to determine, though I'm a little rusty at the moment.  I've not looked much at the metric in non-Euclidean geometries, though these can be deduced from the hyperboloid model.  What do you mean by "making lines from chords"?  There is a one-to-one correspondence between lines of the geometry and chords of the model.  Extending the chords beyond the boundary adds the ideal and ultraideal points, which is useful for reasoning about properties of points inside the geometry.  (The ultraideal points form another type of geometry, which is nicely modelled by the exterior of the Klein disk.  But this is diverting.)  —Quondum 14:08, 22 May 2015 (UTC)

Confused metric
The article currently says:


 * Then the hyperbolic distance between p and q is expressed as
 * $$g (x, dx) = \frac{4 (x \cdot dx)^2}{(1 - \left\Vert x \right\Vert^2)^2} + \frac{4 \left\Vert dx \right\Vert^2}{(1 - \left\Vert x \right\Vert^2)} $$
 * or
 * $$ds^2 \;=\; \frac{ \left\| \mathbf{dx} \right\| ^2 }{ 1 - \left\| \mathbf{x} \right\| ^2 } + \frac{ ( \textbf{x} \cdot \textbf{dx} ) ^2 }{\bigl( 1 - \left\| \mathbf{x} \right\| ^2 \bigr) ^2 } .$$

This seems to be a confusion, inasmuch as the factor of four is different, but otherwise these are identical. There should be no difference between the two: we must have ds2 = g(x, dx). The stackexchange answer only seems to be addressing the distinctions between the Klein and Poincaré models. Where was the first equation above sourced? I suspect that it is incorrect. —Quondum 14:59, 26 July 2015 (UTC)


 * The "g (x, dx)" metric was included on 3 februari 2015 (https://en.wikipedia.org/w/index.php?title=Beltrami%E2%80%93Klein_model&diff=prev&oldid=645436480 ) by user:Jvohn I added the second "ds2" metric because I did not understand the first metric, and wanted one that looked like and was comparable with the metric on the poincare disk model , I don't know a source that mentions the "g (x, dx)" metric.
 * Can the ds''2" metric be simplified to
 * $$ds^2 = \frac{ \left\| \mathbf{dx} \right\| ^2( 1 - \left\| \mathbf{x} \right\| ^2 ) + (\textbf{x} \cdot \textbf{dx} ) ^2 } {\bigl( 1 - \left\| \mathbf{x} \right\| ^2 \bigr) ^2 } $$ or even further?
 * Also can you teach me on "inasmuch as the factor of four is different, but otherwise these are identical" you seem to imply that The "g (x, dx)" metric divided by 4 equals the "ds2" metric. even that I could not figure out.WillemienH (talk) 15:34, 26 July 2015 (UTC)
 * Answering you in reverse order:
 * There are formatting differences that may be confusing. Compare the following, where I've reordered a sum and used the same formatting convention (vectors italic rather than bold, but you could choose all x as bold instead) for both, where it should be obvious that $$g (x, dx) = 4 d s^2 $$:
 * $$g (x, dx) = \frac{4 \left\Vert dx \right\Vert^2}{1 - \left\Vert x \right\Vert^2} + \frac{4 (x \cdot dx)^2}{(1 - \left\Vert x \right\Vert^2)^2}$$
 * $$ds^2 \;=\; \frac{ \left\| dx \right\| ^2 }{ 1 - \left\| x \right\| ^2 } + \frac{ ( x \cdot dx ) ^2 }{\bigl( 1 - \left\| x \right\| ^2 \bigr) ^2 } $$
 * One could re-express it that way, but whether it would be considered a simplification is a subjective choice. I would not describe it that way.
 * Thanks for finding the addition of that metric; it means that we will need to regard that as unsourced and independently source it. I suspect that it derived from a model with a normalization different from that in use in the article (i.e. chosen so that the Gaussian curvature is −1); I would suggest removing it.
 * We can look at the detail of the metric tensor (the term is slightly misapplied in the article) and its relationship to the distance function later. —Quondum 16:52, 26 July 2015 (UTC)
 * I've gone ahead and tweaked it, (hopefully) making it clearer where vectors are used. See whether this makes any more sense to you.  The bit about the metric tensor is still not clear; the metric tensor is really a bilinear product that is the dot product on the tangent vector space; what is given (ds2) is just the square of the distance between points separated by an infinitesimal vector dx in the model, and should be easily derivable from the distance function.  As it is given, it is technically a quadratic form and not a bilinear form or metric tensor. (It takes only one vector dx as its input; the position vector x is only to allow the metric tensor g to be determined.  A better notation would have been qx(dx).  The distance along any curve is essentially $∫√q_{x}(dx)$.) —Quondum 21:25, 26 July 2015 (UTC)


 * Thanks for the edit, I was wondering: could we rename the section to "Metric tensor" and move the section down to after the move the whole section to after the "Klein disk model" section? I like to arrange sections from simple to advanced, and the distance function is allready in the lead. I am not sure of the mathematics yet, in metric tensor it also says "Conversely, the metric tensor itself is the derivative of the distance function (taken in a suitable manner)." haven't been able to make a proof of that yet. I think my problems was more with the function itself: why "g" with two parameters?  maybe it just confused me. Other question is there not a scale (map) function of the model? (we also could do with one at the Poincare disk model WillemienH (talk) 22:36, 26 July 2015 (UTC)


 * I don't think that the lead should contain content that is not in the body of the article. I'm not entirely comfortable with the distance formula being in the lead at all; I just have not objected thusfar.  But I would object to its removal from the body.  The distance formula is not "simple", and probably belongs after the metric, for the simple reason that the metric is probably simpler to determine and the distance function must be determined through integration of the metric.  Where else would a transcendental function such as $log$ come from?  It would be reasonable to split the sections into two sections (distance and metric), but I would keep them together.
 * On your question about a scale function, that is essentially what the metric tensor is. You will understand that the scale of a map varies from point to point of a map, and that is why the vector x is needed: to determine the scale at each point (you need to vary the scale by position).  It is to this function that the displacement dx is input.  That is why I used the notation qx(dx).  Your intuition is spot-on here, only you need to tie it to the existing notation.  If you want a true metric tensor, we would call it gx, and it would take two parameters, both infinitesimals: gx(dx, dy).  —Quondum 03:56, 27 July 2015 (UTC)


 * It is usually a bad idea to make a massive change in notation as you did with replacing $$p,q$$, etc with capital and lower case $$U, V, u, v$$. In this case, it also creates inconsistency between the formula for the distance in the lead and in the body of the article.


 * On the question of metric vs distance, clearly, Cayley (and Beltrami and Klein after him) considered the distance, and not the metric, even though the logical order is reversed if you (ahistorically) treat the Beltrami-Klein model as a 2-dimensional Riemannian manifold of constant scalar curvature &minus;1. In other words, let us not turn this article into an exercise in Riemannian geometry. Arcfrk (talk) 04:50, 27 July 2015 (UTC)


 * I am doubttfull about " the distance function must be determined through integration of the metric. " history is really the other way around (As far as i remember) also using vectors so early in the article, I don't like it. I do think the first bit should be about the 2-dimensional case, I will rewrite the article a bit and lets see how it looks WillemienH (talk) 07:12, 27 July 2015 (UTC)


 * I will not argue about historical ordering; I was simply mentioning a logical sequence for deriving the distance, not necessarily how it should be presented. If the sections are separated, there is some flexibility here.
 * The article lacked clarity on the distinction between points of the geometry (or even points of the model) and the vectors to those points. However, if we remove the concept of vectors (motivated by the metric case), we could phrase distances as before, purely in terms of points and distances between them.  I have no objection to this if the metric is split into a separate section, but uppercase letters for point might be more familiar.  The formula should not (IMO) be in the lead. —Quondum 13:40, 27 July 2015 (UTC)