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I was puzzeling with the distance between points in hyperbolic geometry and found that the same formula is used for calculating the length in the Poincare disk model as for the Beltrami-Klein model the formula

$$ d(PQ)=\frac{1}{2} \left| \log \left(\frac{|QA||BP|}{|PA||BQ|}\right)\right| $$

where A and B are the idealpoints (extremities) of the line (in the Beltrami-Klein model ) or the circle or diameter (in the Poincare disk model) that contains P and Q while PA, PB, QA, QB be the euclidean distances between them. (but see below for an extra question)

But let P and Q for simplicity be points on a diameter, then by going from a Beltrami-Klein model to a Poincare disk model the points P and Q get closer to the centre while the end points stay on the same points so the euclidean distances change, and the formula could give a different value.

Therefore (I think) the formula cannot be correct for both models, and so my question for which model is this equation and what is the formula for the other model.

ADDED LATER:

A more worked out example: (Schweikart Constant, altitude of the largest orthogonal isocleses triangle)

Let r be the radius of the disk Then $ A = ( - \frac{1}{2} r \sqrt{2} , - \frac{1}{2} r \sqrt{2} ) $ , $ B = ( \frac{1}{2} r \sqrt{2} , \frac{1}{2} r \sqrt{2} ) $ , P = (0,0) and Q is on the line x=y

and the hypothenuse is the hyperbolic line between (r,0) and (0,r)

The euclidean lengths for PQ are:

For the Poincare Disk model: $ PQ = r ( \sqrt2 - 1 ) $

For the Beltrami-Klein model: $ PQ= \frac{1}{2} r \sqrt{2} $

What gives for the altitude:

For the Poincare Disk model: $ d(PQ)= \frac{1}{2} | \log ( 1 + \sqrt{2} | $

And for the Beltrami-Klein model: $ d(PQ)= \frac{1}{2} | \log ( 3 + 2 \sqrt{2} ) | = \log ( 1 + \sqrt{2}) $

What is right way to calculate the Schweikart Constant?

The Schweikart Constant is $ \log ( 1 + \sqrt{2}) $ , so it looks like the value in the Beltrami-Klein model is correct, but what is the correct formula for the Poincare Disk model?

Additional Question :

For the lengths in the Poincare disk models: If the hyperbolic line is an euclidean circle are the euclidean lengths measured as the segment-lengths or as arc-lengths (along the circle)?

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Let's first concentrate on a single line. I.e. your hyperbolic 1-space is modeled by the open interval $(-1,1)$. You have $A=-1,B=1$. Take two points in the Poincaré model, and compute the cross ratio

$$ (A,B;Q,P) = \frac{\lvert QA\rvert\cdot\lvert BP\rvert}{\lvert PA\rvert\cdot\lvert BQ\rvert} = \frac{(1+Q)(1-P)}{(1+P)(1-Q)} $$

Now transfer these points into the Klein model, at your discretion either via stereographic projection and the hemisphere model, or via the hyperboloid model, or purely algebraically. You obtain

$$ P' = \frac{2P}{1+P^2} \qquad Q' = \frac{2Q}{1+Q^2} $$

Plug these into the cross ratio and you get

$$ (A,B;Q',P')=\frac{(1+Q^2+2Q)(1+P^2-2P)}{(1+P^2+2P)(1+Q^2-2Q)}= \frac{(1+Q)^2(1-P)^2}{(1+P)^2(1-Q)^2} = (A,B;Q,P)^2 $$

So the Kleinian cross ratio is the square of that from the Poincaré model. Therefore the distances will differ by a factor of two. Since cross ratios are invariant under projective transformations (of $\mathbb{RP}^2$ for Klein resp. $\mathbb{CP}^1$ for Poincaré), the above considerations hold for the plane as well.

So which coefficient is the correct one? That depends on your curvature. If you want curvature $-1$, or in other words, if you want an ideal triangle to have area $\pi$ so that angle deficit equals area, then the $\frac12$ in front of the Klein formula is correct as far as I recall. For Poincaré you'd better use coefficient $1$, then the lengths in the two models will match.

If you use coefficient $\frac12$ in the Poincaré model, then you effectively double your unit of length. All length measurements get divided by two, including the imaginary radius of your surface. Since Gaussian curvature is the product of two inverse radii, you get four times the curvature, namely $-4$, just as Post No Bulls indicated.

For the lengths in the Poincare disk models: If the hyperbolic line is an euclidean circle are the euclidean lengths measured as the segment-lengths or as arc-lengths (along the circle)?

Segment lengths (i.e. chord lengths) are certainly correct. I think of the cross ratio as one of four numbers in $\mathbb C$. If you write your differences like this

$$z_{QA}=Q-A=r_{QA}\,e^{i\varphi_{QA}}=\lvert QA\rvert\,e^{i\varphi_{QA}} \in\mathbb C$$

then the cross ratio becomes

$$ (A,B;Q,P)=\frac{(Q-A)(B-P)}{(P-A)(B-Q)}= \frac{r_{QA}\,e^{i\varphi_{QA}}\cdot r_{BP}\,e^{i\varphi_{BP}}} {r_{PA}\,e^{i\varphi_{PA}}\cdot r_{BQ}\,e^{i\varphi_{BQ}}}=\\ =\frac{r_{QA}\cdot r_{BP}}{r_{PA}\cdot r_{BQ}}\, e^{i(\varphi_{QA}+\varphi_{BP}-\varphi_{PA}-\varphi_{BQ})}= \frac{\lvert QA\rvert\cdot\lvert BP\rvert}{\lvert PA\rvert\cdot\lvert BQ\rvert} \in\mathbb R $$

This is because the phases have to cancel out: the cross ratio of four cocircular points in $\mathbb C$ is a real number, so $\varphi_{AQ}+\varphi_{BP}-\varphi_{PA}-\varphi_{BQ}$ has to be a multiple of $\pi$, and in fact I'm sure it will be a multiple of $2\pi$.

This doesn't neccessarily rule out arc lengths, but a simple example using arbitrarily chosen numbers shows that arc lengths result in a different value, so these are not an option.

You do have to use circle arcs instead of chords if you compute lengths as an integral along some geodesic path. So be sure not to mix these two approaches.

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  • $\begingroup$ can you say more about " I think of the cross ratio as one of four numbers in C. Then the phases cancel, and you end up with the absolut values of the four differences mentioned in that fraction. " i did find out that in my practical example the results are the same m but would like tu understand why $\endgroup$
    – Willemien
    Jan 13, 2014 at 16:37
  • $\begingroup$ @Willemien: expanded my answer in that regard. $\endgroup$
    – MvG
    Jan 13, 2014 at 19:07
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There are two common versions of the Poincaré disk model:

  1. Constant curvature $-1$; metric element $\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}$
  2. Constant curvature $-4$; metric element $\frac{dx^2+dy^2}{(1-x^2-y^2)^2}$

The difference is where you stick that annoying factor of $4$. Wikipedia uses version 1, and its description of the relation with Beltrami-Klein model is also based on version 1.

Let's compute the distance from $(0,0)$ to $(x,0)$, $x>0$, in each version:

  1. $\int_0^x \frac{2\,dt}{ 1-t^2 } = \log\frac{1+x}{1-x}$
  2. $\int_0^x \frac{ dt}{ 1-t^2 } = \frac12\log\frac{1+x}{1-x}$

As you can see, it's the second version that has $\frac12$ in front of the logarithm.

The transformation $s=\frac{2u}{1+u^2}$, stated in Wikipedia under "relation to the Poincaré model", exactly doubles the Poincaré model distance from the origin, because $$\log\frac{1+s}{1-s}=2\log\frac{1+u}{1-u}$$

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  • $\begingroup$ thanks that is not the problem i wanted to get attention to, see the example of the Schweikant constant $\endgroup$
    – Willemien
    Dec 27, 2013 at 10:41
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Your formulae are correct for the Beltrami-Klein disc $D$ (the unit open disc on the Euclidean plane) and are completely wrong (in any variant you suggest) for the Poincaré disc $\mathbb D:=\big\{c\in{\mathbb C}\mid|c|<1\big\}$. The isometry between these two discs can be given by the formula ${\mathbb D}\ni c\mapsto\frac{2c}{|c|^2+1}\in D$. (It is easy to see that the indicated map transforms the circles orthogonal to the absolute onto the chords, i.e., geodesics onto geodesics.) Taking distinct points $a,ab\in\partial{\mathbb D}$ on the absolute, you can parametrize the geodesic $\gamma\subset{\mathbb D}$ joining $a,ab$ by its length as $\gamma(t)=a\frac{2(1-b)+be^{2t}}{2(1-b)+e^{2t}}$, $t\in(-\infty,\infty)$. (Now you can experiment with particular values and see what is wrong in your considerations.)

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I wanted to point out that you can have the same metric on diameters without having the same metric all around. The idea is that you can fiddle around with 'angular' distances while preserving radial distances. As an example, you could look at the half plane and the quarter plane. There is a map sending the quarter plane to the hakf plane by doublng the argument of each point. This preserves radial distance but not the whole metric.

A similar thing is happening here.

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  • $\begingroup$ Can you explain more? my major question is now the additional one, "For the lengths in the Poincare disk models: If the hyperbolic line is an euclidean circle are the euclidean lengths measured as the segment-lengths or as arc-lengths (along the circle)? ", i made an experimental setup in GSP but neither seems to be true, it is all very strange to me $\endgroup$
    – Willemien
    Jan 5, 2014 at 14:46
  • $\begingroup$ @Willemien, your experiments gave you the correct answer: neither way is true. In the Poincaré disc model, the (parts of the) circles orthogonal to the absolute are geodesics. But the distance along such a geodesic should be measured in a specific way. Of course, it cannot be the arc-length, nor segment-length, as the whole geodesic should be of infinite length. The analog of the formula that works for the Beltrami-Klein model is simply wrong in any variant, would you measure arc-length or segment-length between the points of the geodesic (= circle). $\endgroup$ Jan 5, 2014 at 20:29
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You write

the same formula is used for calculating the length in the Poincare disk model as for the Beltrami-Klein model

but this is not correct: that formula is not valid in the Poincare disk model, as you can see by following the link "relation to the Poincare mode" provided by @Post No Bulls. The correct distance formula in the Poincare disc is obtained by integrating the length element, also provided by @Post No Bulls, along Euclidean circles that hit the boundary of the Poincare disc model at right angles.

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