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I am learning bits of hyperbolic geometry and the wikipedia page gives two such standard models for it ; the Beltrami Klein (BK) model and the Poincare (P) disk model.

Now as I understand it hyperbolic geometry has exact analogues for every concept of Euclidean geometry except for Euclid's parallel postulate which is not true here.

In particular in the model P the straight lines are given by the diameters and by circular arcs cutting the unit-disk orthogonally. The distances are given by a certain formula, while the angle between two `lines' is the usual Euclidean angle between curves (which corresponds to angle between two circular arcs at the point of intersection of the arcs )

In BK, the straight line between two points are pictured by the usual Euclidean straight lines clipped off at the boundary i.e. the chord containing the points.

But how does one measure an angle between two lines (i.e. the chords of the disk) ? Certainly not via the Euclidean angles between the them (because according to the wiki page on BK )

Two chords are perpendicular if, when extended outside the disk, each goes through the pole of the other

The wiki page does not mention how one measures general angles between two lines in BK. So how does one do that? My class notes and a reference book also seems to skirt around this issue for the BK model.

Do we ``convert'' lines in BK into P via the isomorphism between the models given by the diagram from the wiki page (also included below) ? (i.e. angle between two klein lines is defined as the angle between the corresponding Poincare lines)

enter image description here

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3 Answers 3

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If your lines $a$ and $b$ are given by the equations $a_1 x + a_2 y = a_0$ and $b_1 x + b_2 y = b_0$ where $(x, y)$ are the Klein model coordinates of any point, then the angle $\gamma$ between the two lines satisfies

$$\cos \gamma = \left|\frac{a_1 b_1 + a_2 b_2 - a_0 b_0}{\sqrt{(a_1^2+a_2^2 - a_0^2)(b_1^2 + b_2^2 - b_0^2)}}\right|$$

If the right hand side of this equation is greater than $1$, then $a$ and $b$ do not intersect.

This and similar equations can be more easily derived from the hyperboloid model, where you can use the Minkowski inner product $\langle a, b\rangle := a_1 b_1 + a_2 b_2 - a_0 b_0$, and linear maps preserving this inner product, for all manner of computations and tests. The above equation is then more concisely written as

$$\cos \gamma = \left|\frac{\langle a, b\rangle}{\sqrt{\langle a, a\rangle\langle b, b\rangle}}\right|.$$

In a similar manner, you can compute the distance $d$ between two points $p$, $q$ from $$\cosh d = \left|\frac{\langle p, q\rangle}{\sqrt{\langle p, p\rangle\langle q, q\rangle}}\right|,$$ the distance $d$ between a point $p$ and a line $a$ from $$\sinh d = \left|\frac{\langle p, a\rangle}{\sqrt{-\langle p, p\rangle\langle a, a\rangle}}\right|,$$

and the distance $d$ between two nonintersecting lines $a$ and $b$ from $$\cosh d = \left|\frac{\langle a, b\rangle}{\sqrt{\langle a, a\rangle\langle b, b\rangle}}\right|.$$

Note that if you do not take the absolute value of the right hand side, you get a signed distance depending on the orientation of the lines involved.

These formulas all apply to the Klein model as well if you set the $0$th coordinate of each point to $1$.

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  • $\begingroup$ I find the hyperboloid model most suitable for any actual symbolic or numerical computation in hyperbolic geometry. $\endgroup$
    – Magma
    Jun 9, 2021 at 16:31
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In the Beltrami-Klein model, let the chords ("lines") have respective endpoints ("ideal points") $A$ and $B$, and midpoints $M$ and $N$, as shown. Let them meet at point $T$; and let corresponding circular arcs orthogonal to the bounding circle meet at $T'$. (The arcs are the corresponding "lines" in the Poincaré model.) The BK-modeled angle $\angle ATB$ corresponds to the Poincaré-modeled angle $\theta$, faithfully represented by the Euclidean angle between (tangents to) those arcs.

enter image description here

Defining $\alpha := \angle AOM$, $\beta := \angle BON$, $\gamma := \angle MON$, we have

$$\cos\theta = \frac{\cos\alpha\cos\beta-\cos\gamma}{\sin\alpha\sin\beta} \tag{$\star$}$$

(Note that this can also be viewed as a way to calculate angles in the Poincaré model.)

Proof. Let $P$ and $Q$ be the centers of those orthogonal arcs (aka, the poles of the chords). Taking the bounding circle to have radius $1$, we express $|PQ|^2$ two ways using the Law of Cosines in $\triangle PT'Q$ and $\triangle POQ$:

enter image description here

$$\begin{align} |T'P|^2+|T'Q|^2-2|T'P||T'Q|\cos(180^\circ-\theta) &= |OP|^2+|OQ|^2-2|OP||OQ|\cos\gamma \tag{1} \\[4pt] \tan^2\alpha+\tan^2\beta+2\tan\alpha\tan\beta\cos\theta &= \sec^2\alpha+\sec^2\beta-2\sec\alpha\sec\beta\cos\gamma \tag{2} \\[4pt] \tan\alpha\tan\beta\cos\theta&=1-\sec\alpha\sec\beta\cos\gamma \tag{3} \end{align}$$ and $(\star)$ follows. $\square$


To see that orthogonality in the Beltrami-Klein model occurs when each extended chord meets the other chord's pole (in particular, when $\overleftrightarrow{AM}$ contains $Q$), observe $\triangle OMQ$:

enter image description here

$$\cos\gamma = \frac{\cos\alpha}{\sec\beta}=\cos\alpha\cos\beta \quad\to\quad \cos\theta=0 \tag{4}$$

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  • $\begingroup$ That is a very nice answer. Is it related to the $\cos$ formula in spherical geometry? $\endgroup$
    – orangeskid
    Jun 11, 2021 at 0:31
  • $\begingroup$ @orangeskid: Thanks. :) ... I also noticed that the relation is eerily reminiscent of the Laws of Cosines in spherical trig. It's possible that the hemisphere model provides the appropriate link, but I didn't (and still don't) see an obvious way to leverage it, so I kept my discussion in two flat dimensions. $\endgroup$
    – Blue
    Jun 11, 2021 at 1:51
  • $\begingroup$ When we consider two diameters in the P model, then their hyperbolic angle is exactly the euclidean angle. Since hyperbolic isometries in the P model preserve the angles the hyperbolic angle between any hyperbolic lines in the P model is the euclidean angle. But, this is not the case with BK-model. When the two chordes are diameters their hyperbolic angle is the euclidean. By an hyperbolic isometry these diameters go to two chordes but their euclidean angle is not the initial. So, there are many chordes orthogonal to a given one (obeying the pole rule) but only one is euclidean orthogonal. $\endgroup$
    – Allotrios
    Dec 31, 2021 at 12:22
  • $\begingroup$ @Allotrios: I'm afraid I don't understand the point of your comment. ... I don't say that the angle between hyperbolic lines is given by the Euclidean angle between the chords in the BK-model. I say that the angle is given by the Euclidean angle between the arcs in the P-model. (In my first diagram, I have the $\theta$ between chords in quotation marks to suggest that it doesn't depict the true angle, whereas the $\theta$ between arcs is non-quotationed, because it does. Perhaps that could be clearer.) I use the geometry of both models (P for (*), BK for (4)) to get to the target result. $\endgroup$
    – Blue
    Dec 31, 2021 at 12:45
  • $\begingroup$ @Blue. I didnt say you were wrong, sorry. I was just asking for validation of defining "orthogonality" in BK model by the rule of each line passing through the pole of the other (pole determined by the tangents) we have many perpendicular lines to a given line. And i should have added, "is this correct?" $\endgroup$
    – Allotrios
    Jan 1, 2022 at 9:41
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The correspondence from the lines in the Klein model to lines in the Poincare model preserves the angles. Now, the angle of lines in the Poincare models (which are circle orthogonal to the given unit circle) is the usual angle between two curves, since the model is conformal. So exactly what you said.

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  • $\begingroup$ The correspondence from the lines in the Klein model to lines in the Poincare model preserves the angles. Yes, but how does one define the angle between lines in the Klein model in the first place. For the correspondence to preserve angles we should have an intrinsic way to define a map between two Klein lines without resorting to mapping to the Poincare model, no? As an alternative definition, measuring angle by mapping is fine,but I was wondering if there is something more "intrinsic" $\endgroup$ Jun 9, 2021 at 7:23
  • $\begingroup$ @smilingbuddha: your last sentence in the posting sounded like the main question, so I just confirmed your hypothesis. I don't know another way to geometrically give the angle . Note however, this correspondence between lines also gives a correspondence between points. $\endgroup$
    – orangeskid
    Jun 9, 2021 at 7:39
  • $\begingroup$ What do you mean when you say that "the correspondence from the lines in the Klein model to lines in the Poincare model preserves the angles"? $\endgroup$ Mar 16, 2023 at 14:16

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