Measuring angles in the Beltrami Klein model of Hyperbolic geometry

Question

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)

Answer 1

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.

Answer 2

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 $1$.

Answer 3

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.

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 $\triangle POQ$:

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

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