Hyperboloid and Klein coordinates

Try

The point with Poincaré disk radius 1/3 has Klein radius 3/5 under k=2w/(1+|w|²). Both represent the same hyperbolic point, but the coordinate pictures preserve different features. The Poincaré disk preserves angles; the Klein disk turns geodesics into Euclidean chords.

A third representation, the hyperboloid, makes the geometry accessible to linear algebra. Use coordinates p=(p₀,p₁,p₂) and the Lorentz bilinear form ⟨p,q⟩L=−p₀q₀+p₁q₁+p₂q₂. This form is not a Euclidean inner product on all of R³.

Predict

Predict whether a negative square norm makes the induced tangent metric negative. The upper hyperboloid is defined by ⟨p,p⟩L=−1 and p₀>0, but its tangent vectors are Lorentz-orthogonal to p. On those tangent planes the form is positive definite.

Predict whether a straight chord in a disk must represent Euclidean shortest distance. The Klein metric is not the Euclidean disk metric, so its straight geodesics do not justify Euclidean angle or length calculations.

Experiment

Follow a radial point through three coordinate systems.

Representation	Point on a common radial geodesic
Hyperboloid	(cosh t,sinh t,0)
Klein disk	(tanh t,0)
Poincaré disk	(tanh(t/2),0)

At t=log 2, the hyperboloid point is (5/4,3/4,0), the Klein radius is 3/5, and the Poincaré radius is 1/3. Check each quotient exactly. Different visible radii do not represent a disagreement about hyperbolic distance; the distance from the common origin is t in every model.

Understand

Write p=(p₀,v) on the upper hyperboloid, so p₀²=1+||v||². A tangent vector w=(w₀,u) satisfies w₀=v·u/p₀. Therefore ⟨w,w⟩L=||u||²−(v·u)²/p₀²≥||u||²/p₀². This is positive for every nonzero tangent vector. Thus the restriction supplies a genuine Riemannian metric.

The curve γ(t)=(cosh t,sinh t,0) lies on the hyperboloid and has velocity of Lorentz square one. Lorentz transformations preserving the positive sheet are isometries, and geodesics are intersections with suitable two-dimensional planes through the origin. The distance relation is cosh d(p,q)=−⟨p,q⟩L. Its full global minimizing proof can be obtained by reducing to the displayed radial curve; this structural model equivalence is used here with that reduction identified.

Central projection k=(p₁/p₀,p₂/p₀) gives ||k||²=1−1/p₀²<1. Conversely k lifts uniquely to (1,k₁,k₂)/√(1−||k||²). Origin planes become straight lines in the chart p₀=1, hence geodesics become Klein chords.

Apply

The Poincaré projection is w=(p₁,p₂)/(p₀+1). Eliminating p gives k=2w/(1+|w|²), which explains the opening conversion. It maps the open disk to itself and fixes its ideal boundary only as a limiting description.

Repair “A Klein triangle’s Euclidean angles are its hyperbolic angles because its edges are straight.” Incidence and straightness survive projective projection, but the induced tangent metric is not a scalar multiple of the Euclidean metric everywhere. Angles must be computed with the correct metric or transferred to the conformal disk.

For optional depth, the hyperboloid distance and geodesic statements are manifestations of a homogeneous-space geometry. A full classification of its isometry group is beyond the coordinate checks here.

Continue

Choose the model that simplifies the intended proof: hyperboloid for dot-product identities, Klein for incidence, Poincaré for angles. Continue to hyperbolic isometries to classify motions by their fixed-point behavior.