Is there a way to show the infinite Euclidean plane in a finite way?

Question

I'm a huge fan of hyperbolic geometry. I find the Poincaré disk pretty impressive because it represents an infinite space within a finite space.

Is there a way to show the infinite Euclidean 2D space (aka an Euclidean plane) in a finite way? Is it feasible or just unthinkable?

Related (but not a duplicate): Analogy of spherical and hyperbolic geometry projection. — Andrew D. Hwang, May 21 '17 at 14:21
The beginning of this answer references the article "A Euclidean model for Euclidean geometry", in which Adolf Madur describes (what I called) the "Gans disk" model of Euclidean geometry. Lines there are diameters, as well as half-ellipses with diameters for major axes, and the angle between two "Gans lines" is simply the traditional angle between the corresponding diameters. — Blue, May 21 '17 at 16:00

score 2 · Answer 1 · answered May 21 '17 at 14:49

$\newcommand{\Reals}{\mathbf{R}}$If $\phi:\Reals^{2} \to D$ is a bijection to a bounded set, then $D$ can be viewed as a model of the Cartesian/Euclidean plane. With no further constraints, however, such a model is unlikely to satisfy: Lines map to (effectively) arbitrary curves, rigid motions look "far from rigid" in $D$, and so forth.

Here are a couple of "relatively nice" representations of the Euclidean plane in a bounded spatial or planar region:

Stereographic projection from $(0, 0, 1)$ (below, left) $$ \phi(u, v) = (x, y, z) = \frac{(2u, 2v, u^{2} + v^{2} - 1)}{u^{2} + v^{2} + 1},\qquad \phi^{-1}(x, y, z) = (u, v) = \frac{(x, y)}{1 - z}, $$ maps the plane to the complement of the north pole $N = (0, 0, 1)$ in the unit sphere $S^{2} \subset \Reals^{3}$.

These maps are conformal (angle-preserving), Euclidean lines map to circles through $N$, and Euclidean rigid motions look not unlike hyperbolic rigid motions acting on the Poincaré disk.
Central/gnomonic projection from $(0, 0, 1)$ (below, right) $$ \phi(u, v) = (x, y, z) = \frac{(u, v, -1 + \sqrt{u^{2} + v^{2} + 1})}{\sqrt{u^{2} + v^{2} + 1}},\qquad \phi^{-1}(x, y, z) = (u, v) = \frac{(x, y)}{1 - z} $$ maps the plane to the "southern hemisphere" $x^{2} + y^{2} + (z - 1)^{2} = 1$, $0 \leq z < 1$. (If desired, map the open hemisphere to the open unit disk, either by vertical projection, or by stereographic projection from $(0, 0, 2)$.)

Every Euclidean line determines a unique plane through the center of the sphere, and therefore maps to half a great circle. Again, Euclidean rigid motions look not unlike hyperbolic rigid motions acting on a disk model of the hyperbolic plane.

Craig Reynolds · Answer 2 · 2020-07-16T01:12:15.977

[Updated about a week later]

I was interested in this same issue. Inspired by the Poincaré disk, I wanted to draw a disk on which an infinite plane was mapped. The plane is “decorated” with a procedural texture, so this is like the computer graphics operation of texture mapping.

Rendering the disk uses the inverse of the plane-to-disk map: given a position (of a pixel) on the disk, the inverse transform finds the corresponding point on the infinite “source” plane. If that “destination” point $d$ is on the disk, $\left\lVert d \right\rVert <1$, then the corresponding “source” point $s$ on the input texture is defined by this mapping, parameterized by scale $a$ and exponent $b$.

$$s= \frac{d} {a(1-\left\lVert d \right\rVert^b)}$$

Shown below is a portion of source texture plane. This texture is defined procedurally across all of ℝ², or at least the part spanned by two floating point numbers: