Context: I am an associate professor at a small liberal arts institution in the US.

I am currently preparing to teach geometry this fall. Our course is mostly focused on Euclidean geometry (it's required for all math education majors), but it does have a few weeks of noneuclidean geometry coverage near the end.

We typically study hyperbolic geometry from a "model-based" perspective, either by using half-plane or Poincare disk as our workspace. While students are able to use these models, I feel that they do not truly understand how they connect to/relate to/represent hyperbolic space.

As such, I would like to first show a model for spherical geometry. But spherical geometry is typically simply pictured on the surface of a sphere!

  1. Does anyone know of a "model" for spherical geometry?
  2. Could I simply use a projection of the northern hemisphere in the z-direction to the closed unit disk as my model? That would mean that only the lines or circular arcs connecting antipodal points (or circular arcs along the boundary) would represent "lines" in our original spherical space.
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    $\begingroup$ To clarify, the model of the elliptic plane that relies on the sphere is spherical geometry with antipodal points identified. This level of abstraction is often enough to make this model an interesting learning experience. $\endgroup$
    – user52817
    Jul 31 at 15:01
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    $\begingroup$ If I remember correctly the lines in this geometry are modeled by great circles on the earth. I don't think it would make sense to use less than the whole surface of the earth. $\endgroup$
    – Sue VanHattum
    Jul 31 at 15:57
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    $\begingroup$ @SueVanHattum: If you want the postulate "two distinct points determine a line" to be true, then you need to identify pairs of antipodal points on the sphere. There are many great circles going from the North Pole to the South Pole. But in the identified geometry these two poles represent just one point. $\endgroup$
    – user52817
    Jul 31 at 16:18
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    $\begingroup$ Not responding about spherical geometry, but still possibly of interest: while the upper half-plane and Poincare disk models are immediately isomorphic via a smidgen of complex analysis, the Beltrami disk model of hyperbolic two-space, where geodesics are (Euclidean) straight lines (restricted to the disk), is considerably harder to compare to the Poincare disk model. :) $\endgroup$ Jul 31 at 16:29
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    $\begingroup$ @SueVanHattum Yes, you are correct that the great circles are the geodesics ("lines") on the surface of a sphere. It makes for fairly easy examples that students can clearly see without the use of a "flat model". But I am looking for a flat model of spherical geometry! (I want to use the idea to introduce the more complicated hyperbolic models.) $\endgroup$ Jul 31 at 16:39

5 Answers 5


Here are three projections that result in models of spherical geometry. I think the stereographic model is closest to what you're looking for.

Stereographic projection

One possible stereographic projection projects from a pole onto the equatorial plane, giving a model that is analogous to the Poincaré disk model. (Projecting onto the equatorial plane identifies the equator with the unit circle. Other stereographic projections are scaled differently.)

Diagram of stereographic projection. By Leonid 2, CC-BY-SA 3.0.

Image credit to Leonid 2, CC-BY-SA 3.0.

This stereographic model has many desirable properties for geometry. It is conformal/angle-preserving, it can represent the entire sphere using the plane plus a point at infinity (cf. the Riemann sphere), and the great circles are the lines and circles that intersect the unit circle at two ends of a diameter. (This includes the unit circle itself.) Here's a nice visualization in GeoGebra. The stereographic projection is the only projection that represents all spherical circles as circles. (If we consider lines to be circles with infinite radius.)

This lab assignment from a 2005 Geometry for Teachers course at the University of Washington taught by James King has teachers explore great circles and spherical triangles under stereographic projection. You can find even more of his resources on this page.

We can think of the Poincaré disk model as the projection of one sheet of the unit hyperboloid of two sheets onto the equatorial plane, with the projection center being the opposite vertex.

Projective relation of hyperboloid and Poincare models. By Selfstudier, CC0.

Image credit to Selfstudier, CC0.

To facilitate comparison of the two models, we can restrict the stereographic model to only the projection of the hemisphere opposite the pole. This would make it a disk model. By identifying opposite points on the boundary, we could also consider it to be a model of elliptic geometry, which as this answer explains is distinct from spherical geometry.

Ian Stevenson developed a computer "microworld" for exploring these two models, and you can read more about it in his doctoral thesis or in this article. He attributes the stereographic model to Klein. He calls it the "hot-plate universe," and he calls the Poincaré disk model the "cold-plate universe." This language is from Jeremy Gray's Ideas of Space, and Gray reports getting it from Feynman. The idea is that, if we use a metal ruler as our metric, the ruler will grow as we move away from the origin in the stereographic model, and the ruler will shrink as we move away from the origin in the Poincaré disk model. Thus, paths far from the origin are shorter than they appear in the stereographic model, and they are longer than they appear in the Poincaré disk model.

You can read about pre-service teachers' investigation of the models in Stevenson's microworld in this article, this article, and this article.

Orthographic projection

Flattening the northern hemisphere in the z-direction onto the equatorial plane as you describe in the question is an orthographic projection. The orthographic projection is easy to visualize, but it isn’t conformal, only shows half the sphere, and crucially, an orthographic projection does not map great circles to circular arcs.

To see this, consider the orthographic projection of the globe shown. The edge of the projection is a circle representing the 30W and 150E meridians. The other meridians except for the central meridian (90E) are tangent to the outer circle at both poles, so they cannot be circular arcs. In general, an orthographic projection maps spherical circles to (possibly degenerate) ellipses. This article on spherical geometry by Lodge and Heawood discusses the drawing of spherical circles under orthographic and stereographic projections.

Orthographic projection of the Earth from 30W to 150E. By Planiglobe, CC-BY 2.5.

Image credit to Planiglobe, CC-BY 2.5.

Gnomonic projection

A gnomonic projection projects from the center of the sphere onto a tangent plane. The resulting model of a hemisphere is analogous to the Klein disk model, which can be thought of as a projection of one sheet of the unit hyperboloid from the origin onto a plane tangent to the vertex. Like the Klein disk model, it represents geodesics as straight lines, but it also isn't conformal. Unlike the Klein disk model, it uses the entire plane. And to be a faithful model of elliptic geometry, we need the entire real projective plane.

Gnomonic projection (projection from the center of the sphere). By Marozols. CC-BY-SA 3.0.

Image credit to Marozols, CC-BY-SA 3.0.


I often teach a geometry class like the one you describe and always try to include both elliptic and hyperbolic plane geometry.

As you indicate, elliptic geometry is modeled by spherical geometry, but we must take care to identify antipodal points. The reason why this needs to be done is to guarantee that the postulate "two distinct points determine a unique line" holds true in the model. For example, {North Pole, South Pole} is just one Point in the model. I find this level of abstraction is enough to make for an interesting learning experience in the course.

As a consequence, lines have finite length $\pi R$ and the total area of the plane is $2\pi R^2$. Another interesting thread is regular tilings of the elliptic plane. These correspond to central projections of Platonic solids out to a circumscribing sphere. Four of the five solids have antipodal symmetry so give tilings of the elliptic plane. The exception is the tetrahedron.

Another interesting feature of the geometry resulting from the antipodal identification is that it is not orientable. When you slide a right handed glove around it becomes left handed.

Bringing all these concepts together can make for about three interesting classes/lectures if you also include the formula for the area of a triangle in terms of the angles and the parameter $R$.

To bring more history into the discussion after hyperbolic geometry, one can explain how geometers first started developing hyperbolic geometry, before elliptic. They were assuming that they would find contradictions, thereby proving the parallel postulate. They started finding some strange formulas. But it was realized that any such formula could be transformed into a formula for spherical trigonometry by the transformation $k\rightarrow ik$. In view of the evident consistency of spherical trigonometry, it was realized that the "strange new world" of hyperbolic non-Euclidean geometry must be consistent!

  • $\begingroup$ Could someone please explain: " {North Pole, South Pole} is just one Point in the model." I find this very confusing as to me they are obviously two points. Is this model mapping all opposite points to the same one (sort of like an absolute value in one of the dimensions)? Is there a variable transform that works here? $\endgroup$ Aug 2 at 16:01
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    $\begingroup$ @JosephDoggie A point in the model is the equivalence class of two points on the sphere, where $p$ and $q$ are equivalent if $p=-q$. So if $(x,y,z)$ is a point on the sphere, then it is the same as $(-x,-y,-z)$. If you travel from the North Pole to the South Pole, then in the model of the elliptic plane, you are back where you started. Hilbert said "In geometry we must always be ready to replace 'points', 'lines' and 'planes' by 'tables', 'chairs' and 'beer mugs'." We are replacing 'line' by 'great circle' and 'point' by an equivalence class under $p\to -p$. $\endgroup$
    – user52817
    Aug 2 at 17:13

Imagine the world were a big glass ball, that is, transparent all the way through, and you could see your "shadow" on the antipodal point at all times.

So as you run around the spherical surface, your "shadow" runs around on the surface at the opposite end of the earth in exactly the opposite direction to you at all times.

Extrapolate that out to imagining space to be like that.


One relatively simple projection starts with a disk, like tge Poibcaré disk of hyperbolic geonetry. But we will define the "lines" differently. The interior os the disk will correspond to half the sphere, the exterior to the other half.

Define a line as a Euclidean circular arc passing through two points (which may be on, insideor outside the disk boundary), and intersecting the boundary at two diametrically opposite points. If you draw the three "lines" defined in this way passing pairwise through three points, you see the sides of the formed triangle bulging out as they would when viewing a spherical triangle. So you satisfy the angukar sum exceeding 180°. Indeed, with the right selection of a metric to measure distances, the triangle meets all the trigonometric laws applying to spherical triangles!


Assuming you want to talk about spherical and elliptical geometry, you might want to try this. It's gonna sound kind of crazy, but it's worked (at least slightly) for me; it will definitely get their attention.

  • Take an inflatable sphere-like object. A beach ball might do, or a large yoga ball. Big is good.
  • Inflate (possibly before class) and discuss what axioms do work. Maybe positive curvature and triangle area formulas, if you've covered some of that.
  • Note that not all neutral axioms are satisfied (see @user52817 answer, for instance).
  • Now deflate the ball in such a way that the "bottom half" sort of curves inside the rest of the ball. This is much easier with a large exercise/yoga ball than a beach ball, but in principle both might work. At the least you should be able to get a kind of hemisphere with the "bottom half" smashed under the rest, if you move it in such a way to "keep air in the top half".
  • Discuss with students whether two points might determine a line uniquely now. (It won't because the "equator" will probably still all be visible, but it will be a lot closer, and should get them talking about "identification of antipodes", which implies that each great circle is in some sense the same as the "visible half" of the great circle when we require this axiom.)

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