Elliptic (Spherical) Geometry

If hyperbolic geometry replaces "exactly one parallel" with "infinitely many," elliptic geometry replaces it with "none at all." Every pair of lines intersects. The simplest model is the surface of an ordinary sphere — mathematicians have studied spherical geometry for navigation and astronomy since antiquity, long before anyone noticed it was a legitimate non-Euclidean geometry. This section gives the formal definition, identifies the geodesics, derives the area formula, and notes the subtle modification (identifying antipodes) that turns the sphere into the genuine elliptic plane.

The sphere as a model

Consider the sphere $S^2(R) = {(x,y,z) : x^2 + y^2 + z^2 = R^2}$ in $\mathbb{R}^3$, with the induced (intrinsic) distance: the distance between two points is the length of the shorter great-circle arc connecting them. The geodesics are exactly the great circles — intersections of the sphere with planes through its centre. The Gaussian curvature is constant $K = 1/R^2 > 0$. For the unit sphere $R = 1$, $K = 1$.

No parallels

Any two great circles lie in planes through the origin, and two distinct planes through the origin intersect in a line through the origin. That line meets the sphere in two antipodal points. So every pair of great circles intersects in two points — there are no parallel lines, and the Euclidean parallel postulate fails maximally. Even worse, two "lines" intersect in two points, not just one, which is awkward for axiomatic geometry.

Girard's theorem on triangle area

The spherical analogue of the hyperbolic angular-defect formula is Girard's theorem. For a spherical triangle on $S^2(R)$ with interior angles $\alpha, \beta, \gamma$:

The angle sum is strictly greater than $\pi$; the angular excess $E = \alpha + \beta + \gamma - \pi > 0$ is exactly the area divided by $R^2$. Smaller triangles have smaller excess; you can verify by drawing a small triangle near a single point on the sphere and seeing that its angle sum is only slightly more than $\pi$.

As in hyperbolic geometry, there are no similar triangles in elliptic geometry — angles determine area and hence side lengths.

The real projective plane

The sphere isn't quite the "right" model for elliptic geometry because two lines meet in two points, not one. Identifying antipodal points solves this: define the real projective plane $\mathbb{RP}^2 = S^2 / {\pm 1}$. Now every pair of distinct great-circles-modulo-antipodes meets in exactly one point. The Euler characteristic of $\mathbb{RP}^2$ is $\chi = 1$ (half that of the sphere), and it is non-orientable — you cannot define a continuous unit normal on it. This is the genuine elliptic plane.

Geometry on Earth and in the sky

The Earth's surface is approximately a sphere of radius about $6371$ km, so for problems at continental scale we are doing spherical geometry whether we admit it or not. The angular excess of a triangle formed by three intercontinental airports (say New York, London, Tokyo) is small but measurable, and ignoring it produces systematic errors in great-circle distance estimates.

Spherical triangles also govern celestial navigation: a star's position is a point on the celestial sphere, and the triangle formed by the star, the observer's zenith, and the celestial pole is solved with spherical trigonometry. The same formulas, with $R = 1$, work for any unit sphere.

Pause and think: A triangle drawn on the sphere with all three angles equal to $\pi/2$ has angle sum $3\pi/2$. What is its area on the unit sphere, and what fraction of the total sphere area $4\pi$ does it occupy? (Hint: imagine the triangle bounded by the equator, the prime meridian, and the meridian at longitude $\pi/2$.)

Try it

• Predict first: on a sphere of radius $R = 10$, a triangle has angles $70^\circ, 80^\circ, 70^\circ$. Compute its area.
• Explain in one sentence why a "Euclidean" architect designing a polygonal courtyard on the surface of the moon would find the sides don't close up as predicted if the courtyard is large enough.
• True or false: a spherical polygon with $n$ sides has angle sum exceeding $(n-2)\pi$ by exactly its area over $R^2$. Justify by triangulating the polygon and summing.
• The real projective plane $\mathbb{RP}^2$ has Euler characteristic $\chi = 1$. Apply Gauss-Bonnet (in the form valid for non-orientable surfaces with the half-density convention) to find $\int K,dA$ for a constant-curvature $K = 1$ metric.

A trap to watch for

The sphere $S^2$ and the elliptic plane $\mathbb{RP}^2$ are different — the former has $\chi = 2$ and is orientable, the latter has $\chi = 1$ and is non-orientable. Textbooks sometimes use "elliptic geometry" loosely to mean either. For navigation on Earth, the sphere is the right model (the Earth's surface is orientable). For axiomatic geometry where each pair of lines meets in exactly one point, the projective plane is the right model. Always check which one the source intends.

What you now know

You can describe spherical geometry as a model for the elliptic case $K > 0$, identify great circles as the geodesics, apply Girard's theorem to compute areas from angular excess, and distinguish the sphere from the projective plane obtained by identifying antipodes. The final section ties the three geometries — Euclidean, hyperbolic, and elliptic — together as one curvature-classified family and previews how higher-dimensional analogues unfold from the same idea.

References

• Garrity, T. (2002). All the Mathematics You Missed: But Need to Know for Graduate School. Cambridge University Press, ch. 8.
• Greenberg, M. J. (2008). Euclidean and Non-Euclidean Geometries (4th ed.). W. H. Freeman, ch. 11.
• Stillwell, J. (2008). Mathematics and Its History (3rd ed.). Springer, ch. 18 (Riemann's habilitation lecture).
• Coxeter, H. S. M. (1969). Introduction to Geometry (2nd ed.). Wiley, ch. 6 and 14 (spherical and projective geometry).
• Penrose, R. (2004). The Road to Reality. Knopf, ch. 27 (cosmological models with positive curvature).