Elliptic Geometry
Compare the sphere and antipodal quotient, then measure great-circle distance and spherical excess.
Learning goals
- Distinguish single and double elliptic geometry.
- Identify great circles and their stereographic images.
- Calculate distance and area with the correct model and radius.
The sphere and its great circles
A straight path on a curved surface
A great circle is the intersection of a sphere with a plane through its center.
Its shorter arcs give shortest surface paths between nonantipodal points, making great circles the spherical analogues of straight lines. Small circles such as most lines of latitude do not generally have this property.
Imagine stretching a thread along the surface of a sphere between nearby points. The shortest route follows a great-circle arc, not a straight chord through the interior. This is our first reminder that distance belongs to the surface being studied. We will use that viewpoint when deciding what counts as a line.
Great circle
A plane through the center of a sphere cuts out a great circle. Its shorter arcs give shortest paths between nonantipodal points. A circle cut out by a plane away from the center is a small circle.
On the sphere, antipodal points lie on infinitely many great circles. Distinct nonantipodal points determine one great circle. Any two distinct great circles intersect at two antipodal points, so there are no parallel great circles.
Three right angles
Take the north pole and two equatorial points separated by a quarter-turn. Find the angles of the triangle bounded by the short great-circle arcs.
Show worked solution
Each meridian meets the equator perpendicularly, giving two right angles. The meridians are a quarter-turn apart at the pole, giving the third. The triangle occupies one octant of the sphere.
Single versus double elliptic geometry
The sphere gives us great-circle geometry, but we must decide whether opposite points count as different locations. Identifying antipodal points changes the global geometry and the distance between represented points. Keep that choice explicit before transferring a spherical calculation to the elliptic model.
| Model | Points | Length of a complete line on unit radius | Total area |
|---|---|---|---|
| Double elliptic | Individual sphere points | \(2\pi\) | \(4\pi\) |
| Single elliptic | Antipodal pairs | \(\pi\) | \(2\pi\) |
Identifying antipodal points turns two intersections of great circles into one point. A disk model keeps one hemisphere and identifies opposite points on its boundary. Unlike the hyperbolic disk, that boundary belongs to the geometry.
Two points determine one line in the quotient
Prove line uniqueness for two distinct antipodal pairs.
Show worked solution
Each pair determines a one-dimensional subspace through the sphere center. Distinct pairs determine distinct subspaces, which span exactly one two-dimensional plane. Its great circle, modulo antipodal identification, is the unique line containing both pairs.
Do not mix distances on the sphere with distances between antipodal pairs. They are different metric spaces.
Elliptic transformations
The sphere suggests rotations as natural motions, but our elliptic model also identifies antipodal points. Check that a proposed transformation respects that identification, so the two representatives of one point do not acquire inconsistent images. This connects the spherical picture to a well-defined action on the elliptic plane.
\[T(z)=\frac{az+b}{-\bar b z+\bar a},\qquad |a|^2+|b|^2=1\]These complex formulas represent rotations of the unit sphere under stereographic projection. They preserve the antipodal pairing. The coefficients and their simultaneous negatives represent the same map.
Check a simple rotation
For \(a=e^{i\theta/2},b=0\), identify \(T\).
Show worked solution
It rotates the plane about zero and corresponds to a rotation about the north–south axis of the sphere.
Great-circle equation in the plane
Project a sphere plane \(AX+BY+CZ=0\).
Show worked solution
If C is zero this is a line through the origin; otherwise it is a circle. It contains the antipodal partner of every one of its points.
Distance: keep track of identification
On the sphere, distance follows a great-circle arc. In the elliptic plane, either representative of an antipodal pair describes the same destination. We therefore compare the possible spherical routes before choosing the elliptic distance.
Distance formulas
For unit vectors \(u,v\) on the sphere, their spherical distance is their central angle:
\[\delta=\arccos(u\cdot v),\qquad0\le\delta\le\pi\]For antipodal pairs the shorter of the two choices is used:
\[d([u],[v])=\min(\delta,\pi-\delta)=\arccos|u\cdot v|\]In the complex chart, \(ds=2|dz|/(1+|z|^2)\). Moving one endpoint to zero and integrating a radius gives:
\[\delta(z,w)=2\arctan\left|\frac{z-w}{1+\bar zw}\right|\]A zero denominator gives the limiting value pi, meaning antipodal sphere points. Take the minimum with the supplementary distance for single elliptic geometry.
Same formula, different model
For \(z=0,w=2\), compare the distances.
Show worked solution
The spherical distance is \(2\arctan2\approx2.2143\). The single elliptic distance is \(\pi-2\arctan2\approx0.9273\), because the antipodal representative \(-1/2\) is nearer to zero.
Choose representatives before measuring
A single elliptic point \([u]\) represents the pair of opposite unit vectors \(\{u,-u\}\). Changing either representative must leave a distance unchanged. Since \((-u)\cdot v=-(u\cdot v)\), the expression \(\arccos(u\cdot v)\) alone fails this test, whereas \(\arccos|u\cdot v|\) passes. On a sphere of radius R, multiply this angle by R to obtain length.
Two representatives, one answer
Unit directions u and v satisfy \(u\cdot v=-3/5\). Find the distance between their elliptic classes for sphere radius \(2\). Explain whether changing v to its antipode changes the answer.
Show worked solution
The nearer representative is −v, whose dot product with u is \(3/5\). The distance is \(2\arccos(3/5)\approx1.8546\). Choosing v initially gives the supplementary spherical angle, but taking the shorter representative yields the same result. At zero dot product both choices are equally near: the distance is \(R\pi/2\).
In the quotient, a unique complete line through two distinct points does not always mean a unique shortest arc. Orthogonal representative directions lie halfway around their projective line, giving two equal shortest arcs. Specify the chosen arcs when describing a triangle.
Area from angular excess
Hyperbolic triangles had an angular deficit. On the sphere the comparison goes the other way: a triangle has excess angle, and that excess measures its area. The sign change reflects curvature rather than a different choice of algebraic notation.
Triangle area
For a convex triangle bounded by specified short great-circle arcs on a sphere of radius R:
\[A=R^2(\alpha+\beta+\gamma-\pi)\]On the unit sphere, a lune of angle alpha occupies the fraction alpha over two pi of the whole sphere, hence has area twice alpha. The three lune pairs determined by the triangle’s sides cover the sphere once and the triangle and its antipode an extra two times. Thus four times the sum of the angles equals four pi plus four times the triangle area. Rearranging gives the excess formula.
Area of an octant
Find the area of the three-right-angle triangle on a sphere of radius two.
Show worked solution
The full sphere has area sixteen pi, so this is one eighth as expected.
Practice: a positive excess
A unit-sphere triangle has angles \(80^\circ,70^\circ,60^\circ\). Find the area.
Show worked solution
Circles, area, and circumference
We have measured triangular regions using their angles. Now consider all points at a fixed intrinsic distance from a center. That distance is measured along the surface, so it should not be confused with the radius of the visible circle in space. Keep the spherical or antipodal convention in view when interpreting the formulas.
On the unit sphere, a metric disk of radius R up to pi is a spherical cap. A circle of angular radius R has Euclidean radius sine R in its cutting plane.
\[C=2\pi\sin R,\qquad A=2\pi(1-\cos R)=4\pi\sin^2(R/2)\]Derive cap area
Integrate the spherical area element to find the area within angular radius R of a pole.
Show worked solution
The angular variable runs around the pole and the polar variable runs from the pole to the boundary. For single elliptic disks use radii below pi over two; at the maximal radius the quotient cut locus needs separate treatment.
A great circle is a metric circle on the sphere with radius pi over two. Its role as a line does not prevent it from being a distance locus.
Proof practice: spherical Pythagoras
A right angle still carries useful information on a sphere, but the Euclidean squared-length formula cannot simply be assumed. Track how side lengths correspond to central angles and use the spherical relationship. The behavior of a very small triangle offers a check on the result, not a substitute for its proof.
The right-triangle identity
For a unit-sphere right triangle with legs \(a,b\) and opposite side \(c\), prove \(\cos c=\cos a\cos b\).
Show worked solution
Rotate the right-angle vertex to \((0,0,1)\) and its perpendicular tangent directions to the coordinate axes. The other vertices are \((\sin a,0,\cos a)\) and \((0,\sin b,\cos b)\). Their dot product is \(\cos a\cos b\); by the central-angle definition it also equals \(\cos c\).
At very small lengths, expanding cosine to second order recovers the Euclidean Pythagorean relation to leading order.
Elliptic Geometry: discussion A–E
Work through each question before revealing its solution. For proof questions, compare both the reasoning and the required hypotheses.
Question A
Compare the formulas for elliptic and hyperbolic transformations.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Both are fractional linear and contain rotations. Their normalized matrices differ by a sign: elliptic matrices are \(\begin{pmatrix}a&b\\-\bar b&\bar a\end{pmatrix}\) with \(|a|^2+|b|^2=1\), while disk-isometry matrices have \(\begin{pmatrix}a&b\\\bar b&\bar a\end{pmatrix}\) with \(|a|^2-|b|^2=1\). Correspondingly the centered formulas have denominators \(1+\bar z_0z\) and \(1-\bar z_0z\). The disk parameter requires \(|z_0|<1\); spherical transformations have no disk restriction. The centered spherical formula with finite \(z_0\) omits maps whose zero is infinity, so the matrix formula is the complete description.
Question B
In the sphere and plane versions of Figure 11.1, identify the lines through the labeled points and the pairs lying on more than one line.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
The two drawn great circles through \(z\) also pass through its antipode \(z^d\), and the three through \(w\) pass through \(w^d\). In the sphere drawing these are intersections with planes through the corresponding diameter. The pairs \((z,z^d)\) and \((w,w^d)\) each lie on infinitely many great circles, although only a few are drawn. A nonantipodal pair determines one plane through the origin and hence one great circle.
Question C
Give two distinct repeating decimal representations that are identified as the same rational number.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
For example \(0.5000\ldots=0.4999\ldots\). Indeed \(0.0999\ldots=\sum_{n=2}^\infty9\cdot10^{-n}=0.1\), so adding \(0.4\) proves the equality. The representations differ, but the real number does not.
Question D
Name a familiar reentrant curve in Euclidean and hyperbolic geometry.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
A circle is reentrant: following it once returns to the starting point. Hyperbolic circles have finite circumference \(2\pi\sinh R\), even though hyperbolic geodesics have infinite length.
Question E
Is a horocycle reentrant?
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
No. In the upper half-plane it is isometric to \(y=1\), whose intrinsic length coordinate is \(x\in\mathbb R\). It never closes. Its disk image may look like a Euclidean circle, but the ideal tangency point is excluded and is infinitely far along the horocycle.
Elliptic Geometry: exercises 1–6
Work through each question before revealing its solution. For proof questions, compare both the reasoning and the required hypotheses.
Exercise 1
Prove the elliptic transformations form a group.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Let \(J(z)=-1/\bar z\) be the antipodal involution on the sphere, with \(J(0)=\infty\). The elliptic group consists precisely of Möbius transformations commuting with \(J\). The identity commutes with \(J\); if \(S,T\) do, then \((ST)J=SJT=JST\); and \(TJ=JT\) implies \(T^{-1}J=JT^{-1}\). Thus it is a group of bijections. Equivalently the normalized matrices \(\begin{pmatrix}a&b\\-\bar b&\bar a\end{pmatrix}\) of determinant one are closed under multiplication and inverse.
Exercise 2
Show a nonidentity elliptic-group transformation is elliptic in the Möbius classification and lifts to a rotation of the sphere.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
A Möbius transformation has a fixed point \(p\) on the sphere. Commutation with the antipodal map makes \(J(p)\) another fixed point, distinct from \(p\). A spherical transformation \(U\) can send this antipodal pair to \(0,\infty\): for finite \(p\), use \((z-p)/(1+\bar pz)\) up to a unit scalar, and handle infinity by a rotation. Then \(UTU^{-1}(z)=\lambda z\). Preserving antipodes gives \(\lambda=1/\bar\lambda\), hence \(|\lambda|=1\). Because \(T\) is not the identity, \(\lambda\ne1\).
The lift of \(z\mapsto e^{i\theta}z\) is rotation through \(\theta\) about the polar axis. The lift of \(U\) is a sphere isometry, as follows either from the preserved line element \(2|dz|/(1+|z|^2)\) or the stereographic dot-product formula. Conjugation rotates the axis to the diameter through \(p\). The identity must be treated separately; it fixes every point and is the zero-angle rotation.
Exercise 3
Prove stereographic projection carries sphere great circles exactly to elliptic straight lines.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
A great circle is the sphere's intersection with a plane through its center. It contains the antipode of every point on it. Stereographic projection maps circles to clines and antipodes to \(J(z)=-1/\bar z\), so its image has the defining elliptic-line property.
Conversely lift an antipode-invariant cline to a sphere circle. If its plane is \(n\cdot u=c\), both \(u\) and \(-u\) lie in it, so \(c=-c\) and \(c=0\). Thus its plane passes through the center and it is a great circle. Here “circle” in the planar definition must include generalized circles, or the meridians would be excluded.
Exercise 4
Prove elliptic lines through zero are Euclidean straight lines and are the only elliptic lines that are Euclidean straight lines.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
An elliptic line through zero contains its antipode infinity, so it is a Euclidean line. Conversely a Euclidean line that is elliptic contains infinity and therefore \(J(\infty)=0\). Every Euclidean line through zero is indeed antipode-invariant, since \(-1/\bar z\) is a real multiple of \(z\) for nonzero \(z\).
Exercise 5
Discuss the first four Euclidean postulates in the single elliptic model, with the necessary interpretation of extension and radius.
Qualifies the source assertion: the compact projective plane has finite diameter, so arbitrary metric radii are impossible.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
(1) Two distinct projective points are distinct diameters of the sphere; their span is a unique plane and therefore determines one projective line. (2) A geodesic can be continued for any parameter length, but it closes after length \(\pi\) on the unit model; indefinite travel is not unbounded distance. (3) Metric circles exist for \(0<R<\pi/2\). At \(R=\pi/2\) the locus is a projective line, and for \(R>\pi/2\) it is empty. Thus the unrestricted Euclidean postulate about every radius is not literally valid with shortest-path distance. (4) Rotations preserve angles and act transitively on orthogonal tangent frames, so all right angles are congruent.
Exercise 6
Prove directly in the disk model that two single-elliptic lines cannot be parallel.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Rotate one line to the real diameter. The other line is represented by an antipode-invariant cline. If it is a Euclidean line, it also passes through zero and intersects the first there. Otherwise let its circle have center \(a\) and radius \(r\). Antipodal invariance forces \(r^2=1+|a|^2\): substituting \(-1/\bar z\) into its equation proves this relation. On the real axis its intersections solve \(x^2-2\operatorname{Re}(a)x-1=0\), with two real roots of product \(-1\). One lies in the closed unit disk; if both are boundary points they are identified. Thus the two projective lines intersect.
Elliptic Geometry: exercises 7–12
Work through each question before revealing its solution. For proof questions, compare both the reasoning and the required hypotheses.
Exercise 7
Find the length of a line, the area of an angle-\(\alpha\) two-gon, and the total area in double elliptic geometry.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Double elliptic geometry is the unit sphere without antipodal identification. A great circle has length \(2\pi\). A lune of angle \(\alpha\) occupies the fraction \(\alpha/(2\pi)\) of the sphere when measured between its two meridians, and has area \(2\alpha\); alternatively integrate \(\int_0^\alpha\int_0^\pi\sin\theta\,d\theta\,d\phi\). The whole sphere has area \(4\pi\). Do not double the single-projective two-gon formula blindly: the standard spherical two-gon is one lune, while its antipodal lune is a second region.
Exercise 8
Show a fixed-distance cycle in the single elliptic plane is represented by portions of one or two Euclidean circles after the disk identifications.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Let \(n\) be a unit sphere representative of the center. Projective distance is \(d([u],[n])=\arccos|u\cdot n|\). For \(0<r<\pi/2\), the locus lifts to the two small circles \(u\cdot n=\pm\cos r\), exchanged by the antipodal map. Their stereographic images are clines, and restricting representatives to the disk may show one circle or two portions. A circle through the north pole projects to a straight line, so “Euclidean circles” must allow generalized circles. At \(r=\pi/2\) the two lifts coincide in a great circle.
Exercise 9
At which radii is an elliptic cycle a line, and at which is it not a curve?
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
In the single elliptic plane, the maximum distance is \(\pi/2\). At \(r=\pi/2\), the lift is \(u\cdot n=0\), a great circle, so the cycle is a projective line. At \(r=0\) it is the center point, not a curve; for \(r>\pi/2\) it is empty. On the unquotiented sphere instead, \(r=\pi\) is the antipodal point. The distinction prevents contradictory answers.
Exercise 10
Explain why every Euclidean circle in the plane represents an elliptic cycle or an elliptic straight line.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Its inverse stereographic image is a sphere circle in a plane \(n\cdot u=c\), with \(|n|=1\) and \(|c|<1\). If \(c=0\), it is a great circle. Otherwise it is a spherical distance circle centered at \(\operatorname{sgn}(c)n\) of radius \(\arccos|c|<\pi/2\). Under antipodal identification its image is the corresponding single-elliptic cycle. Its other antipodal lift represents the same projective points. “Represents” matters: as subsets of the unquotiented plane, the full projective distance locus may include the antipodal companion circle as well.
Exercise 11
Do elliptic triangles have incircles?
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Yes for a nondegenerate convex geodesic triangle with the chosen interior. Lift its triangular region to the sphere and choose unit inward normals \(n_1,n_2,n_3\) to its side planes. There is a unit vector \(u\) in the interior with \(n_1\cdot u=n_2\cdot u=n_3\cdot u>0\): take the internal angle bisector from a vertex to the opposite side and use continuity of its distance to that side minus its equal distances to the other two. The endpoint signs are opposite, exactly as in the hyperbolic proof.
The common distance to the great-circle sides is \(r=\arcsin(n_j\cdot u)\). The disk of radius \(r\) about \(u\) lies in all three side hemispheres and is tangent to each. Projecting gives an incircle in the chosen elliptic triangle. Choosing a different triangular region formed by the same three projective lines may give a different incenter.
Exercise 12
Do elliptic triangles have circumcycles?
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Choose unit sphere representatives \(u,v,w\) of the three vertices. Their affine plane has equation \(n\cdot x=c\) with \(|n|=1\). For a nondegenerate projective triangle the vectors are linearly independent, so \(c\ne0\); otherwise the three points would lie on one great circle. Thus \[d([n],[u])=d([n],[v])=d([n],[w])=\arccos|c|.\] This cycle passes through all three projective vertices. Different choices of signs for the sphere representatives can yield different circumcycles; existence is the claim, not a blanket uniqueness statement.
Elliptic Geometry: exercises 13–18
Work through each question before revealing its solution. For proof questions, compare both the reasoning and the required hypotheses.
Exercise 13
What is the locus at fixed perpendicular distance from an elliptic line?
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Write the line as \(u\cdot n=0\) on the sphere. Distance from \([u]\) to the line is \(\arcsin|u\cdot n|\). Setting it equal to \(d\), \(0<d<\pi/2\), gives \(|u\cdot n|=\sin d\). This is the cycle centered at \([n]\) with radius \(\pi/2-d\). At \(d=0\) it is the original line; at \(d=\pi/2\) it collapses to the pole \([n]\).
Exercise 14
Prove an elliptic triangle's angles sum to more than \(180^\circ\).
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
For a nondegenerate convex triangle on the unit spherical/projective model, the area theorem gives \(A=\alpha+\beta+\gamma-\pi\). Since its interior has positive area, the angle sum exceeds \(\pi\). Degenerate triples or an unspecified choice of long arcs are not covered by this triangle convention.
Exercise 15
State and prove the elliptic Pythagorean relation.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
For a convex spherical right triangle with minor-arc legs \(a,b\) and opposite minor arc \(c\), \[\cos c=\cos a\cos b.\] Put the right vertex at \(u=(0,0,1)\), and the others at \(v=(\sin a,0,\cos a)\) and \(w=(0,\sin b,\cos b)\). Their tangent directions at \(u\) are perpendicular. Since sphere distance is the central angle, \(\cos c=v\cdot w=\cos a\cos b\).
For single-elliptic shortest sides \(a,b\le\pi/2\), these representatives give \(v\cdot w\ge0\), so the same equation holds for projective distance \(c=\arccos|v\cdot w|\). Other choices of long spherical arcs require the explicitly chosen spherical lengths.
Exercise 16
Derive the distance formula \(2\arctan|(z_1-z_2)/(1+\bar z_1z_2)|\) and specify which elliptic model it measures.
The displayed source formula is the double-elliptic (spherical) distance.
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
A spherical isometry sends \(z_1\) to \(0\) and \(z_2\) to a point of modulus \(r=|(z_2-z_1)/(1+\bar z_1z_2)|\). The radial line element integrates to \[d_{S^2}=\int_0^r\frac{2\,dt}{1+t^2}=2\arctan r.\] If the denominator is zero, \(r=\infty\) and the distance is \(\pi\), because the points are antipodal.
In the single elliptic plane antipodes represent the same point, so the correct distance is \[d_{\mathrm{proj}}=\min\{d_{S^2},\pi-d_{S^2}\}.\] For example \(z_1=1,z_2=-1\) have spherical distance \(\pi\) but projective distance zero. The unqualified source formula must not be used as a projective metric.
Exercise 17
Find the area enclosed by an elliptic cycle of radius \(R\).
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
Center it at zero and put \(r=\tan(R/2)\). Integrating yields \[A=\int_0^{2\pi}\int_0^r\frac{4t}{(1+t^2)^2}\,dt\,d\theta=\frac{4\pi r^2}{1+r^2}=4\pi\sin^2(R/2)=2\pi(1-\cos R).\] On the sphere this is the cap area for \(0\le R\le\pi\). In the projective plane it is the metric-ball area for \(0\le R\le\pi/2\), with the endpoint giving the whole area \(2\pi\) (the boundary has zero area). A projective line does not bound an ordinary disk globally, so describe this endpoint as a metric ball limit.
Exercise 18
Find the circumference of an elliptic cycle in terms of radius \(R\).
Show worked solution
Approach. Normalize the figure using an isometry, then apply the definition.
At zero, \(r=\tan(R/2)\) and \[C=\int_0^{2\pi}\frac{2r}{1+r^2}\,dt=2\pi\sin R.\] This holds for sphere distance circles with \(0<R<\pi\), and projective cycles with \(0<R<\pi/2\). At the projective endpoint \(R=\pi/2\), antipodal identification folds the boundary circle two-to-one, so the resulting projective line has length \(\pi\), not the limiting circumference \(2\pi\).
References and further study
Primary source: Michael Henle, Modern Geometries: Non-Euclidean, Projective, and Discrete, 2nd edition, Prentice Hall, 2001, pp. 115–124. ISBN 9780130323132. The supplied chapter scans determine the discussion sequence and source questions. Solutions are worked explanations; qualifications are noted where needed.
Book bibliographic record. Supplementary course notes provide supporting examples and model conventions.
Continue exploring
Synthesis and proof challenges
A numerical result is meaningful only after specifying whether points are individual sphere points or antipodal pairs. The same vectors can give different distances in these two models.
Interpretation · Antipodal identification changes distance
Unit vectors u,v satisfy \(u\cdot v=-1/2\). Find their spherical distance and the single elliptic distance between their antipodal classes. What happens if v=-u instead?
Hint
Compare arccos of the dot product with arccos of its absolute value.
Show worked solution
The spherical distance is \(2\pi/3\), while the quotient distance is \(\pi/3\). Replacing v by its antipode selects the nearer representative. When v=-u, the spherical distance is \(\pi\), but the quotient distance is zero because [u]=[v]. Zero distance is not a failure of the quotient metric: these are the same quotient point.
Advanced / Honors · Scale a spherical triangle correctly
A convex short-arc triangle on the unit sphere has angles \(\pi/2,\pi/2,\pi/3\). The same angular configuration is drawn on a sphere of radius R>0. Find its area and its fraction of the full sphere. Explain which quantity changes with R.
Hint
Use angular excess and compare with the sphere's total area.
Show worked solution
The excess is \(\pi/3\), so the area is \(\pi R^2/3\). Dividing by \(4\pi R^2\) gives fraction \(1/12\), independent of R. Corresponding lengths scale by R, areas by R squared, while angles and the area fraction remain unchanged. The excess formula applies to the specified convex short-arc region; selecting its complementary region would change the area.
