Mathematics 138
 Spherical Geometry

Review of 3-dimensional Euclidean geometry (Ryan pp. 84-88)
Denote the "standard basis" of R3 by e1 =    e2 =    e3 = 
Given three vectors u, v, w, we can define the determinant
det(u, v, w) by the properties
1. It is linear in each of its three arguments.
2. It changes sign if two arguments are interchanged.
3. det (e1 ,e2 ,e3) = 1
(This leads to the familar rule for evaluating the determinant of a matrix)
Given an inner product <x, y> on R3, we can define a vector product u x v
by the property that
<y, u x v> = det (y, u, v) for all vectors y in R3.
This leads to the usual rules for computing the vector product; for example if
u = e2  and v = e3 , then taking y = ae1 + be2 + ce3 we have
det (y, e2,e3) = a, so u x v = e1.
The advantage of this approach is that it will work also for hyperbolic geometry, where
a different inner product is used.
Properties of the vector product that you should know:

Directly from the definition
u x v = - v x u    (since det changes sign when u and v are interchanged)
< u x v, w> = <u, v x w>    (since det (w, u, v) = det (v, w, u))
< u x v, u> = <u x v, v> = 0 (since det = 0 if two arguments are equal)

This means that u x v is perpendicular to both u and v.  If u and v are linearly independent, 
then {u, v, u x v} is a basis for R3.

Starting with e1x e2 = e3,  e3 x e1 =e2 ,   e2x e3 =e1 , we can use linearity to show that
(u x v) x w = <u, w> v - <v, w> u    (the "vector in the middle" gets the + sign)

Further calculation gives
<u x v, w x z> = <u , v x (w x z)> = <u,w>,<v,z> - <u, z> <v, w>
and as a special case of this
<u x v, u x v> = |u|2 |v|2 - <u,v>2
If u and v are unit vectors, then
<u x v, u x v> = 1 - <u,v>2 = 1 - cos2 a = sin2 a
where a is the angle between vectors u and v.

Planes 
A plane P is a set of points in E3 characterized by
The line joining any two points in P lies in P
P is not contained in any line (so P is more than a line)
Not every point in E3 is in P (so P is not all of E3 )
Three noncollinear points P, Q, and R determine a plane.
Let v = Q - P,   w = R - P,   N = v x w/|v x w| (a unit normal)

Then the plane a determined by P, Q, and R is
{P + tv + sw} for all real values of s and t

Equivalently, because N was constructed to be perpendicular to v and w,
for any point X = P + tv + sw,

X - P  = tv + sw   and <X - P, N> = t <v, N> + s <w, N> = 0.
Incidence geometry of the sphere

Denote by S2 the set of points in E3 that are at unit distance from the origin
 (equivalently, the set of all unit vectors).

A point P in S2 is a unit vector.  The point -P is called the antipode of P.

A line in S2 is a "great circle" on the sphere: the intersection with the sphere of a plane through the origin.  Such a plane can be characterized by a "unit normal" x, so a line m can be described as the set of all unit vectors that are perpendicular to x.

x is called the "pole" of m, and m is the "polar line" of x .

Duality is spoiled by the fact that a given line m has two poles: x and -x.  This problem is fixed in elliptic geometry by identifying pairs of antipodal points with one another.

Two distinct, non-antipodal points P and Q determine a unique line PQ.

To construct the line, choose the pole x = P x Q/ |P x Q|
Since both P and Q are perpendicular to x, they lie on its polar line.
Consider any other possible pole h with < h, P> = <h, Q> = 0
Then h x (P x Q) = < h, Q> P - <h, P> Q = 0 so h is a multiple of P x Q,
 and since it is a unit vector it must equal either x or -x.

Two distinct lines m and n intersect in a pair of antipodal points.
Let x be a pole of m and h a pole of n.  Construct the point P = x x h / |x x h|.  
This is a unit vector, and it is perpendicular to both x and h, so it lies on both m and n.  
The same is true of the antipodal point -P.

This means there are no parallel lines.  Having two points of intersection is a problem that is fixed in elliptic geometry.

Parametric representation of a line

In affine geometry we are accustomed to representing a line in the form {P + tv} where v is a vector
that is conveniently chosen as v = Q - P where Q is some other point of the line

A similar approach will work in spherical geometry if we choose the vector to be a point Q on the line that is perpendicular to the point P.  You can either think of Q as another point on the line that is 1/4 of the way around the sphere, or you can think of it as a unit vector that is tangent to the line at the point P.

Since Q is a point of the line, it is perpendicular to the pole x.  Since it is perpendicular also to P, the vector Q = x x P (which is a unit vector) will do the job.  Then {x, P, Q} is an orthonormal set.

Now, for any value of t between 0 and 2p, the vector
a(t) = (cos t) P + (sin t) Q
is a unit vector perpendicular to x, hence a point on the line whose pole is x.

If |t2 - t1| < p, the measure of the angle between the vectors a(t2) and a(t1), |t2 - t1|,  is an appropriate measure of the distance between the two points on the line.

As in the Euclidean case, this parameterization of a line is not unique.  Any point on the line
 is an acceptable choice for P, and once P has been chosen
 there are two possible (antipodal) choices for Q.

Measuring distance and angle in spherical geometry
Given points P and Q (think of them as points on the equator), measure the distance by the angle q subtended at the center of the sphere (the difference in longitude). Since P and Q are unit vectors,
<P , Q> = cos q, so we define

d(P, Q) = arc cos <P, Q>
Since the arc cos function has values from 0 to p, distance is non-negative, and it corresponds to measuring "the short way round" from P to Q (rather than "the long way round" which would give a value between p and 2p).

We measure the angle between two lines as "the distance between their poles."  Unfortunately, this is ambiguous because each line has two antipodal poles, and, except in the case where the lines are perpendicular we can choose either a pair of poles that give a value less than p/2 or a pair of poles that give a value greater than p/2.  
(This same ambiguity exists if we try to measure the angle between lines (rather than rays) in Euclidean geometry -- do we choose the acute angle or the obtuse angle?

The ambiguity goes away in Euclidean geometry if we consider the interior angles of a triangle.
Similarly, we can consider a triangle ABC on the sphere, where each side has length less than p.

Spherical trigonometry and the triangle inequality

Let g be the pole of line AB, so g = A x B / |A x B|
But |A x B|2 = A2 B2 - <A, B>2 = 1 - cos2 c
so |A x B| = sin c and 
g = A x B / sin c.
Similarly, let b be the pole of line AC, and b = A x C / sin b

The most natural definition of the angle at A is the Euclidean angle between the unit tangent vector g x A to the great circle AB at A and the unit tangent vector b x A to the great circle AC at A.  This suggests
cos A = <g x A, b x A> = <g, b> <A, A> - <g, A> <b, A>
But A is a unit vector and is perpendicular to both g and b so
cos A = <g, b> = <A x B, A x C> / (sin b sin c)   Note -- A comes first in both cross products.
This agrees with the definition on p. 108 of Ryan.
Now expand the inner product of cross products:
<A x B, A x C> = <A, A><B, C> - <A,B><A,C> = cos a - cos c cos b
so cos A = 
This is the Law of Cosines for spherical trigonometry, well known to navigators and astronomers.
If the angle at A is a right angle, so that cos A = 0, we have
cos a = cos c cos b. which is the spherical counterpart of the Pythagorean Theorem.
Now we can easily prove the triangle inequality.
Since 
cos a - cos c cos b <= sin b sin c
cos a - (cos c cos b + sin c sin b) <= 0
cos a - cos (c - b) <= 0
cos a <= cos (c - b)  (assume c > b, otherwise write cos (b - c))
But a and (c - b)  are both positive and less than p, and the cosine function is decreasing on[0,p], so a >= c - b
Thus a + b >= c, which is the triangle inequality

More spherical trigonometry: the Law of Sines

Let g be the pole of AB and b the pole of AC.
Since cos A = <g, b> and <g, b>2 + |g x b|2 = 1 for unit vectors,
sin A = = |<g x b>| =  |(A x B) x (A x C)| / (sin c sin b)
Now use the triple cross product formula:

(A x B) x (A x C) = <A x B, C> A - <A x B, A> C
The second term is zero, so |(A x B) x (A x C)| = |<A x B, C>|

Thus sin A =  |<A x B, C>| / (sin c sin b)
and 
sin A/ sin a = |<A x B, C>| / (sin a sin c sin b)
But the quantity on the right is invariant under interchange of A, B, and C, so
sin A/ sin a = sin B / sin b = sin C / sin c

This is the Law of Sines for spherical trigonometry.

If the angle at A is a right angle, then

1 / sin a = sin C / sin c
or 
sin C = sin c / sin a
Similarly,
sin B = sin b / sin a
In this case side a is opposite the right angle, so
cos a = cos b cos c.
So the Law of Cosines
cos C = (cos c - cos b cos a)/ (sin b sin a)
becomes
cos C = (cos c - cos2 b cosc)/ (sin b sin a)
or cos C = cos c sin2 b / (sin b sin a)

or cos C = cos c sin b / sin a 
cos C = (cos a / cos b)(sin b / sin a) = tan b / tan a

Isometries of the Sphere

As with Euclidean geometry, we build everything from reflections.
To reflect in a line m with pole x, we simply reflect all of R3 in the plane perpendicular to x.

Tx = x - 2 <x, x> x,    where x is a unit vector.

If x is a unit vector, so is Tx.

Proof: <Tx, Tx> = <x, x> - 4 <x, x>2 + 4 <x, x>2 <x, x> = <x,x> since <x, x> = 1
So T maps the sphere S2 onto itself.
It also preserves distance, because it preserves the inner product

<Tx, Ty> = <x, y> - 4 <x, x><y, x> + 4 <x, x><y, x> <x, x> = <x, y> 

When restricted to S2, T is called Wm, the reflection in line m.

Consider the product of two reflections in lines m (pole x) and n (pole h).  The lines m and n
intersect in points P and -P, where P = x x h/ |x x h|. Choose an orthonormal basis where P 
is the basis vector e3 (the "north pole").  Then x and h, being perpendicular to P, are linear combinations of basis vectors e1 and e2.

In terms of matrices, this is the same as the case of Euclidean plane geometry!  If x makes an angle q with the basis vector e2, then Wm is represented by

which is a familiar reflection matrix.  The calculations that we have already done for the Euclidean case then confirm that if the angle between vectors x and h is y, then the product of reflections Wm Wn is a rotation through angle 2y about the basis vector e3.  The signs are handled carefully on pp. 93-94 of Ryan, but what you should remember is that if a rotation through angle y carries h (the pole for the first reflection) into x (the pole for the second reflection) then the product Wm Wn is a rotation through angle 2y in the same direction.  The rotation
Wn Wm is of course the inverse of Wm Wn.
As in the Euclidean case, the product of three reflections through lines that intersect at P is equivalent to a single reflection, and any rotation about P can be represented as a product of two reflections where the line through P for either the first or second reflection may be chosen arbitrarily.  (Formal statements on p. 94 of Ryan)

What about translations?
Although there are no parallel lines, we can define a translation along a line m as the product of two reflections in lines perpendicular to m.  But this is nothing new: these translations are simply rotations about the pole of m, and they are products of two reflections in lines that intersect at the pole of m.
What about glide reflections?
As in the Euclidean case, we can follow a reflection in line m by a translation along the same line m to obtain a glide reflection.  (In terms of the Earth, think of reflecting in the Equator, then rotating about the North Pole.)  This is a glide reflection.

The following results (proof left to the homework) are the same as in the Euclidean case, but their proofs are simpler because there are no special cases involving parallel lines.

The product of three reflections in lines that do not intersect at a single point is a glide reflection.

The product of a glide reflection and a reflection is a rotation.

The product of any number of reflections can be expressed as the product of three or fewer reflections.
Representing isometries of the sphere by matrices:

An isometry of the sphere is the restriction to S2 of a linear transformation of E3 that preserves the inner product <x. y>. The set of such isometries forms a group called O(3).  The transformation resulting from any product of reflections is an element of this group.  If the number of reflections is even, then the transformation is an element of the subgroup SO(3), the special orthogonal group.

If you choose an orthonormal basis, then an orthogonal transformation T is represented by an orthogonal matrix A.  Such a matrix can be characterized in two equivalent ways:

1. Its columns are orthogonal unit vectors (so it carries an orthonormal basis into another orthonormal basis)
2. Its inverse is equal to its transpose.

If the determinant of A is +1, it represents an element of SO(3).  In general the determinant will be either +1 or -1.

Proofs (which you have probably seen, but for which you are not explicitly responsible) are on pp. 97-98 of Ryan

We now know that every "motion" (product of reflections) is an isometry and that it may be represented by an orthogonal matrix.  We still need to show that every isometry is a motion.
This is done in two steps.

First, show that any orthogonal transformation arises from a motion.
Second, show that every isometry of S2 arises from an orthogonal transformation.
To start, consider an orthogonal transformation T in SO(3), so that the determinant is +1.
To find the eigenvalues, we set det (T - lI) = 0.  This leads to a cubic equation for l, which may have three real roots or which may have only one real root. 
 In either case, there is at least one eigenvector x such that Tx = lx. 
 But since <Tx, Tx> = <x,x>, l2 = 1 and so l = +1 or -1.
The product of the eigenvalues equals the determinant, which is +1.

If there are three real eigenvalues (each +1 or -1), at least one of them must be +1, otherwise their product would be -1 instead of +1.
If there is only one real eigenvalue, the other two roots of the cubic are complex conjugates and their product is a positive number.  Since the product of all three roots is +1, the real root must be +1 rather than -1.

This establishes that any T in SO(3) has an eigenvector x with eigenvalue +1.  (It could have three, but only if it is the identity).  The transformation is a rotation , and the eigenvector x is its pole.  If we choose a basis in which the third basis vector is x, then the transformation has 
as its third column.  The other two columns must be orthogonal unit vectors, and the determinant must be +1, so the matrix of T is of the form
, which represents a rotation.
If T in in O(3) but not in SO(3), then its determinant is -1.  Define the antipodal map E:
E = = ,
which is the product of three reflections.  Now ET has determinant +1, so it is a rotation r. Thus
T = E-1r = Er and T is shown to be a motion.
Finally, consider an arbitrary isometry To of S2.  For any two basis vectors, 
<Toei, Toej> = <ei, ej>
So if we represent To in terms of a transformation T of E3, T must carry an orthonormal basis into another orthonormal basis -- it must be an orthogonal transformation.  For fussy and unenlightening details, see pp 100-101 of Ryan).

Our conclusion: Every isometry of S2 is an orthogonal transformation and can be represented as the product of three or fewer reflections.

Half turns
The case in which a nontrivial rotation T has three real eigenvalues and three eigenvectors is a half-turn.  If the pole of the rotation is the third basis vector, the matrix of T is

Remarkably, any rotation can be represented as the product of two half turns!
First represent the rotation as a product of reflections, T = Wl Wm.  Let n be the polar line of the rotation. Lines l and n intersect at right angles at P, while m and n intersect at right angles at Q.
(there is a nice diagram on p. 102 of Ryan.)
Now write T = Wl Wn Wn Wm. But Wl Wn is a half-turn about P, while Wn Wm is a half-turn about Q.
So T = HP HQ.

A half turn HP about P has the particularly simple representation

HP x = -x + 2 <x, P> P.

Proof: chose an orthonormal basis {x, h, P}

If x = P. then HP P = -P + 2 <P, P> P = P
If x = x, then HP x = -x + 2 <x, P> P = -x
If x = h, then HP h = -h + 2 <h, P> P = -h
So HP has P as an eigenvector with eigenvalue +1 and has x and h as eigenvectors with
 eigenvalue -1, and its matrix with respect to the basis {x, h, P} is .
Since a glide reflection with axis l and pole P consists of a reflection in l followed by a translation along l (rotation about P), we can write it as
T = Wm Wn Wl, where lines m and n are perpendicular to line l.  Let Q be the point where lines l and n intersect.  Then Wn Wl is the half-turn HQ and T = Wm HQ.
Thus we can represent any motion as a product of half turns and no more than one reflection. Since it is easy to construct the matrix for a half turn or reflection, this is a fairly efficient way to construct the matrix that represents an motion.

Segments and rays:
The Euclidean notion of the segment AB as the set of all points between A and B fails in spherical geometry because the line AB is really a great circle and there is no unambiguous notion of betweenness.
However, the parametric representation of a line makes possible some reasonable definitions.
These are discussed in detail on pp. 103-107 of Ryan. This can be done better in the more interesting (and slightly different) case of elliptic geometry.

Here are the definitions:
A line m with pole x is represented parametrically as
a(t) = (cos t) P + (sin t) Q,
where {x, P, Q} are an orthonormal set.

Choosing any two numbers t1 and t2 with t1 < t2 and t2 - t1 < 2p determines a segment.
s =  {(cos t) P + (sin t) Q | }
If  t2 - t1 < p, the seqment is a "minor segment" and its length is t2 - t1.
Otherwise it is a "major segment" and its length is 2p - (t2 - t1).

The line through P and Q is the union of a minor segment PQ and a major segment PQ.

If we have a minor segment PQ represented as
 {(cos t) P + (sin t) Q | }  where L < p
then we can define the ray through P with origin Q as

 {(cos t) P + (sin t) Q | }. 
Given point P on line m, the union of the two rays with origin P is the entire line m with the exception of the point -P, so the two rays determine a straight angle just as in Euclidean geometry.

As in Euclidean geometry, the union of two rays is an angle.  The set of points in the interior of an angle is called a lune; think of it as the portion of the Earth's surface that is bounded by two meridians of longitude less than 180 degrees apart.

To be more precise about the interior of an angle, consider the angle PQR defined by the rays with origin Q through P and R respectively.  A point X is in the interior of the angle if the minor segment XP does not intersect the line QR and the minor segment XR does not intersect the line QP.  The diagram below does not really do justice to this definition; try it with a sphere.

Perpendicular bisectors and midpoints

Given two points P and Q, construct the unit vector x = (P - Q)/ |P - Q|
and let m be the line whose pole is x. Then reflection in m is the unique reflection that interchanges P and Q (proof on p. 111 of Ryan).

This line m is the perpendicular bisector of either the minor segment PQ or the major segment PQ. Its intersection with either segment is the midpoint of that segment.

Given two lines m and n, there are two reflections that interchange them (this is also the case in Euclidean geometry).  To construct them, find the poles +P and -P for m and +Q and -Q for n. The reflection that interchanges +P and +Q interchanges their polar lines m and n; so does the reflection that interchanges +P and -Q.

For a given angle A there is a unique reflection that interchanges its arms (which are rays).  The easiest way to construct similar to the standard construction of Euclidean geometry.

Construct P and Q on the two arms of the angle so that <P,Q> = <R,Q> = 0.
Now PR is the polar line of Q
Construct X = (P + R)/ |P + R|, which is the midpoint of the minor segment PR.
X is in the interior of the angle PQR.

Now the ray with origin Q through X bisects the angle PQR, and reflection in the line m through
Q and X interchanges the arms of the angle.  As in Euclidean geometry, the angles PQX and XQR are congruent.

Concurrence and congruence theorems

Most of these are proved just as in Euclidean geometry.

The perpendicular bisectors of the sides of a triangle are concurrent.

The bisectors of the angles of a triangle are concurrent.

SSS congruence: If triangles PQR and P'Q'R' have d(P,Q) = d(P',Q')
d(P,R) = d(P', R') and d(Q,R) = d(Q',R'), then the triangles are congruent.

SAS congruence:If triangles PQR and P'Q'R' have d(P,Q) = d(P',Q')
 d(Q,R) = d(Q',R'), and angle PQR = angle P'Q'R', then the triangles are congruent.

But there is one congruence theorem that has no Euclidean counterpart.  
The length of any side of a triangle can be expressed in terms of angles by

cos a = (cos A + cos B cos C)/ (sin B sin C)

So if two triangles have their respective angles congruent, their respective sides are also congruent,
and AAA congruence follows from SSS congruence.

This is the best theorem in all of spherical geometry, but somehow it's not mentioned in Ryan

Area of a Spherical Triangle
The triangle ABC has two points
A and C on the "equator" and an
arbitrary third point B.
We know that the entire "northern
hemisphere" has area 2p.
The lune bounded by lines AB 
and AC is bounded by rays that
make angle a; so its area is 2a

and A1 + A2 = 2a.

Similarly, for the lune bounded by
lines AC and BC,

A1 + A4 = 2g. 

The lune bounded by AB and BC
extends into the southern 
hemisphere, where it includes
a triangle congruent to the one
with area A3.  This lune has
area 2b; so 

A1 + A3 = 2b
Adding everything up,
3 A1 + A2 + A3 + A4 = 2(a+b+g)
But the entire northern hemisphere has area 2p, so
A1 + A2 + A3 + A4 = 2p

Subtract to obtain

2A1 = 2(a+b+g-p)

Thus the area A1 of triangle ABC is a+b+g-p

Dual triangles.
As long as every side of a triangle is less than p in length, we can assume that the entire triangle lies in the northern hemisphere and that the north pole is in the interior of the triangle. If we label the vertices A, B, C going counterclockwise around the triangle as seen from above, the cross products AxB, BxC, CxA will all give vectors with a positive z component and lead to poles in the northern hemisphere.
Now construct the poles:
A' = B x C/ |B x C|
B' = C x A/ |C x A|
C' = A x B/ |A x B|

The triangle A' B' C' is the dual triangle.
Now, for example, B' x C' is 
proportional to 
(C x A) x (A x B) = <CxA,B>A,
so that when we make it into a unit 
vector we get back A.  So the dual of
the dual triangle is the original 
triangle.

If the original triangle is all near the 
pole and has short sides, the dual
triangle will have vertices near the
equator and its angles will all be
large.

A triangle with three angles of p/2
and three sides of p/2 will be its own
dual.
Calculate the length of a side of the dual triangle:

d(A',B') = arccos <A', B'> = arccos (<B x C, C x A>/ (|B x C||C x A|))

On the other hand, the angle at C is (Ryan, p. 108)

angle BCA = arccos (<C x B, C x A>/ (|B x C||C x A|))

This has C x B instead of B x C, so we are taking the arc cos of the negative of the same quantity.
and the result is the supplement of the original distance.
You can think of this as cos C = - cos c' = -cos d(A', B')
or as C + c' = p.

A similar calculation shows that cos C' = - cos c = -cos d(A', B')
or. equivalently, that C' + c = p.

Any side of the dual triangle is the supplement of the corresponding angle of the original triangle.
Any angle of the dual triangle is the supplement of the corresponding side of the original triangle