<<

. 9
( 13)



>>

conic projections is radial, to use the radius r and the angle χ is more convenient:

dr = f (θ ) dθ dχ = n d•

The differentials of arc length and area for a conic projection are:

sin 2θ
dr 2
ds = dθ + sin θ d• = dχ 2
+
2 2 2 2

[ f (θ )]2 2
n

sin θ
dA = sinθ dθ § d• = dr § dχ
n f (θ )

Let us see as before the three special cases: equidistant, conformal and
equivalent projections. The differential of area for polar coordinates r, χ is
dA = r dr § dχ . If the projection is equivalent, we must identify both dA to find :
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 185


® sin θ  θ θ
1 2
d
f (θ ) =
 n f (θ ) = f (θ ) ’ r=
cos and sin
dθ 2 2
n
n
° »

r2 n
1’
n 4 r 2 dχ 2
ds = dr +
2 2
Figure 14.9
2
n
nr
1’
4

If the projection is equidistant, the meridians
have zero distortion so dθ = dr / n and:

«2 
«r
1
¬ dr + sin 2 ¬ · dχ 2 ·
ds 2 =
n2 n
 

If the projection is conformal then
ds 2 ∝ dr 2 + r 2 dχ 2 so:

sin 2θ
ds = 2 2 ( dr 2 + r 2 dχ 2 )
2

nr

Solving the differential equation:

sinθ
dθ = dr
nr
with the boundary condition tg θ 0 = r0 as shown by the figure 14.9 we find the
Lambert™s conformal projection:

n
«θ 
¬ tg ·
r = tg θ 0 ¬ 2 ·
¬ θ0 ·
¬ tg ·
2 


Exercises

14.1 We have seen the gnomonic projection of great circles being always straight lines.
Then, why does the shadow of a gnomon of a sundial follow a hyperbola on a plane
surface instead of a line during a day? Why does this hyperbola become a straight line
the March 21 and September 23?

14.2 Three points on the sphere are projected by means of the stereographic projection
into a circle of unity radius with the coordinates A = (’0.5, 0.5), B = (0, ’2/3) and C =
(2/3, 0). Calculate the sides, angles and area of the triangle which they form.

14.3 Built at Belfast, the Titanic begun its first and last travel in Southampton the April
10th 1912. After visiting Cherbourg the Titanic weighed anchor in the Cork harbour
186 RAMON GONZALEZ CALVET


with bearing New York the 11th April. Calculate the length of the shortest trajectory
from Fastnet (Ireland) at 51º30™N 9º35™W to Sandy Hook (New York) at 40º30™N
74ºW. From January 15th to July 15th the ships had to follow the orthodrome (great
circle) from Fastnet to the point 42º30™N 47ºW and from this point to Sandy Hook
passing at twenty miles from the floating lighthouse of Nantucket. Calculate the length
of the route the Titanic should have followed. Take as an averaged radius of the Earth
6366 km. The Titanic sank the 15th April at the position 41º46™N 50º14™W. Is this point
on the obliged track?

Figure 14.10




14.4 The figure 14.11 is a photograph of the Hale-Bopp comet. The orientation of the
camera is unknown, but the W of Cassiopeia constellation appears in the photograph.
The declination D of a star is the angular distance from the celestial equator to the star
(measured on the great circle passing through the celestial pole and the star). The right
ascension A is the arc of celestial equator measured eastward from the vernal equinox
(one of the intersections of the celestial equator with the ecliptic, also called Aries
point) to the foot of the great circle passing through the star and the pole. The right
ascension and declination of the stars are constant while those of the comets, planets,
the Sun and the Moon are variable. The data of Cassiopeia constellation are:

star Magnitude A D
± Cassiopeia (Schedir) 2.2 0h 40min 6.4s = 10.0267º 56º 29™ 57” N
β Cassiopeia (Caph) 2.3 0h 8min 48.1s = 2.2004º 59º 6™ 40” N
γ Cassiopeia (Tsih) 2.4 0h 56min 16.9s = 14.0704º 60º 40™ 44” N
δ Cassiopeia (Ruchbah) 2.7 1h 25min 21.2s = 21.3383º 60º 11™ 57” N

In your system of coordinates, take the x-axis as the Aries point and the z-axis as the
north pole. Then D = π/2’θ and A=• .
a) Calculate the focal distance of the photograph, that is, the distance from the point of
view to the plane of the photograph. Knowing that the negative was universal
(24—36 mm), calculate the focal distance of the camera.
b) Calculate the orientation (right ascension and declination) of the photographic
camera.
c) Calculate the coordinates of the Hale-Bopp comet.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 187




Figure 14.11
188 RAMON GONZALEZ CALVET


15. HYPERBOLOIDAL GEOMETRY IN THE PSEUDO-EUCLIDEAN SPACE
(LOBACHEVSKY™S GEOMETRY)

The geometric algebra of the pseudo-Euclidean space

A vector of the pseudo-Euclidean space is an oriented segment in this space with
direction and sense. The set of all segments (vectors) together with their addition
(parallelogram rule) and the product by real numbers (dilation of vectors) has a structure
of vector space, symbolised here with W3. Every vector in W3 is of the form:

v = v1 e1 + v 2 e 2 + v 3 e3

where ei are three unitary perpendicular vectors, which constitute the base of the space.
The modulus of a vector is:

2
= ’ v12 ’ v 2 + v 3
2 2
v

It determines the geometric properties of the space, very different from the Euclidean
space. Now we define an associative product (geometric or Clifford product) being a
generalisation of those defined for the Euclidean and hyperbolic planes. Imposing the
condition that the square of the modulus be equal to the square of the vector, we find:

2
= v2
v

e12 = ’1 e 2 = ’1 e3 = 1 ei e j = ’e j ei for i ≠ j
2 2
and

From the base vectors one deduces that the geometric algebra generated by the
space W3 has eight components:

Cl (W3 ) = Cl1, 2 = 1, e1 , e 2 , e3 , e 23 , e31 , e12 , e123

Let us see with more detail the product of two vectors:

v w = (v1 e1 + v 2 e 2 + v 3 e3 ) (w1 e1 + w2 e 2 + w3 e 3 ) = ’v1 w1 ’ v 2 w2 + v 3 w3

+ (v 2 w3 ’ v 3 w2 ) e 23 + (v 3 w1 ’ v1 w3 ) e31 + (v1 w2 ’ v 2 w1 ) e12

I shall call the product (or quotient) of two vectors a tetranion. The tetranions are the
even subalgebra of Cl1, 2 that generalises the complex and hyperbolic numbers to the
pseudo-Euclidean space. Splitting a tetranion in the real and bivector parts, we obtain
the inner (or scalar) product and the outer (or exterior) product respectively:

v · w = ’ v1 w1 ’ v 2 w2 + v 3 w3

v § w = (v 2 w3 ’ v 3 w2 ) e 23 + (v 3 w1 ’ v1 w3 ) e31 + (v1 w2 ’ v 2 w1 ) e12
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 189


Here also, the bivectors are oriented plane surfaces indicating the direction of planes in
the pseudo-Euclidean space. As before, vectors and bivectors are different things. This
fact has been experienced also by physicists: in the Minkowski™s space, the
electromagnetic field is a bivector while the tetrapotential is a vector. On the other hand,
the oriented area is, of course, a bivector. As a criterion to distinguish both kind of
magnitudes one also uses the reversal of coordinates, which changes the sense of
vectors while leaves bivectors invariant.
Two vectors are said to be orthogonal if their inner product vanishes:

v⊥w ” v·w=0

So the outer product is the product by the orthogonal component and the inner product
is the product by the proportional component:

v · w = v w|| v § w = v w⊥

The product of two bivectors yields a tetranion. The real and bivector parts can
be separated as the symmetric and antisymmetric product. Since:

e 23 = e31 = 1 e12 = ’1
2 2 2
and

the symmetric product is a real number whose negative value will be denoted here with
the symbol •, whereas the antisymmetric product, denoted here with the symbol —, is
also a bivector:

1
(v w + w v ) = ’v 23 w23 ’ v 31 w31 + v12 w12
v•w=’
2

1
(v w ’ w v )
v—w=’
2

= ’(v 31 w12 ’ v12 w31 ) e 23 ’ (v12 w23 ’ v 23 w12 ) e31 + (v 23 w31 ’ v 31 w23 ) e12

v w = ’v • w ’ v — w

The outer product of three vectors has the same expression as for Euclidean
geometry and this is a natural outcome of the extension theory: the product u § v § w is
the oriented volume generated by the surface represented by the bivector u § v when it
is translated parallelly along the segment w:

ux vx wx
u § v § w = uy vy w y e123
uz vz wz

Finally, let us see how the product of three vectors u, v and w is. The vector v
can be resolved into a component coplanar with u and v and another component
perpendicular to the plane u-v:
190 RAMON GONZALEZ CALVET



u v w = u v || w + u v ⊥ w

Now let us analyse the permutative property. In both Euclidean and hyperbolic planes
we found u v w ’ w v u = 0. In the pseudo-Euclidean space the permutative property
becomes:

u v w ’ w v u = u v ⊥ w ’ w v ⊥ u = v ⊥ (’ u w + w u )

= ’2 v ⊥ u § w = ’ 2 v § u § w = 2 u § v § w

I take the same algebraic hierarchies as in the former chapter: all the abridged
products must be operated before the geometric product, convention adequate to the fact
that in many algebraic situations, the abridged products must be developed in sums of
geometric products.


The hyperboloid of two sheets

According to Hilbert (Grundlagen der Geometrie, Anhang V) the complete
Lobachevsky™s “plane” cannot be represented by a smooth surface with constant
curvature as proposed by Beltrami. But this
result only concerns surfaces in the Euclidean
space. The surface whose points are placed at a
fixed distance from the origin in a pseudo-
Euclidean space (the two-sheeted hyperboloid) is
the surface searched by Hilbert which realises
the Lobachevsky™s geometry1. It is known that it
has a characteristic distance like the radius of the
sphere. Since all the spheres are similar, we need
only study the unitary sphere. Likewise, all the
Figure 15.1
hyperboloidal surfaces z2’x2’y2 = r2 are similar
and the hyperboloid with unity radius (figure
15.1):

z2 ’ x2 ’ y2 =1

where (x, y, z) are Cartesian although not Euclidean2 coordinates, suffices to study the
whole Lobachesky™s geometry.

1
The reader will find a complete study of the hyperboloidal surface in Faber, Foundations of
Euclidean and Non-Euclidean Geometry, chap. VII. “The Weierstrass model”.
2
For me, Cartesian coordinates are orthogonal coordinates in a flat space (with null curvature),
independently of the Euclidean or pseudo-Euclidean nature of the space. Lines and planes
(linear relations between Cartesian coordinates) can always be drawn in both spaces, although
the pseudo-Euclidean orthogonality is not shown as a right angle, and the pseudo-Euclidean
distance is not the viewed length. The incompleteness of Cartesian coordinates was already
criticised by Leibniz in a letter to Huygens in 1679 proposing a new geometric calculus: «Car
premierement je puis exprimer parfaitement par ce calcul toute la nature ou d©finition de la
figure (ce que l™algèbre ne fait jamais, car disant que x2 + y2 aeq. a2 est l™equation du cercle, il
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 191


The fact that the hyperboloidal surface owns the Lobachevsky™s geometry will
be evident through the different projections here reviewed. If we followed the same
order as in the former chapter, we should firstly deal with the Lobachevskian
trigonometry. However, the concept of arc length on the hyperboloid is much less
intuitive than on the sphere. Therefore we must see the concept and determination of the
arc length before studying the hyperboloidal trigonometry.


The central projection (Beltrami disk)

In this projection the centre of the hyperboloid is the centre of projection. Thus,
we project every point on the upper sheet of the hyperboloid z2’x2’y2 = 1 into another
point on the plane z = 1 taking the origin x = 0, y = 0, z = 0 as centre of projection
(figure 15.1). Let u and v be Cartesian coordinates on the plane z =1, which touches the
hyperboloid in the vertex. By similar triangles3:

x y
u= v=
z z

From where we obtain:

u v 1
x= y= z=
1 ’ u2 ’ v2 1 ’ u2 ’ v2 1’ u2 ’ v2

The differential of the arc length upon the hyperboloid is:

ds = dx e1 + dy e 2 + dz e3

(1 ’ v )du + 2 u v du dv + ( 1 ’ u 2 )dv 2
2 2
ds = ’dx ’ dy + dz =’
2 2 2 2

(1 ’ u ’ v2 )
2
2




which is the negative value of the usual metric of the Lobachevsky™s surface in the
Beltrami disk. The negative sign indicates that the arc length is comparable with lengths
in the x-y plane. The whole upper sheet of the hyperboloid is projected inside the
Beltrami™s circle u 2 + v 2 ¤ 1 with unity radius. However, if we project the one-sheeted
hyperboloid z 2 ’ x 2 ’ y 2 = ’1 into the plane z = 1 we find the same metric but
positive, what indicates the lengths being comparable to those measured on the z-axis.
All the upper half of the one-sheeted hyperboloid is projected outside the circle of unity
radius, giving rise to a new geometry not studied up till now.


faut expliquer par la figure ce que c™est x et y.» (Josep Manel Parra, “Geometric Algebra versus
numerical cartesianism” in F. Brackx, R. Delanghe, H. Serras, Clifford Algebras and Their
Applications in Mathematical Physics).
3
Now to consider that the legs of both right angle triangles are horizontal and vertical, and
therefore perpendicular is enough for the deduction. Notwithstanding, in a pseudo-Euclidean
space one must use the algebraic definition of similitude: a triangle with sides a and b is directly
similar to a triangle with sides c and d if and only if a c ’1 = b d ’1. This condition implies that
both quotients of the modulus are equal and both arguments also.
192 RAMON GONZALEZ CALVET


Let us consider a plane passing through the origin of coordinates (central plane).
Its intersection with the two-sheeted hyperboloid is the hyperbola determined by the
equations system:

± z2 ’ x2 ’ y2 =1

z=ax+by a , b real


We search the Frenet's trihedron. By differentiation we find a linear differential
equations system :

± z dz = x dx + y dy

 dz = a dx + b dy

where dx and dy can be isolated:

y ’bz az’x
dx = dy =
dz dz
a y ’bx a y ’bx

Then the differential of the arc length of this hyperbola is:

« y ’bz 
az’x
ds = dx e1 + dy e 2 + dz e3 = ¬ e 2 + e3 · dz
e1 +
¬a y ’bx ·
a y ’bx
 

1 ’ a 2 ’ b2
ds = ’dx ’ dy + dz =
2 2 2 2
dz 2
and its square:
(a y ’ b x )2
Now we obtain the unitary vector t tangent to the hyperbola:

( y ’ b z ) e1 + ( a z ’ x ) e 2 + ( a y ’ b x ) e3
dx dy dz
t= e1 + e 2 + e3 =
ds ds ds 1 ’ a 2 ’ b2

Observe that t and ds have imaginary components since a plane only cuts the two-
sheeted hyperboloid if a2 + b2 > 1. Anyway, the geometric vector can be taken with real
components, although then its modulus is ’1.
The derivative of the tangent vector with respect to the arc length is not only
proportional but equal to the normal vector n of the hyperbola:

dt dt
= ’ ( x e1 + y e 2 + z e3 ) κ=
n= ’ =1
ds ds

because the position vector of every point on the two sheet hyperboloid has unitary
modulus. Hence the curvature κ of the hyperbola is constant and equal to 1. All the
hyperbolas being intersections of the hyperboloid with planes passing through the origin
and a given point on the hyperboloid have the same radius of curvature r:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 193


’1
« dt 
= ’( x e1 + y e 2 + z e3 )
r=¬ ·
 ds 

Two consequences are deduced from this fact: firstly their common radius of curvature
is perpendicular to the surface, and secondly, these hyperbolas are geodesics of the
hyperboloid. Because all the hyperbolas lie on planes passing through the origin, their
central projections are straight lines. In other words, every geodesic hyperbola is
projected into a segment on the Beltrami disk. Taking as equation of the line:

v=ku+l

with k and l constant, the substitution in ds gives4:

1+ k2 ’ l2
ds
= du
e123 ’ u 2 (1 + k 2 ) ’ 2 k l u + 1 ’ l 2

By integration we arrive at the following primitive:

k l + 1+ k2 ’ l2
u (1 + k 2 ) + k l
u+
s 1 1+ k2
= arg tgh + const = log + const
e123 2
1+ k ’ l k l ’ 1+ k ’ l
2 2 2 2
u+
1+ k2

The line v = k u + l cuts the Beltrami™s circle u 2 + v 2 = 1 at the points A and D
whose coordinates are the solution of the system of the equations:

±v =k u +l ’ k l ’ 1+ k2 ’ l2 ’ k l + 1+ k2 ’ l2
’ uA = uD =
2
u + v2 =1 k2 +1 k2 +1


Thus the primitive may be rewritten:
Figure 15.2
u ’ uA
s 1
= + const
log
u ’ uD
e123 2

Then, the arc length between two points B and C on this
geodesic hyperbola is the difference of this primitive
between both points:

(u ’ u A ) (u D ’ u B ) 1
sC ’ s B 1
= log( A B C D )
= log C
(u D ’ u A ) (u C ’ u B ) 2
e123 2

4
As said before, the arc length s on the two-sheeted hyperboloid has imaginary values. I™m
sorry by the inconvenience of that reader accustomed to take real values. However the
coherence of all the geometric algebra force to take account of the imaginary unity e123. This
also explains why Lobachevsky found the hyperboloidal trigonometry by choosing imaginary
values for the sides in the spherical trigonometry.
194 RAMON GONZALEZ CALVET



that is, the half of the logarithm of the cross ratio of the four points (figure 15.2) on the
Beltrami disk.
However it is more advantageous to write it using cosines instead of tangents by
means of the trigonometric identity:

s
sB
1 ’ tgh tgh C
sC ’ s B e123 e123

cosh
e123 s
sB
1 ’ tgh 2 1 ’ tgh 2 C
e123 e123

After removing k and l using the equation of the line in the primitive, we arrive at:

sC ’ s B 1 ’ u B uC ’ v B vC
=
cosh
e123 2 2 2 2
1 ’ u B ’ v B 1 ’ uC ’ vC

where the expression is real for all the points on the hyperboloid, whose projections lie
on the Beltrami disk u2 + v2 ¤ 1. This expression is equivalent to write:

(u B e1 + v B e2 + e3 ) · (u C e1 + v C e 2 + e3 )
sC ’ s B
=
cosh
(u B e1 + v B e 2 + e3 ) (u C e1 + v C e 2 + e3 )
e123

Since the vector u A e1 + v A e 2 + e 3 is proportional to the position vector, we have:

(x B e1 + y B e 2 + z B e3 ) · (x C e1 + y C e 2 + z C e3 )
sC ’ s B
= cos(s C ’ s B ) =
cosh
(x B e1 + y B e 2 + z B e3 ) (x C e1 + y C e 2 + z C e3 )
e123

which is the expected expression to calculate angles between vectors in the pseudo-
Euclidean space. This result must be commented in more detail. Note that every pair of
vectors going from the origin to the two-sheeted hyperboloid always lie on a central
plane having hyperbolic nature, since it cuts the cone z 2 ’ x 2 ’ y 2 = 0 of vectors with
zero length. Because of this, the angle measured on this plane is hyperbolic.
A hyperboloidal triangle is the region of the hyperboloid bounded by three
geodesic hyperbolas. The length of a side of the triangle can be calculated by integration
of the arc length of the hyperbola according to the last result. On the other hand, the
angle of a vertex of a hyperboloidal triangle is the angle between the tangent vectors of
the hyperbolas of each side at the vertex:

’ ( y ’ b z )( y ’ b' z ) ’ (a z ’ x )(a' z ’ x ) + (a y ’ b x )(a' y ’ b' x )
cos ± = t … t ' =
1 ’ a 2 ’ b2 1 ’ a' 2 ’ b' 2

1 ’ a a' ’ b b'
=
1 ’ a 2 ’ b2 1 ’ a' 2 ’ b' 2
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 195


Note that every pair of tangent vectors lie on a plane of Euclidean nature, a plane
parallel to a central plane not cutting the cone of zero length. So the angles measured on
this plane are circular. We arrive at the same conclusion by proving that the absolute
value of the cosine (and also its square) is lesser than the unity:

(1 ’ a a' ’ b b' )2
<1
(1 ’ a ) (1 ’ a' )
22 22
’b ’ b'
2 2




The denominator is positive (although both squares are negative because a 2 + b 2 > 1 )
and it can pass to the right hand side of the inequality without changing the sense.
Removing the denominator and after simplification:

’ 2 a a' ’ 2 b b' + 2 a a' b b' < ’a 2 ’ b 2 ’ a' 2 ’ b' 2 + a 2 b' 2 + a' 2 b 2

and arranging terms we find an equivalent inequality:

a 2 + a' 2 ’ 2 a a' + b 2 + b' 2 ’ 2 b b' < a 2 b' 2 + a' 2 b 2 ’ 2 a a' b b'

(a ’ a' )2 + (b ’ b' )2 < (a b' ’ b a' )2
Just this is the condition for the existence of the intersection point where two central
planes and the hyperboloid meet as I shall show now. The planes with equations a x + b
y = z and a' x + b' y = z are projected into the lines a u + b v = 1 and a' u + b' v = 1 on
the Beltrami disk. Then the point of intersection of both planes and the hyperboloidal
surface is projected into the point on the Beltrami disk given by the system of equations:

± a u + b v =1

a' u + b' v = 1

b' ’ b ’ a' + a
u= v=
whose solution is:
a b' ’ a' b a b' ’ a' b

This point lies inside the Beltrami circle (and the point of intersection of the
hyperboloid and both planes exists) if and only if:

(a ’ a' )2 + (b ’ b' )2
u +v = <1
2 2

(a b' ’ b a' )2
which is the condition above obtained.
Note that the expression for the angle between geodesic hyperbolas coincides
with the angle between the bivectors of the planes containing these hyperbolas:

(a e 23 + b e31 + e12 ) • (a' e 23 + b' e31 + e12 )
cos ± =
a e 23 + b e 31 + e12 a' e 23 + b' e31 + e12
196 RAMON GONZALEZ CALVET


It is not a surprising result, because it happens likewise for the sphere: the angle
between two maximum circles is the angle between the central planes containing them.
The planes passing through the origin only cut the two-sheeted hyperboloid if a2
+ b2 >1. These planes also cut the cone of zero length z2 ’ x2 ’ y2 = 0 so they are
hyperbolic and the angles measured on them are hyperbolic arguments. On the other
hand, the planes with a2 + b2 <1 do not cut the hyperboloid neither the cone and have
Euclidean nature. In the Lobachevskian trigonometry, the planes containing the sides of
a hyperboloidal triangle are always hyperbolic, while the planes touching the
hyperboloid and containing the angles between sides are always Euclidean, and hence
the angles are circular. Resuming, real modulus of a bivector (0<’a2 ’b2 +1) indicates
an Euclidean plane and imaginary modulus (0>’a2 ’b2 +1) a hyperbolic plane.
Now we already have all the formulas to deduce the hyperboloidal trigonometry.


Hyperboloidal (Lobachevskian) trigonometry

The Lobachevskian trigonometry is the study of the trigonometric relations for
the sides and angles of geodesic triangles on two-sheeted hyperboloids in the pseudo-
Euclidean space. We will take for
convenience the hyperboloid with
z2 ’ x2 ’ y2 =1,
unity radius
although the trigonometric identities
are equally valid for a hyperboloid
with any radius.
Consider three points A, B and
C on the unitary hyperboloid (figure
15.3). Then A = B = C = 1 . The
angles formed by each pair of sides
will be denoted by ±, β and γ, and the
real value of the sides respectively
opposite to these angles will be
symbolised by a, b and c respectively.
Figure 15.3
Then a is the real value of the arc of
the geodesic hyperbola passing
through the points B and C, that is, the hyperbolic angle between these vectors:

B §C
sinh a =
e123

As said above, the planes intersecting the hyperboloid have imaginary modulus. So we
must divide by the pseudoscalar imaginary unity, which is also the volume element. In
this equality the hyperbolic sine is taken positive and also a.
± is the circular angle between the sides b and c of the triangle, that is, the angle
between the planes passing through the origin, A and B, and the origin, A and C
respectively. Since the direction of a plane is given by its bivector, which can be
obtained through the outer product, we have:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 197


(A § B) — (A § C )
sin ± = ’
A§B A§C

where the minus sign is due to the imaginary modulus of the outer products of the
denominators. Now we write the products of the numerator using the geometric product:

’ ( A B ’ B A)( A C ’ C A) + ( A C ’ C A)( A B ’ B A)
sin ± = ’
8 A§B A§C

’ A B A C + A B C A + B A2 C ’ B A C A + A C A B ’ A C B A ’ C A2 B + C A B A
sin ± = ’
8 A§B A§C

We extract the vector A as common factor at the left, but without writing it because
A = 1:

’ B A C + B C A + A B C ’ A ’1 B A C A + C A B ’ C B A ’ A C B + A ’1 C A B A
sin ± = ’
8 A§ B A§C


Applying the permutative property to the suitable pairs of products, we have:

6 A § B § C + 2 A ’1 A § B § C A A§ B §C
sin ± = ’ =’
8 A§ B A§C A§ B A§C

since the volume A § B § C is a pseudoscalar, which commutes with all the elements of
the algebra. Now the law of sines for hyperboloidal triangles follows:

A§ B §C
sin ± sin β sin γ
= = = (I)
A § B B § C C § A e123
sinh a sinh b sinh c

Let us see the law of cosines. Since e 23 = e 31 = ’ e12 = 1 then:
2 2 2




(A § B) • (A § C ) = ’ (A § B) • (A § C )
cos ± =
A§ B A§C sinh c sinh b

1
(( A B ’ B A) ( A C ’ C A) + ( A C ’ C A)( A B ’ B A))
sinh b sinh c cos ± =
8

( A B A C ’ A B C A ’ B A 2 C + B A C A + A C A B ’ A C B A ’ C A 2 B + C A B A)
1
=
8

Now taking into account that:
198 RAMON GONZALEZ CALVET


4 A · B A · C = ( A B + B A ) ( A C + C A) = A B A C + A B C A + B A 2 C + B A C A

But also:

4 A · B A · C = ( A C + C A) ( A B + B A) = A C A B + A C B A + C A 2 B + C A B A

and adding the needed terms, we find:

(8 A · B A · C ’ 2 A B C A ’ 2 B A 2 C ’ 2 A C B A ’ 2 C A 2 B )
1
sinh b sinh c cos ± =
8

Extracting common factors and using A2 = B2 =1, we may write:

1
(8 A · B A · C ’ 2 A (B C + C B ) A ’ 2 (B C + C B )) = A · B A · C ’ B · C
sinh b sinh c cos ± =
8

sinh b sinh c cos ± = cosh c cosh b ’ cosh a

cosh a = cosh b cosh c ’ sinh b sinh c cos ± (II)

which is the law of cosines for sides. The substitution of cosh c by means of the law of
cosines gives:

cosh a = cosh b (cosh a cosh b ’ sinh a sinh b cos γ ) ’ sinh b sinh c cos ±

cosh a (1 ’ cosh 2 b ) = ’ cosh b sinh a sinh b cos γ ’ sinh b sinh c cos ±

and the simplification of ’sinh b:

cosh a sinh b = cosh b sinh a cos γ + sinh c cos ±

The substitution of sinh c = sinh a sin γ / sin ± yields:

sinh a sin γ cos ±
cosh a sinh b = cosh b sinh a cos γ +
sin ±

Dividing by sinh a :

coth a sinh b = cosh b cos γ + sin γ cot ± (III)

Figure 15.4
Stereographic projection (Poincar© disk)

We project the hyperboloidal surface into the
plane z = 0 (figure 15.4) taking as centre of
projection the lower pole (0, 0, ’1). The upper sheet
is projected inside the circle of unity radius while the
lower sheet is projected outside. Let u and v be the
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 199


Cartesian coordinates of the projection plane. By similar triangles we have:

x y
= z +1 = z +1
u v

where (x, y, z) are the coordinates of a point on the hyperboloid. Using its equation
z 2 ’ x 2 ’ y 2 = 1 , one arrives at:

2u 2v 2
x= y= z= ’1
1’ u2 ’ v2 1 ’ u2 ’ v2 1 ’ u2 ’ v2

from where the differential of the arc length is obtained:

4 ( du 2 + dv 2 )
ds = ’ dx ’ dy + dz = ’
2 2 2 2

(1 ’ u ’ v2 )
2
2



The geodesic hyperbolas are intersection of the hyperboloid with central planes having
the equation:

z=a x+b y Figure 15.5

The substitution by the stereographic coordinates
yields:

( u ’ a )2 + ( v ’ b )2 = a 2 + b 2 ’ 1
which is the equation of a circle centred at (a, b)
with radius r = a 2 + b 2 ’ 1 . That is, the geodesic
hyperbolas on the hyperboloid are shown as circles
in the stereographic projection (figure 15.5). The
central planes cutting the two-sheeted hyperboloid
fulfil a 2 + b 2 > 1 , so that the radius is real and the centres of these circles are always
placed outside the Poincar© disk.
The angle ± between two circles is obtained through:

r 2 + r' 2 ’ (O ’ O' )
2
a a' + b b' ’ 1
cos ± = =
2 r r' a 2 + b 2 ’ 1 a' 2 + b' 2 ’ 1

Just this is the angle between two geodesic hyperbolas (between the tangent vectors t
and t™) found above. Therefore, the stereographic projection (Poincar© disk) is a
conformal projection of the hyperboloidal surface.
On the other hand, the geodesic circles are always orthogonal to the Poincar©'s
circle u 2 + v 2 = 1 ( r' =1, O' = (0, 0) ) because:

r2 +1’ O2
cos ± = =0
r
200 RAMON GONZALEZ CALVET



The differential of area is easily obtained through the modulus of the differential
bivector:

4 du § dv
(dx § dy )2 ’ (dy § dz )2 ’ (dz § dx )2
dA = =
(1 ’ u ’ v2 )
2
2



When doing an inversion of the
Poincar© disk centred at a point lying on
the limit circle, another projection of the Figure 15.6
Lobachevsky's geometry is obtained: the
Poincar©™s half plane. Here the geodesics
are semicircles orthogonal to the base line
of the half plane. Since the inversion is a
conformal transformation, the upper half
plane is also a conformal projection of the
hyperboloid. The figure 15.6 displays the
projection of a hyperboloidal triangle and
its sides in the Poincar©™s upper half plane.


Azimuthal equivalent projection

This projection preserves the area and is similar to the azimuthal equivalent
projection of the sphere usually used in the polar maps of the Earth.
The differential of the area in the pseudo-Euclidean space is a bivector whose
square is:

dA 2 = (dx § dy ) ’ (dy § dz ) ’ (dz § dx )
2 2 2




The substitution of the hyperboloid equation:

x y
dz = dx + dy
z2 ’ x2 ’ y2 =1 ’
z z

1
(dx § dy )2
dA 2 =
yields: 2
z

Every azimuthal projection has plane coordinates u and v proportional to x and y,
being the proportionality constant only function of z:

u = x f (z ) v = y f (z )

Then by differentiation of these equalities and substitution of dz, which is a linear
combination of dx and dy we arrive at:

« 2 z2 ’1 
du § dv = ¬ f + f f ' · dx § dy
¬ ·
z
 
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 201



Identifying both area differentials we find the following differential equation:

z2 ’1 1
f+ f f' =
2

z z

I introduce the auxiliary function g = f 2 :

z2 ’1 1
g+ g' =
2z z

Rewriting the equation with differentials, we arrive at an exact differential:

« z2 ’1 
z2 ’1
(z g ’ 1) dz + d¬ g ’ z· = 0
dg = 0 ” ¬2 ·
2  

2 (z + const )
2 (z + const )
g (z ) = f (z ) =

z2 ’1 z2 ’1

Next to the pole x and u are coincident and also y and v. Then f(1)=1, which implies the
integration constant be ’1 and then5:

2 (z ’ 1) 2 2 2
f (z ) = ’
= u=x v= y
z +1 z +1 z +1
z2 ’1


Weierstrass coordinates and cylindrical equidistant projection

These are a set of coordinates similar to
the spherical coordinates, but on the
hyperboloid surface in the pseudo-Euclidean
space.
For a hyperboloid with unity radius, the
Figure 15.7
Weierstrass coordinates are:

x = sinh ψ cos •

y = sinh ψ sin •

z = coshψ

Then the differential of arc length is:



5
As a comparison in the central projection f(z) = 1/z while for the stereographic projection f(z) =
1 / (z + 1).
202 RAMON GONZALEZ CALVET


ds = (coshψ cos • dψ ’ sinhψ sin• d• ) e1 + (coshψ sin • dψ + sinhψ cos• d• ) e 2 + sinhψ dψ e 3

Introducing the unitary vectors eψ and e• as:

eψ = coshψ cos • e1 + coshψ sin • e 2 + sinhψ e3

e• = ’sin • e1 + cos • e 2

the differential of arc length in hyperboloidal coordinates becomes:

ds 2 = ’( dψ 2 + sinh 2ψ d• 2 )
ds = dψ eψ + sinh ψ d• e•

Note that eψ and e• are orthogonal vectors, since their inner product is zero. At ψ = 0,
ds only depends on dψ , so that there is a pole. Then ψ is the arc length from the pole to
the given point on the hyperboloid surface (figure 15.7), while • is the arc length over
the equator, the parallel determined by sinhψ = 1 and ψ = log(1+ 2 ). The meridians (•
= constant) are geodesics, while the parallels (ψ = constant) including the equator are
not geodesics because their planes do not pass through the origin of coordinates.
In the cylindrical projections, the hyperboloid is projected onto a cylinder
passing through its equator, whose axis is the z axis. The chart is the Euclidean map
obtained unrolling the cylinder6. Using • and ψ as the coordinates u and v of the chart,
the hyperboloid surface is projected into a rectangle with 2π width and infinite height.
The area on this chart is:

dA = sinhψ d• § dψ = sinh v du § dv

This projection is equidistant for the meridians (• = constant) and for the equator (ψ =
log(1+ 2 )).


Cylindrical conformal projection

If we wish to preserve the angles between curves, we must enlarge the meridians
by the same amount as the parallels are enlarged in a cylindrical projection, that is by a
factor 1/sinh ψ:

dψ ψ
dv = ’ v = log tgh
sinhψ 2

The differential of arc length and area are:




6
Every plane tangent to the two-sheeted hyperboloid has Euclidean nature. So the
Lobachevsky™s surface can be only projected onto a Euclidean chart, and the cylinder of
projection is not hyperbolic and does not belong properly to the pseudo-Euclidean space.
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 203



’ 4 exp(2 v )
ds 2 = ’( dψ 2 + sinh 2ψ d• 2 ) = ’sinh 2ψ (du 2 + dv 2 ) = (du + dv 2 )
2

(1 ’ exp(2 v ) ) 2




4 exp(2 v )
dA = sinhψ d• § dψ = du § dv
(1 ’ exp(2 v ) )2

Cylindrical equivalent projection

If we wish to preserve area, we must shorten the meridians in the same amount
as the parallels are enlarged in the cylindrical projection, what yields:

dv = sinhψ dψ v = coshψ



ds 2 = ’( dψ 2 + sinh 2ψ d• 2 ) = ’(v 2 ’ 1) du 2 ’
1
dv 2
v ’1
2



dA = sinhψ d• § dψ = du § dv

which clearly displays the equivalence Figure 15.8
of the projection. Observe that v =
cosh ψ means the hyperboloid is
projected following planes
perpendicular to the cylinder of
projection (figure 15.8).


Conic projections

These projections are made
into a cone surface tangent to the
hyperboloid. Here to take the cone
z 2 ’ x 2 ’ y 2 = 0 is the most natural
choice, although as before, the cone of
projection has Euclidean nature and does not belong properly to the pseudo-Euclidean
space, where it should be the cone of zero length. The cone surface unrolled is a circular
sector. In a conic projection a parallel is shown as a circle with zero distortion. The
characteristic parameter of a conic projection is the constant of the cone n = cos θ0,
being θ0 the Euclidean angle of inclination of the generatrix of the cone. The relation
with the hyperbolic angle ψ0 of the parallel touching the cone (which is represented
without distortion in the projection) and θ0 is:

1
n = cos θ 0 =
1 + tgh 2 ψ 0
204 RAMON GONZALEZ CALVET


Since the graticule of the conic projections is radial, is more convenient to use
the radius r and the angle χ :

dr = f (ψ ) dψ dχ = n d•

The differentials of arc length and area for a conic projection are:

« dr 2 
sinh 2ψ
ds = ’( dψ + sinh ψ d• ) = ’ ¬ dχ 2 ·
+
2 2 2 2
¬ [ f (ψ )]2 ·
n2
 

sinhψ
dA = sinhψ d• § dψ = dr § dχ
n f (ψ )

Let us see as before the three special cases: equidistant, conformal and
equivalent projections. The differential of area for polar coordinates r, χ is
dA = r dr § dχ . If the projection is equivalent, we must identify both dA to find:

® sinhψ  ψ ψ
1 2
d
f (ψ ) =
 n f (ψ )  = f (ψ ) ’ r=
cosh and sinh
dψ 2 2
n n
° »

« 
n r2
¬ ·
1+
n 4 r 2 dχ 2 ·
ds = ’ ¬ dr +
2 2
¬ ·
n r2 n
¬1 + ·
 
4

If the projection is equidistant, the meridians have zero distortion so dψ = dr / n and:

1« 2 
«r
¬ dr + sinh 2 ¬ · dχ 2 ·
ds 2 = ’
n2  n 

If the projection is conformal then ds 2 ∝ dr 2 + r 2 dχ 2 so:

sinh 2ψ
ds = ’ 2 2 ( dr 2 + r 2 dχ 2 )
2

nr

Then we solve the differential equation:

sinhψ
dψ = dr
nr

with the boundary condition tghψ 0 = r0 to find the conformal projection:
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 205


n
ψ
« 
¬ tgh ·
2
r = tghψ 0 ¬ ·
ψ0
¬ ·
¬ tgh ·
 
2

On the congruence of geodesic triangles

Two geodesic triangles on the hyperboloid having the same angles also have the
same sides and are said to be congruent. This follows immediately from the rotations in
the pseudo-Euclidean space, which are expressed by means of tetranions in the same
way as quaternions are used in the rotations of Euclidean space. However, this falls out
of the scope of this book and will not be treated. The reader must perceive, in spite of
his Euclidean eyes7, that all the points on the hyperboloid are equivalent because the
curvature is always the unity and the surface is always perpendicular to the radius. So
the pole (vertex of the hyperboloid) is not any special point, and any other point may be
chosen as a new pole provided that the new axis are obtained from the old ones through
a rotation.


Comment about the names of the non-Euclidean geometry

Finally it is apparent that the different kinds of non-Euclidean geometry have
been often misnamed. The Lobachevskian surface was improperly called the
“hyperbolic plane” in contradiction with the fact that it has constant curvature and
therefore is not flat. Certainly, in the Lobachevsky™s geometry the hyperbolic functions
are widely used, but in the same way as the circular functions work in the spherical
trigonometry, as Lobachevsky himself showed. In the pseudo-Euclidean plane the
geometry of the hyperbola is properly realised, the hyperbolic functions being also
defined there. Thus the pseudo-Euclidean plane should be named properly the
hyperbolic plane, while the Lobachevsky™s geometry might be alternatively called
hyperboloidal geometry. The plane models of the Lobachevsky™s geometry (Beltrami
disk, Poincar© disk and half plane, etc.) must be understood as plane projections of the
hyperboloid in the pseudo-Euclidean space, which is its proper space, and then the
Lobachevskian trigonometry may be called hyperboloidal trigonometry.
On the other hand, the space-time was the first physical discovery of a pseudo-
Euclidean plane, but one no needs to appeal to relativity because, as the lector has
viewed in this book, both signatures of the metrics (Euclidean and pseudo-Euclidean)
are included in the plane geometric algebra and also in the algebras of higher
dimensions.


Exercises

15.1. According to the Lobachevsky™s axiom of parallelism, at least two parallel lines
pass through an exterior point of a given line. The two lines approaching a given line at

7
In fact, eyes are not Euclidean but a very perfect camera where images are projected on a
spherical surface. The principles of projection are likewise applicable to the pseudo-Euclidean
space. Our mind accustomed to the ordinary space (we learn its Euclidean properties in the first
years of our life) deceives us when we wish perceive the pseudo-Euclidean space.
206 RAMON GONZALEZ CALVET


the infinity are called «parallel lines» while those not cutting and not approaching it are
called «ultraparallel» or «divergent lines».
Lobachevsky defined the angle of parallelism Π(s)
as the angle which forms the line parallel to a given
line with its perpendicular, which is a function of
the distance s between the intersections on the
perpendicular. Prove that Π(s) = 2 arctg(exp(’s))
using the stereographic projection.

15.2. Find the equivalent of the Pythagorean
theorem in the Lobachevskian geometry by taking a
right angle triangle. Figure 15.9
15.3. A circle is defined as the curve which is the intersection of the hyperboloid with a
plane not passing through the origin with a slope less than the unity. Show that:
a) The curve obtained is an ellipse.
b) The distance from a point that we call the centre to every point of this ellipse
is constant, and may be called the radius of the circle.
c) This ellipse is projected as a circle in the stereographic projection.

15.4. A horocycle is the intersection of the hyperboloid and a plane with unity slope
( a 2 + b 2 = 1 ). Show that the horocycles are projected as circles touching the limit circle
x 2 + y 2 = 1 in the stereographic projection. Find the centre of a horocycle.

15.5 a) Show that the differential of area in the Beltrami projection is:

du § dv
dA =
(1 ’ u ’ v2 )
3/ 2
2




b) By integrating dA prove that
the area between two lines
forming an angle • and the
common “parallel” line (figure
15.10-a) is equal to π ’ •.
c) Prove that the area of every
triangle (figure 15.10-b) is equal
Figure 15.10
to its angular defect π ’ ± ’ β ’
γ.

15.6 Use the azimuthal conformal projection to show that the area of a circle
(hyperboloidal segment) with radius ψ is equal to 2 π (cosh ψ ’ 1). This result is
analogous to the area of a spherical segment 2 π (1 ’ cos θ ).
TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 207

16. SOLUTIONS OF THE PROPOSED EXERCISES


1. The vectors and their operations

1.1 Let a, b be the different sides of the parallelogram and c, d both diagonals. Then:

d=a’b
c=a+b

c2 + d2 = ( a + b )2 + ( a ’ b )2 = a2 + 2 a · b + b2 + a2 ’ 2 a · b + b2 =

= 2 a2 + 2 b2

which are the sum of the squares of the four sides of the parallelogram.

(a b + b a )2 (a b ’ b a )2
(a · b ) ’ (a § b ) =
2 2
+
1.2
4 4

a b a b + b a b a + a b2 a + b a 2 b a b a b + b a b a ’ a b2 a ’ b a 2 b
= ’
4 4

= a 2 b2

1.3 Let us prove that a § b c § d = ( a § b c ) · d :

4a§b c§d=(ab’ba)(cd’dc)=abcd’abdc’bacd+badc=

=abcd’dbac’bacd+dabc=(abc’bac)d+d(abc’bac)=

=2(abc’bac)·d=4(a§b c)·d

In the same way the identity a § b c § d = a · ( b c § d ) is proved.

1.4 Using the identity of the previous exercise we have:

a§b c§d+a§c d§b+a§d b§c=

= a·(b c§d)+a·(c d§b)+a·(d b§c)=

= a·(b c§d+c d§b+d b§c)=

=a·(bcd’bdc+cdb’cbd+dbc’dcb)=a·0=0

where the summands are simplified taking into account the permutative property.

1.5 Let us denote the components proportional and perpendicular to the vector a with ||
and ⊥. Then using the fact that orthogonal vectors anticommute and those with the same
direction commute, we have:
RAMON GONZALEZ CALVET
208


a b c = a ( b|| + b⊥ ) ( c|| + c⊥ ) = ( b|| ’ b⊥ ) a ( c|| + c⊥ ) = ( b|| ’ b⊥ ) ( c|| ’ c⊥ ) a =

= ( b|| c|| ’ b|| c⊥ ’ b⊥ c|| + b⊥ c⊥ ) a = ( c|| b|| + c⊥ b|| + c|| b⊥ + c⊥ b⊥ ) a =

= ( c|| + c⊥ ) ( b|| + b⊥ ) a = c b a

1.6 Let a, b, c be the sides of a triangle and h the altitude corresponding to the base b.
Then the altitude divides the triangle in two right triangles where the Pythagorean theorem
may be applied:

b = a2 ’ h2 + c2 ’ h2

Isolating the altitude as function of the sides we obtain:

(a
+ b2 ’ c2 )
2
2
h =a ’
2 2

4 b2
Then the square of the area of the triangle is:

b2h 2 4 a 2b2 ’ ( a 2 + b2 ’ c 2 )
2
’ a 4 ’ b4 ’ c 4 + 2 a 2b2 + 2 a 2c 2 + 2 b2c 2
A= = =
2

4 16 16

1
Introducing the semiperimeter s = (¦a¦+¦b¦+¦c¦) we find:
2

A = s (s ’ a ) ( s ’ b ) ( s ’ c )


<<

. 9
( 13)



>>