The zero blade
From this perspective, the zero blade 0=0 represents UN, as does the pseudoscalar i.
This is unsatisfactory partly because 0 should not imply a particular dimension N. Bouma suggests
interpreting 0 instead as an indeterminate subspace. This is appropriate in a programming context,
since an absolute magnitude measure of a blade then becomes a measure of its "ambiguity",
with large values indicating robust exact representations, small values indicating possibly
inexact results, and zero indicating a total absence of subspace identification
rather than a particular subspace.
Lines and Planes
We now generalise the concepts of lines and planes.
Programmers traditionally represent 3D planes with a normal 1-vector
and either a scalar distance (aka. directance)
from a given origin point, or a particular "base" point in the plane. This approach fails to extend to higher
dimensions where there is no unique normal 1-vector; rather a normal (dual) (N-k)-blade.
A k-plane (aka a displaced subspace or a flat) is a set of the form
{ r : (r-a)Ùbk = 0 }
= { r : rÙbk = aÙbk } where bk is a k-blade.
In particular, a lies in the k-plane.
An (N-1)-plane is known as a hyperplane. The term plane is
traditionally used for a 2-plane.
If bk is invertible (ie. is nondegenerate), we can represent the k-plane by the
mixed-grade multivector
bk + aÙbk
= (1+a)Ùbk
A 0-plane a is the singular pointset { a} .
A 1-plane b + aÙb is the line { r: r = a + lb}.
We call bk the tangent of the k-plane, and aÙbk its moment.
We say the k-plane is parallel to bk.
The dual of the k-plane equation is
(rÙbk)i-1 = (aÙbk)i-1
Û r.(bki-1) = a.(bki-1).
Û r.(bki-1) = a.(bki-1).
Hence for k=2, N=3 the dual of the tangent is the conventional normal vector while
the dual of the moment is the scalar directance a.n of
the conventional 2-plane representation { r : r.n = a.n }.
For N=3, we can accordingly encode lines by a vector and a pseudovector
instead of the usual two vectors; and planes by a pseudovector and a pseudoscalar
instead of the usual vector and scalar.
The k-plane (1+a)Ùbk can be expressed as (1 + ^bk(a))bk ,
the 1-vector ^bk(a) being the vector directance of the k-plane.
Simplexes
Given a frame of k+1 vectors (a0,a1,a2,...ak) where k<N,
their k-simplex is the convex hull defined by the points.
The 1-simplex for
(a0,a1) , for example, is the line segment connecting a0 to a1.
The 2-simplex for (a0,a1,a2) is the flat triangular "surface element"
defined by them.
Writing ak º (a1-a0)Ù(a2-a0)Ù...Ù(ak-a0) ;
where a0 is known as the base point, we note that the simplex
is contained within the k-plane
ak + a0Ùak
which we will refer to as the extended k-simplex.
ak + a0Ùak
has
moment a0Ùa1Ù...Ùak = a0Ùak
which is nonzero only if all k+1 vectors are linearly independant.
The scalar (k!)-1 |ak| is known as the content of
the k-simplex; where |ak| º |ak2|½
is welldefined for any pureblade ak. For k=1 this is the length |a1-a0|; for k=2 it is the area ½|(a2-a0)Ù(a1-a0)|.
Simplexes provide the basic construction element for the multidimensional integration operations of Geoemetric calculus.
One of the fundamental limitations of multivectors is that while they can naturally be used to concisely represent and manipulate extended k-simplexes (the plane containing a given three points for example) and, as we will see below, spherical surfaces with simple blades, they do not provide so ready a means of directly representing and manuipulating bounded k-simplexes such as as the triangular "facet" defined by a particular three points or the line "segment" defined by two. From a programming perspective, this is matter of some frustation.
The representations of general k-planes not containing the origin are not simple blades, so the operators
¯, ^ ,È, Ç are undefined with respect to them. The higher dimensional embeddings discussed below
provide a solution to this.
Frames
The obvious way to represent a k-frame (a1,..,ak) is as k 1-vectors, typically as the columns of a
N×k real matrix.
However alternate seldom discussed possibilities are the mixed multivectors
ak = a1 + a1Ùa2 + ... + a1Ùa2Ù..Ùak
or a1 + a1a2 + ... + a1a2..ak .
[ If we allow the operation <1> (ie. taking only the 1-vector component) then we can clearly recover
a1=ak<1> and thence a1=a1/(a12) . a2 is then available as a1¿ak,
a3 as a2¿(a1¿ak), and so forth
]
Suppose we transform the frame as ai'ºBaiB#-1 where
B#-1B = b .
BakB#-1 is not the correct representor for the transformed frame,
(BakB#-1)<i> requiring scaling by b1-i . However,
provided b>0 we can unambiguously renormalise the ai while reconstructing them
from BakB#-1 so it is a representor for the transformed frame.
Higher Dimensional Embeddings
We can think of k-blades as representing k-dimensional subspaces in
ÂN. For N=3, these correspond to lines and planes through the
origin. We would like to be able to use the operators
¯, ^ ,È, Ç to manipulate general displaced subspaces of ÂN
(ie. ones not containing the origin).
One way of doing this is to "embedd" ÂN points in a higher dimensional
space via a function ¦ : ÂN ®
ÂN+p,q,r and work with the multivectors of that space.
With each point x in ÂN we associate a point x = ¦(x) in the higher space.
With each multivector in ÂN we associate the "same" multivector with regard to an
extended basis.
Multivectors in the higher space are associated with particular structures in ÂN
and we can investigate these structures by working with the multivectors.
We look here at three popular embeddings.
Homogeneous coordinates
We move from ÂN to ÂN+1 by the incorportaion of a new basis vector
e0 with Sig(e0)=1.
The embedding is trivial: ¦(x) = x + e0 .
Some authors favour x ® e0x, mapping 1-vectors to 2-blades (when e0¿x=0), but we do not
take this approach here.
For N=3, our multivectors now require 16 coordinates.
Homogeneous coordinates are so called because they "desingularise" the origin point 0.
In a Euclidean space, all points save 0 have a geometric inverse. By displacing all points by 1-vector e0
outside ÂN we ensure that all points are invertible.
k-planes
Consider the ÂN+1 (k+1)-blade (a0+e0)Ùak where ak is tangent to the ÂN k-simplex for frame { a0,a1,..ak}.
(x+e0) Î (a0+e0)Ùak Û (x+e0)Ù(a0+e0)Ùak = 0
Û xÙa0Ùak = e0(x-a0)Ùak
Û (x-a0)Ùak = 0 which is the condition for membership in the k-plane
containing the k-simplex.
Hence the k-simplex of any k+1 points in ÂN corresponds to a
(k+1)-blade in ÂN+1. The simplex frame is not uniquely recoverable from the blade, only
the extended simplex and measure of the content.
In particular, the ÂN k-plane (1+x)Ùak is represented by ÂN+1 (k+1)-blade (e0+x)Ùak .
For example:
We can accordingly use joins and meets in ÂN+1 to construct and intersect displaced subspaces in ÂN. The advantage of using joins rather than outer products to construct simplex representatives is that we will obtain appropriate results for degenerate simplexes. If p=q, for example, then (e0+p)È(e0+q)=e0+p. If a,b and c are colinear, then (e0+a)È((e0+b)È(e0+c)) represents the line containing them.
If a given ÂN k- and m-planes intersect then the meet of their ÂN+1 representatives represents that intersection, and we can recover it as follows. We have an c Î ÂN+1 which we wish to express as (e0+c0)ÙCn where c0 and Cn are in ÂN. If we have c expressed via coordinates with respect to the extended basis for { e0,e1,...,eN } this is a trivial computation ( Cn[ij..m] = c[0ij..m] ).
Suppose the ÂN k- and m-planes with k+m=N-1 containing given simplexes do not intersect.
The corresponding ÂN+1 blades ((a0+e0)Ùak) and ((b0+e0)Ùbl)
are then complimentary and have a scalar meet equal to
the directed perpendicular Euclidean distance between the ÂN planes.
[ Proof :
((a0+e0)Ùak)Ù((b0+e0)Ùbl)
= a0ÙakÙb0Ùbl
+e0Ù(akÙb0Ùbl-akÙa0Ùbl)
= 0+e0Ù(akÙb0Ùbl-akÙa0Ùbl)
= e0ÙakÙ(b0-a0)Ùbl .
.]
Let us review with regard to N=3. By the adoption of homogeneous coordinates, we can represent 3D points, lines, and planes by blades of Â4 which are implementable as 24=16 dimensional vectors. The use of a sixteen-real multivector to represent a 3D point may seem extravagent, but of course such a multivector has sparsity 13. Similarly, the multivector representations of lines and planes have sparsities 10 and 12 respectively. In compensation for this wastage we have the following advantages:
| ab | |||||
| 1 | e0 | e¥ | e¥0 | ||
| 1 | 1 | e0 | e¥ | e¥0 | |
| b | e0 | e0 | 0 | -1+e¥0 | e0 |
| e¥ | e¥ | -1-e¥0 | 0 | -e¥ | |
| e¥0 | e¥0 | -e0 | e¥ | 1 | |
e0* = e0(e¥0i)-1 = (e0e¥0)i-1 = -e0i-1 ;
e¥* = e¥(e¥0i)-1 = -e¥i-1 .
Geometric Interpretation Overview
We shall see below that for N=3, by the adoption of GHC we
can represent traditional 3D points, lines, planes, bipoints, circles, and spheres by pure blades in
Â4,1 which are implementable as 25=32 dimensional vectors. More generally we can represent N-D
k-spheres and k-planes by particular (k+2)-blades in UN% implimentable as 2N+2 dimensional 1-vectors.
We will associate
UN% point s=l(c+e0+½e¥(c2±r2) ) via its dual with
the UN hyper(anti)sphere of squared radius s2=±r2 and centre c
and can think of e¥ as providing a "squared radius coordinate".
Thus 1-curves in
UN% represent paths through the space of UN hyperspheres and we can represent the trajectory of a hypersphere with
constant radius r as a UN% 1-curve confined to x2=r2.
If points a,b are distinct
then a¿b = (aÙb)2 = -½(a-b)2 and
aÙb represents the bipoint {a,b}
while e¥ÙaÙb represents the extended line through a,b
; with (e¥ÙaÙb)2 = (a-b)2 . Here a is the horoemebdding of
UN point a into UN% described below.
If points a,b,c are noncolinear aÙbÙc represents the
1-sphere (ie. the circle) through a,b,c
with (aÙbÙc)2 = r2 4 Area(a,b,c)2
where r is the circle radius;
while
e¥ÙaÙbÙc represents the 2-plane (ie. the plane) containing {a,b,c}
; with
(e¥ÙaÙbÙc)2 = -2! Area(a,b,c)2 .
If points a,b,c,d are noncoplanar
aÙbÙcÙd represents the
2-sphere (ie. 3D sphere) through a,b,c,d, with
(aÙbÙcÙd)2
= -r2 3! Volume(a,b,c,d) ; while
e¥ÙaÙbÙcÙd represents the 3-plane containing them
with
(e¥ÙaÙbÙcÙd)2 = -3! Volume(a,b,c,d) .
These results generalise for higher N with the k-blade outter product of k£N+1 embedded points representing the (k-2)-sphere through those points and having magnitude r (k-1)! V where r is the radius of the (k-2)-sphere and V is the "volume" content of the (k-1)-simplex formed by the points. And the (k+1)-blade formed by wedging this with e¥ represents the (k+1)-plane containing the points, with magntitude (k-1)! V. What could be more useful?
We will associate more general UN% k-blade s1Ùs2..Ùsk via its (N+2-k)-blade dual (s1Ùs2..Ùsk)* with the space of all planes and spheres that include the (N-k)-sphere meet s1*Çs2*...Çsk* .
This all constitutes a magnificent "payback" for investing in two additional dimensions. A ÂN+1,1 (k+1)-blade ak+1 = a0Ùa1Ù...Ùak derived from ÂN points a0,a1,..,ak expands as a0Ùa1Ù...Ùak = a0Ùa1Ù...ak + e0Ù(a1-a0)Ù(a2-a0)Ù..Ù(ak-a0) + O(terms in e¥ and e¥0)
Generalised homogenised blades thus provide an excellent way to
represent extended lines and planes and circles and spherical surfaces, but they are less
accomodating when it comes to representing finite line segments and the interiors of k-spheres.
Informally, they capture boundaries rather than the regions bound.
But not all ÂN+1,1 blades are of this type, others represent
not k-planes or k-spheres but more general "families" of hyperspheres.
We can form an m-blade meet akÇbl of (k-2)-sphere ak and (l-2)-sphere bl in the usual way, imposing Eucluidean signatures if desired. Any null 1-vector x within their meet corresponds to either a UN point in their geometric intersction, or to ¥x when e0¿x=0 . A nonnull 1-vector will be dual to a (possibly anti) hypersphere whose centre will have geometric relevance to the nature of their nonintersection.
Suppose for example that we have two 3-pointblades in Â4,1 each representing either a circle or an infinite lines in Â3. Two 3-blades must intersect in a 5D space so the meet of the two blades will be a proper blade, ie. have grade ³ 1. If the meet is a null 1-vector, then the two objects intersect in Â3 at a single point. If the meet is a nonnull 1-vector m then the circles do not intersect and the meet is dual to a hypersphere that intersects with both, the sign of m2 indicates whether the line passes through the inside (negative) or outside (postive) of the circle as it crosses its tangent plane.
Hence the meet may not be scalar when the ÂN structures do not intersect, representing hyper(anti)spherical solutions.
The nature of an intersection in ÂN
is indicated by the grade of the delta product and the sign of the square of the meet, which is often most directly
computaable via the square of the delta product. When the meet is scalar valued, it provides a measure of the
seperation such as when provideing the squared radius of a hypersphere tangent to both.
When it is proper-blade valued, it either geometrically represents the ÂN intersection or an "imaginary intersection"
embodying information pertaining to the geometry of the ÂN nonintersection, eg. providing a seperating 2-plane.
The meet Ç acts precisely as programmers might wish it to, essentially providing the biggest possible seperator between a given
two nonintesrsecting pointsets.
The horosphere point embedding
We are interested in the point embedding taking ÂN to the subspace of ÂN+1,1 defined by
HNe¥ = { r Î ÂN+1,1 : r2 = 0 ; e0¿r = 1 }.
This is the horosphere, the intersection of a hyperplane normal to e¥ containing e0,
and the null cone { x Î ÂN+1,1 : x2 = 0 }.
x = x + ae0 + be¥ is null
Û x2 = ab so
HNe¥ = { x + ½x2e¥ + e0 : x Î ÂN } .
Our embeding is accordingly
x = f0(x)
º f0e¥(x) º x + e0 + ½x2e¥
=
x + ½e-(x2+1) + ½e+(x2-1)
though some authors favour
lx + l2e0 + ½x2e¥
= lx + ½e-(x2+l2) + ½e+(x2-l2)
with l a unit length in order to homogenise the "dimension" or "units" of x .
We will refer to such null x of unit e0 coordinate (-e¥¿x=1)
as the horosphere embedding or horoembedding of
x = ¯(e¥0*)(x).
In particular,
e0 corresponds to the ÂN origin 0 while e¥ represents a hypothetical ÂN point at infinity which we will denote ¥.
When x2=1 this coincides with the spherical conformal embedding x = x+e0+½e¥ = x+e- .
Some authors consider an embedding ÂN ® ÂN+1,1
defined by x ® e¥0x, mapping 1-vectors to 3-blades. We do not take this
approach here, although similar correspondances arise as consequences of the
point embedding since we associate the horosphere intersection of e¥0x=e¥Ùe0Ùx with the 1-plane (line)
through 0 and x. The 3-blade e¥0x also spans UN% points of the form
e0+l(x-e0)
= (1-l)((1-l)-1x + e0)
= (1-l)(((1-l)-1x)' - ½(1-l)-2x2e¥)
dual to hyperspheres of centre (1-l)-1x and radius (1-l)-1|x|,
ie. hyperspheres that contain (pass through) 0 with centre on the line through 0 and x.
In Homogenous coordinates, we "de-specified" the origin by replacing 0 with 02=0 by e0 with e02=1, however (when extending a Euclidean space) the embedded origin remains a unique point in that it is to sole point minimally attaining the inequality x2³1 so one might more properly refer to "quasi-homogenised coordinates". With GHC the embedded origin e0 is truly equivalent to every other embedded point in that all nontrivially satisfy x2=0, and in this sence GHC can be regarded as a "true" homogenisation.
By embedding into a nullcone we arrange that x is noninvertible even when x-1 exists.
Although "inverting a point" seldom has a physical interpretation,
we will see that self-inverse 2-blade " flat point" e¥Ù(x+e0)=e¥Ùx
also represents the ÂN point x but in a slightly different way.
The nearest we have to a 1-vector inverse for the embedded point is
x' = 2(x2+1)-2x[+]
where
x[+] = x - ½e-(x2+1) + ½e+(x2-1)
= x - x2e0 - ½e¥
= -x2f0(-x-1) for nonnull x
has xx[+]
= ½(x2 + 1)2 + 2xÙe- + (x2-1)e¥0
so, provided x2¹-1, we have
xx'
= 1 + 4(x2+1)-1((x2+1)-1xÙe- + (x2-1)e¥0)
, though this is of little practical use.
Inverse Point Embedding
The simplest inverse mapping
f0-1(x) º (e0¿x)-1 ^(x, e¥0)
= -( e¥¿x)-1(xÙe¥0)e¥0
= (e¥¿x)-1(e¥0Ùx)e¥0 taking UN% into UN
is unsatisfactory because f0-1(e¥) is undefined and we wish to define f0-1
fully over UN% rather than merely over HNe¥.
Thus we seek a
f0-1 : UN% ® UN È ¥ with f0-1(e¥)=¥.
Some authors favour f0-1(x) =
e+¿(xÙe¥)
= -e-¿(xÙe¥)
= e0¿(xÙe¥)
.e+
= (xe¥ - e¥0).e+ = x + e- but we will take
f0-1 : UN% ® UN% defined by f0-1(x) º e¥¿(e¥Ù((e0¿x)-1x) - e¥0)
which returns the horosphere to the UN% subspace { x : x¿e¥0 = 0 } .
Clearly f0-1f0 is an identity mapping 1 : UN®UN
but f0f0-1 : UN% ® HNe¥ is a many-to-one "projection" into the horosphere
dependant on our choice of f0-1.
All f0-1 coincide over HNe¥, however, and so given any g UN ® HNe¥ we can unambigusously define
f0-1g UN% ® UN .
Embedded Products
Note that (x+e0+le¥)2 = x2 - 2l.
Note also the scaled idempotent 2-versors
(e0x)2 = (-x2)e0x
and
(xe0)2 = (-x2)xe0 with similar reslts for e¥x and xe¥.
Further,
e0xe¥x
= -e0(x2e¥ + 2x) has
(e0xe¥x)2
= e0(x2e¥ + 2x)e0(x2e¥ + 2x)
= e0(x2e¥)e0(x2e¥ + 2x)
= -2x2 (e0xe¥x)
while
(e0xe¥x)§
(e0xe¥x) =
(e0xe¥x)
(e0xe¥x)§ = 0.
Let a,b be the null ÂN+1,1 1-vectors associated with ÂN points a and b.
ab =
-½(a-b)2 + aÙb +½(a2b-b2a)e¥
+ (a-b)e0 - ½(a2-b2)e¥0 .
[ Proof :
(a + ½a2e¥ + e0)(b + ½b2e¥ + e0)
= ab - ½a2be¥ - be0
+ ½ab2e¥ + ½b2e0e¥
+ ae0 + ½a2e¥e0
=
a¿b + aÙb - be0
+ ½(ab2-a2b)e¥ + ½b2(-1-e¥0)
+ ae0 + ½a2(-1+e¥0)
=
-½(a-b)2 + aÙb +½(a2b-b2a)e¥
+ (a-b)e0 - ½(a2-b2)e¥0 .
.]
Hence
a¿b =a.b = -½(a-b)2 = -½(a-b)2
and so for nullseperated points a and b ab=-ba=aÙb is a null 2-blade.
Also (aÙb)2 = (a¿b)2 = 4-1(a-b)4
.
The 1-vector linear combination l1a+l2b represents a point
(ie. is null) only if (a-b)2=0 .
aÙb =
aÙb +½(a2b-b2a)e¥
+ (a-b)e0 - ½(a2-b2)e¥0.
For N=3, our multivectors now require 25=32 coordinates but are frequently sparse.
An embedded point has just five nonzero coordinates, for example.
Dropping to the horosphere
We can express a general 1-vector y Î ÂN+1,1 as
x + ae0 + be¥ where x Î ÂN but this appears less useful than
the
decomposition ae¥xe¥ + be¥e¥ where xe¥2=0 and e0¿xe¥=1 ;
achieved by ae¥ = e0¿y = -e¥¿y ;
be¥ = ½(e¥¿y)-1y2 ; and
xe¥ = - ½(e0¿y)-2 ye¥y
corresponding to reflecting e¥ in y and rescaling.
More generally, for a given unit or null 1-vector e in UN% we can define
null horodrop along e or e-horodrop of y denoted De(y) by removing enough e from y to turn it null.
If this has nonzero e0 coordinate then we can rescale to give the normalised e-horodrop
De~(y).
with unit e0 coordinate.
For e=e¥ we can think of moving from scaled hyper(anti)sphere dual
y=l(c+e0+½(c2±r2)e¥)
to the embedded centre point c as a natural definition of f0-1 f0.
This fails for y=x+be¥
where x Î UN and so e0¿y=0, which we can intrepret as representing a particular "directed infinity"
¥x .
For general null e
then provided e¿y ¹ 0 we can
remove ½(e¿y)-1y2 e
and rescale for unit e0 coordinate equivalent to yey rescaled.
For general unit e with (e¿y)2 ³ e2y2
we can remove
e2(-e¿y ±
((e¿y)2-e2y2)½) e
and reach HNe¥ at two alternative nullvectors having opposite e coordinate and typically
corresponding
to x4 greater and less than 1.
If we obtain unrescalable l(x+be¥) we interpret it as ¥x .
Clearly De¥(y)
= (e0¿y) f0((e0¿y)-1 ^e¥0(y))
= (e0¿y) f0f0-1(y)
if y has nonzero e0 coordinate. f0-1 De¥ for y with nonzero e0 coordinate thus tricvially consists
of dividing the UN compobnent by the e0 coo0rdinate, independant of the e¥ coordinate.s
The question arises of how much
e must be added to a null x to make it unit, ie. we
seek b so that the e-horolift Le±(x) = x + be has
Le±(x) 2 = ±1.
For e=e¥ we solve (x+be¥)2 = ±1 with
b = -/+½(e0¿x)-1 corresponding
(for "normalised" null x with e0¿x=1) to obtain
Le¥±(x) =
x + e0 + ½(x2 -/+ 1)e¥
= x + ½(e--e+) -/+ ½(e-+e+) + ½x2e¥
= x -/+ e± + ½x2e¥ .
which we can think of as moving from scaled null point embedding
lx = lf0(x)
to the (dual of) the unit-radius (anti)hypersphere with centre x.
Hyperspheres
The non degenerate N-D sphere (aka. hypersphere) { x : (x-c)2 = r2 }
where r>0
corresponds to the ÂN+1,1 equation
x¿c = -½ r2.
Writing
s = c-½r2e¥
= c + e0 +½(c2-r2)e¥
= c + ½(1+c2-r2)e-
+ ½(-1+c2-r2)e+
we have
s2 = r2 ;
e0¿s=1
, and x¿s=0 Û x lies on the hypersphere.
[ Proof : x¿s = x¿(c-½r2e¥)
= -½ r2 - ½r2x¿e¥
= 0.
.]
So for any plussquare s Î ÂN+1,1 satisfying e0¿s = 1
the solution set
{ x Î HNe¥ : x¿s = 0 } =
{ x Î HNe¥ : xÙ(s*) = 0 }
corresponds to a hypersphere (ie. a spherical surface) in ÂN having centre
c = ^e¥0(s) and radius r=(s2)½ ;
and we can associate any hyperblade
s*=se¥0i-1
with an N-D hypersphere provided e0¿s¹1
and s2>0.
In particular, (e0-½e¥)*=-e+* represents the unit sphere at 0.
Conversely, any nondegenerate hypersphere in ÂN corrseponds to a solution set
{ x Î HNe¥ : x¿s = 0 } where
s=c+e0 +½(c2-r2)e¥ satisfies
s2=r2 ; e¥¿s = -1.
Further, s = a¿(cÙe¥)
where a is any point on the hypersphere.
Whence s* = aÙ(cÙe¥)* .
[ Proof :
Setting a=c+rb~ for arbitary unit b~ we have
a¿(cÙe¥) = (c+rb~ + e0
+ ½(c2+2rc¿b~ + r2)e¥)
¿(ce¥-e¥0)
= c2e¥ + r(b~¿c)e¥+c
+ e0
- ½(c2+2rc¿b~ + r2)e¥
= c+e0+½(c2-r2)e¥
.]
To what does the horosphere solution set for x¿s = 0 correspond if
s2>0 but e0¿s=0 rather than 1?
s = ^e¥0(s) + ¯e¥0(s)
= s+(s¿e¥)e¥ when e0¿s=0 so
s = |s|(n-(s¿e0/|s|)e¥) where n is a unit vector
in ÂN and |s|=(s2)½=(s2)½.
Thus x¿s=0
Û x¿(n-(s¿e0/|s|)e¥) = 0
Û x¿n - e0¿(s¿e0/|s|)e¥ = 0
Û x¿n + (s¿e0/|s|) = 0
Û x¿n + d = 0
where d = (s¿e0/|s|) = s~¿e0.
This defines a ÂN hyperplane having normal s and
directance -ds.
Since e¥¿s = 0, e¥ lies in the solution set, corrseponding to a
"point at infinity" ¥
attained by the hyperplane. We can accordingly view a hyperplane as a sphere
(of infinite radius) that passes through ¥.
We will refer to s2 = r2 , the squared radius of the UN hypersphere to which s is dual,
as the squadius of plussquare UN% point s. For s2<0 we have a negative squadius
corresponding to the squared radius of an antihypersphere ( { x : x2=-r2 } ) in UN.
The dual equation to x¿s = 0 is
xÙ(s*) = 0
where s* is a pseudovector in ÂN+1,1.
In particular, consider a nondegenerate (N+1)-blade
s*=a0Ùa1Ù....ÙaN .
We have s¿e¥ = 0 Û (e¥Ùs*)2 = 0
so if e¥Ùs* ¹ 0 the blade represents a
sphere in ÂN with x lieing on the sphere iff
xÙs*=0.
In particular, aiÙs*=0 for i=0,1,..N so
a0,a1,...aN lie on the sphere.
[ Proof : (s¿e¥)2 - (sÙe¥)2
= (s¿e¥ + sÙe¥)(s¿e¥ - sÙe¥)
= (se¥)(e¥s)
= s(e¥e¥)s) = 0 .]
If e¥Ùs* = 0 the blade represents a
hyperplane in ÂN which can similarly be shown to contain
a0,a1,...aN.
We can thus represent the hyperplane containing a1,a2,...aN
by the blade e¥Ùa1Ù...ÙaN.
If s1* and s2* represent spheres then
s1¿s2 =
½(r12+r22 - (c1-c2)2 ) .
This may initially seem a disappointing inner product since we might instead have wished for the closest distance between the spheres
, but it has a definite geometric interpretation nonetheless.
If a is a point common to both spheres
then the cosine of the angle subtended by the "radial normals" (a-c1) and (a-c2) is given
(via the traditional triangular cosine rule)
as ½(r12 +r22-(c1-c2)2)(r1r2)-1 .
Thus s1¿s2 is r1r2 times the cosine of the angle subtended by the radial normals at
any common point. This result also holds when one of s1 or s2 represents a hyperplane.
s1Ùs2
= c1Ùc2 + e0d + ½e¥c + ½e¥0( c12-c22+r22-r12)
where c º (c12-r12)c2 - (c22-r22)c1
and d º c2-c1.
(s1Ùs2)2
= (s1¿s2)2 - r12r22
= ¼( (r12-r22)2 - 2(c1-c2)2(r12+r22)
+ (c1-c2)4 )
.
[ Proof :
(s1Ùs2)2
= (c1Ùc2)2 + c¿d + ¼(c12-c22+r22-r12)2
Set c1=0
so that c=-r12c2 and d=c2 .
We then have
(s1Ùs2)2 =
-r12c22 + ¼(c22-r22+r12)2
=
¼c24 + ¼(r12-r22)2 +
-½c22(r12+r22) .
Replacing c2 with c2-c1 gives
¼( (r12-r22)2 - 2(c1-c2)2(r12+r22)
+ (c1-c2)4 )
which rearranges as
¼((r12+r22)2 - 2(c1-c2)2(r12+r22)
+ (c1-c2)4
- 4r12r22)
= (½(r12+r22 - (c1-c2)2 ))2 - r12r22
.]
With s1=x so r1=0 we have x¿s2 = ½(r22-(c2-x)2) which is positive only when x lies outside s2* and zero only (for Euclidean UN) when it lies on it. Also (xÙs2)2= ¼(r22-(c2-x)2)2 is zero if x lies in hypersphere s2* . More generally, the sign of (s1Ùs2)2 indicates where s1* intersects s2*. Neagtive implies an interesction and zero implies tnagential contact.
s+le¥ is dual to a hypersphere with centre c and squadius r2-2l.
s+le0 = (1+l)( (1+l)-1c + e0
+ ½e¥(1+l)-1(c2-r2)) is dual to
a hypersphere with centre (1+l)-1c and squadius
(1+l)-2(r2-lc2).
It is easily verified that (ls1+(1-l)s2)* represents a hypersphare of centre
lc1 + (1-l)c2 and squadius l(l-1)(c1+c2)2 + lr12+(1-l)r22.
[ Proof :
ls1+(1-l)s2 = lc1+(1-l)c2 + e0 +
½e¥(lc12+(1-l)c22-lr12-(1-l)r22)
= c2+e0+½e¥(c2 -
(-lc12-(1-l)c22+lr12+(1-l)r22+c2))
dual to a hypersphere centre c=lc1+(1-l)c2 of squadius
(-lc12-(1-l)c22+lr12+(1-l)r22+c2)
= (l(l-1)c12+l(1-l)c22+2l(1-l)(c1¿c2)
+ lr12+(1-l)r22 )
= l(l-1)(c12+c22+2(c1¿c2))
+ lr12+(1-l)r22 )
= l(l-1)(c1+c2)2 + lr12+(1-l)r22
.]
We saw that the meet Ç can be evaluated with forced Eucliden signatures and when interscting blades in
Âp,q% we can stay in
Âp,q% or "lift" into
Âp+q,0% or Âp+q+2,0
according to choice.
The point x=X+xNeN is represented as x=x+e0+½(X2-(xN)2) in
ÂN,1%
which represents the dual of a hypersphere centre x and radius 2½xN .
If x2<0 this sphere encloses 0, if x2=0 it includes 0.
The ÂN,1 squaredlength x2 represents the ÂN+1,0 squeperation
of 0 from the spherical surface.
k-planes
Let a0,a1,..ak be k+1 1-vector horopoints
in ÂN+1,1 corresponding to points a0,a1,..,ak in ÂN
and set ÂN k-blade
ak º (a1-a0)Ù(a2-a0)Ù...Ù(ak-a0) .
We have
e¥Ùa0Ùa1Ù...Ùak
= e¥Ù(a0 + e0)Ùak
.
[ Proof :
e¥Ù(a0+½a02e¥+e0)Ùa1Ù...Ùak
= (e¥Ùa0+e¥0)Ùa1Ù...Ùak
= (e¥a0+e¥0)Ù(a1+½a12e¥+e0)Ùa3Ù...Ùak
= (e¥a0+e¥0)Ù(a1+e0)Ùa3Ù...Ùak
= ((e¥a0)Ùa1+e¥0Ùa1+e¥Ùa1Ùe0)Ùa3Ù...Ùak
= (e¥a0Ùa1+e¥0Ù(a1-a0))Ùa3Ù...Ùak
= ...
= (e¥Ùa0Ù...Ùak + e¥0Ù(a1-a0)Ù...Ù(ak-a0))
= e¥Ù(a0Ù(a1-a0)Ù...Ù(ak-a0) + e0Ù(a1-a0)Ù...Ù(ak-a0))
.]
Also, (e¥Ùa0Ùa1Ù...Ùak)2
= ak2
= (-1)½k(k-1)(k!Vk)2
where Vk is the volume (content) of the k-simplex.
[ Proof :
For a0=0 we trivially have
(e¥Ùe0Ùak)2 = (-e¥0ak)2 = ak2.
More generally
(e¥Ù(e0+a0)Ùak)2 = (½e¥(a0Ùak)-e¥0ak)2
= (-e¥0(½e¥(a0Ùak)+ak)2
= (½e¥(a0Ùak)+ak)2
= (½(e¥(a0Ùak)ak + ake¥(a0Ùak))+ak2)
= (½e¥((a0Ùak)ak + ak#(a0Ùak))+ak2)
= ak2 since we know result is scalar
.]
Now, x + e0 + ½ x2e¥ Î
e¥Ù(a0 + e0)Ùak
Û (x-a0)Ùak = 0 . Thus
e¥Ùa0Ùa1Ù...Ùak
= e¥Ùa0Ùak
= -i2 (a0¿(e¥ak*))*
represents the ÂN k-plane containing a0,a1,...ak.
.
[ Proof :
(x + e0 + ½ x2e¥)Ùe¥Ù(a0 + e0)Ùak = 0
Û e¥Ù(x + e0)Ù(a0 + e0)Ùak = 0
Û e¥Ù(xÙa0Ùak - e0(x-a0)Ùak) = 0 .
Also, e¥Ùa0Ùak
= -(a0¿((e¥Ùak)*))-*
= -(a0¿((e¥Ak)*))-*
= -(a0¿(e¥Ake¥0i-1))-*
= -(a0¿(e¥aki-1))-*
.]
In particular
e¥0Ùbk
= e¥0bk represents the k-plane through 0 with tangent k-blade bk.
If a0¿ak=0 we have
e¥Ù(a0 + e0)Ùak
= (e¥(a0 + e0)+1)ak and it is sometimes convenient to express the (k+2)-blade in product form
as (e¥(e0+dn)+1)ak where ak is the tangent blade
and dn is the directance 1-vector normal to ak, ie. the ÂN
point in the k-plane closest to the origin.
For example, the 3-blade e¥Ù(a+e0)Ùd represents the 1-plane (extended line) through a and a+d If a is the closest point of the line to the origin so that a¿d=0 we have e¥Ù(a+e0)Ùd = (e¥(a+e0)+1)Ùd = (e¥(a+e0)+1)d represents the 1-plane (line) through a and a+ld
The unit plusquare 2-blade e¥Ùa = e¥Ù(a+e0) = e¥(a+e0) + 1 = e¥0+e¥a represents the 0-plane (point) {a} and so represents the same thing as does the null 1-blade a=a+e0+½a2e¥ . It is this invertible 0-plane representation that is "output" by the meet and join operations, eg. the meet of lines e¥Ù(a+e0)Ùu and e¥Ù(a+e0)Ùv is (e¥Ù(a+e0)Ùu) Ç (e¥Ù(a+e0)Ùv) = e¥Ù(a+e0) .
We can recover the tangent ak from Ak+2 = e¥Ù(a0 + e0)Ùak via ak = e¥0¿