 Multivectors as Geometric Objects
"As long as algebra and geometry have been separated, their progress have been slow and their uses limited; but when these two sciences have been united, they have lent each mutual forces, and have marched together towards perfection." --- Joseph-Louis Lagrange

Introduction
We have covered multivectors and their associated mathematical operations in some considerable depth. Now we turn at last to what they are good for.

Subspaces
We can regard a proper k-blade ak as representing the set of 1-vectors { x : xÙak = 0 } or, equivalently, { x : x¿ak* = 0 } where x is understood to be a 1-vector from a vector space UN so that ak 1-represents the "space" of all 1-vectors (interpreted as points) satisfying a particular geometric condition.
Since xÙak is a pure (k+1)-blade the equivalence with zero condition involves NCk+1 coordinates .
More generally we can regard nonnull ak as spanning the set of multivectors { x : ¯ak(x)=x } = { x : x¿ak=xak }  .

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.
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}.
A 2-plane is the traditional plane.
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). Hence for k=2, N=3 the dual of the tangent is the conventional normal vector n 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.

Since the representations of k-planes not containing the origin point 0 are not pure blades, the operators ¯, ^ ,È, Ç are undefined with respect to them. The higher dimensional embeddings   discussed below provide a solution to this.

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.

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:

• The ÂN point (0-plane) p corresponds to the ÂN+1 vector e0+p.
• The ÂN line (1-plane) through 2 distinct points p,q corresponds to the ÂN+1 2-blade (e0+p)Ù(q-p) = (e0+p)Ù(e0+q) = È(e0+p , e0+q). Any segment of the same length of the extended line through p and q will generate this same ÂN+1 bivector,
• The ÂN 2-plane containing 3 noncolinear points a,b, and c corresponds to the ÂN+1 3-blade
(e0+a)Ù(b-a)Ù(c-a) = (e0+a)Ù(e0+b)Ù(e0+c) = (e0+a)È((e0+b)È(e0+c)) .
Any three points in the a,b,c plane defining a triangle of the the same area will generate this same trivector.

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:

• Uniformity of representation.
• Generally applicable operators for geometric projection, rejection, and intersection/seperation replacing the conventional unholy suite of point-line, line-line, line-plane, point-plane, line-plane, plane-plane geometric primitves.
• Affine invarient representation. Given a linear or affine map g:Â3 ® Â3, we have (g(r)-g(a))Ùg(bk)=g(r-a)Ùg(bk)=g((r-a)Ùbk) so we can represent the image under g of a k-plane by the image under g of its "pseudo representors". We cannot do this for the "normal representation" since g(a.n) ¹ g(a).g(n).
• Ease of extension into higher dimensions. The conventional vector cross product so pivotal in conventional 3D geometry, for example, only exists for N=3 or N=7 (when it is a product of three vectors, see Lounesto Ch 23.4)
Though the above advantages might alone suffice to convince a jaded 3D programmer to rethink his paradigms, there are more to come.

Affine Model
We incorporate a new basis vector e0 with Sig(e0)=0. We denote the space of multivectors for such a basis by ÂN,0,1.
The embedding is ¦(x) = 1 + e0x = 1 + e0Ùx .

Generalised Homogeneous Coordinates
This system, proposed by Li et al provides many advantages including all those of Homogeneous and affine model imbedding, at the cost of a less trivial ¦ and (N+2)-dimensional multivectors. It is now increasingly known as the conformal geometric algbra or the conformal embedding.

e0 and e¥
We extend a basis { e1,...,eN } for ÂN   by two orthogonal vectors e+ and e- satisfying e+2 = 1; e-2 = -1; e+¿e- = 0 to form a Minkowski space ÂN% º ÂN+1,1; but it will be convenient to consider instead the nonorthogonal basis { e0,e¥,e1,...,eN } for ÂN+1,1 where
e0 = ½(e- - e+) ; e¥ = e- + e+     are null vectors with e+=-e0e¥ ; e-=e0e¥ .
We can construct an "inverse" frame with e0=-e¥ ; e¥=-e0 so that e0¿e0 = e¥¿e¥ = 1 but note that e0¿e¥ = -1 rather than zero and e0Ùe0=-e+Ùe- rather than zero.
We write e¥0 for the unit plussuare 2-blade e¥0 º e¥Ùe0 = e¥e0 + 1 = e+Ùe- = e+e- that commutes with everything in ÂN while anticommuting with e0,e¥,e+, and e-.
Note carefully that e¥0 ¹ e¥e0 because e¥ and e0 are not orthogonal. We can usually freely replace extended orthonormal basis blades by versors , eiÙejÙek = eiejek = eijk for example, but more care is needed with blades formed from e0 and e¥ since though 2-blade e¥0=e+e- is a 2-versor, it is not the 2-versor e¥e0 which includes scalar component -1.
We can tabulate the geometric product over e¥0 as
 ab in e¥0 a 1 e0 e¥ e¥0 e+ e- 1 1 e0 e¥ e¥0 e+ e- b e0 e0 0 -1+e¥0 e0 ½(-1+e¥0) ½(-1+e¥0) e¥ e¥ -1-e¥0 0 -e¥ 1+e¥0 -1-e¥0 e¥0 e¥0 -e0 e¥ 1 e- e+ e+ e+ ½(-1-e¥0) 1-e¥0 -e- 1 -e¥0 e- e- ½(-1-e¥0) -1+e¥0 -e+ e¥0 -1

Note the "absorbtion" of e¥0 by the nullvectors e¥0e¥ = -e¥e¥0 = -e¥ and e¥0e0 = -e0e¥0 = e0 as opposed to the more usual dualities e+e¥0=e- and e-e¥0=e+ . Hence e¥0xe¥0§ preserves e0 and e¥ but negates xÎÂN while e¥0× annihilates xÎÂN while preserving xe0 and negating xe¥.
Also e0e¥e0 = -2e0 while e¥e0e¥ = -2e¥ and e0e¥ = -e¥e0 - 2.
If i=e1Ù...ÙeN is a unit pseudoscalar for ÂN then e¥0i = ie¥0 is a unit pseudoscalar for ÂN+1,1.

We will denote the "extended" ÂN+1,1 dual in ie¥0 by a* º ai-1e¥0 and the "unextended" ÂN+1 dual in i by a* º ai-1.
e0* = e0(e¥0i)-1 = (e0e¥0)i-1 = -e0i-1 ; e¥* = e¥(e¥0i)-1 = -e¥i-1 so the extended and unextended duals of the nullextendors differ only in sign. This runs counter to our experience of nonnull basies, which might lead us to expect e0* to exclude e0 and include e¥.

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+e0e¥(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 can 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 s1*Çs2*...Çsk* .

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)

• If   (a1-a0)Ù(a2-a0)Ù..Ù(ak-a0) ¹ 0 , then ak+1 represents the (k-1)-sphere containing the k+1 points.
• Otherwise if a0Ùa1Ù...ak ¹ 0 it represents the extended k-simplex {a0,..ak} .
• In the "doubly degenerate" case of both (a1-a0)Ù(a2-a0)Ù..Ù(ak-a0) and a0Ùa1Ù...ak vanishing, ak represents a (k-1)-plane through the origin.

This all constitutes a magnificent "payback" for investing in two additional dimensions and GHC blades provide an excellent way to represent extended lines and planes and circles and k-spherical surfaces, but they are less accomodating when it comes to representing finite line segments and the interiors of k-spheres. Informally, they embody boundaries rather than the regions bound. 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 Euclidean 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-horoblades 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 indicating 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 "imaginary" 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 providing the squared radius of a hypersphere tangent to both. When the meet 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 nonintersecting pointsets.

We will now substantiate and justify the associations summarised above, a process which will of necessity be somewhat mathematically intensive. The reader should remember that when we speak of associating a particular ÂN% or UN% (k+2)-blade ak+2 with a particular ÂN or UN pointset such as a k-sphere, we should more correctly speak of associating the intersection of ak+2 with the horosphere since ak+2 actually represents a set of hyperspheres in ÂN or UN.

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 = 2ab 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+e0e¥ = 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.

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¥ lies in the horosphere (ie. 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 just enough e from y to make 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 .

Thus we define the centre of UN% 1-blade s by Ú(s)   =   s[e0] + ½s[e0]2e¥ where s[e0] º (e0¿s)-1 s when (e0¿s) ¹ 0 , and e¥ else is s reccaled for unit e0 coordinate when possible.
Unlike the disembedded centre ¯I(s[e0])

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 trivially consists of dividing the UN component by the e0 coordinate, independant of the e¥ coordinate.

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 = cr2e¥ = 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¿(cr2e¥) = -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, (e0e¥)*=-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¥  .]

The UN centre c can be recovered from s as ¯I((e0¿_ve2(s))-1s)   =   (e0¿_ve2(s))-1 (s¿I)I-1 corresponding to recaling for unit e0 coordinate and then discarding the e0 and e¥ (or e+ and e-) coordinates.

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.

The Power Distance

If s1* and s2* represent hyperspheres then  s1¿s2   =   ½(r12+r22 - (c1-c2)2 ) provides a natural metric on spherespace sometimes refered to outside the GA literature (without the ½ factor) as the power distance [ Amenta ]. This apparently disappointing inner product (we might instead have wished for the closest distance between the spheres) has a definite geometric interpretation and has significant utility. If a is a point common to both hyperspheres 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.

If two sepeperate real hyperspheres do not interesect with |c1-c2| > r1+r2 then s1¿s2 is -½ times the square of the length of a tangent to s2* from the meet of a tangent from c2 to s1* with s1*. (c1-c2)2 - r12 - r22 is the average of the squared common external and internal tangent lengths (c1-c2)2 - (r1-r2)2 and (c1-c2)2 - (r1+r2)2 .

(N-1)-spheres s1* and s2* are considered orthogonal if s1¿|s2 = 0 , ie. Û r12+r22 = (c1-c2)2 . For real spheres in a Euclidean space with postive r12 and r22, this coincides with the spheres intersecting at right angles;
When r12 < 0, for othogonality we require r22 ³ |r12| and c1 placed inside s2* such that the intersection of real sphere s1'* centered at c1 with radius |r12|½ is a great circle of s1'* .
When r22 < |r12| and r12 <  0 then hypersheres s1* and s2* in ÂN cannot be orthogonal.

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((r1+r2)2-(c1-c2)2)((r1-r2)2-(c1-c2)2)   =   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  .]

When c1=a+b , c2=a-b, this is  (a+b)Ù(a-b) - 2e0b + ½e¥ ( ((a+b)2-r12)(a-b) - ((a-b)2-r22)(a+b) ) + ½e¥0( (a+b)2-(a-b)2+r22-r12) and when r1 = r2 = 0 we have the 0-sphere   2(bÙa - e0b + ½e¥(aba -b3) + e¥0(a¿b) ) representing the bipoint a±b .
[ Proof : (a+b)Ù(a-b) + 2e0b + ½e¥ ( (a+b)2(a-b) - (a+b)(a-b)2 ) + ½e¥0( (a+b)2-(a-b)2)   =   2bÙa + 2e0b + ½e¥( ( (a+b)(a+b-(a-b))(a-b) + ½e¥0( 4 a¿b)   =   2(bÙa - e0b + ½e¥(a+b)b(a-b) + e¥0(a¿b) )   =   2(bÙa - e0b + ½e¥(aba -b3) + e¥0(a¿b) )  .]

When s1=x and 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 whether s1* intersects s2*. Negative implies an intersection while zero implies tangential 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).
ls1 + ms2 is dual to a sphere of center (_lamba+m)-1(ls1+ms2) and squadius (l+m)-2( ls1 + ms2)2 = (l+m)-2(l2r12 + m2r22 +lm(r12+r22 - (c1-c2)2 ) ) ..
In particular ½(p1+p2) has square -¼(c1-c2)2 and so is dual to a sphere passing through the two points centered on their midpoint. .

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,s..,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(a0a02e¥+e0)Ùa1Ù...Ùak = (e¥Ùa0+e¥0)Ùa1Ù...Ùak = (e¥a0+e¥0)Ù(a1a12e¥+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¥0e¥(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+e0a2e¥ . 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¿Ak+2 = e0¿(e¥¿Ak+2). To retrieve an a0 we form
s = ak¿(e¥¿Ak+2) parallel to ¯e¥¿Ak+2(e0) and rescale so that e0¿s=1. We then have null a0 = s + ½s2e¥ with a0ÙAk+2=0.

We can also represent ak+2 as a translated k-plane through 0 [  See next chapter ] as
ak+2 = (1+½e¥a0)(e¥Ùe0Ùak)(1-½e¥a0)   =   (1+½e¥a0)e¥0ak(1-½e¥a0)   =   (1+½e¥(a0+e0))e¥0ak(1-½e¥(a0+e0)) .

Since the UN hyperplane through point a with normal n is represented by UN% hyperblade e¥Ù(a+e0)Ù(ni-1) , the hyperplane of points equidistant from points a and b is represented by UN% hyperblade (a-b)* where * denotes the GHC extended dual.
[ Proof : e¥Ù(½(a+b)+e0)Ù((a-b)i-1) = e¥Ù((½(a+b)¿(a-b))i-1 + e0(a-b)i-1) = ½(a2-b2)e¥i-1 + e¥0(a-b)i-1 = (½(a2-b2)e¥e¥0i-1 + (a-b)e¥0i-1
= (½(a2-b2)e¥ + a-b)(e¥0i)-1 = (a-b)(e¥0i)-1 . Verification: xÙ((a-b)*)=0 Û x¿a=x¿b Û (x-a)2=(x-b)2  .]

k-spheres
An N-D k-sphere is the intersection of a hypersphere and a (k+1)-plane in N-D. [   A strong case can be made for refering to this as a (k+1)-sphere rather than a k-sphere, as does much topology literature, but the term 2-sphere is so uniformly used in the geometry literature to represent a spherical surface in 3D that we retain the terminology here. We will use the terms "sphere" and "hypersphere" to mean an (N-1)-sphere in the N-D space currently under discussion, favouring "hypersphere" except when N=3.   ]

• A (-1)-sphere is a single point represented by the null 1-blade x=x+e0e¥x2.
• A 0-sphere or bipoint is the intersection of an N-D hypersphere with a 1-plane (line) through its centre, ie. two points {a,b} a distance 2R apart. It is represented by the 2-blade aÙb = aÙ(d + (d2+2d¿a)e¥) when d=b-a or by
(c+d+e0+½(c+d)2e¥)Ù(c-d+e0+½(c-d)2e¥) = (c+d+½(d2+2c¿d) Ù(c-d+½(d2-2c¿d) = -2(cd2e¥)Ù(d+(c¿d)e¥) when a=c+d, b=c-d .
• A 1-sphere or circle is the intersection of an N-D hypersphere with a 2-plane (plane) containing its centre. ie. a circle of "interior area" content pR2 and "length" content  2pR; represented by the 3-blade aÙbÙc where a,b,c are any three distinct points on the circle.
• A 2-sphere or sphere is the intersection of an N-D hypersphere with a 3-plane containing its centre, ie. a sphere (globe) of interior "volume" content 2 ò0Rp(R2-t2) dt = 4/3 pR3 and "surface area" content 4pR2 ; represented by a 4-blade.
• A 3-sphere is a 4-D hypersphere having interior "four-volume" content ½ p2 R4 and "boundary volume" content 2 p2 R3 ; represented by a 5-blade.
[ Proof :  2 ò0R4/3 p(R2-t2)3/2  dt = 8/3 p[ ¼(t(R2-t2)3/2 +3/2 R2t (R2-t2)½ + 3/2 R4 sin-1(t/R)  )]0R = 2/3 p[ 3/2 R4 sin-1(t/R)  )]0R = ½ p2 R4  .]
• A 4-sphere exists in 5-D or higher and has interior content 2 ò0R ½ p2 (R2-t2)4/2  dt =p2 ò0R(R2-t2)2  dt = 8/15 p2R5 and content 8/3 p2 R4. It is represented by a 6-blade in the extended space.

When we speak of a point being "within" a k-sphere, we mean that it lies in the "surface". A circle divides its plane into "inside" and "outside" regions containing c and e¥ respectively. More generally a k-sphere divides its (k+1) dimensional tangent space into two regions, or four in the case of a tsphere in a Minkowski space.

We will denote the "solid interior" of a k-sphere a by a·, although we do not yet have a multivector representation of it.

We will refer to the outter product of k UN% embedded point 1-vectors as a k-pointblade.

When we say a particular UN% (k+2)-blade ak+2 corresponds to a pointset A such as a k-sphere in UN what we actually mean is that the meet of ak+2 and the UN% horosphere { x : x2=0 } maps 1-1 (apart from scale) to the embedded set ¦(A). Furthermore, a UN% (k+2)-blade actually represents a set of hypersphere duals, each UN% point l(c+e0+½(c2±r2)e¥) within ak+2 corresponding dually to a UN hyper(anti)sphere of squadius ±r2. The horosphere ponts are dual to zero radius hyperspheres, ie. to points for Euclidean UN, nullcones for Minkowski.

The (k+2)-blade representation ak+2 of a k-sphere of centre c, radius r and (k+1)-plane  through c parallel to tangent (k+1)-blade ak+1=(a1-a0)Ù(a2-a0)Ù...(ak+1-a0) has four alternatively covenient representational forms:

1. The outer product  ak+2 = a0Ùa1Ù...Ùak+1 of the ÂN+1,1 embeddings of k+2 linearly independant points a0,a1,..,ak+1 on the k-sphere, the choice of points effecting only the scale and orientation of the (k+2)-horoblade .
ak+22 = (-1)kr2ak+12 = (-1)½(k+1)k(r(k+1)! Vk+1)2 where ak+1=(a1-a0)Ù(a2-a0)Ù..(ak+1-a0) , r is the radius of the k-sphere , and   Vk+1 is the content of the (k+1)-simplex.
For example consider (k+2)-blade   representing k-sphere centre c through a0=c+re1 with tangent (k+1)-blade e12..(k+1) (c+re1+e0+½(c+re1)2e¥) Ù(c-re1+e0+½(c-re1)2e¥) Ù(c+re2+e0+½(c+re2)2e¥) Ù... Ù(c+rek+1+e0+½(c+rek+1)2e¥)
=   -2rk+1 ((a0+e0)Ù e1 Ù(e2+½(c¿(e2-e1))e¥) Ù... Ù(ek+1+½(c¿(ek+1-e1))e¥) + (c¿e1) (a0+e0)Ù e¥ Ù(e2-e1) Ù... Ù(ek+1-e1) + ½a02e¥e12..(k+1) )
Choosing e1 perpendicular to c gives -2rk+1 ((a0+e0)Ù e1 Ù(e2+½(c¿e2)e¥) Ù... Ù(ek+1+½(c¿ek+1)e¥) + ½a02e¥e12..(k+1) )

ak+2 splits naturally into unextended 1, e0, e¥, and e¥0 "factors". Observe that the scaled tangent(k+1)-blade is available as the e0 factor and having obtained a unit tangent blade, it can be contracted with the (k+2)-blade 1 factor of ak+2 to recover the (same scaled) a0.

2. The meet of N-k hyperspheres s1* Ç s2* Ç  ... sN-k* . Any frame for the (N-k)-blade s1Ùs2Ù ... ÙsN-k representing the space of hyperspheres containing ak+2 (of which s* is the r minimal) will suffice, the choice effecting only sign and scale.
3. The meet of the hypersphere and (k+1)-plane defining the k-sphere
ak+2 = i2 s¿(e¥ÙsÙak+1)   =   i2 s(e¥ÙsÙak+1)   =   i2 s(e¥Ù(c+e0)Ùak+1)   =   i2( sÙ(s¿(e¥Ùak+1)) - r2e¥Ùak+1 )
This is a particularly fundamental form for ak+2 because s and ak+1 are unambiguously specified (apart from scale) whereas the meet and outter product factored forms are not unique.
[ Proof : The  meet is   s* Ç (e¥ÙcÙak+1)   =   s* Ç (e¥ÙsÙak+1) where s=cr2e¥. Since the join s* È (e¥ÙsÙak+1)   is the full pseudoscalar e¥0i we have ak+2   =   (s* *)¿(e¥ÙsÙak+1)   =   i2 s¿(e¥ÙsÙak+1)   =   i2 s(e¥ÙsÙak+1)  .]
4. The dual form   ak+2 = -s(c¿(e¥ak+1*))* = -s(s¿(e¥ak+1*))*     where unextended dual ak+1*=ak+1i-1 lies withing UN.
[ Proof : ak+2-* = (i2 s(e¥ÙsÙak+1))-* = (s(e¥ÙsÙak+1))* = s((e¥ÙsÙak+1)*) = -s((cÙe¥Ùak+1)*) = -s(c¿((e¥Ùak+1)*)) = -s(c¿(e¥ak+1*))  .]

To recover the enclosing hypersphere (aka. surround) s* (and hence r2 and c) and the tangency ak+1 (independant of c) of a given (k+2)-horoblade ak+2=la0Ùa1Ù...Ùak+1 , we can recover s as the dual of ak+2 in e¥Ùak+2 ie. s = -ak+2¿((e¥Ùak+2)-1) =  ^ak+2(e¥))-1 ( ie. the dual to ak+2 in e¥Ùak+2) . We can neglect the inversion and calculate s = ak+2¿(e¥Ùak+2)   =   ak+2(e¥Ùak+2)     provided we then rescale so that e0¿s=1 . The squared radius r2 is then available as s2; or we can compute it directly as r2=(-1)kak+22(e¥Ùak+2)-2 .
Note that s = ak+2(e¥Ùak+2)-1 = (-1)(k+1)(N+1)(¯(ak+2*)(e¥)-1)*
[ Proof : ak+2(e¥Ùak+2)-1 = ((e¥Ùak+2)ak+2-1)-1 = ((e¥¿ak+2*))-*ak+2-1)-1 = (-1)k(N+1)((e¥¿(ak+2*))ak+2-1-*)-1 = (-1)k(N+1)(¯(ak+2*)(e¥)-*)-1 = (-1)k(N+1)i-1¯(ak+2*)(e¥)-1 = (-1)(k+1)(N+1)¯(ak+2*)(e¥)-1*  .]

Thus we can define the centre of a (k+2)-blade as the centre of the 1-blade dual surround
Ú(ak+2) º Ú(_Akp¿(e¥Ùak+2)), or e¥ if e¥Ùak+2=0.

If s2=r2<0 for Euclidean UN then we interpret the blade as representing an "imaginary" UN pointset.
Once we have s then provided r¹0 we can form ak+1 =e¥0¿(sak+2) with ak+2=2r-2 s¿(e¥ÙsÙak+1) because sak+2 = i2 s2 (e¥ÙsÙak+1) = i2 r2 e¥Ù(c+e0)Ùak+1 .

Note that ak+2e¥ak+2 ¹ ak+2¿(e¥Ùak+2) but is rather a multiple of c , residing entirely within ÂN+1.

If ak+2 is a k-plane rather than a k-sphere we will have e¥Ùak+2=0 and can recover the tangent k-blade and point a0 withing the k-plane from ak+2 as previously described.

If ak+2 is not a pointblade, then s = ak+2¿(e¥Ùak+2) may have e0¿s=0 and so not represent a hypersphere dual. If r2=0 (such as for a nullcone) then bk+1 cannot be recovered in this way.

When s=-e+ for the origin-centred unit hypersphere we have ak+2   =   -i2 e+e¥0ak+1   =   -i2 e-ak+1   =   -i2 e-Ùak+1 .

Contents of k-spheres

It can be inductively shown that a (2k-1)-sphere has interior "volume" content |S2k-1,R|·   =   R2k pk (k!)-1   =   R2k pk (G(k+1))-1 while a 2k-sphere has content |S2k,R|· = (2R)2k+1pk k! ( (2k+1)!)-1   =   R2k+1pk (G(k+1+½))-1 .
The "surface area" boundary content of a k-sphere, is (k+1)/R times the "volume" interior content.
Thus a 5-sphere (existing in 6 or more dimensions) has content R6 p3/6 and boundary R5 p3 ; a 6-sphere has content (2R)7 p3 3!/7! = R7 16p3/105 and boundary R6 16p3/15 ; a 7-sphere has content   R8 p4 / 4! and boundary R7 p4 / 3 ; and so forth.

We can unify the content expressions for odd and even dimensions by recourse to the conventional scalar gamma function ; a k-sphere having interior content
|Sk,R|·   =   Rk+1p½(k+1) G(½(k+3))-1   =   (R2p)½(k+1) G(½(k+3))-1     ; and "surface" content
|Sk,R|   =   (K+1) (R2p)½(k-1)p G(½(k+3))-1   =   2 (R2p)½(k-1)p G(½(k+1))-1 .

This prompts us to define real scalar constant ok º |Sk-1,1| = 2p½k (Gk))-1     as the surface area of a unit (k-1)-sphere [ HS 7-4.12 ].
oN is thus the boundary content of the N-D hypersphere and we have ok -1 = ½pk Gk) . In particular o1-1 = ½ ; o2-1 = ½p-1 ; o3-1 = ¼p-1 ; o4-1 = ½ p-2 ; o5-1 = 3/8 p-2 ; o6-1 = p-3 ; o7-1 = 15/16 p-3 .

k-antispheres

If the j-2 and (k-2)-spheres represented by ÂN+1,1 j and k-blades aj and bk where j£k intersect it will be in a l-sphere for some l£j represented by the meet ajÇbk . If they do not inertsect then the meet may not vanish, but instead represent a l-antisphere, ie. a l-sphere of imaginary radius and so negative real squared-radius.
So, for example, in N=3 the meet of the 4-blades representing hyperspheres of radius ½ centred at points e1 and e2 is a 4-blade representing a 2-antisphere of squared radius ¼ centred at ½(e1+e2) . This 4-blade is dual to a 1-blade rpresenting the 0-sphere bipoint consisting of the two closest points on the hyperspheres.
If we replace the e2 centred hypersphere with the 0-plane e¥Ù(e0+e1) through 0 and e1 then the meet is the bipoint { ½e1, 1½e1} as expected but if we replace it with e¥Ù(e0+e2) through 0 and e2 that does not intersect Oe1then the meet is the 0-antisphere centre 0 tangent e2 of radius ½3½ which can be regharded as the bipoint 0± ½3½ie2 both of which are squared distance ¼ from e1. This is dual to the 1-sphere centre 0 of radius ½3½ , axis vector e2, and tangent e13 .

e--negation

If we flip the e- coordinate of s = cr2e¥ = c + e0 +½(c2-r2)e¥ = c + ½(1+c2-r2)e-   + ½(-1+c2-r2)e+ we obtain
s[e-] = -(e-*)s(e-*)-1 = c - ½(1+c2-r2)e- + ½(-1+c2-r2)e+   =   c - ½(1+c2-r2)(e0e¥) + ½(-1+c2-r2)(-e0e¥)
=   c - (c2-r2)e0 - ½e¥   =   - (c2-r2)-1 (-(c2-r2)-1c + e0 + ½e¥(c2-r2)-1)

-(c2-r2)-1c + e0 + ½e¥(c2-r2) with square r2(c2-r2)2 represents a sphere of centre c'=-(c2-r2)-1c (so c'2 = (c2-r2)-2c2 ) and squared radius r2(c2-r2)2 ; when r=0 we have -c-2 times the horoembedding of -c-2c which is reflection (aka. inversion) in the unit hypersphere combined with negation (reflection in e0).

Note that e0[e-] = -½e¥ and e¥[e-]=-2e0.

s[e-])s   =   c2 + (½(1+c2-r2))2 + (½(-1+c2-r2))2 + (c + ½(-1+c2-r2)e+)Ùe- with positive scalar part ³½.

s[+-]s = (c-e0-½(c2-r2)) (c+e0+½(c2-r2)) =  c2 - (e0+½(c2-r2)e¥)2 = 2c2 - r2 vanishes only for s=e0 or

e+-negation

Flipping the e+ coordinate of s we obtain
s[e+] = -(e+*)s(e+*)-1 = c + ½(1+c2-r2)e- - ½(-1+c2-r2)e+   =   c + ½(1+c2-r2)(e0e¥) - ½(-1+c2-r2)(-e0e¥)
=   c + (c2-r2)e0 + ½e¥   =   (c2-r2)((c2-r2)-1c + e0 + ½(c2-r2)-1e¥   ) with square r2 corresponding dually to a (c2-r2) weighted hypershere centre  (c2-r2)-1c and squared radius r2(c2-r2)-2 . When r=0 this c-2 times the horoembedding of c-2c corresponding to reflection (ie. inversion) in the unit hypersphere.
s[e+]¿s = (c + ½(1+c2-r2)e-)2 - (½(-1+c2-r2)e+)2   =   c2 - ½(1+(c2-r2)2)   =   -½ + c2 - ½(c2-r2)2 which vanishes only for hyperspheres of radius ½½ containing 0.
s[e+]Ùs = 2 (c + ½(1+c2-r2)e-)Ù(½(-1+c2-r2)e+   =   (-1+c2-r2)cÙe+  - ½(1+c2-r2)(-1+c2-r2)e¥0   =   (-1+c2-r2)cÙe+  - ½((c2-r2)2-1)e¥0 .
Hyperspheres containing 0 have
s[e+]s   =   -½ + c2 - cÙe+ .

If we instead eliminate the e+ component we obtain ¯e+*(s)   = c + ½(1+c2-r2)e-    = c + ½(1+c2-r2)(e0e¥ )    = ½(1+c2-r2)( 2(1+c2-r2)-1c+e0e¥) with square c2 - ¼(1+c2-r2)2    which is a ½(1+c2-r2) weighted hypersphere of centre 2(1+c2-r2)-1c and squared radius (c2 - ¼(1+c2-r2)2) (½(1+c2-r2))-2 = 4c2(1+c2-r2))-2 - 1 .

e0[e+] = ½e¥; e¥[e+] = 2e0

Example Geometric Manipulations
Let us define the squeperation (short for squared seperation) of a point from a hypersphere to be the square of the shortest distance from the point to the sphere surface. We can think of this as the "squared height" of points "outside" the hypershere, zero for points "within" (ie. "on") the sphere, and the "squared depth" of points "inside" it.

Leting ei º ei+e0e¥ = ei+e- for i=1,2,..N denote N of the 2N embedded corners of a particular unit UN hypercube we have

• e1Èe2 = e1Ùe2 = (e1-e2)e- + e12 with (e1Ùe2)2=1 represents the 0-sphere (bipoint) {e1,e2}
• e1Èe2Èe3  = e1Ùe2Ùe3  = e123 + (e12-e13+e23)e- with (e1Ùe2Ùe3)2=2 represents the unit 1-sphere centre 1/3(e1+e2+e3) radius (2/3)½ through e1,e2 and e3.
• e1Èe2Èe3Èe0 = e1Ùe2Ùe3Ùe0 = ½(e¥0(e12-e13+e23)-e0e123) represents the 2-sphere with centre ½(e1+e2+e3)of radius 3½ passing though e1,e2,e3 and 0
• e1Èe2Èe3Èe4  = e1Ùe2Ùe3Ùe4 = e1234 - e123e+ + (e134-e234-e124)e-  + (e12-e13+e23)e¥0 . If e42=1 we have (e1Ùe2Ùe3Ùe4)2=-3 and the join represents the 3-sphere through e1,e2,e3, and e4  centred at ¼(e1+e2+e3+e4) . If e42=0 the join is null and represnts the null cone Le4 centred on point e4 . Encorporating the negative signature thus zeroes the join radius and centres it on e4; and the join remains so and null even if we bring in further orthogonal plussquare ei.

Now consider le1+me2 = (l+m)( (l+m)-1(le1+me2) + e0 + ½e¥) = (l+m)( ((l+m)-1(le1+me2))' + e0  + ½e¥(1- (l+m)-2(le1+me2)2)) which for l+m¹0 is dual to the (N-1)-sphere of centre (l+m)-1(le1+me2) and radius (l+m)-1 (l2+m2)½ .
Now (e1-(l+m)-1(le1+me2))2   =   (l+m)-2 ( m(e1-e2))2   =   2 m2(l+m)-2 so |e1-c|=2½|m(l+m)-1| which differs from the radius by (l+m)-1((l2+m2)½-m and we recognise le1+me2 as dual to the hypersphere having centre (l+m)-1(le1+me2) and radius chosen so that the sum of the squared distances of e1 and e2 from the hyperspherical surface is (e1-e2)2 . We can regard this as the hypersphere moving and expanding to keep the "combined squeperation" from points e1 and e2 constant.

Pencils
Consider a ÂN+1,1 2-blade s2 = s1Ùs2 where s1 and s2 represent hyperspheres in ÂN.
The hyperspheres intersect in an (N-2)-sphere s2* = s1* Ç s2* Û s22 < 0. Since e¥Ùs2 ¹ 0 the 1-vector s = s2(e¥Ùs2)-1 = ¯s2(e¥)-1 represents a sphere having the same centre and radius as the intersection.
s2* represents an intersective pencil of all spheres containing s1* Ç s2* as well as the hyperplane containing it. Further, if a is any point in the intersective (N-2)-sphere s1*s2 = (a-c1)*(a-c2) = ½(r1r2)-1(r12+r22-|c1-c2|2) [   where a*b º (a¿b) (|a||b|)-1 is the inversive product ] . The geometric product of the 1-vectors representing two intersecting hyperspheres is thus a 2-blade representing all hyperspheres sharing that intersection and a scalar having value the product of the radii and the scalar cosine of the angle subtended by the hypersphere centres from any point on the intersection.

s=l1s1+l2s2 = (l1+l2) (l1(l1+l2)-1c1 +(l2(l1+l2)-1c2 + e0e¥(...) has
s2 = l12r12 +l22r22 + l1l2(r12+r22-(c1-c2)2) = (l1+l2)(l1r12 +l2r22) - l1l2(c1-c2)2 so the weighted sum of two spheres is another sphere.
When l1+l2=1 for colinearity with s1 and s2 we have s = (l1c1+l2c2 + e0e¥(...) with s2 = (r22 +l1(r12 -r22) - l1(1-l1)(c1-c2)2 .

Setting r1=r2=0 we can now interpret the wieghted sum of non null-seperated points as dual to a hypersphere of centre l1c1+l2c2 and squared radius -l1l2(c1-c2)2 . Thus a-b is dual to a hypersphere of centre a-b and radius (a-b)2 for spacelike seperated a,b.
Spheres s1 and s2 are nonintersecting iff s22 > 0 in which case s2 contains two noncolinear null vectors which represent two particular ÂN points. s2* represents a Poncelet pencil of all spheres and hyperplames with respect to which these two points are inversive.
The spheres are tangent iff s22 = 0. The contact point is then represented by ^s1(s2) and s2* represents a tangent pencil of all spheres tangent to this point.

Thus if c1¹c2 then (s1Ùs2)* represents all hyperspheres that intersect identically with s1 as does s2, while if c1=c2=c but r1¹r2 then s1Ùs2 = ½(r22-r12)e¥Ù(c+e0) and (s1Ùs2)* represents all hyperspheres with centre c.

Suppose now that s2 represents a hyperplane n2 + d2e¥.
Sphere s1 intersects the hyperplane iff s22 < 0 as above. The interesction has the same centre and radius as sphere ^s2(s1)*.
s2* = (s1Ùs2)* represents a concurrent pencil of spheres as above. Further, if a is any point in the intersective (N-2)-sphere
s1*s2 = -(a-c1)*n2 = (c1.n2-d2)/r1 .
The sphere and hyperplane are seperate iff s22 > 0 in which case s2* represents a Poncelet pencil as above.
The sphere and hyperplane are tangent (at ^(s1,s2)) iff s22 = 0 in which case s2* represents a tangent pencil as above.

If both s1 and s2 are hyperplanes then once again they intersect iff s22 < 0. If s1 and s2 both contain 0 then their intersection is the (N-2)-space s2i, otherwise it is (e0¿s2)* reprenting an (N-2)-plane (a line if N=3) having the same normal and distance from 0 as ^(e0,s2)*.
s2* represents a concurrent pencil of hyperplanes containing the (N-2)-plane. Further, s1*s2 = n1*n2 .
Otherwise s22 = 0 and s1,s2 are parallel a distance |e0¿^(s1~,s2)| apart.
s2* represents a parallel pencil of hyperplanes normal to ^(s1,s2).

We can generalise pencils as 2-bunches where a k-bunch is represented by (the dual of) a k-blade s1Ùs2Ù...Ùsk where s1,s2,...,sk are representors of hyperspheres or hyperplanes. Those interested should consult Li et al. A key result is that if the k hyperspheres s1,s2,..sk intersect in an (N-k)-sphere s1*Çs2*Ç...sk*  then (s1Ùs2Ù..sk)* is an intersecting k-bunch representing all m-spheres and m-planes containing s1*Çs2*Ç...sk* where m³N-k.
In particular a point a satisfies aÙ(s*)=0 for any hypersphere s*  containing a in its surface, so a can be regarded as a 1-bunch m-representing the set of (N-m)-spheres and (N-m)-planes containing a.

We now summarise the GHC geometric interpretations of particular ÂN+1,1 or UN% pure blades.
Let bk be a pure k-blade in ÂN ; null 1-vectors a0,a1,... be embedded points ; and 1-vectors s1,s2,... represent (duals of) hyperspheres.
 Blade Grade Square Horosphere intersection represents e0 1 0 Origin point 0 (embedded) e¥ 1 0 Infinity point ¥ e¥0 2 1 Origin point {0} (0-plane) a+e0+½a2e¥ 1 0 Point a   (embedded) e¥Ù(a+e0) 2 1 Point {a}   (0-plane) e¥Ùa 2 0 Direction a e+* N+1 (-1)N+1i2 Unit hypersphere (squared radius=1) centre 0 e-* N+1 (-1)Ni2 Unit antihypersphere (squared radius=-1) centre 0 e0* N+1 0 Null (ie. zero radius) hypersphere centre 0 e¥* N+1 0 No horosphere intersection e¥0bk k+2 bk2 k-plane through 0 with tangent bk (c+e0+½(c2-r2)e¥)* N+1 (-1)N+1i2r2 Hypersphere centre c squared radius r2 a0Ùa1...Ùak k+1 (k!-1 r Volume(a0,a1,...,ak))2 (k-1)-sphere through a0,a1,...and ak (e0+a0)Ùbk k+1 (-1)k(^bk(a0))2bk2 (k-1)-sphere through a0 with tangent bk e-Ùbk k+1 (-1)kbk2 Unit (k-1)-sphere with centre 0 and tangent bk e¥Ù(e0+a0)Ùbk k+2 bk2 k-plane through a0 with tangent bk (a1-a0)* N+1 (a1-a0)2 Bisecting hyperplane between a0 and a1 (s1Ùs2Ù...Ùsk)* N+2-k ? k-bunch of spheres and planes containing s1*Çs2*...Çsk*.

Each such blade b has a variety of "duals". We have the unextended dual * in i which sends e- and e+ to (N+1)-vectors and the ei to (N-1)-vectors; and the extended dual * in i = ie¥0 . For example consider 3-blade b= (e0+8e¥)e12 = ½(8e-+7e+)e12  representing the 1-sphere of radius 4, centre 0 and tangent 2-plane e12 . When N=5, its unextended dual is (e0+8e¥)e345 representing the 2-sphere of radius 4, centre 0, and tangent 3-plane e345 whereas the extended dual is b* =  (e0-8e¥)e345 representing a 2-antisphere of radius -4, centre 0 and tangent volume e345. In Euclidean Â5 this is an empty pointset but if e42=-1 (for example) then b does contain points in the GHC horosphere such as the embedding of Â4,1 point e3 + 15½e4.

In an GHC extended Euclidean space, the extended dual of a k-sphere can be regarded as an "imaginary (N-k)-sphere". Imaginary in that it has negative squared radius and so does not contain any representors of points in ÂN. We can then more safely consider the extended dual b* as representing the same geometric construct as b, considering the multiplication by imaginary i=e¥0i as being an "irrelevant scaling" of b much as we consider multiplication of b by a scalar value a to be irrelevant. Encountering a nonnull 1-vector s, for example, we might consider it to represent the (N-1)-sphere more properly represented by (N+1)-blade s*.
If N is odd so that i is central, we can "normalise away" a complex rather than a real "scale" for a nonnull blade b. All that is required is a reversing conjugation ^ that preserves k-blade b while negating (N+2)-blade i and we have (ab)~ º |(ab)^(ab)| ab.
However in a nonEuclidean space, the dual may represent a distinct "actual" or "real"  geometric pointset.

Point Versors
Conisder the point k-versor p1p2p3...pk where the pi is the GHC embeddings pi = pi + e0 + ½pi2e¥ .
For k=2 we have p1¿p2 + p1Ùp2 which comprises the squared seperation distance and the 0-sphere bipoint { p1,p2 } .
For k=3 we have p1(p2 ¿ p3) + p1¿(p2Ùp3) + p1Ùp2Ùp3 comprising the 3-blade 1-sphere through the points; and the 1-vector S = -½(a2p1 - b2p2 + c2p3 ) with S2 = ¼a2b2c2 , where a2 º (p2-p3)2 ; b2 = (p1-p3)2; c2=(p1-p2)2 .,
The e0 coordinate -½(a2-b2+c2) of S is zero when the pi form a right angled triangle with p2 opposing the hypotenuse; but is otherwise positive in a Euclidean space with s = -2(a2-b2+c2)-1S = (a2-b2+c2)-1(a2p1-b2p2+c2p3) then dual to a hypersphere with center (a2-b2+c2)-1(a2p1-b2p2+c2p3) and squadius a2b2c2(a2-b2+c2)-2 ;   passing through the endpoints p1 and p3 and a squared distance ??? from p2.. .

[ Proof :  S = p1(p2 ¿ p3) + p1¿(p2Ùp3)   =   p1(p2 ¿ p3) + (p1¿p2)p3 - p2(p1¿p3)   =   -½(p1a2 - p2b2 + p3c2) º -½(a2-b2+c2)s
S2 = (p1(p2 ¿ p3) - p2(p1¿p3) + p3(p1¿p2) )2   =   2( -(p1¿p2)(p2 ¿ p3)(p1¿p3) +(p1¿p3)(p2 ¿ p3)(p1¿p2) - (p2¿p3)(p1¿p3)(p1¿p2) )   =   -2(p1¿p2)(p2 ¿ p3)(p1¿p3)   =   ¼a2b2c2 Þ s2 = (a2-b2+c2)-2a2b2c2 . .
Now p1¿S   =   p3¿S = 0 while p2¿S   =   2(p1¿p2)(p2 ¿ p3)   =   ½a2c2 hence p2¿s = -(a2-b2+c2)-1a2c2 so
(p2-c)2 = r2 - 2(p2¿s)   =   r2 + 2(a2-b2+c2)-1a2c2   =   a2c2(a2-b2+c2)-2(b2+2(a2-b2+c2))   =   a2c2(a2-b2+c2)-2(2a2-b2+2c2)) from which we deduce that s is dual to a hyperplane or hypersphere that passes through p1 and p3 and within squared seperation a2c2 (a2-b2+c2)-1 of p2 .  .]

For example p1 = 2e1 + e0 - 2e¥ ; p2=e0 ; p3 = e2+e0e¥ with a2=4; b2=5; c2 = 1; and p2Ùp3 = _e02 + ½ e¥0 genretes the circle centred at e1e2 with squadius 2-25 and tangent e12 .

Convergent Point Projection

The point x = e0 + xe1 + ze3 + ½x2e¥ = xe1 + ze3 + ½(x2+1)e- + ½(x2-1)e+ where x=xe1+ze3 inverts in 3-blade e-12 representing the unit radius circle centered at e0 with tangent e12 to e-12xe-12-1   =   xe1 - ze3 + ½(x2+1)e- - ½(x2-1)e+ and projects to e-12 as
¯e-12(x) = ((xe1 + xe3 + ½(x2+1)e- + ½(x2-1)e+)¿e-12)e-12-1 = ((-xe-2 -½(x2+1)e12 )(-e-12) = ((xe-2 + ½(x2+1)e12 )e-12 = -xe1 - ½(x2+1)e- = -xe1 - ½(x2+1)(e0e¥) = -½(x2+1)( (½(x2+1))-1xe1 + e0e¥ ) , dual to an antihypersphere centre (½(x2+z2+1))-1xe1 with squared radius (½(x2+1))-2 x2 - 1 , which is   -(x2+1)-2 (x2 - 1)2 when z=0.
Note that this projection is also given by ½(x + e-12xe-12-1), The dualed contraction projection provides a more efficiemt and robust grade limited computation, but the averaged inversion can be more useful intuitively.
If z=0 and x=1+d with |d| < 1 then d' = x'-1 has |d'| < d2 so the repeatedly projected centre will converge reasonably rapidly to e1
[ Proof :  (½(x2+1))-1x -1 = (½(x2+1))-1 (x-½(x2+1)) = (½((1+d)2+1))-1 (1+d - ½((1+d)2+1)) = -(½((1+d)2+1))-1 ½d2 = -((1+d)2+1)-1 d2  .]

We accordingly define the convergent point projection of x into (k+2)-blade ak+2 ¯Ú(x)ak+2(x) as the limit of xj+1 = ¯Ú(xj, ak+2), ie. we have ¯Ú º (¯Ú)¥ .

Nonflat Embeddings
Our approach for intersecting planes and spheres in UN is to embed k+2 nondegenerate points in a k-sphere into the UN% horosphere HNe¥ with f0(x) º x+e0x2e¥ and then form their (k+2)-blade outterproduct. We intersect such blades using the meet, and we can do this forcing any signatures we like. The actual signatures (or metric) of UN contribute only to the e¥ coordinate of f0(x), and then arise again when considering where and whether the m-blade meet intersects HNe¥.
Suppose instead we embedd with ¦: UN ® UN% defined by ¦(x) = h(f0(x)) where h : UN% ® UN% is an arbitary function. We can extend linear h to an outtermorphism mapping UN% ® UN% and we are particularly interested in h() with h(e¥)=le¥ for nonzero scalar l so that ¦(¥) = le¥ and k-planes map to k-planes.

If (k+2)-horoblade ak+2 represents a k-sphere through UN points a0,a1,..,ak+1 then h(ak+2) is a more general (k+2)-blade containing nonhoro h(a0), h(a1),..h(ak+1) whose intersection with HNe¥ correspnds to a k-curve in UN of a type dependant on h(). More interestingly, if isomorphic h-1 exists we can form h(h-1(e¥Ùbl+1) Ç ak+2) propotionate to h((e¥Ùh-1(bl+1)) Ç ak+2) to obtain a (nonhoro) m-blade whose intersection with HNe¥ represents the intersection of l-plane e¥Ùbl+1 with the particular k-curve of type dictated by h() .

We can also compose h with a horodrop to obtain Deh : UN%®HNe¥ inducing a UN transformation he º f0-1 Dehf0 : UN ®  UN È {¥}.

k-conics
Conics are naturally present in GA as the medial axies of Dupin Cycides. For example: 3-blade e-12 represents the unit radius circle centered at e0 with tangent e12. Rotating s   =   xe1 + ½(1+x2-r2)e- + ½(-1+x2-r2)e+   =   xe1 + ½(1+D)e- + ½(-1+D)e+, where D = x2-r2, in 3D dual e-12* = e+3 we have (½qe+3)§(s)   =   xe1 + ½(1+D)e- + ½(-1+D) cos(q)e+ - ½(-1+D) sin(q)e3 .
A hypershere dual is "normalised" with unit e0 coordinate if the e- coordinate minus the e+ coordinate is unity, so the centre of the rotated hypersphere is c(q)   =   X(q)e1 +Z(q)e3   =   (½(1+D) - ½(-1+D) cos(q))-1 (xe1 - ½(-1+D) sin(q)e3) which for x2 > r2 is an ellipse centre ½xD-1(1+D)e1 eccentricity x-1D½ passing through (½xD-1(1+D) ± ½xD-1(-1+D))e1 at q=0,p and ½xD-1(1+D)e1  ± ½D (-1+D)e3 at q = ± cos-1((1+D)-1(-1+D)) .

The conic transform y = h(x) = Sa,e0(x) º x + ½(a¿x)Ùe0 for UN 1-vector a with inverse
h-1(y) = S-a,e0 =   y - ½(a¿y)e0 has y2 = (a¿x)x¿e0 = -½(a¿x)x2 .
Sa,e0(x) = x + ½(a¿x)e0 is dual to the hypersphere of radius (-½a¿x)½|x| centred at (1+½a¿x)-1 x .
h(x)=x for all x perpendicular to a (including e¥ and e0) and is linear in x , so only h(la) is nontrivial. y = h(f0(la)) = la + e0(1+½la2) + ½l2a2e¥ has y2 = -½l3a4 and e¥-horodrops to
De¥(h(f0(la))) = (1+½la2) f0(la(1+½la2)-1) .
For l<0, y thus represents a hypersphere with c=la(1+½la2)-1 and radius 2l3/2a . For nonnull a we have f0-1 De¥hf0(la) = la~(1±½al)-1 according as a2=±a2.

h(aÙb) =  h(a)Ùh(b) so h maps k-blades to k-blades but while h(e¥Ùa0Ùa1...Ùak+1) = e¥Ùh(a0)Ùh(a1)...Ùh(ak+1)) is proportionate to e¥Ùf0he¥(a0) Ùf0he¥(a1) Ù...Ùf0he¥(ak) so that h maps k-planes representors to k-planes repersentors, it maps k-sphere representors to more general k-curve representors.

UN transformation he¥(x) = (1+½a¿x)-1x with inverse he¥-1(y) = (1-½a¿y)-1y acts as a sort of "directed dilation" with the 0-centred hypersphere of radius r < 2a-2  mapping under he¥ to an ellipsoid that passes through ±r(1±½ar)-1 a and intersects the orginal hypersphere in a (N-2)-sphere of radius r and (N-2)-plane through 0 perpendicular to a. Taking a=ae1 and N=3 we have an ellipsoid having 0 as its rightmost focus and lefthand focus  at -r2a(1-¼a2r2)-1e1 with eccentricity era, passing through -r(1-½ar)-1e1 ; r(1+½ar)-1e1 ; ± re2; and ± re3 with "semi-major axis" r(1-¼a2r2)-1 and b = r(1-¼a2r2) .
Points with a¿x=-2 map to ¥ provoiding hyperbolids and parabaloids.

To intersect such a 0-focussed k-conic with a ray e¥Ùf0(a0)Ùf0(a1) through points a0 and a1 it thus suffices to intersect the k-sphere with the ray e¥Ùf0he¥-1(a0)Ùf0he¥-1(a1) through points he¥-1(a0) and he¥-1(a1) and apply he¥ to the resultant two points.

More generally, given any k+2 points a0,a1,..ak+1 then for arbitary a we have a k-conic passing through a0,a1,..,ak corresponding to the image under Sa,e0 of the k-sphere through the k+2 points S-a,e0(ai) .
a=0 gives the k-sphere through a0,a1,..,ak+1 and while the k-sphere through the he¥-1(ai) has radius < 2 we will have an k-ellipsoid.
 PolarGrid(f0-1 De(x+½a(x¿a)b)) where De is dropping to the horosphere in direction e a=e1 ;b=e0 ; e=e¥    8×8 cells r=½ a=0 8×8 cells r=½ a=½ 8×8 cells r=½ a=1 8×8 cells r=½ a=8 a=e1 ; b=e0 ; e=e+ ; - in root   4×4 cells r=½ a=½ 4×4 cells r=½ a=1 4×4 cells r=½ a=2 b=e0 ; e=e+ ; + in root   2×2 cells r=½ a=¼ 2×2 cells r=½ a=½ 2×2 cells r=½ a=1

Regeneralised Homogeneous Coordinates
We can repeat the GHC embedding, mapping UN% point x to (N+4)-dimensional UN%% point x =x+e0e¥x2 which in particular takes s=x+e0+½(x2-r2)e¥  with s2 = r2 to null 1-vector s x,r   º s + e0 + ½r2e¥   =   x+e0+e0 + ½x2e¥ + ½r2(e¥-e¥) .

s 1 ¿ s 2   =   s1 ¿ s2 + ½(r12+r22)(e0¿e¥)   =   ½(r12+r22-(c1-c2)2) - ½(r12+r22)   =   -½(c1-c2)2 and hence   ( s 1Ù s 2)2   =   ¼(c1-c2)4 independant of r1 and r2 . .
[ Proof :  s1¿s2 + (e0 + ½r12e¥)¿(e0 + ½r22e¥)   =   ½(r12+r2-(c1-c2)2)   + ½r12e¥¿e0+r22e0¿e¥)  .]

s e1 Ù s e2 Ù s -e1 represents a 1-torus formed by sweeping a sphere of radius ½ along the 1-sphere circular path having centre e0 and tangent plane e12. However if we intersect this with another torus we can obtain only spheres in which they intersect (if any) rather than the more desirable set of shared UN points .
If, for example,  we intersect s 1Ù s 2Ù s 3 with the cone s 0,0 Ù s 2 Ù e¥ we retrieve null 1-vector s 2 but if instead we use s 0,0 Ù s c2,d Ù e¥ we obtain a minussquared 1-vector having positive e0,e¥,e0, and e¥ coordinates represnting a 4-antisphere of radius 0.6i in UN% whose "centre" is a UN 2-antisphere centred at 0.8e1 of radius .331i .

Line Segments
If we add a scaled line representation to the representation of a point on the line g = e¥ÙaÙd  + a we obtain a <1;3> grade pointed line (aka. flag) satisfying gg§ = -d2 .
pg is a <0;2;4> multivector with zero 4-vector component when p lies on the line and scalar component p¿a = -½(p-a)2

Consider the noninvertible 3-versor e¥(a+e0)d . Its 3-vector component e¥Ù(a+e0)Ùd embodies the extended line through a and a+d and we can recover d from this (as the "coefficient" of e¥0) but a is ambiguous since any point on the line will give the same 3-blade. The 1-vector component e¥(a¿d) - d resolves the ambiguity in a by specifying a¿d.
However this is not a very useful form because e¥(a+e0)de¥(b+e0) = -de¥(b+e0) independant of a.

The "One Up" Embeddings

Moving off the horosphere
We saw that adding -/+ ½(e0¿x)-1 e¥ to x gives a 1-vector with square ±1 corresponding to moving from a scaled embedded point lx to the (dual of) the unit (anti)hypersphere centre x, but we can also move off the horosphere in directions e+ and e- and this is the basis for the one-up embeddings proposed by Lasenby.

Recalling that x   =   x + e0 + ½x2e¥   =   x + ½e-(x2+1) + ½e+(x2-1) we define
x± º ¦±(x) º   (^e-/+(x))~   =   -2(x2±1)-1x + e-/+   =   -2(x2±1)-1 (x + ½(x2 -/+ 1)e±)     with x±¿e-/+=0 and x±2=±1 ; but with the caveat that we have null x± º x when x2=-/+1 , ie. when x¿e-/+ = 0. Otherwise, we can recover x from x± via
x   =   -½(x2±1)(x± - e-/+)   =    -(e±¿x± ±1)-1(x± - e-/+) .
[ Proof : e±¿x± = -/+ (x2±1)-1(x2 -/+ 1) Û (e±¿x±)(x2±1) = -/+ (x2 -/+ 1) Û x2(e±¿x± ± 1) = 1 -/+ e±¿x±
Þ x2 = (1 -/+ e±¿x±)(e±¿x± ± 1)-1 Þ x2 ± 1 = (e±¿x± ± 1)-1 ((1 -/+ e±¿x±) ± (e±¿x± ± 1)) = 2(e±¿x± ± 1)-1  .]

Lasenby suggests regarding x with x2=1 as boundary points for the x- embedding. Since e0 ± = e± while e¥ ± = -e± we lack a distinct e¥ ± .

a±¿b±   =   ±1 - 2(a2±1)-1(b2±1)-1(a-b)2 provides our inner product.
[ Proof : (-2(a2±1)-1a + e-/+)¿ (-2(b2±1)-1b + e-/+)   =   4(a2±1)-1(b2±1)-1a¿b - e-/+2  .]

k-planes and k-spheres

Let UN% (k+2)-blade ak+2 = a0Ùa1Ù..Ùak+1 represent a k-sphere in UN in the usual way. Provided no aj2=1 we have
ak+2 = l(a0 --e+)Ù(a1 --e+)Ù...Ù(ak+1 --e+) where l = (-½)k+2 (a02-1) (a12-1)... (ak+12-1) and we can expand this as
ak+2 = l(ak+2 - + e+Ùbk+1 -) where (k+2)-blade ak+2 - º a0 -Ùa1 -Ù...Ùak+1 - lies within e+* and (k+1)-vector bk+1 -   =   åi=0k+1 ai - lies within ak+2 - and so either commutes or anticommutes with it according as k is odd or even. Here (k+1)-blade ai - º (-1)i a0 -Ùa1 -Ù...[i]...Ùak+1 - with the [i] denoting ommission of ai - .
ak+2 - and bk+1 - thus provide a (N+1)-D representation of the k-sphere ak+2 with the k+2CN+1 coordinate condition x-Ùbk+1 -=ak+2 - replacing k+3CN+2 coordinate condition xÙak+2=0 as the criteria for x lieing in the k-sphere.

Note that e-Ùak+2 - = l-1 e-Ùa0Ùa1Ù..Ùak+1 and e-Ùbk+1 - = e-Ùbk+1 where bk+1 º åi=0k+1 (Pj¹i(-2(aj2-1)-1)a0Ùa1Ù...[i]..Ùak+1 .

e¥Ù(a0 --e+)Ù(a1 --e+)Ù...Ù(ak --e+) expands proportionate to e¥Ù(ak+1 - + e+Ùbk)   =   e¥Ùak+1 - - e¥0Ùbk so we can also represent k-planes with (N+1)-D vectors rather than (N+2)-D (k+2)-blades.

k-Planes and k-Spheres
la± + (1-l)b± = -2(l(a2±1)-1(a+½(a2 -/+ 1)e±) +(1-l)(b2±1)-1(b+½(b2 -/+ 1)e±) is a scaled multiple of the unit representor for UN point l(a2±1)-1a + (1-l)(b2±1)-1b which does not in general lie on the line through a and b. Thus a±Ùb± actually represents a particular 1-curve through a, b rather than the 1-plane through them. Lasenby asserts that the tangent at x- for x-Ùa-Ùb- = 0 has direction x-¿(a-Ùb-).
For want of better terms, we will capitalise, and refer to the k-curve   { x : x-Ùa0 -Ù..ak - = 0 } as a k-Plane and the (N-1)-curve { x : x-¿d- = -/+½m2 } as a hyper(anti)Sphere of squadius m2 and centre d. A k-Sphere is of course the intersection of a k-Plane with a hyperSphere.
We have ak+2e+ak+2 = 2ak+2 -bk+1 - + (bk+12+(-1)kak+2 -2)e+ .
[ Proof : (ak+2 - + e+bk+1 -)e+(ak+2 - + e+bk+1 -)   =   (-1)kak+2 -2e+ + (-1)k+1bk+1 -ak+2 - + ak+2 -bk+1 - + e+bk+1 -2  .]
Numerical experimentation confirms Lasenby's assertion [6.2] that ¯e+*(ak+2e+ak+2) = 2ak+2 -bk+1 - is a multiple of d-, the centre of the k-Sphere through a0,a1,...,ak+1.

Spherical Conformal Coordinates
As described by Hestenes et al, the intersection of the ÂN+1,1 = ÂN% plane e-¿x=-1 and the horosphere { x : x2=0 } represents the unit hypersphere in Euclidean space ÂN+1, ie. an N-sphere. Thus, for example, the surface of a 3D globe (a 2-sphere) is represented by the e-¿x=-1 plane in Â3,1 .
There are a variety of equivalent distance measures one can adopt between two points a and b on the unit N-sphere (perhaps the most natural is the subtended angle) but we will favour the chord distance (a-b)2 ½   =   (2(1-a¿b))½ even though ÂN+1 1-vector a-b does not strictly speaking exist inside SN  . This ranges from 0 when a=b through 2½ when a¿b=0 to a maximum of 2 when a=-b corresponmding to a single point.
Our embedding ¦: ÂN+1 ® ÂN+1,1 is  ¦(x) = e- + x . Those x mapping to the horosphere have x2 = 1. This coincides over ÂN with our GHC embedding over the unit origin centred hypersphere -e+* of xÎÂN with x2=1 since then x = x + e0 + ½e¥ = x + ½(e--e+) + ½(e-+e+) = x + e-
Within SN, we can define a k-sphere having centre c (with c2=1) as the set { x: x2=1, (x-c)2=R2 } for any 0£r£2 which we can regard from our SN Ì ÂN+1 perspective as the intersection of SN and a (k-1)-plane in ÂN+1. But spheres of radius greater than 2½ can arguably be said to have two centres c and -c with -c the "better" one, so it is natural to restrict attention to k-spheres of radius r £ 2½ and refer to SN k-spheres having radius 2½ < r £ 2 as a k-antisphere.
There are no k-planes within SN but analagous to k-planes are the great k-spheres of radius 2½.
The dual of 1-vector s   =   c + (1-½r2)e-   =   c - ½r2e- with s2 = r2(1-¼r2) corresponds to the (N-1)-sphere of radius 0£r<2½ when s2 > 0 ; to the great (N-1)-sphere with r=2½ when s2 = 0 ; and to the (N-1)-antisphere with 2½<r£2  when s2 < 0 .

Given k+1 points a0, a1,... ak in SN , the (k+1)-blade ak+1 = a0Ùa1Ù...Ùak represents the (k-1)-sphere containing the k+1 points if e-Ùak+1 ¹ 0 or the great (k-1)-sphere containing them if e-Ùak+1 = 0.
Intersection results similar to those for the GHC emebedding follow, and conformal transformations of SN can be represented by multivectors in like manner to our following discussion of Lorentz transformations in GHC. See Hestenes et al for a full treatment.

Tspherical Conformal Coordinates
The intersection of the ÂN,2 = ÂN-1,1%  plane e+¿x=-1 and the horosphere { x : x2=0 } represents the unit hypertsphere in ÂN,1 corresponding to the unit hypertsphere in ÂN-1,2 .
Our embedding ¦: ÂN,1 ® ÂN+1,1 is  ¦(x) = e+ + x . Those x mapping to the horosphere have x2 = -1. This coincides over ÂN-1,1 with our GHC embedding over the unit origin centred hypersphere -e+* of xÎÂN with x2=-1 since then x = x + e0 - ½e¥ = x + e+ .

Soft Geometry
We can now represent particular geometric pointsets like a (k-2)-spheres in ÂN as a pure k-blade bk in ÂN+1,1 with the understanding that x is in the set iff (k+1)-blade xÙbk=0.
Consider the scalar field ¦(x)= (-(xÙbk)4)   =    (-((xÙbk)*(xÙbk))2) . The product (xÙbk)2 is scalar (being the square of a pure (k+1)-blade) and squaring again ensures a nonnegative value. ¦(x)=1 for all x with xÙbk=0 and lies in (0,1) everywhere else, falling rapidly in magnitude for increasing |(xÙbk)2| .
¦ thus "peaks" at 1 over our geometric pointset and falls rapidly to positive nearzero away from it, with the caveat that it is also 1 at x such that xÙbk is null rather than zero. This includes every null x that commutes or anticommutes with bk. We require a measure that is small everywhere "away" from bk and one (frame dependant) way to ensure this is to force Euclidean signatures using  ¦(x)= (-((xÙbk)¿+(xÙbk)2) .

Next : Multivectors as Transformations

Glossary   Contents   Author
Copyright (c) Ian C G Bell 1998, 2014
Web Source: www.iancgbell.clara.net/maths
Latest Edit: 23 Nov 2015. 