Visualizing Conformal Geometric Algebra

Visualizing
Conformal Geometric Algebra
Bas Fagginger Auer
June 30, 2006

Contents
Abstract
i
Introduction
iii
0.1 About the image on the cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
0.2 Used programs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
0.3 Welcome . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Geometric Algebra 1
1.1 Rationale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Our goal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.2 Current model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.3 Extending R n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.1.4 Operations on our extended space . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.1.5 Use of our extended space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.1.6 Embedding R n in our extended space . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3 Implementation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3.1 Data types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3.2 Extended signature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3.3 Order sign . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.3.4 Kronecker delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.3.5 Multivector addition and scalar multiplication . . . . . . . . . . . . . . . . . . . . 13
1.3.6 ·, ∧ and • products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.3.7 Multivector division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Extended Geometric Algebra 17
2.1 Homogeneous geometric algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2 Conformal geometric algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.3 Object classification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.4 Visualizing Euclidian 3-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3

2.5 Visualizing Hyperbolic 3-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3 Conclusion 31
3.1 Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Appendices 33
3.2 Binary numbers and operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3.3 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4

Abstract
This document is intended as an introduction to implementing a geometric algebra interpreter on a
computer and visualizing elements of 3-dimensional conformal geometric algebras with both Euclidian
and Hyperbolic signatures. It provides implementations of the outer, inner and geometric products as
well as inverting a geometric product in arbitrary dimension.
i

Introduction
0.1 About the image on the cover
The image is a so-called xor pattern. It is made by plotting the value ‘x xor y’ at each pixel (x, y) on
the computer screen in 320x200 standard VGA mode using the standard palette (screen 10011). What
this image (or actually, the ‘xor’ operation itself) has to do with geometric algebra shall become clear in
the first chapter of this paper.
0.2 Used programs
The typesetting of this document was done using L A TEXand (almost all) figures were made with the
program Xfig.
0.3 Welcome
The writing of the program CGAview and this document was supervised by Dr. F.M.C. Witte of the
department of physics and astronomy of the university of Utrecht (Holland). This paper was written as
part of my (third year) Bachelor project which concerned visualization of conformal geometric algebra.
Chapter one
I will start this chapter with a rationale, explaining the reason and use of geometric algebra. Following up
to this rationale I shall give definitions for the geometric, outer and inner products and what it actually
means to represent an object in this algebra. After the definitions there shall be a more technical section
concerned with actually implementing all these operations on a computer. This concludes the first
chapter and with all the necessary algebraic tools at our disposal we can continue to the second part of
the paper.
Chapter two
In the first section of the second chapter we will increase the dimensionality of our model to be able
to represent a greater class of objects. After this we shall consider methods to classify and visualize
represented objects in 3 dimensional space with both Euclidian (+ + +) and Hyperbolic (+ - -) signatures.
This concludes the paper and the reader should now be able to implement a program far prettier than
CGAview without too much trouble.
Appendices
The first appendix deals with binary numbers and associated (bitwise) operations, anyone who is unfamiliar
with binary numbers should first leaf through this before reading the paper.
The program
This paper was written during the development the aforementioned program with which CGA could be
visualized: CGAview. CGAview is very similar to (and much inspired by) the program ‘GA viewer’ (see
iii

[4] in particular), but with the additional capability of visualizing Hyperbolic CGA. It was written pretty
much from scratch, I made no use of the source of GA viewer/Gaigen, because I felt that to understand
CGA, I had to write all routines myself. Nevertheless GA viewer was a great inspiration and I very much
encourage everyone who likes playing around with CGAview to take a look at it.
Bas Fagginger Auer
iv

Chapter 1
Geometric Algebra
1.1 Rationale
1.1.1 Our goal
Our aim is to create a model of space that is as complete as possible (in terms of what aspects of space
it can describe), but can still be readily applied to simple problems, such as arise in physics.
We assume space to be described uniquely by its dimension n ∈ N (the number of independent directions
in which we can move) and the signature function σ : {1, 2, . . . , n} → {−1, +1} describing distance.
The signature can be seen as an indication of how your distance to the origin changes if you travel
along one of the independent directions from the origin. For example, suppose n = 2 with σ(1) = 1
and σ(2) = −1, then if we take a step in direction 1, our distance to the origin increases, while if we
take a step in direction 2 (which is independent of direction 1, meaning that we do not at all move in
direction 1), our distance to the origin decreases. Clearly for ‘usual’ (Euclidian) space, we must have
that σ(k) = 1 for all 1 ≤ k ≤ n, as we always move away from the origin if we take a step, no matter
our direction. Spaces where some directions have a negative signature are particularly useful in physics,
where they are used as a model for relativistic space-time.
1.1.2 Current model
Usually space with these parameters is modelled as the vectorspace R n (the elements of which are called
vectors) together with a pseudo-Euclidian inner product (as we want to include the possibility of negative
signatures):
n∑
< ., . >: R n × R n → R : ((a 1 , a 2 , . . . , a n ), (b 1 , b 2 , . . . , b n )) ↦→ σ(i)a i b i
Apart from the inner product we also have two other operations, vector addition:
+ : R n × R n → R : ((a 1 , a 2 , . . . , a n ), (b 1 , b 2 , . . . , b n )) ↦→ (a 1 + b 1 , a 2 + b 2 , . . . , a n + b n )
and scalar multiplication (elements α ∈ R are called scalars):
i=1
· : R × R n → R n : (α, (b 1 , b 2 , . . . , b n )) ↦→ (αb 1 , αb 2 , . . . , αb n )
All these operations have a very clear geometrical interpretation if we interpret elements a = (a 1 , a 2 , . . . , a n ) ∈
R n as taking a 1 steps in direction 1, a 2 steps in direction 2, . . . , and a n steps in direction n. The sum
c = a + b of two vectors is simply the amount of steps we need to move if we want to go where we would
end up after first moving along a and then along b. Similarly scalar multiplication b = αa is simply
moving along a, but doing so α times. The inner product of a vector with itself < a, a > measures
1

a
a+b
a
2a
b
Figure 1.1: Vector addition and scalar multiplication
the (square of the) distance from the origin we would be if we moved along a and the inner product
of two different vectors < a, b > gives us a measure for their lengths and (in)dependence. Note that
(using vector addition and scalar multiplication) we can write all vectors a = (a 1 , a 2 , . . . , a n ) as a linear
combination of the base vectors e 1 = (1, 0, 0, . . . , 0), e 2 = (0, 1, 0, . . . , 0), . . . , e n = (0, 0, . . . , 0, 1), being
a = ∑ n
i=1 a ie i (the e i are in fact nothing but the n independent directions in which we can move). We
call the collection B := {e i |1 ≤ i ≤ n} the basis of R n and this collection has the property that any
element a ∈ R n can be written as a unique linear combination of elements from B using vector addition
and scalar multiplication (as it is an orthogonal basis with respect to < ., . >).
This description is very simple and easily applied to physical problems (Newtonian dynamics for instance),
but also quite limited in terms of what it can describe. The elements of R n only describe
‘movements’ or 1-dimensional linear subspaces with a certain magnitude (think of movements as the
direction or line along which we move and the magnitude as how far we move). Extended objects as
planes or spheres (which are prevalent in the 3-dimensional space we are familiar with) are not at all
included.
1.1.3 Extending R n
Therefore we seek to extend this model (given its success, R n is not a bad model to start out with) to
encompass as much classes of objects as possible. The first thing we will try is to include k-dimensional
linear subspaces for all 0 ≤ k ≤ n since in R n only 1-dimensional linear subspaces are represented
(vectors), but not planes and their higher dimensional extensions. To do so, we introduce the wedge, ∧,
operator. For now we define it only in a pictorial way, a more precise definition can be found in the next
section.
a
a
a^b
c
a
a^b^c
b
b
Figure 1.2: The wedge operator
For any two linear subspaces a and b, a ∧ b can be seen as being the span of a and b with an associated
size depending on a and b (although we must be careful, just like vectors these linear subspaces only have
a certain orientation and size, but no finite shape). From the pictorial definition, it is easily seen that
we must demand the ∧ operator to have the following properties for all vectors a, b and c and scalars α:
a ∧ b = −b ∧ a
(αa) ∧ b = a ∧ (αb) = α(a ∧ b)
(a ∧ b) ∧ c = a ∧ (b ∧ c)
a ∧ (b + c) = a ∧ b + a ∧ c
Note that because of the first property a ∧ a = −a ∧ a so a ∧ a = 0 for all vectors a.
2

a
a^b
a
b^a
b
b
a
b
a^b
2a
b
(2a)^b
a
2b
a^(2b)
c
a
a^b
c
a
(a^b)^c
b
b
c
a
b^c
c
a
a^(b^c)
b
b
a
c
a^b
a
a^c
+ c
=
a
c
a^(b+c)
b
b
b
b+c
Figure 1.3: Wedge operator properties
3

Let us try to write out a general form of the ∧ product for a collection of vectors using the properties
mentioned above:
a 1 ∧ a 2 ∧ . . . ∧ a k =
=
=
( n ∑
i 1=1
n∑
i 1 =1 i 2 =1
) ( ∑
n
a 1i1 e i1 ∧
n∑
. . .
∑
i 2=1
) ( ∑
n
a 2i2 e i2 ∧ . . . ∧
i k =1
a kik e ik
)
n∑
( )
a 1i1 a 2i2 . . . a kik ei1 ∧ e i2 ∧ . . . ∧ e ik
i k =1
1≤i 1

multiplication and in that sense, nothing at all new. Right now, this is quite correct (the only difference
is the interpretation of the base vectors, but this is not yet of any practical use), but if we manage to
find operations on this space with which we can illustrate the geometrical interpretations of the base
vectors, the model G n may become a lot more appealing.
1.1.4 Operations on our extended space
The first operations we can define on G n are the ∧ product, which we already have, and an extension
of the inner product (currently only defined for vectors). First note that both operations (applied to
vectors) are bilinear and associative. These are properties we would like to have the extensions of these
operations to the entire G n to have as well (as we have seen that they make sense geometrically for
vectors). Because of this, we can use a small trick to make it easier to find out how the operations work
on the entire space.
Let us consider some bilinear, associative operation □ on G n and two arbitrary elements (not necessarily
one-dimensional linear subspaces, just anything in the span of the base elements γ i ) a, b ∈ G n . We will
call such arbitrary elements multivectors to distinguish them from vectors. Because of our basis B ′ of
G n and the bilinearity and associativity of □:
a□b =
=
( 2∑
n −1
i=0
2∑
n −1
i=0
a i γ i
)□
2∑
n −1
j=0
( 2 n −1
∑ )
b j γ j
j=0
a i b j
(
γi □γ j
)
So □ is completely defined in terms of how it operates on the base vectors γ i . Therefore we only need
to consider base vectors if we want to know how a bilinear, associative operator works on the entire G n .
The wedge product
Let’s start with examining the ∧ operator of two arbitrary base vectors γ i and γ j . To do so, decompose
both i and j in binary as usual: i = 2 i 1−1 + 2 i 2−1 + . . . + 2 i k−1 , j = 2 j 1−1 + 2 j 2−1 + . . . + 2 j l−1 with
1 ≤ i 1 < i 2 < . . . < i k ≤ n and 1 ≤ j 1 < j 2 < . . . < j l ≤ n and use the definition of the basis B ′ and
associativity of ∧:
γ i ∧ γ j = ( e i1 ∧ e i2 ∧ . . . ∧ e ik
)
∧
(
ej1 ∧ e j2 ∧ . . . ∧ e jl
)
= e i1 ∧ e i2 ∧ . . . ∧ e ik ∧ e j1 ∧ e j2 ∧ . . . ∧ e jl
We know that the ∧ operator is antisymmetric, so we can order this expression such that i 1 ≤ i 2 ≤ . . . ≤
i k ≤ j 1 ≤ j 2 ≤ . . . ≤ j l by swapping any two adjacent base vectors which have an out of order index.
This induces a sign which depends only on the numers i 1 , i 2 , . . . , i k , j 1 , j 2 , . . . , j l . Since these uniquely
define i and j we can define a new function sgn(i, j) which is precisely the sign change because of the
ordering of the base vectors in the expression. With this new function we can write:
γ i ∧ γ j = e i1 ∧ e i2 ∧ . . . ∧ e ik ∧ e j1 ∧ e j2 ∧ . . . ∧ e jl
= sgn(i, j)e i ′
1
∧ e i ′
2
∧ . . . ∧ e i ′
k+l
Where i ′ 1 ≤ i ′ 2 ≤ . . . ≤ i ′ k+l are simply the original i 1, . . . , i k , j 1 , . . . , j l in a different order.
Now suppose there exists a 1 ≤ m ≤ k + l such that i ′ m = i ′ m+1. Then the same vector occurs twice in
the expression and since the ∧ is antisymmetric the wedge product has to be zero. But this can occur
if and only if there exist 1 ≤ k ′ ≤ k and 1 ≤ l ′ ≤ l such that i k ′ = j l ′ since i 1 < i 2 < . . . < i k and
j 1 < j 2 < . . . < j l . If we now write i and j in binary, then the bits on position i k ′ = j l ′ are set in both
binary forms and therefore the bit on position i k ′ = j l ′ in the number i and j is set as well. Because of
this i and j ≠ 0 (see Appendix A for a definition of the binary ‘and’ operator).
Conversely, if no m can be found such that i ′ m = i ′ m+1, then there cannot be k ′ and l ′ such that i k ′ = j l ′
and therefore all the bits set in the binary form of i are not set in the binary form of j and vice versa.
Therefore i and j = 0.
5

So there are two possibilities: either i and j ≠ 0 and the same vector occurs twice in the wedge expression,
resulting in it being equal to 0 or i and j = 0 and all the vectors in the wedge expression are different,
resulting in i ′ 1 < i ′ 2 < . . . < i ′ k+l . In the latter case this wedge e i 1 ′ ∧ e i ′ ∧ . . . ∧ e 2 i ′ = γ
k+l i ′ for i ′ =
2 i′ 1 −1 + 2 i′ 2 −1 + . . . + 2 i′ k+l −1 , because 1 ≤ i ′ 1 < i ′ 2 < . . . < i ′ k+l
≤ n. But we know that the numbers
i ′ 1, i ′ 2, . . . , i ′ k+l are precisely i 1, . . . , i k , j 1 , . . . , j l , only in a different order, so the set bits of i ′ are precisely
the set bits of i and the set bits of j, therefore i ′ = i or j (if some base vector occurs in either γ i or in
γ j , then it has to occur in γ i ∧ γ j if this expression is nonzero).
With this observation the final expression becomes:
γ i ∧ γ j = sgn(i, j)e i ′
1
∧ e i ′
2
∧ . . . ∧ e i ′
k+l
= sgn(i, j)δ i and j
0 γ i or j
Where δj i is the Kronecker delta, defined as being equal to 1 if i = j and equal to 0 otherwise. The above
expression defines the ∧ operator on the entire G n through bilinearity.
The inner product
Now we will generalize the inner product. We define the inner product a • b to be the orthoplement
of the orthogonal projection of a onto b, linear in both its arguments and coinciding with the inner
product < ., . > for vectors we already were provided with. This definition is called the left-contraction
inner product and leads to the following properties (see [2] and [3]) for scalars α, β, vectors a, b and
multivectors A, B, C:
α • β = αβ
a • α = 0
α • a = αa
a • b = < a, b >
a • (b ∧ C) = (a • b) ∧ C − b ∧ (a • C)
(A ∧ B) • C = A • (B • C)
Now we want to derive a closed expression for γ i • γ j , let us first consider a more simple case:
Consider (using the beforelast rule of •):
( )
• ej1 ∧ e j2 ∧ . . . ∧ e jl
e i
e i • e j =< e i , e j >= δ i jσ(i)
= (e i • e j1 ) ∧ (e j2 ∧ . . . ∧ e jl ) − e j1 ∧ (e i • (e j2 ∧ . . . ∧ e jl ))
= (e i • e j1 ) ∧ (e j2 ∧ . . . ∧ e jl ) −
(
)
e j1 ∧ (e i • e j2 ) ∧ (e j3 ∧ . . . ∧ e jl ) − e j2 ∧ (e i • (e j3 ∧ . . . ∧ e jl ))
= (e i • e j1 ) ∧ (e j2 ∧ . . . ∧ e jl ) + (e i • e j2 ) ∧ (e j1 ∧ e j3 ∧ . . . ∧ e jl ) +
e j1 ∧ e j2 ∧ (e i • (e j3 ∧ . . . ∧ e jl ))
= (e i • e j1 ) ∧ (ê j1 ) + (e i • e j2 ) ∧ (ê j2 ) +
. . .
e j1 ∧ e j2 ∧ (e i • (e j3 ∧ . . . ∧ e jl ))
= (e i • e j1 ) ∧ (ê j1 ) + (e i • e j2 ) ∧ (ê j2 ) + . . . + (e i • e jl ) ∧ (ê jl )
Here (ê j2 ) is used to denote the entire wedge e j1 ∧ e j2 ∧ . . . ∧ e jl without the vector e j2 (the vector with
the hat is left out of the expression).
We are now ready to take care of arbitrary base multivectors, let γ i , γ j ∈ B ′ and write i and j in binary
as usual: i = 2 i1−1 + 2 i2−1 + . . . + 2 ik−1 , j = 2 j1−1 + 2 j2−1 + . . . + 2 jl−1 with 1 ≤ i 1 < i 2 < . . . < i k ≤ n
and 1 ≤ j 1 < j 2 < . . . < j l ≤ n. Use the last rule for •:
)
)
γ i • γ j =
(e i1 ∧ e i2 ∧ . . . ∧ e ik •
(e j1 ∧ e j2 ∧ . . . ∧ e jl
6

(
= e i1 •
= e i1 •
(
e i2 •
(
e i2 •
(
. . . •
(
. . . •
)) ))
(e j1 ∧ e j2 ∧ . . . ∧ e jl . . .
e ik •
(
)
(e ik • e j1 ) ∧ (ê j1 ) + . . . + (e ik • e jl ) ∧ (ê jl )
Suppose that i k ≠ j l ′ for all 1 ≤ l ′ ≤ l, then (as e i • e j = δj iσ(i))
))
. . .
the entire expression equals 0. Otherwise, call r k = l ′ (so i k = j rk )
and all inner products vanish except for (e ik • e jrk ) = σ(i k ).
We will assume that such an j rk can be found,
otherwise the entire expression simply equals 0.
( ( ( (
) ))
= e i1 • e i2 • . . . • e ik−1 • σ(i k ) ∧ (ê jrk ) . . .
( ( (
(
))
= e i1 • e i2 • . . . • 0 − σ(i k ) ∧ e ik−1 • (ê jrk )
( ( (
= e i1 • e i2 • . . . • − σ(i k )σ(i k−1 ) ∧ (ê jrk , ê jrk−1 )
. . . Process i k−2 , i k−3 , . . . , i 1 similarly.
= (−1) k+1 σ(i 1 )σ(i 2 ) . . . σ(i k )(ê jr1 , . . . , ê jrk )
= (−1) k+1 σ(i 1 )σ(i 2 ) . . . σ(i k )(ê i1 , . . . , ê ik )
))
. . .
))
))
. . .
This is under the assumption that for any i k ′ we can find an j l ′ = i k ′, otherwise the expression becomes
zero. So any bit that is set in i, should be set in j if the result is to be nonzero. This is the case if and
only if i and j = i. Furthermore, all base vectors occuring in the wedge γ i are removed from the wedge
γ j (note (ê i1 , . . . , ê ik ) in the expression for γ i • γ j ). So all bits set in i are cleared in j. As these bits also
have to be set in j (otherwise the result is 0) this is a xor operation: (ê i1 , . . . , ê ik ) = γ i xor j. Note that
because all set bits in i also occur in j, we have that sgn(i, j) = (−1) k+1 . At this point it is useful to
introduce the extended signature ˆσ(i) := σ(i 1 )σ(i 2 ) . . . σ(i k ). This allows us to write the final expression
as (in the above expression we assumed that i and j = i, so we need to add the Kronecker delta):
γ i • γ j = δ i and j
i (−1) k+1 σ(i 1 )σ(i 2 ) . . . σ(i k )(ê i1 , . . . , ê ik )
= δ i and j
i sgn(i, j)ˆσ(i)γ i xor j
This expression defines the inner product on the entire G n through bilinearity.
The geometric product
The only thing left to do is to define a suitable geometric product on our space. To make our geometric
product easy to work with, it should be bilinear, associative and, if possible, invertible.
Let us first concentrate on the product of just two vectors a and b. We already have two bilinear products
(being ∧ and •), but neither is invertible as becomes apparent if we consider the solution sets for b of
the equations a ∧ b = C and a • b = α. These are the line {b + βa|β ∈ R} for the ∧ product (as
a ∧ (b + βa) = a ∧ b + βa ∧ a = a ∧ b) and a plane with normal a at distance α/|a| from the origin for
the • product. The intersection of both solution sets however (as noted in [2]) is exactly one point, so
if we define the geometric product as a · b := a • b + a ∧ b then a · b = α + C has a unique solution b.
Furthermore with this definition the geometric product is bilinear and associative just as we desired.
For base vectors this definition results in:
{
σ(i) = ej · e
e i · e j = e i • e j + e i ∧ e j =
i if i = j
e i ∧ e j = −e j · e i if i ≠ j
We can easily extend this to base multivectors, as with deriving the ∧ product:
γ i · γ j = ( e i1 ∧ e i2 ∧ . . . ∧ e ik
)
·
(
ej1 ∧ e j2 ∧ . . . ∧ e jl
)
= ( e i1 · e i2 · . . . · e ik
)
·
(
ej1 · e j2 · . . . · e jl
)
= e i1 · e i2 · . . . · e ik · e j1 · e j2 · . . . · e jl
= sgn(i, j)e i ′
1 · e i ′
2 · . . . · e i ′
k+l
7

We used the expression for base vectors and the associativity of the geometric product. 1 ≤ i ′ 1 ≤ i ′ 2 ≤
. . . ≤ i ′ k+l ≤ n is a re-arrangement of i 1, i 2 , . . . , i k , j 1 , j 2 , . . . , j l (we can exchange the base vectors just as
with the ∧ product). Now we consider all the 1 ≤ k ′ < k+l for which i ′ k = ′ i′ k ′ +1 , then e i ′ ·e k ′ i ′ = σ(i′ k ′ +1 k ′).
This can only occur (as all the i k ′ and j l ′ are distinct) if i ′ k = i ′ k ′′ and i′ k ′ +1 = j l ′ (or vice versa) for some
k ′′ and l ′ . Therefore the same bit at position i k ′′ = j l ′ in both i and j needs to be set. Conversely, if in
both i and j the same bit is set, then the same basevector occurs twice and can hence be replaced with
σ(i ′ k ′). Therefore all base vectors removed from the product in this way are exactly those whose bits are
set in i and j and for each removal the sign changes with σ(i ′ k ′):
γ i · γ j = sgn(i, j)ˆσ(i and j)e i ′′
1 · e i ′′
2 · . . . · e i ′′ N
Where 1 ≤ i ′′
1 < i ′′
2 < . . . < i ′′ N ≤ n and N ≤ k + l. This is again a new base multivector, γ i xor j to
be precise (if a bit is set in both i and j, the same basevector occurs in both wedges γ i and γ j , it is
removed, otherwise the vector is added to the wedge). With this information we can write:
We now have closed formulas for all our products:
γ i · γ j = sgn(i, j)ˆσ(i and j)γ i xor j
γ i ∧ γ j = sgn(i, j)δ i and j
0 γ i or j
γ i • γ j = δ i and j
i sgn(i, j)ˆσ(i)γ i xor j
γ i · γ j = sgn(i, j)ˆσ(i and j)γ i xor j
To symmetrize them, note that if i and j = 0 then i or j = i xor j and ˆσ(i and j) = ˆσ(0) = 1 (empty
product). Furthermore δ i and j = 1 always, so we can also write our products as:
i and j
γ i ∧ γ j = sgn(i, j)δ i and j
0 ˆσ(i and j)γ i xor j
γ i • γ j = sgn(i, j)δ i and j
i ˆσ(i and j)γ i xor j
γ i · γ j = sgn(i, j)δ i and j
i and j ˆσ(i and j)γ i xor j
Hopefully, it is now apparent why I chose the ‘xor’ operation to illustrate the cover of this document: it
is embedded in our algebra in the most fundamental way.
1.1.5 Use of our extended space
Now that we have our new space and its operations, it is time to see its geometrical use. Let us consider G n
with the Euclidian signature (so σ(1) = σ(2) = . . . = σ(n) = 1). Let ι C : C → G 2 : a+bi ↦→ a1+b(e 1 ∧e 2 )
this is an embedding and because (e 1 ∧ e 2 ) · (e 1 ∧ e 2 ) = e 1 · e 2 · e 1 · e 2 = −e 1 · e 1 · e 2 · e 2 = −σ(1)σ(2) = −1
we have that ι C (z · w) = ι C (z) · ι C (w). So this embedding of the complex numbers in our extended
space preserves complex multiplication: C is ‘part’ of our extended space and the geometric product is
(when restricted to ι C (C)) nothing more than complex multiplication. This even gives us a geometrical
interpretation of i = √ −1 being e 1 ∧ e 2 : i is the 2-dimensional linear subspace spanned by the base
vectors e 1 and e 2 with size 1.
Furthermore, the quaternions H are part of G 3 in a similar manner, the embedding ι H : H → G 3 :
a + bi + cj + dk ↦→ a1 + b(e 2 ∧ e 3 ) + c(e 1 ∧ e 3 ) + d(e 1 ∧ e 2 ) preserves the quaternion product. So our
geometrical product is ‘right’ in the sense that in certain linear subspaces of our extended space it reduces
to useful products we are already familiar with.
Another use is mentioned in [3], given some linear subspace A = a 1 ∧ a 2 ∧ . . . ∧ a k we can easily seperate
any vector a ∈ R n in a part a ⊥ perpendicular to A and a ‖ parallel to A. This is such that a = a ⊥ + a ‖
with a ⊥ • A = 0 and a ‖ ∧ A = 0.
Note that a ⊥ · A = a ⊥ • A + a ⊥ ∧ A = 0 + a ⊥ ∧ A + a ‖ ∧ A = a ∧ A (the formula a · A = a • A + a ∧ A
can be derived from the definitions of the products in the previous paragraph and the particular form of
8

A). Also a ‖ · A = a ‖ • A + a ‖ ∧ A = a • A. Now dividing by A yields:
a ⊥ = ( a ∧ A ) /A a ‖ = ( a • A ) /A
In [2] it is shown that we can rotate any multivector by simple conjugating it with the multivector
e A φ 2 (we define e x := ∑ ∞
) where φ is the angle in radians and A is the plane of rotation (e.g.
n=1 xn
n!
A = e 1 ∧ e 2 ). So to rotate any multivector x to x ′ simply use x ′ = e −A φ 2 xe +A φ 2 . This formula applies to
any multivector x of any dimension (we can rotate a plane like x = e 2 ∧ e 3 using the same formula) in
arbitrary G n .
Later, after the introduction of the conformal geometric algebra, it will become clear that the products
on our extended space can be used to do a variety of things: rotations, translations, reflections,
sphere/line/plane intersections and more. It turns out that all these operations can be described with
formulas independent of the chosen basis and dimension of our space. This independence is the true
benefit of using geometric algebra.
1.1.6 Embedding R n in our extended space
Until now we have more or less pretended that the (base)vectors are elements of our space G n , but this
is not entirely true. We view e i ∈ G n ≃ R 2n while (strictly speaking) e i ∈ R n and these are two very
different sets. What we mean of course is (through the ‘definition’ of our base multivectors γ i ) that we
associate e i with the wedge consisting of only one element: γ 2 i−1. To make this mathematically precise,
we say that the R n is embedded in our extended space G n via the map (actually a linear embedding)
ι K : R n → G n : ∑ n
i=1 a ie i ↦→ ∑ n
i=1 a iγ 2 i−1. Later we will see that embedding R n in other ways (in
higher dimensional models of space) can be extremely useful. If left unspecified however, the embedding
is assumed to be the default ι K we have been using implicitly all the time.
We set out extending our space with the ∧ operator such that ‘planes and their higher dimensional
equivalents are part of our model’, but without ever specifying what we actually meant. To make this
precise, we consider some object A ⊆ R n and a linear embedding ι of R n into G m for some m. Suppose
A to be a linear subspace, then A is the span of some vectors a 1 , . . . , a k and we have that a ∈ A if and
only if a is a linear combination of the vectors a 1 , . . . , a k , but (in our extended model G n ) this is the
case if and only if a ∧ (a 1 ∧ . . . ∧ a k ) = 0, so we would like to say that a 1 ∧ . . . ∧ a k represents the linear
subspace A.
This motivates the definition that some multivector B ∈ G m represents an object A ⊆ R n if and only if
x ∈ A ⇐⇒ ι(x) ∧ B = 0 (and in our example of a linear subspace A, ι was just the default inclusion of
R n in G n ).
In R n you can also characterize a plane through the origin with just a normal vector a. To include this
type of representation in our extended model, we say that B dually represents an object A if and only
if x ∈ A ⇐⇒ ι(x) • B = 0 (as for all elements x in a plane through the origin with normal a we have
that < x, a >= 0).
There is a way to switch from one representation to another if we make two additional definitions. For
any geometric algebra G n we define γ 2 n −1 (the wedge of all base vectors: all bits are set) to be the
pseudoscalar. The inner product of some multivector B with the pseudoscalar is called the dual of B:
B ∗ := B • γ 2 n −1. The use of these definitions becomes apparent when we consider the final rule of the
inner product: (x ∧ B) • γ 2 n −1 = x • (B • γ 2 n −1) = x • B ∗ . From this we see that B represents A if and
only if B ∗ dually represents A. So the dual of a plane is its normal and vice versa (as (B ∗ ) ∗ = B). This
operation permits us to switch from direct object representation to dual representation and back again.
Also note that (since we xor the bits with 1111 . . . 111) if B has grade k, then B ∗ has grade n − k as all
bits are inverted because of the xor operation when we take the dual.
Further use of the dual is (as illustrated in [2]) the calculation of the largest linear subspace contained
in two given linear subspaces (i.e. for two planes this is usually the line given by their intersection). For
A, B ∈ G n it turns out that B ∗ • A is the largest common linear subspace of A and B .
9

This concludes this (rather lengthy) introduction to geometric algebra as an extended model of space.
Actually it is nothing more than a justification of the definitions that are about to follow, but a very
important one, as it emphasizes their geometrical motivation.
1.2 Definitions
Let n ∈ N and σ : {1, 2, . . . , n} → {−1, +1} be given.
Definition: The extended signature ˆσ:
ˆσ : {0, 1, . . . , 2 n − 1} → {−1, +1} : i ↦→
k∏
σ(i j )
j=1
for i = 2 i1−1 + 2 i2−1 + . . . + 2 i k−1
1 ≤ i 1 < i 2 < . . . < i k ≤ n
This map is well-defined because every natural number can be written uniquely in binary form.
Definition: The order sign sgn:
sgn : {0, 1, . . . , 2 n − 1} × {0, 1, . . . , 2 n − 1} → {−1, +1}
Let (i, j) ∈ {0, 1, . . . , 2 n −1}×{0, 1, . . . , 2 n −1}, write them both in binary form: i = 2 i1−1 +2 i2−1 +. . .+
2 ik−1 and j = 2 j1−1 + 2 j2−1 + . . . + 2 jl−1 for 1 ≤ i 1 < i 2 < . . . < i k ≤ n and 1 ≤ j 1 < j 2 < . . . < j l ≤ n.
Define sgn(i, j) := (−1) N where N equals the number of nontrivial transpositions (exchanges of two
adjacent, different numbers) to order (i 1 , i 2 , . . . , i k , j 1 , j 2 , . . . , j l ) such that i 1 < i 2 < . . . < i k < j 1 <
j 2 < . . . < j l . This is well-defined because any natural number can be written uniquely in binary form
and because if we have two different ways to order the sequences, they must differ an even number of
transpositions and (−1) N+2K = (−1) N for any integer K.
Definition: The Kronecker delta δ i j :
δ : Z × Z → {0, 1} : (i, j) ↦→ δ i j :=
{
1 if i = j
0 if i ≠ j
Definition: The n-dimensional geometrical algebra:
G n := (R 2n , +, ·, ∧, •, (0, 0, . . . , 0), (1, 0, . . . , 0))
Where for any α ∈ R and (a 1 , a 2 , . . . , a 2 n), (b 1 , b 2 , . . . , b 2 n) ∈ R 2n :
(a 0 , a 1 , . . . , a 2 n −1) + (b 0 , b 1 , . . . , b 2 n −1) := (a 0 + b 0 , a 1 + b 1 , . . . , a 2 n −1 + b 2 n −1)
α(a 0 , a 1 , . . . , a 2 n −1) := (αa 0 , αa 1 , . . . , αa 2 n −1)
(a 0 , a 1 , . . . , a 2 n −1) · (b 0 , b 1 , . . . , b 2 n −1) :=
(a 0 , a 1 , . . . , a 2 n −1) ∧ (b 0 , b 1 , . . . , b 2 n −1) :=
(a 0 , a 1 , . . . , a 2 n −1) • (b 0 , b 1 , . . . , b 2 n −1) :=
2∑
n −1
i,j=0
2∑
n −1
i,j=0
2∑
n −1
i,j=0
a i b j sgn(i, j)ˆσ(i and j)δ i and j
i and j γ i xor j
a i b j sgn(i, j)ˆσ(i and j)δ i and j
0 γ i xor j
a i b j sgn(i, j)ˆσ(i and j)δ i and j
i γ i xor j
Note that G n when considered with just vector addition and scalar multiplication is a vector space, while
considered with vector addition and any of the products it is a nonabelian ring with nilpotent elements
(for n ≥ 1) and isomorphic to R for n = 0.
10

Definition: Canonical embedding
The canonical embedding of R n into G n is defined as:
Definition: Object representation
ι K : R n → G n :
n∑
a i e i ↦→
i=1
Let A ⊆ R n , B ∈ G m and ι : R n → G m an embedding.
n∑
a i γ 2 i−1
We say that B represents A if and only if x ∈ A ⇐⇒ ι(x) ∧ B = 0.
We say that B dually represents A if and only if x ∈ A ⇐⇒ ι(x) • B = 0.
Definition: Dual
Let A ∈ G n , the dual of A is defined as A ∗ := A • γ 2 n −1 and γ 2 n −1 is called the pseudoscalar.
Definition: Grade
Let A = ∑ 2 n −1
i=0
a i γ i ∈ G n then we say that A has grade or dimension k ∈ N if and only if for all
0 ≤ i < 2 n we have that a i ≠ 0 → there are precisely k bits set in the number i written in binary form.
i=1
1.3 Implementation
1.3.1 Data types
Now we will discuss implementing this model in some programming language. We will assume that two
datatypes are available: integers and real numbers.
Given are the dimension as some positive integer:
integer n
And the signature as an array (list) of n integers:
integer sigma[n]
Such that n = n and sigma[0] = σ(1), sigma[1] = σ(2), . . . , sigma[n-1] = σ(n). Multivectors are
nothing more than elements of R 2n , so we characterize them as an array of 1

function sigmahat(integer i)
//return value of our function
integer returnval = 1
integer j
set j = 0 and do while j < n
//check if lowest bit is set, if so multiply
//with appropriate signature
if (i and 1 = 1) then returnval = returnval * sigma[j]
//advance to the next bit by shifting i to the right
i = i >> 1
advance j = j + 1
return returnval
1.3.3 Order sign
For this function, we are given two numbers i, j ∈ N such that 0 ≤ i, j < 2 n . We now need to find out
what the set bit positions of i and j are, add these to a list and find out how often we must exchange
two adjacent, different elements of the list to sort it. Sorting lists can be done in a lot of ways, I chose
to use the bubble sort algorithm because of its simplicity. Extracting the bit positions can be done with
the same bit shift trick as we used before: we keep shifting a number to the right while incrementing
the position and if the lowest bit is set, we add the current position to the list. Note that i and j are
both at most n-bit numbers, so a list that is 2n elements long should be big enough to contain all bit
positions. This leads to the following algorithm:
function sgn(integer i, integer j)
//return value
integer returnval = 1
//the actual bit positions of i and j
integer bitpos[2 * n]
//total number of bit positions in both i and j
integer numbitpos = 0
integer k
integer l
//extract bit positions from i
set k = 0 and do while k < n
if (i and 1 = 1) then
bitpos[numbitpos] = k + 1
numbitpos = numbitpos + 1
end if
i = i >> 1
advance k = k + 1
//extract bit positions from j
set k = 0 and do while k < n
if (j and 1 = 1) then
bitpos[numbitpos] = k + 1
numbitpos = numbitpos + 1
end if
j = j >> 1
advance k = k + 1
//sort bit position list (bubble sort)
set k = 0 and do while k < (numbitpos - 1)
12

set l = 0 and do while l < (numbitpos - 1 - k)
//we need to swap two adjacent numbers to sort the list
if (bitpos[l] > bitpos[l + 1]) then
swap bitpos[l], bitpos[l + 1]
//extra transposition, so sign flips
returnval = -returnval
end if
advance l = l + 1
advance k = k + 1
return returnval
1.3.4 Kronecker delta
This is rather easy to implement:
function delta(integer i, integer j)
if (i = j) return 1
return 0
1.3.5 Multivector addition and scalar multiplication
This is simply done componentwise: we loop through all the multivector’s components and apply the
operation.
function add(real a[1

two multivectors according to some given multiplication table C(i, j) of some product □:
In code:
( 2∑
n −1
i=0
a i γ i
)□
( 2 n −1
∑ )
b j γ j =
j=0
2∑
n −1
i,j=0
a i b j C(i, j)γ i xor j
function mul(real a[1

can be done with Gauss reduction. ( ∑2
First we must ( construct the matrix A however. From the formula
we derived above: c = a□b =
n −1 ∑2
)
i=0
a i γ i
)□
n −1
j=0 b jγ j = ∑ 2 n −1
i,j=0 a ib j C(i, j)γ i xor j and the fact
that for ordinary matrix multiplication c i = ∑ 2 n −1
j=0 A ijb j we see that the matrix A must have elements
A i xor j,j = a i C(i, j). Now we just have to solve the system Ab = (1, 0, 0, . . . , 0):
function rightinverse(real a[1

extract multivector
set i = 0 and do while i < (1

Chapter 2
Extended Geometric Algebra
2.1 Homogeneous geometric algebra
With ι K : R n → G n we have embedded R n in our extended space in which all k-dimensional linear
subspaces through the origin are represented. It would be nice to find another embedding ι which
enables us to embed R n in a slightly bigger space which contains k-dimensional linear subspaces at any
point. Consider R 2 and the embedding ι H : R 2 → R 3 : a ↦→ a + e 3 .
Now let a, b ∈ R 2 be two distinct, arbitrary points. The plane through the origin, ι H (a) and ι H (b) is
represented in our extended space G 3 as l := ι H (a) ∧ ι H (b). It is easily seen that any x ∈ R 2 with
ι H (x) ∧ l = 0 must lie in the preimage through ι H of the intersection of the plane l and the plane ι H (R 2 ).
This is precisely the line through the points a and b, so in that sense l = ι H (a) ∧ ι H (b) represents (in R 2 )
the line through a and b. Therefore this construction enables us to represent lines through any two points
in the plane, which was not possible in G 2 (there all represented lines, 1-dimensional linear subspaces,
passed through the origin).
a
c
b
i_H
a
c
b
l = a^b
points identified
with c
Figure 2.1: 2-dimensional homogeneous geometric algebra
But what do the points not lying in the plane ι H (R 2 ) represent? Consider the line in R 3 from the origin
through some point ι H (c) for c ∈ R 2 . In figure 2.1 we see that if we were to identify all points on this
line with c, it would be consistent with l representing the line in R 2 between a and b (with this point
identification all points on the plane in R 3 represented by l would be identified with points on the line
from a to b). Because of this we say that for a ∈ R 2 , a + αe 3 and 1 α a + e 3 for α ≠ 0 represent the same
point 1 α a ∈ R2 .
But what of points for which α = 0? Let us consider points a + ɛe 3 for 0 < ɛ ≪ 1 a very small number,
if we let ɛ ↓ 0 we get the case that α = 0. a + ɛe 3 is identified with 1 ɛ a + e 3, so the point 1 ɛ a in R2 . If
we now let ɛ ↓ 0 we see that the distance between the origin and 1 ɛ
a becomes arbitrarily large for a ≠ 0.
Therefore we view points (x, y, 0) (arbitrary points with α = 0) in R 3 as points at infinity, as they are
identified with points at an infinite distance from the origin in R 2 .
17

To generalize representing subspaces at arbitrary locations to higher dimensions, consider R n and ι H :
R n → R n+1 : a ↦→ a + e n+1 . Let A = a 1 ∧ a 2 ∧ . . . ∧ a k be the representation in G n of some k-dimensional
linear subspace through the origin in R n . We want to move A to some location b ∈ R n different from
the origin. Consider A ′ := ι H (b) ∧ a 1 ∧ a 2 ∧ . . . ∧ a k ∈ G n+1 and let α 1 , . . . , α k ∈ R. Note that:
ι H (b + α 1 a 1 + . . . + α k a k ) ∧ A ′ = ( b + e n+1 + α 1 a 1 + . . . + α k a k
)
∧
(
b + en+1 ) ∧ a 1 ∧ . . . ∧ a k
= 0
Since this holds for arbitrary α 1 , . . . , α k , the linear subspace spanned by a 1 , . . . , a k at location b in
R n (call this A ′′ ) is a subset of the object represented by A ′ . Conversely, suppose that we have an
x ∈ R n such that ι H (x) ∧ A ′ = 0, then x + e n+1 must be a linear combination of b + e n+1 , a 1 , . . . , a k :
x + e n+1 = α ( )
b + e n+1 + α1 a 1 + . . . + α k a k . Because b, a 1 , . . . , a k ∈ R n and the e 1 , e 2 , . . . , e n+1 are
linearly independent we see that α = 1 and therefore x = b+α 1 a 1 +. . .+α k a k : x is an element of A ′′ . So
the object represented by A ′ is a subset of A ′′ . Therefore we may conclude that the object represented
by ι H (b) ∧ a 1 ∧ . . . ∧ a k is precisely A ′′ , the linear subspace spanned by a 1 , . . . , a k at location b. Because
of this, the embedding ι H enables us to represent any linear subspace at any location in R n .
These considerations motivate the following definition of the homogeneous geometric algebra:
Definition: n-dimensional homogeneous geometric algebra
Let n ∈ N and σ : {1, 2, . . . , n} → {−1, +1} be given. Extend σ to ˜σ : {1, 2, . . . , n, n + 1} → {−1, +1} :
k ↦→ σ(k) if 1 ≤ k ≤ n and 1 otherwise. We define the n-dimensional homogeneous geometric algebra
G n H := Gn+1 the geometric algebra with signature ˜σ and associated embedding ι H : R n → G n H :
∑ n
i=1 a ie i ↦→ γ 2 n + ∑ n
i=1 a iγ 2 i−1.
2.2 Conformal geometric algebra
Now we would like to extend our space such that spheres are represented in the geometrical algebra.
Spheres are defined as the collection of points for which the distance to a given point is equal to a given
number (the radius of the sphere): S = {x ∈ R n |‖x − a‖ 2 = r 2 } is a sphere located at position a ∈ R n
with radius r ∈ R. The abbreviation ‖a‖ 2 is defined as ‖a‖ 2 :=< a, a >= ∑ n
i=1 σ(i)a2 i , the length of a
squared.
A sphere around the origin is nothing but the level set of the map a ↦→ ‖a‖ 2 . This can also be seen as
the projection onto R n of the intersection of the plane with normal e n+1 at a distance r from the origin
with the image of the map a ↦→ a + ‖a‖ 2 e n+1 .
Planes can be represented by multivectors in geometric algebra, so embedding our space R n into R n+1
using a ↦→ a + ‖a‖ 2 e n+1 could be a good start. The only problem is that in ordinary geometric algebra
only linear subspaces through the origin are represented, to be able to have linear subspaces at arbitrary
locations we must extend our extended space to the homogeneous geometric algebra. Therefore we
introduce another independent base vector e n+2 as our homogeneous origin (the same procedure as
when we created the homogeneous extension of R 2 where we added e 3 , see the previous section). This
gives us the embedding a ↦→ e n+2 + a + ‖a‖ 2 e n+1 which we will call ι C .
Now we consider R 2 embedded in R 4 using ι C . Let a, b, c ∈ R 2 and define A = ι C (a) ∧ ι C (b) ∧ ι C (d)
this is the plane through the points ι C (a), ι C (b) and ι C (c) on the parabola. We see that A intersects the
parabola in an ellipsoid and that d ∈ R 2 satisfies ι C (d)∧A = 0 if and only if ι C (d) lies on the ellipsoid. So
A represents this ellipsoid, but we wanted to represent spheres (or circles in this case). This means that
we have to ‘unwarp’ our ellipsoid to a circle by choosing a special signature for our extended space. It
turns out that for sensible results, this signature must be chosen such that e n+1 •e n+1 = e n+2 •e n+2 = 0,
but this is not allowed (this would imply that the signature of our extended space assumes the value 0,
but signatures may only assume the values −1 and +1) . Therefore we choose new extension vectors to
be o 0 and o ∞ (the names of which will become clear later on) which are linear combinations of e n+1 and
e n+2 and of which o 0 plays the role of the homogeneous origin.
18

0
1
2
-1
-2
4
3
2
1
0
-2
-1
0
1
2
Figure 2.2: Level surface of a ↦→ ‖a‖ 2
So we extend our space with two vectors o 0 and o ∞ not part of our original space R n . We take the
embedding to be ι C (a) = o 0 + a + 1 2 ‖a‖2 o ∞ for vectors a ∈ R n into R n+2 with signature ˜σ (currently
unspecified) and generate the associated geometric algebra G n+2 . Time to get to work:
ι C (a) • ι C (b) =
(o 0 + a + 1 )
2 ‖a‖2 o ∞ •
(o 0 + b + 1 )
2 ‖b‖2 o ∞
= o 0 • o 0 + 1 4 ‖a‖2 ‖b‖ 2 o ∞ • o ∞ + a • b +
o 0 • b + a • o 0 + 1 2 ‖b‖2 o 0 • o ∞ + 1 2 ‖a‖2 o ∞ • o 0
Suppose we can choose the signature ˜σ for the space R n+2 such that o 0 • o 0 = o ∞ • o ∞ = 0, o 0 • o ∞ = 1
and o 0 • e i = o ∞ • e i = 0 for 1 ≤ i ≤ n. Then the above simplifies to:
ι C (a) • ι C (b) = 1 2 ‖a‖2 · 1 + a • b + 1 2 ‖b‖2 · 1
Furthermore, we choose ˜σ negative, such that a • b = − < a, b > (here < ., . > is the inner product on
our original space R n ). This leads to:
ι C (a) • ι C (b) = 1 2 < a, a > − < a, b > +1 2 < b, b >
= 1 2 < b − a, b − a >= 1 ‖b − a‖2
2
So the inner product now reflects the metric of our original space. But this was all under the assumption
we could find a signature such that the simplification above holds. For this we pick ˜σ(1) = 1, ˜σ(n + 2) =
−1 and o 0 := 1 √
2
(
e1 + e n+2
)
, o∞ := 1 √
2
(
e1 + e n+2
)
. Then certainly o0 • o 0 = o ∞ • o ∞ = 0 and
o 0 • o ∞ = 1. Furthermore, to ensure that a • b = − < a, b > we choose ˜σ(k) = −σ(k − 1) for
2 ≤ k ≤ n + 1. This defines the entire signature for our extended space in which we embed R n as
∑ n
i=1 a ie i ↦→ o 0 + ∑ n
i=1 a ie i+1 + 1 2 ‖a‖2 o ∞ .
Note that ι(a) • o ∞ = o 0 • o ∞ + a • o ∞ + 1 2 ‖a‖2 o ∞ • o ∞ = 1 + 0 + 0 = 1. With this, for some r ∈ R:
ι C (a) •
(ι C (b) − 1 )
2 r2 o ∞ = 1 (‖b − a‖ 2 − r 2)
2
Now define A = ι C (b) − 1 2 r2 o ∞ , then clearly ι C (a) • A = 0 ⇐⇒ ‖b − a‖ 2 = r 2 , so A dually represents
an n-dimensional sphere at location b with radius r in R n . If we take the dual of A, A ∗ then this is a
19

multivector of grade n + 1 (we take the dual in n + 2-dimensional space of a 1 dimensional multivector
A). Furthermore, the object represented by A ∗ is an n-dimensional sphere (as A dually represents an
n-dimensional sphere). Observe that (dualisation is linear because the inner product is linear in its first
argument):
A ∗ =
(ι C (b) − 1 2 r2 o ∞
) ∗
= o ∗ 0 + b ∗ + 1 2(
‖b‖ 2 − r 2) o ∗ ∞
It is not hard to see that we can find vectors a 1 , . . . , a n+1 ∈ R n (this is a grade n + 1 multivector) such
that A ∗ = αι C (a 1 ) ∧ . . . ∧ ι C (a n+1 ) for some constant α. Just as with the homogeneous extension, we
identify multivectors that are scalar multiples of each other, so we may suppose that α = 1. For these
vectors a i , we have that ι C (a i ) ∧ A ∗ = 0, so a i is for 1 ≤ i ≤ k an element of the object represented
by A ∗ : the sphere. Therefore the direct representation of the n-dimensional sphere is a wedge of n + 1
vectors on the sphere and we have achieved our goal.
Now we formalize this with the following definition:
Definition: n-dimensional conformal geometric algebra
Let n ∈ N and σ : {1, 2, . . . , n} → {−1, +1} be given. Extend σ to ˜σ : {1, 2, . . . , n, n+1, n+2} → {−1, +1}
by letting ˜σ(1) = 1, ˜σ(k) = −σ(k − 1) for 2 ≤ k ≤ n + 1 and ˜σ(n + 2) = −1. We define the
n-dimensional conformal geometric algebra G n C := Gn+2 the geometric algebra with signature ˜σ and
associated embedding:
ι C : R n → G n C :
n∑
a i e i ↦→ o 0 +
i=1
n∑
a i γ 2 i + 1 ( ∑
n )
σ(i)a 2 i o ∞
2
i=1
Where o 0 := 1 √
2
(γ 1 + γ 2 n+1) and o ∞ := 1 √
2
(γ 1 − γ 2 n+1).
i=1
2.3 Object classification
Within the conformal geometric algebra, it turns out (see [1]) that there are simple (independent of
dimension and represented object to which they are applied) expressions for reflections and intersections
of multivectors.
Reflection
Let us suppose that A ∈ G n C represents some object (say a plane or a sphere), then we can reflect any
multivector x ∈ G n C in this object A to x′ using x ′ = A · x · A −1 .
Intersection
The intersection C of the representations of two objects A, B ∈ G n C is given by the largest common
subspace of both A and B, through the formula C = A ∗ • B (this formula was also mentioned in the
rationale, for planes it is rather obvious, but in CGA it also holds for representations of more complicated
objects).
Classification of represented objects
Just as with the homogeneous geometric algebra we choose to identify elements x of the form x =
αo 0 + ∑ n
i=1 a ∑n
)
iγ 2 i + 2( 1 i=1 σ(i)a2 i o ∞ with points 1 α x ( 1 α x equals ι C(p) for some p ∈ R n ) for α ≠ 0 and
say that x is at infinity if α = 0. We call (for α ≠ 0) replacing x with the point 1 αx the normalization
of x. Because o ∞ • ( αo 1 + ∑ n
i=1 a iγ 2 i + 1 2 (∑ n
i=1 σ(i)a2 i )o (
∞)
= α we can use the formula x/ o∞ • x ) to
normalize x.
Note that according to these definitions o ∞ in particular is a point at infinity. We can really interpret
this point as being the infinity, after all it is the limit of 1 α o 0 + 0 + o ∞ for α ↑ ∞. This point is for
20

all α > 0 identified with o 0 + 0 + αo ∞ : a point at an ever increasing distance from the origin with no
preferred direction. Similarly, o 0 can be interpreted as the origin, for it equals ι C (0).
Having incorporated infinity as an element of our algebra, we can also represent linear subspaces at any
location (just as with the homogeneous extension of our space). Consider G 3 C where we represent a
sphere with A = ι C (a) ∧ ι C (b) ∧ ι C (c) ∧ ι C (d). If we now let the point d move farther and farther away
from the origin, we see that the sphere represented by A becomes larger and larger and appears more
and more like a plane through a, b and c. For large distances it also does not matter in which direction
we let d move to infinity, since the plane is determined only by a, b and c. Therefore, if we take the
limit of ‖d‖ 2 ↑ ∞ we may just as well take d → o ∞ . In this limit , the sphere represented by A will turn
into a plane through a, b and c, while A → ι C (a) ∧ ι C (b) ∧ ι C (c) ∧ o ∞ . So ι C (a) ∧ ι C (b) ∧ ι C (c) ∧ o ∞
represents of a plane through a, b and c. Similarly a line can be seen as a circle through infinity in G 2 C :
ι C (a) ∧ ι C (b) ∧ o ∞ . In general this enables us represent any k-dimensional linear subspace of R n in G n C
by ι C (a 1 ) ∧ ι C (a 2 ) ∧ . . . ∧ ι C (a k ) ∧ o ∞ for vectors a 1 , . . . , a k inside the subspace. So our new model
represents linear subspaces as well as spheres.
Now we will start making a rough classification of objects represented by multivectors in G n C . Let A ∈ Gn
be a multivector with a well-defined grade k.
There are two possibilities, either o ∞ • A = 0 or o ∞ • A ≠ 0. For our reasoning to be compatible with
what we decided for points, we say that A is located at infinity if o ∞ • A = 0. In this case A cannot
have a well-defined position, only some orientation (remember that points at infinity can have a certain
direction: 0 · o 0 + 4 · e 2 + o ∞ for example). Therefore we call objects without a well-defined position
orientations. Otherwise (if o ∞ • A ≠ 0) we say that A has a well-defined position, we only need to find
it.
If A has a well-defined position, there are again two possibilities: o ∞ ∧ A = 0 or o ∞ ∧ A ≠ 0 (infinity is
either part of our represented object, or it isn’t). In the first case we say the object is flat, as it extends
to infinity. These are the lines, planes, etc.: k-dimensional linear subspaces of our n-dimensional space
with 0 ≤ k ≤ n at a certain position. In the second case, we call A round: it somehow bends such that
it does not extend to infinity. These rounds must be k-dimensional spheres for 0 ≤ k ≤ n with respect
to the metric induced by the signature of our space.
For rounds it is easy to find their position: simply reflect infinity in the given object: A · o ∞ · A −1
(consider a sphere, spherically inverting the sphere’s position gives you infinity and vice versa).
For flats it is slightly more difficult. Consider the homogeneous case in two dimensions G 2 H . A line
through two points certainly classifies as a flat object. For the position of the line, any point on the line
will do (together with the direction, we can reconstruct the line uniquely). In homogeneous space, the
line is represented by the plane l through the origin and two points on the line. To recover a point on
the line, we simply take the orthogonal projection of e 3 on this plane: (e 3 • l) · l −1 . When we extended
e3
e3 parallel
plane representing line
Figure 2.3: Homogeneous position of a flat
our space to conformal geometric algebra, we introduced o 0 as our homogeneous origin (just like e 3 in
the homogeneous extension of R 2 ) to be able to be able to represent linear subspaces at any position,
therefore the position of a flat A in conformal geometric algebra is given by (o 0 • A) · A −1 .
21

Now we only need to determine the orientations of the objects, but what do we actually mean by this?
Consider a flat A located containing the points a 1 , a 2 , . . . , a k , this is A = ι C (a 1 )∧ι C (a 2 )∧. . .∧ι C (a k )∧o ∞ .
Because we have that a ∧ a = 0 for all multivectors a, note that:
A = ι C (a 1 ) ∧ ι C (a 2 ) ∧ . . . ∧ ι C (a k ) ∧ o ∞
(
= o 0 + ∑ a 1,i γ 2 i + 1 )
2 ‖a 1‖ 2 o ∞ ∧
i
(
o 0 + ∑ a 2,i γ 2 i + 1 )
2 ‖a 2‖ 2 o ∞ ∧
i
. . .
(
o 0 + ∑ a k,i γ 2 i + 1 )
2 ‖a k‖ 2 o ∞ ∧ o ∞
i
( ∑
) ( ∑
)
= ι C (a 1 ) ∧ (a 2,i − a 1,i )γ 2 i ∧ . . . ∧ (a k,i − a 1,i )γ 2 i ∧ o ∞
= ι C (a 1 ) ∧ b 2 ∧ . . . ∧ b k ∧ o ∞
i
Where the b j are defined as b j = ∑ i (a j,i −a 1,i )γ 2 i. Clearly the b j can be viewed as the spanning vectors
i
A
a2
b2
a3
b3
a1
Figure 2.4: Orientation of a flat
of the linear space A at position a 1 and therefore it would be sensible to define the orientation of this
object to be b 2 ∧ . . . ∧ b k ∧ o ∞ (we include o ∞ because our usual conception of an orientation is a flat).
But how to extract these vectors b j ? For this we consider a more simple case: A = ι C (a) ∧ b ∧ o ∞ where
b = ∑ i b iγ 2 i. Note that:
( ∑ )
A = ι C (a) ∧ b i γ 2 i ∧ o ∞
=
= ∑ i
(
o 0 + ∑ j
i
) ( ∑ )
a i γ 2 j + ‖a‖ 2 o ∞ ∧ b i γ 2 i ∧ o ∞
b i
(
o 0 ∧ γ 2 i ∧ o ∞ + ∑ j
i
a i γ 2 j ∧ γ 2 i ∧ o ∞
)
If we were to take the inner product with o ∞ (remember that o ∞ • o ∞ = 0 and o ∞ • γ i = 0 for all
2 ≤ i ≤ 2 n ):
o ∞ • A = ∑ (
b i o ∞ • o 0 ∧ γ 2 i ∧ o ∞ + ∑ )
a i γ 2 j ∧ γ 2 i ∧ o ∞
i
j
= (o ∞ • o 0 ) ∑ i
b i γ 2 i ∧ o ∞ + 0
= b ∧ o ∞
22

This is exactly the orientation of A. From this it is easy to see that o ∞ • ( ι C (a) ∧ b 2 ∧ . . . ∧ b k ∧ o ∞
)
=
b 2 ∧ . . . ∧ b k ∧ o ∞ (just add the other b j to the expression above). Therefore the orientation of a flat A
is given by o ∞ • A.
The orientation of an orientation A is much simpler: if A does not yet contain infinity (this is the case
if o 0 • A = 0) the orientation equals A ∧ o ∞ and otherwise (if o 0 • A ≠ 0) the orientation is just A.
For rounds we use the expression for the orientation of a flat. First we convert the given round A to the
flat A ∧ o ∞ (i.e. from a circle through a, b, c we would obtain the plane through a, b, c in which the circle
lies) and calculate the orientation of this flat using the expression we derived above: o ∞ • (A ∧ o ∞ ).
There is one final remark to be made about rounds: they represent objects of a certain finite size (they
do not extend to infinity). Just as with vectors, we see that if we square the dual representation of a
sphere at position a ∈ R n with radius r: A = ι C (a) − 1 2 r2 o ∞ that A · A = r 2 , we retrieve the size of the
sphere squared. This also works for the direct representation (see [4]) if we normalize A before doing so.
So in a nutshell, for a multivector A:
o ∞ • A
=0
Orientation
(2.1)
≠0
o ∞ ∧ A =0
Flat
≠0
Round
Type Position Orientation Size
Orientation - A or A ∧ o ∞ -
Flat (o 0 • A) · A −1 o ∞ • A -
Round A · o ∞ · A −1 o ∞ • (A ∧ o ∞ ) A · A
Remember that the results should be normalized afterwards. These expressions allow us to classify all
multivectors in G n C with a certain grade k.
2.4 Visualizing Euclidian 3-space
We will now use the expressions from the previous paragraph to visualize a given multivector A of grade
k in G 3 C with signature σ(1) = σ(2) = σ(3) = 1 (Euclidian, + + +).
We will assume that we are able to display parameterized objects with our program (i.e. perhaps through
polygonal approximations as with CGAview) and have use of a transformation stack S on which we can
push transformations of the space of the displayed objects. S is also assumed to be resolved last in first
out (i.e. if S = (T 1 , T 2 ) we first apply T 2 and then T 1 ).
Let A ∈ G 3 C be given. If A does not have a certain grade we reject A and do not draw anything, as
it is a linear combination of represented objects and we cannot draw (for example) the sum of a line
and a sphere. If A does have a certain grade k we classify A using scheme 2.1 to determine its type:
orientation, flat or round. Depending on this type we can use table 2.1 to retrieve the properties of A.
If left unspecified, we will assume the position to be (0, 0, 0), the origin, and the size equal to 1. So now
we know the type of object type(A), its grade k, orientation ori(A), position pos(A) and size size(A).
We can assume that the object is positioned at the origin if we push a transformation ‘Translate(0 →
pos(A))’ on the stack S. Now that the object is located at the origin we can assume it has size with
absolute value 1 by pushing ‘Scale( √ |size(A)|)’ onto S if size(A) ≠ 0. This operation scales the entire
space uniformly with √ |size(A)| (note that the size in table 2.1 is given squared, just as with vectors
a · a = ‖a‖ 2 so we have to take the square root). If the size of A equals 0, we do not push the scale
operation (then nothing would remain). What we would like to do now is add a rotation to S such that
23

we can assume A to be oriented along the e 1 axis in R 3 . For this we first have to convert the multivector
ori(A) into a ‘normal’ vector n(A) ∈ R 3 (for a line this is the direction of the line, for a plane this is the
plane normal). It turns out that we can extract this vector directly from the components of ori(A).
Note that if the orientation of A has grade 0 or 1 that it must be of the form (we extended orientations
to infinity in 2.1) α1 or αo ∞ . So for these grades, A does not really have a certain orientation. Therefore
we simply assume it to be n(A) = e 1 . It becomes more interesting if ori(A) has grade 2. Then it is of
the form ( ∑ 3
i=1 a iγ 2 i)
∧ o∞ . Clearly (see the previous paragraph about the spanning vectors b j ) a is
the normal vector we seek. Since
( 3∑ )
a i γ 2 i ∧ o ∞ = √ −1
2
i=1
3∑
a i γ 1 ∧ γ 2 i + a i γ 2 i ∧ γ 2 3+1
i=1
= −1 √
2
3∑
a i γ 1+2 i + a i γ 2 i +16
we see that the third (3 = 1 + 2 1 ) component of ori(A) is equal to −1 √
2
a 1 , the fifth (5 = 1 + 2 2 ) equal
to −1 √
2
a 2
and the ninth (9 = 1 + 2 3 ) equal to −1 √
2
a 3 . So the normal vector simply equals n(A) =
− √ 2(ori(A) 3 , ori(A) 5 , ori(A) 9 ).
Now suppose that ori(A) has grade 3, then it is of the form ( ∑ 3
i=1 a iγ 2 i)
∧
( ∑ 3
j=1 b jγ 2 j
)
∧ o∞ . We can
save ourselves a lot of trouble by noting that n(A) and ori(A) are linear operations and considering three
simple cases. Suppose a = (1, 0, 0) and b = (0, 1, 0), then ori(A) is the orientation of the e 1 -e 2 plane
which has normal (0, 0, 1), on the other hand
ori(A) = γ 2 1 ∧ γ 2 2 ∧ o ∞ = √ 1
)
(γ 1+2
2 1 +2 2 + γ 2 1 +2 2 +16 = √ 1 ( )
γ 7 + γ 22
2
i=1
Similarly a = (1, 0, 0), b = (0, 0, 1), n(A) = (0, −1, 0) yields ori(A) = 1 √
2
(
γ11 + γ 26
)
and a = (0, 1, 0),
b = (0, 0, 1), n(A) = (1, 0, 0) yields ori(A) = 1 √
2
(
γ13 + γ 28
)
. Now use linearity of n(A) (note that
we can swap a and b at the cost of a minus sign, so through taking linear combinations of the cases
we have considered we can really create all possible orientations ori(A) of grade 3) to conclude that
n(A) = √ 2(ori(A) 13 , −ori(A) 11 , ori(A) 7 ).
If ori(A) has grade 4, then A must be of grade 5. Since we are working in a five (3 + 2) dimensional
geometric algebra A has to be a multiple of just γ 31 (the pseudoscalar). Therefore (just as with ordinary
scalars) we say for grade 4 orientations that n(A) = (1, 0, 0). It is impossible that ori(A) has grade 5
(for this would imply that A has grade 6 which is impossible in 5-dimensional geometric algebra) so in
this case we also assign n(A) = (1, 0, 0). Let us summarize this result:
Grade Form Normal vector
0, 1 - (1, 0, 0)
2 a ∧ o ∞ − √ 2(ori(A) 3 , ori(A) 5 , ori(A) 9 )
3 a ∧ b ∧ o ∞
√
2(ori(A)13 , −ori(A) 11 , ori(A) 7 )
4, 5 - (1, 0, 0)
Now that we have this normal vector, we push ‘Rotate((1, 0, 0) → n(A))’ onto S and may assume that
the orientation of A is along the e 1 axis.
So if our transformation stack looks like (do not include the Scale operation if size(A) = 0):
S = (Translate(0 → pos(A)), Scale( √ |size(A)|), Rotate((1, 0, 0) → n(A)))
and we may assume A to be positioned at the origin, have size −1, 0 or +1 and be oriented along e 1 .
The fact that we are now able to make these assumptions will greatly simplify the parameterizations of
the represented objects.
Suppose A has grade k = 4 and is a round, then A is the wedge of 4 = 3+1 vectors (so A = a∧b∧c∧d of
vectors a, b, c, d through ι C ) and must therefore represent a sphere. There are three possibilities, either
size(A) = 0, in which case A represents the set {x ∈ R 3 |‖x‖ 2 = 0} (remember that we can assume A to be
positioned in the origin, have default orientation, etc.), which is just the point (0, 0, 0). If size(A) = −1
24

then the represented object is empty, since ‖x‖ 2 = x 2 1 + x 2 2 + x 2 3 ≥ 0 for all x ∈ R 3 because of our
signature. If size(A) = 1 then the represented object is the unit sphere at the origin: x 2 1 + x 2 2 + x 2 3 = 1.
This object can be parameterized by (use cos(α) 2 + sin(α) 2 = 1 for all α ∈ R):
for 0 ≤ α < π and 0 ≤ β < 2π.
x 1 = cos(α) cos(β), x 2 = cos(α) sin(β), x 3 = sin(α)
Suppose A has grade k = 4 and is a flat, then A = a ∧ b ∧ c ∧ o ∞ which represents the plane through
vectors a, b and c (A can be seen as a sphere a ∧ b ∧ c ∧ d of which ‖d‖ 2 ↑ ∞). We may assume this plane
to be positioned in the origin and have normal vector e 1 , so it is parameterized by:
for −∞ < α, β < ∞.
x 1 = 0, x 2 = α, x 3 = β
Suppose A has grade k = 3 and is a round. Then A = a ∧ b ∧ c for vectors a, b, c. Because of this A lies
within (in terms of linear subspaces represented by wedge expressions in ‘ordinary’ geometric algebra,
there a∧b∧c represents a linear subspace of the subspace represented by a∧b∧c∧d) the plane a∧b∧c∧o ∞
through a, b and c. A also lies within the sphere represented by a ∧ b ∧ c ∧ d for some vector d. So A
should lie within the intersection. The intersection of a plane with a sphere is a circle (for radius > 0),
therefore it is not unreasonable to assume that A represents a circle through a, b and c. If size(A) = 0
then A represents a circle with radius 0, which is just the point (0, 0, 0). If size(A) = −1 then (circles
with negative radii do not exist) A represents the empty set. If size(A) = 1 then A represents a circle
with radius 1. We can assume the orientation of A to be along e 1 , so the circle should lie within a plane
with normal e 1 . It is furthermore located at the origin and has radius 1, so we can parameterize it using:
for 0 ≤ α < 2π.
x 1 = 0, x 2 = cos(α), x 3 = sin(α)
Suppose A has grade k = 3 and is a flat, then A = a ∧ b ∧ o ∞ . This can (just as the grade 4 flat) be seen
as a circle a ∧ b ∧ c of which ‖c‖ 2 ↑ ∞. In the limiting procedure the circle becomes larger and larger
until it represents a line through a and b. This line is positioned at the origin and oriented along e 1 .
Therefore we can parameterize it with:
for −∞ < α < ∞.
x 1 = α, x 2 = 0, x 3 = 0
Suppose A has grade k = 2 and is a round, then A = a ∧ b. This lies within the circle a ∧ b ∧ c for
some vector c and the line a ∧ b ∧ o ∞ . If size(A) = 0 then A represents just (0, 0, 0) (circle consists of
just one point). If size(A) = −1 then A represents the empty set (circle is empty for negative radii). If
size(A) = 1 then A represents a pair of points at the origin, oriented along e 1 : (−1, 0, 0) and (1, 0, 0).
Suppose A has grade k = 2 and is a flat, then A = a ∧ o ∞ . Again this is the limit of a ∧ b for ‖b‖ 2 ↑ ∞,
so it is a pair of points of which one point of the pair is moved to infinity. Therefore A represents just
the point a which is ‘connected’ (as with the point pair) to infinity. We call this a flat point. Since we
may assume A to be positioned at the origin, this is just (0, 0, 0).
Suppose A has grade k = 1 and is a round, then A = a, just a single point (0, 0, 0) (note: just a point a
dually represents a sphere at position a with radius 0, so points really are round objects, just very small
ones).
Suppose A has grade k = 1 and is a flat, then A = o ∞ , this is the point at infinity which does not
represent any subset of R 3 since all elements of R 3 have a finite distance from the origin. Similarly
rounds of grade 0 represent scalars, but these are not subsets of R 3 either and therefore will not be
parameterized.
Note that grade 0 objects (scalars) are always rounds and grade 5 objects are always flats (since they
always include o ∞ ).
We will now summarize these results, for rounds (with size(A) > 0, otherwise it is either the empty set
for size(A) < 0 or just (0, 0, 0) for size(A) = 0):
25

Grade Form Description Parameterization Parameters
0 α scalar - -
1 a point (0, 0, 0) -
2 a ∧ b point pair (±1, 0, 0) -
3 a ∧ b ∧ c circle (0, cos(α), sin(α)) 0 ≤ α < 2π
4 a ∧ b ∧ c ∧ d sphere (cos(α) cos(β), cos(α) sin(β), sin(α)) 0 ≤ α < π, 0 ≤ β < 2π
And flats:
Grade Form Description Parameterization Parameters
1 o ∞ infinity - -
2 a ∧ o ∞ flat point (0, 0, 0) -
3 a ∧ b ∧ o ∞ line (α, 0, 0) −∞ < α < +∞
4 a ∧ b ∧ c ∧ o ∞ plane (0, α, β) −∞ < α, β < +∞
5 a ∧ b ∧ c ∧ d ∧ o ∞ pseudoscalar - -
With transformation stack S:
S = (Translate(0 → pos(A)), Scale( √ |size(A)|), Rotate((1, 0, 0) → n(A)))
Orientations are positioned at infinity and can therefore not truly represent subspaces of R 3 . However,
it is not unreasonable to visualize them as flats (given our definition of orientations in the previous
paragraph) located at the origin using the table for flats above.
There is one more thing I would like to add about Euclidian 3-space. It turns out that (see [4]) rotations
and translations can be done using conjugation with multivectors of the form V = l 1 /l 2 which are called
versors. Here l 1 = ι C (a 1 ) ∧ ι C (a 2 ) ∧ o ∞ and l 2 = ι C (b 1 ) ∧ ι C (b 2 ) ∧ o ∞ represent lines through the
points a1, a2 and b1, b2 in R 3 . Any object in R 3 represented by some multivector x ∈ G 3 C can now be
rotated/translated to x ′ ∈ G 3 C by using x′ = V · x · V −1 . x ′ is equal to x rotated over twice the angle
between the lines l 1 and l 2 and translated over twice the shortest distance between l 1 and l 2 .
This simple expression is yet another example of how the geometrical foundations of our algebra permit
us to do a complex operation (rotation/translation) through the simple use of an object V which has a
clear geometrical interpretation (the ratio of two lines ‘representing’ the rotation/translation).
2.5 Visualizing Hyperbolic 3-space
Things become more challenging if we consider G 3 C with signature: σ(1) = 1, σ(2) = σ(3) = −1 (Hyperbolic,
+ - -). Luckily we have already done a lot of the work necessary to visualize multivectors in the
previous paragraph.
Let A ∈ G 3 C be given and have grade k. Using the same procedure as before, we can determine the type
of A, as well as position, size, orientation and associated normal vector n(A). With the transformation
stack S we can assume that A is positioned at the origin, oriented along e 1 and has size −1, 0 or +1.
Flats can be treated in the exact same way as with the Euclidian signature (identical parameterizations
as the signature does not influence the shape of flats), so for these simply use table for flats from the
previous paragraph. Rounds however change quite a bit with this new signature.
Suppose A is a round and has grade k = 4, so A = a ∧ b ∧ c ∧ d for vectors a,b,c and d. Then we have
three possibilities.
If size(A) = 0, A represents all x ∈ R 3 for which ‖x‖ 2 = x 2 1 − x 2 2 − x 2 3 = 0. Contrary to the Euclidian
case, this is not just (0, 0, 0), but the cone parameterized by:
for −∞ < α < ∞ and 0 ≤ β < 2π.
x 1 = α, x 2 = α cos(β), x 3 = α sin(β)
26

If size(A) = 1, we have x 2 1 − x 2 2 − x 2 3 = 1. This is a hyperboloid which can be parameterized with (use
cosh(α) 2 − sinh(α) 2 = 1 for all α ∈ R):
for 0 ≤ α < ∞ and 0 ≤ β < 2π.
x 1 = ± cosh(α), x 2 = sinh(α) cos(β), x 3 = sinh(α) sin(β)
If size(A) = −1, we have x 2 1 − x 2 2 − x 2 3 = −1, this can be parameterized using:
for −∞ < α < ∞ and 0 ≤ β < 2π.
x 1 = sinh(α), x 2 = cosh(α) cos(β), x 3 = cosh(α) sin(β)
When compared to the Euclidian grade 4 rounds (which were either nothing, a point or a sphere) these
are more complex, but still easy to parameterize. This becomes more challenging if we consider grade 3
Hyperbolic rounds.
Suppose A has grade k = 3, so A = a ∧ b ∧ c. Just as with the Euclidian case we know that A must be
the intersection of a grade 4 round a ∧ b ∧ c ∧ d for some vector d and a plane a ∧ b ∧ c ∧ o ∞ through the
origin. This gives us different possibilities for the represented objects depending on size(A). We may
assume (using the transformation stack S) that A is positioned in the origin, has size −1, 0 or +1 and
normal vector n(A) along e 1 .
Unfortunately, the shape of the intersection depends heavily on the normal vector n(A). With S we can
only assume that A lies within the plane with normal vector e 1 , but not that the intersection of the plane
containing A and the hyperboloid a ∧ b ∧ c ∧ d is, say, an ellipse or a hyperbole. To figure out the shape
of the intersection we rotate our normal vector n(A) around e 1 such that it lies within the e 1 -e 2 plane.
This can freely be done because all grade 4 rounds are invariant under rotations around e 1 . So we add
‘Rotate((x, y, 0) → n(A))’ to our stack S where x = n(A) 1 and y = √ n(A) 2 2 + n(A)2 3 , chosen such that
(x, y, 0) equals n(A) rotated into the e 1 -e 2 plane. Because we added this rotation to S we may assume
that n(A) lies within the e 1 -e 2 plane when determining the shape of the intersection. We also normalize
x and y such that x 2 + y 2 = 1 as only the direction of n(A) matters, not its length.
Figure 2.5: Grade 3 hyperbolic round with size 0
The simplest case is size(A) = 0, here A is the intersection of a plane and a cone (grade 4 round with
size(A) = 0), which for |x| ≤ |y| (see figure 2.7) consists of two lines parameterized by:
x 1 = 0, x 2 = α, x 3 = ±α
for −∞ < α < ∞. For |x| > |y| the intersection is just the point (0, 0, 0).
Figure 2.6: Grade 3 hyperbolic round with size +1
Now consider size(A) = +1. If |x| ≥ |y| then the intersection is empty. If |x| < |y| then the intersection
of the plane containing A and a ∧ b ∧ c ∧ d is a hyperbole, which is stretched as the angle between n(A)
27

and e 1 becomes smaller. Now we view the e 1 -e 2 plane from above. As n(A) lies within this plane and
is the normal vector of the plane containing A, we can view the plane containing A as a line. Because
e_2
y
r
x
r
e_1
Figure 2.7: Hyperbolic warping for size +1
in the e 1 -e 2 plane n(A) has direction (x, y), this line has direction (y, −x). We will call r the smallest
distance from the origin to the intersection of the plane containing A and the hyperboloid a ∧ b ∧ c ∧ d.
Since the closest point of intersection must lie on both the line and the hyperboloid, we must have
(ry, −rx) = (± cosh(α), sinh(α)). Also note that |x| < |y| and x 2 + y 2 = 1 so |y| > 0, therefore we can
divide x by y:
x
y = −−rx ry
= ± sinh(α)
cosh(α) = ± tanh α
So α = ±arctanh(x/y) and with this (the function cosh is even):
|ry| = r|y| = | ± cosh(α)| = cosh(α)
= cosh(±arctanh( x y )) = cosh(arctanh(x y ))
=
√ √ y − x y + x
y + x + y − x
We can now calculate r:
r = 1 ( √ √
y − x y + x
)
2|y| y + x + y − x
Because of the stack S we must give a parameterization for which n(A) = e 1 . Now that we have r this
can easily be done, simply add ‘Scale 2 (r)’ (scale the e 2 axis with r) to S and we may assume that r = 1.
For r = 1 the intersection of the plane containing A with the hyperboloid is the (unstretched) hyperbole
parameterized by:
x 1 = 0, x 2 = ± cosh(α), x 3 = sinh(α)
for −∞ < α < ∞. Technically for r = 1 we have that y = 1 and x = 0, so n(A) = e 2 and we should
parameterize a hyperbole in the e 1 -e 3 plane, but because of the rotation already on our stack S we must
assume that n(A) = e 1 . That is why we set x 1 = 0 for our parameterization and scale in the e 2 direction
instead of e 1 , otherwise our hyperbole is rotated to a wrong orientation by S.
Now consider size(A) = −1. Again we write n(A) = (x, y, 0) for the normal of A. If |x| < |y| then the
intersection of the plane containing A and the hyperboloid a ∧ b ∧ c ∧ d is again a hyperbole and we
28
(2.2)

or
Figure 2.8: Grade 3 hyperbolic round with size -1
treat this case the same as with size(A) = +1. Only the r = 1 parameterization differs (as we intersect
a different hyperboloid because size(A) = −1 instead of 1), now we have
x 1 = 0, x 2 = sinh(α), x 3 = ± cosh(α)
for −∞ < α < ∞. If |x| > |y| the intersection is not empty, but equal to an ellipse, which really is a circle
e_2
intersects ellipse
e_1
intersects
hyperbole
Figure 2.9: Hyperbolic warping for size -1
stretched with r. Through a calculation almost identical to size(A) = +1 (this time p = (ry, −rx) =
(sinh(α), ± cosh(α)), so simply swap x and y in the calculations) we derive:
r = 1 ( √ √
x − y x + y
)
2|x| x + y + x − y
The stack S remains the same as with size(A) = +1 and the parameterization (the intersection is for
r = 1 simply a circle) becomes:
for 0 ≤ α < 2π.
x 1 = 0, x 2 = cos(α), x 3 = sin(α)
Suppose A has grade k = 2, so A = a ∧ b. Then A is the intersection of a line a ∧ b ∧ o ∞ with a grade 3
round a ∧ b ∧ c for some c. This intersection, just as in the Euclidian case, always consists of a point pair.
It is easily seen that the figures above also apply to this case (we intersect a line with a hyperbole/ellipse),
so we must compensate for the point pair being ‘stretched’ with exactly the same amount r as for the
grade 3 rounds. So we add ‘Scale 1 (r)’ (scale e 1 with r because the point pair lies along e 1 ) to our stack
S where we use expressions 2.2 and 2.3 for the cases |x| < |y| and |x| > |y| respectively. Then the point
29
(2.3)

pair can be ‘parameterized’ as (−1, 0, 0) and (+1, 0, 0) because we may now assume that r = 1 and
n(A) = e 1 .
Suppose A has grade k = 1, so A = a. Then A represents a point and we visualize it as (0, 0, 0) just as
with the Euclidian case.
This concludes the classification of objects represented in Hyperbolic G 3 C . Below is a summary of the
parameterizations.
Hyperbolic rounds of grade 4:
S = (Translate(0 → pos(A)), Scale( √ |size(A)|), Rotate((1, 0, 0) → n(A)))
Size Condition Description Parameterization Parameters
-1 - hyperboloid (sinh(α), cosh(α) cos(β), cosh(α) sin(β)) −∞ < α < ∞, 0 ≤ β < 2π
0 - cone (α, α cos(β), α sin(β)) −∞ < α < ∞, 0 ≤ β < 2π
+1 - hyperboloid (± cosh(α), sinh(α) cos(β), sinh(α) sin(β)) 0 ≤ α < ∞, 0 ≤ β < 2π
Hyperbolic rounds of grade 3:
S = (Translate(0 → pos(A)), Scale( √ |size(A)|), Rotate((1, 0, 0) → n(A)), Rotate((x, y, 0) → n(A)), Scale 2 (r))
For x = n(A) 1 , y = √ n(A) 2 2 + n(A)2 3 and r by 2.2 and 2.3 for |x| < |y| and |x| > |y|. If unspecified, the
represented object is empty.
Hyperbolic rounds of grade 2:
Size Condition Description Parameterization Parameters
-1 |x| < |y| hyperbole (0, sinh(α), ± cosh(α) −∞ < α < ∞
-1 |x| > |y| ellipse (0, cos(α), sin(α)) 0 ≤ α < 2π
0 |x| ≤ |y| line pair (0, ±α, α) −∞ < α < ∞
0 |x| > |y| point (0, 0, 0) -
1 |x| < |y| hyperbole (0, ± cosh(α), sinh(α)) −∞ < α < ∞
S = (Translate(0 → pos(A)), Scale( √ |size(A)|), Rotate((1, 0, 0) → n(A)), Rotate((x, y, 0) → n(A)), Scale 1 (r))
With x,y and r the same as before.
Size Condition Description Parameterization Parameters
-1 - point pair (±1, 0, 0) -
0 - point (0, 0, 0) -
1 - point pair (±1, 0, 0) -
Hyperbolic rounds of grade 1 are just the point (0, 0, 0).
30

Chapter 3
Conclusion
We have now classified and parameterized all represented objects in Euclidian and Hyperbolic 3-space.
Together with the implementations of the geometric algebra operations from the first chapter, this should
enable the reader to write a program similar to CGAview or GA viewer, with which conformal geometric
algebra can be visualized and further explored.
Such a program can for example be used to take a look at the use of geometric algebra in physics: easy
object intersection through inner products, rotations (Euclidian signature) or Lorentz transformations
(Hyperbolic signature) through conjugation with the appropriate versors, interpretation of the sum of
(dually) represented objects, simulation of Newtonian/Relativistic dynamics with rotor equations and
much, much more.
Apart from that the reader could also explore the mathematical background of geometric algebra (which
is essentially a Clifford algebra of R n ) and try to prove the geometrical applications of the defined
products. Furthermore the ring G n is bound to have many interesting properties as it contains both
complex and quaternion multiplication through the geometric product.
With these suggestions this paper is concluded and I hope to have given you a satisfactory introduction
to this subject.
Bas Fagginger Auer
3.1 Acknowledgements
I would like to thank Frank Witte for his kind assistance and our stimulating conversations during the
development of CGAview and L. Dorst and D. Fontijne for their previous work concerning geometric
algebra, without which this project could not have progressed quite as much as it did.
31

Appendices
3.2 Binary numbers and operations
Let n ∈ N be an arbitrary number. Usually we write numbers with respect to base-10 (i.e. 324 =
3 · 10 2 + 2 · 10 1 + 4 · 10 0 ), this means that we write n = n k n k−1 . . . n 2 n 1 where the 0 ≤ n i ≤ 9 are
called the digits of the number n, and n = ∑ k
i=1 n i10 i . We can also do this in another basis, say b ∈ N
with b > 1: n = ∑ k
i=1 n ib i . It should be stressed that although the digits will change as we change our
base (both the number of digits and their values), they still represent the same number n. To guarantee
uniqueness of the digit representation of n, we require that (in the representation of n with respect to
basis b) digits lie between 0 ≤ n i < b for all 1 ≤ i ≤ k and that n k ≠ 0 if k > 1. It is easy to see that
now every n is uniquely characterized by a certain number of digits.
Example
Consider the number n = 1023 in base 10 (this is called decimal), we have that 1023 = 1·10 3 +0·10 2 +2·
10 1 +3·10 0 . We can also write n in base 16 (hexadecimal): 1023 = 3·16 2 +15·16 1 +15·16 0 , so n = 3 15 15 =
3FF in hexadecimal (in hexadecimal we use A = 10, B = 11, C = 12, D = 13, E = 14, F = 15 to be able to
write all digits as just one character). Or in base 8 (octal): 1023 = 1·8 3 +7·8 2 +7·8 1 +7·8 0 , so n = 1777
in octal. Or in base 2 (binary): 1023 = 1·2 9 +1·2 8 +1·2 7 +1·2 6 +1·2 5 +1·2 4 +1·2 3 +1·2 2 +1·2 1 +1·2 0 ,
so n = 1111111111 in binary.
We will now write n in base-2: n = n k n k−1 . . . n 2 n 1 . The digits n 1 , n 2 , . . . , n k are called bits and
n k n k−1 . . . n 2 n 1 a k-bit binary number. The bit on the i-th position, n i , can have only two values (since
we are working in base-2): 0 and 1. If n i = 1 we say that the bit on the i-th position or the i-th bit is
set. Otherwise (if n i = 0) we say that the i-th bit is not set or clear.
Example
Let n = 5 then if we write n in binary it is equal to 101. We see that the first and third bits are set,
while the second bit is not.
For binary numbers we can easily define a few operations that work bitwise on their operands. Let
a, b ∈ N and a 1 , a 2 , . . . , a k , b 1 , b 2 , . . . , b k be their binary digits (if they have a different number of bits,
we pad the binary form with the smallest number of bits with 0-valued bits). We define c = a and b as
the binary number c k c k−1 . . . c 2 c 1 where c i = 1 if both bits a i and b i are equal to 1 and c i = 0 otherwise
(it should now be clear why this is called the and operation). Similarly we define c = a or b: c i = 1 if
either a i = 1 or b i = 1 (or both) and 0 otherwise and c = a xor b: c i = 1 if either a i = 1 or b i = 1 (but
not both) and c i = 0 otherwise. This can neatly be summarized in a table (just remember to apply the
operation to every bit of a and b):
a b a and b a or b a xor b
0 0 0 0 0
1 0 0 1 1
0 1 0 1 1
1 1 1 1 0
33

Example Let a = 103 and b = 418, then a = 001100111, b = 110100010 in binary (note that we padded
a with cleared bits to ensure that a and b have an equal number of bits). Using the table above, we see
that a and b = 000100010, a or b = 111100111 and a xor b = 111000101. So a and b = 34, a or b = 487
and a xor b = 453. Note that if we calculate (a xor b) xor b = 00100111 = 103, so (a xor b) xor b = a.
It is not difficult to see (use the table for the xor operation) that this must hold for all a, b ∈ N.
Because bits can only be equal to 0 and 1 we can also uniquely characterize a binary number by the
positions of the bits that are set (since we know for certain that if a bit is not set, that is has to be
clear and cannot assume any other value): a = 10011 = 2 1−1 + 2 2−1 + 2 5−1 , a is characterized by
the bits on positions 1, 2 and 5. More general, let i ∈ N, then we can decompose i in a unique way:
i = 2 i1−1 + 2 i2−1 + . . . + 2 i k−1 with certain 1 ≤ i 1 < i 2 < . . . < i k . Given such a decomposition, we know
that i is an i k -bit number of which k bits are set at exactly the positions i 1 , i 2 , . . . , i k . All unspecified
positions are occupied by cleared bits. This decomposition will be very useful when we start doing base
multivector operations.
There are two more operations that I would like to discuss: the left and right bit shift. Let i = 2 i 1−1 +
2 i 2−1 + . . . + 2 i k−1 be some binary number. Note that if we multiply i by 2, we get 2i = 2 (i 1+1)−1 +
2 (i2+1)−1 + . . . + 2 (i k+1)−1 : all bit positions have shifted one to the left. This is called the left bit shift
of i with 1 bit position. By multiplying i with 2 l for some l ∈ N we get 2 l i = 2 (i1+l)−1 + 2 (i2+l)−1 +
. . . + 2 (i k+l)−1 : all bits are shifted l positions to the left. We write i > l as 2 −l i rounded down: now all bit
positions are shifted to the right l times.
Example Let a = 11001 in binary, then a > 2 = 110, a >> 3 = 11. Furthermore the value 2 l = 1

Bibliography
[1] D. Fontijne and L. Dorst: Performance and elegance of five models of 3D Euclidean geometry in a
ray tracing application, University of Amsterdam 2003
[2] L. Dorst and S. Mann: Geometric algebra: a computational framework for geometrical applications,
IEEE Computer Graphics and Applications, 2002 (part I May/June 2002, part II July/August 2002)
[3] J. Suter: Geometric Algebra Primer, available at http://www.jaapsuter.com/, 2003
[4] L. Dorst and D. Fontijne: 3D Euclidean Geometry Through Conformal Geometric Algebra, University
of Amsterdam 2005
35