Dorst L., Fontijne D., Mann S. Geometric Algebra for Computer Science. An Object Oriented Approach to Geometry
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148 THE FUNDAMENTAL PRODUCT OF GEOMETRIC ALGEBRA CHAPTER 6
process of invention by Clifford. We then claimed it was actually more fundamental than
either. If that is indeed the case, we should be able to start with the geometric product
and define it algebraically without reference to the other products. We do that now, for it
allows us to extend the geometr ic product properly beyond vector terms and use it as the
foundation of our geometric algebra.
6.2.1 ALGEBRAIC DEFINITION OF THE GEOMETRIC PRODUCT
We start with a metr ic vector space R
n
and its linear space of subspaces
R
n
. Its metric is
characterized by a bilinear form Q (see Appendix A), or, e quivalently, by a inner product
of vectors.
We define the geometric product from
R
n
×
R
n
to
R
n
by the following properties:
•
Scalars. The geometric product is an extension of the usual scalar multiplication in
R
n
, so the expressions
αβ and α x, α,β ∈
R
can from now on be read as involving the geometric product. We will explicitly
define this multiplication by a scalar to be commutative with any element A:
α A = A α for all α ∈
R, A ∈
R
n
.
•
Scalar Squares.Thegeometricproductx
2
≡ xxofavectorx with itself is defined to
be a scalar, equivalent to the metric quantity x · x = Q[x,x]. This ties the geometric
product to the metric of the vector space
R
n
.
•
Distributivity and Linearity. The geometric product is defined to be distributive
over the addition of elements:
A (B + C) = AB+ AC and (A + B) C = AC+ BC
This also defines the general linearity of the geometric product (since A can be a
scalar).
•
Associativity. The geometric product is defined to be associative:
A (BC) = (AB) C
so that we may write ABCwithout confusion about the result.
•
Commutativity Not Required!.Thegeometricproductisnot defined to be either
purely commutative or purely anticommutative (although it may be either, for
suitably chosen factors). This is essential, as it permits the geometr ic product to
SECTION 6.2 THE GEOMETRIC PRODUCT OF MULTIVECTORS 149
unite the commutative properties of metric computations with the anticommuta-
tive properties of spanning to produce a product that is complete in its geometric
properties.
Note that our original definition (6.3) of the geometric product as a sum of inner
and outer product is not part of these algebr aic defining properties. We will actually
rederive (6.3), and with it will define the outer product and contraction in terms of the
geometric product. Such a procedure demonstrates that it is indeed the more fundamen-
tal of the three, our order of treatment in this book notwithstanding.
The geometric product makes our algebra of
R
n
into a true geometric algebra, and we
will use that term from now on. It transcends the subspace algebra we have used so far,
and has a much more rich, powerful, and consistent structure. While subspace algebra was
similar to Grassmann-Cayley algebra, geometric algebra closely resembles a nongeomet-
rical mathematical construction called Clifford algebra. The terms are often used inter-
changeably by others, but we will make a distinction. For the moment, you can think of
geometric algebra as the geometrically significant par t of Clifford algebra. We will be able
to make this distinction more precise in Section 7.7.2.
6.2.2 EVALUATING THE GEOMETRIC PRODUCT
Since the above are all the properties of the geometric product, they should enable the
expansion of arbitrary expressions of the geometric product of multivectors. Let us do
some of this before we proceed—we will develop faster techniques to compute with
the geometric product, but it is important to realize that these definitions are indeed
complete.
The completeness is most easily shown when we have an orthonormal basis for the metric
vector space
R
n
. We should demonstrate that we can compute the geometric product
of any two basis vectors. This should of course agree with our special case for vectors,
computed in Section 6.1.4, but we are not allowed to use those. We should work fully
from the algebraic definition just given.
The geometric product of a basis vector with itself is easy, since by definition it can be
expressed in terms of the bilinear form or the inner product that is part of the definition
of the metric vector space:
e
i
e
i
= e
i
· e
i
= Q[e
i
, e
i
]
But the geometric product of two different basis vectors e
i
e
j
does not appear to allow
simplification by the axioms:
e
i
e
j
= ?
We know from the vector computations that it cannot be simplified, but we would at
least like to be able to show that e
j
e
i
= −e
i
e
j
to be in correspondence with the geometric
product as we defined it earlier for vectors.
150 THE FUNDAMENTAL PRODUCT OF GEOMETRIC ALGEBRA CHAPTER 6
There is a neat trick called polarization that comes to the rescue. The bilinear form or
inner product of the metric vector space can be evaluated on any two vectors. The bilinear
nature gives an identity for Q[x, y] or x·y that can be manipulated into a sy mmetric shape:
x · y =
1
2
(x + y) · (x + y) − (x · x) − (y · y)
=
1
2
(x + y)(x + y) − (xx) − (yy)
=
1
2
(xy+ yx) (6.9)
Therefore the inner product x · y of two general vectors is the symmetric part of their
geometric product. We are thus able to derive part of our motivating definitions of
Section 6.1.2 from the algebraic definition above.
This symmetry property gives us the idea to manipulate our different basis vectors by
splitting the product in its symmetric and antisymmetric parts:
e
i
e
j
=
1
2
(e
i
e
j
+ e
j
e
i
) +
1
2
(e
i
e
j
− e
j
e
i
)
= e
i
· e
j
+
1
2
(e
i
e
j
− e
j
e
i
)
= 0 +
1
2
(e
i
e
j
− e
j
e
i
),
since in the orthonormal basis e
i
· e
j
= 0 for i = j. It follows that
e
i
e
j
= −e
j
e
i
.
So the new definition indeed permits us to derive this important property of the multi-
plication of basis vectors.
With the multiplication of the basis elements established, we can use associativit y,
linearity, and distributivity to compute the geometric product of any two elements in
the linear space
R
n
.
6.2.3 GRADES AND THE GEOMETRIC PRODUCT
At this point you may object that we have not really shown that the geometric product,
constructing elements starting from a metric vector space
R
n
, really generates the same
linear structure of elements of different grades that we had before. We are therefore for-
mally not allowed to write
R
n
for the space in which geometric algebra lives.
That is a correct objection; but if the spaces are not the same, they are certainly
algebraically isomorphic, and that is good enough to identify them geometrically. The
reason is that the orthonormal basis of the vector space
R
n
leads to a basis for the prod-
uct structure that satisfies e
i
e
j
= −e
j
e
i
. That is the essence in generating the higher-
grade elements. This property is identical to the outer product antisymmetric property
e
i
∧ e
j
= −e
j
∧ e
i
. The identity of these properties means that we can use the geometric
product to at least faithfully reconstruct the basis of the ladder of subspaces
R
n
that we
originally made using the outer product.
SECTION 6.3 THE SUBSPACE PRODUCTS RETRIEVED 151
And even though the geometric product has richer properties than the outer product,
we cannot make other elements beyond the ladder of subspaces. Consider for example
e
1
(e
1
e
2
) as an attempt to make something new. Had we used the outer product, the
construction e
1
∧ (e
1
e
2
) = e
1
∧ (e
1
∧ e
2
) would have been zero. For the geometric product,
the result is not zero, but it reverts to something we already have:
e
1
(e
1
e
2
) = e
1
e
1
e
2
= (e
2
1
) e
2
,
so this is ±e
2
, depending on the metric. You can generalize this argument to show that
nothing beyond the elements of
R
n
can be made; the scalar squares foil any such
attempt. Therefore, the geometric product of a metric vector space
R
n
“lives” in precisely
the same structure
R
n
as the outer product of the same space R
n
.
However, this analysis brings out an important difference between the geometric product
and the outer product. When multiplying the extended basis elements of grade k and
grade l by the outer product, we are left with a single element of grade k + l (or zero).
With the geometr ic product, the product of two basis elements of gr ade k and l may have
any of the grades
|k − l|, |k − l| + 2,..., (k + l − 2), (k + l).
The highest grade (k + l) occurs when all basis vectors in the elements are different. (The
geometric product is then essentially the same as the outer product of those elements.)
But each vector in common between the two basis elements reduces the grade by two as it
combines to produce a scalar. The extreme case is when all the vectors in one are contained
in the other, leaving only |k − l| factors as a result. (The geometric product is then the left
or right contraction of one argument onto the other.)
If we now have arbitrary elements A
k
and B
l
of grade k and l, respectively, these can be
decomposed on the bases of
k
R
n
and
l
R
n
. When we multiply them using the geo-
metric product, any or all of the possible grades between |k − l| and (k + l) may occur.
Therefore the geometric product produces multivectors of mixed grade.Thegrade() opera-
tion no longer has a single integer value in geometric algebra.
The algebraic invertibility of the geometric product can now be understood in principle.
The series of terms in the geometric product of the two elements A
k
and B
l
apparently
give us a complete inventory of their relative geometric relationship, allowing full recon-
struction of one when we are given the other.
6.3 THE SUBSPACE PRODUCTS RETRIEVED
The geometric product is the fundamental product in geometric algebra—you will
not need any other product, since it contains all geometric relationships between
its arguments. Yet we have seen that the subspace products (by which we mean the
outer product, scalar product, and contraction) are also useful geometrically. In fact,
152 THE FUNDAMENTAL PRODUCT OF GEOMETRIC ALGEBRA CHAPTER 6
the whole geometrical concept of subspace requires the outer product to be encoded
in our algebra.
Since we want to have those products to “do geometry,” we should show that the y are
included in our geometric algebra based on the geometric product. There are two routes:
•
We could use the symmetries of the geometric product to retrieve outer product
and contraction (basically reversing the construction that motivated the geomet-
ric product in Section 6.1.2). This is actually only partly successful. It does not
define the subspace products fully, but it does show that they are consistent with
the symmetry structure of the geometric product. When we perform the analysis
in Section 6.3.1 below, we obtain many useful relationships between the various
products.
•
We can identify the subspace products of blades as certain well-defined grades of
their geometric product. This indeed defines them fully, though it gives us less alge-
braic insight in their relationships. We do this in Section 6.3.2.
In the practice of applying the subspace products, both approaches are useful. Depending
on the geometrical problem that one tries to solve computationally, either may feel like
the more direct route. That is why we present both.
6.3.1 THE SUBSPACE PRODUCTS FROM SYMMETRY
The familiar outer product of a vector a with a blade B can be related to the geometric
product by the following two expressions:
a ∧ B =
1
2
(aB+
Ba), (6.10)
B ∧ a =
1
2
(Ba+ a
B), (6.11)
where
B = (−1)
grade(B)
B is the grade involution of B introduced in Section 2.9.5. (Writing
the equations in this form makes it easier to lift them to general multivectors B.)
The proof of these statements may be found in Section C.1 of Appendix C. It demonstrates
that the two equations above indeed identify the same outer product structure that we
had before, at least when one of the factors is a vector, and it proves the associativity
(x ∧ y) ∧ z = x ∧ (y ∧ z) of the product thus defined. Because of that associativity, this
outer product in terms of the geometric product can be extended to general blades, and
by linearity to general multivectors. Only the case of two scalars is formally not included,
but other than that this outer product is isomorphic to the outer product we had before.
The contractions can be related to the geometric product in similar fashion when they
involve a vector factor a:
aB =
1
2
(aB−
Ba), (6.12)
Ba =
1
2
(Ba− a
B). (6.13)
SECTION 6.3 THE SUBSPACE PRODUCTS RETRIEVED 153
The proof of these statements may be found in Section C.2 of Appendix C.
Unfortunately, because of lack of associativity, we cannot prove the defining equation
(3.11) A(BC) = (A ∧ B)C (nor its counterpart for the right contraction). Neither
can we define the contraction result on two scalar arguments in this manner. So although
the products defined by (6.12) and (6.13) are consistent with the earlier contractions, the
contractions based on the symmetries of the geometric product are not pinned down
very precisely. This algebraic freedom partly explains the variation in inner products in
the geometric algebra literature exposed in Appendix B.
But at least for a vector factor, the constructions above define the subspace products
uniquely. Conversely, this means that the geometric product of a vector with an arbitrary
multivector can be decomposed using the contraction and the outer product.
aB= aB + a ∧ B,
(6.14)
Ba=
Ba +
B ∧ a = −aB + a ∧ B, (6.15)
where we used (3.19) to convert the right contraction to a left contraction. These equa-
tions subsume and generalize (6.3).
The subspace product definitions permit us to change the order of multiplications in a
geometric product, which is often convenient in evaluating expressions:
Ba= aB− 2aB = −aB+ 2a ∧ B
(6.16)
and
aB=
Ba+ 2 aB = −
Ba+ 2 a ∧ B.
(6.17)
In all these equations, you may note that right-multiplication of B by a is
always
accom-
panied by a gra de involution the formulas become simplest or most symmetrical when
you define them in terms of aB, aB, and a ∧ B, combined with
Ba,
Ba (= −aB), and
B∧a (= a ∧ B). This grade involution is apparently the natural geometric sign when mov-
ing a vector to the right of a blade. We will see this phenomenon reappear throughout the
equations of geometric algebra.
You may be puzzled by a paradox in the associativity of the various products. According to
(6.14), the geometric product is the sum of the outer product and contraction. Both the
geometric product and the outer product are associative, whereas the contraction is not.
How can the sum of something associative and something nonassociative ever be asso-
ciative itself ? The solution is: don’t look at it that way. Instead, start with the geometric
product, which is defined to be associative. Then derive the outer product as (6.10) (i.e.,
as half the sum of a geometric product and its swapped version). This is associative since
addition is associative. Then derive the contraction as (6.12) (i.e., as half the difference of
a geometric product and its swapped version). This is nonassociative, since subtraction is
nonassociative. Now there is no paradox.
By linearity, all equations in this section are easily extended from a blade B to a general
multivector B.
154 THE FUNDAMENTAL PRODUCT OF GEOMETRIC ALGEBRA CHAPTER 6
6.3.2 THE SUBSPACE PRODUCTS AS SELECTED GRADES
An alternative way of obtaining the subspace products from the geometric product is as
parts of well-chosen grades using the k-grade selection operator
k
. For the geometric
product on vectors, this is simple:
a · b = ab
0
and a ∧ b = ab
2
.
(6.18)
This generalizes as follows:
A
k
∧ B
l
≡A
k
B
l
k+l
(6.19)
A
k
B
l
≡A
k
B
l
l−k
(6.20)
A
k
B
l
≡A
k
B
l
k−l
(6.21)
A
k
∗ B
l
≡A
k
B
l
0
(6.22)
Blades of negative grade are zero—so the left contraction is zero when k>l, and the
right contraction is zero when k<l. By linearity of the geometric product, all these
definitions can be lifted from blades to k-vectors and then to multivectors as a sum over
the appropriate grades.
In contrast to the symmetry-based approach, these equations are complete definitions for
all arguments. The proofs that these equations give the same products we had in the earlier
chapters may be found in Section C.3 of Appendix C. It involves some new manipulation
techniques that are useful to study.
A surprising property of these definitions is that the selection of certain grades of the
geometric product of blades apparently produces another blade. Beware that this does
not generalize to the selection of other grade parts!
Once these correspondences with the old subspace products have been established, some
of the properties of the subspace products can be used to simplify grade-based expressions
and vice versa. For instance, the symmetry property A ∗ B = B ∗ A of the scalar product
can be easily lifted to multivectors as A ∗ B = B ∗ A. This implies
AB
0
= BA
0
(6.23)
which is a useful cyclic reordering property of the grade-0 symbol.
The grade approach to the subspace products is a feasible way of implementing all
products in geometr ic algebra based on a single implemented product (the geometric
product). This is explored in programming exercise 6.7.1.
6.4 GEOMETRIC DIVISION
With the integration of subspace products with the geometric product, we have a more
powerful algebra to analyze subspaces. We now combine our new capability of div ision
SECTION 6.4 GEOMETRIC DIVISION 155
by a subspace with the earlier techniques. This not only generalizes the projection, but it
also produces the compact representation of a new construction: subspace reflection.
6.4.1 INVERSES OF BLADES
The geometric product is invertible, so dividing by a multivector has a unique meaning,
equivalent to multiplication by the inverse of the multivector. However, not all multivec-
tors have inverses. Fortunately, we are mostly interested in two kinds: blades, and multi-
vectors that can be written as a product of invertible vectors. The latter are called versors,
and we treat them in Chapter 7; they are obviously invertible (their inverse is formed by
the inverses of the vector factors, in reverse order).
Blades are also invertible, if the y have a nonzero norm (i.e., if they are not null-blades; see
Appendix A). The inverse of a blade A is then
inverse of a blade A : A
−1
=
A
A ∗ A
=
A
A ∗
A
=
A
A
2
(6.24)
where
A is the reverse of Section 2.9.5. This formula is based on the property that
the squared norm of a blade is a scalar, which makes the division well defined and
unambiguous (since a scalar commutes with the geometric product, its right-division
and left-division coincide). Its validity is most easily verified using an orthogonal fac-
torization of the blade A as a product of orthogonal vectors: A = a
1
a
2
···a
k
. Such
a factorization can be made for invertible blades by the Gram-Schmidt procedure of
Section 6.7.2. By computing the geometric product A
A vector by vector, you find that
it is equal to A ∗
A, so that the inverse formula is indeed correct.
This inverse of a blade is unique, by the associativity argument also used in Section 6.1.5.
6.4.2 DECOMPOSITION: PROJECTION TO SUBSPACES
We can express a vector x triviallyrelativetoaninvertiblebladeA as x = x (AA
−1
).
Moving the brackets by associativity and invoking (6.14), we get a pleasantly suggestive
rewriting. Let us first explore this for a 1-blade A, the vector a.
x = (xa) a
−1
= (x · a + x ∧ a) a
−1
= (x · a) a
−1
+ (x ∧ a) a
−1
(6.25)
The first term in (6.25) is a vector (since it is a scalar times a vector). We recognize it as
(x · a) a
−1
= (x · a
−1
) a = (xa
−1
)a; that is, as the orthogonal projection of x onto a (see
Section 3.6). We can now write it as a division:
projection of x onto a: (xa)/a.
156 THE FUNDAMENTAL PRODUCT OF GEOMETRIC ALGEBRA CHAPTER 6
It is the component of x in the a-direction. The second term in (6.25) must then be the
component of x that contains no a-components at all (since the two terms must add to
produce x). We follow Hestenes [33] in calling this the rejection of x by a:
rejection of x by a: (x ∧ a)/a
We can imagine its construction visually as in Figure 6.3: span the bivector x ∧ a. This is
a reshapeable area element, and it is equivalent to a rectangle perpendicular to a spanned
by some vector r perpendicular to a. That rectangular element can be written as the outer
product r ∧ a, but because r is perpendicular to a (implying that r · a = 0), we can even
wr ite it as a geomet ric product:
x ∧ a = r ∧ a = r · a + r ∧ a = ra,
This rewriting is helpful because the geometric product is invertible; that makes this equa-
tion for r solvable. Through right-division by a on ra = x ∧ a, we obtain the solution
r = (x ∧ a)/a, which is indeed the rejection of x by a.
W
e thus
see that the identity x = (xa)/a, when written out in terms of the inner product
and the outer product, is actually a decomposition of the vector x relative to a, providing
its a-component and non-a-component. It offers us the possibility of describing a vector
relative to another vector, but does so in a satisfyingly coordinate-free manner.
Returning to the general blade A and again invoking (6.14), we find the decomposition
x = (xA) A
−1
= (xA) A
−1
+ (x ∧ A) A
−1
.
(6.26)
(x
a) /a
x ∧ a
(x ∧ a) /a
a
x
Figure 6.3: Projection and rejection of x relative to a.
SECTION 6.4 GEOMETRIC DIVISION 157
You may expect that the first term is a 1-blade fully contained in A, and that it should be
equal to the projection of x onto A.
projection of x onto A :(xA)/A.
We have encountered this before (in Section 3.6) as (xA)A
−1
,whichcommuteswith
all products. But structural exercise 7 explores why we can replace the contraction by a
geometric division in this formula.
The second term is again called the rejection of x by A,
rejection of x by A: (x ∧ A)/A,
since it is a vector perpendicular to A. To prove that fact compactly, we combine the sub-
space products and the grade selection of the geometric product (to be frank, it took us
about an hour to make it this simple). If (x ∧ A) A
−1
is perpendicular to A, it should be
perpendicular to any vector in A. Let us pick one, a, and compute the inner product:
a ·
(x ∧ A) A
−1
= a (x ∧ A) A
−1
0
=
1
2
axAA
−1
+ a
AxA
−1
0
=
1
2
ax− AaxA
−1
+ 2(a ∧
A) xA
−1
0
=
1
2
ax− axA
−1
A − 0
0
=
1
2
ax− ax
0
= 0.
Identify the steps we took—they are all based on formulas in this chapter and you will
see an instructive example of how the grade approach and the symmetry approach to the
subspace products can be combined. In the rejection, we can substitute the geometric
product in (x ∧ A) A
−1
by a right contraction (see structural exercise 8).
Although we can replace the geometric products by contractions in both projection and
rejection, there is not necessarily an advantage in doing so. The geometric product is
invertible, and this often helps to simplify expressions, so that would plead in favor of
leaving it. On the other hand, the contraction helps remind us of the containment rela-
tionships (subspaces taken out of other subspaces), and makes it easier to apply duality
relationships to convert the subspace products.
Since projection and rejection are linear transformations, we can extend them from vec-
tors to general blades as outermorphisms (and even to multivectors, by linearity). For the
projection, we have done this before in Section 4.2.2, and we derived that it boils down to
substituting the general blade X for the vector x, to obtain
projection of X onto A : X → (XA) A
−1
.
(6.27)
However, the outermorphic extension of the rejection quickly disappoints, since i t
becomes a rather trivial operation (although indeed linear). For (X ∧ A) A
−1
is zero