Dorst L., Fontijne D., Mann S. Geometric Algebra for Computer Science. An Object Oriented Approach to Geometry

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SECTION 4.2 OUTERMORPHISMS: LINEAR TRANSFORMATIONS OF BLADES 107

rather than about linear transformations, and that the determinant is just a property of

a square matrix rather than a fundamental property of a linear transformation. But it

is just that, and it can be deﬁned without referring to matrices. We have just done so in

(4.7), even managing to avoid coordinates altogether. We brieﬂy show how we can use this

geometrical approach to retrieve the determinants of some common transformations.

•

Determinant of a Rotation. The example of the rotation in the Euclidean plane

indicated by the blade u ∧ v demonstrated that R[u ∧ v] = u ∧ v. Since u ∧ v is

proportional to the pseudoscalar I

of the plane, this implies

R[I

] = I

Therefore the determinant of a 2-D rotation equals 1.

If the rotation plane is embedded within an n-dimensional Euclidean metric space

n,0

, then we can span a pseudoscalar for the n-dimensional embedding space

using I

combined with (n − 2) vectors perpendicular to the plane. Each of those

vectors is not affected by the rotation, so for the part of the space they span we

have R[I

n−2

] = I

n−2

(where I

n−2

is a pseudoscalar for the (n − 2)-dimensional

space), by the outermorphism property. We thus ﬁnd: R[I

] = R[I

∧ I

n−2

] =

R[I

] ∧ R[I

n−2

] = I

∧ I

n−2

= I

. So the determinant of a rotation still equals 1,

even in an n-dimensional space (n ≥ 2).

•

Determinant of a Point Reﬂection. We have seen that a point reﬂection satisﬁes

f[I

] = (−1)

. Thus its determinant equals 1 in even dimensions and −1 in odd

dimensions. This suggests that in even dimensions, a reﬂection can be performed

as a rotation, and indeed it can.

•

Determinant of a Projection onto a Line. The projection P onto a line has a deter-

minant that varies with the dimensionality of the space

.Wehaveseenhowany

blade with grade exceeding 1 becomes zero. Therefore any pseudoscalar of

with

n>1 isprojectedto0,anddet(P) = 0. However, for n = 1 the line must necessarily

be the whole space

. Now the projection is the identity, so det(P) = 1.

We can continue the theme of determinants. Applying two linear transformations

f :

→ R

and g : R

→ R

,ﬁrstf then g, we obtain a composite transforma-

tion that is again linear (as you can easily show) and that can therefore also be extended

as an outermorphism. We denote this composite transformation by (g◦f).Wecompute

its determinant:

det(g◦f) I

= (g◦f)[I

] = g[f[I

]] = det (f) g[I

] = det (g) det(f) I

Therefore, we get the composition rule of determinants:

det(g◦f) = det(g) det(f).

(4.8)

This is a well-known result, derived within this context of the outer product in a straight-

forward algebraic manner with satisfying geometrical semantics.

108 LINEAR TRANSFORMATIONS OF SUBSPACES CHAPTER 4

4.3 LINEAR TRANSFORMATION OF THE METRIC

PRODUCTS

The linear transformation f has been extended to an outermorpishm to transform an

outer product completely naturally as

f[A ∧ B] = f[A] ∧f[B].

Of course, we should also understand how the scalar product and the contraction trans-

form; those three combined will then enable us to transform arbitrary geometric com-

positions and will give us the full extent of linear transformations in our subspace

algebra.

The scalar product is easily transformed, but the contraction takes a bit more work and

requires us to introduce an important concept: the adjoint of a linear transformation.

4.3.1 LINEAR TRANSFORMATION OF THE SCALAR PRODUCT

The scalar product returns a scalar, so it transforms under a linear transformation as

f[A ∗ B] = A ∗ B. (4.9)

This looks straightforward, but when you remember that A ∗



A is the squared norm of

the blade A, does this now mean that no linear transformation can change the norm? Or,

a related question since A ∗



B is proportional to the cosine of the angle between A and

B, does this mean that no linear transformation can change the angle? That would imply

that all linear transformations are orthogonal transformations.

Resolve these problems for yourself—or take a hint from structural exercise 6.

4.3.2 THE ADJOINT OF A LINEAR TRANSFORMATION

The transformation f[AB] for the contraction AB will follow from our deﬁnitions so

far, but can only be formulated compactly when we introduce an additional construction:

the adjoint of f. We do this ﬁrst.

The adjoint

f of the linear transformation f :



→



is a linear transformation

f :



→



, deﬁned implicitly for vectors by the equation

adjoint transformation :

f[a] ∗ b = a ∗f[b] (4.10)

for all a and b of

. In nondegenerate metrics, this deﬁnes it fully. In degenerate met-

rics, we have similar incompleteness issues as for (3.6)—you can then use an explicit,

coordinate-based formula. This is explored in structural exercise 9.

It is strange to have the deﬁnition so implicit. To show that the adjoint is actually a familiar

concept from linear algebra, we momentarily convert the equation to matrix notation.

SECTION 4.3 LINEAR TRANSFORMATION OF THE METRIC PRODUCTS 109

Let [[f]] be the matrix of the mappingf and [[ f]] be the matrix off, and convert the inner

product to a matrix product. Vectors like a are represented as a column matrix [[ a]] .By

transferring (4.10) to matrix notation in the case of a Euclidean orthonormal basis and

by using the matrix transpose, we obtain

[[ a]]

[[f]] [[ b]] = ([[

f]] [[ a]] )

[[ b]] = [[ a]]

[[ f]]

[[ b]] ,

so [[

f]] = [[f]]

. This implies that for vectors in a Euclidean orthonormal basis, the adjoint

of f is the transpose mapping, speciﬁed in a coordinate-free manner.

For blades (and even for multivectors), we extend

f as an outermorphism. This leads to

adjoint transformation :

f[A] ∗ B = A ∗f[B], (4.11)

which we could have taken as the deﬁnition of the adjoint for blades. For general f,it

follows easily from the symmetry of the scalar product that

f = f,

since

f[A] ∗ B = A ∗f[B] = f[A] ∗ B, for all b. Another useful property is

−1

= f

−1

which is easily shown from the deﬁnition (4.10).

Some examples:

•

Iffis the uniform scaling deﬁned byf[x] = α x, thenf[x]·a = x·(α a) = (α x) ·a for

all x and a. This yields

f[x] = α x, so in this casef = f, the adjoint equals the original

transformation. As an outermorphism on blades, the adjoint is also

f[X] = f[X].

•

A special case of the uniform scaling is the point-reﬂection into the origin, which

has α = −1. Again,

f = f, but now also f

−1

= f.

•

In Figure 4.2, we met the line projection P[x] = a (x∧b)/(a∧b) = a (x∧b)(a∧b )

−1

Its adjoint is

P[x] = (x · a) b(a ∧ b)

−1

. This, therefore, is proportional to the dual

of b in the ( a ∧ b) plane.

4.3.3 LINEAR TRANSFORMATION OF THE CONTRACTION

To demonstrate how the contraction product transforms under linear transformations,

we ﬁrst show

f[A] B ] = Af[B], (4.12)

which we derive simply using the scalar product deﬁnition:

X ∗ (Af[B]) = (X ∧ A) ∗f[B]

f[X ∧ A] ∗ B

110 LINEAR TRANSFORMATIONS OF SUBSPACES CHAPTER 4

= (f[X] ∧f[A]) ∗ B

f[X] ∗ (f[A] B)

= X ∗f[

f[A] B ].

f is invertible (which happens precisely when f is invertible), we can deﬁne A



−1

[A]. Substituting that in (4.12) and dropping the prime gives the promised trans-

formation law for the contraction product:

contraction transformation : f[AB] =

−1

[A] f[B]. (4.13)

By the linearity of the functions and operators involved, this is of course immediately

extendable to arbitrary multivectors as f[AB] =

−1

[A]f[B].

The result of (4.13) is very powerful, but unfortunately rather abstract. We have not found

an easy geometric picture that will convince you of its necessary truth. But the interpre-

tation of the contraction AB as “the part of B that remains when A is taken out in a per-

pendicular manner” (to paraphrase) helps remember the placement of f and its derived

mappings. Obviously, the transformed result is a part of B, so it should transform asf[B];

and taking A out could explain the inversef

−1

; doing so in an orthogonal manner justiﬁes

the adjoint

4.3.4 ORTHOGONAL TRANSFORMATIONS

If a linear transformation of vectors preserves their inner product, we call it an orthogonal

transformation. It then satisﬁes

orthogonal transformation : f[a] ·f[b] = a · b, for all a, b ∈

Since orthogonal transformations are invertible, we may set a = f

−1

[x], to obtain

x ·f[b] = f

−1

[x] · b for all x, b. It follows that

f = f

−1

so for an orthogonal transformation, the adjoint equals the inverse transformation. You

probably knew this fact from linear algebra, in terms of the inverse of an orthogonal matrix

is its transpose. In the present context, “inverse equals adjoint” is a statement about the

mappings rather than their matrices, and by outermorphism, it also holds for the outer-

morphism extension of the mapping to blades.

With this, the transformation formula of the contraction is much simpler when f is an

orthogonal transformation:

f[AB] = f[A] f[B].

Therefore, for orthogonal transformations, the contraction transforms in a structure-

preserving manner: the contraction of the transformed blades is the transformation

SECTION 4.3 LINEAR TRANSFORMATION OF THE METRIC PRODUCTS 111

of the contraction. Orthogonal transformations are thus “innermorphisms” as well as

outermorphisms (since the contraction is actually an inner product for blades).

Familiar examples are reﬂections and rotations. In fact, they are the prototypical

orthogonal transformations, and we will discover in Chapter 7 that the general case can

always be written as a rotation followed by a reﬂection (or vice versa).

4.3.5 TRANSFORMING A DUAL REPRESENTATION

ThedualofabladeX is X

∗

≡ XI

−1

. When X undergoes a linear transformation f,

the dual is transformed as well. We should deﬁne the transformation f

∗

of the dual by

demanding that it preserve the duality relationship: the f

∗

-transform of the dual should

be the dual of the f-transformed X. In formula:

dual transformation : f

∗

] ≡ (f[X])

∗

That speciﬁes it. We introduce D = X

∗

, and ﬁnd

∗

[D] = (f[D

−∗

])

∗

= f[DI

]I

−1

= (

−1

[D] f[I

]) I

−1

= f

−1

[D] ∧ (f[I

] I

−1

)

= det(f)

−1

[D], (4.14)

as the necessary deﬁnition of the linear transformation on duals. Since this is not gen-

erally equal to f[D], it implies that blades that are intended as dual representations do

not transform in the same way as blades that are intended as direct representations. In

a proper representation of geometry, we therefore need to indicate how a blade is to be

interpreted before we can act on it appropriately with a linear transformation.

Note that the outermorphism transformation law of direct blades is nonmetric, since it

only involves the outer product. By contrast, f

∗

= det(f)

−1

is metric, since it is express-

ible in terms of the adjoint

f whose deﬁnition involves the scalar product. This makes

sense, since in our algebra of subspaces dualization is a metric concept.

An orthogonal transformation has a determinant equal to ±1 (see structural exercise 10),

and has

−1

= f. For such transformations, the dual representation therefore transforms

rather nicely: f

∗

] = ±f[X

∗

]. For rotations, which have determinant +1, this is com-

pletely structure-preserving: the dual of the transform is the transform of the dual. For

rotations it is therefore not necessary to know whether a blade is a dual or direct rep-

resentation before you can transform it. For orthogonal transformations containing a

reﬂection, which have determinant −1, the extra minus sign in the dual is caused by the

reﬂection of the pseudoscalar of the space. It now makes a difference whether you desire

to take the dual of the transform relative to the original pseudoscalar (then you need the

−1), or relative to the transformed pseudoscalar (then there is no sign).

112 LINEAR TRANSFORMATIONS OF SUBSPACES CHAPTER 4

4.3.6 APPLICATION: LINEAR TRANSFORMATION

OF THE CROSS PRODUCT

As we discussed in Section 3.7, the cross product is an inherently 3-D construct, and

we can replace all its uses by elements of our subspace algebra that generalize to

n-dimensional space:

a × b = (a ∧ b)

∗

= (a ∧ b)I

−1

It is interesting to derive an additional argument as to why the cross product should not

be used: the normal vector a×b of a and b, used as a characterization of the homogeneous

plane spanned by a and b, behaves rather awkwardly under a linear transformation f.For

its deﬁnition clearly shows that it is a dual representation of a 2-blade, and therefore it

should transform according to (4.14):

a × b → det(f)

−1

[a × b]. (4.15)

This is not equal tof[a] ×f[b], so the normal vector of the transforms does not equal the

transform of the normal vector.

In computer graphics and engineering applications, it is customary to use “vector” as a

synonym for a tuple of coordinates as well as for a geometrical 1-D direction. The three

coefﬁcients of n = a × b are a vector in the sense of a 3-tuple, but the geometry of

n makes this vector transform unlike a geometrical direction, so it is called a normal

vector. Even though it is given on the same basis as a regular vector, it requires its own

methods to be transformed. The transformation of such a normal vector under the linear

transformation with matrix [[f]] on regular vectors must explicitly be implemented as the

matrix det(f)[[

−1

]] = det(f)[[f]]

−T

,where[[f]]

−T

is the inverse of the transposed matrix.

Within the code, such a different transformation for seemingly similar 3-tuples can only

be invoked if their data type is kept explicitly. Even then, novices easily get confused; this

is a common source of programming error. We provide a programming exercise at the

end of this chapter to explore the differences.

In our subspace algebra, we have two choices. We can introduce normal vectors, which

then of course need to transform according to f

∗

= det(f)

−1

. They do so automatically

when we deﬁne them explicitly in the code. Or we can just characterize the same quantity

using the original 2-blade a ∧ b, which transforms according to the outermorphism f

as f[a ∧ b] = f[a] ∧f[b]. As a characterization of either a plane or a rotation (through

its rotation plane), the 2-blade is just as admissible as the classical normal vector, and

now you don’t need to remember how it transforms. This transformational simplicity

reinforces the practical considerations of Section 3.7, and reafﬁrms our decision to drop

the cross product from sound practice in geometrical representation.

SECTION 4.4 INVERSES OF OUTERMORPHISMS 113

4.4 INVERSES OF OUTERMORPHISMS

With the transformation formula for the contraction, we can derive a closed form,

coordinate-free formula for the inverse of an outermorphism. First, we compute

f[AI

−1

] = f

−1

[A] f[I

−1

] = det (f) f

−1

[A] I

−1

Since det(

f) = det(f) I

−1

∗ I

= f[I

−1

] ∗ I

= I

−1

∗f[I

] = det(f) I

−1

∗ I

= det(f),we

can substitute det(

f) by det(f). Now do the contraction of both sides on I

, giving

−1

[A] =

f[AI

−1

] I

detf

(4.16)

In terms of duals, (4.16) reads

−1

[A]

∗

f[A

∗

]

detf

Mathematicians may object to the use of duality in this formula: taking the inverse of a

mapping does not necessarily require a metric space, whereas duality is of course a metric

concept. It turns out that the two dualities in (4.16) cancel each other, in the sense that

the result does not depend on the precise form of the metric any more, and is therefore

actually nonmetric. We feel that if you have a metric (as you often do), you should use it.

If you don’t have one, you can temporarily introduce a convenient one (e.g., Euclidean)

and compute on.

Here are some examples of evaluation of the inverse on various blades:

1. Pseudoscalar.

−1

] =

f[1] I

detf

showing that det(f

−1

) = (detf)

−1

2. Scalar.

−1

[α] =

f[αI

−1

] I

detf

α det

f (I

−1

I

)

detf

α det

detf

= α,

as expected since the inverse is also a linear transformation.

3. Vectors.

−1

[a] =

f[aI

−1

] I

detf

2 When rereading this book, you may want to replace the contractions in formula (4.16) by the geometric product

of Chapter 6.

114 LINEAR TRANSFORMATIONS OF SUBSPACES CHAPTER 4

For vectors there is no simpliﬁcation of the basic equation, but when you follow

this computation for the matrix off

−1

on some basis, you ﬁnd that you have essen-

tially the minor-based construction of the inverse from classical linear algebra (see

structural exercise 15). But remember, that construction only applies to f acting on

vectors, whereas (4.16) is much more powerful because it also applies to the outer-

morphism extension on arbitrary blades and multivectors.

4.5 MATRIX REPRESENTATIONS

A matrix is a convenient way of representing a linear transformation relative to a given

basis. It is speciﬁc to that basis, and therein lies its strength (it is efﬁcient) and its weakness

(you cannot tell what it does directly from its form). Matrices are used in implementa-

tions of geometric algebra, though usually hidden from the user. You do not need them

to specify your linear transformations, but they are an efﬁcient implementation of the

speciﬁed transformation at the very lowest level.

We introduce our notation for matrices, and show how they can be written completely in

terms of the products from subspace algebra to convey their geometric speciﬁcity. Then

we construct the matrix of the outermorphism of a linear transformation f acting on all

multivectors of



4.5.1 MATRICES FOR VECTOR TRANSFORMATIONS

If you have a basis {b

}

i=1

(not necessarily orthonormal) for the vector space R

, you can

use this to deﬁne a matrix representation of a linear transformation f of that space. This

can then be used to transform arbitrary vectors, since they can be written as a weighted

sum of basis vectors, over which f distributes.

We deﬁne a column representation of the vector x, in which the element in the j

row of

the column is labeled by a superscript as [[ x]]

. Then

[[ x]]

= x

= x · b

using the reciprocal frame from Section 3.8. We deﬁne the matrix through speciﬁcation

of its element ( j , i), which is the j-coordinate of the transformed b

[[f]]

≡f[b

] · b

(4.17)

For this matrix, the superscript index j labels the row of matrix, and the subscript index i

labels the column (so it is consistent with viewing a vector as a matrix with 1 column and

n rows). Equation (4.17) implies that the i

column of the matrix is the image of the i

basis vector b

under the transformation f.

SECTION 4.5 MATRIX REPRESENTATIONS 115

The transformation of an arbitrary vector x can be composed from the matrix

contributions as

f[x] =



i=1

f[(x · b

) b

] =



j=1



i=1

(x · b

)(f[b

] · b

) b



j=1



i=1

[[f]]

[[ x]]

Taking the inner product with b

and using b

· b

= δ

(according to (3.32)), the b

coefﬁcient of the transformed vector is

[[f[x]]]

= f[x] · b



i=1

[[f]]

[[ x]]

We have thus retrieved the familiar multiplication of a vector by a matrix, which may be

denoted in shorthand over all rows as

[[f[x]]] = [[f]] [[ x]] .

Writing out (4.17) using the techniques of (3.32), we ﬁnd the matrix element fully

expressed in geometric algebra (remember that I

= b

∧···∧b

, and

means that

is omitted):

[[f]]

= (−1)

j−1

f[b

]



∧···∧

∧···∧b

)I

−1



= (−1)

j−1

(f[b

] ∧ b

∧···∧

∧···∧b

)I

−1

= (b

∧···∧b

j−1

∧f[b

] ∧ b

j+1

∧···∧b

)I

−1

(4.18)

So the recipe to construct the matrix element is that f[b

] takes the place of b

in the ﬁrst

factor of the expression I

I

−1

The general expression of (4.18) denotes explicitly which geometrical concepts are

involved in the deﬁnition of the matrix of a transformation: a particular basis (not nec-

essarily orthonormal) and a particular choice of pseudoscalar for the space. The latter

is automatically determined once one speciﬁes the basis vectors in the desired order, as

= b

∧···∧b

. But (4.18) demonstrates that there is a rather involved object at the basis

of the common way of doing linear algebra. This has advantages in its compactness; but

taking it fully as the basis of expression, one loses the power to express simpler concepts

when those would sufﬁce.

4.5.2 MATRICES FOR OUTERMORPHISMS

A linear mapping can be represented as a matrix, and this holds not only for the original

mapping on vectors, but also on the outermorphism mappings on each of the k-blades

(or k-vectors). This follows the same procedure as for vectors, but now on a basis for the





-dimensional space



of k-blades (or k-vectors) in R

116 LINEAR TRANSFORMATIONS OF SUBSPACES CHAPTER 4

We ﬁrst need to establish a basis and a reciprocal basis for that space. Let us illustrate

the principle for bivectors; the generalization is straightforward. If the basis of the vector

space is {b

}

i=1

, then we can take as our basis 2-blades:

≡ b

∧ b

There is no natural order among them (except perhaps in 3-D, where you can take them

in order of the missing index: {b

}, with cyclically varying indices). You should

see ij as a single index running through its





possible values. We denote that by a capital

index, so a general bivector would be expressible as X =



, with appropriately

chosen coefﬁcients X

Inametricspace

, the reciprocal basis corresponding to the basis {b

} is then

constructed using the reciprocal basis vectors:



∧ b



∼

The reversion makes the scalar product between the basis vectors behave as if it were an

orthonormal basis under the scalar product (or, if you prefer, the contraction):

∗ b

= b

∗ b

= (b

∧ b

) ∗ (b

∧ b

) = (b

∗ b

)(b

∗ b

) = 1,

and the scalar product is zero for unequal indices. Therefore

∗ b

= δ

The coefﬁcient X

of a bivector X on the basis {b

} is then X ∗ b

, so that we can

write

X =



(X ∗ b

) b

With this preparation, the matrix of the outermorphism of f can be deﬁned by complete

analogy with the vector case as

[[f]]

= f[b

] ∗ b

This is enough to compute the matrix. If the linear transformation fis expensive to evalu-

ate, you may prefer a form that expresses this matrix directly in terms of the vectorsf[b

Let us expand the general term, reverting to the speciﬁc vector indices:

[[f]]

= f[b

] ∗ b

= (f[b

] ∧f[b

]) ∗ (b

∧ b

)

= (f[b

] ∗ b

)(f[b

] ∗ b

) − (f[b

] ∗ b

)(f[b

] ∗ b

)

= [[f]]

[[f]]

− [[f]]

[[f]]

By either method, the outermorphism matrix can be constructed from the matrix acting

on the basis vectors. In any implementation, this trivial computation of the outermor-

phism should be hidden from the user, who should simply be allowed to apply the linear