Chapter 3. The impact of product innovation on firm growth using a multi-stage model: Evidence in a
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The outer product operation ∧ can be defined by the antisymmetry ei∧ej= −ei∧ej, i, j = 0, 1, 2, 3, ∞, independent of the inner product. Hence, we can consider the Grassmann algebra of Chapter 5 in this space and define the “direct representation” of geometric objects in the sense that their equations have the form p ∧ ( · · · ) = 0. In the following, we derive direct representations of lines, planes, spheres, circles, and point pairs. It turns out that circles and spheres are the most fundamental objects and that lines and planes are interpreted to be circles and spheres of infinite radius passing through infinity.
8.3.1 Direct representations of lines
In the 4D homogeneous space of the preceding chapter, a line L passing through two points p1 and p2 was represented by p1∧ p2. In the 5D conformal space, we now show that it becomes
L = p1∧ p2∧ e∞. (8.21)
As in Chapters 5 and 7, we mean by this that its equation is written in terms of point p of Eq. (8.8) in the form
p ∧ L = 0. (8.22)
This is shown as follows. Letting pi = e0+ xi+ kxik2e∞/2, i = 1, 2, we see that p ∧ L = (e0+ x +1
2kxk2e∞) ∧ (e0+ x1+1
2kx1k2e∞) ∧ (e0+ x2+1
2kx2k2e∞) ∧ e∞
= (e0+ x) ∧ (e0+ x1) ∧ (e0+ x2) ∧ e∞. (8.23) Note that the last term ∧e∞of L annihilates the symbol e∞in the expressions of p1and p2. Since e∞is orthogonal to e1, e2, and e3, it is linearly independent of the first three factors.
Hence, p ∧ L = 0 implies
(e0+ x) ∧ (e0+ x1) ∧ (e0+ x2) = 0 (8.24) in the 4D homogeneous space considered in the preceding chapter, which confirms that Eq. (8.21) is the direct representation of line L.
We see that, thanks to the the expression ( · · · ) ∧ e∞in Eq. (8.21), all the results in the 4D homogeneous space of the preceding chapter hold as if the symbol e∞ did not exist . The existence of the extra term ∧e∞in Eq. (8.21) is a natural consequence of our interpretation that a line passes through the infinity e∞. In fact, Eq. (8.22) is satisfied for p = p1, p2, e∞
by the antisymmetry of the outer product (Fig. 8.4). By a similar logic, the line passing through p and extending in the direction u is represented by
L = p ∧ u ∧ e∞. (8.25)
If we let q = e0+ y + kyk2e∞/2, the equality p ∧ q = 0 implies that p is a scalar multiple of q. Hence, p ∧ q = 0 is the “equation of point q.” This means that Eq. (8.8) is the direct representation of point p. At the same time, it is its dual representation, as pointed out at the end of Sec. 8.2.2.
Grassmann algebra in conformal space 123
L p3
Π p2
p1
eοο
FIGURE 8.4 The line L passing through two points p1and p2and the plane Π passing through three points p1, p2, and p3. Both pass through the infinity e∞, so they have respective representations L
= p1∧ p2∧ e∞and Π = p1∧ p2∧ p3∧ e∞, and their equations have the form p ∧ L = 0 and p ∧ Π
= 0, respectively.
8.3.2 Direct representation of planes
By the same logic as above, the direct representation of the plane Π passing through three points p1, p2, and p3 is
Π = p1∧ p2∧ p3∧ e∞, (8.26)
and its equation is written in terms of the point p in Eq. (8.8) as
p ∧ Π = 0. (8.27)
The last term ∧e∞ indicates that a plane passes through the infinity e∞, and Eq. (8.27) is satisfied for p = p1, p2, p3, e∞ (Fig. 8.4). Thanks to the last term ∧e∞ of Eq. (8.26), all the results in the preceding chapter hold for planes, as pointed out earlier. For example, the plane that passes through points p1and p2and contains the direction u is represented by
Π = p1∧ p2∧ u ∧ e∞. (8.28)
This can also be written as Π = −L ∧ u in terms of the line L of Eq. (8.21). Similarly, the plane that passes through point p and contains directions u and v has the representation
Π = p ∧ u ∧ v ∧ e∞, (8.29)
which can also be written as Π = −L ∧ v in terms of the line L of Eq. (8.25).
8.3.3 Direct representation of spheres
For pi = e0+ xi+ kxik2e∞/2, i = 1, 2, 3, 4, representing four positions xi in 3D, the expression
Σ = p1∧ p2∧ p3∧ p4 (8.30)
represents the sphere that passes through them (Fig. 8.5(a)), and
p ∧ Σ = 0 (8.31)
is its equation. This can be shown as follows. Expanding the left side, we see that p ∧ Σ = (e0+ x +1
2kxk2e∞) ∧ (e0+ x1+1
2kx1k2e∞) ∧ (e0+ x2+1
2kx1k2e∞)
∧(e0+ x3+1
2kx3k2e∞) ∧ (e0+ x4+1
2kx4k2e∞)
= ( · · · )e0∧ e1∧ e2∧ e3∧ e∞, (8.32)
p3
Σ
p2
p1
p4 S
p1 p
2
p3
(a) (b)
FIGURE 8.5 (a) The sphere Σ passing through four points p1, p2, p3, and p4has the representation Σ = p1∧ p2∧ p3∧ p4. Its equation is p ∧ Σ = 0. (b) The circle S passing through three points p1, p2, and p3has the representation S = p1∧ p2∧ p3. Its equation is p ∧ S = 0.
where ( · · · ) is a linear expression in x and kxk2. If we let this expression be 0, we obtain the equation of a sphere, and Eq. (8.31) is automatically satisfied if p coincides with any of pi due the properties of the outer product. Thus, Σ represents the sphere that passes through the four points pi, i = 1, 2, 3. Note that if we let p4 = e∞, Eq. (8.30) reduces to Eq. (8.26). This fact provides the interpretation that a plane is a sphere of infinite radius passing through the infinity e∞.
8.3.4 Direct representation of circles and point pairs
For pi = e0 + xi + kxik2e∞/2, i = 1, 2, 3, representing three positions xi in 3D, the expression
S = p1∧ p2∧ p3 (8.33)
represents the circle that passes through them (Fig. 8.5(b)), and
p ∧ S = 0 (8.34)
is its equation. The derivation is a little complicated, but this can be intuitively understood as follows:
It is clear that the object S defined by Eq. (8.34) passes through the three points p1, p2, p3 from the properties of the outer product. As in the case of Eq. (8.32), expansion of the left sides leads to the form
p ∧ S = ( · · · )e1∧ e2∧ e3∧ e∞+ ( · · · )e0∧ e2∧ e3∧ e∞+ ( · · · )e0∧ e3∧ e1∧ e∞
+( · · · )e0∧ e1∧ e2∧ e∞+ ( · · · )e0∧ e1∧ e2∧ e3, (8.35) where ( · · · ) are all linear in x and kxk2. The equation p ∧ S = 0 means p ∧ p1∧ p2∧ p3= 0, implying no sphere that passes through the four points p, p1, p2, and p3 exists. Hence, point p should be on the plane passing through p1, p2, and p3. Thus, S is on a plane and specified by linear equations in x and kxk2. This implies that S is an intersection between a sphere and plane, i.e., a circle. If we let p3 = e∞, Eq. (8.33) reduces to Eq. (8.21). This fact provides the interpretation that a line is a circle of infinite radius passing through the infinity e∞.
Now, consider the two-point version of Eq. (8.33), i.e.,
p1∧ p2. (8.36)
This is a point pair, which can be regarded as a low-dimensional sphere (Fig. 8.6(a)). This is understood by the reasoning that a sphere (2D sphere) is “the set of points equidistant
Dual representation 125
p2
p1
p
eοο
(a) (b)
FIGURE 8.6 (a) A point pair {p1, p2} is regarded as a low-dimensional sphere and hence is repre-sented by p1∧ p2. (b) A flat point p such as the intersection of a plane and a line is a point pair with the infinity e∞and hence is represented by p ∧ e∞.
from a point in 3D” and a circle (1D sphere) is “the set of points equidistant from a point in 2D.” Hence, a point pair, i.e.,“the set of points equidistant from a point in 1D” is a 0D sphere. If one of the two points is replaced by e∞, the point pair p∧e∞represents one point, but this is geometrically distinguished from an isolated point p (= a sphere of radius 0) given by Eq. (8.8) and called a flat point . A flat point p ∧ e∞appears as an intersection of a plane and a line and an intersection of three planes. This is because lines and planes all pass through the infinity e∞, so their intersections should necessarily contain e∞(Fig. 8.6(b)).
8.4 DUAL REPRESENTATION
The volume element of the 5D conformal space is
I5= e0∧ e1∧ e2∧ e3∧ e∞, (8.37) and dual expression is given by
( · · · )∗= −( · · · ) · I5. (8.38) From the rule of the contraction (֒→ Sec. 5.3 of Chapter 5), we observe the following (֒→
Eq. (5.78) in Chapter 5):
(p ∧ ( · · · ))∗= p · ( · · · )∗. (8.39) Hence, if p ∧ ( · · · ) = 0, then p · ( · · · )∗ = 0, and if p · ( · · · ) = 0, then p ∧ ( · · · )∗ = 0 (֒→
Eq. (5.79) in Chapter 5).
In the following, we consider the explicit form of the dual representations for planes, lines, circles, point pairs, and flat points. We will see that the outer product ∧ means “join”
for direct representations and “meet” for dual representations. To distinguish them, we use uppercase letters for direct representations and lowercase letters for dual representation.
8.4.1 Dual representation for planes
Consider the plane Π passing through three points pi = e0+ xi+ kxik2e∞/2, i = 1, 2, 3:
Π = p1∧ p2∧ p3∧ e∞
= (e0+ x1+1
2kx1k2e∞) ∧ (e0+ x2+1
2kx2k2e∞) ∧ (e0+ x3+1
2kx3k2e∞) ∧ e∞
= (e0+ x1) ∧ (e0+ x2) ∧ (e0+ x3) ∧ e∞. (8.40)
As shown in the preceding chapter, this can be written in terms of the Pl¨ucker coordinates in the following form (֒→ Eq. (7.25) in Chapter 7):
Π = (n1e0∧ e2∧ e3+ n2e0∧ e3∧ e1+ n3e0∧ e1∧ e2+ he1∧ e2∧ e3) ∧ e∞. (8.41) Recalling that contraction is defined by consecutive inner products “from inside” with al-ternating signs (֒→ Sec. 5.3 in Chapter 5), we see that the dual of e0∧ e2∧ e3∧ e∞is, from its definition,
(e0∧ e2∧ e3∧ e∞)∗= −e0∧ e2∧ e3∧ e∞· e0∧ e1∧ e2∧ e3∧ e∞
= −e0∧ e2∧ e3· (e∞· e0∧ e1∧ e2∧ e3∧ e∞)
= −e0∧ e2∧ e3· he∞, e0i ∧ e1∧ e2∧ e3∧ e∞
= e0∧ e2∧ e3· e1∧ e2∧ e3∧ e∞= e0∧ e2· (e3· e1∧ e2∧ e3∧ e∞)
= e0∧ e2· e1∧ e2∧ he3, e3ie∞= e0∧ e2· e1∧ e2∧ e∞= e0· (e2· e1∧ e2∧ e∞)
= −e0· e1∧ he2, e2ie∞= −e0· e1∧ e∞= e1he0, e∞i = −e1. (8.42) Similarly,
(e0∧ e3∧ e1∧ e∞)∗= −e2, (e0∧ e1∧ e2∧ e∞)∗= −e3. (8.43) The dual of e1∧ e2∧ e3∧ e∞ is obtained also by consecutive contraction “from inside” in the form
(e1∧ e2∧ e3∧ e∞)∗= −e1∧ e2∧ e3∧ e∞· e0∧ e1∧ e2∧ e3∧ e∞
= −e1∧ e2∧ e3· (e∞· e0∧ e1∧ e2∧ e3∧ e∞)
= −e1∧ e2∧ e3· (he∞, e0ie1∧ e2∧ e3∧ e∞) = e1∧ e2∧ e3· e1∧ e2∧ e3∧ e∞
= e1∧ e2· (e3· e1∧ e2∧ e3∧ e∞) = e1∧ e2· e1∧ e2∧ he3, e3ie∞
= e1· (e2· e1∧ e2e∞) − e1· e1∧ he2, e2ie∞= −e1· e1∧ e∞= −he1, e1ie∞
= −e∞. (8.44)
Hence, the dual of Eq. (8.41) is
Π∗= −n1e1− n2e2− n3e3− he∞= −(n + he∞), (8.45) which agrees with the dual representation π in Eq. (8.13) except for the sign. Since the conformal space is homogeneous, sign change or scalar multiplication does not affect the representation. By similar calculations, it is shown that the dual Σ∗ of the sphere Σ in Eq. (8.30) agrees with the dual sphere representation σ in Eq. (8.18) up to sign and scalar multiplication.
8.4.2 Dual representation for lines
Instead of specifying two points as in Eq. (8.21), a line can be defined as the intersection of two planes. Consider two planes Π1 and Π2. As pointed out in Sec. 7.6.3 in Chapter 7, their intersection Π1∩ Π2 has its dual π1∪ π2(= π1∧ π2), where πi= Π∗i, i = 1, 2. Hence,
l = π1∧ π2 (8.46)
should be the dual representation of the intersection of the two planes. We now confirm this. Using Eq. (8.13) as the dual representation of the plane, let πi = ni+ hie∞, i = 1, 2,
Dual representation 127 where we noted that hp, πii = hni, xi−hifrom Eq. (8.15). Since the intersection exists when the surface normals n1and n2 are linearly independent, the above expression vanishes only when the coefficients of n1, n2, and e∞ are all 0. Hence, p · l = 0 implies hni, xi = hi, i = 1, 2, i.e., x satisfies the equation of the two planes, meaning that it is on their intersection.
Also, the coefficient of e∞ in Eq. (8.47) should be 0. Since the supporting plane of the intersection line has surface normal n = h2n1− h1n2(֒→ Eq. (2.96) in Chapter 2), we have hn, xi = 0 (֒→ Eq. (2.62) in Chapter 2). Thus, p · l = 0 is the equation of the intersection line, and Eq. (8.46) is its dual representation.
We can also use Eq. (8.16) to define dual representations of planes. If we let π1= p1−p2
and π2= p2− p3, Eq. (8.46) gives the dual representation of their intersection, which is the line perpendicular to the triangle △p1p2p3 passing through its circumcenter.
8.4.3 Dual representation of circles, point pairs, and flat points
A circle can be defined as the intersection of two spheres. We can write the dual represen-tations of the two spheres as σ1 = ci− r2ie∞/2, i = 1, 2, in the form of Eq. (8.18). Then, we obtain the dual representation of the intersection circle,
s = σ1∧ σ2, (8.48)
in the same way we define a line as the intersection of two planes. To confirm this, we compute p · s. From Eq. (8.19), we see that Since the centers c1 and c2 of the two spheres are distinct, they are linearly independent, i.e., one is not a scalar multiple of the other. Hence, their coefficients are both 0, and we obtain kx − cik2 = ri2, i = 1, 2, which means that x is on both spheres, defining a circle.
Hence, Eq. (8.48) is its dual representation. Instead of regarding a circle as the intersection of two spheres, we can regard it as the intersection of a sphere σ and a plane π (= a sphere of infinite radius) to define it by σ ∧ π.
Similarly, a point pair is regarded as the intersection of a circle S and a sphere σ or a plane π (= a sphere of infinite radius), and hence its dual representation is given by s ∧ σ or s ∧ π, where s = S∗ is the dual representation of the circle S. Using the same logic, we can regard a flat point whose direct representation is p ∧ e∞as the intersection of a plane Π and a line L and hence its dual representation is given by π ∧ l, where π (= Π∗) and l (= L∗) are the dual representations of the plane Π and the line L, respectively. A flat point can also be regarded as the intersection of three planes, so we can also write its dual representation in the form π1∧ π2∧ π3, where πi (= Π∗i), i = 1, 2, 3, are the dual representations of the three planes Πi.
Table 8.1 summarizes the direct and dual representations in the conformal space de-scribed above. Due to the duality theorem in Sec. 7.6 in Chapter 7, the outer product ∧
TABLE 8.1 Representations in the conformal space. Here, π and πi are the dual representations of planes, l the dual representation of a line, σ and σithe dual representations of spheres, and s the dual representation of a circle.
object direct representation dual representation (isolated) point p = e0+ x + kxk2e∞/2 p = e0+ x + kxk2e∞/2
line p1∧ p2∧ e∞ π1∧ π2
p ∧ u ∧ e∞
plane p1∧ p2∧ p3∧ e∞ n+ he∞
p1∧ p2∧ u ∧ e∞ p1− p2 p ∧ u1∧ u2∧ e∞
sphere p1∧ p2∧ p3∧ p4 c − r2e∞/2
circle p1∧ p2∧ p3 σ1∧ σ2
σ ∧ π
point pair p1∧ p2 s ∧ σ
s ∧ π
flat point p ∧ e∞ π ∧ l
π1∧ π2∧ π3
equation p ∧ ( · · · ) = 0 p · ( · · · ) = 0
indicates join for direct representations and meet for dual representations:
(direct representation) ∧ (direct representation) = (direct representation of their join), (dual representation) ∧ (dual representation) = (dual representation of their meet).