100 Homogeneous Space and Grassmann–Cayley Algebra
Hence, this can be regarded as the “equatio n” of the line L. From Eq. (7.12), we have
p ∧ L = (e
0
+ x) ∧ (e
0
∧ m − n
∗
) = −e
0
∧ n
∗
+ x ∧ e
0
∧ m + x ∧ n
∗
= −e
0
∧ (x ∧ m + n
∗
) + x ∧ n
∗
, (7.15)
so the equation p ∧ L = 0 is equivalent to
x ∧ m = −n
∗
, x ∧ n
∗
= 0. (7.16 )
The dual of the former is (x ∧ m)
∗
= n, but the expression (x ∧ m)
∗
is nothing but the
vector product x ×m (֒→ Eq. (5.73) in Chapter 5). The latter is equivalent to x ·n = 0 (֒→
Eq. (5.79) in Chapter 5). Hence, in terms of the notation of Chapter 2, we can equivalently
write Eq. (7.16) as follows (֒→ Eqs. (2.60) and (2.65) in Chapter 2 ):
x × m = n, hx, ni = 0. (7.17)
This implies that m is the direction vector of the line L and n is the surface normal to the
supporting plane of the line L.
7.3.3 Computation of a line
We now co nsider how to compute the Pl¨ucker coordinates m and n of the line L passing
through points p
1
and p
2
. The bive c tor express ion of L is
L = p
1
∧ p
2
= (e
0
+ x
1
) ∧ (e
0
+ x
2
) = e
0
∧x
2
+ x
1
∧ e
0
+ x
1
∧ x
2
= e
0
∧ (x
2
− x
1
) + x
1
∧x
2
. (7.18)
Comparing this with Eq. (7.12), we find that m = x
2
− x
1
and −n
∗
= x
1
∧ x
2
. The dual
of the latter is n = x
1
× x
2
. Hence, we obtain the following Pl¨ucker coordinates:
m = x
2
− x
1
, n = x
1
× x
2
. (7.19)
Since the Pl¨ucker coordinates are homogeneous coordinates, this result is equivalent to
Eq. (2.68) in Chapter 2.
Next, let L be the line that passes through point p = e
0
+ x and ha s direction u. This
line can be regarded as passing through the finite point p and the point u at infinity. Hence,
can write
L = p ∧ u = (e
0
+ x) ∧ u = e
0
∧ u + x ∧ u. (7.20)
Comparing this with Eq. (7.12), we see that m = u and −n
∗
= x ∧ u. The dual of the
latter is n = x × u. Hence, we o btain the following Pl¨ucker coordinates:
m = u, n = x × u. (7.21)
This coincides with Eq. (7.19) if we let u = x
2
− x
1
.
Finally, consider the line L
∞
passing through two points u
1
and u
2
at infinity:
L
∞
= u
1
∧ u
2
. (7.22)
Such a line is called a line at infinity. Comparing Eq . (7.22) with E q. (7.12), we s e e that
m = 0 and −n
∗
= u
1
∧ u
2
. The dual of the latter is n = u
1
× u
2
. Hence, the Pl¨ucker
coordinates are
m = 0, n = u
1
× u
2
. (7.23)
Pl¨ucker coordinates of planes 101
No finite points are included here. In fact, no finite x exists that satisfies Eq. (7.17) for m
= 0. However, n is the surface normal to the plane spanned by u
1
and u
2
. Hence, the line
L
∞
passing through points u
1
and u
2
at infinity is interpreted to be the “boundary” of the
plane spanned by the directions u
1
and u
2
, surrounding it like a “circle” of infinite radius.
The above results are summarized as follows:
Proposition 7.1 (Description of lines)
1. The line passing through two points p
1
and p
2
is represented by L = p
1
∧ p
2
, where
one or both of p
1
and p
2
can be at infinity.
2. A point at infinity can be identified with the direction toward it.
3. The equation of t he line L has the form p ∧ L = 0.
7.4 PL
¨
UCKER COORDINATES OF PLANES
In this section, we show the following facts about planes in 3D:
• The plane Π passing through three points p
1
, p
2
, and p
3
is repre sented by a trivector
in the 4D homogeneous space in the form Π = p
1
∧ p
2
∧p
3
.
• The equation of the plane Π has the form p ∧ Π = 0.
• A plane is specified by its Pl¨ucker coordinates n a nd h.
• The plane that passes through two given points and contains a given direction is
interpreted to be the plane connecting those two points and the point at infinity
corresponding to that direction.
• The plane that passes through a given point and contains two given directions is
interpreted to be the plane connecting that point and the two points at infinity cor-
responding to those two directions.
• The plane passing through three points at infinity has no direction and is interpreted
to be a plane located infinitely far away.
7.4.1 Representation of a plane
Consider a plane Π passing through three points at x
1
, x
2
, and x
3
(the position vectors
in 3D). In the 4D homogeneous space, these points are represented by p
1
= e
0
+ x
1
, p
2
=
e
0
+ x
2
, and p
3
= e
0
+ x
3
, and the plane Π is represented by their trivector
L = p
1
∧p
2
∧ p
3
. (7.24)
This is based on the following interpretation. From the definition of the o uter product, the
trivector p
1
∧ p
2
∧ p
3
defines a space, i.e., a 3D subspace , spanned by the 4D vectors p
1
,
p
2
, and p
3
. However, we are unable to perceive the entire 4D space; all we can see is its 3D
cross section c ontaining the origin e
0
, with which the 3D subspace p
1
∧ p
2
∧ p
3
intersects
along the plane Π. Since Π is a trivector, it has the following expression with respect to the
basis {e
0
, e
1
, e
2
, e
3
}:
Π = n
1
e
0
∧ e
2
∧ e
3
+ n
2
e
0
∧ e
3
∧ e
1
+ n
3
e
0
∧ e
1
∧e
2
+ he
1
∧ e
2
∧ e
3
. (7.25 )
102 Homogeneous Space and Grassmann–Cayley Algebra
The coefficients n
i
, i = 1, 2, 3, a nd h are called the Pl¨ucker coordinates of this plane. Due to
the homogeneity of the space, p
1
, p
2
, and p
3
represent the same points if they are multiplied
by any nonzero number. Hence, only the ratios among n
i
and h have a geometric meaning.
The Pl¨ucker coordinates of a plane are homogeneous coordinates in this s e nse.
Let n = n
1
e
1
+ n
2
e
2
+ n
3
e
3
by regarding the Pl¨ucker coordinates n
i
as a vector. Then,
Eq. (7.25) is written as
Π = −e
0
∧n
∗
+ hI, (7.26)
where we let n
∗
= −n · I, a bivector dual to n in 3D with I = e
1
∧ e
2
∧ e
3
the volume
element in 3D.
7.4.2 Equation of a plane
A point p = e
0
+ x is on the plane Π = p
1
∧ p
2
∧ p
3
if and only if p ∧ p
1
∧ p
2
∧ p
3
= 0,
namely,
p ∧ Π = 0. (7.27)
Hence, this can be regarded as the “equatio n” of the plane Π. Fr om Eq. (7.2 6), we have
p ∧ Π = (e
0
+ x) ∧ (−e
0
∧ n
∗
+ hI) = −x ∧ e
0
∧ n
∗
+ he
0
∧I
= e
0
∧ x ∧ n
∗
+ he
0
∧ I = e
0
∧ (x ∧ n
∗
+ hI) = (h − hn, xi)e
0
∧I, (7.28)
where we have noted that x ∧n
∗
= −hn, xiI from Eq. (7.13). Hence, the equation p ∧Π =
0 is equivalent to
hn, xi = h . (7.2 9)
As we saw in Chapter 2, the vector n is the surface normal to this plane and h/ knk is the
distance of this plane from the origin e
0
.
7.4.3 Computation of a plane
We now consider how to compute the Pl¨ucke r coordinates n and h of the plane Π passing
through points p
1
, p
2
, and p
3
. The trivector expression of Π is
Π = p
1
∧p
2
∧ p
3
= (e
0
+ x
1
) ∧ (e
0
+ x
2
) ∧ (e
0
+ x
3
)
= e
0
∧ x
2
∧x
3
+ x
1
∧ e
0
∧ x
3
+ x
1
∧ x
2
∧ e
0
+ x
1
∧ x
2
∧ x
3
= e
0
∧ (x
2
∧ x
3
+ x
3
∧x
1
+ x
1
∧ x
2
) + x
1
∧ x
2
∧ x
3
. (7.30)
Comparing this with Eq. (7.26), we find that −n
∗
= x
2
∧x
3
+ x
3
∧x
1
+ x
1
∧x
2
and hI =
x
1
∧x
2
∧x
3
. Replacing both sides with their duals and r e c alling I
∗
= 1 and (x
1
∧x
2
∧x
3
)
∗
= |x
1
, x
2
, x
3
|, we obtain the following Pl¨ucker coo rdinates (֒→ Eqs. (5.64) and (5.73) of
Chapter 5):
n = x
2
× x
3
+ x
3
× x
1
+ x
1
× x
2
, h = |x
1
, x
2
, x
3
|. (7.31)
This is equivalent to Eq. (2.53) in Chapter 2 except for scale normaliza tion.
The plane Π passing through points p
1
, p
2
, and p
3
can be regarded as the plane passing
through point p
1
and the line L connecting points p
2
and p
3
. If we write p
1
as p = e
0
+ p
and let L = e
0
∧ m − n
∗
L
, we can write the trivector Π as follows:
Π = (e
0
+ p) ∧ (e
0
∧ m − n
∗
L
) = −e
0
∧ n
∗
L
+ p ∧ e
0
∧m − p ∧ n
∗
L
= −e
0
∧ (n
∗
L
+ p ∧ m) − p ∧ n
∗
L
= −e
0
∧ (n
L
− (p ∧ m)
∗
)
∗
− p ∧ n
∗
L
= −e
0
∧ (n
L
− p × m)
∗
− p ∧ n
∗
L
= −e
0
∧(n
L
− p × m)
∗
+ hp, n
L
iI, (7.32)
Pl¨ucker coordinates of planes 103
where we have noted that p ∧n
∗
L
= −hp, n
L
iI from Eq. (7.13). Comparing Eq. (7.32) with
Eq. (7.26), we obtain the following Pl¨ucker coordina tes :
n = n
L
− p × m, h = hp, n
L
i. (7.33)
This coincides with Eq. (2.83) in Chapter 2 up to scale normalization.
Next, let Π be the plane that passes through points p
1
= e
0
+ x
1
and p
2
= e
0
+ x
2
at
x
1
and x
2
, respectively, and contains a direction vector u. This plane can be regarded as
passing through the finite points p
1
and p
2
and the point u at infinity. Hence,
Π = p
1
∧p
2
∧ u = (e
0
+ x
1
) ∧ (e
0
+ x
2
) ∧ u = e
0
∧ x
2
∧ u + x
1
∧ e
0
∧ ux
1
∧ x
2
∧ u
= e
0
∧(x
2
− x
1
) ∧ u + x
1
∧ x
2
∧ u. (7.34)
Comparing this with Eq. (7.26), we see that −n
∗
= (x
2
− x
1
) ∧ u and hI = x
1
∧ x
2
∧ u.
Replacing b oth sides with their duals, we obtain the Pl¨ucker coordinates:
n = (x
2
− x
1
) × u, h = |x
1
, x
2
, u|, (7.35 )
which is e quiva lent to the plane in Fig. 2.10 in Chapter 2 if u is replaced by x
3
− x
1
. If
we write the line pa ssing through p
1
and p
2
as L = e
0
∧ m − n
∗
L
, the above res ult can be
rewritten in the form
Π = (e
0
∧ m − n
∗
L
) ∧ u = e
0
∧ m ∧ u − n
∗
L
∧ u
= −e
0
∧ (m ∧ u)
∗∗
+ hn
L
, uiI = −e
0
∧ (m × u)
∗
+ hn
L
, uiI, (7.36)
where we have used n
∗
L
∧ u = −hn
L
, uiI obtained from Eq. (7.13). Comparing Eq. (7.36)
with Eq. (7.26), we obtain the following expression of the Pl¨ucker coordinates:
n = m × u, h = hn
L
, ui. (7.37)
This coincides with the result of Exe rcise 2.14 in Chapter 2 up to scale nor malization.
Now, let Π be the plane that passes through a point p = e
0
+ x at x and contains
direction vectors u and v. This plane can be r e garded as passing through the finite point p
and the two points u and v at infinity. Hence, we c an write
Π = p ∧ u ∧v = (e
0
+ x) ∧ u ∧ v = e
0
∧ u ∧ v + x ∧ u ∧ v. (7.38)
Comparing this with Eq. (7.26), we see that −n
∗
= u ∧ v and hI = x ∧ u ∧ v. Replacing
both sides with their duals, we obtain the following Pl¨ucker c oordinates:
n = u × v, h = |x, u, v|. (7.39)
This coincides with the result of Exercise 2.15 in Chapter 2 up to scale normalization. In
terms of the L
∞
in Eq. (7.22), Eq. (7.38) can also be written as Π = p ∧ L
∞
.
Finally, consider the plane passing through three points u, v, and w at infinity:
Π
∞
= u ∧ v ∧w. (7.40)
Such a plane is called a plane at infinity. Comparing Eq. (7.40) with Eq. (7.26), we see that
−n
∗
= 0 a nd hI = u ∧ v ∧ w. Replacing these with their duals, we obtain the following
Pl¨ucker coordinates:
n = 0, h = |u, v, w|. (7.41)
104 Homogeneous Space and Grassmann–Cayley Algebra
Since the distance of this plane from the origin e
0
is h/ knk, the condition n = 0 means
that the plane is infinitely far away. Moreover, it has no surface normal, so its orientation is
not defined. Due to the homogeneity of the Pl¨ucker coordinates, only the ra tio between n
and h has a meaning, so if n = 0, the absolute value h is irrelevant as long as it is nonzero.
Hence, we can let h = 1 and write Eq. (7.40) as
Π
∞
= I (= e
1
∧ e
2
∧ e
3
), (7.42)
without losing generality. Thus, the existence of the plane Π
∞
at infinity is unique irre-
sp e c tive of the points u, v, a nd w at infinity that define Π. We can interpret it as the
“boundary” of the entire 3D space, surrounding it like a “sphere” of infinite radius. The
fact that Π
∞
does not contain a ny finite points is e asily seen if we note that no finite x can
satisfy Eq. (7.29) for n = 0 and h 6= 0. The a bove results are summarized as follows:
Proposition 7.2 (Description of planes)
1. The plane passing through three points p
1
, p
2
, and p
3
is represented by Π = p
1
∧p
2
∧p
3
,
where any or all of p
1
, p
2
, and p
3
can be at infinity.
2. The plane passing through point p and line L is represented by Π = p ∧ L, where p
can be a point at infinity and L can be a line at infinity.
3. The equation of t he plane Π has the form p ∧ Π = 0.
7.5 DUAL REPRESENTATION
As pointed out in Cha pter 5, a subspace can also be specified by its orthogonal complement.
In the 4D ho mogeneous space, a line is represented by a bivector, i.e., a 2D subspac e , so
its orthogonal complement is also 2D, which repres e nts a line in 3D. Similarly, a plane is
represented by a trivector in the 4D homogeneous space, i.e., a 3D subspace, so its or thogonal
complement is 1D, which represe nts a point in 3D. In describing such duality relationships,
the volume element
I
4
= e
0
∧ e
1
∧ e
2
∧e
3
(7.43)
of the 4 D homogeneous space plays a fundamental role. We define the dual of a k-vector
( ···), k = 0, 1, 2, 3, 4, by
( ···)
∗
= ( ···) · I
4
, (7.44 )
which is a (4 − k)-vector. Note that we defined the dual in 3D by ( ···)
∗
= −( ···) · I (֒→
Eq. (5.54) in Chapter 5). In the general nD, the dual is defined by (−1)
n(n−1)/2
( ···) · I
n
,
and Eq. (7.44) is the expression for n = 4. From this definition, we obtained the following
results:
Proposition 7.3 (Duality of basis) The outer products of the basis elements e
0
, e
1
, e
2
,
and e
3
have the following dual expressions:
1
∗
= e
0
∧ e
1
∧ e
2
∧e
3
, e
∗
0
= e
1
∧ e
2
∧ e
3
, (7.45)
e
∗
1
= −e
0
∧ e
2
∧ e
3
, e
∗
2
= −e
0
∧e
3
∧e
1
, e
∗
3
= −e
0
∧ e
1
∧ e
2
, (7.46)
(e
0
∧e
1
)
∗
= −e
2
∧ e
3
, (e
0
∧ e
2
)
∗
= −e
3
∧e
1
, (e
0
∧ e
3
)
∗
= −e
1
∧e
2
, (7.47)
(e
2
∧e
3
)
∗
= −e
0
∧ e
1
, (e
3
∧ e
1
)
∗
= −e
0
∧e
2
, (e
1
∧ e
2
)
∗
= −e
0
∧e
3
, (7.48)
(e
0
∧ e
2
∧ e
3
)
∗
= −e
1
, (e
0
∧ e
3
∧ e
1
)
∗
= −e
2
, (e
0
∧ e
1
∧ e
2
)
∗
= −e
3
, (7.49)
(e
1
∧ e
2
∧ e
3
)
∗
= e
0
, (e
0
∧e
1
∧ e
2
∧ e
3
)
∗
= 1. (7.50)
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