The Kinematic Image of RR, PR, and RP Dyads
Tudor-Dan Rad, Daniel F. Scharler, Hans-Peter Schröcker
Unit Geometry and CAD, University of Innsbruck
July 3, 2018
Summary
We provide necessary and sufficient conditions for admissible trans-
formations in the projectivised dual quaternion model of rigid body
displacements and we characterise constraint varieties of dyads
with revolute and prismatic joints in this model. Projective trans-
formations induced by coordinate changes in moving and/or fixed
frame fix the quadrics of a pencil and preserve the two families
of rulings of an exceptional three-dimensional quadric. The con-
straint variety of a dyad with two revolute joints is a regular ruled
quadric in a three-space that contains a “null quadrilateral”. If a
revolute joint is replaced by a prismatic joint, this quadrilateral col-
lapses into a pair of conjugate complex null lines and a real line but
these properties are not sufficient to characterise such dyads. We
provide a complete characterisation by introducing a new invariant,
the “fiber projectivity”, and we present examples that demonstrate
its potential to explain hitherto not sufficiently well understood phe-
nomena.
KEYWORDS: Kinematic map, dual quaternion, Study quadric,
null cone, revolute joint, prismatic joint, fiber projectivity, vertical
Darboux motion.
1. Introduction
A common technique in theoretical and applied kinematics is the
use of a point model for the group SE(3) of rigid body displace-
ments. One prominent example is the projectivised dual quaternion
model of SE(3) which exhibits particularly nice geometric and al-
gebraic properties1;2;3;4. In this article, we revisit some fundamen-
tal concepts related to this model, the transformation group gener-
ated by coordinate changes in the moving and the fixed frame and
the kinematic images of dyads with revolute and prismatic joints.
While numerous necessary conditions on these objects are well-
known, we contribute sets of provably sufficient conditions.
Our characterisation of the transformation group generated by
coordinate changes in fixed and moving frame (Section 3) is based
on the pencil of quadrics spanned by the Study quadric and null
cone. These are quadrics corresponding to dual quaternions of real
norm and purely dual norm, respectively. Admissible transforma-
tions fix each member of this pencil and, in addition, preserve the
two families of rulings on a further quadric in a subspace of dimen-
sion three. This leads to the important distinction between “left”
and “right” rulings. At this point we also introduce a further in-
variant, the “fiber projectivity”, which will be crucial in our latter
characterisation of dyads with prismatic joints in Section 5.
The relative position of two rigid bodies can be constrained by a
link or a sequence of links. Fixing one of the two bodies, the col-
lection of all possible poses (position and orientation) of the other
is called a constrained variety. These are important objects in the
study of open and closed kinematic chains, in linkage synthesis or
analysis and other fields. In Section 4, we characterise constraint
varieties generated by dyads of two revolute joints (“RR dyads”) as
regular ruled quadrics in the Study quadric that contain four com-
plex rulings of the null cone.
In Section 5 we extend this result to dyads containing one pris-
matic and one revolute joint (“RP dyads” and “PR dyads”). It is
tempting to view them as limiting cases of RR dyads where one
joint axis becomes “infinite” (lies in the plane at infinity). Indeed,
their kinematic image is a regular ruled quadric in the Study quadric
that intersects the null cone in two complex lines and a real transver-
sal line. Nonetheless, this viewpoint is not complete because of the
possibility of commuting R and P joints (“cylindrical joints”). A
closer investigation leads us to a more refined concept involving the
fiber projectivity which allows to distinguish between the RP, the
PR, and the cylindrical case.
We conclude this paper with an application of our results to a re-
cently presented non-injective extension of the classical kinematic
map5. Here, commuting RP dyads appear naturally as kinematic
images of straight lines. We use this to prove that the extended
kinematic image of a straight line is, in general, a vertical Darboux
motion.
Some parts of this paper, mostly Section 4 and the computations
in the appendix, overlap with a previously published conference pa-
per6. The investigation on the group of admissible transformation
in Section 3, the characterisation of RP and PR dyads in Section 5,
and the relation of straight lines in extended kinematic image space
to vertical Darboux motions in Section 6 are new.
1
arXiv:1607.08119v1  [cs.RO]  27 Jul 2016
2. Preliminaries
This article’s scene is the projectivised dual quaternion model of
spatial kinematics. Here, we give a very brief introduction to this
model for the purpose of settling our notation. More details will be
introduced in the text as needed. For more thorough introductions
to dual quaternions and there relations to kinematics we refer to
Section 3 in Klawitter (2015)4 or Section 11 of Selig (2005)2.
The dual quaternions, denoted by DH, form an associative alge-
bra in R8 where multiplication of the basis elements 1, i, j, k, ε, εi,
εj, and εk is defined by the rules
i2 = j2 = k2 = ijk = −1,
ε2 = 0,
iε = εi,
jε = εj,
kε = εk.
An element q ∈DH may be written as q = p+εd with quaternions
p,d ∈H := ⟨1,i,j,k⟩(angled brackets denote linear span). In this
case the quaternions p and d are referred to as primal and dual part
of q, respectively. The conjugate dual quaternion is q = p+εd and
conjugation of quaternions is done by multiplying the coefficients
of i, j, and k with −1. It satisfies the rule qr = rq for any q,r ∈DH.
The dual quaternion norm is defined as ∥q∥= qq. We readily verify
that it is a dual number, that is, an element of D := ⟨1,ε⟩.
We identify linearly dependent non-zero dual quaternions and
thus arrive at the projective space P7 = P(R8). Writing [q] for the
point in P7 that is represented by q ∈DH, the Study quadric is de-
fined as S := {[q] ∈P7 : ∥q∥∈R}. With q = p+εd, the algebraic
condition for [q] ∈S is pd +dp = 0.
Identifying P3 with the projective subspace generated by
⟨1,εi,εj,εk⟩, a point [q] = [p+εd] ∈S with non-zero primal part
acts on [x] ∈P3 via
[x] 7→[y] := [(p+εd)x(p−εd)].
(1)
The map (1) is the projective extension of a rigid body displacement
in R3. Composition of displacements corresponds to dual quater-
nion multiplication.
The map that takes a point [q] = [p + εd] ∈S \[εH] to the rigid
body displacement (1) is an isomorphism between the factor group
of dual quaternions of non-zero real norm modulo the real multi-
plicative group and SE(3). We refer to it as Study’s kinematic map
or simply as kinematic map.. It provides a rich and solid algebraic
and geometric environment for investigations in kinematic.
Remark 1. Provided p ̸= 0, (1) always describes a rigid body dis-
placement, even if the Study condition is not fulfilled5. In this case,
the map that sends p + εd to the rigid body displacement (1) is
no longer a group isomorphism but a homomorphism. We call it
extended kinematic map. In Section 6 we will characterise the ex-
tended kinematic image of straight lines in P7.
3. Characterisation of the transformation
group
The geometry in kinematic image space P7 is invariant with respect
to coordinate changes in three-dimensional Euclidean space. More
precisely, it is invariant with respect to coordinate changes in both,
the frame of [x] in (1) (the moving frame) and the frame of [y] in (1)
(the fixed frame). The former correspond to right-multiplications,
the latter to left-multiplications with dual quaternions of real norm
and non-zero primal part. Both transformations induce projective
transformations in P7 and our aim in this section is a geometric
characterisation of the transformation group they generate. Nec-
essary geometric conditions on this group are already known but
we are not aware of a formal proof of sufficiency for a set of these
conditions. This we will provide in this section. It will refer to
an important new geometric invariant, the fiber projectivity, whose
usefulness we demonstrate in an example. Later, it will re-appear
in our characterisation of the kinematic image of cylinder spaces.
For a quaternion p = p0 + p1i+ p2j+ p3k we define
ρ(p) :=


p0
−p1
−p2
−p3
p1
p0
−p3
p2
p2
p3
p0
−p1
p3
−p2
p1
p0

,
λ(p) :=


p0
−p1
−p2
−p3
p1
p0
p3
−p2
p2
−p3
p0
p1
p3
p2
−p1
p0

.
(2)
With this notation, the projective maps [x] 7→[px] and [x] 7→[xp] on
[H] = P3 may be written as [x0,x1,x2,x3]⊺7→ρ(p) · [x0,x1,x2,x3]⊺
and [x0,x1,x2,x3]⊺7→λ(p)·[x0,x1,x2,x3]⊺, respectively. Both leave
invariant the quadric E : xx = x2
0 +x2
1 +x2
2 +x2
3 = 0 whence the ge-
ometry in [H], induced by coordinate changes in moving and fixed
frame, is that of elliptic three-space. In fact, the matrices in (2) de-
scribe the well-known Clifford right and left translations that gen-
erate the transformation group of this space, see Coxeter (1998),
p. 140.7 Clifford right (left) translations leave fixed every member
of one family of (complex) rulings of E and we call those rulings
right (left) rulings, respectively.
Similarly, left-multiplication by a dual quaternion ℓ1 + εℓ2 and
right-multiplication by a dual quaternion r1 + εr2 can be effected
by multiplication with matrices L(ℓ1+εℓ2) and R(r1+εr2), respec-
tively. These 8×8 matrices are conveniently described in terms of
blocks of dimension 4×4:
L(ℓ1 +εℓ2) =

λ(ℓ1)
O
λ(ℓ2)
λ(ℓ1)

,
R(r1 +εr2) =

ρ(r1)
O
ρ(r2)
ρ(r1)

.
Here, O denotes the zero matrix of dimension 4×4. Two matrices
of above shape commute and their product
T = L(ℓ1 +εℓ2)·R(r1 +εr2)
=

λ(ℓ1)·ρ(r1)
O
λ(ℓ2)·ρ(r1)+λ(ℓ1)·ρ(r2)
λ(ℓ1)·ρ(r1)

(3)
2
is the matrix of a general coordinate transformation which we want
to characterise geometrically. For that purpose, we introduce two
further invariant quadrics:
Definition 1. The null cone N is the quadric defined by the
quadratic form q = p+εd 7→pp.
The points of the null cone N are characterised by having purely
dual norm (qq ∈εR). The null cone is a singular quadric with three-
dimensional vertex space [εH] := {[εd]: d ∈H} which we call the
exceptional generator. It is contained in the Study quadric S and
the real points of N are precisely those of [εH]. By Y we denote the
regular quadric in [εH] defined by the quadratic form εd ∈[εH] 7→
dd. The quadrics E ⊂[H], N, S, and Y ⊂[εH] are all invariant
under transformations of the shape (3).
Theorem 1. The transformation group described by matrices of
shape (3) where ℓ1 +εℓ2 and r1 +εr2 satisfy the Study condition is
characterised by the following properties: 1) It fixes any quadric in
the pencil spanned by Study quadric S and null cone N and 2) the
restriction to [εH] preserves the two families of (complex) rulings
of the quadric Y.
Proof. To begin with, it is elementary to verify that the conditions
of the theorem are necessary for transformations given by (3). For
the proof of sufficiency, we again employ block matrix notation.
Assume that
T =

A
B
C
D

is the matrix of a projective transformation τ : P7 →P7 with 4× 4
blocks A, B, C, and D. The matrices of the pencil spanned by S and
N are of the shape

νI
σI
σI
O

(4)
where I and O denote the 4 × 4 identity and zero matrix, respec-
tively, and ν and σ are real numbers. The only singular quadric in
this pencil is N and τ must fix its vertex space [εH] (the only three-
dimensional space contained in N). Hence B = O and τ transforms
the matrix of N to

A⊺
C⊺
O
D⊺

·

I
O
O
O

·

A
O
C
D

=

A⊺A
O
O
O

whence A is the scalar multiple of an orthogonal matrix. The trans-
formed matrix of S is

A⊺
C⊺
O
D⊺

·

O
I
I
O

·

A
O
C
D

=

C⊺A+A⊺C
A⊺D
D⊺A
O

.
From this we infer that D = αA for some α ∈R and C⊺A+A⊺C =
O. This latter matrix equation has the general solution C = A−⊺S
with an arbitrary skew-symmetric matrix S of dimension 4×4. Fi-
nally, we study the action on a third quadric in the pencil:

A⊺
C⊺
O
αA⊺

·

I
I
I
O

·

A
O
C
αA

=

A⊺A+C⊺A+A⊺C
αA⊺A
αA⊺A
O

=

A⊺A
αA⊺A
αA⊺A
O

.
This implies α = 1 whence
T =

A
O
A−⊺S
A

.
(5)
The restriction of τ to [εH] transforms the matrix [I] of Y ⊂[εH]
to [A⊺· I · A] = [I] and hence fixes Y. The condition on the rulings
implies detA > 0. By well-known results of three-dimensional el-
liptic geometry, A is uniquely expressible as product of a Clifford
left translation and a Clifford right translation (Page 140 of Coxeter
19987). In our notation, this means that there exist quaternions ℓ1,
r1 such that A = λ(ℓ1)·ρ(r1). Hence, the upper left and lower right
corners of (5) and (3) match. For given A, the set of possible matri-
ces C = A−⊺S is a real vector space of dimension six. The same is
true for the 4×4 sub-matrices in the lower left corner of the matrix
(3) (recall that ℓ1ℓ2 + ℓ2ℓ1 = r1r2 + r2r1 = 0). Hence, there exist
unique quaternions ℓ2, r2 such that C = λ(ℓ2)·ρ(r1)+λ(ℓ1)·ρ(r2)
and the theorem is proved.
Examining the proof, it is easy to see that the system of invariants
in Theorem 1 is minimal.
Invariance of left and right rulings, respectively, of Y is an im-
portant property that is, for example, responsible for different kine-
matic properties of motion and inverse motion. For the planar case,
this is mentioned in Bottema and Roth (1990), Chapter 11, §141.
The map χ : P7 →P7, [q] 7→[q] (quaternion conjugation) is a pro-
jective transformation of P7 leaving invariant the quadrics of the
pencil spanned by S and N (lets call it the absolute pencil) but it
interchanges the left and right rulings of Y. Given a motion γ (a
curve in S), the inverse motion is γ := χ(γ). It is of the same al-
gebraic degree and has the same relative position to absolute pencil
(in projective sense) but not necessarily the same position to the left
and right rulings of Y. In order to demonstrate this at hand of an
example, we introduce a further invariant concept.
Definition 2. The projective map
ϕ : P7 →[εH],
[x′ +εx′′] 7→[εx′]
(6)
that assigns to each point in P7 the projection on its primal part
times ε is called the fiber projectivity.
The name “fiber projectivity” is motivated by a recently intro-
duced non-injective extensions of Study’s kinematic map5 which
reads as in (1) but drops the Study condition pd +dp = 0. Its fibers
are obtained by connecting a point [x] with ϕ([x]). The fiber pro-
jectivity is invariant with respect to transformations τ given by (3)
in the sense that τ ◦ϕ([x]) = ϕ ◦τ([x]) for every point [x] ∈P7. The
restriction of the fiber projectivity to [H] is a bijection in which E
and Y correspond so that we may speak of right (left) rulings of Y
as well.
Example 1. The Darboux motion is the only spatial motion with
planar trajectories, see Bottema and Roth, 1990, Equation (3.4)1;8.
A parametric equation in dual quaternions reads
C(t) = (cε +k)t3 +(1+ε(b−ai−ck))t2 +(k−ε(aj+bk))t +1
3
(see Li et al., 20158). The motion parameter is t while a, b, and c
are constant real numbers. This rational cubic curve intersects [εH]
in the two points
[d1,2] := [ε(±ci+b−ai±aij+(±bi−c)k)]
which even lie on Y. (Note that “i” denotes a complex number
which must not be confused with the quaternion unit “i”.) The fiber
projection of [C(t)] is the point
ϕ([C(t)]) = [(1+t2)ε(kt +1)] = [ε(kt +1)].
For varying t, these points vary on a straight line which intersects
Y in the two points [f1,2] := [ε(1 ± ik)]. It is now easy to verify
that the lines [d1] ∨[f1] and [d2] ∨[f2] are rulings of Y. Moreover,
[d1] = [p f1] and [d2] = [pf2] where p = −b + ai + ck. This means
that a Darboux motion intersects Y in two points [d1], [d2] and its
fiber projection intersects Y in two points [f1], [f2] such that the
lines [d1] ∨[f1] and [d2] ∨[f2] are left rulings of Y. The inverse
motion
C(t) = (cε −k)t3 +(1+ε(b+ai+ck))t2 −(k−ε(aj+bk))t +1
(Mannheim motion) has similar properties as curve in P7 but dif-
ferent kinematic properties. Most notably, the trajectories of C(t)
are rational of degree two while those of C(t) are rational of de-
gree four. The deeper reasons for this is the non-invariance of left
and right rulings with respect to the quaternion conjugation map
χ. Hence, the intersection points of C and ϕ(C) with Y span right
rulings. The case of a vertical Darboux motion (a = 0) is special.
Here we have [d1] = [f1] and [d2] = [f2] and the distinction between
left and right rulings vanishes. Indeed, the inverse motion is again
a vertical Darboux motion.
4. A characterisation of 2R spaces
Given two non co-planar lines ℓ1, ℓ2 in Euclidean three-space, we
consider the set of all displacements obtained as composition of a
rotation around ℓ2, followed by a rotation about ℓ1. Its kinematic
image is known to lie in a three-space (Selig 2005, Section 11.4)2
which we call a 2R space. It always contains the displacement cor-
responding to zero rotation angle around both axes. It is no loss
of generality to view it as identity displacement [1] which we will
often do in proofs (and consistently did in a previous publication6).
However, we avoid this in our theorems because it is not invari-
ant with respect to the transformations characterised in Theorem 1.
Fixing one rotation angle and varying the other yields a straight line
in S. The thus obtained two families of lines form the rulings of a
quadric surface in S. In order to characterise 2R spaces among all
three-dimensional subspaces of P7 with these properties, we intro-
duce the following notions:
Definition 3. A spatial quadrilateral is a set of four different lines
{ℓ0,ℓ1,ℓ2,ℓ3} in projective space such that the intersections ℓ0 ∩ℓ1,
ℓ1 ∩ℓ2, ℓ2 ∩ℓ3, and ℓ3 ∩ℓ0 are not empty. The lines ℓ0, ℓ1, ℓ2, and ℓ3
are called the quadrilateral’s edges. A null line is a straight line con-
tained in the Study quadric S and the null cone N. A null quadri-
lateral is a spatial quadrilateral whose elements are null lines.
Theorem 2. A three-space U ⊂P7 is a 2R space if and only if it
• intersects the Study quadric in a regular ruled quadric Q,
• does not intersect the exceptional three-space [εH], and
• contains a null quadrilateral.
The first and second item in Theorem 2 exclude exceptional cases
with co-planar, complex, or “infinite” revolute axes. The latter cor-
respond to prismatic joints and will be treated later. The crucial
point is existence of a null quadrilateral. We split the proof of The-
orem 2 into a series of lemmas. It will be finished by the end of this
section.
Lemma 1. The straight line [x] ∨[y] is contained in S ∩N if and
only if xx = yy = xy+yx = 0.
We omit the straightforward computational proof of Lemma 1.
Note that the left-hand sides of each of the three conditions in this
lemma are dual numbers. Hence, they give six independent linear
equations for the real coefficients of x and y.
Lemma 2. The conditions of Theorem 2 are necessary for 2R
spaces.
Proof. The quadric Q contains a regular point [u] and, without loss
of generality, we may assume [u] = [1]. Then, the constraint variety
of a 2R chain can be parameterised as
R(t1,t2) = (t1 −h1)(t2 −h2)
(7)
with two dual quaternions h1,h2 that satisfy h1h1 = h2h2 = 1,
h1 +h1 = h2 +h2 = 0 and h1h2 ̸= h2h1. The first condition ensures
that the dual quaternions are suitably normalised and [h1],[h2] ∈S.
The second condition means that h1 and h2 describe half-turns (ro-
tations through an angle of π).
The third condition is true be-
cause the axes of these half turns are not co-planar.
(Two half
turns commute if and only if their axes are identical or orthogo-
nal and co-planar.) Equation (7) describes a composition of two
rotations for all values of t1, t2 in R∪{∞} with ∞corresponding to
zero rotation angle. Even if their kinematic meaning is unclear,
we also allow complex parameter values.
Expanding (7) yields
R(t1,t2) = t1t2 −t1h2 −t2h1 + h1h2. We see that the kinematic im-
age of the 2R dyad lies in the three-space spanned by [1], [h1], [h2],
[h1h2]. In a suitable projective coordinate frame with these points
as base points, we use projective coordinates [x0,x1,x2,x3]. Then,
the surface parameterisation (7) reads x0 = t1t2, x1 = −t1, x2 = −t2,
x3 = 1. It describes the quadric with equation x0x3−x1x2 = 0 which
is indeed regular and ruled.
The intersection of U with the exceptional three-space [εH] is
non empty if and only if the primal part of R(t1,t2) vanishes for cer-
tain parameter values t1, t2. This can only happen if the primal parts
4
of h1 and h2 are linearly dependent over R but then the revolute axes
are parallel and U, contrary to our assumption, is contained in S.
The other possibility for U ⊂S, intersecting revolute axes, has been
excluded as well. Clearly, U is not contained in the null cone N ei-
ther. We claim that the intersection of U and N consists of the four
lines given by t1 = ±i, t2 = ±i. Indeed, they are null lines. Set, for
example, z(t2) := (i−h1)(t2 −h2), x := z(0) and y := z(1). In view
of Lemma 1, we have to verify xx = yy = 0. But this follows from
z(t2)z(t2) = (i−h1)(t2 −h2)(t2 −h2)(i−h1)
= (t2 −h2)(t2 −h2)(i−h1)(i−h1)
= (t2 −h2)(t2 −h2)(−i2 −i(h1 +h1)+h1h1)
= (t2 −h2)(t2 −h2)(−1−0+1) = 0.
Here, we used the fact that (t2 −h2)(t2 −h2) is a real number and
thus commutes with all other factors. The cases t1 = −i, t2 = ±i are
similar so that we have verified all conditions of Theorem 2.
The proof of sufficiency is more involved. We need two addi-
tional lemmas from projective geometry which are formulated and
proved in Appendix A.
Lemma 3. A three-space U that satisfies all conditions of Theo-
rem 2 is a 2R space.
Proof. Denote the vertices of the null quadrilateral by [u1], [v1],
[u2], [v2] such that consecutive points span the quadrilateral’s edges.
Again, it is no loss of generality to assume [1] ∈Q. We denote the
tangent hyperplane of S at [1] by τ1. A point [p′ +ε p′′] is contained
in τ1 if and only if p′′ + p′′ = 0. Since [1] does not lie on the null
quadrilateral, there exist (possibly not unique) points
[m1] ∈[u1]∨[v1],
[n1] ∈[v1]∨[u2],
[m2] ∈[u2]∨[v2],
[n2] ∈[v2]∨[u1]
in the intersection of U and τ1 such that
• [m1], [m2] and [n1], [n2] are pairs of complex conjugate points
(whence their respective joins are real lines) and
• the quadrilateral with these points as vertices is planar and
non-degenerate (Figure 1).
The real points of [m1]∨[m2] and [n1]∨[n2] correspond to rotations
about fixed axes. Pick real rotation quaternions h1, h2 ̸= 1 such that
[h1] ∈[m1]∨[m2] and [h2] ∈[n1]∨[n2]. The axes of these rotations
are the only candidates for our 2R dyad. Hence, we have to show
that either [h1h2] ∈U or [h2h1] ∈U. It is easy to see that this is the
case if and only if either [m1n1] ∈U or [n1m1] ∈U. In fact, we will
even show that one of these products equals v1.
We claim that both M1 := [m1]∨[m1n1] and M2 := [m1]∨[n1m1]
are null lines. In order to show this, we have to verify the conditions
[1]
[m1]
[m2]
[n1]
[n2]
[h1]
[h2]
[u1]
[u2]
[v1]
[v2]
[n2m1]
[n1m1]
[n1m2]
[n2m2]
[m2n1]
[m1n1]
[m1n2]
[m2n2]
Figure 1: Null lines and null quadrilaterals in the proof of Lemma 3.
of Lemma 1:
(m1n1)(m1n1) = m1(n1n1
|{z}
=0
)m1 = 0,
(n1m1)(n1m1) = n1(m1m1
| {z }
=0
)n1 = 0,
m1(m1n1)+(m1n1)m1 = m1(n1 +n1
| {z }
∈C
)m1 = (n1 +n1)(m1m1
| {z }
=0
) = 0,
m1(n1m1)+(n1m1)m1 = (m1m1
| {z }
=0
)n1 +n1(m1m1
| {z }
=0
) = 0.
Similarly, we see that also N1 := [n1] ∨[m1n1] and N2 := [n1] ∨
[n1m1] are null lines.
Thus, we are in the situation depicted in
Figure 1 where we have three null quadrilaterals with respective
vertices
[u1],[v1],[u2],[v2];
[m1n1],[m2n1],[m2n2],[m1n2];
[n1m1],[n1m2],[n2m2],[n2m1].
The second and third quadrilateral are different because m1 and n1
do not commute (otherwise they would lie on the same line through
[1] which contradicts the regularity of the quadric Q :=U ∩S). Our
proof will be finished as soon as we have shown that the first and
the second or the first and the third quadrilateral are equal. For this,
it is sufficient to show that [v1] = [m1n1] or [v1] = [n1m1].
At first, we argue that the primal part of [v1] equals the primal
part of [m1n1] or of [n1m1]. Because U does not intersect [εH], the
projection on the primal part is a regular projectivity U →[H] with
centre [εH]. We denote projected objects by a prime, that is, we
write u′
1, v′
1, m′
1 etc., for the primal parts of u1, v1, m1 etc. The
quadric Q is regular and ruled and so is its primal projection Q′.
Hence, the point [m′
1] ∈Q′ is incident with precisely two lines con-
tained in Q′, say M′
1 and M′
2. But [m′
1] ∨[v′
1] is contained in Q′.
Hence [v′
1] ∈M′
1 or [v′
1] ∈M′
2. Similarly, [v′
1] is also incident with
one of the two lines contained in Q′ and incident with [n′
1]. Thus,
[v′
1] = [m′
1n′
1] or [v′
1] = [n′
1m′
1].
Now we have to lift this result to the dual part and show that
[v1] = [m1n1] or [v1] = [n1m1]. The alternative being similar, we
5
assume [v′
1] = [m′
1n′
1]. This means that we have two null quadri-
laterals, one with vertices [u1], [v1], [u2], [v2] and one with vertices
[m1n1], [m2n1], [m2n2], [m1n2] such that their primal projections are
equal and corresponding sides intersect in the vertices [m1], [n1],
[m2], [n2] of a planar quadrilateral. Now we wish to apply Lemma 7
in Appendix A with E = [εH], F = [H], and Q = S in order to
conclude [v1] = [m1n1]. This is admissible if and only if the plane
L := [1] ∨[h1] ∨[h2] = [1] ∨[m1] ∨[n1] does not intersect the four-
spaces [u1]∨[εH], [v1]∨[εH], [u2]∨[εH], and [v2]∨[εH]. We con-
sider the four-space [v1]∨[εH] = [m1n1]∨[εH]. It does not intersect
L if and only if the linear combination
α ·1+βm1 +γn1 +δm′
1n′
1 +
3
∑
ℓ=0
ϕℓwℓ= 0
(8)
with some basis (w0,w1,w2,w3) of εH is trivial. By considering
only primal parts, we see that (8) implies
α ·1+βm′
1 +γn′
1 +δm′
1n′
1 = 0.
But the points [1], [m′
1], [n′
1], [m′
1n′
1] are vertices of a quadrilateral
contained in a regular ruled quadric and therefore not co-planar.
Thus α = β = γ = δ = 0 and ϕ0 = ϕ1 = ϕ2 = ϕ3 = 0 follows. The
arguments for the remaining four-spaces are similar.
5. A characterisation of RP, PR, and
cylinder spaces
Now we proceed with a geometric characterisation of the images
of RP and PR dyads. It is tempting to view them as limiting case
of RR dyads where one of the revolute axes becomes “infinite”.
However, this is not entirely justified because of the possibility of
commuting joints – a phenomenon which is trivial for RR dyads but
relevant for RP or PR dyads. It happens precisely if the direction
of the revolute axis and translation direction are linearly dependent.
These dyads are special enough to deserve a name of their own.
Since their transformations are usually modelled by a “cylindrical
joint”, we define:
Definition 4. The projective span of the kinematic image of an RP
or PR dyad is called an RP or PR space respectively. If rotation and
translation commute, we speak of a cylinder space or C space.
The following example demonstrates that RP and PR spaces can-
not be characterised by a limiting configuration of Theorem 3.
Example 2. Consider the three-space U spanned by [1], [m1], [n1]
and [s1] where
m1 = εi,
n1 = i+i,
s1 = ε(i+i+j+ik).
It is straightforward to verify that the line [m1] ∨[s1] is a null line
contained in [εH] and the line [n1]∨[s1] and its conjugate complex
line are null lines not contained in [εH]. The three-space U con-
tains all rotations about i and all translations in direction of i. The
[1]
[m1]
[m2]
[n1]
[n2]
[s1]
[s1]
[s2]
ϕ([n1])
ϕ([n1])
e1
ℓ1
ℓ2
Figure 2: Lines in the proof of Theorem 4 (Lemma 4 and Lemma 5).
corresponding RP space is generated by the regular quadric with
parametric equation
R(u,v) = (u−i)(v−εi)
where u and v both range in R∪{∞}. This quadric is not contained
in U which is hence neither an RP nor a PR spac.
It is crucial to above example that m1 and n1 commute. This can
also be seen in our characterisation of RP and PR spaces where C
spaces require a special treatment.
Theorem 3. A three-space U ⊂P7 is an RP but not a C space if
and only if it
• intersects the Study quadric in a regular ruled quadric Q,
• intersects the exceptional three-space [εH] in a straight
line e1, and
• contains a pair of conjugate complex null lines ℓ1, ℓ2 that in-
tersect e1 in points [s1], [s2] respectively, such that ϕ(ℓ1)∨[s1]
and ϕ(ℓ2)∨[s2] are right rulings of Y.
RP spaces that are not C spaces are characterised by the same con-
ditions but with “left rulings” instead of “right rulings”.
Necessity of the first and second condition of Theorem 3 have
been proved in the master thesis by Stigger9 by means of Gröbner
basis computations. Here, we give an independent proof.
Lemma 4. The conditions of Theorem 3 are necessary for an RP
or PR space that is not a C space.
Proof. Assuming without loss of generality [1] ∈U, the constraint
variety of an RP chain can be parameterised as
R(u,v) = (u−h)(v−ε p),
u ∈R∪{∞}, v ∈R∪{∞}
(9)
with a non-zero quaternion p with p+ p = 0 and a dual quaternion h
that satisfies hh = 1, h+h = 0. Since it is not the constraint variety
of a PR chain, the primal part h′ of h and p do not commute, that is
[h′] ̸= [p].
By the same arguments as in Lemma 2, this constraint variety is
a regular quadric: The parameters u and v range in R ∪{∞} (with
∞corresponding to zero rotation angle or translation distance) but
we will also admit complex values for them. Expanding (9) yields
6
R(u,v) = uv −vh −uε p + εhp. We see that the kinematic image
of this RP dyad lies in the three-space spanned by [1], [h], [ε p] and
[εhp] = [εh′p] and is a regular quadric.
Clearly, this quadric it contains the line e1 := [ε p]∨[εhp] ⊂[εH].
Moreover, the lines ℓ1, ℓ2 parameterised by R(±i,v) are conjugate
complex null lines. The intersection points [n1], [n2] of these lines
with τ1 correspond to v = ∞, their intersection points [s1], [s2] with
[εH] correspond to v = 0:
n1 = i−h, n2 = −i−h, s1 = −ε(i−h)p, s2 = −ε(−i−h)p
(Figure 2). In view of Lemma 1 the null line property follows from
s1n1 +n1s1 = −ε(i−h)p(i−h)+(i−h)p(−ε(i−h))
= −(i−h)ε(p+ p
| {z }
=0
)(i−h) = 0
and similar with i replaced by −i. Moreover ϕ([ℓ1]) = ϕ([n1]) =
[ε(i−h′)] and [s1] = [ε(i−h′)p]. Because of [h′] ̸= [p], these points
do not coincide and, indeed, span a right ruling of Y.
The statements for PR spaces follow from Theorem 1 by applica-
tion of the conjugation map χ. It interchanges RP spaces with PR
spaces and right rulings with left rulings but fixes the Study quadric
and the null cone.
The proof of sufficiency of the conditions in Theorem 3 is, in
large parts, a copy of Lemma 3. However, at one point the third
condition comes into play.
Lemma 5. A three-space U that satisfies all conditions of Theo-
rem 3 is a PR or RP space but not a C space.
Proof. Once more, we assume [1] ∈U. The intersection of U and S
is a regular quadric Q that contains the straight line e1 = U ∩[εH]
and the two conjugate complex null lines ℓ1, ℓ2. There exist points
[m1] ∈e1 and [n1] ∈ℓ1 such that [1]∨[m1] and [1]∨[n1] are rulings of
Q. The points of the line [1]∨[m1] correspond to translations with
fixed direction and the points of [1]∨[n1] to rotations around a fixed
axis. There exists an intersection point [s1] := e1∩ℓ1 and we have to
show that either [s1] = [m1n1] or [s1] = [n1m1]. In general, this can
be done just as in our proof of Lemma 3: Denote by m2 the com-
plex conjugate to m1 and by n2 the complex conjugate of n1. Then
[1] = ([m1]∨[m2])∩([n1]∨[n2]), the quadrilaterals with respective
vertices [m1n1], [m2n1], [m2n2], [m1n2] and [n1m1], [n1m2], [n2m2],
[n2m1] are contained in Q, and the points [m1], [n1], [m2], and [n2]
form a planar quadrilateral. Because the intersection of [εH] and
U is not empty, we have to use a projection with three-dimensional
centre Z that does not intersect U (and hence differs from [εH])
and is such that the plane [1] ∨[m1] ∨[n1] is complementary to the
four-spaces
[u1]∨Z,
[v1]∨Z,
[u2]∨Z,
[v2]∨Z
when appealing to Lemma 7. Such a choice is possible and then
the differences to the proof of Lemma 3 are irrelevant unless m1
and n1 commute. In this case our argument fails because the two
spatial quadrilaterals coincide and we may no longer conclude that
the projection of [s1] equals the projection of [m1n1] or of [n1m1].
Fortunately, this case is already special enough to allow a straight-
forward computational treatment.
The dual quaternions m1 and n1 commute (m1n1 = n1m1) if the
translation direction of m1 and the direction of the rotation axis of n1
are linearly dependent. Without loss of generality, we may set m1 :=
εi and n1 := i + i so that m1n1 = n1m1 = ε(ii −1). Now U equals
the span of [1], [m1], [n1], and [s1] = [m1n1] = [n1m1] but ϕ(ℓ1) =
ϕ([n1]) = [ε(i+i)] = [m1n1] – a contraction to the assumption that
[s1] and ϕ([n1]) span a straight line.
Above proof already contains the basic idea for the still missing
characterisation of C spaces.
Theorem 4. A three-space U ⊂P7 is a C space if and only if it
• intersects the Study quadric in a regular ruled quadric Q,
• contains a pair of conjugate complex null lines ℓ1, ℓ2 and a
real null line e1, and
• satisfies ϕ(Q) = e1.
Proof. Lets assume that U is a C space. Without loss of generality,
we may assume that it is spanned by [1], [m1] = [εi], [n1] = [i + i]
and [s1] = [m1n1] = ϕ([n1]). Clearly, it contains the real line e1 =
[m1] ∨[s1], the null line ℓ1 = [s1] ∨[n1], and its complex conjugate
ℓ2. A parametric equation for the intersection quadric Q of U and
S is
R(u,v) = (u−n1)(v−m1),
u ∈R∪{∞}, v ∈R∪{∞}.
It is a regular ruled quadric and its fiber projection equals ϕ(Q) is
parameterized by ε(u−s1). This is indeed the straight line e1.
Conversely, assume that the condition of the theorem hold and,
without loss of generality, [1] ∈U. As in the proof of Lemma 5,
we can find points [m1], [n1], [s1] ∈U such that e1 = [m1] ∨[s1],
ℓ1 = [n1] ∨[s1], and the lines [1] ∨[m1] and [1] ∨[n1] are rulings of
Q. After a suitable choice of coordinates, we may assume n1 = i+i
whence
e1 = ϕ([1])∨ϕ([n1]) = [ε]∨[ε(i+i)] = [ε]∨[εi].
But then [m1] = [εi] because [m1] ∈e1 and [1]∨[m1] is contained in
S. In particular, [m1n1] = [n1m1] = [ε(ii −1)] = [ε(i + i)]. More-
over, there exist scalars α and β such that s1 = ε(α + βi). The
condition n1s1 +s1n1 = 0 implies α = iβ so that [s1] = [ε(i+i)] =
[m1n1] = [n1m1] and U is indeed a C space.
6. Extended Kinematic Mapping and
Straight Lines
In this section we illustrate the usefulness of Theorem 4 at hand
of an example. In Pfurner et al.5 the authors consider the map
7
(1) but without imposing the Study condition (“extended kinematic
map”). The action of a dual quaternion on points is still a rigid
body transformation but the map is no longer an injection. Pfurner
et al. (2016) show that the fibers of (1) (the set of pre-images of
a fixed rigid body displacement) is a straight line. More precisely,
the fiber incident with a point [p] = [p′ + ε p′′] ∈P7 \ [εH] is the
straight line [p′ + ε p′′] ∨[ε p′′] = ϕ([p]). This line intersects the
Study quadric in [ε p] and one further point which gives the corre-
sponding displacement in the classical kinematic map (which also
the underlying concept in this paper).
Pfurner et al. (2016) also mention that the motion correspond-
ing to a generic straight line in P7 is a Darboux motion1;8, that is,
the motion obtained by composing a planar elliptic motion with a
harmonic oscillation in orthogonal direction. This is fairly straight-
forward to see. By virtue of (1), all trajectories are rational curves
degree two or less which already limits possible candidates to Dar-
boux motions, rotations with fixed axis, and curvilinear translations
along rational curves of degree two or less.
We will now prove that the motion of a generic straight line ℓ
is actually a vertical Darboux motion. In this limiting case of the
generic Darboux motion the elliptic translation is replaced by a ro-
tation with fixed axis. Our proof is based on the fact that the three-
space spanned by ℓand its fiber ϕ(ℓ) is a C space.
Theorem 5. If a straight line ℓ⊂P7 contains a point [p] /∈[εH] and
is neither contained in the null cone N nor in the four-dimensional
subspace [T] := [p]∨[εH], the span of ℓand ϕ(ℓ) is a C space.
Proof. Without loss of generality we assume [p] = [1]. Then [T] is
the space of all translations. The straight line ℓis neither contained
in nor tangent to the null cone N (because it is real and does not
intersect [εH]). Hence, it intersects N in two points [a] = [a′+εa′′],
[b] = [b′+εb′′]. They satisfy the null cone condition a′a′ = b′b′ = 0.
Because [1] ∈[a]∨[b], we may write b = a′ −εa′′. Note that a and
b are complex conjugates but we will not make direct use of this,
thus avoiding possible confusion arising from mixing of complex
and quaternion conjugation.
If the vectors a′ and a′ are linearly dependent, there exists a linear
combination εc′′ of a and b with c′′ ̸= 0. But then [1] ̸= [εc′′] ∈[T]
and, contrary to our assumption, ℓ= [1] ∨[εc′′] ⊂[T]. Hence, a′
and a′ are linearly independent. This implies that the points [a] =
[a′ + εa′′], [b] = [a′ −εa′′], ϕ([a]) = [εa′], and ϕ([b]) = [εa′] span
a three-dimensional space U.
Now we compute the intersection of U with S and N.
We
set x = αa + βa + ε(γa′ + δa′) and solve the equation xx = 0 for
[α,β,γ,δ]. The left-hand side is a dual number. Equating its primal
part with zero yields
αβ(a′2 +a′2) = 0.
Note that f := a′2 +a′2 is a real number and it is different from zero
because of linear independence of a′ and a′. Hence α = 0 or β = 0.
We further abbreviate
g1 := a′a′′ +a′′a′,
g2 := a′ a′′ +a′′a′
[1]
[a]
[b]
ϕ([a])
ϕ([b])
[n1]
[n2]
[s1]
[s2]
ℓ
n
e1
ℓ1
ℓ2
Figure 3: Lines in the proof of Theorem 5.
(these are real numbers as well) and equate the dual part of xx = 0
with zero:
(αδ +βγ) f +α(α −β)g1 +β(α −β)g2 = 0
In case of β = 0, this boils down to
α(δ f +αg1) = 0
while we get
β(γ f −βg2) = 0
if α = 0. From these considerations we infer thatU ∩S ∩N consists
of precisely three straight lines: The line e1 given by [α,β,γ,δ] =
[0,0,γ,δ] which is contained in the exceptional generator [εH], and
the two lines ℓ1, ℓ2 given by
[α,β,γ,δ] = [−f,0,γ,g1],
and
[α,β,γ,δ] = [0, f,g2,δ]
respectively. Because of f ̸= 0, the lines ℓ1 and ℓ2 do not inter-
sect and they are complex conjugates by construction. This already
implies that U ∩S is a ruled quadric but neither a quadratic cone
nor a double plane (but we still have to argue that it is not a pair of
planes).
For α = β = 0 and γ = ∞or δ = ∞we obtain the intersection
points
[s1] = [εa′]
and
[s2] = [εa′]
(10)
of ℓ1 and ℓ2, respectively, with e1. There is a unique line n through
[1] that intersects both ℓ1 and ℓ2. Its intersection points [n1], [n2]
with ℓ1 and ℓ2, respectively, are given by
[α,β,γ,δ] = [−f,0,g2,g1]
and
[α,β,γ,δ] = [0, f,g2,g1].
The line [n1]∨[n2] is a fourth ruling in the intersection of U and S
and does not intersect [e1]. Hence U and S cannot intersect in a pair
of planes.
The quadric Q = S ∩U is given by the equations x = αa+βa+
ε(γa′ + δa′) where the parameters α, β, γ, and δ satisfy αδ +
βγ = 0. From this we see that that its fiber projection is given by
[ε(αa′ + βa′)]. It, indeed, coincides with e1 so that also the last
condition of Theorem 4 is fulfilled.
Corollary 1. A straight line ℓsatisfying the assumptions of Theo-
rem 5 corresponds, via (1), to a vertical Darboux motion.
8
Proof. Since the map (1) generically doubles the degree, the trajec-
tories of the motion in question are at most quadratic and the motion
itself is either a curvilinear translation along a curve of degree two
or less, the rotation about a fixed axis or a Darboux motion. Pure
translations are excluded by the assumptions of Theorem 5, non-
vertical Darboux motions by its conclusion.
Remark 2. Corollary 1 is also valid for straight lines contained in
the Study quadric S unless they intersect the exceptional generator
[εH]. These lines are known to describe rotations about a fixed axis
(Selig 2005, Section 11.2)2, that is, a vertical Darboux motion with
zero amplitude. In this sense, we may say that all straight lines in
extended kinematic image space that intersect the null cone in two
distinct points describe a vertical Darboux motion.
7. Conclusion
We provided necessary and sufficient geometric conditions on the
transformation group induced by coordinate changes in the fixed
and moving frame and on the kinematic image of 2R, RP, and PR
dyads. While many properties of these objects are already known,
sufficiency of certain collections of conditions seems to be new. We
payed proper attention to the case of commuting joints (which is
trivial for RR dyads), highlighted the role of left and right rulings,
and introduced the fiber projectivity. It is important in our charac-
terisation of RP and PR spaces but also for the kinematic images of
Darboux motions, as illustrated in Example 1 and Section 6.
A. Auxiliary results
In this appendix we prove two technical results of purely projective
nature and with no direct relations to kinematics. The formulation
of Lemma 6 could be simplified but in its present form its applica-
tion in the proof of Lemma 7 is apparent.
Lemma 6. Given a three-space E ⊂P7 and four points [u′
1], [v′
1],
[u′
2], [v′
2] that span a three-space F ⊂P7 with E ∩F = ∅, set U1 :=
[u′
1]∨E, V1 := [v′
1]∨E, U2 := [u′
2]∨E, V2 := [v′
2]∨E and consider
four projections
ζ1 : U1 →V1,
η1 : V1 →U2,
ζ2 : U2 →V2,
η2 : V2 →U1
with respective centres [m1], [n1], [m2], and [n2]. If the centres are
pairwise different and span a plane L that is complementary to U1,
V1, U2, and V2, respectively, the composition η2 ◦ζ2 ◦η1 ◦ζ1 of the
four projections is the identity.
Proof. Note that dimU1 ∨V1 = 5 so that the projection ζ1 (and also
η1, ζ2, η2) from a suitable point is well defined. Take an arbitrary
point [u1] ∈U1 but not in E and set [v1] := ζ1([u1]), [u2] := η1([v1]),
[v2] := ζ2([u2]). Then the projective space G := [u1] ∨[v1] ∨[u2] ∨
[1,1,1,1]
[1,1,1,1]
[m1] = [1,1,0,0]
[m1] = [1,1,0,0]
[m2] = [0,0,1,1]
[m2] = [0,0,1,1]
[n1] = [0,1,1,0]
[n1] = [0,1,1,0]
[n2] = [1,0,0,1]
[n2] = [1,0,0,1]
[u1] = [1,0,0,0]
[u1] = [1,0,0,0]
[u2] = [0,0,1,0]
[u2] = [0,0,1,0]
[v1] = [0,1,0,0]
[v1] = [0,1,0,0]
[v2] = [0,0,0,1]
[v2] = [0,0,0,1]
Figure 4: Projective coordinates in the proof of Lemma 6 (four trail-
ing zeros are omitted).
[v2] = [u1]∨[m1]∨[n1]∨[m2]∨[n2] (Figure 4) is of dimension three
and does not intersect E because of the dimension formula
dim(G∩E) = dimG+dimE −dim(G∨E)
= 3+3−dim([u1]∨L∨E) = 3+3−7 = −1.
Therefore, we may take [u1], [v1], [u2] and [v2] in that order as base
points of a projective coordinate system in G and complement them
by four further base points in E and a suitable unit point to a projec-
tive coordinate system of P7. We select the unit point such that its
projection into G from E gives ([m1] ∨[m2]) ∩([n1] ∨[n2]). Then,
the projection centres are
[m1] = [1,1,0,0,0,0,0,0],
[n1] = [0,1,1,0,0,0,0,0],
[m2] = [0,0,1,1,0,0,0,0],
[n2] = [1,0,0,1,0,0,0,0].
We pick an arbitrary point h2 := [x0,0,0,0,x4,x5,x6,x7] ∈U1 and
compute
ζ1(h2) = [0,−x0,0,0,x4,x5,x6,x7] =: z1,
η1(z1) = [0,0,x0,0,x4,x5,x6,x7] =: h1,
ζ2(h1) = [0,0,0,−x0,x4,x5,x6,x7] =: z2,
η2(z2) = [x0,0,0,0,x4,x5,x6,x7] = h2.
This concludes the proof.
Lemma 7. Let E, F ⊂P7 be non-intersecting three-spaces, the
latter spanned by points [u′
1], [v′
1], [u′
2], [v′
2]. Consider further a
regular quadric Q containing E and four vertices [m1], [n1], [m2],
[n2] ∈Q of a quadrilateral in a plane L that is complementary to
E ∨[u′
1],
E ∨[v′
1],
E ∨[u′
2],
and
E ∨[v′
2].
Then there exists a unique spatial quadrilateral with vertices [u1],
[v1], [u2], [v2] that is contained in Q, whose sides [u1] ∨[v1], [v1] ∨
[u2], [u2]∨[v2], [v2]∨[u1] are, in that order, incident with [m1], [n1],
[m2], [n2] and whose projection from E into F is the quadrilateral
with vertices [u′
1], [v′
1], [u′
2], and [v′
2].
9
Proof. Denote the quadratic form associated to Q by ω. Given [u′
1],
[v′
1], [u′
2], [v′
2], we have to reconstruct [u1], [v1], [u2], [v2] subject to
the constraints
ω(ui,ui) = ω(vi,vi) = ω(ui,vj) = 0,
i, j ∈{1,2}
(11)
and
[m1] ∈[u1]∨[v1],
[n1] ∈[v1]∨[u2],
[m2] ∈[u2]∨[v2],
[n2] ∈[v2]∨[u1].
(12)
If [u1] is given, we find [v1] by projecting [u1] from centre [m1] onto
E ∨[v′
1]. Lemma 6 tells us that we can find [u2] and [v2] in similar
manner such that (12) is satisfied. Then, some of the conditions in
(11) become redundant and it is sufficient to consider only
ω(u1,u1) = ω(v1,v1) = ω(u2,u2) = ω(v2,v2) = 0.
(13)
In a projective coordinate system with base points [u′
1], [v′
1], [u′
2],
[v′
2] ∈F and further base points in E, the quadratic form ω of the
quadric Q is described by a matrix of the shape

A
B
B
O

where A, B and O are matrices of dimension 4 × 4, O is the zero
matrix and B is regular. Now (13) gives rise to a linear system for
the unknown coordinates of [u1], [v1], [u2], and [v2]. We have
v1 = u1 +ζ1m1,
u2 = v1 +η1n1 = u1 +ζ1m1 +η1n1,
v2 = u2 +ζ2m2 = u1 +ζ1m1 +η1n1 +ζ2m2
with certain scalars ζ1, η1, ζ2 that can be computed from u′
1,
v′
1, u′
2, v′
2 and m1, n1, m2, n2 alone.
Now we write [u1] =
[1,0,0,0,x0,x1x2,x3] with unknown x0, x1, x2, x3. Then, (13) yields
0 = [1,0,0,0]·B[x0,x1,x2,x3]⊺,
0 = [0,1,0,0]·B([x0,x1,x2,x3]+ζ1m′′
1)⊺,
0 = [0,0,1,0]·B([x0,x1,x2,x3]+ζ1m′′
1 +η1n′′
1)⊺,
0 = [0,0,0,1]·B([x0,x1,x2,x3]+ζ1m′′
1 +η1n′′
1 +ζ2m′′
2)⊺
(14)
where the double prime denotes projection on the last four coordi-
nates. (14) is a linear system for x0, x1, x2, x3 with regular coeffi-
cient matrix B. Hence, the solution is unique.
Acknowledgement
This work was supported by the Austrian Science Fund (FWF):
P 26607 (Algebraic Methods in Kinematics: Motion Factorisation
and Bond Theory).
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