SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
PRZEMYS LAW DOBROWOLSKI
Abstract. The swing-twist decomposition is a standard routine in motion
planning for humanoid limbs. In this paper the decomposition formulas are
derived and discussed in terms of Clifford algebra. With the decomposition one
can express an arbitrary spinor as a product of a twist-free spinor and a swing-
free spinor (or vice-versa) in 3-dimensional Euclidean space. It is shown that in
the derived decomposition formula the twist factor is a generalized projection
of a spinor onto a vector in Clifford algebra.
As a practical application of
the introduced theory an optimized decomposition algorithm is proposed. It
favourably compares to existing swing-twist decomposition implementations.
1. Introduction
Swing-twist decomposition of rotations is commonly used in context of humanoid
motion planning. Consider movement of an arm reaching some predefined position.
In order to displace it properly, a controller calculates twist factor of the corre-
sponding rotation. Having this factor computed, a controller is then able to apply
some corrections to the motion so that the arm is not unnaturally twisted during
the motion. Swing-twist decomposition is an inherent part of a correction al-
gorithm. It allows one to decompose an arbitrary rotation into a swing part (tilt
of a given axis) and a twist part (rotation around a given axis). In this paper the
decomposition is derived and discussed in terms of Clifford algebra.
Swing-twist decomposition has already been considered for a few decades. Differ-
ent authors have obtained equivalent formulas, in particular for quaternion algebra.
Current literature tends to neglect a deeper consideration on the spin-twist decom-
position. There is wide range of publications on humanoid motion planning. Most
of these works relate to swing-twist decomposition in some way. In the paper, the
most revelant approaches to swing-twist decomposition are quoted and compared.
Starting from the recent PhD dissertation by Huyghe ([1], 2011), the swing-twist
decomposition is introduced in quaterion algebra by a proposed therein projection
operator. Unfortunately, the origin of the projection operator is not enough ex-
plained. In two previous papers, Baerlocher ([2], 2001) and Baerlocher, Boulic ([3],
2000) investigate joint boundaries for ball-and-socket joints using swing-twist de-
composition. Grassia ([4], 1998) discusses features of a swing-twist decomposition
in terms of an exponential map. However, the author does not present any re-
lated formula for the decomposition. A classic work by Korein ([5], 1984) contains
most of the initial results on body positioning and joint motion. Among others
the work uses swing-twist decomposition of rotations to constrain movement of an
elbow.
None of these are generalized to Clifford or geometric algebras which has deeper
consequences than previous results. In this paper the swing-twist decomposition
1
arXiv:1506.05481v1  [cs.RO]  17 Jun 2015
2
PRZEMYS LAW DOBROWOLSKI
is derived as an inverse of a formula expressing the set of rotations which move a
given initial vector to cover a given terminal one.
2. Existing and related solutions
Swing-twist decomposition splits a given rotation in two parts: a swing part and
a twist part. A schematic view of a limb rotating with a ball joint is presented in
figure 1.
Figure 1. A limb with a ball joint
By zero twist reference vector one refers to a base vector with respect to
which the swing-twist decomposition is performed. Usually, this is the bone of a
rotating limb.
There are several existing approaches. These solutions differ in terms of perfor-
mance, complexity and result exactness.
2.1. Direct method. For quaternions one can make the following argument. Let
q ∈H be a quaternion and v be a zero-twist reference vector. In case of twist-after-
swing type of decomposition (see section 6) the initial vector becomes the given v
vector and the terminal vector is w = qvq∗. Schematic view is presented in figure
2.
Figure 2. A direct decomposition of a quaternion
Swing quaternion can be calculated with an axis angle representation of the
quaternion. Here, the axis is a normalized vector perpendicular to v and w and the
angle is equal to the angle between v and w so:
n =
v × w
∥v × w∥
cos(α) = v · w
qs = cos(α/2) + sin(α/2)(nxi + nyj + nzk)
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
3
Twist quaternion is calculated by the following inversion formula:
q = qtqs =⇒qt = qq−1
s
qt = q(cos(α/2) −sin(α/2)(nxi + nyj + nzk))
In this formulation both trigonometric and inverse trigonometric functions are
used.
It is a practical disadvantage since it requires computationally expensive
functions which can also face some accuracy issues.
2.2. Huyghe’s method. Huyghe uses swing-after-twist decomposition type. First
a simplified formula is derived which is a decomposition of an arbitrary quaternion
q into a product
q = qsqt
with respect to a constant Z-axis aligned zero-twist reference vector (called there
”a twist axis”). With the following coordinates
q = w + xi + yj + zk
qt = wt + ztk
qs = ws + xsi + ysj
the author shows using some algebraic transformations that the decomposition is
wt =
±w
√
w2 + z2
zt =
±z
√
w2 + z2
ws = wtw + ztz
xs = wtx −zty
ys = wty + ztx
It must be noted that some of the results were not rigorously stated. It particular,
scenarios when some coefficients are equal to zero are not discussed separately which
leads to possible division by zero.
In the second part of the thesis, Huyghe derives a generalized formula for swing-
twist decomposition. Assume that axis-angle representation of quaternions is
qs = [ws, vs] = [cos(σ
2 ), us sin(σ
2 )]
qt = [wt, vt] = [cos(τ
2), ut sin(τ
2)]
Multiplying both quaternions one writes
(1)
q = qsqt = [wswt −vt · vs, wsvt + wtvs + vs × vt]
Huyghe notes that the axes vs and vt are perpendicular so formula (1) simplifies to
(2)
q = [wswt, wsvt + wtvs + vs × vt]
Next, the author introduces a new quaternion qp (no origin is provided) which is
”a projected version of the initial quaternion q onto the twist axis” ([1])
qp = [w, (v · ut)ut]
= [wswt, (wsvt · ut + wtvs · ut + (vs × vt) · ut)ut]
= [wswt, ws∥ut∥2 sin(τ
2)ut]
4
PRZEMYS LAW DOBROWOLSKI
= [wswt, wsvt]
Normalization of quaternion qp gives
qp
∥qp∥=
[wswt, wsvt]
p
w2sw2
t + w2s∥vt∥2 =
ws[wt, vt]
ws
p
w2
t + ∥vt∥2 = qt
which in result turns out to be twist quaternion. Note that in the above equation
Huyghe does not consider ws = 0 which is a drawback. Remaining swing quaternion
is calculated from
qs = qq∗
t
which completes the decomposition.
3. Preliminaries
The presented results relate to Cℓ3 := Cℓ3,0(R) - Clifford algebra of 3-dimensional
real space. Herein, a spinor is given by s = a + be12 + ce23 + de31 ∈Spin(3) and a
vector is given by v = xe0 + ye1 + ze2. A rotation of a vector is given by Clifford
multiplication:
v′ = svs−1
for an arbitrary vector v and a spinor s.
In this paper □−1 denotes the inverse of a spinor. In case of Spin(3) it is equiv-
alent to conjugation:
s−1 = ˜s = a −be12 −ce23 −de31
A spinor can be written as a sum of its scalar and bivector parts:
s = [s]0 + [s]2
where
[s]0 ∈V
0 R3 and [s]2 ∈V
2 R3. In several places the Hodge star ⋆operator
is used. In an orthonormal basis it is defined as
⋆(e1 ∧e2 ∧· · · ∧ek) = ek+1 ∧ek+2 ∧· · · ∧en
but in this paper it is used only in relation to the bivector part of a spinor. In this
case the formula can be simplified to:
⋆[s]2 = −e123[s]2 ∈
^
1
R3
A convenient notation is used for normalized vectors. For a given non-zero vector
v ̸= 0, the normalization function is defined as:
n(v) =
v
∥v∥
where ∥v∥= √v · v = √vv is the length of vector v. Note that it is impossible
to define a normalized zero vector. Let s be a spinor and v be a vector in Cℓ3.
The rotation svs−1 of the vector v by the spinor s gives a rotated vector v′ =
x′e0 + y′e1 + z′e2 which is equal to:





x′ = (a2 −b2 + c2 −d2)x + 2y(ab + cd) + 2z(bc −ad)
y′ = (a2 −b2 −c2 + d2)y + 2x(cd −ab) + 2z(bd + ac)
z′ = (a2 + b2 −c2 −d2)z + 2x(bc + ad) + 2y(bd −ac)
(3)
This paper if organized as follows: first we obtain the complete set of spinors
which do not rotate a given non-zero vector. Next we derive a spinor which rotates
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
5
a given initial vector to cover a given terminal vector. By combining these two
formulas we derive a formula which represents an arbitrary rotation by initial and
terminal vectors (swing factor) together with axis rotation (twist factor). Finally
swing-twist representation is inversed for an arbitrary rotation and a unique swing-
twist decomposition of a spinor is obtained. In conclusion some applications of the
proposed decomposition are presented.
4. The set of spinors which do not rotate a given vector
In this section we assume that v is a non-zero vector. The set of all spinors
s ∈Spin(3) which do not rotate the given vector v will be called an invariant set
of spinors for a given vector. The following theorem will be proved:
Proposition 1 (The set of spinors which do not rotate a given vector). Let v be
a non-zero vector in Cℓ3. The complete set of spinors s such that svs−1 = v is
determined by:
s = exp(e123α n(v))
for all α ∈[0; 2π). For a given element a ∈Cℓ3 exponential of a is defined as:
exp(a) :=
∞
X
k=0
ak
k!
Proof. The proof will use the coordinate expansion of spinor rotation formula and
the normalization identity giving the following set of equations:











(a2 −b2 + c2 −d2 −1)x + 2y(ab + cd) + 2z(bc −ad) = 0
(a2 −b2 −c2 + d2 −1)y + 2x(cd −ab) + 2z(bd + ac) = 0
(a2 + b2 −c2 −d2 −1)z + 2x(bc + ad) + 2y(bd −ac) = 0
a2 + b2 + c2 + d2 −1 = 0
(4)
The key idea is to extract simple relations from 4 yet avoiding high order equations.
Denote Qi as the left side of ith equation of 4. A valid solution (a, b, c, d) must
satisfy the following equation:
(5)
xQ1 + yQ2 + zQ3 −(x2 + y2 + z2)Q4 = 0
formula (5) can be expanded:
(a2 −b2 + c2 −d2 −1)x2 + 2xy(ab + cd) + 2xz(bc −ad)
+ (a2 −b2 −c2 + d2 −1)y2 + 2xy(cd −ab) + 2yz(bd + ac)
+ (a2 + b2 −c2 −d2 −1)z2 + 2xz(bc + ad) + 2yz(bd −ac)
−(x2 + y2 + z2)(a2 + b2 + c2 + d2 −1) = 0
(a2 −b2 + c2 −d2 −1 −a2 −b2 −c2 −d2 + 1)x2 + 2xycd + 2xzbc
+ (a2 −b2 −c2 + d2 −1 −a2 −b2 −c2 −d2 + 1)y2 + 2xycd + 2yzbd
+ (a2 + b2 −c2 −d2 −1 −a2 −b2 −c2 −d2 + 1)z2 + 2xzbc + 2yzbd = 0
(b2 + d2)x2 + (b2 + c2)y2 + (c2 + d2)z2 −2xycd −2xzbc −2yzbd = 0
(bx −cz)2 + (dx −cy)2 + (by −dz)2 = 0
which implies that the three identities must hold:
(6)
bx = cz,
dx = cy,
by = dz
6
PRZEMYS LAW DOBROWOLSKI
Next, identities (6) are plugged into the first equation of 4 and simplified using the
identity a2 + b2 + c2 + d2 = 1:
2(a2 + c2 −1)x + 2yab + 2ycd + 2zbc −2zad = 0
2(a2 + c2 −1)x + 2zad + 2ycd + 2zbc −2zad = 0
(a2 + c2 −1)x + ycd + zbc = 0
When x ̸= 0 there is
(a2 + c2 −1)x + ycd + zbc = 0
(a2 + c2 −1)x + yccy
x + z cz
x c = 0
(a2 + c2 −1)x2 + y2c2 + z2c2 = 0
c2(x2 + y2 + z2) = x2(1 −a2)
c = σx
√
1 −a2
p
x2 + y2 + z2
(7)
for σ ∈{−1, 1}. In the other case, when x = 0 is is easy to observe from (6) that
cy = 0,
cz = 0
plugging the above to (4):











yab −zad = 0
(a2 −b2 −c2 + d2 −1)y + 2zbd = 0
(a2 + b2 −c2 −d2 −1)z + 2ybd = 0
a2 + b2 + c2 + d2 −1 = 0
(8)
replace in the second and the third equation of (8) with by = dz:
(
(a2 −b2 −c2 + d2 −1)y + 2b2y = 0
(a2 + b2 −c2 −d2 −1)z + 2d2z = 0
(9)
(
(a2 + b2 −c2 + d2 −1)y = 0
(a2 + b2 −c2 + d2 −1)z = 0
(10)
it is impossible that both y = 0 and z = 0 since in this case x = 0 and ∥v∦= 0 by
assumption. Hence, from any of the above equations there must be:
a2 + b2 −c2 + d2 −1 = 0
subtracting from both sides of this equation the fourth equation of (8) one gets
immediately:
(a2 + b2 −c2 + d2 −1) −(a2 + b2 + c2 + d2 −1) = −2c2 = 0
(11)
=⇒c = 0
(12)
which is also covered by the general solution (7) thus it can be assumed that (7)
is the only solution. Remaining spinor components are derived as follows. When
x ̸= 0 one writes using identities (6):
b = z
xσx
√
1 −a2
p
x2 + y2 + z2 = σz
√
1 −a2
p
x2 + y2 + z2
(13)
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
7
d = y
xσx
√
1 −a2
p
x2 + y2 + z2 = σy
√
1 −a2
p
x2 + y2 + z2
(14)
in the case when x = 0 from (11) there is c = 0. To calculate b and d one rewrites
(8):











yab −zad = 0
(a2 −b2 + d2 −1)y + 2zbd = 0
(a2 + b2 −d2 −1)z + 2ybd = 0
a2 + b2 + d2 −1 = 0
(15)
Consider the case y ̸= 0 then from (6) there is b = z
yd. Plugging this identity into
the fourth equation of (15) one obtains:
a2 −1 + (z2
y2 + 1)d2 = 0
(a2 −1)y2 + (x2 + y2 + z2)d2 = 0
d = σy
√
1 −a2
p
x2 + y2 + z2
and from b = z
yd there is:
b = z
y σy
√
1 −a2
p
x2 + y2 + z2 = σz
√
1 −a2
p
x2 + y2 + z2
In the case when y = 0 there must be z ̸= 0 since ∥v∦= 0. Thus, from (6) there is
d = y
z b. Plugging this identity into the fourth equation of (15) one obtains:
a2 −1 + (y2
z2 + 1)b2 = 0
(a2 −1)z2 + (x2 + y2 + z2)b2 = 0
b = σz
√
1 −a2
p
x2 + y2 + z2
and from b = z
yd there is:
d = y
z σz
√
1 −a2
p
x2 + y2 + z2 = σy
√
1 −a2
p
x2 + y2 + z2
In all cases a general solution to (4) in coordinates is:
a ∈[−1; 1],
σ ∈{−1, 1}
(16)
b = σz
√
1 −a2
p
x2 + y2 + z2 ,
c = σx
√
1 −a2
p
x2 + y2 + z2 ,
d = σy
√
1 −a2
p
x2 + y2 + z2
(17)
There are two parametrized solutions to the set of equations (4):
s =a + σz
√
1 −a2
p
x2 + y2 + z2 e12+
σx
√
1 −a2
p
x2 + y2 + z2 e23 + σy
√
1 −a2
p
x2 + y2 + z2 e31
8
PRZEMYS LAW DOBROWOLSKI
s =a + σ
√
1 −a2
p
x2 + y2 + z2 (ze12 + xe23 + ye31)
s =a + σ
√
1 −a2
p
x2 + y2 + z2 e123(ze3 + xe1 + ye2)
s =a + σ
√
1 −a2
∥v∥
e123v
s =a + σe123
p
1 −a2 n(v)
Finally, since many of the operations were reductions, all solutions are plugged into
the original set of equations to check their validity. It can be seen that:
svs−1 = (a + σe123
p
1 −a2 n(v))v(a + σe123
p
1 −a2 n(v))−1
= (a + σe123
p
1 −a2 n(v))v(a −σe123
p
1 −a2 n(v))
= (av + σe123
p
1 −a2 n(v)v)(a −σe123
p
1 −a2 n(v))
= (av + σe123
p
1 −a2∥v∥)(a −σe123
p
1 −a2 n(v))
= ava + av(−σe123
p
1 −a2 n(v)) + (σe123
p
1 −a2∥v∥)a+
(σe123
p
1 −a2∥v∥)(−σe123
p
1 −a2 n(v))
= ava + av(−σe123
p
1 −a2 n(v)) + av(σe123
p
1 −a2 n(v))+
σ(−σ)e123e123
p
1 −a2p
1 −a2∥v∥n(v)
= ava + (1 −a2)v = a2v + v −a2v = v
which confirms that all the solutions are valid. Now we obtain the following formula:
s = a + σe123
p
1 −a2 n(v)
for a ∈[−1; 1]. Now it is further simplified by using trigonometric and exponential
series converging for all arguments. We substitute a = cos(α) to obtain:
a + σe123
p
1 −a2 n(v) = cos(α) + sin(α)e123 n(v)
where α ∈[0; 2π). Then, we express the trigonometry using exponential function.
We use series expansions which are valid for all α. Note that (−1)k = (e123e123)k =
(e123)2k and (n(v))2k = 1.
cos(α) + sin(α)e123 n(v) =
∞
X
k=0
(−1)k
(2k)! α2k + e123 n(v)
∞
X
k=0
(−1)k
(2k + 1)!α2k+1
=
∞
X
k=0
(e123)2k
(2k)! α2k(n(v))2k + e123 n(v)
∞
X
k=0
(e123)2k
(2k + 1)!α2k+1(n(v))2k
=
∞
X
k=0
(e123)2k
(2k)! α2k(n(v))2k +
∞
X
k=0
(e123)2k+1
(2k + 1)! α2k+1(n(v))2k+1
=
∞
X
k=0
(e123α n(v))2k
(2k)!
+
∞
X
k=0
(e123α n(v))2k+1
(2k + 1)!
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
9
=
∞
X
k=0
(e123α n(v))k
k!
= exp(e123α n(v))
which is precisely the stated general formula. This completes the proof.
□
5. A spinor which rotates a given vector
The following lemma states what is a spinor which rotates a given vector to be
equal to a target one.
In literature it is sometimes called direct rotation, as in
[3]. Schematic view is presented in figure 3. The lemma is required to prove a
Figure 3. A direct rotation of AB to AB′
general theorem about the set of spinors which rotate a given initial vector to a
given target vector. We start with the following initial lemma which applies to the
Clifford algebra of R2. At first the following basic property of Cℓ2 is proved:
Lemma 1. The external product of two given vectors v and w in Cℓ2 can be written
as:
w ∧v = 1
2 sgn(v ∧we12)∥v −w∥∥v + w∥e12
Proof. Assume that v = xe1 +ye2 and w = pe1 +qe2. In special case when w = ±v
both sides of the equation are equal to zero therefore the equation holds. In general
case w ̸= ±v and one can rewrite both sides in coordinates. Left-hand side is equal
to:
w ∧v = 1
2(wv −vw) = 1
2 ((pe1 + qe2)(xe1 + ye2) −(xe1 + ye2)(pe1 + qe2))
= 1
2 (2pye12 −2qxe12) = (py −qx)e12
Right-hand side is equal to:
1
2 sgn(v ∧we12)∥v −w∥∥v + w∥e12
= 1
2 sgn(1
2(vw −wv)e12)
p
(v −w)2p
(v + w)2e12
= 1
2 sgn(((xe1 + ye2)(pe1 + qe2)
−(pe1 + qe2)(xe1 + ye2))e12)
p
(v −w)2(v + w)2e12
= 1
2 sgn((2qxe12 −2pye12)e12)
p
((x −p)2 + (y −q)2)((x + p)2 + (y + q)2)e12
= 1
2 sgn(py −qx)
p
4(py −qx)2e12 = sgn(py −qx)|py −qx|e12
= (py −qx)e12
Both sides are equal so it completes the proof.
□
10
PRZEMYS LAW DOBROWOLSKI
Lemma 2. Let v and w be a pair of non-zero vectors of equal length in Cℓ2 such
that v ̸= −w. Then there exist a pair of spinors ±s ∈Spin(2) which rotate v so
that the rotated vector is equal to w:
svs−1 = w
then the pair of spinors is equal to:
s = ± n(v + w) n(v)
Proof. Let s = a + be12 ∈Spin(2) be a spinor and v = xe1 + ye2, w = pe1 + qe2
be a pair of given vectors. The equation can be expressed in coordinates with the
following set of equations:
(
(a2 −b2)x + 2aby = p
(a2 −b2)y −2abx = q
(18)
From the assumption it is impossible that both x and y are simultaneously zero.
When x ̸= 0 we solve the set of equations:





a2 −b2 = p −2aby
x
p −2aby
x
y −2abx = q
The second equation is then simplified:
(p −2aby)y −2abx2 −qx = 0
2ab = py −qx
x2 + y2
Which is then plugged into the first equation of (18):
(a2 −b2)x + py −qx
x2 + y2 y = p
(2a2 −1)x = p −py2 −qxy
x2 + y2
2a2 = px2 + py2 −py2 + qxy
x(x2 + y2)
+ 1 = px + qy
x2 + y2 + 1
a = ±
s
x2 + y2 + px + qy
x2 + y2
The same result can be obtained when we consider the other case y ̸= 0:







a2 −b2 = q + 2abx
y
q + 2abx
y
x + 2aby = p
The second equation can be simplified:
(q + 2abx)x + 2aby2 −py = 0
2ab = py −qx
x2 + y2
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
11
Which is then plugged into the second equation of (18):
(a2 −b2)y −py −qx
x2 + y2 x = q
(2a2 −1)y = q + pxy −qx2
x2 + y2
2a2 = qx2 + qy2 + pxy −qx2
y(x2 + y2)
+ 1 = px + qy
x2 + y2 + 1
a = ±
s
x2 + y2 + px + qy
2(x2 + y2)
We obtained one general solution which is correct for all x, y. Now, for each
σ ∈{−1, 1} of a = σ
q
x2+y2+px+qy
2(x2+y2)
we have exactly one corresponding solution
b. It can be calculated by plugging it into 2ab =
py−qx
2(x2+y2) valid for all x, y. When
a ̸= 0 then:
b =
py −qx
2a(x2 + y2)
b =
(py −qx)
p
2(x2 + y2)
2σ
p
x2 + y2 + px + qy(x2 + y2)
b = σ
py −qx
p
x2 + y2 + px + qy
p
2(x2 + y2)
b = σ
(py −qx)
p
x2 + y2 −px −qy
p
(x2 + y2)2 −(px + qy)2p
2(x2 + y2)
b = σ
(py −qx)
p
x2 + y2 −px −qy
p
(x2 + y2)2 −(px + qy)2p
2(x2 + y2)
b = σ
(py −qx)
p
x2 + y2 −px −qy
p
x4 + y4 + 2x2y2 −p2x2 −q2y2 −2pqxy
p
2(x2 + y2)
b = σ
(py −qx)
p
x2 + y2 −px −qy
p
x4 + y4 + 2x2y2 −(x2 + y2 −q2)x2 −(x2 + y2 −p2)y2 −2pqxy
·
1
p
2(x2 + y2)
b = σ
py −qx
p
(py −qx)2
s
x2 + y2 −px −qy
2(x2 + y2)
note that (py−qx)e12 = w∧v but it is assumed that v is not parallel to w therefore
py −qx ̸= 0. Since e2
12 = −1 the following applies:
py −qx
p
(py −qx)2 = sgn(v ∧we12)
and b is equal to:
b = σ sgn(v ∧we12)
s
x2 + y2 −px −qy
2(x2 + y2)
12
PRZEMYS LAW DOBROWOLSKI
When a = 0 then from b2 = 1 −a2 we know that b = ±1 which is also a special
case of the general solution. Finally, we simplify the formula to a coordinate-free
solution. For each σ ∈{−1, 1} there is a solution:
s = σ
s
x2 + y2 + px + qy
2(x2 + y2)
+ σ sgn(v ∧we12)
s
x2 + y2 −px −qy
2(x2 + y2)
e12
s =
σ
p
2(x2 + y2)
(
p
x2 + y2 + px + qy+
sgn(v ∧we12)
p
x2 + y2 −px −qye12)
s =
σ
p
2(x2 + y2)
(
r
1
2(x2 + y2 + p2 + q2 + 2(px + qy))+
sgn(v ∧we12)
r
1
2(x2 + y2 + p2 + q2 −2(px + qy))e12)
s =
σ
2
p
x2 + y2 (
p
x2 + y2 + p2 + q2 + 2(px + qy)+
sgn(v ∧we12)
p
x2 + y2 + p2 + q2 −2(px + qy)e12)
s =
σ
2∥v∥(
√
vv + ww + vw + wv+
sgn(v ∧we12)
√
vv + ww −vw −wve12)
s =
σ
2∥v∥(
p
(v + w)2 + sgn(v ∧we12)
p
(v −w)2e12)
s =
σ
2∥v∥(∥v + w∥+ sgn(v ∧we12)∥v −w∥e12)
From the assumptions, there is ∥v + w∦= 0 so:
s =
σ
2∥v∥∥v + w∥(∥v + w∥2 + sgn(v ∧we12)∥v −w∥∥v + w∥e12)
Next, the basic identity from lemma 1 is used so the formula can be rewritten with
only external product:
w ∧v = 1
2 sgn(v ∧we12)∥v −w∥∥v + w∥e12
It reads that the external product is equal to a bivector whose area is equal to the
half of the area of the parallelogram spanned by the both vectors and with the sign
adjusted to the sign of the external product. Using the above identity, the following
further simplifications are possible:
s =
σ
2∥v∥∥v + w∥(∥v + w∥2 + 2v ∧w)
s =
σ
2∥v∥∥v + w∥(vv + ww + 2v · w + 2v ∧w)
s =
σ
∥v∥∥v + w∥(vv + wv)
s =
σ
∥v∥∥v + w∥(v + w)v
s = ± n(v + w) n(v)
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
13
Which is precisely the stated formula.
□
We observe that the same vector formula applies to any Clifford algebra of n-
dimensional real space. Thus, we propose the following:
Proposition 2. Let v and w be a pair of non-zero vectors of equal length in Cℓn
such that v ̸= −w. Then there exist a pair of spinors each of which rotate v so that
it is equal to w:
svs−1 = w
the pair of spinors is equal to:
s = ± n(v + w) n(v)
Proof. Instead of the original equation svs−1 = w an equivalent equation will be
proved:
sv = ws
Let v and w are the given vectors. Start with:
v −v = v −v
since vv = ww > 0 each element can be multiplied by the scalar vv or ww:
vvv −wwv = wvv −wvv
vvv + wvv = wvv + wwv
(v + w)vv = w(v + w)v
it is assumed that v ̸= −w so ∥v + w∥> 0 and both sides can be divided by
∥v + w∥∥v∥:
n(v + w) n(v)v = w n(v + w) n(v)
To ensure that s is a spinor it is sufficient to check whether it is a direct product
of a scalar and a bivector and whether its norm is equal to 1. Indeed:
s = n(v + w) n(v) =
vv + wv
∥v + w∥∥v∥= ∥v∥2 + wv
∥v + w∥∥v∥∈
^
0
R3 ⊗
^
2
R3
and since
s−1 = n(∥v∥2 + w · v + w ∧v)−1
= n(∥v∥2 + v · w + v ∧w) = n(vv + vw) = n(v) n(v + w)
the norm is:
ss−1 = n(v + w) n(v) n(v) n(v + w) = ∥n(v + w)∥2∥n(v)∥2 = 1
Which completes the proof.
□
Since the theorem is valid for Cℓn it is also valid for Cℓ3.
In more general
conclusion, for any dimension there exists a simple formula giving a pair of spinors
rotating a given vector so that it covers another given vector of the same length.
14
PRZEMYS LAW DOBROWOLSKI
6. The set of spinors which rotate a given initial vector to a given
target vector
There are two ways of defining a swing-twist representation (composition or
decomposition):
• twist q is done before swing p; for a given spinor r we have: s = pq.
This is the swing-after-twist representation.
• twist q is done after swing p; for a given spinor r we have: s = qp.
This is the twist-after-swing representation.
The swing-after-twist representation is used in [1] while the twist-after-swing rep-
resentation is usually used in the direct method. Since in practice both conventions
are used (the first one is slightly less common) in this paper both decompositions
will be presented and proved. The following theorem is a general solution to Clifford
product equation:
svs−1 = w
We prove the following
Proposition 3. Let v and w be a pair of non-zero vectors of equal length in Cℓ3
such that v ̸= −w. The set of spinor solutions s ∈Spin(3) to
svs−1 = w
in the case of swing-after-twist representation is equal to:
s = ± n(v + w) n(v) exp(e123α n(v))
and in the case of twist-after-swing representation is equal to:
s = ± exp(e123α n(w)) n(v + w) n(v)
Proof. The solution is the set of all possible rotations which swing axis from initial
v to terminal w with any possible twist during the movement. Using propositions 1
and 2 one composes swing and twist according to the order used in a given represen-
tation. In the case of swing-after-twist representation twist q = exp(e123α n(v)) fac-
tor (around v axis) is applied at first and only after it, swing factor p = n(v+w) n(v)
is applied (swinging the axis from v to w). In the other case of twist-after-swing
representation swing p = n(v + w) n(v) is applied first (swinging the axis from v to
w) and after that twist factor q = exp(e123α n(w)) is applied with respect to the
terminal axis w.
□
7. Swing-twist decomposition of a spinor
This is the main result of this paper. In this section the inverse of formulas given
in proposition 3 is derived. Given a spinor it is possible to calculate its decomposi-
tion into twist and swing factors in respect to a given non-zero vector. Since there
are two different swing-twist representations, there are also two swing-twist decom-
positions for swing-after-twist and twist-after-swing representation respectively. In
this section, the initial vector will be called a base vector.
Theorem 1 (Swing-twist decomposition of a spinor in swing-after-twist represen-
tation). Assume that s ∈Spin(3) is a spinor. For any non-zero base vector v ∈Cℓ3
such that svs−1 ̸= −v there exist a unique up to the sign swing-twist decompo-
sition in swing-after-twist representation
s = ±pq
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
15
where swing spinor p and twist spinor q are equal to:
p = ±s˜σv(s)
q = ±σv(s)
where σv(s) : Spin(3) −→Spin(3) is a function of spinor s:
σb(s) = n(v(v · s))
the reversion ˜σv(s) is equal to:
˜σv(s) = n(v(v · ˜s))
The function σv(s) will be called a twist projection function.
Proof. The proof is divided into several steps. At first the problem is reformulated
in coordinates. Then, twist angle is calculated and twist spinor and finally swing
spinor. Denote the following:
s = a + be12 + ce23 + de31
v = vxe1 + vye2 + vze3
w = wxe1 + wye2 + wze3
The angle α will be described implicitly by:
cos(α) = k
sin(α) = l
k2 + l2 = 1
Vector w exists on the assumption that svs−1 ̸= −v. What does this assumption
require about the base vector and the spinor is explained in the appendix of this
paper. At first, the equation is rewritten in coordinates. The swing factor is equal
to:
p = n(w + v) n(v)
= ((wx + vx)e1 + (wy + vy)e2 + (wz + vz)e3)(vxe1 + vye2 + vze3)
p
(wx + vx)2 + (wy + vy)2 + (wz + vz)2q
v2x + v2y + v2z
= ((wx + vx)2 + (wy + vy)2 + (wz + vz)2)−1
2 (v2
x + v2
y + v2
z)−1
2 ·
[(wx + vx)vx + (wy + vy)vy + (wz + vz)vz+
((wx + vx)vy −(wy + vy)vx)e12 + ((wy + vy)vz −(wz + vz)vy)e23+
((wz + vz)vx −(wx + vx)vz)e31]
Twist factor is equal to:
(19)
q = cos(α) + e123 n(v) sin(α) = k +
l
q
v2x + v2y + v2z
(bze12 + bxe23 + bye31)
Combining the above formulas there is:
pq = ((wx + vx)2 + (wy + vy)2 + (wz + vz)2)−1
2 (v2
x + v2
y + v2
z)−1
2 ·
[(wx + vx)vx + (wy + vy)vy + (wz + vz)vz+
((wx + vx)vy −(wy + vy)vx)e12 + ((wy + vy)vz −(wz + vz)vy)e23+
((wz + vz)vx −(wx + vx)vz)e31]·
16
PRZEMYS LAW DOBROWOLSKI
[k +
l
q
v2x + v2y + v2z
(bze12 + bxe23 + bye31)]
= ((wx + vx)2 + (wy + vy)2 + (wz + vz)2)−1
2 (v2
x + v2
y + v2
z)−1
2 ·
[k((wx + vx)vx + (wy + vy)vy + (wz + vz)vz)+
(k(wxvy −wyvx) + l
q
v2x + v2y + v2z(wz + vz))e12+
(k(wyvz −wzvy) + l
q
v2x + v2y + v2z(wx + vx))e23+
(k(wzvx −wxvz) + l
q
v2x + v2y + v2z(wy + vy))e31]
Two spinors are equal if and only if corresponding coefficients are equal. There-
fore the following set of equations determines the solution:
(20)



























a = [k((wx + vx)vx + (wy + vy)vy + (wz + vz)vz)]∆
b = [k(wxvy −wyvx) + l
q
v2x + v2y + v2z(wz + vz)]∆
c = [k(wyvz −wzvy) + l
q
v2x + v2y + v2z(wx + vx)]∆
d = [k(wzvx −wxvz) + l
q
v2x + v2y + v2z(wy + vy)]∆
k2 + l2 = 1
w2
x + w2
y + w2
z = v2
x + v2
y + v2
z
a2 + b2 + c2 + d2 = 1
Where ∆= ((wx+vx)2+(wy+vy)2+(wz+vz)2)−1
2 (v2
x+v2
y+v2
z)−1
2 . By substituting
variables in the set of equations (20) it is easy to rise unfavourably the degree of
involved polynomials. Thus the following careful operations are performed. First,
note that it is easy to obtain variable k from the first equation:
(21)
a = k(w + v) · v
∥w + v∥∥v∥
Observe that:
∥w + v∥2 = (w + v)(w + v) = ww + vv + wv + vw = 2(v · v + v · w) = 2v · (w + v)
Using this, one simplifies (21) to:
a = k(w + v) · v
∥w + v∥∥v∥= k∥w + v∥2
∥w + v∥∥v∥= k ∥w + v∥
2∥v∥
From that one obtains the value of k:
(22)
k = a
2∥v∥
∥w + v∥
The value of k is now put into the set of equations (20) which then can be simplified
to:
(23)





b∥w + v∥= 2a
1
∥w+v∥(wxvy −wyvx) + l(wz + vz)
c∥w + v∥= 2a
1
∥w+v∥(wyvz −wzvy) + l(wx + vx)
d∥w + v∥= 2a
1
∥w+v∥(wzvx −wxvz) + l(wy + vy)
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
17
It is a set of three linear equations with three unknowns M[wx, wy, wz]T = N. The
characteristic matrix M is:
M =

2a
1
∥w+v∥vy
−2a
1
∥w+v∥vx
l
l
2a
1
∥w+v∥vz
−2a
1
∥w+v∥vy
−2a
1
∥w+v∥vz
l
2a
1
∥w+v∥vx

and the vector N is equal to:
N = [b∥w + v∥−lvz, c∥w + v∥−lvx, d∥w + v∥−lvy]T
Any known method can be used to compute the determinant of M which is equal
to:
det(M) = l
The case l = 0 needs special care and will be now discussed separately. In this case,
from (19) twist factor is equal to:
q = k +
l
q
v2x + v2y + v2z
(bze12 + bxe23 + bye31) = k
Since q is a spinor, its component k must be equal to:
k = ±1
which then implies that twist and swing factors are equal to:
p = ±s
q = ±1
It is a special case of the general formula. It holds for all vector v and spinors s
such that α = zπ, z ∈Z. In the case of l ̸= 0 there exist exactly one solution to
(23). After having the matrix inverted and solution calculated, one gets:
wx = [4a2(v2
x + v2
y + v2
z)l + ∥w + v∥2l3]−1
[2a(bvy −dvz)∥w + v∥2l+
∥w + v∥2l2(c∥w + v∥−vxl) + 4a2vx((bvz + vxc + vyd)∥w + v∥−(v2
x + v2
y + v2
z)l)]
wy = [4a2(v2
x + v2
y + v2
z)l + ∥w + v∥2l3]−1
[2a(−bvx + cvz)∥w + v∥2l+
∥w + v∥2l2(d∥w + v∥−vyl) + 4a2vy((bvz + vxc + vyd)∥w + v∥−(v2
x + v2
y + v2
z)l)]
wz = [4a2(v2
x + v2
y + v2
z)l + ∥w + v∥2l3]−1
[2a(cvy + dvx)∥w + v∥2l+
∥w + v∥2l2(b∥w + v∥−vzl) + 4a2vz((bvz + vxc + vyd)∥w + v∥−(v2
x + v2
y + v2
z)l)]
All three denominators are equal and can be simplified:
4a2(v2
x + v2
y + v2
z)l + ∥w + v∥2l3
= l(4a2∥v∥2 + ∥w + v∥2(1 −a2
4∥v∥2
∥w + v∥2 )) = l∥w + v∥2
18
PRZEMYS LAW DOBROWOLSKI
A compact form is achieved with the following simplifications:

wx
wy
wz

=
1
l∥w + v∥2 (2a

bvy −dvz
cvz −bvx
dvx −cvy

∥w + v∥2l + ∥w + v∥2l2

c∥w + v∥−vxl
d∥w + v∥−vyl
b∥w + v∥−vzl

+
4a2

vx
vy
vz

((bvz + cvx + dvy)∥w + v∥−∥v∥2l))
= 2a

vx
vy
vz

×

c
d
b

+ l∥w + v∥

c
d
b

−l2

vx
vy
vz

+
4a2
l∥w + v∥2

vx
vy
vz

(∥w + v∥

vx
vy
vz

·

c
d
b

−∥v∥2l)
Vector [c, d, b]T can be formulated with coefficients of spinor s:
⋆[s]2 = −e123[s]2 = −e123(be12 + ce23 + de31) = ce1 + de2 + be3
With this substitution further simplification is possible:
w = 2av × ⋆[s]2 + l∥w + v∥⋆[s]2 −l2v +
4a2
l∥w + v∥2 v(∥w + v∥v · ⋆[s]2 −∥v∥2l)
but form previous equations there is:
(24)
l2 = 1 −a2
4∥v∥2
∥w + v∥2
so after substitution one obtains:
w = 2av × ⋆[s]2 + l∥w + v∥⋆[s]2 −v +
4a2v
l∥w + v∥v · ⋆[s]2
The above formula is useful to compute ∥w + v∥. Note that one can move −v to
the left hand side of the equation and compute lengths of both sides.
∥w + v∥2 = (2av × ⋆[s]2 + l∥w + v∥⋆[s]2 +
4a2v
l∥w + v∥2 v · ⋆[s]2)2
= (2av × ⋆[s]2)2 + (l∥w + v∥⋆[s]2)2 + (
4a2v
l∥w + v∥2 v · ⋆[s]2)2
+ 2[(2av × ⋆[s]2) · (l∥w + v∥⋆[s]2) + (2av × ⋆[s]2) · (
4a2v
l∥w + v∥2 v · ⋆[s]2)
+ (l∥w + v∥⋆[s]2) · (
4a2v
l∥w + v∥2 v · ⋆[s]2)]
= 4a2(b × ⋆[s]2)2 + l2∥w + v∥2(⋆[s]2)2 + 16a4∥v∥2
l2∥w + v∥2 (v · ⋆[s]2)2 + 8a2(v · ⋆[s]2)2
From the property that (b×⋆[s]2)2 +(b·⋆[s]2)2 = ∥b∥2(⋆[s]2)2 the following formula
is obtained:
∥w+v∥2 = 4a2∥b∥2(⋆[s]2)2+l2∥w+v∥2(⋆[s]2)2+4a2(v·⋆[s]2)2+ 16a4∥v∥2
l2∥w + v∥2 (v·⋆[s]2)2
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
19
There is:
(25)
l2∥w + v∥2 =

1 −4a2∥v∥2
∥w + v∥2

∥w + v∥2 = ∥w + v∥2 −4a2∥v∥2
So one can write the following equivalent formula:
∥w + v∥2 = 4a2∥b∥2(⋆[s]2)2 + (∥w + v∥2 −4a2∥v∥2)(⋆[s]2)2+
4a2(v · ⋆[s]2)2 + 16a4∥v∥2
l2∥w + v∥2 (v · ⋆[s]2)2
= ∥w + v∥2(⋆[s]2)2 + 4a2(v · ⋆[s]2)2 + 16a4∥v∥2
l2∥w + v∥2 (v · ⋆[s]2)2
Substituting (⋆[s]2)2 = 1 −a2 and once again (25) one obtains:
∥w + v∥2 = ∥w + v∥2(1 −a2) + (v · ⋆[s]2)2
4a2(∥w + v∥2 −4a2∥v∥2) + 16a4∥v∥2
l2∥w + v∥2

Which can be simplified to:
a2∥w + v∥2 = (v · ⋆[s]2)2 4a2
l2
a2(∥w + v∥2 −4a2∥v∥2) = 4a2(v · ⋆[s]2)2
There are two cases. When a ̸= 0 the following equation applies:
∥w + v∥2 = 4(v · ⋆[s]2)2 + 4a2∥v∥2
Using the above identity, there is:
l2∥w + v∥2 = ∥w + v∥2 −4a2∥v∥2 = 4(v · ⋆[s]2)2 + 4a2∥v∥2 −4a2∥v∥2
l2∥w + v∥2 = 4(v · ⋆[s]2)2
Modifying the identity (24) there is:
l2 = 1 −a2
4∥v∥2
∥w + v∥2
∥w + v∥2 −4a2∥v∥2 = l2∥w + v∥2
∥w + v∥2 −4a2∥v∥2 = 4(v · ⋆[s]2)2
∥w + v∥2 = 4(v · ⋆[s]2)2 + 4a2∥v∥2
∥w + v∥= 2
p
(v · ⋆[s]2)2 + a2∥v∥2
Using the formula (22) one obtains the value of k:
(26)
k = a
2∥v∥
∥w + v∥= a
2∥v∥
2
p
(v · ⋆[s]2)2 + a2∥v∥2 =
a∥v∥
p
(v · ⋆[s]2)2 + a2∥v∥2
Again from (24) it is possible to calculate l:
l2 = 1 −a2
4∥v∥2
∥w + v∥2 = 1 −
4a2∥v∥2
4[(v · ⋆[s]2)2 + a2∥v∥2]
= (v · ⋆[s]2)2 + a2∥v∥2
(v · ⋆[s]2)2 + a2∥v∥2 −
a2∥v∥2
(v · ⋆[s]2)2 + a2∥v∥2 =
(v · ⋆[s]2)2
(v · ⋆[s]2)2 + a2∥v∥2
It is possible to take the square root and abandon sign. This comes from the fact
that when s is a given spinor then the solution represented by variable l must
20
PRZEMYS LAW DOBROWOLSKI
include the solution for −s as well. In that case the numerator is negated, so also
−l is a solution automatically.
(27)
l =
v · ⋆[s]2
p
(v · ⋆[s]2)2 + a2∥v∥2
In the case when a = 0 from (24) there is:
l2 = 1
k2 = 1 −l2 = 0
The same argumentation as above can be used to select any solution of l2 = 1.
Both cases are special cases of (26) and (27) so it is sufficient to consider only those
general solutions. The twist factor equal to:
q = k + e123 n(v)l =
=
a∥v∥
p
(v · ⋆[s]2)2 + 4a2∥v∥2 +
e123 n(v)(v · ⋆[s]2)
p
(v · ⋆[s]2)2 + a2∥v∥2
=
a∥v∥+ n(v)(v · ⋆[s])
p
(v · ⋆[s]2)2 + a2∥v∥2 = a∥v∥+ n(v)(v · s) −n(v)(v · [s]0)
p
(v · ⋆[s]2)2 + a2∥v∥2
= a∥v∥+ n(v)(v · s) −a∥v∥
p
(v · ⋆[s]2)2 + a2∥v∥2
=
n(v)(v · s)
p
(v · ⋆[s]2)2 + a2∥v∥2
Define a norm of a pinor p = p0 + p12e12 + p23e23 + p31e31 ∈Pin(3):
n(p) :=
p
√p˜p = p0 + p12e12 + p23e23 + p31e31
p
p2
0 + p2
12 + p2
23 + p2
31
∈Spin(3)
where ˜p = p0 −p12e12 −p23e23 −p31e31.
Using the fact that s = a+e123⋆[s]2 the numerator of twist factor can be written
as:
n(v)(v · s) = n(v)(v · (a + e123 ⋆[s]2) = n(v)(va + e123v · ⋆[s]2)
= a∥v∥+ e123 n(v)v · ⋆[s]2
(28)
the norm of the numerator of twist factor is equal to:
p
(a∥v∥+ e123 n(v)v · ⋆[s]2)(a∥v∥−e123 n(v)v · ⋆[s]2)
=
p
a2∥v∥2 + (e123 n(v)v · ⋆[s]2)2 =
p
a2∥v∥2 + (v · ⋆[s]2)2
one notices that it is equal to the denominator. In result, it is possible to write
simply:
q = n(v(v · s))
The swing factor is calculated by modifying the original equation:
s = pq
p = sq−1
by using (28) twist inverse can be easily calculated:
q−1 = ˜q = ˜σv(s) =
^
a∥v∥+ e123 n(v)v · ⋆[s]2 = a∥v∥−e123 n(v)v · ⋆[s]2
= a∥v∥+ e123 n(v)v · ⋆[−s]2 = n(v)(va + e123v · ⋆[−s]2)
= n(v)(v · (a + e123 ⋆[−s]2)) = n(v)(v · ˜s)
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
21
so
p = s˜σv(s)
where ˜σv(s) = n(v)(v · ˜s). Both swing and twist factors can be negated as a given
spinor and its negation define the same rotation:
p = ±s˜σv(s)
q = ±σv(s)
which completes the proof.
□
The following dual theorem can be easily proved with the previous theorem (1):
Theorem 2 (Swing-twist decomposition of a spinor in twist-after-swing represen-
tation). Assume that s ∈Spin(3) is a spinor. For any non-zero base vector v ∈Cℓ3
such that svs−1 ̸= −v there exist a unique up to the sign swing-twist decompo-
sition in twist-after-swing representation
s = ±qp
where swing spinor p and twist spinor q are equal to:
p = ±˜σv(s)s
q = ±σv(s)
where σv(s) is a twist projection function.
Proof. Assume that u = s−1. According to theorem 2 there exists a swing-twist
decomposition of spinor u in swing-after-twist representation:
u = ±pq
where p is a swing factor and q is a twist factor of u in respect to the base vector
v. Taking inverse of both sides one obtains:
u−1 = s = ±q−1p−1
which is a twist-after-swing decomposition of spinor s. Swing and twist factors can
be rewritten as:
p−1 = (u˜σv(u))−1 = (s−1˜σv(s−1))−1 = ˜σv(s−1)−1s = ˜σv(s)s
q−1 = σv(u)−1 = σv(s−1)−1 = σv(s)
which completes the proof.
□
8. Twist projection function
In this section basic properties of the twist projection function are discussed.
Intuitively, twist projection function takes a spinor and returns its twist factor in
respect to a given vector. In this paper the following definition was assumed
σv(s) := n(v(v · s))
This is algebraically equivalent to projecting it onto a vector in the sense of Clif-
ford algebra. This an improvement over the work of Huyghe [1] where the author
introduces similar projection operator artificially in quaternion algebra. Twist pro-
jection function is a projection due to the following
Proposition 4. A twist projection function is a projection.
22
PRZEMYS LAW DOBROWOLSKI
Proof. It is enough to prove that σv(σv(s)) = σv(s) for any vector v and any spinor
w:
σv(σv(s)) = n(v(v · n(v(v · s)))) = n(v(v · a∥v∥+ e123 n(v)(v · ⋆[s]2)
p
(v · ⋆[s]2)2 + a2∥v∥2
))
= n(v v · a∥v∥+ e123(v · n(v))(v · ⋆[s]2)
p
(v · ⋆[s]2)2 + a2∥v∥2
) = n(v v · a∥v∥+ e123∥v∥v · ⋆[s]2
p
(v · ⋆[s]2)2 + a2∥v∥2
)
= n(v
(∥v∥v) · (a + ⋆[s]2)
p
(v · ⋆[s]2)2 + a2∥v∥2 ) = n(∥v∥2
n(v)(v · s)
p
(v · ⋆[s]2)2 + a2∥v∥2 )
= n(v(v · s)) = σv(s)
which completes the proof.
□
9. Applications
In this section an exemplary application of the proposed theoretical results is
presented.
It is a fast, concise and numerically stable algorithm for calculating
swing-twist decomposition of a spinor.
9.1. An efficient algorithm for swing-twist decomposition of a spinor. To
propose an efficient method of calculating swing-twist decomposition the following
formula is used for twist projection function:
σv(s) = n(v(v · s)) = n(v(v · (a + e123 ⋆[s]2)) =
= n(v(va + e123v · ⋆[s]2)) = n(a∥v∥2 + e123v(v · ⋆[s]2))
If a rotation is represented by a pinor normalization can be omitted. In case of
spinor representation normalization is needed and requires square root computa-
tion or equivalently requires arithmetic with square root extension. The proposed
method uses swing-after-twist representation and is presented in algorithm 1.
Algorithm 1 Compute swing-twist decomposition of a spinor s = pq
Require: v = xe1 + ye2 + ze3, s = a + be12 + ce23 + de31
Ensure: ∥v∦= 0
u ←xc + yd + zb
n ←x2 + y2 + z2
m ←an
l ←
√
m2 + u2n
q ←m
l + zu
l e12 + xu
l e23 + yu
l e31
p ←s˜q
return p, q
Because of its simplicity the proposed method favourably compares to existing
methods for computing swing-twist decomposition (as presented in section 2.
Appendix A. Which decompositions are impossible
When a combination of a spinor and a base vector cannot be decomposed into
swing and twist. From theorem 1 or 2 a decomposition is impossible when:
svs−1 = −v
SWING-TWIST DECOMPOSITION IN CLIFFORD ALGEBRA
23
the above condition can be rewritten in terms of spinor coordinates. Using (3) one
writes











(a2 −b2 + c2 −d2 + 1)x + 2y(ab + cd) + 2z(bc −ad) = 0
(a2 −b2 −c2 + d2 + 1)y + 2x(cd −ab) + 2z(bd + ac) = 0
(a2 + b2 −c2 −d2 + 1)z + 2x(bc + ad) + 2y(bd −ac) = 0
a2 + b2 + c2 + d2 −1 = 0
(29)
Denote as Wi the left side of ith equation of 29. A valid solution (a, b, c, d) must
satisfy the following equation:
xW1 + yW2 + zW3 + (x2 + y2 + z2)W4 = 0
the above formula can be expanded:
(a2 −b2 + c2 −d2 + 1)x2 + 2xy(ab + cd) + 2xz(bc −ad)
+ (a2 −b2 −c2 + d2 + 1)y2 + 2xy(cd −ab) + 2yz(bd + ac)
+ (a2 + b2 −c2 −d2 + 1)z2 + 2xz(bc + ad) + 2yz(bd −ac)
+ (x2 + y2 + z2)(a2 + b2 + c2 + d2 −1) = 0
(a2 −b2 + c2 −d2 + 1 + a2 + b2 + c2 + d2 −1)x2 + 2xycd + 2xzbc
+ (a2 −b2 −c2 + d2 + 1 + a2 + b2 + c2 + d2 −1)y2 + 2xycd + 2yzbd
+ (a2 + b2 −c2 −d2 + 1 + a2 + b2 + c2 + d2 −1)z2 + 2xzbc + 2yzbd = 0
(cx + dy + bz)2 + a2(x2 + y2 + z2) = 0
which implies that the two identities must hold:
a2(x2 + y2 + z2)
∧
cx + dy + bz = 0
but since ∥v∦= 0 it must be
a = 0
∧
v · ⋆[s]2 = 0
These conditions define when a spinor can be written as a swing-twist decomposition
in respect to a given base vector.
References
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ing, Ph.D. thesis, Ghent University (2011).
[2] P. Baerlocher, Inverse Kinematics Techniques for the Interactive Posture Control of Articulated
Figures, Ph.D. thesis, Swiss Federal Institute of Technology (2001).
[3] P. Baerlocher, R. Boulic, Parametrization and Range of Motion of the Ball-and-Socket Joint,
in: Proceedings of the IFIP TC5/WG5.10 DEFORM’2000 Workshop and AVATARS’2000
Workshop on Deformable Avatars, DEFORM ’00/AVATARS ’00, Kluwer, B.V., Deventer,
The Netherlands, The Netherlands, 2001, pp. 180–190.
[4] F. S. Grassia, Practical Parameterization of Rotations Using the Exponential Map, J. Graph.
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[5] J. Korein, A Geometric Investigation of Reach, The MIT Press, 1985.
Faculty of Mathematics and Information Science, Warsaw University of Technol-
ogy, Poland