Spherical-vectors and geometric interpretation ofunit quaternions

Lahcen Lamgouni∗

August 31, 2019

Abstract

The purpose of this work is to introduce and study the notion of spherical-vectors, which we can consider as a natural generalization of the arguments of com-plex numbers in the case of quaternions. After having established some elemen-tary properties of these particular vectors, we show by transport of structure thatspherical-vectors form a non-abelian additive group, isomorphic to the group of unitquaternions. This identification allows us, first, to present a new polar form of quater-nions, then to represent the unit quaternions on the unit sphere of R3 and to interprettheir multiplications geometrically.

Keywords. quaternions, spherical-vectors, argument, geometric interpretation, polarform, exponential notation, spherical form.

1 Introduction

The algebraic properties of the additive group of oriented angles of two vectors in an Eu-clidean plane have allowed to define unambiguously the argument of a complex numberand to take full advantage of the algebraic properties of the exponential form. However,if one places oneself in a three Euclidean space, most of the preceding results collapse.Thus, we cannot define the measurement of an oriented angle of two vectors (in the casewhere we consider this measurement is a real number); the quaternions arguments arenon-oriented and unstructured angles that no longer obey the Chasles relation.

The present work aims to generalize, in a way naturally compatible with complexnumbers, the notions of argument, polar form, and exponential notation to quaternions.

E-mail: lahcen.lamgouni@gmail.comMathematics Subject Classification (2010): Primary 51N30; Secondary 14L35

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2 Lahcen Lamgouni∗

Thus, we obtain a manageable tool allowing to establish algebraically and geometricallyall the properties related to the multiplication of quaternions in their exponential forms.In order to do so, we first need to define and study the notion of spherical-vectors. Bythe way, this goes back to Hamilton’s model for the group of unit quaternions (Hamilton[1], Book II, Chapter I, Section 9). In his model, Hamilton interpreted unit quaternions asvector-arcs on the unit sphere S2 of R3 .

Section 2 is devoted to spherical-vectors. First, we present the definition and mainproperties related to this notion. Then we give a geometric representation of a spherical-vector on S2. In order to establish a relation between spherical-vectors and unit quater-nions, we introduce in Section 3 a new form of quaternions, which we call spherical form.In Section 4, we construct a bijection map µ from the set V of spherical-vectors, to themultiplicative group G of unit quaternions. This bijection then allows us to transport thestructure of G into V . So we define in V the following additive internal composition lawα+ β = µ−1 (µ(β)µ(α)). Equipped with this operation, V is a non-commutative additivegroup and µ is an anti-isomorphism of groups. Thanks to spherical-vectors, we present inSection 5 a new definition of quaternion argument and polar form. Then we demonstratethe algebraic properties of the argument and exponential notation. Finally, in Section 6,we give some examples of applications.

In the sequel, the space R3 is endowed with its standard oriented Euclidean structureand its standard basis (ex, ey, ez). We designate respectively by “ · ”, “ × ” and “ ‖ · ‖ ”,the dot product, the cross product, and the Euclidean norm.

2 Spherical-Vectors

In this section, we define a spherical-vector, as well as its components and support, thenwe give its geometric representation on the unit sphere S2.

2.1 Basic Notions

Definition 2.1. Let (u, v) be an ordered pair of non-zero vectors in R3. We call scalarcomponent and vector component of (u, v), the real number λ = u·v

‖u‖‖v‖ and the vectorn = u×v

‖u‖‖v‖ , respectively.

Definition 2.2. A spherical-vector of R3 is an equivalence class of ordered pairs of non-zero vectors. Two ordered pairs represent the same spherical-vector if and only if theyhave the same scalar component and the same vector component.

Definition 2.3. Let α be a spherical-vector such that its vector component n is non-zero.We call support of α, and denote by Pα, the vector plane whose n is a normal vector.

Spherical-vectors 3

Remark 2.4. Let α = (u, v) be a spherical-vector such that its vector component n is non-zero. Let Pα be the support of α. As n = u×v

‖u‖‖v‖ , the two vectors u and v are orthogonalto n. But n is normal to Pα, so u and v belong to Pα.

The following theorem gives a characteristic property of the components of a spherical-vector.

Theorem 2.5. Let x be a real number and let w be a vector in R3. We have x2+‖w‖2 = 1

if and only if there exists a unique spherical-vector whose components are x and w.

Proof. − Suppose there exists a spherical-vector (u, v) whose components are x and w.Then x = u·v

‖u‖‖v‖ and w = u×v‖u‖‖v‖ (Definition 2.1). Let θ ∈ [0, π] be the non-oriented angle

of the two vectors u and v. So x = cos θ and ‖w‖ = sin θ. Then x2 + ‖w‖2 = 1.− Conversely, suppose now that x2+ ‖w‖2 = 1 and show there exists a unique spherical-vector (u, v) whose components are x and w. The uniqueness comes from the fact thattwo spherical-vectors are equal if and only if they have the same components (Definition2.2.). It remains now to show the existence. So, it suffices to choose u to be a non-zerovector orthogonal to w and take v = xu + w × u. Let’s first show that v is non-zero. Wehave

‖v‖2 = ‖xu+ w × u‖2

= x2‖u‖2 + ‖w × u‖2 (u ⊥ (w × u))= x2‖u‖2 + ‖w‖2‖u‖2 (w ⊥ u)

= ‖u‖2 (x2 + ‖w‖2 = 1).

So ‖v‖ = ‖u‖ 6= 0. Thus, the ordered pair (u, v) represents a spherical-vector whosecomposants are

u · v‖u‖‖v‖

=u · (xu+ w × u)

‖u‖2= x,

andu× v‖u‖‖v‖

=u× (xu+ w × u)

‖u‖2=u× (w × u)‖u‖2

= w.

This according to the double product formula:

u× (w × u) = (u · u)w − (u · w)u = ‖u‖2w.

2.2 Geometric interpretation of a spherical-vector

Let u and v be two non-zero vectors in R3. The two ordered pairs (u, v) and ( u‖u‖ ,

v‖v‖)

have the same components (Definition 2.1), so they represent the same spherical-vector

4 Lahcen Lamgouni∗

(Definition 2.2). The vectors of the second ordered pair are unitary, so there exists a uniqueordered pair (A,B) of points in the unit sphere such that u

‖u‖ = A and v‖v‖ = B. Thus, we

write (u, v) = (A,B). Hence the following remark.

Remark 2.6. − Any spherical-vector can be represented by an ordered pair (u, v) of unitvectors in R3. In this case, the components of α are written λ = u · v and n = u× v.− Any spherical-vector can be represented by an ordered pair (A,B), which we denote

byyAB, of points in the unit sphere S2 (Figure 1).

P : Support

of a

S2

A

B

α

α

u

n

vO

θ

Vector component of α

Figure 1. Geometric representation of a spherical-vector α (α = (u, v) =yAB). The components of α are

λ = u · v = cos θ and n = u× v with ‖n‖ = sin θ and θ is the non-oriented angle of the vectors u and v.

3 Spherical form of a quaternion

In this section, we introduce a new form of quaternions, which will allow, in section 4,to establish a relation between unit quaternions and spherical-vectors and will showcaseinteresting results.

Definition 3.1. We denote by N the set of quaternions written in the form ja + kb + c,where (a, b, c) ∈ R3.

Theorem 3.2. The map (a, b, c) ∈ R3 7−→ ja+ kb+ c ∈ N is an isomorphism of vectorspaces.

Proof. immediate.

Remark 3.3. In quaternion theory, we are used to identifying a vector with a pure quater-nion. So in everything that follows and to avoid any confusion, by theorem 3.2 above, wewill identify each vector w = (a, b, c) in R3 with the quaternion ja+ kb+ c in N , whichwe also denote by w. The reason for this choice, which will be clear in section 5, is to be

Spherical-vectors 5

able to define a new exponential notation of quaternions that obeys algebraic propertiesidentical to those of the exponential notation of complex numbers.

Let q = x + iy + jz + kt, (x, y, z, t) ∈ R4, be a quaternion. We factorize by i

its imaginary part in the following way, q = x + i(jt − kz + y). Thus we may writeq = x + iw, where w is the quaternion jt − kz + y ∈ N identified with the vector(t,−z, y) in R3 (Remark 3.3 below). We call spherical form of q the writing x + iw. Wecall spherical components of q, the real number x (scalar component) and the vector w(vector component).

Property 3.4. (i) Two quaternions are equal if and only if they have the same sphericalcomponents.(ii) If x and w are the spherical components of a quaternion q then |q|2 = x2 + ‖w‖2.

Proof. immediate.

We summarize in the following lemma some useful properties of the spherical form,which we will use later in this article.

Lemma 3.5. For all non-zero vectors u and v in R3, we have the following properties

(i) iu = ui,

(ii) uv = v · u+ i(v × u),

(iii) If u is a unit vector then u−1v = uv = u · v − i(u× v).

Proof. (i) Let u = (x, y, z) be a vector in R3 that we identify with the quaternion jx +

ky + z (Remark 3.3). We have then

iu = i(jx+ ky + z) = kx− jy + iz = (−jx− ky + z)i = ui.

(ii) Let u = (x, y, z) and v = (x′, y′, z′) be two vectors in R3. We have

uv = (jx+ ky + z)(jx′ + ky′ + z′)

= (−xx′ − yy′ + zz′) + i(xy′ − x′y) + j(xz′ + x′z) + k(yz′ + y′z)

= (−xx′ − yy′ + zz′) + i [(xy′ − x′y)− k(xz′ + x′z) + j(yz′ + y′z)]

= (−xx′ − yy′ + zz′) + i [j(yz′ + y′z)− k(xz′ + x′z) + (xy′ − x′y)]

= (−xx′ − yy′ + zz′) + i

yz′ + y′z−xz′ − x′zxy′ − x′y

= v · u+ i(v × u).

(iii) By (ii), we easily see that uv = v · u + i(v × u) = u · v − i(u × v). For any unitvector u, its inverse is equal to its conjugate; therefore, u−1v = uv.

6 Lahcen Lamgouni∗

4 Algebraic structure of spherical-vectors

In this section, we define and study an additive group structure on the set of spherical-vectors. This is thanks to a bijective map from the set of spherical-vectors to the multi-plicative group of unit quaternions.

In all that follows, we denote by V the set of spherical-vectors and by G the multiplica-tive group of unit quaternions.

4.1 Additive group of spherical-vectors

Let α be a spherical-vector whose components are λ and n; we write α(λ, n). We caneasily associate α with the quaternion qα defined by its spherical form qα = λ + in. ByTheorem 2.5, λ2 + ‖n‖2 = 1, so the quaternion qα is unit.

Theorem 4.1. The map below, which associates each spherical-vector α(λ, n) to the unitquaternion λ+ in, is bijective.

µ : V −→ G, α(λ, n) 7−→ λ+ in.

Proof. Let q = x+ iw be a unit quaternion written in its spherical form. As q is unit,

x2 + ‖w‖2 = 1.

So, by Theorem 2.5, there exists a unique spherical-vector α whose components are x andw. That is, µ(α) = q and hence the result.

The bijective map µ allows us to transport the structure of G, as a noncommutativemultiplicative group, to V . We then define an additive internal composition law on V by:

α + β = µ−1 (µ(β)µ(α)) .

That is,µ(α + β) = µ(β)µ(α).

So µ is an anti-isomorphism of groups, and we have therefore the following theorem.

Theorem 4.2. The map µ is an anti-isomorphism of groups, and the set V of spherical-vectors is a non-commutative additive group isomorphic to the group G of unit quater-nions.

Property 4.3. Let α = (u, v) be a spherical-vector represented by an ordered pair of unitvectors of R3. We have

(i) µ(α) = u · v + i(u× v).

Spherical-vectors 7

(ii) µ(α) = v−1u.

Proof. The first expression follows from the definition of µ (Theorem 4.1) and the factthat u · v and u× v are the components of α (Definition 2.1). Now, according to Lemma3.5.(iii), we have v−1u = v·u−i(v×u) = u·v+i(u×v). Hence the second expression.

4.2 Algebraic properties of spherical-vectors

In this section, using the anti-isomorphism µ established in Paragraph 4.1, we transportthe algebraic properties of the group G of unit quaternions into the group V of spherical-vectors.

4.2.1 Zero spherical-vector (neutral element of V)

Property 4.4. The zero spherical-vector, which we denote byy0 , is represented by any

ordered pair (u, u) where u is a unit vector of R3.

Proof. The map µ being a group anti-isomorphism, the zero spherical-vectory0 is defined

by µ(y0) = 1. So, let u and v be two unit vectors in R3 such that µ ((u, v)) = 1. Then

v−1u = 1 (Property 4.3.(ii)), so u = v. Conversely, we have µ(u, u) = u−1u = 1. Hencethe result.

4.2.2 Opposite of a spherical-vector

Property 4.5. Let u and v be two unit vectors in R3. The opposite of the spherical-vector(u, v), which we denote by −(u, v), is the spherical-vector (v, u).

Proof. Let (u, v) be a spherical-vector where u and v are two unit vectors in R3. Theopposite of (u, v) is the spherical-vector −(u, v) defined by

µ (−(u, v)) = (µ ((u, v)))−1 .

Thus, by Property 4.3.(ii)

µ (−(u, v)) =(v−1u

)−1= u−1v

= µ ((v, u)) .

Hence the result.

8 Lahcen Lamgouni∗

4.2.3 Chasles relation

Property 4.6. Let u, v and w be three unit vectors in R3. We have

(u, v) + (v, w) = (u,w).

Proof. As µ is an anti-isomorphism of groups,

µ ((u, v) + (v, w)) = µ ((v, w))µ ((u, v)) .

Then, by Property 4.3.(ii),

µ ((u, v) + (v, w)) = w−1vv−1u

= w−1u

= µ ((u,w)) .

Hence the result.

4.2.4 Straight spherical-vector

Let u be a unit vector in R3. The components of the spherical-vector (u,−u) are −1 and~0 (Definition 2.1) and do not depend on u. Hence the following definition.

Definition 4.7. We call straight spherical-vector and denote byyπ, the spherical-vector

whose components are −1 and ~0, represented by any ordered pair (u,−u) such that u is aunit vector in R3.

Let α be a spherical-vector represented by an ordered pair (u, v) of unit vectors in R3.Let nα be the vector component of α. By Definition 2.1, nα = u × v. Hence u and v areunit vectors, then the vector nα is zero if and only if v = u or v = −u (that is to say

α =y0 or α =

yπ). Hence the following remark.

Remark 4.8. The vector component of a spherical-vector is zero if and only if thisspherical-vector is zero or straight.

4.3 Sum of two spherical-vectors

In this section, we return, with more details, to the sum of two spherical-vectors definedby their components.

Theorem 4.9. Let α be a spherical-vector whose components are λ and n such that n 6= ~0,and whose support is Pα. For any unit vector u in Pα, we have

(i) The vector v = u(λ − in) = λu + n × u is the unique unit vector in Pα such thatα = (u, v),

Spherical-vectors 9

(ii) The vector w = u(λ + in) = λu − n × u is the unique unit vector in Pα such thatα = (w, u).

u

n

v

w

O

Figure 2. For all α(λ, n) ∈ V (α 6=y0 , α 6= y

π ), and for all unit vector u in Pα, there exist two uniquevectors v = λu+ n× u and w = λu− n× u such that α = (u, v) = (w, u).

Proof. (i) It suffices to solve the equation (u, v) = α where the unknown is a unitvector v.

(u, v) = α⇐⇒ µ((u, v)) = µ(α)

⇐⇒ v−1u = λ+ in (Property 4.3.(ii))

⇐⇒ v = u(λ− in).

Let’s show now that u(λ− in) = λu+ n× u.

u(λ− in) = λu− uin= λu− iun (Lemma 3.5.(i))

= λu− i(u · n− i(u× n)) (Lemma 3.5.(ii))

= λu+ n× u. (u ⊥ n,Remark 2.4)

(ii) In the same way as in (i), we solve the equation (w, u) = α where the unknownnow is a unit vector w. We have

(w, u) = α⇐⇒ µ((w, u)) = µ(α)

⇐⇒ u−1w = λ+ in (Property 4.3.(ii))

⇐⇒ w = u(λ+ in).

10 Lahcen Lamgouni∗

On the other hand, we have

u(λ+ in) = λu+ uin

= λu+ iun (Lemma 3.5.(i))

= λu+ i(u · n− i(u× n)) (Lemma 3.5.(ii))

= λu− n× u. (u ⊥ n,Remark 2.4)

Hence the proof is complete.

Theorem 4.10. Let α and β be two spherical-vectors whose supports are respectively Pαand Pβ and whose components are respectively (λα, nα) and (λβ, nβ) such that nα 6= ~0

and nβ 6= ~0.

(i) If nα × nβ 6= ~0 (Figure 3), then there are exactly two combinations of three unitvectors u, v and w such that α = (u, v) and β = (v, w). These combinations are

v =nα×nβ‖nα×nβ‖

u = v(λα + inα) = λαv − nα × vw = v(λβ − inβ) = λβv + nβ × v

and

v′ = −vu′ = −uw′ = −w.

(ii) If nα × nβ = ~0 (Figure 4), then α and β have the same support P (Pα = Pβ = P ).In this case, for any unit vector v in P , we have− The vector u = v(λα + inα) = λαv − nα × v is the unique unit vector in P suchthat α = (u, v).− The vector w = v(λβ − inβ) = λβv + nβ × v is the unique unit vector in P suchthat β = (v, w).

u

v

w

O

P

P

O

-u

-v

-w

P

P

Figure 3. nα × nβ 6= ~0 : The supports of the two spherical-vectors α and β intersect in a vector linespanned by the vector nα × nβ .

Spherical-vectors 11

α

αβ

β

αn

nβ

v

u

w

αP = Pβ

O

P : Support of α and β

Figure 4. nα × nβ = ~0 : The two spherical-vectors α and β have the same support.

Proof. (i) We have to solve the system{(u, v) = α

(v, w) = β,

where the unknowns are three unit vectors u, v and w. Suppose that

(u, v) = α and (v, w) = β,

then v is a common vector between the two vector planes Pα and Pβ (Remark 2.4). Sov belongs to the vector line Pα ∩ Pβ spanned by the non-zero vector nα × nβ , nα andnβ being respectively normal vectors to the vector planes Pα and Pβ . Therefore, v andnα × nβ are parallel. But v is unitary, so either v =

nα×nβ‖nα×nβ‖

or v = − nα×nβ‖nα×nβ‖

. Thus,

{(u, v) = α

(v, w) = β⇐⇒

v =

nα×nβ‖nα×nβ‖

(u, v) = α

(v, w) = β

or

v = − nα×nβ

‖nα×nβ‖

(u, v) = α

(v, w) = β

.

Suppose now that v ∈ Pα. So by Theorem 4.9.(ii),

(u, v) = α if and only if u = v(λα + inα) = λαv − nα × v.

Similarly, if v ∈ Pβ then we have by Theorem 4.9.(i),

(v, w) = β if and only if w = v(λβ − inβ) = λβv + nβ × v.

Therefore{(u, v) = α

(v, w) = β⇐⇒

v =

nα×nβ‖nα×nβ‖

u = λαv − nα × vw = λβv + nβ × v

or

v = − nα×nβ

‖nα×nβ‖

u = λαv − nα × vw = λβv + nβ × v

.

(ii) This result is an immediate consequence of Theorem 4.9.

12 Lahcen Lamgouni∗

We have then the necessary tools to build the sum of two spherical-vectors. Let α andβ be two spherical-vectors whose vector components are respectively nα and nβ .• If nα and nβ are not both zero (Figure 5), then there are three unit vectors u, v andw such that α = (u, v) and β = (v, w) (Theorem 4.10). Thus, according to the Chaslesrelation (Property 4.6),

α + β = (u, v) + (v, w)

= (u,w).

• If one of the two vectors nα or nβ is zero, nβ for example (the other case is treated

in the same way), so either β =y0 , or β =

yπ (Remark 4.8). Therefore, we have:

− If β =y0 , then α +

y0 = α.

− If β =yπ. So let u and v be two unit vectors such that α = (u, v). We can then write

the straight spherical-vector asyπ = (v,−v) (Definition 4.7). So,

α +yπ = (u, v) + (v,−v)= (u,−v).

u

v

w

α

α

β

β

O

αP

Pβ

α+βA

B

C

α

α

β

β

Figure 5. We have (u, v) + (v, w) = (u,w). That is,yAB +

yBC =

yAC.

5 Argument and polar form of a quaternion

In this section, we introduce, through spherical-vectors, a definition of the quaternionargument.

Spherical-vectors 13

5.1 Argument of a quaternion

Definition 5.1. Let q be a non-zero quaternion. The map µ is bijective from V in G andq/|q| ∈ G. We call argument of q, and denote by arg(q), the spherical-vector defined by

arg(q) = µ−1(q

|q|

).

That is to say,α = arg(q) if and only if µ(α) =

q

|q|.

Remark 5.2. Let q be a unit quaternion and α its argument. Since q = µ(α) (Definition5.1 above), then the spherical components of q and the components of α are the same(Theorem 4.1).

5.2 Cosine and sine of a spherical-vector

Definition 5.3. Let α be a spherical-vector whose components are λ and n.−We call cosine of α, and denote cos(α), the scalar component λ.−We call sine of α, and denote sin(α), the vector component n identified with a quater-nion in N (Remark 3.3).Thus, we define the following two applications:

cos : V −→ R sin : V −→ Nα 7−→ λ α 7−→ n.

Remark 5.4. Let α be a spherical-vector represented by an ordered pair (u, v) of unitvectors in R3.(i) The components of α are u · v and u× v (Definition 2.1). So, by Definition 5.3 above,cos(α) = u · v and sin(α) = u× v. Thus by Property 4.3.(i)

µ(α) = cos(α) + i sin(α).

(ii) Let θ be the non-oriented angle of the two vectors u and v (θ ∈ [0, π]). We have| cos(α)| = |u · v| = | cos(θ)| and | sin(α)| = ‖u× v‖ = sin(θ), thus

cos2(α) + | sin(α)|2 = 1.

Example 5.5. Consider the two unit vectors u = (√22,√22, 0) and v = (

√33

√33

√33). The

components of the spherical-vector α represented by the ordered pair (u, v) are

λ = u · v =

√6

3and n = u× v = (

√6

6,−√6

6, 0).

That is, by Remark 3.3,

cos(α) =

√6

3and sin(α) =

√6

6j −√6

6k.

14 Lahcen Lamgouni∗

5.3 Polar form of a quaternion

Theorem 5.6. Let q be a non-zero quaternion, r a strictly positive real number, and α aspherical-vector. We have q = r(cosα + isinα) if and only if |q| = r and arg(q) = α.

Proof. − Suppose that q = r(cos(α) + isin(α)). Then, by Remark 5.4.(ii)

|q|2 = r2(cos2(α) + | sin(α)|2) = r2.

Thus |q| = r. On the other hand, q = |q|(cos(α) + isin(α)) = |q|µ(α) (Remark 5.4.(i)),so µ(α) = q

|q| . That is, arg(q) = α (Definition 5.1).− Conversely, suppose that |q| = r and arg(q) = α. Then q/r = µ(α) (Definition 5.1).Thus q/r = cos(α) + i sin(α) (Remark 5.4.(i)).

5.4 Exponential notation of a quaternion

Definition 5.7. Let α be a spherical-vector. The notation eiα denotes the unit quaternionwhose argument is α; that is:

eiα = cosα + isinα.

Therefore, if q is a quaternion whose modulus and argument are respectively r and α, then

q = r(cosα + isinα) = reiα.

Remark 5.8. Let α be a spherical-vector. According to Remark 5.4.(i), we have

µ(α) = eiα.

5.5 Algebraic properties of the exponential notation

Property 5.9. Let α and β be two spherical-vectors and m an integer. We have the fol-lowing properties.

(1) ei(α+β) = eiβeiα (2)(eiα)−1

= eiα = e−iα

(3) ei(α−β) =(eiβ)−1

eiα (4) eiyπ = −1

(5) − eiα = ei(yπ+α) = ei(α+

yπ ) (6)

(eiα)m

= eimα.

Proof. (1)− Since µ is an anti-isomorphism, then µ(α+β) = µ(β)µ(α). Thus, accordingto Remark 5.8, we have ei(α+β) = eiβeiα.(2)− Similarly, we have µ(−α) = µ(α)−1, so e−iα = (eiα)

−1 and since eiα is a unitquaternion, then eiα = (eiα)

−1.(3)− Simply combine (1) and (2).

Spherical-vectors 15

(4)− According to Definition 4.7, the components ofyπ are −1 and ~0. So µ(

yπ) = −1

(Theorem 4.1). Thus, eiyπ = −1.

(5)− It suffices to combine (1) and (4).(6)− This result is easily proved by recursion on m ∈ N, of course using (1). To extendit to Z, we apply (2).

We can rewrite the above properties as follows.

Property 5.10. Let p and q be two non-zero quaternions and m an integer. We have thefollowing properties.

(1) arg(pq) = arg(q) + arg(p) (2) arg(q−1) = arg(q) = −arg(q)(3) arg(p−1q) = arg(q)− arg(p) (4) arg(−1) = y

π

(5) arg(−q) = yπ + arg(q) = arg(q) +

yπ (6) arg(qm) = m arg(q).

6 Examples of applications

We have seen in the previous sections that the anti-isomorphism µ allows us to iden-tify every unit quaternion q to its argument, which is none other than a spherical-vector.Thus, we can easily interpret geometrically unit quaternions and their algebraic propertiesthrough spherical-vectors.

6.1 Practical method for determining the argument of a non-zerounit quaternion

Let q = x + iω be a non-zero unit quaternion written in its spherical form (Section 3)and α be its argument. The components of α are x and ω (Remark 5.2). The methodpresented here consists of representing the spherical-vector α by an ordered pair (u, v) ofunit vectors.− Suppose that ω = ~0. Then α =

y0 = (u, u) or α =

yπ = (u,−u), for all unit vector u

(Remark 4.8).− Suppose now that ω 6= ~0. The ordered pair (u, v) is characterized by u · v = x andu × v = ω. The vector ω is normal to the support P of α. So, to have (u, v) = α, itsuffices, by Theorem 4.9, to choose u orthogonal to ω (that is, u ∈ P ), then to determinatev by one of the two formulas v = xu+ω×u or v = uq. Hence the following proposition.

Proposition 6.1. Let q = x + iω be a unit quaternion, where x and ω are its sphericalcomponents. Let α be a spherical-vector represented by an ordered pair (u, v) of unitvectors. We have :

α = arg(q) if and only if

{u ⊥ ω

v = xu+ ω × u = uq.

16 Lahcen Lamgouni∗

Examples: Argument of quaternions i, j and k

• Argument of i − Since the spherical form of i is (Section3)

i = i(0j + 0k + 1)

= i(0, 0, 1) (Remark 3.3)

= iez,

then the spherical components of i are 0 and ez. By Proposition 6.1 above, for an orderedpair (u, v) of unit vectors to represent the argument of i, it is necessary and sufficient thatu ⊥ ez and v = ez×u. As ex ⊥ ez, we can take for example u = ex. So v = ez× ex = ey

and the argument of i is the spherical-vector (ex, ey). That is,

i = ei(ex,ey).

• Argument of j − As j = −ik = −i(0j + 1k + 0) = −i(0, 1, 0) = −iey, thespherical components of j are 0 and−ey. An ordered pair (u, v) of unit vectors representsthe argument of j if and only if u ⊥ ey and v = −ey × u. As ex ⊥ ey, we choose thenu = ex. So v = −ey × ex = ez. Therfore, arg(j) = (ex, ez) and

j = ei(ex,ez).

• Argument of k − We have k = ij = i(1j + 0k + 0) = i(1, 0, 0) = iex. So thespherical components of k are 0 and ex. An ordered pair (u, v) of unit vectors representsthe argument of k if and only if u ⊥ ex and v = ex×u. As ey ⊥ ex, we take u = ey. Thusv = ex × ey = ez. Therfore, arg(k) = (ey, ez) and

k = ei(ey ,ez).

6.2 Geometric interpretation of the relation ki = j

In Section 6.1, we identified the unit quaternions i, j and k to their respective argumentsαi = (ex, ey), αj = (ex, ez) and αk = (ey, ez). So, we have

ki = j ⇐⇒ arg(ki) = arg(j)

⇐⇒ arg(i) + arg(k) = arg(j) (Property 4.11)

⇐⇒ (ex, ey) + (ey, ez) = (ex, ez).

That is,ki = j ⇐⇒ αi + αk = αj.

Spherical-vectors 17

xe

ze

ye

kα

jα

iαxe

ze

ye

kα

jα

iα

O

Figure 6. Geometric interpretation of the relation ki = j via the Chasles relation on the arguments of thequaternions i, j and k, αi + αk = αj .

6.3 Geometric interpretation of the non-commutativity of the multi-plication of unit quaternions

• General case − Let p and q be two unit quaternions whose arguments are respectivelyα and β. According to Property 5.10, we have arg(pq) = β + α and arg(qp) = α + β.Thus, the non-commutativity of the multiplication of p and q can be interpreted geomet-rically by the non-commutativity of the addition of their arguments α and β (Figure 7).

Support of α+β

Support of β+α

α+β

β+α

α

α

β

β

α+β

β+α

α

α

β

β

O

Figure 7. The two arguments α+ β and β + α have the same scalar component, but they do not have thesame support (pq 6= qp⇐⇒ β + α 6= α+ β).

• Example − Consider the following two unit quaternions

p =

√6

6(2− j − k) and q =

√2

2(1 + i).

18 Lahcen Lamgouni∗

In this last example, thanks to the spherical-vectors, we give a geometrical aspect to thequaternions p and q, as well as their products h = qp =

√33(1 + i − k) and h′ = pq =

√33(1 + i − j). Let us designate by αp, αq, αh and αh′ the respective arguments of the

quaternions p, q, h and h′. Thus αh = αp + αq and αh′ = αq + αp (Property 5.10). Torepresent these four spherical-vectors on the unit sphere S2, we must first write p and q intheir spherical forms (Section 3):

p =

√6

3+ i

√6

6(−j + k + 0) =

√6

3+ i

(−√6

6,

√6

6, 0

),

and

q =

√2

2+ i

(0j + 0k +

√2

2

)=

√2

2+ i

(0, 0,

√2

2

).

Thus, the spherical components of p are

λp =

√6

3and np =

(−√6

6,

√6

6, 0

),

and those of q are

λq =

√2

2and nq =

(0, 0,

√2

2

).

Since the components of αp are the spherical components of p, and the componentsof αq are the spherical components of q (Remark 5.2), then according to Theorem 4.10,np × nq being non-zero, there are only two combinations of three unit vectors u, v and wsuch that αp = (u, v) and αq = (v, w). We choose, for example, the combination

v = np×nq‖np×nq‖

u = λpv − np × vw = λqv + nq × v.

After calculation we get v =

(√22,√22, 0)

u =(√

33,√33,√33

)w = (0, 1, 0) = ey.

Thus, we represent the product qp = h, where

p =

√6

6(2− j − k), q =

√2

2(1 + i) and h =

√3

3(1 + i− k),

through the sum αp + αq = αh (Figure 8), where

αp = (u, v), αq = (v, ey) and αh = (u, ey)

Spherical-vectors 19

with

u =

(√3

3,

√3

3,

√3

3

)and v =

(√2

2,

√2

2, 0

).

Similarly, we have nq×np 6= ~0, so there are two combinations of three unit vectors u′, v′

and w′ such that αq = (u′, v′) and αp = (v′, w′). We choose for example the combinationv′ = np×nq

‖np×nq‖ = v

u′ = λqv − nq × vw′ = λpv + np × v.

That is v′ = v =

(√22,√22, 0)

u′ = (1, 0, 0) = ex

w′ =(√

33,√33,−√33

).

We represent now the product pq = h′, where

p =

√6

6(2− j − k), q =

√2

2(1 + i) and h′ =

√3

3(1 + i− j)

, through the sum αq + αp = αh′ (Figure 8), where

αq = (ex, v), αp = (v, w′) and αh′ = (ex, w′)

with

v =

(√2

2,

√2

2, 0

)and w′ =

(√3

3,

√3

3,−√3

3

).

ze

u

v

ye

xev

ze

ye

xe

αp

αp

αh’α

h’

αqα

q

αhα

h

αp

αp

αqα

q

u

w’w’

O O

Figure 8. qp = h⇐⇒ αp + αq = αh and pq = h′ ⇐⇒ αq + αp = αh′ .

20 Lahcen Lamgouni∗

By identification between unit quaternions and their arguments, which are spherical-vectors, it is legitimate to write p = αp, q = αq, qp = αp + αq, and pq = αq + αp

(Figure 9).

ze

ye

xe

O

qp

p

p

qq

pq

v

u

w’

Figure 9. p =√66 (2− j − k) , q =

√22 (1 + i), qp =

√33 (1 + i− k) and pq =

√33 (1 + i− j).

References

[1] W.R. Hamilton and W.E. Hamilton. Elements of Quaternions. Longmans, Green, &Company, 1866.