Self-Reverse Elements and Lines in an Algebra for 3D Space

A geometric algebra provides a single environment in which geometric entities can be represented and manipulated and in which transforms can be applied to these entities. A number of versions of geometric algebra have been proposed and the aim of the paper is to investigate one of these as it has a number of advantageous features. Points, lines and planes are presented naturally by element of grades 1, 2, and 3 respectively. The self-reverse elements in the algebra form a field. This allows an equivalence relation between elements of grade 2 to be defined so that, although not every grade 2 element corresponds to a line, each equivalence class does, and vice versa. Examples are given to illustrate the ease in which geometric objects are represented and manipulated.


Introduction
Geometry is concerned with the properties of objects such as points and lines and whether these remain invariant under various transforms. In the 1800s, work by researchers such as Clifford, Grassmann and Hamilton sought means to represent transforms (particularly rigid-body transforms) such as rotations and translations. What emerged were various approaches including the quaternions and the Clifford and Grassmann algebras [1,[7][8][9]18,19]. However, these ideas lay dormant for the early part of the 1900s and transforms were studied using 4 × 4 matrices and homogeneous coordinates [29]. The algebra-based ideas started to reappear towards the end of the 1900s as they provided a more robust means for dealing with transforms in such applications as computer games [20].
Of the algebra-based ideas, the quaternions and dual quaternions can be used as a means for handling rigid-body transforms of 3D space in a compact way [9,17,18,22]. While they allow points to be defined, they are limited in their ability to represent other geometric objects (lines, planes, etc.).
The algebras of Clifford and Grassmann have evolved into the ideas of geometric algebra which provides a single framework for dealing with both geometric objects (such as points, line and planes) and the transformations acting upon them. A number of versions of geometric algebra have been proposed [27] which have somewhat different properties and are therefore better suited to specific applications. Perhaps the version most frequently used is the conformal geometric algebra (CGA) [2,7,33].
Other versions include the homogeneous model [10,11,30,31] and projective geometric algebra (PGA) [12,13] which are created by extending a real vector space of dimension 4, with the introduction of a multiplication in which the square of one of the basis vectors is zero. This has the effect of relating vectors in the algebra to planes in the geometry, which seems unnatural, rather than to points which seems more natural. An alternative version, G 4 , of this algebra has been proposed [3,23,24] in which the square of the particular basis element is made (effectively) infinite. This has the effect of associating vectors in the algebra with points. Both the homogeneous model and G 4 use an underlying vector space of dimension 4 to describe 3D space, the CGA uses dimension 5. The differences in these dimensions is considered in [21].
The purpose of this paper is to study geometric attributes of G 4 in more detail. In particular, three significant properties are dealt with. The first is that points, lines and planes are represented naturally in the algebra by elements of grades 1, 2 and 3 respectively. A self-reverse element is a linear combination of elements of grade 0 and grade 4, and the second property is that the set of self-reverse elements forms a field. While lines are represented by elements of grade 2, not every such element corresponds to a line. However, the second property leads to the third: there is an equivalence relation between elements of grade 2, and every line corresponds to a unique equivalence class and vice versa.
To make this paper more self-contained, Sect. 2 provides an overview of how the geometric algebra G 4 is formed, and Sect. 3 discusses self-reverse elements and their properties. Sect. 4 shows how geometric objects (points, lines and planes in three dimensions) are represented in G 4 . Lines are considered further in Sect. 5 where, as noted above, not all elements of grade 2 represent lines, but every equivalence class of such elements does. This section also discusses results relating geometric properties to relations between the elements representing them. Sect. 6 applies these results to two geometric applications [4]: proofs of Desargues's theorem and of Fermat's triangle theorem. Finally, Sect. 7 draws conclusions: the geometric algebra G 4 provides a single environment in which three-dimensional geometric entities (such as points, line and planes) can be represented in a natural way and rigid-body transforms can be applied to them. Elements representing geometric objects Vol. 30 (2020) Self-Reverse Elements and Lines in an Algebra for 3D Space Page 3 of 26 50 are shown to have a particular "grade" and the algebra is rich enough to allow use of the grade as a means of distinguishing object types.
The version used here is called G 4 . This is discussed in greater detail elsewhere [3,23,24] and what follows here is an overview of its construction. The G 4 version is used here since the purpose of this paper is to investigate its properties which are different from other versions of geometric algebra: in particular and importantly, as discussed in Sect. 3, the subalgebra of selfreverse elements forms a field.
The algebra G 4 can be regarded as a real vector space of dimension 16 which has basis elements denoted by e σ where σ is an ordered subset of the set of subscripts {0, 1, 2, 3}. The basis vectors are the elements e 0 , e 1 , e 2 , e 3 . A multiplication is created on the basis elements by defining the following products of basis vectors Additionally the following definition is made The squares of the basis vectors are defined to be where ε is a symbol which can be regarded as a small real quantity. The typical element a of G 4 is a linear combination of basis elements where the a σ are real coefficients. The multiplication is extended to the product of two such general elements by multiplying out on a term-by-term basis. The basis vector e φ corresponding to the empty set acts as unity with respect to multiplication and is identified with the real number 1. The basis element e 0123 is denoted by ω: The quantity ε is carried through all multiplications. This means that the coefficients of the basis elements become polynomials in ε, and, as a result of other operations, they can also become power series in ε. So an alternative view is that the typical element a in (2.3) is a linear combination of basis elements with coefficients a σ lying in the field R((ε)) of formal (Laurent) where the α i are real numbers and m is a finite integer (possibly negative). Assuming that α m is non-zero, it is called the leading coefficient, and m is the leading power. An element a ∈ G 4 is said to be of order ε i if the leading power of each of its non-zero coefficients is at least i, and one is equal to i. This is written The grade of the basis element e σ is the size of the subset σ. More generally, if an element is a linear combination of basis elements of a single grade, then this is also the grade of the element. Elements of grade 1 are called vectors; those of grade 2 are bivectors; and grade 3 elements are trivectors. An element of the form α + εβω where α and β are ε-scalars is called a self-reverse element.
The reverse of a basis element is obtained by reversing the order of its subscripts. For example The reverse of the general element of (2.3) is obtained by taking the reverse of each of its summands. An inner and outer product of any two elements x, y ∈ G 4 are defined by equations x y = 1 2 (xy + yx), x y = 1 2 (xy − yx). Note that this is different from other definitions [7,25] which rely upon finding components of a particular grade within a product. The above definitions deal entirely with addition, subtraction and multiplication of elements within the algebra.
In addition, if a, b, c ∈ G 4 are any three elements, their triproduct is defined to be the following.
[a, b, c] = 1 6 (abc + bca + cab − acb − bac − cba). The following result is immediate. The next two results look to relating the triproduct to the inner and outer products when the three elements involved have the same grade. Proof. Firstly note that x y has even grade and is equal to its own reverse. Hence it is a self-reverse element. Since ω commutes with all even-grade elements of G 4 , this proves the result when k is 0, 2 or 4.
If k is 1 or 3, there is not enough scope for any ω component to appear in the products forming x y and so it is a pure ε-scalar and the result is trivial. Lemma 2.3. If a, b, c ∈ G 4 are three elements of the same grade, then Proof. Multiplying out the first three product gives since a c commutes with b by Lemma 2.2. So and, by symmetry, these are also equal to the third of the products; call this common value t. Then adding the expressions for the three products gives Also, from Eq. (2.4), it is seen that By symmetry, this common values is also equal to abc − cba and the last part of the lemma follows.
As discussed in Sect. 4, a point in three-dimensional space is represented by the vector where W is the additional coordinate and the cartesian coordinates of the point are (X/W, Y /W, Z/W ). Products of vectors representing points are now considered.
Proof. The first equality is from Lemma 2.3. The second follows by multiplying out the triproduct.
where the W i are the additional coordinates of the vectors p i , and V is the volume of the tetrahedron formed by the points corresponding to the four vectors.
Proof. This is derived from Lemma 2.4 by taking the outer product with p 4 .
where V is the volume of the tetrahedron formed by the p i .
with similar expressions for the other triple products.
Combining the determinants gives Vol. 30 (2020) Self-Reverse Elements and Lines in an Algebra for 3D Space Page 7 of 26 50 and the result follows from the last lemma.

Lemma 2.9. For any elements
Proof. Expanding the right hand side gives the following

Self-Reverse Elements
The only elements of G 4 that have grade 4 are non-zero multiples of ω = e 0123 . It anticommutes with every element of odd grade, commutes with every element of even grade, ω 2 = ε −1 , and ω † = ω. More generally, an element of the form where α and β are ε-scalars, is called a self-reverse element. Such elements are the only ones of even grade equal to their own reverse.
The conjugate of a self-reverse element is the result of changing the sign of ω.
Theorem 3.1. The self-reverse elements form a field.
Proof. The only property that is not immediately obvious is the existence of multiplicative inverses. If γ = α + εβω is a non-zero self-reverse element, then it has a multiplicative inverse in the element Consideration of powers of ε in the denominator shows that it cannot be zero since α and β are not both zero.
The next results consider when a self-reverse element can have a square root. Such roots are used in other forms of geometric algebra including the CGA [6].

Lemma 3.2. Suppose that α is an ε-scalar with a positive leading coefficient. Then
(i) if its leading power is even, α has a square root which is an ε-scalar; (ii) more generally, α has a square root which is a self-reverse element.
Proof. In case (i), the element has the form where the α i are real numbers, α 0 > 0, and β has non-negative leading power. Then, using the binomial expansion, α has two square roots, namely , if the leading power of α is even, case (i) applies. If it is odd, then, since εω 2 = 1, α can be written as α = ω 2 (εα) and its square roots are ±ω √ [εα] which exist by case (i).

Lemma 3.3.
Suppose that γ = α + εβω is a self-reverse element where α has zero leading power and positive leading coefficient, and the leading power of β is non-negative. Then γ has a square root.
Proof. Since α is non-zero, the self-reverse element can be expressed as where φ = β/α. Here φ is an ε-scalar whose leading power is non-negative. Lemma 3.2 shows that α has a square root. Hence, using the binomial series, the following are square roots of γ: Since the power series being used are treated as "formal", the question of their convergence is not important. However, it can be noted that the series appearing in the last two lemmas are derived from standard convergent series and hence are themselves convergent for suitably small values of ε.

Points and Planes
The typical vector in G 4 is the element This is used to represent the point (W, X, Y, Z) in the projective space R((ε)) 3 , and the point ( 3 . By letting ε become zero, it also represents a point in Euclidean space R 3 , assuming that the components remain finite. An element of G 4 is said to be in standard form if the coefficients of its components have non-negative leading powers and at least one leading power is zero. This definition is only used here with respect to the particular choice of basis vectors. However, it is independent of that choice provided the new set of basis vectors satisfies (2.1) and (2.2). Clearly, any element can be put into standard form by multiplying by a power of ε, and any ε-scalar multiple of a point p corresponds to the same point in Euclidean space.
Conversely, the point (x, y, z) is represented by the vector W (e 0 + xe 1 + ye 2 + ze 3 ) ∈ G 4 for any non-zero choice of the ε-scalar W .
A vector in G 4 is also called a point. When required to avoid confusion, the corresponding entity in Euclidean space is called a geometric point. The component W of the vector p is the additional coordinate, and p, in standard Vol. 30 (2020) Self-Reverse Elements and Lines in an Algebra for 3D Space Page 9 of 26 50 form, is said to be finite if W has zero leading power, and normalized if W = 1.
A vector of the form is regarded as a direction.
In Euclidean geometry, three geometric points which are not collinear define a plane. If p 1 , p 2 , p 3 ∈ G 4 represent three such points, then the plane is represented by the triproduct Π = [p 1 , p 2 , p 3 ]. It is assumed that Π is nonzero and this happens if the points are indeed not collinear as shown in the following lemma Then for ε-scalars λ i and μ i . Column operations on the determinant in Lemma 2.4 show that Π is a non-zero ε-scalar multiple of Similarly, [q 1 , q 2 , q 3 ] is also a (possibly different) non-zero ε-scalar multiple of this determinant as required.
This means that the triproduct Π for a plane is independent of the choices of points chosen within the plane up to multiplication by a non-zero ε-scalar.  Proof. Suppose that Π = [p 1 , p 2 , p 3 ] where p 1 , p 2 , p 3 are three points defining the plane. If q is a point in the plane, then Corollary 2.7 says that Π q = 0 and so Π and q commute. The extension to a combination of points in the plane is immediate.

Lines
The previous section shows that points and planes correspond precisely to elements of grades 1 and 3 respectively in G 4 . This section considers lines and the situation is not as simple. A line is defined as the outer product of two points and as such is an element of grade 2. However not every element of grade 2 in G 4 is a line. It is shown that an equivalence relation can be established between elements. This relies upon the result, Theorem 3.1, that the self-reverse elements form a field. It can then be shown that lines correspond precisely to the associated equivalence classes (Theorem 5.13). Additionally, this section provides some results relating geometric properties of lines to expressions involving elements of G 4 . Lines appear in other forms of geometric algebra (e.g. [11,13,14,16]).
A line in G 4 is defined to be the outer product p 1 p 2 of two vectors. It is a finite line if it is non-zero and the two vectors are finite and represent different geometric points. Proof. In case (i), q = λ 1 p 1 + λ 2 p 2 where λ 1 and λ 2 are ε-scalars, so that p 1 q = λ 2 and p 2 q = −λ 1 . Case (ii) is an obvious extension of (i).
Thus any line in Euclidean space is represented by the element = p 1 p 2 where p 1 and p 2 correspond to geometric points on the line. Except for multiplication by a non-zero ε-scalar, the element is independent of the choice of the points.

Lemma 5.2.
If p 1 and p 2 are vectors with p i = W i e 0 + X i e 1 + Y i e 2 + Z i e 3 for i = 1, 2, then = p 1 p 2 is given as Proof. The proof is by multiplying out the product.
Note that the coefficients in p 1 p 2 are the Plücker coordinates of the line joining the points showing that representation of a line as a bivector is closely related to other representations and also to the associated theory of screws [26,28,31].
Vol. 30 (2020) Self-Reverse Elements and Lines in an Algebra for 3D Space Page 11 of 26 50

Lemma 5.3. Every line is a bivector and † is an ε-scalar.
Proof. Suppose that = p 1 p 2 where p 1 and p 2 are vectors. Then Lemma 5.2 shows that is a bivector, and, using the notation of that lemma, where d is the distance between the points p 1 and p 2 , and W 1 and W 2 are their additional coordinates.
Proof. From the proof of the Lemma 5.3, where (x 1 , y 1 , z 1 ) and (x 2 , y 2 , z 2 ) are the cartesian coordinates of the points. This proves the result. The following question is now considered. If b is a bivector, is it necessarily a line: that is, can it be expressed as the outer product of two vectors?
Any bivector b ∈ G 4 can be written as where v and b 1 are a vector and a bivector respectively which do not involve e 0 . A bivector is called finite if it is in standard form and v is non-zero and has leading power zero.
where q 1 , q 2 are vectors independent of e 0 , and (e 0 +q 1 ) and (e 0 +q 2 ) represent different geometric points. Then and it is seen that is a finite bivector.
Lemma 5.7. If b is a finite bivector, then εb † b is a self-reverse element with a square root.
Proof. Firstly, εb † b is certainly a self-reverse element since it has even grade and is equal to its own reverse.
Writing b = e 0 v + b 1 as in (5.1), it follows that The first summand here is v 2 which is a non-zero ε-scalar with zero leading power (from the definition of b as a finite bivector). The other summands have a factor of ε. So Lemma 3.3 applies and completes the proof.
If b is a finite bivector, then its size is the self-reverse element defined by The square root is an ε-scalar of the form in (3.1) with coefficients α and β. The plus sign means choosing the root whose α has strictly positive leading coefficient if α is non-zero; otherwise choosing the β with strictly positive leading coefficient. (If both are zero, then b is itself zero.) A finite bivector is said to be normalized if it has unit size. Note that in stating some of the following results the reverse of the bivector b is used. This is to show the symmetry involved. The results could be restated simply in terms of b itself since Lemma 5.8. Suppose that b ∈ G 4 is a non-zero bivector and that p ∈ G 4 is a point. Set Then q is a point if and only if b  † b is an ε-scalar. Vol. 30 (2020) Self-Reverse Elements and Lines in an Algebra for 3D Space Page 13 of 26 50 Proof. Note that γ = b † b is certainly a self-reverse element, and that q has odd grade. The reverse of q is given by and only if γ * = γ, and this is the case if and only if γ is an ε-scalar. a finite bivector with b  † b being an ε-scalar, and  p ∈ G 4 is a finite vector, then Proof. The last lemma shows that q is a vector, so it remains to show that it is finite. Suppose that where p 1 , v are vectors, b 1 is a bivector and all three are independent of e 0 . Then where x and y are independent of e 0 . Hence q α 0 v 2 e 0 + 1 2 ε(x − y) and the coefficient of e 0 has leading power zero.

Lemma 5.10. Suppose that b is a non-zero bivector for which b † b is an ε-scalar and that p is a point. Set
Proof. Note that b † = −b and that b 2 commutes with all elements of G 4 since it is an ε-scalar. So Lemma 5.11. Any non-zero finite bivector b ∈ G 4 for which b † b is an ε-scalar is a finite line, that is b is the outer product of two finite vectors. Proof. Choose two finite points p 1 , p 2 and set Since v i has odd grade and is equal to its own reverse, v i is a vector.
Consider the outer product of q 2 and q 1 .
Considering part of the second summand here In the first summand, α = (v 1 p 2 ) is the inner product of two vectors and hence is an ε-scalar. Thus Since p 1 and p 2 are open to choice, p 1 can be chosen arbitrarily, and then p 2 chosen such that α is non-zero. Then εb 2 and εα are non-zero ε-scalars with zero leading powers. Thus q 1 /(εb 2 ) and q 2 /(εα) are finite vectors whose outer product is b, as required.
Two elements of G 4 are said to be equivalent if one is the product of the other and a self-reverse element. Since the self-reverse elements form a field, this is an equivalence relation. Proof. It is clear that B is equivalent to b, and Vol. 30 (2020) Self-Reverse Elements and Lines in an Algebra for 3D Space Page 15 of 26 50 Theorem 5.13. Every non-zero finite bivector b ∈ G 4 is equivalent to a finite line, and, if b † b is an ε-scalar, then b is itself that line.
Proof. If b † b is an ε-scalar, then Lemma 5.11 says that b is a finite line. If not, then by the last lemma, b is certainly equivalent to a bivector B such B † B = ε −1 , and B is a finite line by Lemma 5.11.

Now consider four finite points in
whereâ,b,ĉ,d do not involve e 0 . Let A, B, C, D denote the corresponding geometric points. Then The first term in the square brackets is essentially the scalar product of two vectors. Lemma 2.6 and Corollary 2.8 can be applied to the second term. The third term can be ignored because of the ε multiplier. Hence where Δ(p, q) is used to denote the Euclidean distance between the geometric points corresponding to points p, q ∈ G 4 , α is the angle between lines AB and CD (when viewed along their common normal), and V is the volume of the tetrahedron formed by the four points in Euclidean space. Using traditional geometry, the volume of the tetrahedron is given by the following as shown in Lemma 8.1 in the Appendix.
This gives the inner product of two lines. In particular, taking c = a, d = b and setting = a b, the inner product of a line with itself is given by so that, using (5.2),  (5.4) provide the proof of the following result which extends the idea of the scalar product of two ordinary vectors. Not only does it gives the angle between two lines, but also it provides the distance between them.
There are related results in other forms of geometric algebra (e.g. [5,11]).
Theorem 5.14. Suppose that a, b, where α is the angle between the lines (when viewed along their common normal), and h is the distance between them.
This result allows more insight to be provided into the choice of points made in the proof of Lemma 5.11. These points are the feet of perpendiculars dropped from the chosen points onto the line represented by the bivector in that result. This is shown by the following lemma. Proof. Set q = ε (p ) = 1 2 ε p 2 − p , which has odd grade. Since 2 = − † is an ε-scalar (Lemma 5.3), it commutes with all elements of G 4 , and hence q † = q. Thus q is indeed a point (vector).
The line joining p and q is m = p q = 1 4 ε(p 2 2 − p p − p 2 p + p p) = 1 4 ε( p p − p p ), which is the expression in (ii). Then m = 1 8 ε( p p − p p 2 + 2 p p − p p ) = 1 8 ε −p p 2 + 2 p p = 0, so that, by Theorem 5.14, the lines are perpendicular (and zero distance apart). Proof. First note that q is an element of odd grade and is equal to its own reverse. Hence q is a point.
Since is the outer product of two points in Π, Lemma 4.2 says that Π commutes with . Hence, using Lemmas 2.3 and 2.1, Vol. 30 (2020) Self-Reverse Elements and Lines in an Algebra for 3D Space Page 17 of 26 50 So, by Lemma 5.5, point q lies on line . Similarly, q lies also on m and so q is the point of intersection of the lines.
Finally, in this section, suppose that 1 and 2 are two lines, and consider their outer product. It certainly has even grade and is equal to minus its own reverse. Hence it has grade 2. So it is equivalent to a line. This raises the question: what line is it? The following result provides the answer.
Theorem 5.17. The outer product of two distinct lines is equivalent to the line which is their common normal.
Proof. As before, let 1 and 2 be the lines. Let , where γ is the appropriate self-reverse element, be the line equivalent to their outer product.
Since γ commutes with all even-grade elements, it is seen that m 1 = γ [ 1 , 2 , 1 ] = 0, using Lemmas 2.3 and 2.1. Theorem 5.14 now shows that line m is normal to line 1 . Similarly, it is normal to line 2 and hence is the required common normal.

Geometry
This section shows how the geometric algebra G 4 can be used to deal with geometrical applications. It makes use of the fact that G 4 has both a model of projective space (and hence Euclidean space) and the ability to generate rigid-body transforms. These ideas are used to present a proof of Desargues's theorem and Fermat's triangle theorem [4]. Different proofs using others form of geometric algebra are given in [15,32].
Some preliminary ideas and results about G 4 and transforms of Euclidean space are required for the Fermat result. These are dealt with first and are discussed in the case of two-dimensional space. They are particular cases of more general results [24].
Any element S ∈ G 4 defines a map F S of G 4 to itself given by If S has even-grade and x is a vector, then F S (x) is an element of odd grade equal to its own reverse and so is also a vector. Hence F S generates a transform of Euclidean space. Further, this is a linear transform [24]. There are two particular cases of importance. The even-grade elements Adv. Appl. Clifford Algebras of these elements by multiplying out the appropriate products.) Further, these elements are normalized in the sense that A rotation about the Euclidean point Q corresponding to the vector e 0 + ue 1 + ve 2 can be formed by translating this point to the origin, performing the rotation about the origin, and then translating back again. Thus the even-grade element which generates such a rotation is R Q = T † R O T and multiplying out gives the following result.
Application to the proofs of the two theorems are now given.

Desargues's Theorem
Desargues's theorem is stated below and is illustrated in Fig. 1. Note that it is not a requirement that the points are all coplanar. In the version given here, it is assumed that the six points A 1 , A 2 , B 1 , B 2 , C 1 , C 2 are distinct. This is the version usually illustrated in text books. The result holds in some other cases (possibly trivially) and the proof given here can be modified appropriately. Proof. Regard the points indicated in the theorem as being normalized vectors in G 4 . The collinearity of points I, A 1 , A 2 means that A 2 is a linear combination of I and A 1 . Similar considerations apply to the other two original lines and so where the coefficients are ε-scalars. (If I coincides with one of the other six points, then assume, without loss, that the other point has subscript 2. Then the corresponding expression above remains valid with the coefficient of I being unity and the other coefficient zero.) The two lines IB 1 B 2 and IC 1 C 2 form a plane which can be represented by the triproduct which is non-zero as the lines are distinct. It follows that Since points B 2 and C 2 commute with Π (Lemma 4.2), and εΠ 2 is an ε-scalar, Lemma 5.16 yields By cyclic symmetry, the following also hold Hence X Y = γ 2 α 2 (B 2 C 2 ) + β 2 γ 2 (C 2 A 2 ) + γ 2 2 (A 2 B 2 ), and then [X, Y, Z] = (X Y ) Z = −γ 2 α 2 β 2 [B 2 , C 2 , A 2 ] + β 2 γ 2 α 2 [C 2 , A 2 , B 2 ] = 0.
Thus the points X, Y , Z are linearly dependent and form a line (Lemma 4.1).

Fermat's Triangle Theorem
Fermat's triangle theorem relates to the construction shown in Fig. 2. Here A, B, C are three non-collinear points in the plane. On each side of the triangle ABC an equilateral triangle is constructed lying outside triangle ABC. The additional vertices are points X, Y , Z. Proof. Let A, B, C ∈ G 4 be normalized vectors representing the vertices of the original triangle. For simplicity, assume that these vertices lie in the xyplane. This is the plane Π = e 012 , and then A, B, C depend only on e 0 , e 1 , e 2 .
Let R A , R B , R C ∈ G 4 be even-grade elements which generate anticlockwise rotations through angle π/3 about the points A, B, C respectively. As