Solving some quadratic Diophantine equa- tions with

Solving some quadratic Diophantine equations with Clifford algebra G. Aragón-González, J. L. Aragón and M. A. Rodr´ıguez-Andrade Abstract. In this work, the equivalence class representatives of integer solutions of the Diophantine equation of the type a1 x21 + · · · + ap x2p = ap+1 x2p+1 +· · ·+ap+q x2p+q +a1 x2n+1 (ai > 0, i = 1, . . . , p+q, xn+1 6= 0) are found using simple reflections of orthogonal vectors, manipulated using the Clifford algebra over orthogonal spaces Rp,q . These solutions are obtained from the application of an useful Lemma: given two different non-zero vectors of the same norm, we can map one onto the other, or its negative, by means of a simple reflection. With this Lemma, we extend and improve a previous work [1] concerning generalized Pythagorean numbers, which now can be obtained as a Corollary. We also show that our technique is promising for solving others Diophantine equations. Mathematics Subject Classification (2010). Primary 15A66; Secondary 11D09. Keywords. Clifford algebras, Diophantine equations.

1. Introduction In a recent work [1], rotations and reflections of canonical vectors, as elements of a Clifford algebra, were used to find all the Pythagorean numbers, i. e., x1 n ∈ all the equivalence class representatives of vectors x = xn+1 , . . . , xxn+1 Qn ∩ S n−1 were found. This approach is used and improved here to find all integer solutions of the Diophantine quadratic equation: a1 x21 + · · · + an x2n = a1 x2n+1 ,

(1.1)

and, more generally a1 x21 + · · · + ap x2p = ap+1 x2p+1 + · · · + ap+q x2p+q + a1 x2n+1 .

(1.2)

where ai > 0 for all i. Our procedure is based on an useful Lemma (Lemma 4), which states that given two different non-zero vectors of the same norm, we

2

G. Arag´ on-Gonz´ alez, J. L. Arag´ on and M. A. Rodr´ıguez-Andrade

can map one onto the other, or its negative, by means of a simple reflection. This Lemma establishes a fundamental difference with the approach in Ref. [1], since rotations are now not needed. Actually, the Pythagorean numbers can be obtained as a Corollary of Lemma 4. In order to set main ideas and notation, in Section 3 we first solve a particular case of the general equation: find (x1 , x2 , x3 ) ∈ Z3 satisfying the equation a1 x21 + a2 x22 = a1 x23 ,

x3 6= 0,

for positive integers a1 and a2 . Next, in Section 4 we move on to a more general case of finding (n + 1)tuples of integers (x1 , . . . , xn , xn+1 ) such that a1 x21 + a2 x22 + · · · + an x2n = a1 x2n+1 ,

xn+1 6= 0,

(1.3)

for positive integers a1 , a2 , . . . , an . Finally, in Section 5 the algorithm developed to solve (1.3) is extended mutatis mutandis to find n + 1-tuples of integers (x1 , . . . , xn , xn+1 ) satisfying a1 x21 + · · · + ap x2p = ap+1 x2p+1 + · · · + ap+q x2p+q + a1 x2n+1 ,

(1.4)

where xn+1 6= 0 and n = p + q. ai , i = 1, 2, ..., n, are positive integers. √ a1 To solve (1.3) we use simple reflections of certain vectors in S n−1 in the Euclidean space. Similarly, solutions to (1.4) are generated by means of simple reflections of vectors belonging to the set n √ o n−1 √ Sp,q ( a1 ) = x ∈ Rp,q kxkp,q = a1 , where Rp,q (n = p+q) is the orthogonal space with signature (p, q) and kxkp,q is the pseudo norm associated to their corresponding bilinear form. We add the further comment that the trivial solution xi = 0, i = 1, . . . , n + 1 is not obtained with the proposed algorithm. Also that with our approach, we can find the equivalence class representatives of vectors (x1 , . . . , xn+1 ) that satisfy the equations (1.3) or (1.4), which are denoted by h(x1 , . . . , xn+1 )i. We say that (y1 , . . . , yn+1 ) ∈ h(x1 , . . . , xn+1 )i, that is (x1 , . . . , xn+1 ) ∼ (y1 , . . . , yn+1 ) , if and only if yi xi = , gcd (x1 , . . . , xn+1 ) gcd (y1 , . . . , yn+1 )

i = 1, . . . , n + 1.

Thus, given an arbitrary solution (x1 , . . . , xn+1 ) of (1.3) or (1.4), the algorithm proposed here allows us to find a solution (y1 , . . . , yn+1 ) ∈ h(x1 , . . . , xn+1 )i. This is what is meant when we say that all the solutions of (1.3) or (1.4) are found.

Solving some quadratic Diophantine equations with Clifford algebra

3

2. Simple reflections with Clifford algebra Clifford algebra provide us with a convenient way of representing simple reflections and in this Section we present the very basics just to fix motation. With the exception of the new Lemma mentioned in the previous Section, proofs are referred to Refs. [1] and [7]. From the several ways to construct Clifford Algebras[4, 5, 2, 6] we use generators. The Clifford algebra of the orthogonal non-degenerate space Rp,q , p + q = n, is defined as [2, 3]: Definition 1. The real associative and distributive algebra, generated by the space Rp,q , p + q = n, with the following product rules e2i e2i ei ej + ej ei

= = =

1, −1, 0,

i = 1, 2, . . . , p, i = p + 1, p + 2, . . . , p + q i 6= j, i, j = 1, 2, . . . , n,

where {e1 , . . . , en } is the canonical basis of Rp,q , is called universal Clifford algebra of the space Rp,q and is denoted by Rp,q . The particular case of Rn is obtained by setting q = 0 and, consequently p = n. The inverse of a nonzero vector can be defined in a Clifford algebra [2]. A vector a ∈ Rp,q is said to be invertible if a2 6= 0 and the element a (2.1) a−1 = 2 , a is called the inverse of a. Simple reflections are obtained by means of geometric products between vectors belonging to Rp,q : Proposition 2. The simple reflection φx : Rp,q → Rp,q is given by φx (y) = −xyx−1 . The following result will be useful below. Proposition 3. Let a and b be invertible elements of Rp,q , with a2 = b2 . Then a + b and a − b are mutually orthogonal and either a + b or a − b is invertible. The following Lemma is central for the results that follow; it states that given two different non-zero vectors of the same norm, we can map one onto the other, or its negative, by means of a simple reflection. This Lemma makes unnecessary the use of rotations to find Pythagorean numbers, as it was proposed in Lemma 10 of Ref. [1]. It also complements the Proposition 5.14 of Ref. [3]. Lemma 4. Consider any two non-zero invertible vectors a, b ∈ Rp,q , such that a 6= b and a2 = b2 . Then either one or both of the following conditions are fulfilled: 1. a can be mapped onto b by means of a simple reflection.

4

G. Arag´ on-Gonz´ alez, J. L. Arag´ on and M. A. Rodr´ıguez-Andrade 2. −a can be mapped onto b by means of a simple reflection.

Proof. From Proposition 3 we have that either a + b or a − b is invertible. Assume that a − b is invertible. In this case, we have that φa−b (a) = b. 2

2

Indeed, since a = b

−1

− (a − b) a (a − b)

−1 = − a2 − ba (a − b) , −1 = − b2 − ba (a − b) , −1

= −b (b − a) (a − b)

= b.

Now, if a + b is invertible then φa+b (−a) = b. Indeed, −1

− (a + b) (−a) (a + b)

−1 = − −a2 − ba (a + b) , −1 = b2 + ba (a + b) −1

= b (b + a) (a + b)

= b. −1

Finally, an alternative algebraic expression of φx (y) = −xyx is obtained, that will be useful in the next Sections. For a given y, such that y2 6= 0, there exists λ ∈ R and z ∈ Rp,q , such that x = λy + z, 2 where λ = (x, y) /x , (·, ·) is the inner product in Rp,q and z is orthogonal to y. Thus yz = −zy, and φx (y)

= = = =

−xyx−1 , 1 − 2 (λy + z) y (λy + z) , x 1 − 2 λ2 y2 y + λy2 z + λy2 z − z2 y x 1 − 2 λ2 y2 − z2 y + 2λy2 z . x

(2.2)

3. The equation ax21 + bx22 = ax23 In order to introduce our main procedure, we first consider the problem of finding integer triples (x1 , x2 , x3 ) satisfying ax21 + bx22 = ax23 ,

x3 6= 0,

(3.1)

where a and b are positive integers. √ 2 Let √ B = {w1 , w2 } be an2 orthogonal basis of R , where w1 = ae1 and w2 = be2 . Now, let u ∈ R be a vector such that its representation with


5

respect to the basis B is [u]B = (α1 , α2 ) ∈ Z2 , that is u = α1 w1 + α2 w2 . Then u2 = aα12 + bα22 . As it will be shown in what follows, from the vector v = φu (w1 ) , we can generate all the integer solutions of (3.1). Indeed

v

= −uw1 u−1 , 1 = − 2 (α1 w1 + α2 w2 ) w1 (α1 w1 + α2 w2 ) , u 1 = − 2 α12 w12 − α22 w22 w1 + 2α1 α2 w12 w2 , u = β1 w1 + β2 w2 ,

where p aα12 − bα22 = , 2 2 aα1 + bα2 r 2aα1 α2 q − 2 = . aα1 + bα22 r

β1

= −

β2

=

Since φu is an orthogonal transformation, we have that v2 = w12 = a. Therefore a = v2 = β12 w12 + β22 w22 = a

p2 q2 +b 2, 2 r r

or ap2 + bq 2 = ar2 . Consequently (p, q, r) ∈ Z3 is a solution of (3.1) given by: − aα12 − bα22 , − (2aα1 α2 ) , aα12 + bα22 . Solutions of the equation (3.1) can be thus generated by sampling non-zero relatively prime integers (α1 , α2 ). It should be noticed that since φu (x) = φλu (x) for some non-zero λ ∈ Z, it is enough to consider the set A = −uw1 u−1 | u = α1 w1 + α2 w2 ; gcd (α1 , α2 ) = 1 . As we will prove in the next Section, where a more general equation is considered, at least one solution of (3.1), (y1 , y2 , y3 ) ∈ h(x1 , x2 , x3 )i, can be obtained from the elements of A.

6


4. The equation a1 x21 + · · · + an x2n = a1 x2n+1 Now consider the problem of finding (n + 1)-tuples (x1 , . . . , xn , xn+1 ) of nonzero integers satisfying a1 x21 + · · · + an x2n = a1 x2n+1 ,

(4.1)

where ai > 0, i = 1, . . . , n, are integers. √ Let {e1 , . . . , en } be the canonical basis of Rn and define wi = ai ei , i = 1, . . . , n. Clearly B = {w1 , . . . , wn } is an orthogonal basis Rn . Also let u=

n X

αi wi ,

i=1

such that (α1 , . . . , αn ) ∈ Zn and gcd (α1 , . . . , αn ) = 1. Since wi2 = ai , i = 1, . . . , n, we have n X αi2 ai . u2 = i=1

Now, u can be decomposed as u = α1 w1 + z, n P

where z =

αi wi . The integer solutions of (4.1) can be generated from the

i=2

simple reflection: v1 = φu (w1 ) . This can be seen as follows. Since from (2.2) v1 = φu (w1 ) = − clearly v1 ∈

n L

1 u2

α12 w12 − z2 w1 + 2α1 w12 z ,

Qwi and, since φu is an orthogonal transformation, v12 = w12 .

i=1

Now, if we take v = u2 v1 ∈

n L

Zwi , then

i=1

v=− that is v =

n P

α12 w12 − z2 w1 + 2α1 w12 z ,

(4.2)

pi wi , where (p1 , . . . , pn ) ∈ Zn . By squaring this vector we find

i=1

on the one hand that: n X

v2 =

ai p2i .

(4.3)

i=1

Since, on the other hand v2 = u2

2

w12 = u2

2

a1 ,

we set r = u2 , so that (4.3) becomes a1 p21 + · · · + an p2n = a1 r2 .


7

Therefore, the (n + 1)-tuple (p1 , . . . , pn , r) ∈ Zn+1 , given by (see Eq. 4.2) p1

a1 α12

= −

−

n X

! ai αi2

,

i=2

pi r

= −2a1 α1 αi , n X = αi2 ai ,

i = 2, 3, ..., n,

i=1

is a solution of (4.1). Solutions of (4.1) can be thus generated from the above expressions by sampling non-zero integers (α1 , . . . , αn ). Since for a given integer λ 6= 0 we have that φu (x) = φλu (x), in what follows we prove that with the elements of the set ) ( n X −1 αi wi ; gcd (α1 , . . . , αn ) = 1 , A = −uw1 u u= i=1

we can generate at least one solution (p1 , . . . , pn+1 ) ∈ h(x1 , . . . , xn+1 )i of (4.1). Suppose that (x1 , . . . , xn , xn+1 ), xn+1 6= 0, is a solution of (4.1). By n n P P xi 1 2 x2i ai = a1 . defining x = xn+1 wi , we have that x = x2 n+1

i=1

i=1

Let u1 = x − w1 =

n X x1 − xn+1 xi w1 + wi . xn+1 x i=2 n+1

By Lemma 4 we have that φu1 (w1 ) = x. and if we take u0

=

xn+1 u1 gcd (x1 − xn+1 , x2 , . . . , xn )

=

n X (x1 − xn+1 ) xi w1 + wi , gcd (x1 − xn+1 , x2 , . . . , xn ) gcd (x1 − xn+1 , x2 , . . . , xn ) i=2

=

n X

βi wi ,

i=1

we have that (β1 , . . . , βn ) ∈ Zn and gcd (β1 , . . . , βn ) = 1. Therefore φu0 (w1 ) ∈ A.

8


Thus, with the vector w = φu0 (w1 ), we can generate the solution (x1 , . . . , xn , xn+1 ). Even more, if ! n X 2 2 y1 = − a1 α1 − ai αi ; i=2

yi yn+1

= −2a1 α1 αi , n X = αi2 ai ,

i = 2, . . . , n,

i=1

then (y1 , . . . , yn+1 ) ∈ h(x1 , . . . , xn+1 )i. With all these results, we have proved the following Theorem: Theorem 5. Let {e1 , . . . , en } be the canonical basis of Rn and define wi = √ ai ei , for i = 1, . . . , n. Let [x]B be the representation of the vector x with respect to the basis B = {w1 , . . . , wn }. Then, the vector x1 xn [x]B = ,..., ∈ Qn , xn+1 6= 0, xn+1 xn+1 yields a (n + 1)-tuple (x1 , . . . , xn , xn+1 ) satisfying a1 x21 + · · · + an x2n = a1 x2n+1 , for integers ai > 0, if an only if there exists a vector a, represented in the basis B as [a]B = (α1 , . . . , αn ) ∈ Zn , gcd (α1 , . . . , αn ) = 1, such that −aw1 a−1 = x. Even more, by considering y1

= −

a1 α12

−

n X

! ai αi2

(4.4a)

i=2

yi yn+1

= −2a1 α1 αi , n X = αi2 ai ,

i = 2, . . . , n

(4.4b) (4.4c)

i=1

we obtain a (n + 1)-tuple (y1 , . . . , yn+1 ) ∈ h(x1 , . . . , xn , xn+1 )i.

5. The equation a1 x21 + · · · + ap x2p = ap+1 x2p+1 + · · · + ap+q x2p+q + a1 x2n+1 If the algorithm of the previous section is now implemented for to the space Rp,q , we can solve the equation (1.4).


9

√ Let B = {w1 , . . . , wn }, where wi = ai ei , be an orthogonal basis of Rp,q . Now, let u ∈ Rp,q be an invertible vector represented in this basis as [u]B = (α1 , . . . , αn ) ∈ Zn , that is u=

p+q X

αi wi .

i=1

Since wi2 = ai e2i , then u2 =

p X

p+q X

ai αi2 −

i=1

ai αi2 .

(5.1)

i=p+1

If u is decomposed as u = α1 w1 + z, with z =

n P

αi wi , then, as in Section 4, the integers solutions of (1.4) are

i=2

generated by the vector v1 = φu (w1 ) .

(5.2)

Indeed, since from (2.2) φu (w1 ) = − then v1 ∈

n L

1 u2

α12 w12 − z2 w1 + 2α1 w12 z ,

(5.3)

Qwi and v12 = w12 = a1 .

i=1

If we take v = u2 v1 ∈

n L

Zwi , then from (5.3) we have

i=1

v=−

α12 w12 − z2 w1 + 2α1 w12 z ,

that is v can be written as v =

p+q P

(5.4)

ri wi , where (r1 , . . . , rn ) ∈ Zn . By squaring

i=1

this vector we have, on the one hand, that v2 =

p X

p+q X

ai ri2 −

i=1

ai ri2 .

i=p+1

On the other hand v2 = u2

2

w12 = u2

2

a1 ,

then if we take s = u2 , Eq. (5.5) becomes 2 a1 r12 + · · · + ap rp2 − ap+1 rp+1 − · · · − an rn2 = a1 s2 .

(5.5)

10


Therefore, the (n + 1)-tuple (r1 , . . . , rn , s) ∈ Zn+1 is a solution of (4.1) and, from (5.4), it is given by    p p+q X X r1 = − a1 α12 −  αi2 ai − αi2 ai  , i=2

i=p+1

= −2a1 α1 αi , i = 2, 3, ..., n, p p+q X X s = ai αi2 − ai αi2 .

ri

i=1

i=p+1

Thus (n + 1)-tuples satisfying (1.4) are obtained from the above equations by sampling non-zero integers (α1 , . . . , αn ). Notice that there could be some integers producing s = u2 = xn+1 = 0 which in fact implies a non invertible u (see Eq. 5.1). Despite that this possibility is ruled out by the fact that the problem is posed specifying xn+1 6= 0, its occurrence does not represent a problem since the (n + 1)tuple (r1 , . . . , rn , 0) is still a solution of (1.4). Consider the sets ( ) p+q X A = −uw1 u−1 u = αi wi , gcd (α1 , . . . , αn ) = 1 i=1

and ( B=

−1

−u (−w1 ) u

) p+q X αi wi , gcd (α1 , . . . , αn ) = 1 . u= i=1

In what follows it will be shown that if (x1 , . . . , xn xn+1 ) ∈ Zn+1 is a solution of (1.4), then we can always find a solution, a representative in h(x1 , . . . , xn , xn+1 )i, in the set A ∪ B. Assume that x=

1

p+q X

xn+1

i=1

xi wi ,

xi ∈ Qp,q , i = 1, . . . , n, xn+1

satisfies x2 = a1 = w12 , that is, the (n + 1)-tuple (x1 , . . . , xn , xn+1 ) ∈ Zn+1 is a solution of (1.4). Now, by considering p+q X x1 − xn+1 xi x − w1 = w1 + wi , xn+1 x i=2 n+1

two possibilities arise: 1. x − w1 is invertible.

(5.6)

Solving some quadratic Diophantine equations with Clifford algebra 11

u0

Let u1 = x − w1 ; from Lemma 4, we have that φu1 (w1 ) = x. If we define xn+1 u1 = gcd (x1 − xn+1 , x2 , . . . , xn ) =

=

p+q X (x1 − xn+1 ) xi w1 + wi , gcd (x1 − xn+1 , x2 , . . . , xn ) gcd (x1 − xn+1 , x2 , . . . , xn ) i=2 p+q X

βi wi ,

i=1

we have that (β1 , . . . , βn ) ∈ Zn and gcd (β1 , . . . , βn ) = 1. Therefore φu0 (w1 ) ∈ A. That is, with a element of the set A we can generate a solution in h(x1 , . . . , xn , xn+1 )i. 2. x − w1 6= 0 is non-invertible. According to Proposition 3, in this case we have that x + w1 is invertible, so define u1 = x + w1 =

p+q X x1 + xn+1 xi w1 + wi . xn+1 x i=2 n+1

From Lemma 4 we have that φu1 (−w1 ) = x. If we define u0

=

xn+1 u1 gcd (x1 + xn+1 , x2 , . . . , xn )

=

p+q X (x1 + xn+1 ) xi w1 + wi , gcd (x1 + xn+1 , x2 , . . . , xn ) gcd (x + x 1 n+1 , x2 , . . . , xn ) i=2

=

p+q X

βi wi ,

i=1

we have that (β1 , . . . , βn ) ∈ Zn and gcd (β1 , . . . , βn ) = 1. Therefore φu0 (−w1 ) ∈ B. Therefore, with a element of the set B we can generate a solution of (1.4) in h(x1 , . . . , xn , xn+1 )i. p+q P To sum up: given u = αi wi , where u2 6= 0, to generate solutions of i=1

(1.4) we use φu (w1 ) and φu (−w1 ). But since φu (−w1 ) = −φu (w1 ) , if φu (w1 ) yields a solution (x1 , . . . , xn , xn+1 ), then φu (−w1 ) generates the solution (−x1 , . . . , −xn , xn+1 ). Both solutions belong to the same equivalence class h(x1 , . . . , xn , xn+1 )i.

12


Clearly A∩B 6= ∅ but nevertheless any solution of (1.4) can be obtained from an element of A ∪ B. All these results can be summarized in the following Theorem: Theorem 6. Let {e1 , . . . , en } be the canonical basis of the orthogonal nondegenerate space Rp,q , p + q = n, and let B = {w1 , . . . , wn } be an orthogonal √ basis with wi = ai ei . If (x1 , . . . , xn , xn+1 ), xn+1 6= 0, is an integer solution of the equation a1 x21 + · · · + ap x2p = ap+1 x2p+1 + · · · + ap+q x2p+q + a1 x2n+1 , then there exists an equivalent solution (y1 , . . . , yn , yn+1 ) ∈ Rn+1 , which is given by    p p+q X X y1 = − a1 α12 −  αi2 ai − αi2 ai  , i=2

yi

=

yn+1

=

i=p+1

−2a1 α1 αi , i = 2, 3, . . . , n = p + q, p p+q X X 2 ai αi − ai αi2 . i=1

i=p+1

Here [u]B = (α1 , . . . , αn ) is a n-tuple of integers such that gcd (α1 , . . . , αn ) = 1 and u2 6= 0. 5.1. Some special cases Let us close this Section discussing some special cases that can arise when we considered the vectors x − w1 and x + w1 in the analysis of equation (5.6). Two possibilities should be taken into account: 2

1. (x − w1 ) = 0. That is 2

(x − w1 ) = x2 − 2 (x, w1 ) + w12 = 0. where (·, ·) is the canonical inner product in Rp,q . As x2 = w12 = a1 > 0 and (x, w1 ) = x1 /xn+1 , we have that x1 = xn+1 . 2

2. (x + w1 ) = 0. That is 2

(x + w1 ) = x2 + 2 (x, w1 ) + w12 = 0, which yields x1 = −xn+1 . Both cases are mutually exclusive but nevertheless can help us to solve equations of the type a1 x21 + · · · + ap x2p = ap+1 x2p+1 + · · · + ap+q x2p+q . Indeed, if we extend this equation to x20 + a1 x21 + · · · + ap x2p = ap+1 x2p+1 + · · · + ap+q x2p+q + x2n+1 ,

(5.7)

Solving some quadratic Diophantine equations with Clifford algebra 13 then we can already find solutions with xn+1 6= 0. Therefore, we could obtain solutions of (5.7) with an algorithm for the cases x0 = xn+1 6= 0 or x0 = −x1 . This work is under way.

6. Conclusions We have used reflections of certain vectors in Rn and Rp,q to find integer solutions to quadratic Diophantine equations of the type a1 x21 + · · · + an x2n = a1 x2n+1 ,

xn+1 6= 0,

for positive integers a1 , a2 , . . . , an , and a1 x21 + · · · + ap x2p = ap+1 x2p+1 + · · · + ap+q x2p+q + a1 x2n+1 ,

xn+1 6= 0,

for positive integers a1 , a2 , . . . , an , respectively. We show that Clifford algebra turns out to be an adequate language to handle reflections and provides useful insights when applied to the problem of solving quadratic equations. Acknowledgements G.A.G. Acknowledges financial support from the Program for the Professional Development in Automation (grant from the Universidad Autónoma Metropolitana and Parker Haniffin-México). J.L.A. is indebted to DGAPAUNAM and CONACYT for sabbatical grants and CONACyT for research grants Nos. 50368 and 79641. M.A.R.A. acknowledges financial suport from COFAA-IPN.

References [1] G. Arag´ on-Gonz´ alez, J.L. Arag´ on, M.A. Rodr´ıguez-Andrade, Pythagorean vectors and Clifford numbers, submitted to Adv. Appl. Clifford. Al. [2] M. Riesz, Clifford Numbers and Spinors: with Riesz’s Private Lectures to E. Folke Bolinder and a Historical Review by Pertti Lounesto, Springer, New York, 1993. [3] I. Porteous, Clifford algebras and the classical groups. Cambridge University Press, Cambridge, 1995. [4] C.C. Chevalley, The Algebraic Theory of Spinors, Columbia University Press, Cambridge, 1954. [5] E. Artin, Geometric Algebra, Interscience Publishers, New York, 1957. [6] D. Hestenes, Clifford Algebra to Geometric Calculus. A Unified Language for Mathematics and Physics, D. Reidel Publishing, The Netherlands, 1985. [7] G. Arag´ on-Gonz´ alez, J.L. Arag´ on, M.A. Rodr´ıguez-Andrade, The decomposition of an orthogonal transformation as a product of reflections, J. Math. Phys. 47 (2006), Art. No. 013509.

14


G. Arag´ on-Gonz´ alez Programa de Desarrollo Profesional en Automatizaci´ on, Universidad Aut´ onoma Metropolitana, Azcapotzalco, San Pablo 180, Colonia Reynosa-Tamaulipas, 02200 D. F. México. e-mail: [email protected] J. L. Arag´ on Centre for Mathematical Biology, Mathematical Institute, University of Oxford. 24-29 St Giles, Oxford OX1 3LB, U.K. Centro de F´ısica Aplicada y Tecnolog´ıa Avanzada, Universidad Nacional Aut´ onoma de México, Apartado Postal 1-1010, 76000 Querétaro, México. e-mail: [email protected] M. A. Rodr´ıguez-Andrade Departamento de Matem´ aticas, Escuela Superior de F´ısica y Matem´ aticas, Instituto Politécnico Nacional. Unidad Profesional Adolfo L´ opez Mateos, Edificio 9. 07300 D. F. México. e-mail: [email protected]