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#117 Inheritance of Monoid Structure in Direct Product ←	↑	→ #119 solve returns wrong answers and multiple answers to same trig problem

#118 Quaternion restriction

Edit detail for #118 Quaternion restriction revision 2 of 3

	1 2 3
Editor: test1 Time: 2014/04/15 17:19:47 GMT+0
Note:

changed:
-q**2
q^2

added:

Yes, use GeneralQuaternion:
\begin{axiom}
qD := GeneralQuaternion(Fraction(Polynomial(Integer)), a, b)
quatern(0, 1, 0, 0)$qD^2
quatern(0, 0, 1, 0)$qD^2
quatern(0, 0, 0, 1)$qD^2
\end{axiom}

changed:
-H := CliffordAlgebra(2, K, quadraticForm m)
H := CliffordAlgebra(2, K, m)

originally posted by Anonymous

I've been checking quaternion support in Axiom. And I've found out that it is a bit too restricted. More exactly it seems to be aimed on Hamiltonian quaternions only:

fricas

q := quatern(0,1,0,0)

$\label{eq1}i$

(1)

Type: Quaternion(Integer)

fricas

q^2

$\label{eq2}- 1$

(2)

Type: Quaternion(Integer)

This is fine over the reals. But only there. On any field with bigger group of square classes (e.g. rationals, algebraic rationals,...) we have far more quaternion algebras. Namely for any two square classes , there exist a quaternion algebra with and .

Is there a way to declare such a quaternions in Axiom?

Yes, use GeneralQuaternion?:

fricas

qD := GeneralQuaternion(Fraction(Polynomial(Integer)), a, b)

$\label{eq3}\hbox{\axiomType{GeneralQuaternion}\ } (\hbox{\axiomType{Fraction}\ } (\hbox{\axiomType{Polynomial}\ } (\hbox{\axiomType{Integer}\ })) , a , b)$

(3)

Type: Type

fricas

quatern(0, 1, 0, 0)$qD^2

$\label{eq4}a$

(4)

Type: GeneralQuaternion?(Fraction(Polynomial(Integer)),a,b)

fricas

quatern(0, 0, 1, 0)$qD^2

$\label{eq5}b$

(5)

Type: GeneralQuaternion?(Fraction(Polynomial(Integer)),a,b)

fricas

quatern(0, 0, 0, 1)$qD^2

$\label{eq6}-{a \ b}$

(6)

Type: GeneralQuaternion?(Fraction(Polynomial(Integer)),a,b)

... --Bill Page, Mon, 07 Mar 2005 10:33:49 -0600 reply

From the description of quatern on in section

9.64 Quaternion, page 734 of the Axiom Book I think that what you want is not possible with this domain. But since the quaternions can be thought of as a Clifford algebra, please refer to section 9.10.2 The Quaternion Numbers as a Clifford Algebra, page 483 of the Book For example

fricas

K := Fraction Polynomial Integer

$\label{eq7}\hbox{\axiomType{Fraction}\ } (\hbox{\axiomType{Polynomial}\ } (\hbox{\axiomType{Integer}\ }))$

(7)

Type: Type

fricas

m := matrix [ [a,0],[0,b] ]

$\label{eq8}\left[ \begin{array}{cc} a & 0 \ 0 & b$

(8)

Type: Matrix(Polynomial(Integer))

fricas

H := CliffordAlgebra(2, K, m)

$\label{eq9}\hbox{\axiomType{CliffordAlgebra}\ } (2, \hbox{\axiomType{Fraction}\ } (\hbox{\axiomType{Polynomial}\ } (\hbox{\axiomType{Integer}\ })) , [ [ a , 0 ] , [ 0, b ] ])$