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SandBoxCliffordAlgebra

Edit detail for SandBoxCliffordAlgebra revision 9 of 9

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Editor: test1 Time: 2013/07/31 14:43:30 GMT+0
Note:

changed:
-We want to test some properties of the CliffordAlgebra domain implemented in AXIOM(TM)
We want to test some properties of the CliffordAlgebra domain implemented in FriCAS

We want to test some properties of the CliffordAlgebra? domain implemented in FriCAS

fricas

diagMat:=matrix[[1,0,0],[0,1,0],[0,0,1]]

$\label{eq1}\left[ \begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1$

(1)

Type: Matrix(NonNegativeInteger?)

fricas

CLDQ := CliffordAlgebra(3, Fraction(Integer), diagMat)

$\label{eq2}\hbox{\axiomType{CliffordAlgebra}\ } (3, \hbox{\axiomType{Fraction}\ } (\hbox{\axiomType{Integer}\ }) , [ [ 1, 0, 0 ] , [ 0, 1, 0 ] , [ 0, 0, 1 ] ])$

(2)

Type: Type

CLDQ contains now the Clifford algebra constructor for the Clifford algebra CL(R^3,Q) corresponding to the diagonal quadratic form in the basis of generators ei=e(i)

fricas

basGen: List CLDQ :=[1, e(1), e(2), e(1)*e(2), e(3), e(1)*e(3), e(2)*e(3), e(1)*e(2)*e(3)]

$\label{eq3}\left[ 1, \:{e_{1}}, \:{e_{2}}, \:{{e_{1}}\ {e_{2}}}, \:{e_{3}}, \:{{e_{1}}\ {e_{3}}}, \:{{e_{2}}\ {e_{3}}}, \:{{e_{1}}\ {e_{2}}\ {e_{3}}}\right]$

(3)

Type: List(CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]]))

And we can now compute within this basis

fricas

elem1: CLDQ :=2*e(1)+e(2)

$\label{eq4}{2 \ {e_{1}}}+{e_{2}}$

(4)

Type: CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]])

fricas

elem2: CLDQ :=1/2+e(2)+3*e(1)*e(3)

$\label{eq5}{1 \over 2}+{e_{2}}+{3 \ {e_{1}}\ {e_{3}}}$

(5)

Type: CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]])

fricas

elem1*elem2

$\label{eq6}1 +{e_{1}}+{{1 \over 2}\ {e_{2}}}+{2 \ {e_{1}}\ {e_{2}}}+{6 \ {e_{3}}}-{3 \ {e_{1}}\ {e_{2}}\ {e_{3}}}$

(6)

Type: CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]])

which is fine.

Lets now see what happens if we change the basis. We define new generators

fricas

f1:CLDQ := e(1)-e(2)

$\label{eq7}{e_{1}}-{e_{2}}$

(7)

Type: CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]])

fricas

f2:CLDQ := e(2)-e(3)

$\label{eq8}{e_{2}}-{e_{3}}$

(8)

Type: CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]])

fricas

f3:CLDQ := e(1)+e(2)+e(3)

$\label{eq9}{e_{1}}+{e_{2}}+{e_{3}}$

(9)

Type: CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]])

fricas

lstFGen: List CLDQ := [f1, f2, f3]

$\label{eq10}\left[{{e_{1}}-{e_{2}}}, \:{{e_{2}}-{e_{3}}}, \:{{e_{1}}+{e_{2}}+{e_{3}}}\right]$

(10)

Type: List(CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]]))

and check what the new defining relations are

fricas

matrix [[y*x for x in lstFGen] for y in lstFGen]

$\label{eq11}\left[ \begin{array}{ccc} 2 &{- 1 +{{e_{1}}\ {e_{2}}}-{{e_{1}}\ {e_{3}}}+{{e_{2}}\ {e_{3}}}}&{{2 \ {e_{1}}\ {e_{2}}}+{{e_{1}}\ {e_{3}}}-{{e_{2}}\ {e_{3}}}} \ {- 1 -{{e_{1}}\ {e_{2}}}+{{e_{1}}\ {e_{3}}}-{{e_{2}}\ {e_{3}}}}& 2 &{-{{e_{1}}\ {e_{2}}}+{{e_{1}}\ {e_{3}}}+{2 \ {e_{2}}\ {e_{3}}}} \ {-{2 \ {e_{1}}\ {e_{2}}}-{{e_{1}}\ {e_{3}}}+{{e_{2}}\ {e_{3}}}}&{{{e_{1}}\ {e_{2}}}-{{e_{1}}\ {e_{3}}}-{2 \ {e_{2}}\ {e_{3}}}}& 3$

(11)

Type: Matrix(CliffordAlgebra?(3,Fraction(Integer),[[1,0,0],[0,1,0],[0,0,1]]))

However, let us do the same calculations with another, symmetric but not diagonal quadratic form

fricas

offDiagMat:=matrix[[0,0,1],[0,1,0],[1,0,0]]

$\label{eq12}\left[ \begin{array}{ccc} 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0$

(12)

Type: Matrix(NonNegativeInteger?)

fricas

CLODQ := CliffordAlgebra(3, Fraction(Integer), offDiagMat)

$\label{eq13}\hbox{\axiomType{CliffordAlgebra}\ } (3, \hbox{\axiomType{Fraction}\ } (\hbox{\axiomType{Integer}\ }) , [ [ 0, 0, 1 ] , [ 0, 1, 0 ] , [ 1, 0, 0 ] ])$

(13)

Type: Type

fricas

basGenO: List CLODQ :=[1, e(1), e(2), e(1)*e(2), e(3), e(1)*e(3), e(2)*e(3), e(1)*e(2)*e(3)]

$\label{eq14}\left[ 1, \:{e_{1}}, \:{e_{2}}, \:{{e_{1}}\ {e_{2}}}, \:{e_{3}}, \:{1 +{{e_{1}}\ {e_{3}}}}, \:{{e_{2}}\ {e_{3}}}, \:{-{e_{2}}+{{e_{1}}\ {e_{2}}\ {e_{3}}}}\right]$

(14)

Type: List(CliffordAlgebra?(3,Fraction(Integer),[[0,0,1],[0,1,0],[1,0,0]]))

fricas

basO: List CLODQ :=[e(1), e(2), e(3)]

$\label{eq15}\left[{e_{1}}, \:{e_{2}}, \:{e_{3}}\right]$

(15)

Type: List(CliffordAlgebra?(3,Fraction(Integer),[[0,0,1],[0,1,0],[1,0,0]]))

and let us check the multiplication table on the basis and in general:

fricas

matrix [[y*x for x in basO] for y in basO]

$\label{eq16}\left[ \begin{array}{ccc} 0 &{{e_{1}}\ {e_{2}}}&{1 +{{e_{1}}\ {e_{3}}}} \ -{{e_{1}}\ {e_{2}}}& 1 &{{e_{2}}\ {e_{3}}} \ {1 -{{e_{1}}\ {e_{3}}}}& -{{e_{2}}\ {e_{3}}}& 0$

(16)

Type: Matrix(CliffordAlgebra?(3,Fraction(Integer),[[0,0,1],[0,1,0],[1,0,0]]))

fricas

matrix [[y*x for x in basGenO] for y in basGenO]

$\label{eq17}\left[ \begin{array}{cccccccc} 1 &{e_{1}}&{e_{2}}&{{e_{1}}\ {e_{2}}}&{e_{3}}&{1 +{{e_{1}}\ {e_{3}}}}&{{e_{2}}\ {e_{3}}}&{-{e_{2}}+{{e_{1}}\ {e_{2}}\ {e_{3}}}} \ {e_{1}}& 0 &{{e_{1}}\ {e_{2}}}& 0 &{1 +{{e_{1}}\ {e_{3}}}}& 0 &{-{e_{2}}+{{e_{1}}\ {e_{2}}\ {e_{3}}}}& 0 \ {e_{2}}& -{{e_{1}}\ {e_{2}}}& 1 & -{e_{1}}&{{e_{2}}\ {e_{3}}}&{{e_{2}}-{{e_{1}}\ {e_{2}}\ {e_{3}}}}&{e_{3}}&{- 1 -{{e_{1}}\ {e_{3}}}} \ {{e_{1}}\ {e_{2}}}& 0 &{e_{1}}& 0 &{-{e_{2}}+{{e_{1}}\ {e_{2}}\ {e_{3}}}}& 0 &{1 +{{e_{1}}\ {e_{3}}}}& 0 \ {e_{3}}&{1 -{{e_{1}}\ {e_{3}}}}& -{{e_{2}}\ {e_{3}}}&{{e_{2}}+{{e_{1}}\ {e_{2}}\ {e_{3}}}}& 0 &{2 \ {e_{3}}}& 0 &{2 \ {e_{2}}\ {e_{3}}} \ {1 +{{e_{1}}\ {e_{3}}}}&{2 \ {e_{1}}}&{{e_{2}}-{{e_{1}}\ {e_{2}}\ {e_{3}}}}&{2 \ {e_{1}}\ {e_{2}}}& 0 &{2 +{2 \ {e_{1}}\ {e_{3}}}}& 0 &{-{2 \ {e_{2}}}+{2 \ {e_{1}}\ {e_{2}}\ {e_{3}}}} \ {{e_{2}}\ {e_{3}}}&{{e_{2}}+{{e_{1}}\ {e_{2}}\ {e_{3}}}}& -{e_{3}}&{1 -{{e_{1}}\ {e_{3}}}}& 0 &{2 \ {e_{2}}\ {e_{3}}}& 0 &{2 \ {e_{3}}} \ {-{e_{2}}+{{e_{1}}\ {e_{2}}\ {e_{3}}}}&{2 \ {e_{1}}\ {e_{2}}}&{- 1 -{{e_{1}}\ {e_{3}}}}&{2 \ {e_{1}}}& 0 &{-{2 \ {e_{2}}}+{2 \ {e_{1}}\ {e_{2}}\ {e_{3}}}}& 0 &{2 +{2 \ {e_{1}}\ {e_{3}}}}$

(17)

Type: Matrix(CliffordAlgebra?(3,Fraction(Integer),[[0,0,1],[0,1,0],[1,0,0]]))

Note: Before Martin Baker fixed CliffordAlgebra? the results with off diagonal form were wrong.