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SandBoxCliffordAlgebra

Edit detail for SandBoxCliffordAlgebra revision 8 of 9

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Editor: test1 Time: 2013/03/18 00:58:21 GMT+0
Note:

changed:
-dq := quadraticForm diagMat
-CLDQ := CliffordAlgebra(3, Fraction(Integer), dq)
CLDQ := CliffordAlgebra(3, Fraction(Integer), diagMat)

changed:
-where dq is the diagonal quadratic form in the basis of generators ei=e(i)
corresponding to the diagonal quadratic form in the basis of generators ei=e(i)

changed:
-odq := quadraticForm offDiagMat
-CLODQ := CliffordAlgebra(3, Fraction(Integer), odq)
CLODQ := CliffordAlgebra(3, Fraction(Integer), offDiagMat)

changed:
-which is irritating, since the Clifford product is just 'unevaluated', hence the
-basis elements are generated by Cliffrord products. We would like to see this more
-explicit in the Grassmann basis, which we define as follows:
-\begin{axiom}
-f12: CLODQ := 1/2*( e(1)*e(2)-e(2)*e(1) )
-f13: CLODQ := 1/2*( e(1)*e(3)-e(3)*e(1) )
-f23: CLODQ := 1/2*( e(2)*e(3)-e(3)*e(2) )
-f123: CLODQ := 1/3*( f12*e(3)$CLODQ- f13*e(2)$CLODQ + f23*e(1)$CLODQ ) -- need this od $CLODQ
-fbasis:List CLODQ :=[ 1, e(1), e(2), f12 , e(3), f13 , f23, f123 ]
-\end{axiom}
-and this is plainly wrong. Eg
-\begin{equation}
-f_{13} = \frac{1}{2}\ e_1 e_3 - \frac{1}{2} \ (e_3 \wedge e_1 + B(e_3,e_1))
-\end{equation}
-and further simplification leads to
-\begin{eqnarray}
-f_{13} &=& \frac{1}{2} \ e_1 e_3 + \frac{1}{2} \ ( e_1 \wedge e_3 - B(e_1,e_3)) \\
-&=& e_1 e_3 - \frac{1}{2} \ B(e_1,e_3)
-\end{eqnarray}
-and all these off diagonal contraction
-terms are missing in the basis. Hence the $\text{AXIOM}^{TM}$ CliffordAlgebra domain works only in the basis of generators
-which diagonalized the quadratic form!
-
-Actually this calls for a new domain CliffordAlgebra which does it right ;-). 
-

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

We want to test some properties of the CliffordAlgebra? domain implemented in AXIOM(TM)

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.