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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.




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