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last edited 13 years ago by Bill Page |
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Editor: Bill Page
Time: 2011/05/24 20:22:13 GMT-7 |
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changed: -q0:=1 -- not split-complex q0:=σ::Symbol --q0:=1 -- not split-complex changed: -q1:=1 -- quaternion q1:=μ::Symbol --q1:=1 -- quaternion changed: -factor Ů -\end{axiom} factor(numer Ů)/factor(denom Ů) \end{axiom} changed: -Frobenius Condition -\begin{axiom} Frobenius Condition (fork) \begin{axiom} removed: - -Bi-algebra conditions -\begin{axiom} -ΦΦ:= _ - ( λ λ ) / _ - ( I X I ) / _ - ( Y Y ) ; - -solve(equate(ΦΦ=H),Ξ(sb('p,[i]), i,1..#Ñ)) -\end{axiom} changed: -ck3:=[p[1]=1,p[2]=1,p[3]=1] ck3:=[p[1]=1,p[2]=1,p[3]=1,σ=1,μ=1] changed: -i = Unit of the algebra -\begin{axiom} Handle \begin{axiom} Φ:𝐋 := λ / Y \end{axiom} The Cartan-Killing form makes Φ of the identity \begin{axiom} test( eval(Φ,ck)=I ) \end{axiom} but it can be the identity in many ways. For example, \begin{axiom} solve(equate(eval(Φ,[p[1]=1,p[2]=1,p[3]=1,σ=1,μ=1])=I),[p[4]]) \end{axiom} If handle is identity then fork is idempotent but the converse is not true \begin{axiom} Φ1:=map(numer,ravel(Φ-I)::List FRAC POLY INT); Φ2:=groebner Φ1 in?(ideal h2, ideal Φ2) in?(ideal Φ2, ideal h2) \end{axiom} Bi-algebra conditions \begin{axiom} ΦΦ:= _ ( λ λ ) / _ ( I X I ) / _ ( Y Y ) ; bi1:=map(numer,ravel(ΦΦ-H)::List FRAC POLY INT); bi2:=groebner bi1 solve(bi2,[p[1],p[2],p[3],p[4]]) \end{axiom} Definition 5 i = Unit of the algebra \begin{axiom} changed: -Handle -\begin{axiom} - -Φ:𝐋 := - λ / - X / - Y - -\end{axiom} - -Definition 5 - - <center>Co-unit<pre> - i - U - </pre></center> - -\begin{axiom} Co-unit <center><pre> i U </pre></center> \begin{axiom} changed: -ex1:=[q[1]=1,p[1]=1,p[2]=1,p[3]=1,p[4]=1] ex1:=[σ=1,μ=1,p[1]=1,p[2]=1,p[3]=1,p[4]=1]
Quaternion Algebra Is Frobenius In Many WaysLinear operators over a 4-dimensional vector space representing quaternion algebra
Ref:
$S_3$-permuted Frobenius Algebras
Zbigniew Oziewicz (UNAM), Gregory Peter Wene (UTSA)
Frobenius algebras and 2D topological quantum field theories
Joachim Kock
We need the Axiom LinearOperator library.
(1) -> )library CARTEN ARITY CMONAL CPROP CLOP CALEY
CartesianTensor is now explicitly exposed in frame initial CartesianTensor will be automatically loaded when needed from /var/aw/var/LatexWiki/CARTEN.NRLIB/CARTEN Arity is now explicitly exposed in frame initial Arity will be automatically loaded when needed from /var/aw/var/LatexWiki/ARITY.NRLIB/ARITY ClosedMonoidal is now explicitly exposed in frame initial ClosedMonoidal will be automatically loaded when needed from /var/aw/var/LatexWiki/CMONAL.NRLIB/CMONAL ClosedProp is now explicitly exposed in frame initial ClosedProp will be automatically loaded when needed from /var/aw/var/LatexWiki/CPROP.NRLIB/CPROP ClosedLinearOperator is now explicitly exposed in frame initial ClosedLinearOperator will be automatically loaded when needed from /var/aw/var/LatexWiki/CLOP.NRLIB/CLOP CaleyDickson is now explicitly exposed in frame initial CaleyDickson will be automatically loaded when needed from /var/aw/var/LatexWiki/CALEY.NRLIB/CALEY
Use the following macros for convenient notation
-- summation macro Σ(x,i, n)==reduce(+, [x for i in n])
-- list macro Ξ(f,i, n)==[f for i in n]
-- subscript macro sb == subscript
𝐋 is the domain of 4-dimensional linear operators over the rational functions ℚ (Expression Integer), i.e. ratio of polynomials with integer coefficients.
dim:=4\begin{equation} \label{eq1}4\end{equation}
macro ℒ == List
macro ℂ == CaleyDickson
macro ℚ == Expression Integer
𝐋 := ClosedLinearOperator(OVAR ['1,\begin{equation*} \label{eq2}\hbox{\axiomType{ClosedLinearOperator?}\ } \left({{\hbox{\axiomType{OrderedVariableList?}\ } \left({\left[ 1, \: i , \: j , \: k \right]?}\right)}, \:{\hbox{\axiomType{Expression}\ } \left({\hbox{\axiomType{Integer}\ }}\right)}}\right)\end{equation*}'i, 'j, 'k], ℚ)
𝐞:ℒ 𝐋 := basisOut()\begin{equation*} \label{eq3}\left[{|_{\ 1}}, \:{|_{\ i}}, \:{|_{\ j}}, \:{|_{\ k}}\right]?\end{equation*}
𝐝:ℒ 𝐋 := basisIn()\begin{equation*} \label{eq4}\left[{|^{\ 1}}, \:{|^{\ i}}, \:{|^{\ j}}, \:{|^{\ k}}\right]?\end{equation*}
I:𝐋:=[1] -- identity for composition\begin{equation} \label{eq5}{|_{\ 1}^{\ 1}}+{|_{\ i}^{\ i}}+{|_{\ j}^{\ j}}+{|_{\ k}^{\ k}}\end{equation}
X:𝐋:=[2,\begin{equation} \label{eq6}\begin{array}{@{}l} \displaystyle {|_{\ 1 \ 1}^{\ 1 \ 1}}+{|_{\ i \ 1}^{\ 1 \ i}}+{|_{\ j \ 1}^{\ 1 \ j}}+{|_{\ k \ 1}^{\ 1 \ k}}+{|_{\ 1 \ i}^{\ i \ 1}}+ \ \ \displaystyle {|_{\ i \ i}^{\ i \ i}}+{|_{\ j \ i}^{\ i \ j}}+{|_{\ k \ i}^{\ i \ k}}+{|_{\ 1 \ j}^{\ j \ 1}}+{|_{\ i \ j}^{\ j \ i}}+{|_{\ j \ j}^{\ j \ j}}+ \ \ \displaystyle {|_{\ k \ j}^{\ j \ k}}+{|_{\ 1 \ k}^{\ k \ 1}}+{|_{\ i \ k}^{\ k \ i}}+{|_{\ j \ k}^{\ k \ j}}+{|_{\ k \ k}^{\ k \ k}} \end{array} \end{equation}1] -- twist
V:𝐋:=ev(1) -- evaluation\begin{equation} \label{eq7}{|^{\ 1 \ 1}}+{|^{\ i \ i}}+{|^{\ j \ j}}+{|^{\ k \ k}}\end{equation}
Λ:𝐋:=co(1) -- co-evaluation\begin{equation} \label{eq8}{|_{\ 1 \ 1}}+{|_{\ i \ i}}+{|_{\ j \ j}}+{|_{\ k \ k}}\end{equation}
equate(eq)==map((x,y)+->(x=y), ravel lhs eq, ravel rhs eq);
Now generate structure constants for Quaternion Algebra
The basis consists of the real and imaginary units. We use quaternion multiplication to form the "multiplication table" as a matrix. Then the structure constants can be obtained by dividing each matrix entry by the list of basis vectors.
Split-complex and co-quaternions can be specified by Caley-Dickson parameters (q0 = -1, q1 = -1)
--q0:=sb('q, [0])
q0:=σ::Symbol
\begin{equation*}
\label{eq9}��\end{equation*}
--q0:=1 -- not split-complex
--q1:=sb('q, [1])
q1:=μ::Symbol
\begin{equation*}
\label{eq10}��\end{equation*}
--q1:=1 -- quaternion QQ := ℂ(ℂ(ℚ,'i, q0), 'j, q1);
Basis: Each B.i is a quaternion number
B:ℒ QQ := map(x +-> hyper x,\begin{equation*} \label{eq11}\left[ 1, \: i , \: j , \:{ij}\right]?\end{equation*}1$SQMATRIX(dim, ℚ)::ℒ ℒ ℚ)
-- Multiplication table: M:Matrix QQ := matrix Ξ(Ξ(B.i*B.j,\begin{equation*} \label{eq12}\left[ \begin{array}{cccc} 1 & i & j &{ij} \ i & - �� & -{ij}&{�� j} \ j &{ij}& - �� &{- �� i} \ {ij}&{- �� j}&{�� i}& -{�� \ ��} \end{array} \right]\end{equation*}i, 1..dim), j, 1..dim)
-- Function to divide the matrix entries by a basis element S(y) == map(x +-> real real(x/y),M)
-- The result is a nested list ѕ :=map(S,B)::ℒ ℒ ℒ ℚ
Compiling function S with type CaleyDickson(CaleyDickson(Expression(
Integer), i, σ), j, μ) -> Matrix(Expression(Integer))-- structure constants form a tensor operator Y := Σ(Σ(Σ(ѕ(i)(k)(j)*𝐞.i*𝐝.j*𝐝.k,\begin{equation*} \label{eq14}\begin{array}{@{}l} \displaystyle {|_{\ 1}^{\ 1 \ 1}}+{{\frac{��}{\overline ��}}\ {|_{\ i}^{\ 1 \ i}}}+{{\frac{��}{\overline ��}}\ {|_{\ j}^{\ 1 \ j}}}+{|_{\ k}^{\ 1 \ k}}+{{\frac{��}{\overline ��}}\ {|_{\ i}^{\ i \ 1}}}- \ \ \displaystyle {�� \ {|_{\ 1}^{\ i \ i}}}+{|_{\ k}^{\ i \ j}}-{{\frac{�� \ ��}{\overline ��}}\ {|_{\ j}^{\ i \ k}}}+{{\frac{��}{\overline ��}}\ {|_{\ j}^{\ j \ 1}}}-{|_{\ k}^{\ j \ i}}- \ \ \displaystyle {�� \ {|_{\ 1}^{\ j \ j}}}+{{\frac{�� \ ��}{\overline ��}}\ {|_{\ i}^{\ j \ k}}}+{|_{\ k}^{\ k \ 1}}+{{\frac{�� \ ��}{\overline ��}}\ {|_{\ j}^{\ k \ i}}}- \ \ \displaystyle {{\frac{�� \ ��}{\overline ��}}\ {|_{\ i}^{\ k \ j}}}-{�� \ �� \ {|_{\ 1}^{\ k \ k}}} \end{array} \end{equation*}i, 1..dim), j, 1..dim), k, 1..dim)
arity Y\begin{equation} \label{eq15}\frac{{+}^{2}}{+}\end{equation}
matrix Ξ(Ξ((𝐞.i*𝐞.j)/Y,\begin{equation*} \label{eq16}\left[ \begin{array}{cccc} {|_{\ 1}}&{{\frac{��}{\overline ��}}\ {|_{\ i}}}&{{\frac{��}{\overline ��}}\ {|_{\ j}}}&{|_{\ k}} \ {{\frac{��}{\overline ��}}\ {|_{\ i}}}& -{�� \ {|_{\ 1}}}& -{|_{\ k}}&{{\frac{�� \ ��}{\overline ��}}\ {|_{\ j}}} \ {{\frac{��}{\overline ��}}\ {|_{\ j}}}&{|_{\ k}}& -{�� \ {|_{\ 1}}}& -{{\frac{�� \ ��}{\overline ��}}\ {|_{\ i}}} \ {|_{\ k}}& -{{\frac{�� \ ��}{\overline ��}}\ {|_{\ j}}}&{{\frac{�� \ ��}{\overline ��}}\ {|_{\ i}}}& -{�� \ �� \ {|_{\ 1}}} \end{array} \right]\end{equation*}i, 1..dim), j, 1..dim)
Multiplication of arbitrary quaternions $a$ and $b$
a:=Σ(sb('a, [i])*𝐞.i, i, 1..dim)
\begin{equation}
\label{eq17}{{a_{1}}\ {|_{\ 1}}}+{{a_{2}}\ {|_{\ i}}}+{{a_{3}}\ {|_{\ j}}}+{{a_{4}}\ {|_{\ k}}}\end{equation}
b:=Σ(sb('b, [i])*𝐞.i, i, 1..dim)
\begin{equation}
\label{eq18}{{b_{1}}\ {|_{\ 1}}}+{{b_{2}}\ {|_{\ i}}}+{{b_{3}}\ {|_{\ j}}}+{{b_{4}}\ {|_{\ k}}}\end{equation}
(a*b)/Y\begin{equation*} \label{eq19}\begin{array}{@{}l} \displaystyle {{\left({{\left(-{{a_{4}}\ {b_{4}}\ ��}-{{a_{2}}\ {b_{2}}}\right)}\ ��}-{{a_{3}}\ {b_{3}}\ ��}+{{a_{1}}\ {b_{1}}}\right)}\ {|_{\ 1}}}+ \ \ \displaystyle {{\frac{{\left({{\left({{a_{3}}\ {b_{4}}}-{{a_{4}}\ {b_{3}}}\right)}\ ��}+{{a_{1}}\ {b_{2}}}+{{a_{2}}\ {b_{1}}}\right)}\ ��}{\overline ��}}\ {|_{\ i}}}+ \ \ \displaystyle {{\frac{{{\left(-{{a_{2}}\ {b_{4}}}+{{a_{4}}\ {b_{2}}}\right)}\ �� \ ��}+{{\left({{a_{1}}\ {b_{3}}}+{{a_{3}}\ {b_{1}}}\right)}\ ��}}{\overline ��}}\ {|_{\ j}}}+ \ \ \displaystyle {{\left({{a_{1}}\ {b_{4}}}+{{a_{2}}\ {b_{3}}}-{{a_{3}}\ {b_{2}}}+{{a_{4}}\ {b_{1}}}\right)}\ {|_{\ k}}} \end{array} \end{equation*}
Multiplication is Associative
test( ( I Y ) / _ ( Y ) = _ ( Y I ) / _ ( Y ) )\begin{equation} \label{eq20} \mbox{\rm false} \end{equation}
A scalar product is denoted by the (2,0)-tensor $U = \{ u_{ij} \}$
U:=Σ(Σ(script('u, [[], [i, j]])*𝐝.i*𝐝.j, i, 1..dim), j, 1..dim)
\begin{equation}
\label{eq21}\begin{array}{@{}l}
\displaystyle
{{u^{1, \: 1}}\ {|^{\ 1 \ 1}}}+{{u^{1, \: 2}}\ {|^{\ 1 \ i}}}+{{u^{1, \: 3}}\ {|^{\ 1 \ j}}}+{{u^{1, \: 4}}\ {|^{\ 1 \ k}}}+
\
\
\displaystyle
{{u^{2, \: 1}}\ {|^{\ i \ 1}}}+{{u^{2, \: 2}}\ {|^{\ i \ i}}}+{{u^{2, \: 3}}\ {|^{\ i \ j}}}+{{u^{2, \: 4}}\ {|^{\ i \ k}}}+{{u^{3, \: 1}}\ {|^{\ j \ 1}}}+
\
\
\displaystyle
{{u^{3, \: 2}}\ {|^{\ j \ i}}}+{{u^{3, \: 3}}\ {|^{\ j \ j}}}+{{u^{3, \: 4}}\ {|^{\ j \ k}}}+{{u^{4, \: 1}}\ {|^{\ k \ 1}}}+
\
\
\displaystyle
{{u^{4, \: 2}}\ {|^{\ k \ i}}}+{{u^{4, \: 3}}\ {|^{\ k \ j}}}+{{u^{4, \: 4}}\ {|^{\ k \ k}}}
\end{array}
\end{equation}
We say that the scalar product is associative if the tensor equation holds:
Y = Y
U U
In other words, if the (3,0)-tensor: $$ \scalebox{1} % Change this value to rescale the drawing. { \begin{pspicture}(0,-0.92)(4.82,0.92) \psbezier[linewidth=0.04]?(2.2,0.9)(2.2,0.1)(2.6,0.1)(2.6,0.9) \psline[linewidth=0.04cm]?(2.4,0.3)(2.4,-0.1) \psbezier[linewidth=0.04]?(2.4,-0.1)(2.4,-0.9)(3.0,-0.9)(3.0,-0.1) \psline[linewidth=0.04cm]?(3.0,-0.1)(3.0,0.9) \psbezier[linewidth=0.04]?(4.8,0.9)(4.8,0.1)(4.4,0.1)(4.4,0.9) \psline[linewidth=0.04cm]?(4.6,0.3)(4.6,-0.1) \psbezier[linewidth=0.04]?(4.6,-0.1)(4.6,-0.9)(4.0,-0.9)(4.0,-0.1) \psline[linewidth=0.04cm]?(4.0,-0.1)(4.0,0.9) \usefont{T1}{ptm}{m}{n} \rput(3.4948437,0.205){-} \psline[linewidth=0.04cm]?(0.6,-0.7)(0.6,0.9) \psbezier[linewidth=0.04]?(0.0,-0.1)(0.0,-0.9)(1.2,-0.9)(1.2,-0.1) \psline[linewidth=0.04cm]?(0.0,-0.1)(0.0,0.9) \psline[linewidth=0.04cm]?(1.2,-0.1)(1.2,0.9) \usefont{T1}{ptm}{m}{n} \rput(1.6948438,0.205){=} \end{pspicture} } $$
\begin{equation} \label{eq22} \Phi = \{ \phi^{ijk} = {y^e}_{ij} u_{ek} - u_{ie} {y_e}^{jk} \} \end{equation} (three-point function) is zero.
Using the LinearOperator? domain in Axiom and some carefully chosen symbols we can easily enter expressions that are both readable and interpreted by Axiom as "graphical calculus" diagrams describing complex products and compositions of linear operators.
ω:𝐋 :=
( Y I ) /
U -
( I Y ) /
U
\begin{equation*}
\label{eq23}\begin{array}{@{}l}
\displaystyle
{{\frac{{{u^{1, \: 2}}\ {\overline ��}}-{{u^{1, \: 2}}\ ��}}{\overline ��}}\ {|^{\ 1 \ 1 \ i}}}+{{\frac{{{u^{1, \: 3}}\ {\overline ��}}-{{u^{1, \: 3}}\ ��}}{\overline ��}}\ {|^{\ 1 \ 1 \ j}}}+
\
\
\displaystyle
{{\frac{{\left({u^{2, \: 1}}-{u^{1, \: 2}}\right)}\ ��}{\overline ��}}\ {|^{\ 1 \ i \ 1}}}+{{\frac{{{u^{1, \: 1}}\ �� \ {\overline ��}}+{{u^{2, \: 2}}\ ��}}{\overline ��}}\ {|^{\ 1 \ i \ i}}}+
\
\
\displaystyle
{{\frac{-{{u^{1, \: 4}}\ {\overline ��}}+{{u^{2, \: 3}}\ ��}}{\overline ��}}\ {|^{\ 1 \ i \ j}}}+{{\frac{{{u^{1, \: 3}}\ �� \ �� \ {\overline ��}}+{{u^{2, \: 4}}\ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ 1 \ i \ k}}}+
\
\
\displaystyle
{{\frac{{\left({u^{3, \: 1}}-{u^{1, \: 3}}\right)}\ ��}{\overline ��}}\ {|^{\ 1 \ j \ 1}}}+{{\frac{{{u^{1, \: 4}}\ {\overline ��}}+{{u^{3, \: 2}}\ ��}}{\overline ��}}\ {|^{\ 1 \ j \ i}}}+
\
\
\displaystyle
{{\frac{{{u^{1, \: 1}}\ �� \ {\overline ��}}+{{u^{3, \: 3}}\ ��}}{\overline ��}}\ {|^{\ 1 \ j \ j}}}+{{\frac{{{u^{3, \: 4}}\ �� \ {\overline ��}}-{{u^{1, \: 2}}\ �� \ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ 1 \ j \ k}}}+
\
\
\displaystyle
{{\left({u^{4, \: 1}}-{u^{1, \: 4}}\right)}\ {|^{\ 1 \ k \ 1}}}+{{\frac{{{u^{4, \: 2}}\ {\overline ��}}-{{u^{1, \: 3}}\ �� \ ��}}{\overline ��}}\ {|^{\ 1 \ k \ i}}}+
\
\
\displaystyle
{{\frac{{{u^{4, \: 3}}\ {\overline ��}}+{{u^{1, \: 2}}\ �� \ ��}}{\overline ��}}\ {|^{\ 1 \ k \ j}}}+{{\left({{u^{1, \: 1}}\ �� \ ��}+{u^{4, \: 4}}\right)}\ {|^{\ 1 \ k \ k}}}+
\
\
\displaystyle
{{\frac{-{{u^{2, \: 1}}\ {\overline ��}}+{{u^{2, \: 1}}\ ��}}{\overline ��}}\ {|^{\ i \ 1 \ 1}}}+
\
\
\displaystyle
{{\frac{-{{u^{2, \: 3}}\ �� \ {\overline ��}}+{{u^{2, \: 3}}\ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ i \ 1 \ j}}}+
\
\
\displaystyle
{{\frac{-{{u^{2, \: 4}}\ {\overline ��}}+{{u^{2, \: 4}}\ ��}}{\overline ��}}\ {|^{\ i \ 1 \ k}}}+{{\frac{-{{u^{1, \: 1}}\ �� \ {\overline ��}}-{{u^{2, \: 2}}\ ��}}{\overline ��}}\ {|^{\ i \ i \ 1}}}+
\
\
\displaystyle
{{\left({u^{2, \: 1}}-{u^{1, \: 2}}\right)}\ �� \ {|^{\ i \ i \ i}}}+{{\left(-{{u^{1, \: 3}}\ ��}-{u^{2, \: 4}}\right)}\ {|^{\ i \ i \ j}}}+
\
\
\displaystyle
{{\frac{-{{u^{1, \: 4}}\ �� \ {\overline ��}}+{{u^{2, \: 3}}\ �� \ ��}}{\overline ��}}\ {|^{\ i \ i \ k}}}+{{\frac{{{u^{4, \: 1}}\ {\overline ��}}-{{u^{2, \: 3}}\ ��}}{\overline ��}}\ {|^{\ i \ j \ 1}}}+
\
\
\displaystyle
{{\left({u^{4, \: 2}}+{u^{2, \: 4}}\right)}\ {|^{\ i \ j \ i}}}+{{\left({{u^{2, \: 1}}\ ��}+{u^{4, \: 3}}\right)}\ {|^{\ i \ j \ j}}}+
\
\
\displaystyle
{{\frac{{{u^{4, \: 4}}\ {\overline ��}}-{{u^{2, \: 2}}\ �� \ ��}}{\overline ��}}\ {|^{\ i \ j \ k}}}+{{\frac{-{{u^{2, \: 4}}\ {\overline ��}}-{{u^{3, \: 1}}\ �� \ ��}}{\overline ��}}\ {|^{\ i \ k \ 1}}}+
\
\
\displaystyle
{{\frac{{\left(-{u^{3, \: 2}}-{u^{2, \: 3}}\right)}\ �� \ ��}{\overline ��}}\ {|^{\ i \ k \ i}}}+
\
\
\displaystyle
{{\frac{-{{u^{3, \: 3}}\ �� \ �� \ {\overline ��}}+{{u^{2, \: 2}}\ �� \ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ i \ k \ j}}}+
\
\
\displaystyle
{{\frac{{{u^{2, \: 1}}\ �� \ �� \ {\overline ��}}-{{u^{3, \: 4}}\ �� \ ��}}{\overline ��}}\ {|^{\ i \ k \ k}}}+{{\frac{-{{u^{3, \: 1}}\ {\overline ��}}+{{u^{3, \: 1}}\ ��}}{\overline ��}}\ {|^{\ j \ 1 \ 1}}}+
\
\
\displaystyle
{{\frac{{{u^{3, \: 2}}\ �� \ {\overline ��}}-{{u^{3, \: 2}}\ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ j \ 1 \ i}}}+{{\frac{-{{u^{3, \: 4}}\ {\overline ��}}+{{u^{3, \: 4}}\ ��}}{\overline ��}}\ {|^{\ j \ 1 \ k}}}+
\
\
\displaystyle
{{\frac{-{{u^{4, \: 1}}\ {\overline ��}}-{{u^{3, \: 2}}\ ��}}{\overline ��}}\ {|^{\ j \ i \ 1}}}+{{\left({{u^{3, \: 1}}\ ��}-{u^{4, \: 2}}\right)}\ {|^{\ j \ i \ i}}}+
\
\
\displaystyle
{{\left(-{u^{4, \: 3}}-{u^{3, \: 4}}\right)}\ {|^{\ j \ i \ j}}}+{{\frac{-{{u^{4, \: 4}}\ {\overline ��}}+{{u^{3, \: 3}}\ �� \ ��}}{\overline ��}}\ {|^{\ j \ i \ k}}}+
\
\
\displaystyle
{{\frac{-{{u^{1, \: 1}}\ �� \ {\overline ��}}-{{u^{3, \: 3}}\ ��}}{\overline ��}}\ {|^{\ j \ j \ 1}}}+{{\left(-{{u^{1, \: 2}}\ ��}+{u^{3, \: 4}}\right)}\ {|^{\ j \ j \ i}}}+ \
\
\displaystyle
{{\left({u^{3, \: 1}}-{u^{1, \: 3}}\right)}\ �� \ {|^{\ j \ j \ j}}}+{{\frac{-{{u^{1, \: 4}}\ �� \ {\overline ��}}-{{u^{3, \: 2}}\ �� \ ��}}{\overline ��}}\ {|^{\ j \ j \ k}}}+
\
\
\displaystyle
{{\frac{-{{u^{3, \: 4}}\ {\overline ��}}+{{u^{2, \: 1}}\ �� \ ��}}{\overline ��}}\ {|^{\ j \ k \ 1}}}+
\
\
\displaystyle
{{\frac{-{{u^{3, \: 3}}\ �� \ �� \ {\overline ��}}+{{u^{2, \: 2}}\ �� \ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ j \ k \ i}}}+
\
\
\displaystyle
{{\frac{{\left({u^{3, \: 2}}+{u^{2, \: 3}}\right)}\ �� \ ��}{\overline ��}}\ {|^{\ j \ k \ j}}}+{{\frac{{{u^{3, \: 1}}\ �� \ �� \ {\overline ��}}+{{u^{2, \: 4}}\ �� \ ��}}{\overline ��}}\ {|^{\ j \ k \ k}}}+
\
\
\displaystyle
{{\frac{{{u^{4, \: 2}}\ {\overline ��}}-{{u^{4, \: 2}}\ ��}}{\overline ��}}\ {|^{\ k \ 1 \ i}}}+{{\frac{{{u^{4, \: 3}}\ {\overline ��}}-{{u^{4, \: 3}}\ ��}}{\overline ��}}\ {|^{\ k \ 1 \ j}}}+
\
\
\displaystyle
{{\frac{{{u^{3, \: 1}}\ �� \ �� \ {\overline ��}}-{{u^{4, \: 2}}\ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ k \ i \ 1}}}+{{\frac{{{u^{4, \: 1}}\ �� \ {\overline ��}}+{{u^{3, \: 2}}\ �� \ ��}}{\overline ��}}\ {|^{\ k \ i \ i}}}+
\
\
\displaystyle
{{\frac{-{{u^{4, \: 4}}\ {\overline ��}}+{{u^{3, \: 3}}\ �� \ ��}}{\overline ��}}\ {|^{\ k \ i \ j}}}+{{\frac{{\left({u^{4, \: 3}}+{u^{3, \: 4}}\right)}\ �� \ ��}{\overline ��}}\ {|^{\ k \ i \ k}}}+
\
\
\displaystyle
{{\frac{-{{u^{4, \: 3}}\ �� \ {\overline ��}}-{{u^{2, \: 1}}\ �� \ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ k \ j \ 1}}}+
\
\
\displaystyle
{{\frac{{{u^{4, \: 4}}\ {\overline ��}}-{{u^{2, \: 2}}\ �� \ ��}}{\overline ��}}\ {|^{\ k \ j \ i}}}+{{\frac{{{u^{4, \: 1}}\ �� \ {\overline ��}}-{{u^{2, \: 3}}\ �� \ ��}}{\overline ��}}\ {|^{\ k \ j \ j}}}+
\
\
\displaystyle
{{\frac{{\left(-{u^{4, \: 2}}-{u^{2, \: 4}}\right)}\ �� \ ��}{\overline ��}}\ {|^{\ k \ j \ k}}}+{{\left(-{{u^{1, \: 1}}\ �� \ ��}-{u^{4, \: 4}}\right)}\ {|^{\ k \ k \ 1}}}+
\
\
\displaystyle
{{\frac{-{{u^{1, \: 2}}\ �� \ �� \ {\overline ��}}-{{u^{4, \: 3}}\ �� \ ��}}{\overline ��}}\ {|^{\ k \ k \ i}}}+
\
\
\displaystyle
{{\frac{-{{u^{1, \: 3}}\ �� \ �� \ {\overline ��}}+{{u^{4, \: 2}}\ �� \ ��}}{\overline ��}}\ {|^{\ k \ k \ j}}}+{{\left({u^{4, \: 1}}-{u^{1, \: 4}}\right)}\ �� \ �� \ {|^{\ k \ k \ k}}}
\end{array}
\end{equation*}
An algebra with a non-degenerate associative scalar product is called a [Frobenius Algebra]?.
The Cartan-Killing Trace
Ú:=
( Y Λ ) / _
( Y I ) / _
V
\begin{equation*}
\label{eq24}\begin{array}{@{}l}
\displaystyle
{{\frac{{{\left({2 \ {\overline ��}}+ �� \right)}\ {\overline ��}}+{�� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ 1 \ 1}}}+{{\frac{{{\left(-{2 \ �� \ {\overline ��}}-{�� \ ��}\right)}\ {\overline ��}}-{{{��}^{2}}\ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ i \ i}}}+ \
\
\displaystyle
{{\frac{{{\left(-{2 \ �� \ {\overline ��}}-{{��}^{2}}\right)}\ {\overline ��}}-{�� \ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ j \ j}}}+{{\frac{{{\left(-{2 \ �� \ �� \ {\overline ��}}-{{{��}^{2}}\ ��}\right)}\ {\overline ��}}-{�� \ {{��}^{2}}\ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ k \ k}}}
\end{array}
\end{equation*}
Ù:=
( Λ Y ) / _
( I Y ) / _
V
\begin{equation*}
\label{eq25}\begin{array}{@{}l}
\displaystyle
{{\frac{{{\left({2 \ {\overline ��}}+ �� \right)}\ {\overline ��}}+{�� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ 1 \ 1}}}+{{\frac{{{\left(-{2 \ �� \ {\overline ��}}-{�� \ ��}\right)}\ {\overline ��}}-{{{��}^{2}}\ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ i \ i}}}+ \
\
\displaystyle
{{\frac{{{\left(-{2 \ �� \ {\overline ��}}-{{��}^{2}}\right)}\ {\overline ��}}-{�� \ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ j \ j}}}+{{\frac{{{\left(-{2 \ �� \ �� \ {\overline ��}}-{{{��}^{2}}\ ��}\right)}\ {\overline ��}}-{�� \ {{��}^{2}}\ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ k \ k}}}
\end{array}
\end{equation*}
test(Ù=Ú)\begin{equation} \label{eq26} \mbox{\rm true} \end{equation}
forms a non-degenerate associative scalar product for Y
Ũ := Ù\begin{equation*} \label{eq27}\begin{array}{@{}l} \displaystyle {{\frac{{{\left({2 \ {\overline ��}}+ �� \right)}\ {\overline ��}}+{�� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ 1 \ 1}}}+{{\frac{{{\left(-{2 \ �� \ {\overline ��}}-{�� \ ��}\right)}\ {\overline ��}}-{{{��}^{2}}\ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ i \ i}}}+ \ \ \displaystyle {{\frac{{{\left(-{2 \ �� \ {\overline ��}}-{{��}^{2}}\right)}\ {\overline ��}}-{�� \ �� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ j \ j}}}+{{\frac{{{\left(-{2 \ �� \ �� \ {\overline ��}}-{{{��}^{2}}\ ��}\right)}\ {\overline ��}}-{�� \ {{��}^{2}}\ {\overline ��}}}{{\overline ��}\ {\overline ��}}}\ {|^{\ k \ k}}} \end{array} \end{equation*}
test
( Y I ) /
Ũ =
( I Y ) /
Ũ
\begin{equation}
\label{eq28} \mbox{\rm false} \end{equation}
determinant Ξ(Ξ(retract((𝐞.i * 𝐞.j)/Ũ),\begin{equation*} \label{eq29}\frac{{{\left(-{{16}\ {{��}^{2}}\ {{��}^{2}}\ {{\overline ��}^{4}}}-{{32}\ {{��}^{3}}\ {{��}^{2}}\ {{\overline ��}^{3}}}-{{24}\ {{��}^{4}}\ {{��}^{2}}\ {{\overline ��}^{2}}}-{8 \ {{��}^{5}}\ {{��}^{2}}\ {\overline ��}}-{{{��}^{6}}\ {{��}^{2}}}\right)}\ {{\overline ��}^{4}}}+{{\left(-{{32}\ {{��}^{2}}\ {{��}^{3}}\ {{\overline ��}^{4}}}-{{48}\ {{��}^{3}}\ {{��}^{3}}\ {{\overline ��}^{3}}}-{{24}\ {{��}^{4}}\ {{��}^{3}}\ {{\overline ��}^{2}}}-{4 \ {{��}^{5}}\ {{��}^{3}}\ {\overline ��}}\right)}\ {{\overline ��}^{3}}}+{{\left(-{{24}\ {{��}^{2}}\ {{��}^{4}}\ {{\overline ��}^{4}}}-{{24}\ {{��}^{3}}\ {{��}^{4}}\ {{\overline ��}^{3}}}-{6 \ {{��}^{4}}\ {{��}^{4}}\ {{\overline ��}^{2}}}\right)}\ {{\overline ��}^{2}}}+{{\left(-{8 \ {{��}^{2}}\ {{��}^{5}}\ {{\overline ��}^{4}}}-{4 \ {{��}^{3}}\ {{��}^{5}}\ {{\overline ��}^{3}}}\right)}\ {\overline ��}}-{{{��}^{2}}\ {{��}^{6}}\ {{\overline ��}^{4}}}}{{{\overline ��}^{4}}\ {{\overline ��}^{4}}}\end{equation*}j, 1..dim), i, 1..dim)
General Solution
We may consider the problem where multiplication Y is given, and look for all associative scalar products $U = U(Y)$
This problem can be solved using linear algebra.
)expose MCALCFN
MultiVariableCalculusFunctions is now explicitly exposed in frame initial J := jacobian(ravel ω,concat map(variables, ravel U)::ℒ Symbol);
u := transpose matrix [concat map(variables,ravel U)::ℒ Symbol];
J::OutputForm * u::OutputForm = 0\begin{equation*} \label{eq30}\begin{array}{@{}l} \displaystyle {{\left[ \begin{array}{cccccccccccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 &{\frac{{\overline ��}- ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 &{\frac{{\overline ��}- ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & -{\frac{��}{\overline ��}}& 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ �� & 0 & 0 & 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & - 1 & 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & -{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 \ �� & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 \ 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 \ 0 & 0 & 0 & - 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ {�� \ ��}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 &{\frac{-{\overline ��}+ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 &{\frac{-{�� \ {\overline ��}}+{�� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{-{\overline ��}+ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ - �� & 0 & 0 & 0 & 0 & -{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & - �� & 0 & 0 & �� & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & - �� & 0 & 0 & 0 & 0 & - 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & - �� & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & �� & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & - 1 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 &{�� \ ��}& 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{-{\overline ��}+ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{{�� \ {\overline ��}}-{�� \ {\overline ��}}}{{\overline ��}\ {\overline ��}}}& 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{-{\overline ��}+ ��}{\overline ��}}& 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{��}{\overline ��}}& 0 & 0 & - 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & �� & 0 & 0 & 0 & 0 & - 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 0 & - 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & - 1 \ - �� & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 \ 0 & - �� & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & - �� & 0 & 0 & 0 & 0 & 0 & �� & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & - �� & 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & - 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}&{�� \ ��}& 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{{\overline ��}- ��}{\overline ��}}& 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{{\overline ��}- ��}{\overline ��}}& 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & -{\frac{��}{\overline ��}}& 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & �� & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & - 1 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 \ 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{��}{\overline ��}}& 0 \ 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \ 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & �� & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 & 0 \ -{�� \ ��}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & - 1 \ 0 & -{�� \ ��}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -{\frac{�� \ ��}{\overline ��}}& 0 \ 0 & 0 & -{�� \ ��}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{\frac{�� \ ��}{\overline ��}}& 0 & 0 \ 0 & 0 & 0 & -{�� \ ��}& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &{�� \ ��}& 0 & 0 & 0 \end{array} \right]}\ {\left[ \begin{array}{c} {u^{1, \: 1}} \ {u^{1, \: 2}} \ {u^{1, \: 3}} \ {u^{1, \: 4}} \ {u^{2, \: 1}} \ {u^{2, \: 2}} \ {u^{2, \: 3}} \ {u^{2, \: 4}} \ {u^{3, \: 1}} \ {u^{3, \: 2}} \ {u^{3, \: 3}} \ {u^{3, \: 4}} \ {u^{4, \: 1}} \ {u^{4, \: 2}} \ {u^{4, \: 3}} \ {u^{4, \: 4}} \end{array} \right]}}= \ \ \displaystyle 0 \end{array} \end{equation*}
nrows(J),\begin{equation*} \label{eq31}\left[{64}, \:{16}\right]?\end{equation*}ncols(J)
The matrix J transforms the coefficients of the tensor $U$
into coefficients of the tensor $\Phi$. We are looking for
the general linear family of tensors $U=U(Y,p_i)$ such that
J transforms $U$ into $\Phi=0$ for any such $U$.
If the null space of the J matrix is not empty we can use
the basis to find all non-trivial solutions for U:
Ñ:=nullSpace(J)\begin{equation*} \label{eq32}\left[{\left[ -{\frac{1}{�� \ ��}}, \: 0, \: 0, \: 0, \: 0, \:{\frac{\overline ��}{�� \ ��}}, \: 0, \: 0, \: 0, \: 0, \:{\frac{\overline ��}{�� \ ��}}, \: 0, \: 0, \: 0, \: 0, \: 1 \right]?}\right]\end{equation*}
ℰ:=map((x,\begin{equation*} \label{eq33}\begin{array}{@{}l} \displaystyle \left[{{u^{1, \: 1}}= -{\frac{p_{1}}{�� \ ��}}}, \:{{u^{1, \: 2}}= 0}, \:{{u^{1, \: 3}}= 0}, \:{{u^{1, \: 4}}= 0}, \:{{u^{2, \: 1}}= 0}, \: \right. \ \ \displaystyle \left.{{u^{2, \: 2}}={\frac{{p_{1}}\ {\overline ��}}{�� \ ��}}}, \:{{u^{2, \: 3}}= 0}, \:{{u^{2, \: 4}}= 0}, \:{{u^{3, \: 1}}= 0}, \:{{u^{3, \: 2}}= 0}, \: \right. \ \ \displaystyle \left.{{u^{3, \: 3}}={\frac{{p_{1}}\ {\overline ��}}{�� \ ��}}}, \:{{u^{3, \: 4}}= 0}, \:{{u^{4, \: 1}}= 0}, \:{{u^{4, \: 2}}= 0}, \:{{u^{4, \: 3}}= 0}, \: \right. \ \ \displaystyle \left.{{u^{4, \: 4}}={p_{1}}}\right] \end{array} \end{equation*}y)+->x=y, concat map(variables, ravel U), entries Σ(sb('p, [i])*Ñ.i, i, 1..#Ñ) )
This defines a family of pre-Frobenius algebras:
zero? eval(ω,\begin{equation} \label{eq34} \mbox{\rm true} \end{equation}ℰ)
In general the pairing is not symmetric!
Ų:𝐋 := eval(U,\begin{equation*} \label{eq35}-{{\frac{p_{1}}{�� \ ��}}\ {|^{\ 1 \ 1}}}+{{\frac{{p_{1}}\ {\overline ��}}{�� \ ��}}\ {|^{\ i \ i}}}+{{\frac{{p_{1}}\ {\overline ��}}{�� \ ��}}\ {|^{\ j \ j}}}+{{p_{1}}\ {|^{\ k \ k}}}\end{equation*}ℰ)
matrix Ξ(Ξ((𝐞.i 𝐞.j)/Ų,\begin{equation*} \label{eq36}\left[ \begin{array}{cccc} -{\frac{p_{1}}{�� \ ��}}& 0 & 0 & 0 \ 0 &{\frac{{p_{1}}\ {\overline ��}}{�� \ ��}}& 0 & 0 \ 0 & 0 &{\frac{{p_{1}}\ {\overline ��}}{�� \ ��}}& 0 \ 0 & 0 & 0 &{p_{1}} \end{array} \right]\end{equation*}i, 1..dim), j, 1..dim)
This is the most general form of the "dot product" of two quaternions
(a*b)/Ų\begin{equation*} \label{eq37}\frac{{{a_{2}}\ {b_{2}}\ {p_{1}}\ {\overline ��}}+{{a_{3}}\ {b_{3}}\ {p_{1}}\ {\overline ��}}+{{a_{4}}\ {b_{4}}\ {p_{1}}\ �� \ ��}-{{a_{1}}\ {b_{1}}\ {p_{1}}}}{�� \ ��}\end{equation*}
(a*a)/Ų\begin{equation*} \label{eq38}\frac{{{{a_{2}}^{2}}\ {p_{1}}\ {\overline ��}}+{{{a_{3}}^{2}}\ {p_{1}}\ {\overline ��}}+{{{a_{4}}^{2}}\ {p_{1}}\ �� \ ��}-{{{a_{1}}^{2}}\ {p_{1}}}}{�� \ ��}\end{equation*}
Cartan-Killing is a special case
ck:=solve(equate(Ũ=Ų),Ξ(sb('p, [i]), i, 1..#Ñ)).1
Compiling function equate with type Equation(ClosedLinearOperator(
OrderedVariableList([1, i, j, k]), Expression(Integer))) -> List(
Equation(Expression(Integer)))
>> Error detected within library code:
index out of rangeThe scalar product must be non-degenerate:
Ů:=determinant Ξ(Ξ(retract((𝐞.i * 𝐞.j)/Ų),\begin{equation*} \label{eq39}-{\frac{{{p_{1}}^{4}}\ {\overline ��}\ {\overline ��}}{{{��}^{3}}\ {{��}^{3}}}}\end{equation*}j, 1..dim), i, 1..dim)
factor(numer Ů)/factor(denom Ů)\begin{equation*} \label{eq40}-{\frac{{{p_{1}}^{4}}\ {\overline ��}\ {\overline ��}}{{{��}^{3}}\ {{��}^{3}}}}\end{equation*}
Co-pairing
Solve the [Snake Relation]? as a system of linear equations.
Ω:𝐋:=Σ(Σ(script('u, [[i, j]])*𝐞.i*𝐞.j, i, 1..dim), j, 1..dim)
\begin{equation}
\label{eq41}\begin{array}{@{}l}
\displaystyle
{{u_{1, \: 1}}\ {|_{\ 1 \ 1}}}+{{u_{1, \: 2}}\ {|_{\ 1 \ i}}}+{{u_{1, \: 3}}\ {|_{\ 1 \ j}}}+{{u_{1, \: 4}}\ {|_{\ 1 \ k}}}+
\
\
\displaystyle
{{u_{2, \: 1}}\ {|_{\ i \ 1}}}+{{u_{2, \: 2}}\ {|_{\ i \ i}}}+{{u_{2, \: 3}}\ {|_{\ i \ j}}}+{{u_{2, \: 4}}\ {|_{\ i \ k}}}+{{u_{3, \: 1}}\ {|_{\ j \ 1}}}+
\
\
\displaystyle
{{u_{3, \: 2}}\ {|_{\ j \ i}}}+{{u_{3, \: 3}}\ {|_{\ j \ j}}}+{{u_{3, \: 4}}\ {|_{\ j \ k}}}+{{u_{4, \: 1}}\ {|_{\ k \ 1}}}+
\
\
\displaystyle
{{u_{4, \: 2}}\ {|_{\ k \ i}}}+{{u_{4, \: 3}}\ {|_{\ k \ j}}}+{{u_{4, \: 4}}\ {|_{\ k \ k}}}
\end{array}
\end{equation}
d1:=(I*Ω)/(Ų*I);
d2:=(Ω*I)/(I*Ų);
eq1:=equate(d1=I);
eq2:=equate(d2=I);
snake:=solve(concat(eq1,eq2), concat Ξ(Ξ(script('u, [[i, j]]), i, 1..dim), j, 1..dim));
if #snake ~= 1 then error "no solution"
Ω:=eval(Ω,\begin{equation*} \label{eq42}-{{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ 1}}}+{{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ i \ i}}}+{{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ j \ j}}}+{{\frac{1}{p_{1}}}\ {|_{\ k \ k}}}\end{equation*}snake(1))
matrix Ξ(Ξ(Ω/(𝐝.i*𝐝.j),\begin{equation*} \label{eq43}\left[ \begin{array}{cccc} -{\frac{�� \ ��}{p_{1}}}& 0 & 0 & 0 \ 0 &{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}& 0 & 0 \ 0 & 0 &{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}& 0 \ 0 & 0 & 0 &{\frac{1}{p_{1}}} \end{array} \right]\end{equation*}i, 1..dim), j, 1..dim)
Check "dimension" and the snake relations.
d:𝐋:=
Ω /
X /
Ų
\begin{equation}
\label{eq44}4\end{equation}
test
( I Ω ) /
( Ų I ) = I
\begin{equation}
\label{eq45} \mbox{\rm true} \end{equation}
test
( Ω I ) /
( I Ų ) = I
\begin{equation}
\label{eq46} \mbox{\rm true} \end{equation}
Co-algebra
Compute the "three-point" function and use it to define co-multiplication.
W:=(Y,\begin{equation*} \label{eq47}\begin{array}{@{}l} \displaystyle -{{\frac{p_{1}}{�� \ ��}}\ {|^{\ 1 \ 1 \ 1}}}+{{\frac{p_{1}}{��}}\ {|^{\ 1 \ i \ i}}}+{{\frac{p_{1}}{��}}\ {|^{\ 1 \ j \ j}}}+ \ \ \displaystyle {{p_{1}}\ {|^{\ 1 \ k \ k}}}+{{\frac{p_{1}}{��}}\ {|^{\ i \ 1 \ i}}}+{{\frac{p_{1}}{��}}\ {|^{\ i \ i \ 1}}}+{{p_{1}}\ {|^{\ i \ j \ k}}}- \ \ \displaystyle {{p_{1}}\ {|^{\ i \ k \ j}}}+{{\frac{p_{1}}{��}}\ {|^{\ j \ 1 \ j}}}-{{p_{1}}\ {|^{\ j \ i \ k}}}+{{\frac{p_{1}}{��}}\ {|^{\ j \ j \ 1}}}+ \ \ \displaystyle {{p_{1}}\ {|^{\ j \ k \ i}}}+{{p_{1}}\ {|^{\ k \ 1 \ k}}}+{{p_{1}}\ {|^{\ k \ i \ j}}}-{{p_{1}}\ {|^{\ k \ j \ i}}}+{{p_{1}}\ {|^{\ k \ k \ 1}}} \end{array} \end{equation*}I)/Ų
λ:=(Ω,\begin{equation*} \label{eq48}\begin{array}{@{}l} \displaystyle -{{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ 1}^{\ 1}}}+{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ i}^{\ 1}}}+{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ j}^{\ 1}}}+ \ \ \displaystyle {{\frac{1}{p_{1}}}\ {|_{\ k \ k}^{\ 1}}}-{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {\overline ��}}}\ {|_{\ 1 \ i}^{\ i}}}-{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {\overline ��}}}\ {|_{\ i \ 1}^{\ i}}}- \ \ \displaystyle {{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ j \ k}^{\ i}}}+{{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ k \ j}^{\ i}}}-{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ 1 \ j}^{\ j}}}+ \ \ \displaystyle {{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ i \ k}^{\ j}}}-{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ j \ 1}^{\ j}}}-{{\frac{�� \ ��}{{p_{1}}\ {\overline ��}}}\ {|_{\ k \ i}^{\ j}}}- \ \ \displaystyle {{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ k}^{\ k}}}-{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ j}^{\ k}}}+{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ i}^{\ k}}}- \ \ \displaystyle {{\frac{�� \ ��}{p_{1}}}\ {|_{\ k \ 1}^{\ k}}} \end{array} \end{equation*}I, Ω)/(I, W, I)
test
( I Ω ) /
( Y I ) = λ
\begin{equation}
\label{eq49} \mbox{\rm true} \end{equation}
test
( Ω I ) /
( I Y ) = λ
\begin{equation}
\label{eq50} \mbox{\rm true} \end{equation}
Co-associativity
test( ( λ ) / _ ( I λ ) = _ ( λ ) / _ ( λ I ) )\begin{equation} \label{eq51} \mbox{\rm false} \end{equation}
Frobenius Condition (fork)
H :=
Y /
λ
\begin{equation*}
\label{eq52}\begin{array}{@{}l}
\displaystyle
-{{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ 1}^{\ 1 \ 1}}}+{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ i}^{\ 1 \ 1}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ j}^{\ 1 \ 1}}}+{{\frac{1}{p_{1}}}\ {|_{\ k \ k}^{\ 1 \ 1}}}-
\
\
\displaystyle
{{\frac{�� \ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ i}^{\ 1 \ i}}}-{{\frac{�� \ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ 1}^{\ 1 \ i}}}-
\
\
\displaystyle
{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ k}^{\ 1 \ i}}}+{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ j}^{\ 1 \ i}}}-
\
\
\displaystyle
{{\frac{{{��}^{3}}\ ��}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ j}^{\ 1 \ j}}}+{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ k}^{\ 1 \ j}}}-
\
\
\displaystyle
{{\frac{{{��}^{3}}\ ��}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ 1}^{\ 1 \ j}}}-{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ i}^{\ 1 \ j}}}-
\
\
\displaystyle
{{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ k}^{\ 1 \ k}}}-{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ j}^{\ 1 \ k}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ i}^{\ 1 \ k}}}-{{\frac{�� \ ��}{p_{1}}}\ {|_{\ k \ 1}^{\ 1 \ k}}}-
\
\
\displaystyle
{{\frac{�� \ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ i}^{\ i \ 1}}}-{{\frac{�� \ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ 1}^{\ i \ 1}}}-
\
\
\displaystyle
{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ k}^{\ i \ 1}}}+{{\frac{�� \ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ j}^{\ i \ 1}}}+
\
\
\displaystyle
{{\frac{�� \ {{��}^{2}}}{p_{1}}}\ {|_{\ 1 \ 1}^{\ i \ i}}}-{{\frac{�� \ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ i}^{\ i \ i}}}-
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ j}^{\ i \ i}}}-{{\frac{��}{p_{1}}}\ {|_{\ k \ k}^{\ i \ i}}}-{{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ k}^{\ i \ j}}}-
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ j}^{\ i \ j}}}+{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ i}^{\ i \ j}}}-
\
\
\displaystyle
{{\frac{�� \ ��}{p_{1}}}\ {|_{\ k \ 1}^{\ i \ j}}}+{{\frac{{{��}^{3}}\ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ j}^{\ i \ k}}}-
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ k}^{\ i \ k}}}+{{\frac{{{��}^{3}}\ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ 1}^{\ i \ k}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ i}^{\ i \ k}}}-{{\frac{{{��}^{3}}\ ��}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ j}^{\ j \ 1}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ k}^{\ j \ 1}}}-{{\frac{{{��}^{3}}\ ��}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ 1}^{\ j \ 1}}}-
\
\
\displaystyle
{{\frac{{{��}^{2}}\ ��}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ i}^{\ j \ 1}}}+{{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ k}^{\ j \ i}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ j}^{\ j \ i}}}-{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ i}^{\ j \ i}}}+
\
\
\displaystyle
{{\frac{�� \ ��}{p_{1}}}\ {|_{\ k \ 1}^{\ j \ i}}}+{{\frac{{{��}^{2}}\ ��}{p_{1}}}\ {|_{\ 1 \ 1}^{\ j \ j}}}-{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ i}^{\ j \ j}}}-
\
\
\displaystyle
{{\frac{{{��}^{3}}\ ��}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ j}^{\ j \ j}}}-{{\frac{��}{p_{1}}}\ {|_{\ k \ k}^{\ j \ j}}}-
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ i}^{\ j \ k}}}-{{\frac{{{��}^{2}}\ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ 1}^{\ j \ k}}}- \
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ k}^{\ j \ k}}}+{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ j}^{\ j \ k}}}-
\
\
\displaystyle
{{\frac{�� \ ��}{p_{1}}}\ {|_{\ 1 \ k}^{\ k \ 1}}}-{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ j}^{\ k \ 1}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ i}^{\ k \ 1}}}-{{\frac{�� \ ��}{p_{1}}}\ {|_{\ k \ 1}^{\ k \ 1}}}-
\
\
\displaystyle
{{\frac{{{��}^{3}}\ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ j}^{\ k \ i}}}+{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ i \ k}^{\ k \ i}}}-
\
\
\displaystyle
{{\frac{{{��}^{3}}\ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ 1}^{\ k \ i}}}-{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ i}^{\ k \ i}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ 1 \ i}^{\ k \ j}}}+{{\frac{{{��}^{2}}\ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ 1}^{\ k \ j}}}+ \
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ j \ k}^{\ k \ j}}}-{{\frac{{{��}^{2}}\ {{��}^{2}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k \ j}^{\ k \ j}}}+
\
\
\displaystyle
{{\frac{{{��}^{2}}\ {{��}^{2}}}{p_{1}}}\ {|_{\ 1 \ 1}^{\ k \ k}}}-{{\frac{{{��}^{2}}\ {{��}^{3}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ i \ i}^{\ k \ k}}}-
\
\
\displaystyle
{{\frac{{{��}^{3}}\ {{��}^{2}}}{{p_{1}}\ {{\overline ��}^{2}}}}\ {|_{\ j \ j}^{\ k \ k}}}-{{\frac{�� \ ��}{p_{1}}}\ {|_{\ k \ k}^{\ k \ k}}}
\end{array}
\end{equation*}
test
( λ I ) /
( I Y ) = H
\begin{equation}
\label{eq53} \mbox{\rm false} \end{equation}
test
( I λ ) /
( Y I ) = H
\begin{equation}
\label{eq54} \mbox{\rm false} \end{equation}
The Cartan-Killing form makes H of the Frobenius condition idempotent
test( eval(H,ck)=eval(H/H, ck) )
There are 10 exposed and 6 unexposed library operations named eval having 2 argument(s) but none was determined to be applicable. Use HyperDoc Browse,or issue )display op eval to learn more about the available operations. Perhaps package-calling the operation or using coercions on the arguments will allow you to apply the operation.
Cannot find a definition or applicable library operation named eval with argument type(s) ClosedLinearOperator(OrderedVariableList([1,i, j, k]), Expression(Integer)) Variable(ck)
Perhaps you should use "@" to indicate the required return type,or "$" to specify which version of the function you need.
But it is not unique. E.g. other idempots
h1:=map(numer,ravel(H-H/H)::List FRAC POLY INT);
Cannot convert the value from type List(Expression(Integer)) to List (Fraction(Polynomial(Integer))) .
Handle
Φ:𝐋 :=
λ /
Y
\begin{equation*}
\label{eq55}\begin{array}{@{}l}
\displaystyle
{{\frac{{{\left(-{2 \ �� \ �� \ {{\overline ��}^{2}}}-{{{��}^{3}}\ ��}\right)}\ {{\overline ��}^{2}}}-{�� \ {{��}^{3}}\ {{\overline ��}^{2}}}}{{p_{1}}\ {{\overline ��}^{2}}\ {{\overline ��}^{2}}}}\ {|_{\ 1}^{\ 1}}}+
\
\
\displaystyle
{{\frac{-{2 \ {{��}^{2}}\ {{��}^{2}}\ {\overline ��}}-{2 \ �� \ {{��}^{3}}\ {\overline ��}}}{{p_{1}}\ {\overline ��}\ {{\overline ��}^{2}}}}\ {|_{\ i}^{\ i}}}+
\
\
\displaystyle
{{\frac{-{2 \ {{��}^{3}}\ �� \ {\overline ��}}-{2 \ {{��}^{2}}\ {{��}^{2}}\ {\overline ��}}}{{p_{1}}\ {{\overline ��}^{2}}\ {\overline ��}}}\ {|_{\ j}^{\ j}}}+
\
\
\displaystyle
{{\frac{-{2 \ �� \ �� \ {\overline ��}\ {\overline ��}}-{2 \ {{��}^{2}}\ {{��}^{2}}}}{{p_{1}}\ {\overline ��}\ {\overline ��}}}\ {|_{\ k}^{\ k}}}
\end{array}
\end{equation*}
The Cartan-Killing form makes Φ of the identity
test( eval(Φ,ck)=I )
There are 10 exposed and 6 unexposed library operations named eval having 2 argument(s) but none was determined to be applicable. Use HyperDoc Browse,or issue )display op eval to learn more about the available operations. Perhaps package-calling the operation or using coercions on the arguments will allow you to apply the operation.
Cannot find a definition or applicable library operation named eval with argument type(s) ClosedLinearOperator(OrderedVariableList([1,i, j, k]), Expression(Integer)) Variable(ck)
Perhaps you should use "@" to indicate the required return type,or "$" to specify which version of the function you need.
but it can be the identity in many ways. For example,
solve(equate(eval(Φ,\begin{equation*} \label{eq56}\left[ \right]?\end{equation*}[p[1]=1, p[2]=1, p[3]=1, σ=1, μ=1])=I), [p[4]])
If handle is identity then fork is idempotent but the converse is not true
Φ1:=map(numer,ravel(Φ-I)::List FRAC POLY INT);
Cannot convert the value from type List(Expression(Integer)) to List (Fraction(Polynomial(Integer))) .
Bi-algebra conditions
ΦΦ:= _ ( λ λ ) / _ ( I X I ) / _ ( Y Y ) ;
bi1:=map(numer,ravel(ΦΦ-H)::List FRAC POLY INT);
Cannot convert the value from type List(Expression(Integer)) to List (Fraction(Polynomial(Integer))) .
i = Unit of the algebra
i:=𝐞.1\begin{equation} \label{eq57}|_{\ 1}\end{equation}
test
i /
λ = Ω
\begin{equation}
\label{eq58} \mbox{\rm false} \end{equation}
Co-unit
i U
ι:𝐋:=
( i I ) /
Ų
\begin{equation*}
\label{eq59}-{{\frac{p_{1}}{�� \ ��}}\ {|^{\ 1}}}\end{equation*}
Y=U ι
test
Y /
ι = Ų
\begin{equation}
\label{eq60} \mbox{\rm false} \end{equation}
For example:
ex1:=[σ=1,\begin{equation*} \label{eq61}\left[{�� = 1}, \:{�� = 1}, \:{{p_{1}}= 1}, \:{{p_{2}}= 1}, \:{{p_{3}}= 1}, \:{{p_{4}}= 1}\right]?\end{equation*}μ=1, p[1]=1, p[2]=1, p[3]=1, p[4]=1]
Ų0:𝐋 :=eval(Ų,\begin{equation} \label{eq62}-{|^{\ 1 \ 1}}+{|^{\ i \ i}}+{|^{\ j \ j}}+{|^{\ k \ k}}\end{equation}ex1)
Ω0:𝐋 :=eval(Ω,\begin{equation} \label{eq63}-{|_{\ 1 \ 1}}+{|_{\ i \ i}}+{|_{\ j \ j}}+{|_{\ k \ k}}\end{equation}ex1)$𝐋
λ0:𝐋 :=eval(λ,\begin{equation} \label{eq64}\begin{array}{@{}l} \displaystyle -{|_{\ 1 \ 1}^{\ 1}}+{|_{\ i \ i}^{\ 1}}+{|_{\ j \ j}^{\ 1}}+{|_{\ k \ k}^{\ 1}}-{|_{\ 1 \ i}^{\ i}}-{|_{\ i \ 1}^{\ i}}- \ \ \displaystyle {|_{\ j \ k}^{\ i}}+{|_{\ k \ j}^{\ i}}-{|_{\ 1 \ j}^{\ j}}+{|_{\ i \ k}^{\ j}}-{|_{\ j \ 1}^{\ j}}-{|_{\ k \ i}^{\ j}}-{|_{\ 1 \ k}^{\ k}}- \ \ \displaystyle {|_{\ i \ j}^{\ k}}+{|_{\ j \ i}^{\ k}}-{|_{\ k \ 1}^{\ k}} \end{array} \end{equation}ex1)$𝐋
Φ0:𝐋 :=eval(Φ,\begin{equation} \label{eq65}-{4 \ {|_{\ 1}^{\ 1}}}-{4 \ {|_{\ i}^{\ i}}}-{4 \ {|_{\ j}^{\ j}}}-{4 \ {|_{\ k}^{\ k}}}\end{equation}ex1)$𝐋
! Argument of �has an extra }.
<inserted text>
\par
l.179 ...t[{\left[ 0, \:{\frac{��}{\overline ��}
}, \: 0, \: 0 \right]}, \:...
Runaway argument?
Paragraph ended before �was complete.
<to be read again>
\par
l.179 ...t[{\left[ 0, \:{\frac{��}{\overline ��}
}, \: 0, \: 0 \right]}, \:...
Package utf8x Error: Character131appearedalone.
See the utf8x package documentation for explanation.
Type H <return> for immediate help.
...
l.179 ...t[{\left[ 0, \:{\frac{��}{\overline ��}
}, \: 0, \: 0 \right]}, \:...
Argument of �has an extra }.
<inserted text>
\par
l.179 ...ht]}, \:{\left[{\frac{��}{\overline ��}
}, \: 0, \: 0, \: 0 \right...
Runaway argument?
Paragraph ended before �was complete.
<to be read again>
\par
l.179 ...ht]}, \:{\left[{\frac{��}{\overline ��}
}, \: 0, \: 0, \: 0 \right...
Package utf8x Error: Character131appearedalone.
See the utf8x package documentation for explanation.
Type H <return> for immediate help.
...
l.179 ...ht]}, \:{\left[{\frac{��}{\overline ��}
}, \: 0, \: 0, \: 0 \right...
Argument of �has an extra }.
<inserted text>
\par
l.179 ... 0, \: -{\frac{�� \ ��}{\overline ��}
}\right]}, \:{\left[ 0, \:...
Runaway argument?
Paragraph ended before �was complete.
<to be read again>
\par
l.179 ... 0, \: -{\frac{�� \ ��}{\overline ��}
}\right]}, \:{\left[ 0, \:...
Package utf8x Error: Character131appearedalone.
See the utf8x package documentation for explanation.
Type H <return> for immediate help.
...
l.179 ... 0, \: -{\frac{�� \ ��}{\overline ��}
}\right]}, \:{\left[ 0, \:...
Argument of �has an extra }.
<inserted text>
\par
l.179 ...\: 0, \:{\frac{�� \ ��}{\overline ��}
}, \: 0 \right]}\right]
Runaway argument?