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Edit detail for Frobenius Algebra, Vector Spaces and Polynomial Ideals revision 1 of 4

1 2 3 4
Editor: Bill Page
Time: 2011/04/06 21:14:06 GMT-7
Note: turtles are symmetric

changed:
-
References

- "Frobenius algebras and 2D topological quantum field theories":http://mat.uab.es/~kock/TQFT.html
  by Joachim Kock

  Especially "Tensor calculus (linear algebra in coordinates)"
  in section 2.3.31, page 123.

See also:

- http://ncatlab.org/nlab/show/Frobenius+algebra
  by John Baez, et al.

- "Ideals, Varieties, and Algorithms":http://www.cs.amherst.edu/~dac/iva.html#MR5
  by David A. Cox, et al.

An n-dimensional algebra is represented by a (2,1)-tensor
$Y=\{ {y^k}_{ij} \ i,j,k =1,2, ... n \}$
viewed as a linear operator with two inputs $i,j$ and one
output $k$. For example in 2 dimensions
\begin{axiom}
n:=2
T:=CartesianTensor(1,n,FRAC POLY INT)
Y:T := unravel(concat concat
  [[[script(y,[[i,j],[k]])
    for i in 1..n]
      for j in 1..n]
        for k in 1..n]
          )
\end{axiom}
Given two vectors $P=\{ p^i \}$ and $Q=\{ q^j \}$
\begin{axiom}
P:T := unravel([script(p,[[],[i]]) for i in 1..n])
Q:T := unravel([script(q,[[],[i]]) for i in 1..n])
\end{axiom}
the tensor $Y$ operates on their tensor product to
yield a vector $R=\{ r_k = {y^k}_{ij} p^i q^j \}$
\begin{axiom}
R:=contract(contract(Y,3,product(P,Q),1),2,3)
\end{axiom}
Pictorially::

  P Q
   Y
   R

  or more explicitly

  Pi Qj
   \/
    \
     Rk

In Axiom we may use the more convenient tensor inner
product denoted by '*' that combines tensor product with
a contraction on the last index of the first tensor and
the first index of the second tensor.
\begin{axiom}
R:=(Y*P)*Q
\end{axiom}
An algebra is said to be *associative* if::

  Y    =    Y
   Y       Y

**Note:** the right hand side of the equation above is
implicitly the mirror image of the left hand side::

  i   j   k   i  j     k   i     j  k
   \  |  /     \/     /     \     \/
    \ | /       \    /       \    /
     \|/    =    e  k    -    i  e
      |           \/           \/
      |            \           /
      l             l         l

This requires that the following (3,1)-tensor
\begin{equation}
\Psi  = \{ {\psi_l}^{ijk} =  {y^e}_{ij} {y^l}_{ek} - {y^l}_{ie} {y^e}_{jk} \}
\end{equation}
(associator) is zero.
\begin{axiom}
YY := reindex(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),[1,4,3,2])-Y*Y; ravel(YY)
\end{axiom}

The algebra $Y$ is *commutative* if::

  Y = Y

  i   j     i  j     j  i
   \ /   =   \/   -   \/
    |         \       /
    k          k     k

This requires that the following (2,1)-tensor
\begin{equation}
\mathcal{C} = \{ {c^k}_{ij} = {y^k}_{ij} - {y^k}_{ji} \}
\end{equation}
(commutator) is zero.
\begin{axiom}
YC:=Y-reindex(Y,[1,3,2])
\end{axiom}
A basis for the ideal defined by the coefficients of the
commutator is given by:
\begin{axiom}
groebner(ravel(YC))
\end{axiom}

The algebra $Y$ is *anti-commutative* if::

  Y = -Y

  i   j     i  j     j  i
   \ /   =   \/   =   \/
    |         \       /
    k          k     k

This requires that the following (2,1)-tensor
\begin{equation}
\mathcal{A} = \{ {a^k}_{ij} = {y^k}_{ij} + {y^k}_{ji} \}
\end{equation}
(anti-commutator) is zero.
\begin{axiom}
YA:=Y+reindex(Y,[1,3,2])
\end{axiom}
A basis for the ideal defined by the coefficients of the
commutator is given by:
\begin{axiom}
groebner(ravel(YA))
\end{axiom}

The *Jacobi identity* is::

            X
  Y =  Y + Y
   Y  Y     Y

  i     j     k  i      j     k  i     j      k   i  j   k
   \    |    /    \    /     /    \     \    /     \  \ /
    \   |   /      \  /     /      \     \  /       \  0
     \  |  /        \/     /        \     \/         \/ \
      \ | /          \    /          \    /           \  \
       \|/     =      e  k      -     i  e       -     e  j
        |              \/              \/               \/
        |               \              /                /
        l                l            l                 l

An algebra satisfies the Jacobi identity if and only if
the following (3,1)-tensor
\begin{equation}
\Theta = \{ {\theta^l}_{ijk} =  {y^l}_{ek} {y^e}_{ij} - {y^l}_{ie} {y^e}_{jk} - {y^l}_{ej} {y^e}_{ik} \}
\end{equation}
is zero.

\begin{axiom}
YX := YY - reindex(contract(Y,1,Y,2),[3,1,4,2]); ravel(YX)
\end{axiom}
A scalar product is denoted by the (2,0)-tensor
$U = \{ u_{ij} \}$
\begin{axiom}
U:T := unravel(concat
  [[script(u,[[],[j,i]])
    for i in 1..n]
      for j in 1..n]
        )
\end{axiom}
Definition 1

  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::

    i  j  k   i  j  k   i  j  k
     \ | /     \/  /     \  \/
      \|/   =   \ /   -   \ /
       0         0         0

  \begin{equation}
  \Phi = \{ \phi^{ijk} = {y^e}_{ij} u_{ek} - u_{ie} {y_e}^{jk} \}
  \end{equation}
  (three-point function) is zero.

\begin{axiom}
YU := reindex(reindex(U,[2,1])*reindex(Y,[1,3,2]),[3,2,1])-U*Y
\end{axiom}
Definition 2

  An algebra with a non-degenerate associative scalar product
  is called *pre-Frobenius*.

We may consider the problem where multiplication Y is given,
and look for all associative scalar products $U = U(Y)$ or we
may consider an scalar product U as given, and look for all
algebras $Y=Y(U)$ such that the scalar product is associative. 

This problem can be solved using linear algebra.
\begin{axiom}
)expose MCALCFN
K := jacobian(ravel(YU),concat(map(variables,ravel(Y)))::List Symbol);
yy := transpose matrix [concat(map(variables,ravel(Y)))::List Symbol];
K::OutputForm * yy::OutputForm = 0
\end{axiom}
The matrix 'K' transforms the coefficients of the tensor $Y$
into coefficients of the tensor $\Phi$. We are looking for
coefficients of the tensor $U$ such that 'K' transforms the
tensor $Y$ into $\Phi=0$ for any $Y$.

A necessary condition for the equation to have a non-trivial
solution is that the matrix 'K' be degenerate.

Theorem 1

  All 2-dimensional pre-Frobenius algebras are symmetric.

Proof: Consider the determinant of the matrix 'K' above.
\begin{axiom}
Kd := factor(determinant(K)::DMP(concat map(variables,ravel(U)),FRAC INT))
\end{axiom}
The scalar product must also be non-degenerate
\begin{axiom}
Ud:DMP(concat map(variables,ravel(U)),FRAC INT) := determinant [[U[i,j] for j in 1..n] for i in 1..n]
\end{axiom}
therefore U must be symmetric.
\begin{axiom}
nthFactor(Kd,1)
US:T := unravel(map(x+->subst(x,U[2,1]=U[1,2]),ravel U))
\end{axiom}

Theorem 2

  All 2-dimensional algebras with associative scalar product
  are commutative.

Proof: The basis of the null space of the symmetric 
'K' matrix are all symmetric

\begin{axiom}
YUS:T :=  reindex(reindex(US,[2,1])*reindex(Y,[1,3,2]),[3,2,1])-US*Y
KS := jacobian(ravel(YUS),concat(map(variables,ravel(Y)))::List Symbol);
NS:=nullSpace(KS)
SS:=map((x,y)+->x=y,concat map(variables,ravel Y),
  entries reduce(+,[p[i]*NS.i for i in 1..#NS]))
YS:T := unravel(map(x+->subst(x,SS),ravel Y))
\end{axiom}
This defines a 4-parameter family of 2-d pre-Frobenius algebras
\begin{axiom}
test(unravel(map(x+->subst(x,SS),ravel YUS))$T=0*YU)
\end{axiom}

Alternatively we may consider
\begin{axiom}
J := jacobian(ravel(YU),concat(map(variables,ravel(U)))::List Symbol);
uu := transpose matrix [concat(map(variables,ravel(U)))::List Symbol];
J::OutputForm * uu::OutputForm = 0
\end{axiom}
The matrix 'J' transforms the coefficients of the tensor $U$
into coefficients of the tensor $\Phi$. We are looking for
coefficients of the tensor $Y$ such that 'J' transforms the
tensor $U$ into $\Phi=0$ for any $U$.

A necessary condition for the equation to have a non-trivial
solution is that all 70 of the 4x4 sub-matrices of 'J' are
degenerate. To this end we can form the polynomial ideal of
the determinants of these sub-matrices.
\begin{axiom}
JP:=ideal concat concat concat
  [[[[ determinant(
    matrix([row(J,i1),row(J,i2),row(J,i3),row(J,i4)]))
      for i4 in (i3+1)..maxRowIndex(J) ] 
        for i3 in (i2+1)..(maxRowIndex(J)-1) ]
          for i2 in (i1+1)..(maxRowIndex(J)-2) ]
            for i1 in minRowIndex(J)..(maxRowIndex(J)-3) ];
#generators(%)
\end{axiom}
Theorem 3

  If a 2-d algebra is associative, commutative, anti-commutative
  or if it satisfies the Jacobi identity then it is a
  pre-Frobenius algebra.

Proof

Consider the ideals of the associator, commutator, anti-commutator
and Jacobi identity
\begin{axiom}
YYI:=ideal ravel YY;
in?(JP,YYI)  -- associative
YCI:=ideal ravel YC;
in?(JP,YCI)  -- commutative
YAI:=ideal ravel YA;
in?(JP,YAI)  -- anti-commutative
YXI:=ideal ravel YX;
in?(JP,YXI) -- Jacobi identity
\end{axiom}

Y-forms

Three traces of two graftings of an algebra gives six
(2,0)-forms.

Left snail and right snail::

  LS                    RS

  Y /\                    /\ Y
   Y  )                  (  Y
    \/                    \/

  i  j                        j  i
   \/                          \/
    \    /\              /\    /
     e  f  \            /  f  e
      \/    \          /    \/
       \    /          \    /
        f  /            \  f
         \/              \/

\begin{equation}
LS = \{ {y^e}_{ij} {y^f}_{ef} \} \
RS = \{ {y^f}_{fe} {y^e}_{ji} \}
\end{equation}

\begin{axiom}
LS:=contract(Y*Y,1,2)
RS:=reindex(contract(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),1,2),[2,1])
test(LS=RS)
\end{axiom}

Left and right deer::

   RD                 LD

   \ /\/              \/\ /
    Y /\              /\ Y
     Y  )            (  Y
      \/              \/

   i            j    i            j
    \    /\    /      \    /\    /
     \  f  \  /        \  /  f  /
      \/    \/          \/    \/
       \    /\          /\    /
        e  /  \        /  \  e
         \/    \      /    \/
          \    /      \    /
           f  /        \  f
            \/          \/

\begin{equation}
RD = \{ {y^e}_{if} {y^f}_{ej} \} \
LD = \{ {y^f}_{ie} {y^e}_{fj} \}
\end{equation}
Left and right deer forms are identical but different from snails.
\begin{axiom}
RD:=contract(Y*Y,1,3)
LD:=reindex(contract(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),1,3),[2,1])
test(LD=RD)
test(RD=RS)
test(RD=LS)
\end{axiom}

Left and right turtles::

  RT                   LT

   /\ / /               \ \ /\
  (  Y /                 \ Y  )
   \  Y                   Y  /
    \/                     \/

           i     j      i     j
    /\    /     /        \     \    /\
   /  f  /     /          \     \  f  \
  /    \/     /            \     \/    \
  \     \    /              \    /     /
   \     e  /                \  e     /
    \     \/                  \/     /
     \    /                    \    /
      \  f                      f  /
       \/                        \/

\begin{equation}
RT = \{ {y^e}_{fi} {y^f}_{ej} \} \
LT = \{ {y^f}_{ie} {y^e}_{jf} \}
\end{equation}

\begin{axiom}
RT:=contract(Y*Y,1,4)
LT:=reindex(contract(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),1,4),[2,1])
test(LT=RT)
\end{axiom}
The turles are symmetric
\begin{axiom}
test(RT=reindex(RT,[2,1]))
test(LT=reindex(LT,[2,1]))
\end{axiom}
Five of the six forms are independent.
\begin{axiom}
test(RT=RS)
test(RT=LS)
test(RT=RD)
test(LT=RS)
test(LT=LS)
test(LT=RD)
\end{axiom}
Associativity implies right turtle equals right snail
and left turtle equals left snail.
\begin{axiom}
in?(ideal ravel(RT-RS),YYI)
in?(ideal ravel(LT-LS),YYI)
\end{axiom}
If the Jacobi identity holds then both snails are zero
\begin{axiom}
in?(ideal ravel(RS),YXI)
in?(ideal ravel(LS),YXI)
\end{axiom}
and right turtle and deer have opposite signs
\begin{axiom}
in?(ideal ravel(RT+RD),YXI)
\end{axiom}

References

See also:

An n-dimensional algebra is represented by a (2,1)-tensor Y=\{ {y^k}_{ij} \ i,j,k =1,2, ... n \} viewed as a linear operator with two inputs i,j and one output k. For example in 2 dimensions

axiom
n:=2

\label{eq1}2(1)
Type: PositiveInteger?
axiom
T:=CartesianTensor(1,n,FRAC POLY INT)

\label{eq2}\hbox{\axiomType{CartesianTensor}\ } (1, 2, \hbox{\axiomType{Fraction}\ } (\hbox{\axiomType{Polynomial}\ } (\hbox{\axiomType{Integer}\ })))(2)
Type: Type
axiom
Y:T := unravel(concat concat
  [[[script(y,[[i,j],[k]])
    for i in 1..n]
      for j in 1..n]
        for k in 1..n]
          )

\label{eq3}\begin{array}{@{}l}
\displaystyle
\left[{\left[ 
\begin{array}{cc}
{y_{1, \: 1}^{1}}&{y_{2, \: 1}^{1}}
\
{y_{1, \: 2}^{1}}&{y_{2, \: 2}^{1}}
(3)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

Given two vectors P=\{ p^i \} and Q=\{ q^j \}

axiom
P:T := unravel([script(p,[[],[i]]) for i in 1..n])

\label{eq4}\left[{p^{1}}, \:{p^{2}}\right](4)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
Q:T := unravel([script(q,[[],[i]]) for i in 1..n])

\label{eq5}\left[{q^{1}}, \:{q^{2}}\right](5)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

the tensor Y operates on their tensor product to yield a vector R=\{ r_k = {y^k}_{ij} p^i q^j \}

axiom
R:=contract(contract(Y,3,product(P,Q),1),2,3)

\label{eq6}\begin{array}{@{}l}
\displaystyle
\left[{{{p^{2}}\ {q^{2}}\ {y_{2, \: 2}^{1}}}+{{p^{2}}\ {q^{1}}\ {y_{2, \: 1}^{1}}}+{{p^{1}}\ {q^{2}}\ {y_{1, \: 2}^{1}}}+{{p^{1}}\ {q^{1}}\ {y_{1, \: 1}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{{{p^{2}}\ {q^{2}}\ {y_{2, \: 2}^{2}}}+{{p^{2}}\ {q^{1}}\ {y_{2, \: 1}^{2}}}+{{p^{1}}\ {q^{2}}\ {y_{1, \: 2}^{2}}}+{{p^{1}}\ {q^{1}}\ {y_{1, \: 1}^{2}}}}\right] 
(6)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

Pictorially:

  P Q
   Y
   R

  or more explicitly

  Pi Qj
   \/
    \
     Rk

In Axiom we may use the more convenient tensor inner product denoted by * that combines tensor product with a contraction on the last index of the first tensor and the first index of the second tensor.

axiom
R:=(Y*P)*Q

\label{eq7}\begin{array}{@{}l}
\displaystyle
\left[{{{p^{2}}\ {q^{2}}\ {y_{2, \: 2}^{1}}}+{{p^{2}}\ {q^{1}}\ {y_{2, \: 1}^{1}}}+{{p^{1}}\ {q^{2}}\ {y_{1, \: 2}^{1}}}+{{p^{1}}\ {q^{1}}\ {y_{1, \: 1}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{{{p^{2}}\ {q^{2}}\ {y_{2, \: 2}^{2}}}+{{p^{2}}\ {q^{1}}\ {y_{2, \: 1}^{2}}}+{{p^{1}}\ {q^{2}}\ {y_{1, \: 2}^{2}}}+{{p^{1}}\ {q^{1}}\ {y_{1, \: 1}^{2}}}}\right] 
(7)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

An algebra is said to be associative if:

  Y    =    Y
   Y       Y

Note: the right hand side of the equation above is implicitly the mirror image of the left hand side:

  i   j   k   i  j     k   i     j  k
   \  |  /     \/     /     \     \/
    \ | /       \    /       \    /
     \|/    =    e  k    -    i  e
      |           \/           \/
      |            \           /
      l             l         l

This requires that the following (3,1)-tensor


\label{eq8}
\Psi  = \{ {\psi_l}^{ijk} =  {y^e}_{ij} {y^l}_{ek} - {y^l}_{ie} {y^e}_{jk} \}
(8)
(associator) is zero.
axiom
YY := reindex(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),[1,4,3,2])-Y*Y; ravel(YY)

\label{eq9}\begin{array}{@{}l}
\displaystyle
\left[{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{1}}}}, \:{{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}-{{y_{2, \: 1}^{1}}\ {y_{2, \: 1}^{2}}}}, \right.
\
\
\displaystyle
\left.\:{{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{2}}}+{{\left(-{y_{1, \: 2}^{2}}+{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 1}^{1}}}-{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{-{{y_{2, \: 1}^{1}}\ {y_{2, \: 2}^{2}}}+{{\left({y_{2, \: 1}^{2}}-{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 2}^{1}}}+{{y_{2, \: 1}^{1}}^2}}, \: \right.
\
\
\displaystyle
\left.{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{1, \: 2}^{2}}}}, \:{{\left(-{y_{2, \: 1}^{2}}+{y_{1, \: 2}^{2}}\right)}\ {y_{2, \: 2}^{1}}}, \: \right.
\
\
\displaystyle
\left.{{{y_{1, \: 2}^{1}}\ {y_{2, \: 2}^{2}}}+{{\left(-{y_{1, \: 2}^{2}}+{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 2}^{1}}}-{{y_{1, \: 2}^{1}}^2}}, \: \right.
\
\
\displaystyle
\left.{{\left({y_{2, \: 1}^{1}}-{y_{1, \: 2}^{1}}\right)}\ {y_{2, \: 2}^{1}}}, \:{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{2}}}}, \: \right.
\
\
\displaystyle
\left.{{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}-{{y_{2, \: 1}^{2}}^2}+{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{2}}}-{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}-{{y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{1}}}}, \:{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{2, \: 1}^{1}}\ {y_{2, \: 1}^{2}}}}, \right.
\
\
\displaystyle
\left.\:{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 2}^{2}}^2}-{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{{{\left(-{y_{2, \: 1}^{2}}+{y_{1, \: 2}^{2}}\right)}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{2}}}-{{y_{1, \: 2}^{2}}\ {y_{2, \: 1}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}-{{y_{1, \: 2}^{1}}\ {y_{1, \: 2}^{2}}}}, \:{{\left({y_{2, \: 1}^{2}}-{y_{1, \: 2}^{2}}\right)}\ {y_{2, \: 2}^{1}}}\right] 
(9)
Type: List(Fraction(Polynomial(Integer)))

The algebra Y is commutative if:

  Y = Y

  i   j     i  j     j  i
   \ /   =   \/   -   \/
    |         \       /
    k          k     k

This requires that the following (2,1)-tensor


\label{eq10}
\mathcal{C} = \{ {c^k}_{ij} = {y^k}_{ij} - {y^k}_{ji} \}
(10)
(commutator) is zero.
axiom
YC:=Y-reindex(Y,[1,3,2])

\label{eq11}\begin{array}{@{}l}
\displaystyle
\left[{\left[ 
\begin{array}{cc}
0 &{{y_{2, \: 1}^{1}}-{y_{1, \: 2}^{1}}}
\
{-{y_{2, \: 1}^{1}}+{y_{1, \: 2}^{1}}}& 0 
(11)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

A basis for the ideal defined by the coefficients of the commutator is given by:

axiom
groebner(ravel(YC))

\label{eq12}\left[{{y_{2, \: 1}^{2}}-{y_{1, \: 2}^{2}}}, \:{{y_{2, \: 1}^{1}}-{y_{1, \: 2}^{1}}}\right](12)
Type: List(Polynomial(Integer))

The algebra Y is anti-commutative if:

  Y = -Y

  i   j     i  j     j  i
   \ /   =   \/   =   \/
    |         \       /
    k          k     k

This requires that the following (2,1)-tensor


\label{eq13}
\mathcal{A} = \{ {a^k}_{ij} = {y^k}_{ij} + {y^k}_{ji} \}
(13)
(anti-commutator) is zero.
axiom
YA:=Y+reindex(Y,[1,3,2])

\label{eq14}\begin{array}{@{}l}
\displaystyle
\left[{\left[ 
\begin{array}{cc}
{2 \ {y_{1, \: 1}^{1}}}&{{y_{2, \: 1}^{1}}+{y_{1, \: 2}^{1}}}
\
{{y_{2, \: 1}^{1}}+{y_{1, \: 2}^{1}}}&{2 \ {y_{2, \: 2}^{1}}}
(14)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

A basis for the ideal defined by the coefficients of the commutator is given by:

axiom
groebner(ravel(YA))

\label{eq15}\left[{y_{2, \: 2}^{2}}, \:{y_{2, \: 2}^{1}}, \:{{y_{2, \: 1}^{2}}+{y_{1, \: 2}^{2}}}, \:{{y_{2, \: 1}^{1}}+{y_{1, \: 2}^{1}}}, \:{y_{1, \: 1}^{2}}, \:{y_{1, \: 1}^{1}}\right](15)
Type: List(Polynomial(Integer))

The Jacobi identity is:

            X
  Y =  Y + Y
   Y  Y     Y

  i     j     k  i      j     k  i     j      k   i  j   k
   \    |    /    \    /     /    \     \    /     \  \ /
    \   |   /      \  /     /      \     \  /       \  0
     \  |  /        \/     /        \     \/         \/ \
      \ | /          \    /          \    /           \  \
       \|/     =      e  k      -     i  e       -     e  j
        |              \/              \/               \/
        |               \              /                /
        l                l            l                 l

An algebra satisfies the Jacobi identity if and only if the following (3,1)-tensor


\label{eq16}
\Theta = \{ {\theta^l}_{ijk} =  {y^l}_{ek} {y^e}_{ij} - {y^l}_{ie} {y^e}_{jk} - {y^l}_{ej} {y^e}_{ik} \}
(16)
is zero.

axiom
YX := YY - reindex(contract(Y,1,Y,2),[3,1,4,2]); ravel(YX)

\label{eq17}\begin{array}{@{}l}
\displaystyle
\left[{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}-{{y_{1, \: 1}^{1}}^2}}, \: \right.
\
\
\displaystyle
\left.{{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{\left(-{y_{2, \: 1}^{1}}-{y_{1, \: 2}^{1}}\right)}\ {y_{2, \: 1}^{2}}}-{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{2}}}-{{y_{1, \: 2}^{2}}\ {y_{2, \: 1}^{1}}}-{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}, \right.
\
\
\displaystyle
\left.\:{-{{y_{2, \: 1}^{1}}\ {y_{2, \: 2}^{2}}}-{{y_{1, \: 1}^{1}}\ {y_{2, \: 2}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{-{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}-{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{-{{y_{1, \: 2}^{1}}\ {y_{2, \: 2}^{2}}}+{{\left(-{y_{2, \: 1}^{2}}+{y_{1, \: 2}^{2}}-{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 2}^{1}}}}, \: \right.
\
\
\displaystyle
\left.{
\begin{array}{@{}l}
\displaystyle
{{y_{1, \: 2}^{1}}\ {y_{2, \: 2}^{2}}}+{{\left(-{2 \ {y_{1, \: 2}^{2}}}+{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 2}^{1}}}-{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{1}}}- 
\
\
\displaystyle
{{y_{1, \: 2}^{1}}^2}
(17)
Type: List(Fraction(Polynomial(Integer)))

A scalar product is denoted by the (2,0)-tensor U = \{ u_{ij} \}

axiom
U:T := unravel(concat
  [[script(u,[[],[j,i]])
    for i in 1..n]
      for j in 1..n]
        )

\label{eq18}\left[ 
\begin{array}{cc}
{u^{1, \: 1}}&{u^{1, \: 2}}
\
{u^{2, \: 1}}&{u^{2, \: 2}}
(18)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

Definition 1

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:

    i  j  k   i  j  k   i  j  k
     \ | /     \/  /     \  \/
      \|/   =   \ /   -   \ /
       0         0         0


\label{eq19}
  \Phi = \{ \phi^{ijk} = {y^e}_{ij} u_{ek} - u_{ie} {y_e}^{jk} \}
  (19)
(three-point function) is zero.

axiom
YU := reindex(reindex(U,[2,1])*reindex(Y,[1,3,2]),[3,2,1])-U*Y

\label{eq20}\begin{array}{@{}l}
\displaystyle
\left[{\left[ 
\begin{array}{cc}
{{\left({u^{2, \: 1}}-{u^{1, \: 2}}\right)}\ {y_{1, \: 1}^{2}}}&{-{{u^{1, \: 2}}\ {y_{2, \: 1}^{2}}}-{{u^{1, \: 1}}\ {y_{2, \: 1}^{1}}}+{{u^{2, \: 2}}\ {y_{1, \: 1}^{2}}}+{{u^{1, \: 2}}\ {y_{1, \: 1}^{1}}}}
\
{{{u^{2, \: 1}}\ {y_{2, \: 1}^{2}}}+{{u^{1, \: 1}}\ {y_{2, \: 1}^{1}}}-{{u^{1, \: 2}}\ {y_{1, \: 2}^{2}}}-{{u^{1, \: 1}}\ {y_{1, \: 2}^{1}}}}&{-{{u^{1, \: 2}}\ {y_{2, \: 2}^{2}}}-{{u^{1, \: 1}}\ {y_{2, \: 2}^{1}}}+{{u^{2, \: 2}}\ {y_{2, \: 1}^{2}}}+{{u^{1, \: 2}}\ {y_{2, \: 1}^{1}}}}
(20)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

Definition 2

An algebra with a non-degenerate associative scalar product is called pre-Frobenius.

We may consider the problem where multiplication Y is given, and look for all associative scalar products U = U(Y) or we may consider an scalar product U as given, and look for all algebras Y=Y(U) such that the scalar product is associative.

This problem can be solved using linear algebra.

axiom
)expose MCALCFN
MultiVariableCalculusFunctions is now explicitly exposed in frame initial K := jacobian(ravel(YU),concat(map(variables,ravel(Y)))::List Symbol);
Type: Matrix(Fraction(Polynomial(Integer)))
axiom
yy := transpose matrix [concat(map(variables,ravel(Y)))::List Symbol];
Type: Matrix(Polynomial(Integer))
axiom
K::OutputForm * yy::OutputForm = 0

\label{eq21}\begin{array}{@{}l}
\displaystyle
{{\left[ 
\begin{array}{cccccccc}
0 & 0 & 0 & 0 &{{u^{2, \: 1}}-{u^{1, \: 2}}}& 0 & 0 & 0 
\
{u^{1, \: 2}}& -{u^{1, \: 1}}& 0 & 0 &{u^{2, \: 2}}& -{u^{1, \: 2}}& 0 & 0 
\
0 &{u^{1, \: 1}}& -{u^{1, \: 1}}& 0 & 0 &{u^{2, \: 1}}& -{u^{1, \: 2}}& 0 
\
0 &{u^{1, \: 2}}& 0 & -{u^{1, \: 1}}& 0 &{u^{2, \: 2}}& 0 & -{u^{1, \: 2}}
\
-{u^{2, \: 1}}& 0 &{u^{1, \: 1}}& 0 & -{u^{2, \: 2}}& 0 &{u^{2, \: 1}}& 0 
\
0 & -{u^{2, \: 1}}&{u^{1, \: 2}}& 0 & 0 & -{u^{2, \: 2}}&{u^{2, \: 2}}& 0 
\
0 & 0 & -{u^{2, \: 1}}&{u^{1, \: 1}}& 0 & 0 & -{u^{2, \: 2}}&{u^{2, \: 1}}
\
0 & 0 & 0 &{-{u^{2, \: 1}}+{u^{1, \: 2}}}& 0 & 0 & 0 & 0 
(21)
Type: Equation(OutputForm?)

The matrix K transforms the coefficients of the tensor Y into coefficients of the tensor \Phi. We are looking for coefficients of the tensor U such that K transforms the tensor Y into \Phi=0 for any Y.

A necessary condition for the equation to have a non-trivial solution is that the matrix K be degenerate.

Theorem 1

All 2-dimensional pre-Frobenius algebras are symmetric.

Proof: Consider the determinant of the matrix K above.

axiom
Kd := factor(determinant(K)::DMP(concat map(variables,ravel(U)),FRAC INT))

\label{eq22}{{\left({u^{1, \: 2}}-{u^{2, \: 1}}\right)}^4}\ {{\left({{u^{1, \: 1}}\ {u^{2, \: 2}}}-{{u^{1, \: 2}}\ {u^{2, \: 1}}}\right)}^2}(22)
Type: Factored(DistributedMultivariatePolynomial?([*002u11,*002u12,*002u21,*002u22],Fraction(Integer)))

The scalar product must also be non-degenerate

axiom
Ud:DMP(concat map(variables,ravel(U)),FRAC INT) := determinant [[U[i,j] for j in 1..n] for i in 1..n]

\label{eq23}{{u^{1, \: 1}}\ {u^{2, \: 2}}}-{{u^{1, \: 2}}\ {u^{2, \: 1}}}(23)
Type: DistributedMultivariatePolynomial?([*002u11,*002u12,*002u21,*002u22],Fraction(Integer))

therefore U must be symmetric.

axiom
nthFactor(Kd,1)

\label{eq24}{u^{1, \: 2}}-{u^{2, \: 1}}(24)
Type: DistributedMultivariatePolynomial?([*002u11,*002u12,*002u21,*002u22],Fraction(Integer))
axiom
US:T := unravel(map(x+->subst(x,U[2,1]=U[1,2]),ravel U))

\label{eq25}\left[ 
\begin{array}{cc}
{u^{1, \: 1}}&{u^{1, \: 2}}
\
{u^{1, \: 2}}&{u^{2, \: 2}}
(25)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

Theorem 2

All 2-dimensional algebras with associative scalar product are commutative.

Proof: The basis of the null space of the symmetric K matrix are all symmetric

axiom
YUS:T :=  reindex(reindex(US,[2,1])*reindex(Y,[1,3,2]),[3,2,1])-US*Y

\label{eq26}\begin{array}{@{}l}
\displaystyle
\left[{\left[ 
\begin{array}{cc}
0 &{-{{u^{1, \: 2}}\ {y_{2, \: 1}^{2}}}-{{u^{1, \: 1}}\ {y_{2, \: 1}^{1}}}+{{u^{2, \: 2}}\ {y_{1, \: 1}^{2}}}+{{u^{1, \: 2}}\ {y_{1, \: 1}^{1}}}}
\
{{{u^{1, \: 2}}\ {y_{2, \: 1}^{2}}}+{{u^{1, \: 1}}\ {y_{2, \: 1}^{1}}}-{{u^{1, \: 2}}\ {y_{1, \: 2}^{2}}}-{{u^{1, \: 1}}\ {y_{1, \: 2}^{1}}}}&{-{{u^{1, \: 2}}\ {y_{2, \: 2}^{2}}}-{{u^{1, \: 1}}\ {y_{2, \: 2}^{1}}}+{{u^{2, \: 2}}\ {y_{2, \: 1}^{2}}}+{{u^{1, \: 2}}\ {y_{2, \: 1}^{1}}}}
(26)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
KS := jacobian(ravel(YUS),concat(map(variables,ravel(Y)))::List Symbol);
Type: Matrix(Fraction(Polynomial(Integer)))
axiom
NS:=nullSpace(KS)

\label{eq27}\begin{array}{@{}l}
\displaystyle
\left[{\left[{{{u^{1, \: 1}}^2}\over{{u^{1, \: 2}}^2}}, \:{{u^{1, \: 1}}\over{u^{1, \: 2}}}, \:{{u^{1, \: 1}}\over{u^{1, \: 2}}}, \: 1, \: 0, \: 0, \: 0, \: 0 \right]}, \: \right.
\
\
\displaystyle
\left.{\left[ -{{u^{2, \: 2}}\over{u^{1, \: 2}}}, \: 0, \: 0, \: 0, \: 1, \: 0, \: 0, \: 0 \right]}, \: \right.
\
\
\displaystyle
\left.{\left[{{-{{u^{1, \: 1}}\ {u^{2, \: 2}}}+{{u^{1, \: 2}}^2}}\over{{u^{1, \: 2}}^2}}, \: -{{u^{2, \: 2}}\over{u^{1, \: 2}}}, \: -{{u^{2, \: 2}}\over{u^{1, \: 2}}}, \: 0, \: 0, \: 1, \: 1, \: 0 \right]}, \: \right.
\
\
\displaystyle
\left.{\left[{{u^{1, \: 1}}\over{u^{1, \: 2}}}, \: 1, \: 1, \: 0, \: 0, \: 0, \: 0, \: 1 \right]}\right] 
(27)
Type: List(Vector(Fraction(Polynomial(Integer))))
axiom
SS:=map((x,y)+->x=y,concat map(variables,ravel Y),
  entries reduce(+,[p[i]*NS.i for i in 1..#NS]))

\label{eq28}\begin{array}{@{}l}
\displaystyle
\left[{
\begin{array}{@{}l}
\displaystyle
{y_{1, \: 1}^{1}}={{\left(
\begin{array}{@{}l}
\displaystyle
{{u^{1, \: 1}}\ {u^{1, \: 2}}\ {p_{4}}}+{{\left(-{{u^{1, \: 1}}\ {u^{2, \: 2}}}+{{u^{1, \: 2}}^2}\right)}\ {p_{3}}}- 
\
\
\displaystyle
{{u^{1, \: 2}}\ {u^{2, \: 2}}\ {p_{2}}}+{{{u^{1, \: 1}}^2}\ {p_{1}}}
(28)
Type: List(Equation(Fraction(Polynomial(Integer))))
axiom
YS:T := unravel(map(x+->subst(x,SS),ravel Y))

\label{eq29}\begin{array}{@{}l}
\displaystyle
\left[{\left[ 
\begin{array}{cc}
{{{{u^{1, \: 1}}\ {u^{1, \: 2}}\ {p_{4}}}+{{\left(-{{u^{1, \: 1}}\ {u^{2, \: 2}}}+{{u^{1, \: 2}}^2}\right)}\ {p_{3}}}-{{u^{1, \: 2}}\ {u^{2, \: 2}}\ {p_{2}}}+{{{u^{1, \: 1}}^2}\ {p_{1}}}}\over{{u^{1, \: 2}}^2}}&{{{{u^{1, \: 2}}\ {p_{4}}}-{{u^{2, \: 2}}\ {p_{3}}}+{{u^{1, \: 1}}\ {p_{1}}}}\over{u^{1, \: 2}}}
\
{{{{u^{1, \: 2}}\ {p_{4}}}-{{u^{2, \: 2}}\ {p_{3}}}+{{u^{1, \: 1}}\ {p_{1}}}}\over{u^{1, \: 2}}}&{p_{1}}
(29)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))

This defines a 4-parameter family of 2-d pre-Frobenius algebras

axiom
test(unravel(map(x+->subst(x,SS),ravel YUS))$T=0*YU)

\label{eq30} \mbox{\rm true} (30)
Type: Boolean

Alternatively we may consider

axiom
J := jacobian(ravel(YU),concat(map(variables,ravel(U)))::List Symbol);
Type: Matrix(Fraction(Polynomial(Integer)))
axiom
uu := transpose matrix [concat(map(variables,ravel(U)))::List Symbol];
Type: Matrix(Polynomial(Integer))
axiom
J::OutputForm * uu::OutputForm = 0

\label{eq31}\begin{array}{@{}l}
\displaystyle
{{\left[ 
\begin{array}{cccc}
0 & -{y_{1, \: 1}^{2}}&{y_{1, \: 1}^{2}}& 0 
\
-{y_{2, \: 1}^{1}}&{-{y_{2, \: 1}^{2}}+{y_{1, \: 1}^{1}}}& 0 &{y_{1, \: 1}^{2}}
\
{{y_{2, \: 1}^{1}}-{y_{1, \: 2}^{1}}}& -{y_{1, \: 2}^{2}}&{y_{2, \: 1}^{2}}& 0 
\
-{y_{2, \: 2}^{1}}&{-{y_{2, \: 2}^{2}}+{y_{2, \: 1}^{1}}}& 0 &{y_{2, \: 1}^{2}}
\
{y_{1, \: 2}^{1}}& 0 &{{y_{1, \: 2}^{2}}-{y_{1, \: 1}^{1}}}& -{y_{1, \: 1}^{2}}
\
0 &{y_{1, \: 2}^{1}}& -{y_{2, \: 1}^{1}}&{-{y_{2, \: 1}^{2}}+{y_{1, \: 2}^{2}}}
\
{y_{2, \: 2}^{1}}& 0 &{{y_{2, \: 2}^{2}}-{y_{1, \: 2}^{1}}}& -{y_{1, \: 2}^{2}}
\
0 &{y_{2, \: 2}^{1}}& -{y_{2, \: 2}^{1}}& 0 
(31)
Type: Equation(OutputForm?)

The matrix J transforms the coefficients of the tensor U into coefficients of the tensor \Phi. We are looking for coefficients of the tensor Y such that J transforms the tensor U into \Phi=0 for any U.

A necessary condition for the equation to have a non-trivial solution is that all 70 of the 4x4 sub-matrices of J are degenerate. To this end we can form the polynomial ideal of the determinants of these sub-matrices.

axiom
JP:=ideal concat concat concat
  [[[[ determinant(
    matrix([row(J,i1),row(J,i2),row(J,i3),row(J,i4)]))
      for i4 in (i3+1)..maxRowIndex(J) ] 
        for i3 in (i2+1)..(maxRowIndex(J)-1) ]
          for i2 in (i1+1)..(maxRowIndex(J)-2) ]
            for i1 in minRowIndex(J)..(maxRowIndex(J)-3) ];
Type: PolynomialIdeal?(Fraction(Integer),IndexedExponents?(Symbol),Symbol,Polynomial(Fraction(Integer)))
axiom
#generators(%)

\label{eq32}51(32)
Type: PositiveInteger?

Theorem 3

If a 2-d algebra is associative, commutative, anti-commutative or if it satisfies the Jacobi identity then it is a pre-Frobenius algebra.

Proof

Consider the ideals of the associator, commutator, anti-commutator and Jacobi identity

axiom
YYI:=ideal ravel YY;
Type: PolynomialIdeal?(Fraction(Integer),IndexedExponents?(Symbol),Symbol,Polynomial(Fraction(Integer)))
axiom
in?(JP,YYI)  -- associative

\label{eq33} \mbox{\rm true} (33)
Type: Boolean
axiom
YCI:=ideal ravel YC;
Type: PolynomialIdeal?(Fraction(Integer),IndexedExponents?(Symbol),Symbol,Polynomial(Fraction(Integer)))
axiom
in?(JP,YCI)  -- commutative

\label{eq34} \mbox{\rm true} (34)
Type: Boolean
axiom
YAI:=ideal ravel YA;
Type: PolynomialIdeal?(Fraction(Integer),IndexedExponents?(Symbol),Symbol,Polynomial(Fraction(Integer)))
axiom
in?(JP,YAI)  -- anti-commutative

\label{eq35} \mbox{\rm true} (35)
Type: Boolean
axiom
YXI:=ideal ravel YX;
Type: PolynomialIdeal?(Fraction(Integer),IndexedExponents?(Symbol),Symbol,Polynomial(Fraction(Integer)))
axiom
in?(JP,YXI) -- Jacobi identity

\label{eq36} \mbox{\rm true} (36)
Type: Boolean

Y-forms

Three traces of two graftings of an algebra gives six (2,0)-forms.

Left snail and right snail:

  LS                    RS

  Y /\                    /\ Y
   Y  )                  (  Y
    \/                    \/

  i  j                        j  i
   \/                          \/
    \    /\              /\    /
     e  f  \            /  f  e
      \/    \          /    \/
       \    /          \    /
        f  /            \  f
         \/              \/


\label{eq37}
LS = \{ {y^e}_{ij} {y^f}_{ef} \} \
RS = \{ {y^f}_{fe} {y^e}_{ji} \}
(37)

axiom
LS:=contract(Y*Y,1,2)

\label{eq38}\left[ 
\begin{array}{cc}
{{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{2}}}+{{y_{1, \: 1}^{1}}^2}}&{{{y_{2, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{2, \: 1}^{1}}\ {y_{2, \: 1}^{2}}}+{{\left({y_{1, \: 2}^{2}}+{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 1}^{1}}}}
\
{{{y_{1, \: 2}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 2}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{1, \: 2}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}&{{{y_{2, \: 2}^{2}}^2}+{{y_{2, \: 1}^{1}}\ {y_{2, \: 2}^{2}}}+{{\left({y_{1, \: 2}^{2}}+{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 2}^{1}}}}
(38)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
RS:=reindex(contract(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),1,2),[2,1])

\label{eq39}\left[ 
\begin{array}{cc}
{{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{1}}}+{{y_{1, \: 1}^{1}}^2}}&{{{y_{2, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{\left({y_{2, \: 1}^{1}}+{y_{1, \: 2}^{1}}\right)}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{1}}}}
\
{{{y_{1, \: 2}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 2}^{1}}\ {y_{1, \: 2}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}&{{{y_{2, \: 2}^{2}}^2}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 2}^{2}}}+{{\left({y_{2, \: 1}^{2}}+{y_{1, \: 1}^{1}}\right)}\ {y_{2, \: 2}^{1}}}}
(39)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
test(LS=RS)

\label{eq40} \mbox{\rm false} (40)
Type: Boolean

Left and right deer:

   RD                 LD

   \ /\/              \/\ /
    Y /\              /\ Y
     Y  )            (  Y
      \/              \/

   i            j    i            j
    \    /\    /      \    /\    /
     \  f  \  /        \  /  f  /
      \/    \/          \/    \/
       \    /\          /\    /
        e  /  \        /  \  e
         \/    \      /    \/
          \    /      \    /
           f  /        \  f
            \/          \/


\label{eq41}
RD = \{ {y^e}_{if} {y^f}_{ej} \} \
LD = \{ {y^f}_{ie} {y^e}_{fj} \}
(41)
Left and right deer forms are identical but different from snails.
axiom
RD:=contract(Y*Y,1,3)

\label{eq42}\left[ 
\begin{array}{cc}
{{{y_{1, \: 2}^{2}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{1}}}+{{y_{1, \: 1}^{1}}^2}}&{{{y_{2, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{2, \: 1}^{1}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{1}}}}
\
{{{y_{1, \: 2}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{1, \: 2}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}&{{{y_{2, \: 2}^{2}}^2}+{{\left({y_{2, \: 1}^{2}}+{y_{1, \: 2}^{2}}\right)}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{1}}}}
(42)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
LD:=reindex(contract(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),1,3),[2,1])

\label{eq43}\left[ 
\begin{array}{cc}
{{{y_{1, \: 2}^{2}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{1}}}+{{y_{1, \: 1}^{1}}^2}}&{{{y_{2, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{2, \: 1}^{1}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{1}}}}
\
{{{y_{1, \: 2}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{1, \: 2}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}&{{{y_{2, \: 2}^{2}}^2}+{{\left({y_{2, \: 1}^{2}}+{y_{1, \: 2}^{2}}\right)}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{1}}}}
(43)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
test(LD=RD)

\label{eq44} \mbox{\rm true} (44)
Type: Boolean
axiom
test(RD=RS)

\label{eq45} \mbox{\rm false} (45)
Type: Boolean
axiom
test(RD=LS)

\label{eq46} \mbox{\rm false} (46)
Type: Boolean

Left and right turtles:

  RT                   LT

   /\ / /               \ \ /\
  (  Y /                 \ Y  )
   \  Y                   Y  /
    \/                     \/

           i     j      i     j
    /\    /     /        \     \    /\
   /  f  /     /          \     \  f  \
  /    \/     /            \     \/    \
  \     \    /              \    /     /
   \     e  /                \  e     /
    \     \/                  \/     /
     \    /                    \    /
      \  f                      f  /
       \/                        \/


\label{eq47}
RT = \{ {y^e}_{fi} {y^f}_{ej} \} \
LT = \{ {y^f}_{ie} {y^e}_{jf} \}
(47)

axiom
RT:=contract(Y*Y,1,4)

\label{eq48}\left[ 
\begin{array}{cc}
{{{y_{2, \: 1}^{2}}^2}+{2 \ {y_{1, \: 1}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 1}^{1}}^2}}&{{{y_{2, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}
\
{{{y_{2, \: 1}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{2}}\ {y_{2, \: 1}^{1}}}+{{y_{1, \: 1}^{1}}\ {y_{1, \: 2}^{1}}}}&{{{y_{2, \: 2}^{2}}^2}+{2 \ {y_{1, \: 2}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}^2}}
(48)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
LT:=reindex(contract(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),1,4),[2,1])

\label{eq49}\left[ 
\begin{array}{cc}
{{{y_{1, \: 2}^{2}}^2}+{2 \ {y_{1, \: 1}^{2}}\ {y_{1, \: 2}^{1}}}+{{y_{1, \: 1}^{1}}^2}}&{{{y_{1, \: 2}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{1}}}}
\
{{{y_{1, \: 2}^{2}}\ {y_{2, \: 2}^{2}}}+{{y_{1, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{1, \: 2}^{1}}\ {y_{2, \: 1}^{2}}}+{{y_{1, \: 1}^{1}}\ {y_{2, \: 1}^{1}}}}&{{{y_{2, \: 2}^{2}}^2}+{2 \ {y_{2, \: 1}^{2}}\ {y_{2, \: 2}^{1}}}+{{y_{2, \: 1}^{1}}^2}}
(49)
Type: CartesianTensor?(1,2,Fraction(Polynomial(Integer)))
axiom
test(LT=RT)

\label{eq50} \mbox{\rm false} (50)
Type: Boolean

The turles are symmetric

axiom
test(RT=reindex(RT,[2,1]))

\label{eq51} \mbox{\rm true} (51)
Type: Boolean
axiom
test(LT=reindex(LT,[2,1]))

\label{eq52} \mbox{\rm true} (52)
Type: Boolean

Five of the six forms are independent.

axiom
test(RT=RS)

\label{eq53} \mbox{\rm false} (53)
Type: Boolean
axiom
test(RT=LS)

\label{eq54} \mbox{\rm false} (54)
Type: Boolean
axiom
test(RT=RD)

\label{eq55} \mbox{\rm false} (55)
Type: Boolean
axiom
test(LT=RS)

\label{eq56} \mbox{\rm false} (56)
Type: Boolean
axiom
test(LT=LS)

\label{eq57} \mbox{\rm false} (57)
Type: Boolean
axiom
test(LT=RD)

\label{eq58} \mbox{\rm false} (58)
Type: Boolean

Associativity implies right turtle equals right snail and left turtle equals left snail.

axiom
in?(ideal ravel(RT-RS),YYI)

\label{eq59} \mbox{\rm true} (59)
Type: Boolean
axiom
in?(ideal ravel(LT-LS),YYI)

\label{eq60} \mbox{\rm true} (60)
Type: Boolean

If the Jacobi identity holds then both snails are zero

axiom
in?(ideal ravel(RS),YXI)

\label{eq61} \mbox{\rm true} (61)
Type: Boolean
axiom
in?(ideal ravel(LS),YXI)

\label{eq62} \mbox{\rm true} (62)
Type: Boolean

and right turtle and deer have opposite signs

axiom
in?(ideal ravel(RT+RD),YXI)

\label{eq63} \mbox{\rm true} (63)
Type: Boolean