Edit detail for SandBox Grassmann Algebra Is Frobenius In Many Ways revision 2 of 8

-dimensional vector space representing Grassmann algebra with generators

An algebra is represented by a (2,1)-tensor viewed as a linear operator with two inputs and one output . For example:

n:=2

(1)

dim:=2^n

(2)

T:=CartesianTensor(1,dim,FRAC POLY INT)

(3)

Y:T := unravel(concat concat
  [[[script(y,[[i,j],[k]])
    for i in 1..dim]
      for j in 1..dim]
        for k in 1..dim]
          )

(4)

A scalar product is denoted by the (2,0)-tensor

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

(5)

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

(6)

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

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 or we may consider an scalar product U as given, and look for all algebras such that the scalar product is associative.

This problem can be solved using linear algebra.

)expose MCALCFN

   MultiVariableCalculusFunctions is now explicitly exposed in frame 
      initial 
K := jacobian(ravel(YU),concat(map(variables,ravel(Y)))::List Symbol);

The matrix K transforms the coefficients of the tensor into coefficients of the tensor . We are looking for coefficients of the tensor such that K transforms the tensor into for any .

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

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..dim] for i in 1..dim]
  \end{axiom}

The basis of the null space of the K matrix:

  \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 family of pre-Frobenius algebras:

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

Alternatively we may consider

J := jacobian(ravel(YU),concat(map(variables,ravel(U)))::List Symbol);

The matrix J transforms the coefficients of the tensor into coefficients of the tensor . We are looking for coefficients of the tensor such that J transforms the tensor into for any .

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}