Edit detail for SandBoxFrobeniusAlgebra revision 15 of 26

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

)library DEXPR

 DistributedExpression is now explicitly exposed in frame initial 
 DistributedExpression will be automatically loaded when needed from 
 /var/zope2/var/LatexWiki/DEXPR.NRLIB/DEXPR
n:=2

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

--T:=CartesianTensor(1,n,HDMP(concat[concat concat
-- [[[script(y,[[k],[j,i]])
-- for i in 1..n]
-- for j in 1..n]
-- for k in 1..n],
-- [script(u,[[i]]) for i in 1..n],
-- [script(v,[[i]]) for i in 1..n] ],FRAC INT))
Y:T := unravel(concat concat
 [[[script(y,[[k],[j,i]])
 for i in 1..n]
 for j in 1..n]
 for k in 1..n]
 )

Given two vectors U=\{ u_i \} and V=\{ v_j \}

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

V:T := unravel([script(v,[[i]]) for i in 1..n])

the tensor Y operates on their tensor product to yield a vector W=\{ w_k = {y_k}^{ji} u_i v_j \}

W:=contract(contract(Y,3,product(U,V),1),2,3)

Diagram:

 U V
 2i 3j
 \ /
 |
 1k
 W

or in a more convenient notation:

W:=(Y*U)*V

The algebra Y is commutative if the following tensor (the commutator) is zero

K:=Y-reindex(Y,[1,3,2])

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

C:=groebner(ravel(K))

An algebra is associative if:

 Y I = I Y
 Y Y
 Note: right figure is mirror image of left!
 2 3 6 2 5 6 2 3 4
 \/ / \ \/ \ | /
 \/ = \/ = \|/
 \ / |
 4 1 1

In other words an algebra is associative if and only if the following (3,1)-tensor A=\{ {a_s}^{kji} = {y_s}^{kr} {y_r}^{ji} - {y_r}^{kj} {y_s}^{ri} \} is zero.

test(Y*Y = contract(product(Y,Y),3,4))

test(Y*Y = contract(Y,3,Y,1))

test(reindex(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),[1,4,3,2]) = reindex(contract(product(Y,Y),1,5),[3,1,2,4]))

test(reindex(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),[1,4,3,2]) = reindex(contract(Y,1,Y,2),[3,1,2,4]))

AA := reindex(reindex(Y,[1,3,2])*reindex(Y,[1,3,2]),[1,4,3,2])-Y*Y; ravel(AA)

AB:=groebner(ravel(AA))

#AB

The Jacobi identity requires the following tensor to be zero:

 2 3 6 2 5 6 2 6 3 
 \ / / \ \ / \ \/ 
 \/ / \ \/ \/\ 
 \/ - \/ - \/ 
 \ / \ 
 4 1 4 

BA := AA - reindex(contract(Y,1,Y,2),[3,1,4,2]); ravel(BA)

BB:=groebner(ravel(BA));

#BB

A scalar product is denoted by U = \{ u^{ij} \}

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

Definition 1

We say that the scalar product is "associative" if the following tensor equation holds:

 Y I = I Y
 U U

UA := 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 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.

)expose MCALCFN

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

YY := transpose matrix [concat(map(variables,ravel(Y)))::List Symbol];

K::OutputForm * YY::OutputForm = 0

The matrix K transforms the coefficients of the tensor Y into coefficients of the tensor UA. We are looking for coefficients of the tensor U such that K transforms Y into UA=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

The scalar product of all 2-dimensional pre-Frobenius algebras is symmetric.

Proof: Consider the determinant of the matrix K above.

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

The scalar product must also be non-degenerate

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

therefore U must be symmetric.

nthFactor(Kd,1)

US:T := unravel(map(x+->subst(x,U[2,1]=U[1,2]),ravel U))

Theorem 2

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

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

UAS:T := unravel(map(x+->subst(x,U[2,1]=U[1,2]),ravel UA))

--solve(ravel(UAS),removeDuplicates concat map(variables,ravel(US)))
KS := jacobian(ravel(UAS),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))

This is a 4-parameter family of 2-d pre-Frobenius algebras with a given admissible (i.e. symmetric) scalar product.

UASS:T := unravel(map(x+->subst(x,SS),ravel UAS))