Difference between revisions of "ApCoCoA-1:BBSGen.TraceSyzStep"

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  BBSGen.TraceSyzStep(Mon,X,OO,BO,N);
 
  BBSGen.TraceSyzStep(Mon,X,OO,BO,N);
 
    
 
    
   c[1,2]t[1,2,3,1] + c[2,2]t[1,2,3,2] + c[3,2]t[1,2,3,3] + c[4,2]t[1,2,3,4] + c[1,4]t[1,2,4,1] + c[2,4]t[1,2,4,2] + c[3,4]t[1,2,4,3] + c[4,4]t[1,2,4,4] + t[1,2,1,3] + t[1,2,2,4]
+
   c[1,2]t[1,2,3,1] + c[2,2]t[1,2,3,2] + c[3,2]t[1,2,3,3] +
 +
c[4,2]t[1,2,3,4] + c[1,4]t[1,2,4,1] + c[2,4]t[1,2,4,2] +  
 +
c[3,4]t[1,2,4,3] + c[4,4]t[1,2,4,4] +
 +
t[1,2,1,3] + t[1,2,2,4]
  
 
-------------------------------
 
-------------------------------

Revision as of 18:10, 31 May 2012

BBSGen.TraceSyzStep

This function only computes the trace syzygy for the degree of the given monomial.

Syntax

TraceSyzLin(OO,BO,N);
TraceSyzLin(OO:LIST,BO:LIST,N:INTEGER):LIST

Description


Let Tau^kl_ij :=t[k,l,i,j] be the (i,j) ^th entry of matrix the operation [A_k,A_l]. The result of the Trace Syzygy computation is K[c]-linear combination of Tau^kl_ij that is equal to 0. This function only computes the trace syzygy for the degree of the given monomial.


  • @param The Monomial Mon, the distinguished indterminate of choice, order ideal OO, border BO, the number of Indeterminates of the Polynomial.

  • @return Trace syzygy of the degree of the given monomial.


Example

Use R::=QQ[x[1..2]];

OO:=BB.Box([1,1]);
BO:=BB.Border(OO);
Mu:=Len(OO);
Nu:=Len(BO);

Mon:=x[1]^2x[2];--------Target Monomial

X:=x[1];   ------------Choice of the Indeterminate

Use XX::=QQ[c[1..Mu,1..Nu],t[1..N,1..N,1..Mu,1..Mu]]; 
 
 BBSGen.TraceSyzStep(Mon,X,OO,BO,N);
  
  c[1,2]t[1,2,3,1] + c[2,2]t[1,2,3,2] + c[3,2]t[1,2,3,3] +
 c[4,2]t[1,2,3,4] + c[1,4]t[1,2,4,1] + c[2,4]t[1,2,4,2] + 
c[3,4]t[1,2,4,3] + c[4,4]t[1,2,4,4] +
 t[1,2,1,3] + t[1,2,2,4]

-------------------------------

BB.Border

BB.Box

BBSGen.Wmat

BBSGen.TraceSyzLin

BBSGen.TraceSyzFull