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D.4.3.1 cup
Procedure from library homolog.lib (see homolog_lib).
- Usage:
cup(M,[,any,any]); M=module
- Compute:
cup-product Ext^1(M’,M’) x Ext^1(M’,M’) —> Ext^2(M’,M’), where M’:=R^m/M, if M in R^m, R basering (i.e. M’:=coker(matrix(M))).
If called with >= 2 arguments: compute symmetrized cup-product- Assume:
all Ext’s are finite dimensional
- Return:
- if called with 1 argument: matrix, the columns of the output present the coordinates of b_i&b_j with respect to a kbase of Ext^2, where b_1,b_2,... is a kbase of Ext^1 and & denotes cup product;
- if called with 2 arguments: matrix, the columns of the output present the coordinates of (1/2)(b_i&b_j + b_j&b_i) with respect to a kbase of Ext^2;
- if called with 3 arguments: list,L[1] = matrix see above (symmetric case, for >=2 arguments) L[2] = matrix of kbase of Ext^1 L[3] = matrix of kbase of Ext^2- Note:
printlevel >=1; shows what is going on.
printlevel >=2; shows result in another representation.
For computing cupproduct of M itself, apply proc to syz(M)!
Example:
LIB "homolog.lib"; int p = printlevel; ring rr = 32003,(x,y,z),(dp,C); ideal I = x4+y3+z2; qring o = std(I); module M = [x,y,0,z],[y2,-x3,z,0],[z,0,-y,-x3],[0,z,x,-y2]; print(cup(M)); → 0,1,0, 0, 0,0,0,0,0,0,0, 0,0,0,0,0,0, → 0,0,-1,0, 0,1,0,0,0,0,0, 0,0,0,0,0,0, → 0,0,0, -1,0,0,0,0,0,1,0, 0,0,0,0,0,0, → 0,0,0, 0, 1,0,0,1,0,0,-1,0,0,1,0,0,0 print(cup(M,1)); → 0,1,0,0,0,0,0,0,0,0,0, → 0,0,0,0,0,0,0,0,0,0,0, → 0,0,0,0,0,0,0,0,0,0,0, → 0,0,0,0,1,0,0,0,0,0,0 // 2nd EXAMPLE (shows what is going on) printlevel = 3; ring r = 0,(x,y),(dp,C); ideal i = x2-y3; qring q = std(i); module M = [-x,y],[-y2,x]; print(cup(M)); → // vdim (Ext^1) = 2 → // kbase of Ext^1(M,M) → // - the columns present the kbase elements in Hom(F(1),F(0)) → // - F(*) a free resolution of M → -1,0, → 0, y, → 0, 1, → -1,0 → // lift kbase of Ext^1: → // - the columns present liftings of kbase elements into Hom(F(2),F(1)) → // - F(*) a free resolution of M → 1,0, → 0,y, → 0,1, → 1,0 → // vdim (Ext^2) = 2 → // kbase of Ext^2(M,M) → // - the columns present the kbase elements in Hom(F(2),F(0)) → // - F(*) is a a free resolution of M → -1,0, → 0, y, → 0, 1, → -1,0 → // matrix of cup-products (in Ext^2) → 0,-1,0, 0,y, → 0,0, -y,y,0, → 0,0, -1,1,0, → 0,-1,0, 0,y → ////// end level 2 ////// → // the associated matrices of the bilinear mapping 'cup' → // corresponding to the kbase elements of Ext^2(M,M) are shown, → // i.e. the rows of the final matrix are written as matrix of → // a bilinear form on Ext^1 x Ext^1 → //-----component 1: → 0,1, → 0,0 → //-----component 2: → 0, 0, → -1,1 → ////// end level 3 ////// → 0,1,0, 0,0, → 0,0,-1,1,0 printlevel = p; |