If $C$ is a graded (homogeneous) complex over a ring $R$, and $d$ is a degree, this method computes the degree $d$ part of the complex over the coefficient ring of $R$.
Taking parts of a graded (homogeneous) complex commutes with taking homology.
i1 : R = QQ[a,b,c,d]; |
i2 : I = ideal(a*b, a*c, b*c, a*d) o2 = ideal (a*b, a*c, b*c, a*d) o2 : Ideal of R |
i3 : C = freeResolution I
1 4 4 1
o3 = R <-- R <-- R <-- R
0 1 2 3
o3 : Complex
|
i4 : D = part(4,C)
35 40 16 1
o4 = QQ <-- QQ <-- QQ <-- QQ
0 1 2 3
o4 : Complex
|
i5 : prune HH D == part(4, HH C) o5 = true |
i6 : prune HH D == part(4, complex(R^1/I)) o6 = true |
Given a squarefree monomial ideal corresponding to a simplicial complex, in a polynomial ring equipped with the fine grading, parts of the dual of the free resolution of the monomial ideal are the chain complexes of the induced simplicial subcomplexes.
i7 : S = QQ[a..d, DegreeRank=>4]; |
i8 : I = intersect(ideal(a,b), ideal(c,d)) o8 = ideal (b*d, a*d, b*c, a*c) o8 : Ideal of S |
i9 : C = dual freeResolution I
1 4 4 1
o9 = S <-- S <-- S <-- S
-3 -2 -1 0
o9 : Complex
|
i10 : prune HH (part({-1,-1,-1,-1}, C)) -- empty quadrilateral
1
o10 = QQ
-3
o10 : Complex
|
i11 : prune HH part({-1,-1,0,0}, C) -- 2 points
1
o11 = QQ
-2
o11 : Complex
|
i12 : prune HH part({0,0,-1,-1}, C) -- 2 points
1
o12 = QQ
-2
o12 : Complex
|
i13 : prune HH part({0,0,0,0}, C) -- solid quadrilateral
o13 = 0
0
o13 : Complex
|