# Type [000], CY of degree 20 -- lifting to a 3-fold with two singular points

We construct a singular Calabi-Yau of dimension 3 with 2 nodes, of codimension 4, of degree 20. This example cannot be smoothed. However, there is the example of the determinantal ($3 \times 3$ minors of a $4 \times 4$ matrix of linear forms, which is a smooth Calabi-Yau 3-fold of degree 20.

 i1 : kk = ZZ/31 o1 = kk o1 : QuotientRing i2 : U = kk[y0,y1,y2,y3,y4,y5,y6,y7]; i3 : setRandomSeed 8327432 o3 = 8327432 i4 : U1 = vars U o4 = | y0 y1 y2 y3 y4 y5 y6 y7 | 1 8 o4 : Matrix U <--- U i5 : P6 = matrix{{y1,y2,y3,y4,y5,y6,y7}} -- The hyperplane y0 == 0 o5 = | y1 y2 y3 y4 y5 y6 y7 | 1 7 o5 : Matrix U <--- U

The original example code

 i6 : F1 = P6 * random(kk^7,kk^5); 1 5 o6 : Matrix U <--- U i7 : F2 = P6 * random(kk^7,kk^5); 1 5 o7 : Matrix U <--- U i8 : F3 = P6 * random(kk^7,kk^5); 1 5 o8 : Matrix U <--- U i9 : fg = transpose F1 | transpose F2 | transpose F3 ; 5 3 o9 : Matrix U <--- U i10 : ff = matrix{{y0,0,0,0,0},{0,0,0,0,0},{0,0,0,0,0}}; 3 5 o10 : Matrix U <--- U i11 : fg = transpose fg + ff; --3x5 matrix of rank 1 at (1:0:..:0), of rank 2 on fourfold Y 3 5 o11 : Matrix U <--- U i12 : Y = fg3 = minors(3,fg); o12 : Ideal of U i13 : fg2 = minors(2,fg); o13 : Ideal of U i14 : codim fg3 o14 = 3 i15 : codim fg2, degree fg2 o15 = (7, 1) o15 : Sequence i16 : saturate fg2 o16 = ideal (y7, y6, y5, y4, y3, y2, y1) o16 : Ideal of U i17 : sf1 = fg^{0,1} o17 = | y0+12y1+3y2+7y3+4y4+13y5+5y6-y7 -13y1+3y2-9y3-3y4-y5+2y6-2y7 | 9y1-7y2-13y3-15y4-2y5+8y6+15y7 -9y1-8y2+14y3-3y4-5y5+4y6+15y7 ----------------------------------------------------------------------- 7y1+5y3-y4-6y5+5y6-6y7 -7y1-3y2+3y3-11y4-3y5-y6-5y7 7y1+9y2-5y3-2y4+6y5+15y6+6y7 14y1+12y2+5y3-6y4+15y5+13y6+4y7 ----------------------------------------------------------------------- 9y1+4y2-5y3+3y4+4y5-14y6-13y7 | -4y1+8y2+12y3+14y4-13y5+3y6-10y7 | 2 5 o17 : Matrix U <--- U i18 : sf2 = fg^{0,2} o18 = | y0+12y1+3y2+7y3+4y4+13y5+5y6-y7 -13y1+3y2-9y3-3y4-y5+2y6-2y7 | -6y1-7y2+14y3-7y4-9y5+2y6+9y7 9y1+9y2-2y3+12y4+7y5-2y6 ----------------------------------------------------------------------- 7y1+5y3-y4-6y5+5y6-6y7 -7y1-3y2+3y3-11y4-3y5-y6-5y7 -7y1+8y2+12y3+6y4-15y5+5y6-y7 8y1-10y2+5y3-11y4+9y5-11y6+3y7 ----------------------------------------------------------------------- 9y1+4y2-5y3+3y4+4y5-14y6-13y7 | -5y1-15y2-8y3-4y4+5y5-12y6+3y7 | 2 5 o18 : Matrix U <--- U i19 : Z1 = sf12 = minors(2,sf1);--threefold rational scroll of degree 5 in Y o19 : Ideal of U i20 : Z2 = sf22 = minors(2,sf2);--threefold rational scroll of degree 5 in Y o20 : Ideal of U i21 : sf = trim intersect(Z1, Z2);-- threefold rational scroll over reducible conic of degree 10 in Y o21 : Ideal of U i22 : (dim Z1, degree Z1) o22 = (4, 5) o22 : Sequence i23 : betti sf --in 11 cubics, so sf is in a cubic hypersurface section of Y 0 1 o23 = total: 1 54 0: 1 . 1: . . 2: . 11 3: . 43 o23 : BettiTally i24 : k3 = ideal select(sf_*, f -> degree f == {3}); -- the cubics in the ideal o24 : Ideal of U i25 : k3=submatrix(gens sf,{0,1,2,3,4,5,6,7,8,9,10}); --the cubics in the ideal of sf 1 11 o25 : Matrix U <--- U i26 : k3=ideal flatten k3; o26 : Ideal of U i27 : X = cy = (k3:Z1):Z2; --the CY threefold, residual to sf in a cubic hypersurface section o27 : Ideal of U i28 : betti X 0 1 o28 = total: 1 16 0: 1 . 1: . . 2: . 16 o28 : BettiTally i29 : betti res X 0 1 2 3 4 o29 = total: 1 16 30 16 1 0: 1 . . . . 1: . . . . . 2: . 16 30 16 . 3: . . . . . 4: . . . . 1 o29 : BettiTally i30 : (dim X, degree X) o30 = (4, 20) o30 : Sequence

Now we analyze the singularities of this Calabi-Yau. First, note that $X$ is singular at the point$(1, 0, \ldots, 0)$.

 i31 : J = jacobian X; 8 16 o31 : Matrix U <--- U i32 : minors(4, sub(J, matrix{{1_kk,0,0,0,0,0,0,0}})) == 0 o32 = true i33 : sub(X, matrix{{1_kk,0,0,0,0,0,0,0}}) o33 = ideal (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0) o33 : Ideal of kk

Now we find other singular points. We would like to just compute the singular locus as the $4 \time 4$ minors of the $8 \times 16$ Jacobian matrix, which is $127,400$ determinants, but this is pretty large.

We find that there is exactly one more singular point, and that it is of multiplicity 2.

 needsPackage "FastMinors" -- "FastLinAlg" in versions of M2 before 1.19 setRandomSeed 500 elapsedTime chooseGoodMinors(200, 4, J); J1 = cy + oo; elapsedTime gbJ1 = groebnerBasis(J1, Strategy => "F4"); inJ1 = monomialIdeal leadTerm gbJ1; codim inJ1 == 7 -- so singular locus is finite degree inJ1 == 2 (L1, d) = divideByVariable(gbJ1, U_7); L1 = ideal L1; pt2 = trim L1

This implies that the singular locus is a finite set of points, and is at most 2 points, and we know one of the points, and the other pointis off of the hyperplane $x_7 = 0$.

However, we must check that this point is singular on $X$, as our ideal is only a subset of the ideal of the singular locus.

 isSubset(cy, pt2) minors(4, J % pt2) == 0

We now know that the singular locus consists of these 2 points, as the degree of $J_1$ is 2, so there can be no other components of dimension 0.