Given a variety X in projective r-space Pr, the Fano scheme Fanok(X) is the natural parameter space for the linear k-planes lying on X. In this tutorial we explore the methods for computing it. The tutorial is in three parts
A. The twenty-seven lines
B. General methods
C. Surfaces of degree 4 in P5
In section A, we treat by hand the Fano variety of lines on a nonsingular cubic surface in P3, and find that there are indeed 27 lines lying on the surface.
In section B, we explain a general purpose function, written to compute Fano schemes.
There are (up to linear transformations) just 3 nondegenerate smooth surfaces of degree 4 in P5: the Veronese embedding of P2 and the rational normal scrolls S(1,3) and S(2,2). Can they be distinguished by their Fano varieties of lines? To find out, read section C!
First make the homogeneous coordinate ring of the ambient projective 3-space
i1 : R = ZZ/32003[a,b,c,d] o1 = R o1 : PolynomialRing |
and the ideal of a nonsingular cubic
i2 : X = ideal(a^3+b^3+c^3+d^3) 3 3 3 3 o2 = ideal(a + b + c + d ) o2 : Ideal of R |
We make a parametrized indeterminate line in our projective space, adding parameters s,t for the line and two points p0..p3 and q0..q3 representing the points 0 and infinity on the line.
i3 : KK = coefficientRing R o3 = KK o3 : QuotientRing |
i4 : S = KK [s,t,p_0..p_3,q_0..q_3] o4 = S o4 : PolynomialRing |
Then we make a map F from R to the new ring sending the variables to the coordinates of the general point on the line
i5 : F = map(S,R, s*matrix{{p_0..p_3}} + t*matrix{{q_0..q_3}} ) o5 = map(S,R,{s*p + t*q , s*p + t*q , s*p + t*q , s*p + t*q }) 0 0 1 1 2 2 3 3 o5 : RingMap S <--- R |
We now apply F to the ideal of X
i6 : FX = F X 3 3 3 3 3 3 3 3 2 2 2 2 3 3 o6 = ideal(s p + s p + s p + s p + 3s t*p q + 3s*t p q + t q + 0 1 2 3 0 0 0 0 0 ------------------------------------------------------------------------ 2 2 2 2 3 3 2 2 2 2 3 3 2 2 3s t*p q + 3s*t p q + t q + 3s t*p q + 3s*t p q + t q + 3s t*p q 1 1 1 1 1 2 2 2 2 2 3 3 ------------------------------------------------------------------------ 2 2 3 3 + 3s*t p q + t q ) 3 3 3 o6 : Ideal of S |
and the condition we want becomes the condition that FX vanishes identically in s,t. The following line produces the coefficients:
i7 : cFX = last coefficients(gens FX, Variables => {s,t}) o7 = {3} | p_0^3+p_1^3+p_2^3+p_3^3 | {3} | 3p_0^2q_0+3p_1^2q_1+3p_2^2q_2+3p_3^2q_3 | {3} | 3p_0q_0^2+3p_1q_1^2+3p_2q_2^2+3p_3q_3^2 | {3} | q_0^3+q_1^3+q_2^3+q_3^3 | 4 1 o7 : Matrix S <--- S |
The interface to the coefficients routine is a bit baroque, and might change in the future. For now, the {0,1} says to find the coefficients of each column of the matrix, with respect to the first two variables. The routine returns a list of two matrices, the second one being the one we need (index 1, since all indices start at 0 in Macaulay2)
We can get rid of some of the variables of S, to ease the computation:
i8 : S1 = KK[p_0..p_3,q_0..q_3] o8 = S1 o8 : PolynomialRing |
i9 : cFX = substitute(cFX, S1) o9 = {3} | p_0^3+p_1^3+p_2^3+p_3^3 | {3} | 3p_0^2q_0+3p_1^2q_1+3p_2^2q_2+3p_3^2q_3 | {3} | 3p_0q_0^2+3p_1q_1^2+3p_2q_2^2+3p_3q_3^2 | {3} | q_0^3+q_1^3+q_2^3+q_3^3 | 4 1 o9 : Matrix S1 <--- S1 |
The ring we want is the quotient
i10 : S1bar = S1/ideal cFX o10 = S1bar o10 : QuotientRing |
Now we want to move to the Grassmannian, so we take a new polynomial ring in 6 variables that will correspond to the minors of the matrix with rows p0..p3 and q0..q3,
i11 : GR = coefficientRing R[x_0..x_5] o11 = GR o11 : PolynomialRing |
We define a map sending the xi to the minors, regarded as elements of S1bar
i12 : M = substitute( exteriorPower(2, matrix{{p_0..p_3},{q_0..q_3}}), S1bar) o12 = | -p_1q_0+p_0q_1 -p_2q_0+p_0q_2 -p_2q_1+p_1q_2 -p_3q_0+p_0q_3 ----------------------------------------------------------------------- -p_3q_1+p_1q_3 -p_3q_2+p_2q_3 | 1 6 o12 : Matrix S1bar <--- S1bar |
i13 : gr = map (S1bar, GR, M) o13 = map(S1bar,GR,{- p q + p q , - p q + p q , - p q + p q , - p q + p q , - p q + p q , - p q + p q }) 1 0 0 1 2 0 0 2 2 1 1 2 3 0 0 3 3 1 1 3 3 2 2 3 o13 : RingMap S1bar <--- GR |
i14 : fano = trim ker gr 2 2 2 o14 = ideal (x x - x x + x x , x x x , x x x , x x + x x , x x x , x x + 2 3 1 4 0 5 3 4 5 1 2 5 0 4 1 5 0 2 4 0 4 ----------------------------------------------------------------------- 2 3 3 3 2 2 2 2 2 2 2 x x , x + x + x , x x + x x , x x - x x , x x + x x , x x x , x x 1 5 3 4 5 1 3 2 4 0 3 2 5 1 3 2 4 0 1 3 0 3 ----------------------------------------------------------------------- 2 2 2 2 2 2 2 2 2 3 3 + x x , x x + x x , x x + x x , x x + x x , x x - x x , x + x - 2 5 1 2 3 4 0 2 3 5 1 2 3 4 0 2 3 5 1 2 ----------------------------------------------------------------------- 3 2 2 2 2 3 3 3 x , x x - x x , x x - x x , x - x - x ) 5 0 1 4 5 0 1 4 5 0 2 4 o14 : Ideal of GR |
trim replaces the given set of generators with a minimal set of generators. We get an ideal representing points:
i15 : codim fano o15 = 5 |
and the number of these points -- the number of the corresponding lines - is 27:
i16 : degree fano o16 = 27 |
It is interesting to note that the ideal of the Fano scheme that we have produced is NOT saturated, as the number (25) of cubics it contains is less than 56 - 27 = 29:
i17 : betti fano 0 1 o17 = total: 1 20 0: 1 . 1: . 1 2: . 19 o17 : BettiTally |
Possible next steps in this computation would be to compute the Jacobian ideal of Fano to show that we really got 27 distinct lines, etc.
The first step in writing a program is to decide what the program should do, and it is just as well to write the documentation at this point .
The documentation has the following form:
i18 : needsPackage "Text" o18 = Text o18 : Package |
If we take the variety that is the whole of Pr, we get the Grassmannian. It is useful to be able to make the ring representing the ambient space of the Grassmannian beforehand by hand, so the ideals of several Fano varieties can be compared. But often we won’t need this. Thus we make the function capable of accepting this ambient ring as an argument, or of fending for itself if no ambient ring is given.
i19 : document { Key => Fano2, TT "Fano2(k,X,GR) or Fano2(k,X)", " -- computes the ideal of a Fano scheme in the Grassmannian.", PARA{}, "Given an ideal X representing a projective variety in P^r, a positive integer k<r, and optionally a ring GR with (exactly) r+1 choose k+1 variables, representing the ambient space of the Grassmannian of k-planes in P^r, this routine returns the ideal in GR of the Fano scheme that parametrizes the k-planes lying on X. If the optional third argument is not present, the routine fabricates its own local ring, and returns an ideal over it." } |
In order to make Fano2 handle an optional number of arguments, we make it a method instead of a function, as follows
i20 : document { Key => symbol Grassmannian2, TT "Grassmannian2(k,r,R) or Grassmannian2(k,r)", "-- Given natural numbers k <= r, and optionally a ring R with at least binomial(r+1,k+1) variables, the routine defines the ideal of the Grassmannian of projective k-planes in P^r, using the first binomial(r+1,k+1) variables of R. If R is not given, the routine makes and uses ZZ/31991[vars(0..binomial(r+1,k+1)-1]." } |
Here is the code for the first case, with comments interspersed:
i21 : Fano2 = method() o21 = Fano2 o21 : MethodFunction |
The second case reduces to the first:
i22 : Fano2(ZZ,Ideal,Ring) := (k,X,GR) -> ( -- Get info about the base ring of X: -- The coefficient ring (to make new rings of -- the same characteristic, for example) -- and the number of variables KK:=coefficientRing ring X; r := (numgens ring X) - 1; -- Next make private variables for our -- intermediate rings, to avoid interfering -- with something outside: t:=symbol t; p:=symbol p; -- And rings S1 := KK[t_0..t_k]; S2 := KK[p_0..p_(k*r+k+r)]; S := tensor(S1,S2); -- Over S we have a generic point of a generic -- line, represented by a row vector, which -- we use to define a map from the base ring -- of X F := map(S,ring X, genericMatrix(S,S_0,1,k+1)* genericMatrix(S,S_(k+1),k+1,r+1) ); -- We now apply F to the ideal of X FX := F X; -- and the condition we want becomes the condition -- that FX vanishes identically in the t_i. -- The following line produces the matrix of -- coefficients of the monomials in the -- variables labelled 0..k: cFX := last coefficients (gens FX, Variables => toList apply(0..k, i -> S_i)); -- We can get rid of the variables t_i -- to ease the computation: cFX = substitute(cFX, S2); -- The ring we want is the quotient S2bar := S2/ideal cFX; -- Now we want to move to the Grassmannian, -- represented by the ring GR -- We define a map sending the variables of GR -- to the minors of the generic matrix in the -- p_i regarded as elements of S1bar gr := map(S2bar,GR, exteriorPower(k+1, genericMatrix(S2bar,S2bar_0,k+1,r+1) ) ); -- and the defining ideal of the Fano variety is ker gr ) o22 = {*Function[stdio:52:34-100:7]*} o22 : FunctionClosure |
With the 0 ideal we get the Grassmannian of projective k-planes in Pr:
i23 : Fano2(ZZ, Ideal) := (k,X) -> ( KK:=coefficientRing ring X; r := (numgens ring X) - 1; -- We can specify a private ring with binomial(r+1,k+1) -- variables as follows GR := KK[Variables => binomial(r+1,k+1)]; -- the work is done by Fano2(k,X,GR) ) o23 = {*Function[stdio:102:27-109:13]*} o23 : FunctionClosure |
i24 : Grassmannian2 = method() o24 = Grassmannian2 o24 : MethodFunction |
i25 : Grassmannian2(ZZ,ZZ,Ring) := (k,r,R) ->( KK := coefficientRing R; RPr := KK[Variables => r+1]; Pr := ideal(0_RPr); Fano2(k,Pr) ) o25 = {*Function[stdio:112:38-116:17]*} o25 : FunctionClosure |
As a first example we can try the Fano of lines on the nonsingular quadric in P3
i26 : Grassmannian2(ZZ,ZZ) := (r,k) -> ( R := ZZ/31991[ vars(0..(binomial(r+1,k+1)-1)) ]; Grassmannian2(k,r,R) ) o26 = {*Function[stdio:118:31-122:27]*} o26 : FunctionClosure |
i27 : KK = ZZ/31991 o27 = KK o27 : QuotientRing |
i28 : R = KK[a,b,c,d] o28 = R o28 : PolynomialRing |
i29 : X = ideal(a*b-c*d) o29 = ideal(a*b - c*d) o29 : Ideal of R |
we investigate by checking its dimension and degree
i30 : I = Fano2(1,X) 2 o30 = ideal (p p , p p , p p + 15995p p - 15995p , p p + p p , p p - 3 4 2 4 1 4 0 5 5 0 4 4 5 2 3 ----------------------------------------------------------------------- 2 15995p p - 15995p , p p , p p - p p , p p , p p - p p , p p + p p , 0 5 5 1 3 0 3 3 5 1 2 0 2 2 5 0 1 1 5 ----------------------------------------------------------------------- 2 2 p - p ) 0 5 o30 : Ideal of KK[p , p , p , p , p , p ] 0 1 2 3 4 5 |
The answer “2” means that I is the ideal of a curve in P5, the ambient space of the Grassmannian of lines.
i31 : dim I o31 = 2 |
The answer is 4. In fact, the ideal I represents the union of two conics.
We now turn to the three surfaces of degree 4 in P5, and make their ideals:
The ring of P5
i32 : degree I o32 = 4 |
i33 : KK = ZZ/31991 o33 = KK o33 : QuotientRing |
It happens that the ideals of all three surfaces are generated by minors of suitable matrices:
The Veronese embedding of P2:
i34 : P5 = KK[a..f] o34 = P5 o34 : PolynomialRing |
i35 : MVero = genericSymmetricMatrix(P5,a,3) o35 = | a b c | | b d e | | c e f | 3 3 o35 : Matrix P5 <--- P5 |
The other scrolls are defined by the minors of matrices that are made from “catalecticant” blocks, that is, from matrices such as
|
which are manufactured by
i36 : Vero = minors(2,MVero) 2 2 o36 = ideal (- b + a*d, - b*c + a*e, - c*d + b*e, - b*c + a*e, - c + a*f, - ----------------------------------------------------------------------- 2 c*e + b*f, - c*d + b*e, - c*e + b*f, - e + d*f) o36 : Ideal of P5 |
for example
i37 : catalecticant = (R,v,m,n) -> map(R^m,n,(i,j)-> R_(i+j+v)) o37 = catalecticant o37 : FunctionClosure |
produces the example above. The rational normal scroll S13, which is the union of lines joining a line with the corresponding points of a twisted cubic in a disjoint subspace of P5
i38 : catalecticant(P5,1,2,4) o38 = | b c d e | | c d e f | 2 4 o38 : Matrix P5 <--- P5 |
i39 : M13 = catalecticant(P5,0,2,1) | catalecticant(P5,2,2,3) o39 = | a c d e | | b d e f | 2 4 o39 : Matrix P5 <--- P5 |
Finally, the rational normal scroll S22, which is made by a similar construction starting with two conics in P5
i40 : S13 = minors(2,M13) 2 o40 = ideal (- b*c + a*d, - b*d + a*e, - d + c*e, - b*e + a*f, - d*e + c*f, ----------------------------------------------------------------------- 2 - e + d*f) o40 : Ideal of P5 |
i41 : M22 = catalecticant(P5,0,2,2) | catalecticant(P5,3,2,2) o41 = | a b d e | | b c e f | 2 4 o41 : Matrix P5 <--- P5 |
It is interesting to note that the numerical invariants of these surfaces are very hard to distinguish. In particular, the graded betti numbers
i42 : S22 = minors(2, M22) 2 o42 = ideal (- b + a*c, - b*d + a*e, - c*d + b*e, - b*e + a*f, - c*e + b*f, ----------------------------------------------------------------------- 2 - e + d*f) o42 : Ideal of P5 |
i43 : Verores = res coker gens Vero 1 6 8 3 o43 = P5 <-- P5 <-- P5 <-- P5 <-- 0 0 1 2 3 4 o43 : ChainComplex |
i44 : S22res = res coker gens S22 1 6 8 3 o44 = P5 <-- P5 <-- P5 <-- P5 <-- 0 0 1 2 3 4 o44 : ChainComplex |
i45 : S13res = res coker gens S13 1 6 8 3 o45 = P5 <-- P5 <-- P5 <-- P5 <-- 0 0 1 2 3 4 o45 : ChainComplex |
i46 : betti Verores 0 1 2 3 o46 = total: 1 6 8 3 0: 1 . . . 1: . 6 8 3 o46 : BettiTally |
i47 : betti S22res 0 1 2 3 o47 = total: 1 6 8 3 0: 1 . . . 1: . 6 8 3 o47 : BettiTally |
coincide, so the three cannot be distinguished on the basis of these or on the basis of the (weaker) invariants the Hilbert series or Hilbert polynomials. But the Fano varieties are more obviously different:
We compute the Fano varieties of lines on each of our surfaces.
i48 : betti S13res 0 1 2 3 o48 = total: 1 6 8 3 0: 1 . . . 1: . 6 8 3 o48 : BettiTally |
i49 : FVero = Fano2(1, Vero) 2 o49 = ideal (p , p p , p p , p p , p p , p p , p p , p p , p p , 14 13 14 12 14 11 14 10 14 9 14 8 14 7 14 6 14 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p p , p , p p , p p , p p , 5 14 4 14 3 14 2 14 1 14 0 14 13 12 13 11 13 10 13 ----------------------------------------------------------------------- p p , p p , p p , p p , p p , p p , p p , p p , p p , p p , 9 13 8 13 7 13 6 13 5 13 4 13 3 13 2 13 1 13 0 13 ----------------------------------------------------------------------- 2 p , p p , p p , p p , p p , p p , p p , p p , p p , p p , 12 11 12 10 12 9 12 8 12 7 12 6 12 5 12 4 12 3 12 ----------------------------------------------------------------------- 2 p p , p p , p p , p , p p , p p , p p , p p , p p , p p , 2 12 1 12 0 12 11 10 11 9 11 8 11 7 11 6 11 5 11 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p , p p , p p , p p , p p , 4 11 3 11 2 11 1 11 0 11 10 9 10 8 10 7 10 6 10 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p p , p , p p , p p , p p , p p , 5 10 4 10 3 10 2 10 1 10 0 10 9 8 9 7 9 6 9 5 9 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p , p p , p p , p p , p p , p p , p p , 4 9 3 9 2 9 1 9 0 9 8 7 8 6 8 5 8 4 8 3 8 2 8 ----------------------------------------------------------------------- 2 2 p p , p p , p , p p , p p , p p , p p , p p , p p , p p , p , p p , 1 8 0 8 7 6 7 5 7 4 7 3 7 2 7 1 7 0 7 6 5 6 ----------------------------------------------------------------------- 2 2 p p , p p , p p , p p , p p , p , p p , p p , p p , p p , p p , p , 4 6 3 6 2 6 1 6 0 6 5 4 5 3 5 2 5 1 5 0 5 4 ----------------------------------------------------------------------- 2 2 2 p p , p p , p p , p p , p , p p , p p , p p , p , p p , p p , p , p p , 3 4 2 4 1 4 0 4 3 2 3 1 3 0 3 2 1 2 0 2 1 0 1 ----------------------------------------------------------------------- 2 p ) 0 o49 : Ideal of KK[p , p , p , p , p , p , p , p , p , p , p , p , p , p , p ] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
The ideal contains all 120 quadrics, and represents the empty set: The Veronese surface contains no lines!
i50 : betti gens FVero 0 1 o50 = total: 1 120 0: 1 . 1: . 120 o50 : BettiTally |
It turns out that the dimension (1) and degree (4) of these varieties coincide! Moreover, since the ideals are not saturated, one cannot directly compare the Hilbert series or free resolutions (of course one could first compute a saturation). But there is the arithmetic genus, that is, 1-H(0), where H is the Hilbert polynomial.
i51 : FS13 = Fano2(1, S13) 2 o51 = ideal (p , p p , p p , p p , p p , p p , p p , p p , p p , 14 13 14 12 14 11 14 10 14 9 14 8 14 7 14 6 14 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p p , p , p p , p p , p p , 5 14 4 14 3 14 2 14 1 14 0 14 13 12 13 11 13 10 13 ----------------------------------------------------------------------- p p , p p , p p , p p , p p , p p , p p , p p , p p , p p , 9 13 8 13 7 13 6 13 5 13 4 13 3 13 2 13 1 13 0 13 ----------------------------------------------------------------------- 2 p , p p , p p , p p , p p , p p , p p , p p , p p , p p , 12 11 12 10 12 9 12 8 12 7 12 6 12 5 12 4 12 3 12 ----------------------------------------------------------------------- p p , p p , p p , p p , p p , p p - p p , p p , p p - 2 12 1 12 0 12 9 11 8 11 7 11 10 11 5 11 4 11 ----------------------------------------------------------------------- 2 p p , p p - p p , p p , p - p p , p p , p p , p p - p p , 6 11 2 11 3 11 0 11 10 6 11 9 10 8 10 7 10 6 11 ----------------------------------------------------------------------- p p - p p , p p , p p - p p , p p - p p , p p - p p , 6 10 3 11 5 10 4 10 3 11 3 10 1 11 2 10 1 11 ----------------------------------------------------------------------- 2 2 p p , p , p p , p p , p p , p p , p p , p p , p p , p p , p p , p , 0 10 9 8 9 7 9 6 9 5 9 4 9 3 9 2 9 1 9 0 9 8 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p p , p p , p p , p - p p , p p - 7 8 6 8 5 8 4 8 3 8 2 8 1 8 0 8 7 6 11 6 7 ----------------------------------------------------------------------- p p , p p , p p - p p , p p - p p , p p - p p , p p - p p , 3 11 5 7 4 7 3 11 3 7 1 11 2 7 1 11 1 7 1 10 ----------------------------------------------------------------------- 2 p p , p - p p , p p , p p - p p , p p - p p , p p - p p , p p , 0 7 6 1 11 5 6 4 6 1 11 3 6 1 10 2 6 1 10 0 6 ----------------------------------------------------------------------- 2 2 p , p p , p p , p p , p p , p p , p - p p , p p - p p , p p - 5 4 5 3 5 2 5 1 5 0 5 4 1 11 3 4 1 10 2 4 ----------------------------------------------------------------------- 2 2 p p , p p - p p , p p , p - p p , p p - p p , p p , p - p p , p p 1 10 1 4 1 6 0 4 3 1 6 2 3 1 6 0 3 2 1 6 1 2 ----------------------------------------------------------------------- - p p , p p , p p ) 1 3 0 2 0 1 o51 : Ideal of KK[p , p , p , p , p , p , p , p , p , p , p , p , p , p , p ] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
The output, 4 P1 - 2 P0, means “four times the Hilbert polynomial of the projective line minus 2”; that is, the polynomial is H(d) = 4d + 2; so arithmetic genus is -1.
i52 : hilbertPolynomial coker gens FS13 o52 = - 2*P + 4*P 0 1 o52 : ProjectiveHilbertPolynomial |
i53 : FS22 = Fano2(1, S22) 2 o53 = ideal (p , p p , p p , p p , p p , p p , p p , p p , p p , 14 13 14 12 14 11 14 10 14 9 14 8 14 7 14 6 14 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p p , p , p p , p p , p p , 5 14 4 14 3 14 2 14 1 14 0 14 13 12 13 11 13 10 13 ----------------------------------------------------------------------- p p , p p , p p , p p , p p , p p , p p , p p , p p , p p , 9 13 8 13 7 13 6 13 5 13 4 13 3 13 2 13 1 13 0 13 ----------------------------------------------------------------------- p p , p p - p p , p p - p p , p p - p p , p p - p p , 9 12 8 12 11 12 7 12 10 12 5 12 10 12 4 12 6 12 ----------------------------------------------------------------------- 2 p p , p p , p p , p - p p , p p - p p , p p , p p - 2 12 1 12 0 12 11 10 12 10 11 6 12 9 11 8 11 ----------------------------------------------------------------------- p p , p p - p p , p p - p p , p p - p p , p p - p p , 10 12 7 11 6 12 6 11 3 12 5 11 6 12 4 11 3 12 ----------------------------------------------------------------------- 2 p p , p p , p p , p - p p , p p , p p - p p , p p - p p , 2 11 1 11 0 11 10 3 12 9 10 8 10 6 12 7 10 3 12 ----------------------------------------------------------------------- 2 p p - p p , p p - p p , p p - p p , p p , p p , p p , p , 6 10 3 11 5 10 3 12 4 10 3 11 2 10 1 10 0 10 9 ----------------------------------------------------------------------- 2 p p , p p , p p , p p , p p , p p , p p , p p , p p , p - p p , p p 8 9 7 9 6 9 5 9 4 9 3 9 2 9 1 9 0 9 8 10 12 7 8 ----------------------------------------------------------------------- - p p , p p - p p , p p - p p , p p - p p , p p - p p , p p , 6 12 6 8 3 12 5 8 6 12 4 8 3 12 3 8 3 11 2 8 ----------------------------------------------------------------------- 2 p p , p p , p - p p , p p - p p , p p - p p , p p - p p , p p 1 8 0 8 7 3 12 6 7 3 11 5 7 3 12 4 7 3 11 3 7 ----------------------------------------------------------------------- 2 - p p , p p , p p , p p , p - p p , p p - p p , p p - p p , 3 10 2 7 1 7 0 7 6 3 10 5 6 3 11 4 6 3 10 ----------------------------------------------------------------------- 2 p p , p p , p p , p - p p , p p - p p , p p - p p , p p , p p , 2 6 1 6 0 6 5 3 12 4 5 3 11 3 5 3 10 2 5 1 5 ----------------------------------------------------------------------- 2 2 p p , p - p p , p p - p p , p p , p p , p p , p p , p p , p p , p , 0 5 4 3 10 3 4 3 6 2 4 1 4 0 4 2 3 1 3 0 3 2 ----------------------------------------------------------------------- 2 2 p p , p p , p , p p , p ) 1 2 0 2 1 0 1 0 o53 : Ideal of KK[p , p , p , p , p , p , p , p , p , p , p , p , p , p , p ] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
The output, 4 P1 - 3 P0, means H(d) = 4d + 1, arithmetic genus -1. In fact, the Fano variety of S22 consists of a projective line, embedded as a smooth rational quartic; while the Fano variety of S13 consists of a smooth rational quartic (corresponding to the rulings of the surface S13 and an isolated point, corresponding to the section of negative self-intersection on the surface).