Rationality proofs by curve counting

We propose an approach for showing rationality of an algebraic variety $X$. We try to cover $X$ by rational curves of certain type and count how many curves pass through a generic point. If the answer is $1$, then we can sometimes reduce the question of rationality of $X$ to the question of rationality of a closed subvariety of $X$. This approach is applied to the case of the so-called Ueno-Campana manifolds. Our experiments indicate that the previously open cases $X_{4,6}$ and $X_{5,6}$ are both rational. However, this result is not rigorously justified and depends on a heuristic argument and a Monte Carlo type computer simulation. In an unexpected twist, existence of lattices $D_6$, $E_8$ and $\Lambda_{10}$ turns out to be crucial.


Introduction
In November 2014 F. Catanese gave a talk at ICTP, Trieste about Ueno-Campana varieties. In particular he spoke about the following open problem. Let E be the elliptic curve over C with complex multiplication by 1+ √ −3 2 or the curve with complex multiplication by √ −1. Let Γ ≃ Z/cZ be the group of automorphisms of E or its subgroup with c ≥ 3. So we have 1 c = 3, c = 4 or c = 6 and c determines E uniquely. Let X n,c = E n /Γ, the quotient of E n by the diagonal action of Γ. It is well-known that E n /Γ is rational for n = 1, 2. Ueno first studied these varieties in [Uen75] and showed that E n /Γ cannot be rational for n ≥ c = |Γ|. Campana asked ( [Cam11]) Problem. For which c, n is X n,c rational?
An introduction to the problem and the state of the art is given in [COV15]. In particular, unirationality of X 3,4 was proved in [COT13]. Then rationality of X 3,4 was proved in [Col15]. Rationality of X 3,6 was proved in [OT13]. Then [COV15] E-mail address: mellit@gmail.com. 1 We include the case c = 3 for completeness and because it helps to illustrate our techniques. established unirationality of X 4,6 . Rationality of X 4,6 and unirationality of X 5,6 are still open.
In this paper we give evidence towards rationality of X 4,6 and X 5,6 . Below we will explain certain curve counting problem. We could only solve this problem by a certain computer-based heuristic approach and our answer is not rigorously justified. So we formulate results of these computations as Conjectures 1, 2, 3.
Theorem. If Conjecture 1 is true, then X 4,6 is rational.
Theorem. If Conjecture 2 is true, then X 5,6 is unirational. If moreover Conjecture 3 is true and X 4,6 is rational, then X 5,6 is rational.
As a summary of all the known results we conclude: Corollary. Suppose Conjectures 1, 2, 3 are true. Let E be an elliptic curve over C and let Γ be a subgroup of the automorphism group of E. Let n be an integer such that 0 < n < |Γ|. Then the quotient of E n by the diagonal action of Γ is rational.
It would be interesting to try to apply our methods to some other abelian varieties or other group actions.
Hopefully, the corresponding curve counting can be achieved by some clever enumerative geometry techniques. This would turn our "heuristic proofs" into real proofs.

The main idea
Mori program teaches us that birational properties of varieties are very much controlled by rational curves on them. Let us try to be not too precise and make a guess, how existence of curves (or rather families of curves) would prove rationality of X 5,6 for us? It would be a good situation if some family of rational curves {C s } s∈S existed such that the base S is rational and such that exactly one curve passes through a generic point of X 5,6 . It turns out that just having the latter property is enough for establishing unirationality of X 5,6 . To see this, consider an embedding ι : X 4,6 ֒→ X 5,6 . If through a generic point of the image of ι we have exactly one curve from our family we are done, because then the curves can be parametrized by X 4,6 , so we obtain a dominant rational map X 4,6 → S and unirationality of X 5,6 as a consequence. Now notice that the union of images of all embeddings X 4,6 ֒→ X 5,6 is Zariski dense in X 5,6 , so ι with the required property exists. A more careful analysis leads to the following general Lemma 2.1. Let X be an irreducible algebraic variety of dimension n over C, and let C = {C s } s∈S be an algebraic family of rational curves in X. Suppose for a generic point x ∈ X there is exactly one curve from C containing x. Let Z ⊂ X be an irreducible closed subvariety of dimension n − 1 such that for a generic point x ∈ Z there is exactly one curve from C containing x. Then the following holds: If moreover Z is rational and there exists open V ⊂ Z such that any curve from C intersects V in no more than one point, then X is rational.
Proof. Denote the total space of the family of curves also by C. It comes with maps π : C → S and f : C → X. Let L be the locus of points x ∈ X such that there is exactly one curve from C containing x. This is a constructible algebraic subset of X.
Let s : U → S be the algebraic map which sends a point x ∈ U to the unique s(x) ∈ S such that x ∈ C s(x) . The pullback s * C of the original family of curves to U ∩ Z has a natural section: for any x ∈ U ∩ Z the curve C s(x) contains x. Therefore over a non-empty open subset W ⊂ U ∩ Z this family is trivial. We obtain a map If Z is unirational, then W is unirational. Hence W ×P 1 is unirational. The image of f ′ is irreducible and contains W . Thus it is either contained in W = Z, or has dimension n. The former is not possible because curves from C are not contained in Z. Thus the image of f ′ has dimension n. Therefore f ′ is dominant and X is unirational. The first statement is proved.
If Z is rational, then W , and hence also W ×P 1 is rational. So it is enough to show that a generic point of X has not more than 1 preimage under f ′ . Without loss of generality we assume W ⊂ V . Suppose x ∈ U has at least 2 preimages. This means there are (v 1 , t 1 ), (v 2 , t 2 ) ∈ W × P 1 that go to x. Since there is exactly one curve from C passing through x, and that curve can intersect W in at most one point, we obtain v 1 = v 2 . On the other hand, for each v ∈ W there is at most finitely many values of t such that there exist t ′ such that f ′ (v, t) = f ′ (v, t ′ ). So the dimension of such pairs (v, t) is at most n − 1, and therefore the dimension of the space of such x is also at most n − 1. So a generic x has no more than 1 preimage.
2.1. Counting curves on a computer. The families of curves we will be dealing with are such that one can write down explicitly a system of equations whose solutions correspond to curves passing through a given point. So we can implement the following strategy. Pick a big prime number, for instance p = 1000003 or p = 1000033. We will work over F = GF(p). Generate a random point x ∈ X(F ). Compute the number of curves passing through x by counting solutions overF of the corresponding system of equations by the standard Gröbner basis techniques 2 . If this number is k, p is large and x is "sufficiently random", then we expect x to behave like a generic point, so the number of curves for a generic point over the complex numbers should also be k.

Rational curves
There are exactly 3 pairs E, Γ where E is an elliptic curve over C and Γ is a subgroup of the group of automorphisms of E with |Γ| > 2. Consider an elliptic curve E of the form x 2 − y 3 = z 6 in P(3, 2, 1) or x 2 − y 4 = z 4 in P(2, 1, 1) or x 3 − y 3 = z 3 in P(1, 1, 1) = P 2 . The equation of the curve in all cases is x a − y b = z c in P( c a , c b , 1). We choose (1, 1, 0) as the zero point on E. There are gcd(a, b) points with z = 0, which we call "points at infinity". Let ζ be a primitive root of unity of order c. The group Γ of the roots of unity of order c acts on E by ζ(x, y, z) = (x, y, ζz).
We construct rational curves in E n /Γ as follows. Let k ≥ 1 be an integer, and let R(t, u) be a homogeneous polynomial of degree ck. For each i = 1, 2, . . . , n let P i (u, v), Q i (u, v) be relatively prime homogeneous polynomials of degrees kc a , kc b respectively satisfying LetC be the curve given by equation R(u, v) = w c in P(1, 1, k). The group Γ acts onC by ζ(u, v, c) = (u, v, ζc) andC/Γ = P 1 . For each i we have a Γ-equivariant Quotienting out by Γ we obtain a commutative diagram 3.1. Discrete invariants. To every such curve we associate discrete invariants as follows. For each i = j we have Thus we have Using the assumption that P i and Q i are relatively prime and considering contribution of an arbitrary linear form in u, v to various M l,r i,j we establish the following: Note that gcd(G l,r i,j , G l ′ ,r ′ i,j ) = 1 whenever l = l ′ and r = r ′ because otherwise all the 4 polynomials P i , P j , Q i , Q j have a common divisor.

Cohomology classes.
It is useful to match the discrete invariants M to the homology classes of the strict pullbacks of our curves in H 2 (Ẽ n /Γ, Z), whereẼ n /Γ is the blowup of E n /Γ in the fixed points of Γ. It is possible to describe this homology group explicitly, but we will not do this. Instead we will think of the homology class of a rational curve as above consisting of two pieces of data: (i) The homology class ofC in H 2 (E n , Z).
(ii) For each Γ-fixed point x ∈ E n the intersection number of the strict pullback ofC to the blowup of E n in x with the exceptional divisor. This, roughly speaking, counts how may points onC go to x.
Furthermore, the homology class ofC in H 2 (E n , Z) can be specified by the following data.
Proposition 3.1. For each curveC ⊂ E n there exists a unique n × n Hermitian matrix where D v is the divisor class given by the pullback of 0 ∈ E to E n via the map We have Since this holds for all v, v ′ the map H(C) must be Q[ζ]-linear. So it can be represented by a matrix with entries in Q[ζ], and that matrix must be Hermitian because the form B was symmetric.
It is clear that the diagonal entries of H(C) are simply the degrees of the components f i of f , f i :C → E. Let us calculate the degree of these components for our construction. Consider the function This is a rational function of degree b on E because for a generic t ∈ C there are exactly b solutions to x z c a = t corresponding to the b-th roots of t a − 1. Its pullback toC is the function A recipe to calculate the off-diagonal entries from the matrices M i,j will be given in the next section on a case-by-case basis.
3.3. Calculating k. Finally, we calculate the value of k as a function of n for which we expect to have finite number of our curves passing through a generic point of E n . The first coefficient of R(u, v) can be normalized to 1, and we have kc remaining coefficients. A generic point is given by pairs x i , y i satisfying x a i −y b i = 1, and we can parametrize our curve so that the point (u, v, w) = (1, 0, 1) goes to (x i , y i , 1). This fixes the first coefficient of P i and Q i . Then the condition for a polynomial R to be of the form . We want this number to be equal to 2 because there is a 2-dimensional group of translations an rotations acting on solutions that needs to be gauged out. Thus we have ).
• Calculate k = 2 c−n . Suppose it is an integer 3 . • For each n × n matrix H list the degrees M l,r i,j . • For a point p = (p 1 , p 2 , . . . , p n ) ∈ E n , p i = (x i , y i , 1) try to compute how many curves with discrete invariants M l,r i,j pass through p. A curve is determined by a sequence of homogeneous polynomials G l,r i,j (u, v) with first coefficient 1 of degrees M l,r i,j . These polynomials must satisfy gcd(G l,r i,j , G l ′ ,r ′ i,j ) = 1 whenever l = l ′ and r = r ′ , and the equations obtained by elimination of P 1 , . . . , P n and Q 1 , . . . , Q n from the following (i, j = 1, . . . , n, l = 0, . . . , a−1, 3 The only cases with n > 1 when this number is not an integer are (a, b, c) = (2, 3, 6) with n = 2, 3.
In these cases the method can still be applied. The curveC should pass through Γ-fixed points of orders different from 6, which implies a slightly different general shape of the equations (1). We do not include these situations here because it would complicate the notations, and because these cases are already known to be rational anyway. r = 0, . . . , b − 1) system of main equations: • If we are lucky and the answer to the previous step is 1 for a generic point p, then try to construct a vector v ∈ Z[ζ] n such that for a generic point p ∈ D v the number of curves is also one, and the number of intersection points of D v /Γ ∩ C outside the set of fixed points of Γ is at most 1.
In this case the group Γ has order c = 3, so we have only one case n = 2, k = 2. The discrete invariant has the form of a matrix of non-negative integers with all the row and column sums equal 2. To calculate the 2 × 2 matrix H(C) we already know that the diagonal entries are 2. Let One can relate h 1,2 to M by the following. Let ∆ ⊂ E × E be the diagonal. Then we have It turns out, that E 0,0 , E 1,2 , E 2,1 do not intersect. Therefore they have the same homology class. So, by counting the intersection points Analogously, E 0,1 ·C ≥ M 0,1 + M 1,0 + M 2,2 + 2, where D is the divisor at infinity of E, which has degree 3. Thus we obtain Hence the inequalities are equalities.
So we have shown that the curve is unique provided This means we have to remove the divisors given by vectors (1, ±ζ i ), (0, 1), (1, 0). Taking any other divisor class we will satisfy conditions for part (i) of Lemma 2.1.
To show rationality we need to satisfy the assumptions of part (ii). So we need a divisor with small intersection number withC, i.e. a vector not of the form (±ζ i , 0), (0, ±ζ i ), (±ζ i , ±ζ j ) whose length is small with respect to the form H.
Take v = (1, 2 + ζ), which corresponds to the divisor D v consisting of (p 1 , p 2 ) ∈ E 2 such that p 1 + 2p 2 + ζp 2 = 0. We have v * Hv = 5. So there is at most 5 points of intersection in D v ∩C. Going down to E 2 /Γ we obtain at most ⌊ 5 3 ⌋ = 1 of points of intersection (D v /Γ) ∩ C satisfying z i = 0. Clearly, D v /Γ is rational. So the conditions of Lemma 2.1 are satisfied.

H 4 curves.
In this case computer experiments showed that there are 3 curves passing through a generic point for each of the last 3 matrices M. However these curves can be distinguished by their incidence information with the Γ-fixed points, so probably it is possible to use these curves for an alternative rationality proof. 4.3. Total curve count. In total we obtain 3 curves for H 3 and 9 curves for H 4 . However, these curves can be distinguished by our discrete invariants and by their intersections with z 1 = z 2 = 0.
Proof. We have curves E l,r ⊂ E × E given by equations (l = 0, 1, r = 0, 1, 2, 3) The pairs representing the same homology class are listed as follows (E 0,0 , E 1,2 ), (E 0,1 , E 1,3 ), (E 0,2 , E 1,0 ), (E 0,3 , E 1,1 ). So This is because each root of gcd(P 1 − P 2 , Q 1 − Q 2 ) has multiplicity 4 in E 0,0 ·C, and there are further kc = 4k points with w = 0 onC which map to the points with z 1 = z 2 = 0. Producing similar inequality for E 1,0 and adding to the one above we obtain On the other hand, E 1,0 + E 0,0 is equivalent to E × D + D × E, where D is the divisor at infinity consisting of 2 points. So the intersection equals 8k. Therefore our inequalities must be equalities.
This allows us to compute h i,j as a function of the entries of M i,j . The diagonal corresponds to the vector e i − e j , so we have The vector e i − ζe j corresponds to the curve E 1,3 , so the corresponding intersection number is We obtain Im Thus 5.1. The case k = 1, n = 2. There are two H-matrices (up to automorphisms) for n = 2, k = 1, of determinants 2 and 4: For H 2 there is only 1 matrix M: M = 1 1 0 0 0 0 1 1 .  for (a, b, c) = (2, 4, 4), n = 3, k = 2. Counting curves on a computer produces Table 1.
The H-matrices with 0 curves are the following matrices with determinants 8 resp. 16:  It turns out that non-existence of these curves is explained by the fact that the matrices can be conjugated to which contain forbidden off-diagonal entries ζ + 3. Note that for each matrix H there are several triples of matrices M i,j . The table was obtained by adding the point counts for all triples. In some situations the total number of curves can be greater than 1, but for some individual triples M i,j the number is 1. We will work with the matrix of determinant 16 which gives 1 curve. The matrix is The matrices M i,j are as follows: M 1,2 = 0 1 2 1 2 1 0 1 , M 1,3 = 1 2 1 0 1 0 1 2 , M 2,3 = 0 1 2 1 2 1 0 1 .
Computer experiments show that exactly one curve passes through a generic point of E 3 . To apply Lemma 2.1 in full generality we need to choose a divisor. So we look for a vector v whose H-norm v * Hv is small, but not too small. All vectors of norm 4 do not produce good divisors: through a generic point of such divisor there are no curves of our type. There are no vectors of norms between 4 and 8. There are 252 vectors of norm 8. Let Aut(H) be the group of matrices g ∈ GL 3 (Z[ζ]) such that g * Hg = H. The vectors of norm 8 form 3 Aut(H)-orbits. Some of these vectors are also such that through a generic point of the corresponding divisor there are no curves. In the orbit of v = (1, −2, 1), which consists of 192 vectors, for 168 vectors 4 the curve count is 1 and for the remaining 24 it is zero. This vector produces a divisor D v /Γ satisfying the conditions of Lemma 2.1. We have D v ·C = 8, so if we show that at least one intersection point is at infinity, we obtain the number of finite intersection points of D v /Γ with C is at most ⌊ 7 4 ⌋ = 1. The points at infinity of D v are 4 points out of the total 2 3 = 8 points at infinity on E 3 . These are the points (p 1 , p 2 , p 3 ) satisfying p 1 − 2p 2 + p 3 = 0. The points at infinity are of order 2, so this condition is equivalent to p 1 = p 3 . Let C ′ be the projection ofC ⊂ E × E × E to E × E using coordinates 1, 3. So it is enough to show that C ′ intersects ∆ at infinity. The intersection number is 8, but there are only 4 finite intersection points because M 0,0 1,3 = 1. Thus there must be intersections at infinity.
6. Examples for (a, b, c) = (2, 3, 6) Finally, we turn to the most interesting example, which includes open cases. We have (a, b, c) = (2, 3, 6) and n = 4 or n = 5. Here ζ = e 2πi 6 . For n = 4 we obtain k = 1. For n = 5 we obtain k = 2. The matrices M i,j are 2 × 3 with column sums 2k and row sums 3k. The matrices H have 6k on the diagonal. Some things are simpler because there is only 1 point at infinity, and the correspondence between M-matrices and the off-diagonal entries of the H-matrix are bijective.
On the other hand, the divisor of the rational function y 1 where D is the point at infinity. Therefore Therefore our inequalities must be equalities.
This allows us to compute h i,j as a function of the entries of M i,j . The diagonal corresponds to the vector e i − e j , so we have The vector e i − ζe j corresponds to the curve E 1,2 , so the corresponding intersection number is 12k − ζh i,j −ζh i,j = 6k + 6M 1,2 i,j . So we can recover h i,j : 6.1. The case k = 1, n = 4. The diagonal entries of H are 6 and the possible off-diagonal entries are in the set We classified all matrices H up to GL 4 (Z[ζ])-equivalence satisfying the following conditions: It turns out there are 5 matrices with determinants 144, 432, 576, 864, 1296: Note that the off-diagonal values 0 resp. 2ζ − 4 correspond to M i,j = 1 1 1 1 1 1 , M i,j = 2 1 0 0 1 2 . The curve counts are given in Table 2. It is not clear why curves corresponding to H 864 do not pass through generic points. We turn our attention to the matrix H = H 144 , which already implies unirationality of X 4,6 and will also imply rationality if we find a "good" divisor class. det (H) 144 432 576 864 1296  #C  1  6  12  0  72  Table 2. Curve counts for (a, b, c) = (2, 3, 6), n = 4, k = 1.

The group
Aut(H) = {g ∈ GL 4 (Z(ζ)) | g * Hg = H} has order 155520 and acts transitively on the 240 vectors of H-norm 6 and on the 2160 vectors of H-norm 12. Vectors of norm 6 intersect C only at infinity, so we pick a vector of norm 12. Some of the vectors of norm 12 correspond to the "diagonals", for instance v = (1, 0, 1, 0). For this vector we obtained 0 curves. However picking v = (0, 1, 2, 1), and any other vector not of the form (0, 0, ζ i , ζ j ) for some i, j or a permutation of such, we obtain 1 curve. Note that for any v of norm 12 and any curve C of our kind the number of intersection points #(C ∩ D v /Γ) outside of the Γ-fixed points is at most 1. This is true becauseC · D v = 12, and the intersection D v ∩ C contains at least one point at infinity.
We verified this conjecture by testing the statement on 10000 random points on D 0,1,2,1 and 10000 random points on E 4 over the field GF 1000003 . Only 1 point got "unlucky" and the number of curves was 0. For every other point the number was 1. Counting the curves took ≈ 0.05 seconds per point on an ordinary laptop.
We could not classify all such matrices H up to GL 5 (Z[ζ])-equivalence because the set of possibilities is too big. However the following matrix seems to be the For such a vector v we have D v ·C = 18, and at least 12 points of intersection are at infinity. Therefore |D v /Γ ∩ C| ≤ 1. Some vectors represent "generalized diagonals", for instance (0, 0, 1, 0, −ζ). We found that our curves do not pass through generic points on the corresponding divisors. Taking any vector different from those do seem to produce good divisors, for instance we take v = (1, 0, ζ, 0, −1).
The following conjecture implies unirationality of X 5,6 by part (i), Lemma 2.1: Conjecture 2. For x ∈ E 5 denote by #(x) the number of curvesC of our type corresponding to the matrix H 13824 and containing x. Then for a generic point x ∈ E 5 we have #(x) = 1.
6.3. Computations for n = 5. The computations in these cases take much more time than in the n = 4 case. It can probably be explained by the fact that the set of divisors where the number of curves is not 1 is huge: for instance, it must contain all the 336 divisors D v corresponding to vectors v of H-norm 12. Another issue is that when we create the ideal parametrizing our curves, we have besides equations also inequalities of the form Each inequality is imposed by adding an extra variable J i and an extra equation Note that the degrees of P i and Q i are 6 and 4 respectively, so the resultant has degree 24 and these extra equations are very long. On the other hand, when we tried to keep only the equations without the inequalities the length of the scheme of solutions grew up to 99. The scheme turned out to contain a single isolated point and several very fat points failing the conditions gcd(P i , Q i ) = 1. The computation with the inequalities (3) takes ≈ 1 hour 15 minutes on an ordinary laptop (for each random point on E 5 ). It turns out, it is better to extend the set of inequalities that translate to equations (4) by a much larger set of 32 inequalities (5) resultant (G l,r i,j , G l ′ ,r ′ i,j ) = 0 (1 ≤ i < j ≤ 5, 0 ≤ l < l ′ ≤ 1, 0 ≤ r, r ′ ≤ 2 : r = r ′ , M l,r i,j = 0, M l ′ ,r ′ i,j = 0). .
For each inequality we have to create a new variable and a new equation as in (4). These inequalities formally follow from (3) as explained in Section 3.1, but their degrees are much smaller. On the other hand, inequalities 5 do not seem to imply 3. Thus we must additionally test that every solution we find satisfies 3. It turns out, that it is faster to build the ideal step-by step. On each step we add some new equations and recompute the Gröbner basis. In the very beginning we choose a cell of the cell decomposition of the weighted projective space we do computations in. The total number of variables is 120 (we have 10 pairs 1 ≤ i < j ≤ 5 and for each pair i, j we have 6 polynomials G l,r i,j whose degrees are given by the entries of M i,j ). Among these variables 42 have weight 1, 38 have weight 2, 22 have weight 3 and 18 have weight 4. We order the variables by weight, and if the weights agree by the degree of the polynomial they are coefficients of. Because we should consider the solutions up to translation, we set the very first variable to 0. The choice of a cell in the weighted projective space means we set the first r variables to 0, the r + 1-st variable to 1. We need to do this for every r, 1 ≤ r ≤ 119. Then we have 3 steps (for each r): (i) Add equations coming from elimination of P i , Q i from the main equations (2). (ii) Add variables and equations representing (5). (iii) Add variables and equations representing (3).
Then we compute the dimension over the base field of the quotient ring with respect to the ideal obtained in the final step. This number divided by the weight of the variable we made equal to 1 is the number of points in the given cell. If after some step we obtain ideal (1), this means there is no solutions in a given cell, so we abort and pass to the next cell, i.e. next value of r. In all situations we encountered, all the solutions belonged to the biggest cell.
Complete computation for each point p ∈ E 5 (GF 1000003 ) took ≈ 6 minutes on an ordinary laptop. We made 10 trials for each of the Conjectures 2, 3 and obtained exactly 1 curve in all cases.
Remark 6.1. The quadratic form induced by H 13824 on the rank 10 lattice Z[ζ] 5 is proportional to the so-called laminated lattice Λ 10 , see [CS93]. We discovered this fact with the help of OEIS ( [Oei], sequence A006909) by searching for the sequence of numbers of vectors of given norm, which begins as follows: 1, 0, 336, 768. In fact, the matrix H 144 from Section 6.1 in a similar way corresponds to the lattice E 8 . The matrix H 16 from Section 5.2 corresponds to the lattice D 6 .