Systems of cubic forms in many variables
Simon L. Rydin Myerson

TL;DR
This paper proves an asymptotic formula for counting integer solutions to systems of smooth cubic forms in many variables, significantly reducing the variable count needed compared to previous results.
Contribution
It establishes an asymptotic count for solutions when the number of variables is at least 25 times the number of forms, improving upon the prior requirement of much larger variable counts.
Findings
Asymptotic formula for solutions in systems with n ≥ 25R variables.
Reduction of variable count needed from R^2 to linear in R.
Application of Davenport's method to bound solutions of auxiliary inequalities.
Abstract
We consider a system of cubic forms in variables, with integer coefficients, which define a smooth complete intersection in projective space. Provided , we prove an asymptotic formula for the number of integer points in an expanding box at which these forms simultaneously vanish. In particular we can handle systems of forms in variables, previous work having required that . One conjectures that should be sufficient. We reduce the problem to an upper bound for the number of solutions to a certain auxiliary inequality. To prove this bound we adapt a method of Davenport.
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Taxonomy
TopicsLimits and Structures in Graph Theory · Analytic Number Theory Research · Algebraic Geometry and Number Theory
Systems of cubic forms in many variables
S. L. Rydin Myerson
Abstract
We consider a system of cubic forms in variables, with integer coefficients, which define a smooth complete intersection in projective space. Provided , we prove an asymptotic formula for the number of integer points in an expanding box at which these forms simultaneously vanish. In particular we can handle systems of forms in variables, previous work having required that . One conjectures that should be sufficient. We reduce the problem to an upper bound for the number of solutions to a certain auxiliary inequality. To prove this bound we adapt a method of Davenport.
Contents
1 Introduction
1.1 Main result
Let be homogeneous cubic forms in variables , with integer coefficients. We treat the simultaneous Diophantine equations
[TABLE]
and the corresponding projective variety in which we call . We assume throughout that the generate the ideal of and are linearly independent. The cubic case of a classic result of Birch gives us:
Theorem 1.1** (Birch [2]).**
Let be a box in contained in the box and having sides of length at most 1 which are parallel to the coordinate axes. For each write
[TABLE]
If the variety is a smooth complete intersection, and we have
[TABLE]
then for some and all the bound
[TABLE]
holds, where the implicit constant depends only on the forms and is a positive real number depending only on and . If the variety has a smooth point over for each prime and a real point whose homogeneous co-ordinates lie in then and are positive.
In particular this follows from Theorem 1 of Birch [2], on inserting the bound for the dimension of the variety occurring in that result. This bound follows from Lemma 3.1 of Browning and Heath-Brown [4] whenever is a smooth complete intersection. See [17, Lemma LABEL:1.lem:nonsing_case] for details.
We sharpen (1.1) as soon as . In §1.3 we prove:
Theorem 1.2**.**
In Theorem 1.1 we may replace (1.1) with the condition
[TABLE]
For example when and is a smooth complete intersection, Theorem 1.2 applies when whereas Birch’s theorem requires .
The “square-root cancellation” heuristic suggests that in place of (1.1) the condition should suffice, see for example the discussion around formula (1.5) in Browning [3]. By handling systems of forms in variables we come within a constant factor of this conjecture.
Our strategy is an extension of our previous work [17]. In forthcoming papers we further generalise this approach to treat systems of forms with degree with rational or real coefficients.
1.2 Related work
We begin with the case when the forms are diagonal.
In the case of a single diagonal form Baker [1] proves that has a rational point whenever .
Brüdern and Wooley [8, 7, 11] treat diagonal systems in variables. This is the best value of possible with the classical circle method. In particular they prove the Hasse principle for whenever the are diagonal, is smooth and . They also prove an asymptotic formula of the type (1.2) whenever holds, or when and holds [5, 6, 9]. In the case they prove a Hasse principle for certain pairs of diagonal cubics in as few as 11 variables [10].
Returning to the case of general (not necessarily diagonal) forms, we consider the case . Let be a cubic form. Hooley [16] proves that if the variety is smooth, and the box does not contain a point at which the Hessian determinant of vanishes, then the asymptotic formula (1.2) holds. In this work he assumes a Riemann hypothesis for a certain modified Hasse-Weil -function. When he proves the same result unconditionally, the weaker error term in place of the from (1.2). Heath-Brown [15] proves that if then always has a rational point, regardless of whether it is singular.
In the case Dietmann and Wooley [14] have shown that always has a rational point when whether or not it is smooth.
In the general case Schmidt [18] shows that always has a rational point if . Recent work of Dietmann [13] improves this condition to .
1.3 Reduction to an auxiliary inequality
To prove Theorem 1.2 we will use Theorem LABEL:1.thm:manin from the author’s previous work [17]. This will reduce the problem to proving an upper bound for the number of solutions to the following auxiliary inequality.
Definition 1.1**.**
For any and we write for the supremum norm. When is a real cubic form in variables with real coefficients, we define a symmetric matrix
[TABLE]
where \lVert c\rVert_{\infty}=\frac{1}{6}\max_{i,j,k\in\{1,\dotsc,n\}}\big{\lvert}\frac{\partial^{3}c(\bm{x})}{\partial x_{i}x_{j}x_{k}}\big{\rvert}. Thus is the Hessian of the cubic form which has been normalised so that 1 is the absolute value of its largest coefficient. For each we put for the number of pairs of vectors with
[TABLE]
We show that this definition of above agrees with the one given in [17, Definition LABEL:1.def:aux_ineq]. There we consider a degree polynomial and a system of multilinear forms and when and we see that
[TABLE]
It follows that the definition above agrees with Definition LABEL:1.def:aux_ineq from [17]. The case of Theorem LABEL:1.thm:manin in [17] therefore states that:
Theorem 1.3**.**
Let the counting function be as in Theorem 1.1. Suppose that for some and we have
[TABLE]
for all and where we write for . Then for some we have
[TABLE]
for all where the implicit constant depends at most on and the and the positive constant depends at most on and . The constants and are positive under the same conditions as in Theorem 1.1.
We give the following bound for the counting function . The proof occupies the bulk of this paper and is completed in §6.
Proposition 1.1**.**
We call a set of real cubic forms in variables a closed cone if (i) for all and we have and (ii) is closed in the real linear space of cubic forms in variables.
Let be a closed cone as above, and let be as in Definition 1.1. If we set
[TABLE]
so that then for all and we have
[TABLE]
We will outline the proof after deducing Theorem 1.2.
Proof of Theorem 1.2.
Suppose that (1.3) holds. We claim that for all and we have
[TABLE]
where is as in Theorem 1.3. If we set and let be sufficiently large in terms of the forms we can then apply Theorem 1.3. For (1.8) implies (1.5) on setting in (1.8). Moreover we have by (1.3). So the hypotheses of Theorem 1.3 are satisfied, and Theorem 1.2 follows.
Setting in Proposition 1.1, we see that (1.8) follows from (1.7) unless holds. Suppose for a contradiction that we have .
By the definition (1.6) there must be with
[TABLE]
We may assume that holds, after permuting the if necessary. Since is a smooth complete intersection, we then have
[TABLE]
and so holds. Thus (1.9) implies that is singular, which is false by assumption. ∎
1.4 Outline of remaining steps
To prove Proposition 1.1 we adapt the argument used to prove Lemma 3 in Davenport [12], and subsequently a somewhat more general result in §5 of Schmidt [18]. These authors consider the counting function defined by
[TABLE]
for some cubic form with integer coefficients. Davenport proves that either is small, or there is a large rational linear space on which vanishes. In order to briefly sketch his line of reasoning, we define some additional notation.
Definition 1.2**.**
Let and let be the eigenvalues of the real symmetric matrix listed with multiplicity and in order of decreasing absolute value. Observe that
[TABLE]
For each let be the vector of all minors of arranged in some order. This is a vector of degree homogeneous forms in the variables with real coefficients. Let be the Jacobian matrix .
Davenport’s argument runs as follows.
- (1)
Let . Suppose that for some sufficiently large implicit constant. The contribution to this count from any one vector is at most . So there must be an integer in the set such that at least integer points satisfy both and . 2. (2)
If are as in 1, then it follows that there is an integer point such that holds and the tangent space to the affine variety at the point has dimension or more. Equivalently, and both hold. This follows from Lemma 2 of Davenport [12]. 3. (3)
If there exists a vector as in 2, then it follows that there exist linear subspaces of with dimensions and respectively, such that for all and the equality holds. See Lemma 4 in Schmidt [18] or the proof of Lemma 3 in Davenport [12]. 4. (4)
We conclude that if then there are spaces as in 3. So the space defined by is a rational linear space, with dimension at least such that for all the equality holds.
Our setting differs in three ways from that of Schmidt and Davenport. First, we consider the inequality rather than the equation . Second, for us the cubic form may have real coefficients. And third, rather than concluding that has a rational linear space of zeroes, we seek to show that the variety is very singular.
1.5 Structure of this paper
In §2 and the three sections §§4-6 we will modify each of the four steps 1-4 above to accommodate the three changes described at the end of §1.4. In the remaining section, §3, we prove some technical lemmas relating the minors and eigenvalues of real matrices.
1.6 Notation
Throughout, we let and be as in Definition 1.1, and we let and be as in Definition 1.2. We do not require algebraic varieties to be irreducible, and we adopt the convention that . We use Vinogradov’s notation and big- notation in the usual way.
2 The eigenvalues of the Hessian matrix
We show that if the counting function from Definition 1.1 is large, then there are many integer points for which the eigenvalues of lie in some fixed dyadic ranges. This corresponds to step 1 from §1.4.
Lemma 2.1**.**
Let be a real symmetric matrix and let be the eigenvalues of the matrix listed with multiplicity and in order of decreasing absolute value. Let and and suppose that holds. Set
[TABLE]
Then we have
[TABLE]
Proof.
The integral vectors counted by are all contained in the box and in the ellipsoid
[TABLE]
which has principal radii . Hence
[TABLE]
It follows that
[TABLE]
Since the inequalities hold, we deduce that
[TABLE]
Definition 2.1**.**
Suppose that and that such that the inequalities hold. Then we define to be the set of all vectors in satisfying the following conditions: the inequality holds, and we have
[TABLE]
whenever holds, and we have
[TABLE]
whenever holds.
Corollary 2.1**.**
Let be as in Definition 1.1, let and be as in Definition 1.2, and let be as in Definition 2.1. For any one of the following alternatives holds. Either
[TABLE]
or there is and there are such that the inequalities hold and
[TABLE]
or there are satisfying and
[TABLE]
Proof.
Note that in the case that there are no values of satisfying so the last condition in the definition of is vacuously true and can be omitted. In particular, if then (2.3) follows from (2.2), because
[TABLE]
So it is enough to prove that either (2.1) holds or there exist integers and satisfying the inequalities and such that (2.2) holds.
Now the set together with the sets partition the box into disjoint pieces. So, if we let
[TABLE]
then we have
[TABLE]
The total number of terms on the right-hand side of (2.4) is at most, so it follows that either
[TABLE]
holds, or else there are and such that
[TABLE]
If (2.5) holds then the trivial bound implies (2.1). Suppose instead that (2.6) holds.
By (1.10), for each real vector the bound holds. So we may apply Lemma 2.1 with the choice and some depending on only. This shows that
[TABLE]
Substituting this into (2.6) we see that (2.2) holds, as claimed. ∎
3 Intermission: Eigenvalues and minors
Here we collect some elementary facts about the eigenvalues and minors of real matrices which will be needed in §§4-5.
Lemma 3.1**.**
For each let
[TABLE]
This set has members. For each such that and each real matrix define an real matrix by
[TABLE]
so that the are the minors of . For all matrices all matrices and all we have . That is, we have
[TABLE]
Proof.
Let be the standard basis of . Fix ; then each side of (3.1) is an alternating multilinear form in those columns of whose indices appear in the vector . This is some -tuple of -vectors.
If one is given the value of an alternating multilinear form at the -tuple for each one can extend by linearity and the alternating property to find its value at any -tuple of -vectors. In other words, it suffices to check (3.1) when, for some the submatrix is the identity and all other entries of are zero. In this case both sides of (3.1) are equal to . ∎
Lemma 3.2**.**
Let be a real matrix. Recall that is positive semidefinite and symmetric. Let the eigenvalues of be in decreasing order, where the are nonnegative and in decreasing order. That is, the are the singular values of listed in decreasing order.
In particular, if is a symmetric matrix, then the are exactly the absolute values of the eigenvalues of by diagonalisation.
Given a natural number with let be the vector of minors of arranged in some order. Then we have:
- (i)
The maximum norm satisfies
[TABLE] 2. (ii)
There is a -dimensional linear space such that for all ,
[TABLE]
We may take to be a span of standard basis vectors in . 3. (iii)
For any either there is an -dimensional linear subspace of such that
[TABLE]
or there is a -dimensional linear subspace of spanned by standard basis vectors of such that
[TABLE]
Proof. Part i.
First we prove the result on the assumption that is diagonal. Let the sets and the matrices be as in Lemma 3.1. Since is diagonal with diagonal entries we have
[TABLE]
by (3.1). The left-hand side of (3.5) is and the right-hand side is so this proves (3.2).
Let be an orthogonal matrix such that is diagonal. Let be the vector of minors of . We claim that the norms and are of comparable size.
Lemma 3.1 shows that and since and we see that is orthogonal. Hence the maximum norm of the entries satisfies
[TABLE]
So in proving (3.2) we may assume that is diagonal. The result follows. ∎
Part ii.
By permuting the rows and columns of we may assume that
[TABLE]
Let be in the span of the first basis vectors. If (3.3) holds for all such then we have proved the lemma. Since for one finds that
[TABLE]
where we have divided each matrix into three blocks, and stands for a block of zeroes. By (3.6) we have as the determinant of the left-hand side. Expanding the determinant of the right-hand side in the first column, we find that it is equal to
[TABLE]
Note the determinant in which runs over with the value omitted, and runs over .
By part i, this implies that so provided that then (3.3) holds.
If we apply the same permutation both to the and to the first rows of then both sides of our claim (3.3) and our assumption (3.6) remain the same. By applying such a permutation we may assume and so we have proved (3.3). ∎
Part iii.
Let be the span of the -eigenvectors of where runs from up to . As the matrix is symmetric, we have for all and so
[TABLE]
for all Therefore either this space satisfies (3.4), or the bound holds and the existence of the space follows by part ii. ∎
4 Counting points in the sets
In this section our goal is to estimate the number of integer points in the sets from Definition 2.1. We give the following result.
Lemma 4.1**.**
Let and be as in Definition 1.1, let be as in Definition 1.2, and let be as in Definition 2.1. Suppose that and and that . Then at least one of the following holds:
- (I)k
The set may be covered by a collection of at most
[TABLE]
boxes in of side . Such a box contains integral points, so it follows that
[TABLE] 2. (II)k
There exist an integer a point and a -dimensional linear subspace of such that
[TABLE]
- (III)
There is a -dimensional linear subspace of such that
[TABLE]
with as in §1.4.
We have subscripted the first two items to emphasize their dependence on ; note that item (III) has no such dependence.
In Corollary 5.1 below, we will use Lemma 4.1 to bound the quantities (2.1) and (2.2) from Corollary 2.1. Before proving the lemma, we give a comparison with step 2 in §1.4.
If there are many integer points for which holds, then step 2 gives us a point for which the matrix has a kernel of dimension or more and holds.
If there are many integer points for which then (4.1) is false and so either (II)k or (III) must hold. Of these, case (II)k gives us a point such that is small on a -dimensional space. Moreover it states that and that so that the th eigenvalue of the matrix is about times smaller than the th eigenvalue. Thus (II)k gives us a point for which in some sense is close to having a kernel of dimension at least and is close to having rank .
The third case (III) is less directly comparable to step 2. We suggest that it could correspond to the case of step 2.
Proof of Lemma 4.1.
The proof is by induction on . Let and be fixed. ∎
The case .
Let let and suppose that alternative (III) does not hold. We claim that alternative (I)0 holds, that is is covered by boxes of side .
As (III)0 is false, applying Lemma 3.2ii to the matrix of the linear map shows that there is an -dimensional subspace of with
[TABLE]
For each let be the box in defined by
[TABLE]
Now is contained in the box . It follows that we can cover with a collection of boxes of the form each one of which is centred at a point belonging to . We will show below that for each the intersection is contained in a box of side . It follows that is covered by boxes of side as claimed.
It remains to let and let and to deduce that must hold.
By definition of we have and and the bounds and follow by (1.10). So we have
[TABLE]
Let and let such that holds. Since lies in we have and with (4.5) this implies that
[TABLE]
By (4.4) it follows that and hence that as claimed. ∎
The inductive step.
Let and let . We suppose that (II)k and (III) are both false, and claim that (I)k holds.
By induction, at least one of (I)k-1, (II)k-1, or (III) holds. Note that of these (III) is false by assumption, and (II)k-1 is false since it implies (II)k, and so (I)k-1 must hold.
Suppose for the time being that
[TABLE]
The contrary case is almost trivial and will be dealt with at the end of the proof. We claim that
[TABLE]
where runs over those -dimensional subspaces of which are spanned by standard basis vectors, and we define
[TABLE]
We have assumed that and that the case of (II)k is false. So the case of (4.2) must be false for every -dimensional subspace of and every .
That is, for any and any -dimensional linear subspace of there is some such that
[TABLE]
Applying Lemma 3.2ii with the choice shows that for each there is an -dimensional subspace of spanned by standard basis vectors, such that
[TABLE]
for all . This proves (4.7), and so to prove (I)k it now suffices to show that for each -dimensional space the set (4.8) is covered by a union of boxes of side .
Let be a sufficiently small constant depending at most on and and for each set
[TABLE]
We assumed at the start of this inductive step that (I)k-1 holds, and so is covered by a collection of boxes of side . Subdivide each of these boxes into sub-boxes of the form . Since
[TABLE]
it follows that the set may be covered by a collection of boxes of the form each of which is centred at a point belonging to the set .
We will show below that for each such box the intersection is covered by a box of side . It follows that each set (4.8) is covered by boxes of side and by the comments after (4.9) this proves the lemma.
We suppose that and that and we claim that holds. Let and let such that and note that since we have
[TABLE]
Now the th partial derivatives of the minors are linear combinations of the minors with coefficients of size at most . So we have
[TABLE]
and Taylor expansion shows that
[TABLE]
It follows that
[TABLE]
Since Lemma 3.2 gives us the bounds
[TABLE]
and since it follows from (4.8) that
[TABLE]
Substituting (4.13)-(4.14) into (4.12) yields
[TABLE]
Applying the bounds from (4.11) and the inequalities we deduce that
[TABLE]
Since is assumed to be small in terms of and it follows that holds and hence that holds. By the comments after (4.10), this proves the lemma.
It remains to consider the case when (4.6) is false and so holds. At the start of the inductive step we supposed that (I)k-1 holds, so the set may be covered by boxes of side . We have
[TABLE]
and so the set is also covered by this collection of boxes. Since holds, we can divide each of these boxes into boxes of side . This proves (I)k. ∎
5 Small values of a trilinear form
Part 3 of Davenport’s argument from §1.4 starts from a point for which the matrices and have prescribed ranks, and finds linear spaces such that for all and the equation holds. Our analogue is the following pair of results, which give linear spaces on which the trilinear form is small.
Lemma 5.1**.**
Let be as in Definition 1.1, and let and be as in Definition 1.2. Suppose that and that . Then, provided is nonzero, there exists an -dimensional linear subspace of such that for all and all we have
[TABLE]
By setting and one may recover Davenport’s result. We prove Lemma 5.1 at the end of this section, after deducing
Corollary 5.1**.**
Let and be as in Definition 1.1. For any and any one of the following alternatives holds. Either
[TABLE]
or there exist positive-dimensional linear subspaces and of for which such that
[TABLE]
Proof.
We saw in Lemma 4.1 that for any and any satisfying
[TABLE]
one of (I)k, (II)k, or (III) must hold. Suppose first that in every case alternative (I)k holds. By (4.1), we then have
[TABLE]
for every and every . Now Corollary 2.1 shows that either
[TABLE]
or
[TABLE]
where and the inequalities hold, or
[TABLE]
where the inequalities hold. We may assume that is sufficiently large in terms of so that holds in (5.6)-(5.7). Substituting the bound (5.4) into each of (5.5)-(5.7) proves the conclusion (5.2).
Suppose next that alternative (III) holds in Lemma 4.1. In this case we let and the conclusion (5.3) follows from (4.3).
It remains to treat the case when there exist and such that alternative (II)k holds in Lemma 4.1. This means that there exist an integer a point and a -dimensional linear subspace of such that
[TABLE]
Since the inequalities hold. Therefore (5.8) implies
[TABLE]
Note that (5.10) implies that is nonzero, and so is nonzero, by Lemma 3.2. Hence we may apply Lemma 5.1 to find an -dimensional linear space such that for all and all the bound (5.1) holds. The conclusion (5.3) follows on taking in (5.1) of Lemma 5.1 and substituting in the bounds (5.9) and (5.10). ∎
Proof of Lemma 5.1.
We imitate the proof of Lemma 3 in Davenport [12], which begins by considering the following easy “warm-up” problem. Suppose we were to look for linearly independent vectors at which vanishes. One approach would be as follows. One can construct matrices for with entries in such that the vectors
[TABLE]
That is, the components of are polynomials of the form and the components of are polynomials of the form .
If had exactly linearly independent solutions we would have while would be nonzero. We would then have solutions defined by
[TABLE]
and if we chose our matrices appropriately these would be linearly independent.
We now return to the proof of the lemma. Assume for the time being that and satisfying (5.11)-(5.13) are given, and let be as in the lemma. Let . Let be the directional derivative along defined by and apply to both sides of (5.12). This shows that
[TABLE]
Since is a linear operator and we have
[TABLE]
it follows from (5.14) that
[TABLE]
Now Lemma 3.2 shows that
[TABLE]
and substituting these bounds into (5.15) gives
[TABLE]
where the are as in (5.13).
The idea is now to let be the span of the and deduce (5.1) from (5.16). Since we are looking for an -dimensional space we will need to be linearly independent. In order to prove (5.1) we require the following slightly stronger statement. We claim there are and satisfying (5.11)-(5.13), such that the linear combination defined by satisfies for every vector in real -space. The lemma then follows, with being the span of the on expressing as linear combinations of the and applying (5.16).
For the remainder of the proof we will assume for simplicity that the minor of with largest absolute value is the minor in the lower right-hand corner, that is, we will assume that
[TABLE]
In general (5.17) holds after permuting the rows and columns of the matrix and one can then apply the same permutations throughout the rest of our construction of every time the matrix appears.
Define by
[TABLE]
where means that first takes the value and then runs over the numbers with omitted. Now this is of the form (5.11), and one can check that
[TABLE]
which is of the form (5.12). Define a matrix by
[TABLE]
or equivalently by
[TABLE]
so that the entries have absolute value at most 1. Then one sees from (5.17) that
[TABLE]
where is some matrix. In particular and so the entries of are bounded in terms of . It follows that if then as claimed. ∎
6 Constructing singular points on
Corollary 5.1 shows that either is small, or there are spaces of large dimension on which is small. To prove Proposition 1.1 we show that the second alternative implies that is singular. This is our analogue of Davenport’s step 4, as described in §1.4.
Proof of Proposition 1.1.
Suppose for a contradiction that the result is false. Then for every there is with
[TABLE]
By Corollary 5.1, this implies that there are linear subspaces of such that
[TABLE]
holds and for all and we have
[TABLE]
If we multiply by a constant then the matrix does not change. So we may assume that for each the equality holds. After passing to a subsequence we have as and it follows that there are subspaces of such that and
[TABLE]
Let such that
[TABLE]
Let be a basis of such that is a basis of .
Let be the projective linear space in associated to . Take homogeneous co-ordinates on so that takes values in .
Let be the projective variety cut out in by the equations
[TABLE]
so that
[TABLE]
We claim that is contained in the singular locus of the projective hypersurface . It follows that which is a contradiction, by (1.6).
Now (6.1) implies that for every we have
[TABLE]
So if we let such that (6.2) holds, then we have
[TABLE]
This implies that holds, by the definition (1.4). It follows that every point of is contained in as claimed. ∎
7 Acknowledgements
This work is based on a DPhil thesis submitted to Oxford University, and was supported by EPSRC grants EP/J500495/1 and EP/M507970/1. I would like to thank my DPhil supervisor, Roger Heath-Brown. I am grateful to Rainer Dietmann for helpful conversations.
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