EFFECTIVE RESULTS ON THE SIZE AND STRUCTURE OF SUMSETS ANDREW GRANVILLE, GEORGE SHAKAN, AND ALED WALKER

arXiv:2105.09181v1 [math.CO] 19 May 2021

ANDREW GRANVILLE, GEORGE SHAKAN, AND ALED WALKER

Abstract. Let A ⊂ Z ^d be a finite set. It is known that N A has a particular size (|N A| = P A (N ) for some P A (X) ∈ Q[X ]) and structure (all of the lattice points in a cone other than certain exceptional sets), once N is larger than some threshold. In this article we give the first effective upper bounds for this threshold for arbitrary A. Such explicit results were only previously known in the special cases when d = 1, when the convex hull of A is a simplex or when |A| = d + 2 [3], results which we improve.

1. Introduction

For any given finite subset A of an abelian group G, we consider the sumset NA := {a 1 + a 2 + · · · + a N : a i ∈ A}.

If G is finite and N is sufficiently large then

NA = Na 0 + hA − Ai (1.1)

for any a ₀ ∈ A where hA − Ai is the subgroup of G generated by A − A, so that |NA| is eventually constant. In this article we study instead the case when G = Z ^d is infinite, and ask similar questions about the size and structure of NA when N is large.

The size of NA. Khovanskii’s 1992 theorem [8] states that if A ⊂ Z ^d is finite then there exists P A (X) ∈ Q[X] of degree 6 d such that if N > N Kh (A) then

|NA| = P A (N ).

Although there are now several different proofs of Khovanskii’s theorem [12, 7], the only effective bounds on N Kh (A) have been obtained when d = 1 [11, 14, 5, 6], when the convex hull of A is a d-simplex or when |A| = d + 2 (see [3]).

We will determine an upper bound for N _Kh (A) for any such A in terms of the width of A, w(A) = width(A) := max

a

,a

∈A ka 1 − a 2 k ∞ . (1.2) Theorem 1.1 (Effective Khovanskii). If A ⊂ Z ^d is finite then

|NA| = P _A (N ) for all N > (2|A| · width(A)) ^(d+4)|A| .

The theorem states that N Kh (A) 6 (2ℓ w(A)) ^(d+4)ℓ where ℓ := |A|. We expect that N Kh (A) is considerably smaller (see Section 2); for example, if |A| = d + 2 and A − A generates Z ^d then [3, Theorem 1.2] gives that

N Kh (A) = d! Vol(H(A)) − d − 1, (1.3)

A.G. is funded by the Natural Sciences and Engineering Research Council of Canada (NSERC) under the Canada Research Chairs program.

G.S. is supported by Ben Green’s Simons Investigator Grant 376201.

A.W. was supported by a postdoctoral research fellowship at the Centre de Recherches Math´ematiques and a junior fellowship at Institut Mittag-Leffler, and is a Junior Research Fellow at Trinity College Cambridge.

1

where the convex hull H(A) is defined by H(A) :=

X

a∈A

c a a : Each c a ∈ R >0 , X

a∈A

c a = 1

.

We can replace w(A) in Theorem 1.1 by w ^∗ (A) which is defined to be the minimum of w(A ^′ ) over all A ^′ ⊂ Z ^d that are Freiman isomorphic to A. ¹

Previous proofs of Khovanskii’s theorem [12, 7] relied on the following ineffective principle.

Lemma 1.2 (The Mann-Dickson Lemma). For any S ⊂ Z ^d _>0 there exists a finite subset S _min ⊂ S such that for all s ∈ S there exists x ∈ S _min with s − x ∈ Z ^d _>0 .

For a proof see [5, Lemma 5]. Here we rework the method of Nathanson–Ruzsa from [12]

as a collection of linear algebra problems which we solve quantitatively (see Section 6), and therefore bypass Lemma 1.2 and prove our effective threshold.

The structure of NA. For a given finite set A ⊂ Z ^d with 0 ∈ A we have H(A) =

X

a∈A

c a a : Each c a ∈ R >0 , X

a∈A

c a 6 1

.

We let ex(H(A)) be the set of extremal points of H(A), that is the “corners” of the boundary of A, ² which is a subset of A. We define the lattice generated by A,

Λ A :=

X

a∈A

x a a : x a ∈ Z for all a

.

For a domain D ⊂ R ^d we set N · D := {Nx : x ∈ D} so that N · H(A) = NH(A) as H(A) is convex and so, as 0 ∈ A,

H(A) ⊂ 2H(A) ⊂ 3H(A) · · · ⊂ C A := lim

N →∞ NH (A) =

X

a∈A

c a a : c a ∈ R >0 for all a

,

the cone generated by A. Now, by definition,

0 ∈ A ⊂ 2A ⊂ 3A · · · ⊂ P (A) :=

∞

[

N=1

NA,

and each

NA ⊂ NH (A) ∩ Λ _A so that P (A) ⊂ C _A ∩ Λ _A . Define the set of exceptional elements

E (A) := (C A ∩ Λ A ) \ P (A).

Therefore, for any finite A ⊂ Z ^d and a ∈ A we have

N (a − A) ⊂ (NH (a − A) ∩ Λ a−A ) \ E (a − A), as 0 ∈ a − A. So

NA ⊂ (NH (A) ∩ (aN + Λ a−A )) \ (aN − E (a − A)).

1 That is, there is a map φ : A → A ^′ such that for all a 1 , . . . , a k , b 1 , . . . , b k ∈ A and k > 1, we have a 1 + · · · + a k = b 1 + · · · + b k if and only if φ(a 1 ) + · · · + φ(a ^k ) = φ(b 1 ) + · · · + φ(b ^k ).

2 That is, those points p ∈ H(A) for which there is a vector v ∈ span(A − A) \ {0} and a constant c such

that hv, pi = c and hv, xi > c for all x ∈ H (A) \ {p}; see Appendix A.

Hence, as aN + Λ a−A is independent of the choice of a ∈ A and Λ a−A = Λ A−A , for any fixed a 0 ∈ A we have

NA ⊂ (NH (A) ∩ (a 0 N + Λ A−A )) \ [

a∈ex(H(A))

(aN − E (a − A))

. (1.4)

In [5] the first two authors showed ³ there exists a constant N Str (A) such that we get equality in (1.4) provided N > N _Str (A); that is,

NA = (NH (A) ∩ (a 0 N + Λ A−A )) \ [

a∈ex(H(A))

(aN − E (a − A))

. (1.5)

(Compare this statement to (1.1).) The proof in [5] relied on the ineffective Lemma 1.2 so did not produce a value for N Str (A).

In this article we give an effective bound on N Str (A):

Theorem 1.3 (Effective structure). If A ⊂ Z ^d is finite then

(1.5) holds for all N > (d|A| · width(A)) ^13d

. That is, Theorem 1.3 implies that N Str (A) 6 (dℓ w(A)) ^13d

where |A| = ℓ.

The 1-dimensional case is easier than higher dimensions, since if 0 = min A and Λ A = Z then E (A) is finite, and so has been the subject of much study [11, 14, 5, 6]: We have N Str (A) = 1 if |A| = 3 in [5], and N Str (A) 6 w(A) + 2 − |A| in [6], with equality in a family of examples. There are also effective bounds known when H(A) is a d-simplex, as we will discuss in the next subsection.

Suppose that x belongs to the right-hand side of (1.5). To prove Theorem 1.3 when x is far away from the boundary of NH (A) we develop an effective version of Proposition 1 of Khovanskii’s original paper [8] using quantitative linear algebra. Otherwise x is close to a separating hyperplane of NH (A): Suppose the hyperplane is z d = 0; write each a = (a 1 , . . . , a d ) and x = (x 1 , . . . , x d ), so that every a d > 0 and x d is “small”. Now x = P

a∈A m a a where each m a ∈ Z >0 as x ∈ P (A) and so P

a∈A,a

6=0 m a a d 6 x d is small. The contribution from those a with a _d = 0 is a “smaller dimensional problem”, living in the hyperplane z _d = 0.

Carefully formulated, one can apply induction on the dimension to bound P

a∈A m a , and hence show that x ∈ NA.

The structure (1.5) is evidently related to Khovanskii’s theorem. However, we have not been able to find a precise way to relate Khovanskii’s theorem and Theorem 1.3. Our proofs of Theorem 1.1 and Theorem 1.3 are almost entirely disjoint, and we get a different quality of bound in each theorem.

The size and structure of NA when A is contained in a simplex. If A ⊂ R ^d then the convex hull H(A) is a d-simplex if there exists B ⊂ A with |B| = d + 1 for which B − B spans R ^d and H(A) = H(B) (whence ex(H(A)) = B).

Theorem 1.4 (Effective Khovanskii, simplex case). If A ⊂ Z ^d is finite and H(A) is a d-simplex then |NA| = P A (N ) for all

N > (d + 1)! vol(H(A)) − (d + 1)(|A| − d) + 1. (1.6) Theorem 1.5 (Effective structure, simplex case). If A ⊂ Z ^d is finite and H(A) is a d- simplex then (1.5) holds for all N in the range (1.6), and if |A| = d + 1 or d + 2 then (1.5) holds for all N > 1.

3 The result in [5] was only stated when 0 ∈ A and Λ A = Z ^d , and the union was over all of A rather than

just ex(H (A)), but the methods give the general version (1.5).

Therefore if A ⊂ Z ^d is finite and H(A) is a d-simplex then

N Kh (A), N Str (A) 6 (d + 1)! vol(H(A)) − (d + 1)(|A| − d) + 1.

The hypotheses imply that |A| > d + 1. If d = 1 our bound gives N Str (A) 6 2w(A) −2|A| + 3 which is weaker than the bound N _Str (A) 6 w(A) − |A| + 2 from [6], which suggests that Theorem 1.5 is still some way from being “best possible”.

Curran and Goldmakher [3] gave a similar (but slightly weaker) bound in the simplex case, showing that (1.5) holds provided N > (d + 1)! vol(H(A)) − 2d − 2. In the statement of their result they replace (1.5) by the equivalent

NA = \

b∈ex(H(A))

(bN + P (A − b)).

Their (seemingly different) proof proceeds via generating functions but also makes use of

‘minimally represented elements’ . Our improvement (see the discussion around Lemma 3.5 below) comes from refining an additive combinatorial lemma concerning these minimal elements, related to the Davenport constant of the group Z ^d /Λ B−B .

Even though the main bounds in Theorems 1.4 and 1.5 are the same, we have not been able to find a way to directly deduce one theorem from the other. Instead, we present separate arguments for each theorem (in Sections 4 and 5 respectively), albeit based on the same fundamental lemmas in Section 3.

In the next section we briefly discuss the 1-dimensional case, and in the three subsequent sections, the simplex case. In Section 6, we prove the effective Khovanskii theorem (Theo- rem 1.1). In Section 7 we then prove the effective structure result (Theorem 1.3); this part may be read essentially independently of the previous section, although there is one piece of quantitative linear algebra in common. An appendix collects together some facts from the theory of convex polytopes (which are useful in Section 7).

2. One dimension and speculations

It might well be that for A ⊂ Z ^d , the general case considered in this paper, that

N Str (A) 6 N Kh (A) 6 d! vol(H(A)). (2.1) We refrain from calling this speculation a conjecture since we have not even proved it for d = 1, but we know of no counterexample, so it is certainly worth investigating; we make a few remarks in this section. First we note that if E (A) = E (b − A) = ∅ then N Kh (A) = N Str (A) by the definitions and Theorem 1.3. We obtain the bounds N Kh (A), N Str (A) <

(d + 1)! vol(H(A)) in Theorems 1.4 and 1.5, bigger than in (2.1) by a factor of d + 1 (and one can see where this comes from in the proof). If d = 1 then Vol(H(A)) = w(A), so the inequalities N _Str (A), N _Kh (A) 6 d! vol(H(A)) can be deduced from the following:

Lemma 2.1. If A ⊂ Z with gcd _a∈A a = 1 and |A| > 3 then N Str (A), N Kh (A) 6 w(A) − 1.

Proof. We may translate A so that it has minimal element 0 and largest element b = w(A).

(If |A| = 2 then A = {0, 1} and N Str (A) = N Kh (A) = 1). The main theorem of [6] gives that N Str (A) 6 b − |A| + 2, which is 6 w(A) − 1 for |A| > 3.

If N > N Str (A) then NA = (NH (A) ∩ Z ^d ) \ (E (A) S

(bN − E (b − A))). Let e A denote the largest element of E (A), or e A = −1 if E (A) is empty. If bN > e A + e b−A then E (A) and bN − E (b − A) are disjoint subsets of {0, . . . , bN} so that |NA| = bN − c where c =

|E (A)| + |E (b − A)| − 1. Therefore N Kh (A) 6 max n

N Str (A), 1 + j e A + e b−A

b

ko .

In fact if N Str (A) 6 ⌊ ^e

^+e _b

⌋ then N Kh (A) = ⌊ ^e

^+e _b

⌋ + 1.

In particular if A = {0, a, b} with (a, b) = 1 then N Str (A) = 1 by [5, Theorem 4] and e A = ba − b − a so that N Kh (A) = max(1, b − 2).

Now suppose that |A| > 4. By [4, Theorem 1] we have e A 6 b(b − 1)

|A| − 2 − 1 so that 1 + j e A + e b−A

b

k < 1 + 2(b − 1)

|A| − 2 6 b.

Therefore we have N _Kh (A) 6 b − 1 = w(A) − 1.

Curran and Goldmakher give the precise value of N Kh (A) in (1.3) in certain special cases including the useful example A := {(0, . . . , 0), (1, . . . , 1), m 1 e 1 , . . . , m d e d } ⊂ Z ^d where the m j are pairwise coprime positive integers and the e 1 , . . . , e d are the standard basis vectors.

If all the m j are close to x so that w(A) ≈ x for some large x then N Kh (A) ∼ _x→∞ w(A) ^d , which suggests we might be able to reduce the bound in Theorem 1.1 to w(A) ^d . However d! vol(H(A)) seems preferable to w(A) ^d since it is smaller and more precise in the example where we let m 2 = · · · = m d = 1 and m 1 = x be arbitrarily large so that N Kh (A) ∼ x→∞

w(A).

3. Preparatory lemmas for the simplex case

Throughout this section, 0 ∈ A ⊂ Z ^d and A is finite. Let N A (0) = 0 and for each v ∈ P (A) \ {0} let N _A (v) denote the minimal positive integer N such that v ∈ NA.

Definition 3.1 (B-minimal elements). Suppose that B ∪ {0} ⊂ A ⊂ Z ^d , with A finite.

Let S(A, B) denote the set of B-minimal elements ⁴ , which comprises 0 and those elements u ∈ P (A) \ {0} such that a i 6∈ B ∪ {0} for every i whenever

u = a ₁ + a ₂ + · · · + a _N

_(u) with each a _i ∈ A.

B-minimal elements can be used to decompose NA and P (A) into simpler parts.

Lemma 3.2. If B ^∗ := B ∪ {0} ⊂ A ⊂ Z ^d with A finite then P(A) = S(A, B) + P (B ^∗ ) and NA = [

u∈S(A,B) N

(u) 6 N

(u + (N − N A (u))B ^∗ ).

Proof. The second assertion implies the first by taking a union over all N . That each u + (N − N A (u))B ^∗ ⊂ NA is immediate, so we need only show that if v ∈ NA then v ∈ u + (N − N _A (u))B ^∗ for some u ∈ S(A, B ) with N _A (u) 6 N .

Now, for any v ∈ NA we can write

v = u + w with u = a 1 + · · · + a L and w = b 1 + · · · + b M ,

where L, M > 0, and each a _i ∈ A \ B and b _i ∈ B , with M maximal and L + M = N _A (v).

Then N A (u) = L and N A (w) = M , else we could replace the above expression for u or w by a shorter sum of elements of A, and therefore obtain a shorter sum of elements to give v, contradicting that L + M = N A (v) is minimal. Moreover u ∈ S(A, B) else we could replace the sum a 1 + · · · + a L in the expression for v by a different sum of length L which includes some elements of B, contradicting the maximality of M .

Therefore u ∈ S(A, B ) with N A (u) = L 6 N A (v) 6 N and

v ∈ u + MB ^∗ = u + (N _A (v) − N _A (u))B ^∗ ⊂ u + (N − N _A (u))B ^∗ ,

since 0 ∈ B ^∗ .

It will be useful to control the complexity of the B-minimal elements.

4 We observe that S(A, B) is the set of u ∈ P(A) such that (u, N A (u)) ∈ Z ^d+1 is minimal in the sense of

[3, Section 3].

Definition 3.3. Let B ∪ {0} ⊂ A ⊂ Z ^d , with A finite. If S(A, B) is a finite set, we define K(A, B) := max

u∈S(A,B) N A (u).

In certain circumstances we will bound K(A, B) using results on Davenport’s problem, which asks for the smallest integer D(G) such that any set of D(G) (not necessarily distinct) elements of an abelian group G contains a subsum ⁵ that equals 0 G . It is known that D(G) 6 m(1 + log |G|/m) where m is the maximal order of an element of G.

Definition 3.4. Given a finite abelian group G, if 0 6∈ H ⊂ G let k(G, H ) be the length of the longest sum of elements of H which contains no subsum equal to 0, and no subsum of length > 1 belonging to H.

Lemma 3.5. Given a finite abelian group G, for any 0 ∈ / H ⊂ G we have k(G, H) 6

|G| − |H|. Moreover k(G, H ) 6 m(1 + log |G|/m) − 1, where m is the maximal order of an element of G.

Proof. Suppose we are given a longest sum h ₁ + · · · + h _k of elements of H defining k(G, H ), so that k = k(G, H). Then

0, h 1 + h 2 , h 1 + h 2 + h 3 , . . . , h 1 + · · · + h k ,

are all distinct in G, else subtracting would give a subsum equal to 0, and they are all contained in G \ H. Therefore k + |H| 6 |G| and the first result follows.

By definition k(G, H ) < D(G) so the second result claims from the result noted for D(G)

above.

Curran and Goldmakher’s [3, Lemma 3.1] implies the weaker upper bound k(G, H) 6

|G| − 1. This difference leads in part to the improvements in Theorems 1.4 and 1.5.

3.1. d-dimensional simplices. Let B = {b 1 , . . . , b d } ⊂ A be a basis for R ^d with B ∪ {0} ⊂ A ⊂ H(B ∪ {0}) and A ⊂ Z ^d

so that A is finite. Since C A = C B , and B is a basis, there is a unique representation of every vector r ∈ C A as

r =

d

X

i=1

r i b i where each r i > 0. (3.1) If r ∈ H(A) = H (B ∪ {0}) then P d

i=1 r _i 6 1.

Lemma 3.6. Suppose B = {b 1 , . . . , b d } is a basis for R ^d with B ∪ {0} ⊂ A ⊂ H(B ∪ {0}) and A ⊂ Z ^d is finite. If r ∈ P (A) and r ≡ a (mod Λ B ) with a ∈ A then r − a ∈ P (B ∪ {0}) (where we choose a = 0 if r ∈ Λ B ).

Proof. Since r ∈ P (A) ⊂ C A , we have the representation (3.1) for r. Moreover since a ∈ H(A) = H(B ∪{0}) we have the representation a = P d

i=1 a i b i by (3.1) with P d

i=1 a i 6 1.

If a 6≡ 0 (mod Λ B ) then each a i ∈ [0, 1), and otherwise we choose a = 0 so each a i = 0.

Therefore P d

i=1 r i b i = r ≡ a = P d

i=1 a i b i (mod Λ B ), and each r i ≡ a i (mod 1). As each r _i > 0 we write m _i = r _i − a _i so that each m _i ∈ Z > 0 and r − a = P d

i=1 m _i b _i ∈ P(B ∪ {0}).

We use this lemma to bound K(A, B).

Lemma 3.7. Suppose B = {b ₁ , . . . , b _d } is a basis for R ^d with B ∪ {0} ⊂ A ⊂ H(B ∪ {0}) and A ⊂ Z ^d is finite. If u = a 1 + · · · + a _N

_(u) ∈ S(A, B) is non-zero then any subsum with two or more elements cannot belong to A B := A mod Λ B , and no subsum can be congruent to 0 mod Λ B . Therefore

K(A, B) 6 k(Λ A /Λ B , A B \ {0}).

5 A subsum of a given sum P

i∈I g i is a sum of the form P

i∈I

g i where I ^′ ⊂ I is non-empty, of length |I ^′ |.

Proof. Let r be a subsum of a 1 + · · · + a N

(u) of size ℓ > 1. Then ℓ = N A (r) and r ∈ S(A, B ) as u ∈ S(A, B ). We write r as in (3.1) so that P

i6d r i 6 ℓ = N A (r). Suppose that r ≡ a (mod Λ B ) for some a ∈ A (where we choose a = 0 if r ∈ Λ B ) so that m := r−a ∈ P(B ∪{0}) by Lemma 3.6. Therefore N A (m) > ℓ − N A (a) > ℓ − 1 > 0 (so m 6= 0). On the other hand N _A (m) = P

i6d (r _i − a _i ) = ℓ if a = 0, and < ℓ if a 6= 0, so N _A (m) 6 ℓ − N _A (a). We deduce that r can be represented as a plus the sum of ℓ − N A (a) elements of B, contradicting that

r ∈ S(A, B ).

If the convex hull of A is not a simplex then S(A, B) need not be finite. For example, if B = {(0, 1), (1, 0)} ⊂ A = {(0, 0), (−1, 1), (0, 1), (1, 0)} then S(A, B) = {(−k, k) : k ∈ Z >0 }.

This is one reason why S(A, B) is not used later in Section 7, when dealing with general sets A.

3.2. Translations. We finish by observing that under rather general hypotheses the sets S(A, B), and consequently the quantities K (A, B), are well-behaved under translations.

Lemma 3.8. Let B ∪ {0} ⊂ A ⊂ Z ^d , with A finite. If b ∈ B then S(b − A, b − B) = {bN _A (u) − u : u ∈ S(A, B)}

and if v = bN A (u) − u with u ∈ S (A, B ) then N b−A (v) = N A (u). In particular we have K(b − A, b − B) = K(A, B).

Proof. Let N = N A (u). If u = a 1 + a 2 + · · · + a N then v := bN − u = (b − a 1 ) + · · · + (b − a N ) so that N b−A (v ) 6 N A (u). If N b−A (v) 6 N − 1 say v = (b − a ^′ ₁ ) + · · · + (b − a ^′ _M ) with M < N then u = a ^′ ₁ + · · · + a ^′ _M + (N − M)b contradicting the definition of u ∈ S(A, B). We deduce that there is a 1-to-1 correspondance between the representations of u as the sum of N elements of A, and of v as the sum of N elements of b − A, and the result follows.

4. Structure bounds in the simplex case First we deal with the special cases.

Proof of Theorem 1.5 for |A| = d + 1 and |A| = d + 2. Let ex(H(A)) = B where |B | = d+1 and span(B − B) = R ^d . Write B = {b ₀ , . . . , b _d }, and translate so that 0 = b ₀ ∈ B.

If |A| = d + 1 then B = A and E (b − A) = ∅ for all b ∈ B. We immediately see that NA = NH (B) ∩ Λ _B = NH (A) ∩ Λ _A for all N > 1.

If |A| = d + 2 write A = B ∪ {a}. Since a ∈ H(B) we can write a = P d

i=0 a _i b _i uniquely with each a i > 0 and P d

i=0 a i = 1. We know that the finite group Λ A /Λ B is generated by a.

If a has order M in the group Λ A /Λ B then the classes of Λ A /Λ B can be represented by S(A, B) = {ma : 0 6 m 6 M − 1}.

Now let

v ∈ (NH(A) ∩ Λ A ) \ ( [

b∈B

(bN − E (b − A))). (4.1) Since v ∈ NH (A) = NH (B) we can write v = P d

i=0 v i b i in a unique way with each v i > 0 and P d

i=0 v i = N . This implies that b i N − v = P d

j=0 v j (b i − b j ) ∈ C b

−A , and from (4.1) we have b i N − v ∈ Λ A = Λ b

−A and b i N − v / ∈ E (b i − A). Hence b i N − v ∈ P(b i − A) for all i, in particular v ∈ P (A) (from i = 0).

Suppose that v ≡ ma mod Λ B for some 0 6 m 6 M − 1. This implies that v i − ma i ∈ Z for i = 0, 1, . . . , d, and we now show that v i − ma i ∈ Z >0 if i 6= 0: Since v ∈ P(A) we may write

v = (m + λM )a +

d

X

i=1

(v i − (m + λM )a i )b i

for some λ ∈ Z >0 with v i − (m + λM )a i ∈ Z >0 for i = 1, . . . , d. Therefore we conclude that v i − ma i ∈ Z >0 for all i > 1.

We now give an analogous argument for representations of b j N − v for each j = 1, . . . , d:

For each j we also have b j N − v =

d

X

i=0

v i (b j − b i ) ≡

d

X

i=0

ma i (b j − b i ) = m(b j − a) mod Λ b

−B

Since b j N − v ∈ P (b j − A) we may write b j N − v = (m + λM )(b j − a) +

d

X

i=0

(v i − (m + λM )a i )(b j − b i )

for some λ ∈ Z > 0 with v _i − (m + λM )a _i ∈ Z > 0 for i = 0, . . . , d with i 6= j (we can’t deduce this for i = j since then b j − b i = 0). Therefore v i > ma i for all i 6= j.

Combining these observations, we deduce that v _i − ma _i ∈ Z > 0 for all i, which implies that v = ma +

d

X

i=0

(v i − ma i )b i ∈

m +

d

X

i=0

(v i − ma i )

A = NA.

We now prove the rest of Theorem 1.5. We’ll use our bound on K(A, B ) from Lemma 3.7, combined with the following theorem.

Theorem 4.1. Let A ⊂ Z ^d be a finite set, for which H(A) is a d-simplex and 0 ∈ B :=

ex(H(A)). Then (1.5) holds for all N > (d + 1)(K (A, B ) − 1) + 1.

Proof. The proof follows similar lines to [5]. For all v ∈ (NH(A) ∩ (a 0 N + Λ A−A )) \ ( [

b∈B

(bN − E (b − A))) (4.2) we wish to show that v ∈ NA. Now v ∈ NH (A) = NH (B ), so if B = {0 = b 0 , b 1 , . . . , b d } then v = P d

i=0 v i b i for some v i ∈ R >0 with P d

i=0 v i = N . We will now show that v ∈ N j A for each j, where N j = K(A, B) + P

i6=j ⌊v _i ⌋:

Taking j = d (all other cases are analogous), we observe that b d N − v =

d−1

X

i=0

v i (b d − b i ) ∈ (N − v d ) · H(b d − B),

so that b _d N − v ∈ C _b

−B = C _b

−A . Therefore b _d N − v ∈ P (b _d − A), as b _d N − v / ∈ E (b _d − A) and b _d N − v ∈ Λ _b

−A by (4.2). So we may write

b _d N − v = u + w with u ∈ S(b _d − A, b _d − B) and w ∈ P (b _d − B) by Lemma 3.2. Then w = P d−1

i=0 w _i (b _d − b _i ) for some w _i ∈ Z > 0 , which implies 0 6 w _i 6 v _i so that w _i 6 ⌊v _i ⌋ for each i. But then w ∈ ( P

i6=d ⌊v _i ⌋)B ⊂ (N _d −K(A, B ))A and u ∈ K (A, B)A since K(A, B) = K(b d − A, b d − B) by Lemma 3.8. Therefore v = u + w ∈ N d A as claimed.

The result follows if P

i6=j ⌊v _i ⌋ 6 N − K for some j , where K = K(A, B). If not then P

i6=j v _i > P

i6=j ⌊v _i ⌋ > N − K + 1 for each j, and so N =

d

X

i=0

v i = 1 d

d

X

j=0

X

i6=j

v i > d + 1

d (N − K + 1),

which implies that N 6 (d + 1)(K − 1), a contradiction.

Proof of Theorem 1.5 for |A| > d + 3. Now A \ B is non-empty. Replacing A with A − b (for some b ∈ ex(H(A)) we may assume, without loss of generality, that 0 ∈ ex(H(A)) = B.

Applying Lemma 3.7 and Lemma 3.5, we then have K(A, B) 6 k(Λ A /Λ B , A B \ {0})

6 |Λ _A /Λ _B | − |A| + |B|

6 |Z ^d /Λ B | − |A| + |B| = d! vol(H(A)) − |A| + d + 1.

By Theorem 4.1, we conclude that (1.5) holds for all N in the range (1.6), as required.

5. The Khovanskii polynomial in the simplex case

In this section we prove Theorem 1.4, and make various remarks about the form of the Khovanskii polynomial itself.

Since the size of NA is preserved by translation, we assume that B = {b 1 , . . . , b d } is a basis for R ^d and

B ^∗ := B ∪ {0} ⊂ A ⊂ H(B ∪ {0}) ⊂ Z ^d . For each g ∈ G := Z ^d /Λ _B we have a coset representative g = P d

i=1 g _i b _i where each g _i ∈ [0, 1).

We may partition NA as the (disjoint) union over g ∈ G of (NA) g := {v ∈ NA : v ∈ g + Λ B }, and thus we wish to count the number of elements in each (NA) _g . If

S(A, B) g := {u ∈ S(A, B) : u ∈ g + Λ B } then, by Lemma 3.2,

(NA) g = [

u∈S(A,B)

N

(u)6N

u + (N − N A (u))B ^∗

.

This union is not necessarily disjoint, but we may nonetheless develop a formula for its size by using inclusion-exclusion:

Let S(A, B ) g = {u 1 , . . . , u k }, as S(A, B ) is finite by Lemma 3.7, and so write u j = g +

d

X

i=1

u j,i b i = a 1 + · · · + a N

(u

)

where each u j,i ∈ Z >0 . Expressing each a ℓ in terms of the basis B, we deduce that

d

X

i=1

u _j,i 6 N _A (u _j ) so that ∆ _j := N _A (u _j ) −

d

X

i=1

u _j,i > 0.

Therefore if N > N A (u j ) then u j + (N − N A (u j ))B ^∗ = g +

^d X

i=1

m i b i : Each m i ∈ Z >u

and

d

X

i=1

m i 6 N − ∆ j

(and the set on the right-hand side of the above expression is empty when N < N _A (u _j )).

Therefore for all N and for all non-empty subsets J ⊂ {1, . . . , k} we have

\

j∈J

u _j + (N − N _A (u _j ))B ^∗

= g + d

X

i=1

m _i b _i : Each m _i > u _J,i and

d

X

i=1

m _i 6 N − ∆ _J

,

(5.1) where we understand the m _i to always be integers, and we let

u J,i := max

j∈J u j,i and ∆ J := max

j∈J ∆ j .

Let

N _J := ∆ _J +

d

X

i=1

u _J,i .

To count the number of points in the intersection (5.1) we write each m i = u J,i + ℓ i , and then

\

j∈J

(u _j +(N −N _A (u _j ))B ^∗ )

= #{(ℓ ₁ , . . . , ℓ _d ) ∈ Z ^d _>0 : ℓ ₁ +· · ·+ℓ _d 6 N −N _J } =

N − N _J + d d

,

where we define ^N−N _d

^+d

:= 0 if N < N J . Hence, by inclusion-exclusion we obtain

#(NA) _g = X

J⊂{1,...,k}

|J|>1

(−1) ^|J|−1

\

j∈J

(u _j + (N − N _A (u _j ))B ^∗ )

= X

J⊂{1,...,k}

|J|>1

(−1) ^|J|−1

N − N _J + d d

.

Therefore we have the general formula

#NA = X

g∈G

#(NA) g = X

g∈G S(A,B)

={u

,...,u

}

X

J⊂{1,...,k}

|J|>1

(−1) ^|J|−1

N − N J + d d

. (5.2)

We wish to replace the binomial coefficients in this formula by polynomials in N ; that is, Replacing

N − N J + d d

by 1

d! (N − N J + d) · · · (N − N J + 1), but these are only equal if N > N J − d. Therefore we are guaranteed that

#(NA) g = P g (N ) where P g (T ) := X

J⊂{1,...,k}

|J|>1

(−1) ^|J|−1 (T − N J + d) · · · (T − N J + 1)

d! ,

provided N > max J N J − d = N {1,...,k} − d. Therefore

#NA = P _A (N) where P _A (T ) := X

g∈G

P _g (T )

once N > max _g (N {1,...,k} − d). It remains to bound N {1,...,k} . If k = 1 then N {1} = N _A (u ₁ );

if k > 1 then by definition we have

N {1,...,k} 6 max

j N A (u j ) +

d

X

i=1

max j u j,i . (5.3)

Now

u j,i 6 N A (u j ) 6 k(Λ A /Λ B , A B \ {0})

6 |Λ A /Λ B | − |A| + |B| 6 |Z ^d /Λ B | − |A| + |B | 6 d! vol(H(A)) − |A| + d + 1, by Lemma 3.7 and Lemma 3.5. Therefore

N {1,...,k} − d 6 (d + 1)N A (u j ) − d 6 (d + 1)(d! vol(H(A)) − |A| + d + 1) − d,

and the result follows.

This proof, following (5.3), seems wasteful and hopefully some more careful geometric

consideration might improve the bound on N _Kh (A).

5.1. Smaller N . One may sometimes show that #(NA) g = P g (N ) for more values of N . Proposition 5.1. Define

W (h) := X

J⊂{1,...,k}

|J|> 1 N

=N

−h

(−1) ^{|J |} ,

and let h be the smallest non-negative integer for which W (h) 6= 0. Then #(NA) g = P g (N ) for all N > N {1,...,k} − d − h, but not for N = N {1,...,k} − d − h − 1.

Proof. Letting m > 0 and N = N _{1,...,k} − d − 1 − m we have

#(NA) g − P g (N ) = X

J ⊂{1,...,k}

|J|> 1 N

>N

−m

(−1) ^|J| (N − N J + d) · · · (N − N J + 1) d!

= (−1) ^d

m

X

κ=0

κ + d d

W (m − κ) since if N J = N {1,...,k} − (m − κ) then

(N − N J + d) · · · (N − N J + 1)

d! = (−1) ^d

κ + d d

If m 6 h − 1 then every term on the right-hand side is 0 and so #(NA) g = P g (N). If m = h

then #(NA) _g = P _g (N ) + (−1) ^d W (h).

5.2. Determing W (0). We do not see how to easily determine h in general, though it is sometimes possible to identify whether W (0) = 0.

Proposition 5.2. Let J 0 := {j : ∆ j = ∆ {1,...,k} } and J i := {j : u j,i = u {1,...,k},i } for 1 6 i 6 d, with J ^∗ := ∪ _06i6d J i .

(i) If J ^∗ is a proper subset of {1, . . . , k} then W (0) = 0.

(ii) If J ^∗ = {1, . . . , k} and, for each i, there exists j _i ∈ J _i such that j _i 6∈ J _ℓ for any ℓ 6= i, then W (0) = (−1) ^d+1 6= 0. (For example, when the sets J i are disjoint.)

Proof. We have N _J = N {1,...,k} if and only if J ∩ J _i 6= ∅ for all 0 6 i 6 d. Therefore, by inclusion-exclusion we have

W (0) = X

J⊂{1,...,k}

|J∩J

|>1 for each i

(−1) ^|J| = X

I ⊂{0,...,d}

I 6=∅

(−1) ^|I| X

J ⊂{1,...,k}

J∩J

=∅ for each i∈I

(−1) ^|J|

= X

I⊂{0,...,d}

I 6=∅

(−1) ^{|I |} X

J⊂{1,...,k}\∪

J

(−1) ^|J| = X

I ⊂{0,...,d}

∪

J

={1,...,k}

(−1) ^{|I |}

(i) If J ^∗ is a proper subset of {1, . . . , k} then there are no terms in the sum.

(ii) If ∪ _ℓ∈I J ℓ = {1, . . . , k} then each j i ∈ ∪ _ℓ∈I J ℓ , so we conclude that i ∈ I. Therefore

I = {0, . . . , d} and the result follows.

5.3. Explicitly enumerating the coefficients of P g (T ). The leading two terms of P g (T ) are

1 d!

X

J ⊂{1,...,k}

|J|>1

(−1) ^|J|−1 (T ^d − d(N _J − ^d+1 ₂ )T ^d−1 )

= T ^d d! + 1

2

(d + 1)T ^d−1

(d − 1)! − T ^d−1 (d − 1)!

1 min 6 j 6 k ∆ j +

d

X

i=1

1 min 6 j 6 k u j,i

since N J is a sum of various maximums and we have the identity X

J⊂{1,...,k}

|J|>1

(−1) ^|J| max

j∈J a j = − min

16j6k a j (5.4)

for any sequence {a _j }. The proof of (5.4) is an exercise in inclusion-exclusion.

6. Delicate linear algebra and an effective Khovanskii’s theorem The proof of Theorem 1.1 rests on various principles of quantitative linear algebra. The first is an application of the pigeon-hole principle.

Lemma 6.1. Let M be a non-zero m-by-n matrix with integer coefficients and n > m. Let K be the maximum of the absolute values of the entries of M . Then there is a solution to MX = 0 with X ∈ Z ⁿ \ {0} and

||X|| ∞ 6 (Kn) ^m .

To prove Corollary 7.9 in the next section, we will need the more sophisticated Siegel’s lemma due to Bombieri–Vaaler [1], which gives a basis for ker M rather than just a single vector X; for the results in this section, the elementary result in Lemma 6.1 suffices.

Proof. Suppose first that Kn is odd. If there were two distinct vectors X 1 , X 2 ∈ Z ⁿ for which MX 1 = MX 2 and kX _i k ∞ 6 ¹ ₂ (Kn) ^m , then by choosing X = X 1 − X 2 we would be done. Now, the number of vectors X ∈ Z ⁿ for which kXk ∞ 6 ¹

2 (Kn) ^m is equal to (2( ¹ ₂ ((Kn) ^m − 1)) + 1) ⁿ , which is (Kn) ^mn . For all such X we have kMXk ∞ 6 ¹ ₂ (Kn) ^m+1 and MX ∈ Z ^m . We may further assume that MX 6= 0, since otherwise we would be immediately done. There are exactly (2( ¹ ₂ ((Kn) ^m+1 − 1)) + 1) ^m vectors Y ∈ Z ^m \ {0} with kY k ∞ 6 ¹ ₂ (Kn) ^m+1 , i.e exactly (Kn) ^m(m+1) − 1 such vectors. Since n > m + 1, by the pigeonhole principle we may find distinct X ₁ , X ₂ with MX ₁ = MX ₂ as required.

If Kn is even, then the number of vectors X ∈ Z ⁿ for which kXk ∞ 6 ¹

2 (Kn) ^m is exactly ((Kn) ^m + 1) ⁿ , and there are at most ((Kn) ^m+1 + 1) ^m − 1 vectors Y ∈ Z ^m \ {0} with kY k ∞ 6 ¹

2 (Kn) ^m+1 . Since

((Kn) ^m + 1) ⁿ > ((Kn) ^m+1 + 1) ^m − 1,

we can conclude using the pigeonhole principle as before.

Next, we will consider solutions to the equation My = b in which all the coordinates of y are positive integers.

Lemma 6.2. Let M = (µ ij ) i6m, j6n be an m-by-n matrix with integer coefficients, with m 6 n and rank M = m, and let b ∈ Z ^m . Suppose that max i,j |µ _ij | 6 K 1 and kbk _∞ 6 K 2

(where we choose K 1 , K 2 > 1), and suppose that there is some x ∈ Z ⁿ _>0 for which Mx = b.

Then we may find y ∈ Z ⁿ _>0 for which My = b and

kyk _∞ 6 (n − m)(n ^m m ^m K ₁ ^2m + m ^m K ₁ ^m ) + m ^m K ₁ ^m−1 K 2 .

Proof. We prove this by induction on n. The base case is n = m. In this case we observe that M is invertible, and so x = y = M ⁻¹ b. Since (det M ) ⁻¹ 6 1 (as M has integer entries), we conclude that kyk ∞ 6 m!K ₁ ^m−1 K 2 6 m ^m K ₁ ^m−1 K 2 , which gives the base case.

We proceed to the induction step, assuming that n > m + 1. By Lemma 6.1, there is a vector X ∈ Z ⁿ \ {0} such that MX = 0 and

kXk ∞ 6 n ^m K ₁ ^m .

Replacing X by −X if necessary, we may assume that X has at least one positive coordinate

with respect to the standard basis. Let S ⊂ {1, . . . , n} be the set of indices where the

coordinate of X is positive.

Take x from the hypotheses of the lemma, and write x = (x 1 , . . . , x n ) ^T ∈ Z ⁿ _>0 . By replacing x with x − λX for some λ ∈ Z >0 as appropriate, we may assume that there is some i ∈ S for which 1 6 x i 6 kXk _∞ + 1 6 n ^m K ₁ ^m + 1. Fix such an i and x i , and now consider the m-by-(n − 1) matrix M ^{i} which is M with the i ^th column removed. Similarly define x ^{i} ∈ Z ⁿ⁻¹ _>0 to be the vector x with the i ^th coordinate removed. Then

M ^{i} x ^{i} = b − M (x i e i ), where e i is the i ^th standard basis vector in R ⁿ .

Observe that b − M (x i e i ) ∈ Z ^m with

kb − M (x i e i )k _∞ 6 K 2 + K 1 x i 6 K 2 + K 1 (1 + n ^m K ₁ ^m ) 6 n ^m K ₁ ^m+1 + K 1 + K 2 .

Now rank M ^{i} is either m or m − 1. If rank M ^{i} = m then, by the induction hypothesis (with x replaced by x ^{i} ), there is some y ^{i} ∈ Z ⁿ⁻¹ _>0 for which M ^{i} y ^{i} = b − M (x i e i ) and

ky ^{i} k ∞ 6 (n − m − 1)((n − 1) ^m m ^m K ₁ ^2m + m ^m K ₁ ^m ) + m ^m K ₁ ^m−1 (n ^m K ₁ ^m+1 + K 1 + K 2 ) 6 (n − m − 1)(n ^m m ^m K ₁ ^2m + m ^m K ₁ ^m ) + m ^m K ₁ ^m−1 (n ^m K ₁ ^m+1 + K 1 + K 2 )

= (n − m)(n ^m m ^m K ₁ ^2m + m ^m K ₁ ^m ) + m ^m K ₁ ^m−1 K 2 .

Let y := y ^{i} + x i e i , where we have abused notation by treating y ^{i} also as an element of Z ⁿ _{> 0} by extending by 0 in the i ^th coordinate. Then we have y ∈ Z ⁿ _>0 , My = b, and

kyk _∞ 6 max(ky ^{i} k _∞ , n ^m K ₁ ^m + 1) 6 (n − m)(n ^m m ^m K ₁ ^2m + m ^m K ₁ ^m ) + m ^m K ₁ ^m−1 K 2

since n > m + 1. Thus we have completed the induction in this case.

If rank M ^{i} = m − 1 then there are some further cases. If m = 1 and M ^{i} is the zero matrix, then we can choose any vector y ^{i} ∈ Z ⁿ⁻¹ _>0 . Otherwise, we may replace M ^{i} with m − 1 of its rows. Call this new (m − 1)-by-(n − 1) matrix M res ^{i} , and further we may assume that rank M res ^{i} = m − 1. Denote the analogous restriction of the vector b − M (x _i e _i ) as b _res − M (x _i e _i ) _res . Then by the induction hypothesis as applied to M res ^{i} , there is some y ^{i} ∈ Z ⁿ⁻¹ _>0 for which M res ^{i} y ^{i} = b res − M(x i e i ) res and

ky ^{i} k _∞ 6 (n − m)(n ^m−1 m ^m−1 K ₁ ^2m−2 + m ^m−1 K ₁ ^m−1 ) + m ^m−1 K ₁ ^m−2 (n ^m−1 K ₁ ^m + K 1 + K 2 ) 6 (n − m)(n ^m m ^m K ₁ ^2m + m ^m K ₁ ^m ) + m ^m K ₁ ^m−1 K ₂

since m > 2, thus completing the induction as above.

Corollary 6.3. Let M = (µ ij ) i6m, j6n be an m-by-n matrix with integer coefficients, and let b ∈ Z ^m . Suppose that max i,j |µ _ij | 6 K 1 and kbk _∞ 6 K 2 (where we choose K 1 , K 2 > 1), and suppose that there is some x ∈ Z ⁿ _>0 for which Mx = b. Then we may find y ∈ Z ⁿ _>0 for which My = b and

kyk ∞ 6 2n ^m+1 m ^m K ₁ ^2m + m ^m K ₁ ^m−1 K 2 .

Proof. We restrict M to a maximal linearly independent subset of its rows and so obtain an m ^′ -by-n matrix M ^′ with rank M ^′ = m ^′ 6 n. The result follows by applying Lemma 6.2 to

M ^′ .

We introduce a partial ordering < unif on Z ^d by saying that x ≤ unif y if x i 6 y i for all i 6 d (that is, y − x ∈ Z ^d _>0 as in the Mann-Dickson lemma). The next lemma controls the set of minimal solutions (with respect to the partial ordering < unif ) to a certain kind of linear equation.

Lemma 6.4. Let n = n 1 + n 2 > 2 with n 1 , n 2 ∈ Z _>0 . Let M = (µ ij ) i 6 m, j 6 n be an m-by-n

matrix with integer coefficients, and b ∈ Z ^m . Suppose that max _i,j |µ _ij | 6 K ₁ and kbk ∞ 6 K ₂

(where we choose K 1 , K 2 > 1). Let S = S(M, b) = x

y

∈ Z ⁿ _>0

× Z ⁿ _>0

: M x

y

= b

,

and let S _min = S _min (M, b) be defined as S min =

x ∈ Z ⁿ _>0

: ∃y ∈ Z ⁿ _>0

with x

y

∈ S and 6 ∃ x ∗

y ∗

∈ S with x ∗ < unif x

.

If x min ∈ S min then

kx min k ∞ 6 2 ²ⁿ m ^mn K ₁ ^m(n+3) n ^m+1 + 2 ⁿ m ^mn K ₁ ^mn K 2 .

Proof. We use induction on n 1 . If S is empty then Lemma 6.4 is vacuously true. Otherwise S is non-empty and so is S min .

If n 1 = 1 then S ⊂ Z >0 so |S min | = 1 by the well-ordering principle. Writing S min = {x min } we note that there exists ( ^x y ) ∈ S by Corollary 6.3 with x 6 2n ^m+1 m ^m K ₁ ^2m + m ^m K ₁ ^m−1 K 2 , and so x _min 6 x 6 2n ^m+1 m ^m K ₁ ^2m + m ^m K ₁ ^m−1 K ₂ .

If n 1 > 2 let x min ∈ S min and choose y ∈ Z ⁿ _>0

with ( ^x

y ) ∈ S. By Corollary 6.3 we may choose ( ^x y

) ∈ S with kx ∗ k ∞ 6 2n ^m+1 m ^m K ₁ ^2m + m ^m K ₁ ^m−1 K 2 . Thus there is some i 6 n 1 for which

|x _min,i | 6 2n ^m+1 m ^m K ₁ ^2m + m ^m K ₁ ^m−1 K ₂ ,

as otherwise x ∗ < _unif x _min , in contradiction to the fact that x _min ∈ S _min . Fixing such a coordinate i, as in the proof of Lemma 6.2 we let x ^{i} _min denote the vector x min with the i ^th coordinate removed, and let M ^{i} be the matrix M but with the i ^th column removed (from the initial set of n 1 columns). Then

M ^{i}

x ^{i} _min y

= b − M

x _min,i e _i 0

,

where e _i is the i ^th basis vector in R ⁿ

. We have

kb − M((x min,i e i , 0) ^T )k ∞ 6 K 2 + K 1 |x min,i |

6 K 2 + K 1 (2n ^m+1 m ^m K ₁ ^2m + m ^m K ₁ ^m−1 K 2 )

= 2n ^m+1 m ^m K ₁ ^2m+1 + (m ^m K ₁ ^m + 1)K 2

6 2n ^m+1 m ^m K ₁ ^2m+1 + 2m ^m K ₁ ^m K 2 .

The vector x ^{i} _min ∈ Z ⁿ _>0

⁻¹ is in S min (M ^{i} , b − M ( ^x

₀ ^e

)). Indeed, were there another vector (w, z) ^T ∈ Z ⁿ _>0

⁻¹ × Z ⁿ _>0

with (w, z) ^T ∈ S(M ^{i} , b − M ( ^x

₀

^e

)) and w < unif x ^{i} _min , then w + x i e i < unif x min and ( ^w+x

_z ^e

) ∈ S(M, b), contradicting the minimality of x. (We have abused notation here by treating w as both an element of Z ⁿ⁻¹ _>0 and, by extending by 0, an element of Z ⁿ _>0 .) So by the induction hypothesis we have

kx ^{i} _min k _∞

6 2 ²⁽ⁿ⁻¹⁾ m ^m(n−1) K ₁ ^m(n+2) n ^m+1 + 2 ⁿ⁻¹ m ^m(n−1) K ₁ ^m(n−1) (2n ^m+1 m ^m K ₁ ^2m+1 + 2m ^m K ₁ ^m K 2 ) 6 2 ²ⁿ m ^mn K ₁ ^m(n+3) n ^m+1 + 2 ⁿ m ^mn K ₁ ^mn K 2 .

So

kx min k ∞ = max(kx ^{i} _min k ∞ , |x min,i |) 6 2 ²ⁿ m ^mn K ₁ ^m(n+3) n ^m+1 + 2 ⁿ m ^mn K ₁ ^mn K 2

too, and the induction is completed.

We are now ready to prove an effective version of Khovanskii’s theorem. Our method is

a quantitative adaptation of Nathanson–Ruzsa’s argument from [12].

Proof of Theorem 1.1. Without loss of generality, we may first translate A (which preserves the width w(A)) so that 0 ∈ A. Therefore we can assume that max a∈A kak _∞ 6 w(A). We can also assume that A − A contains d linearly independent vectors: If not we can project the question down to a smaller dimension (by removing some co-ordinate but keeping all the linear dependencies) and the result follows by induction on d. So |A| =: ℓ > d + 1.

Let us now recall the lexicographic ordering on Z ^d . If x = (x ₁ , . . . , x _d ) ^T ∈ Z ^d and y = (y 1 , . . . , y d ) ^T ∈ Z ^d we say that x < lex y if there exists some i 6 d for which x i < y i and x _j = y _j for all j < i. This is a total ordering on Z ^d .

Following Nathanson–Ruzsa, we say that an element x ∈ Z ^ℓ _{> 0} is useless if there exists

y ∈ Z ^ℓ _>0 with y < lex x, kyk 1 = kxk 1 and P

i 6 ℓ x j a j = P

j 6 ℓ y j a j . We say that element

x ∈ Z ^ℓ _>0 is minimally useless if there does not exist a useless x ^′ ∈ Z ^ℓ _>0 for which x ^′ < unif x.

Let U denote the set of useless elements and U min be the set of minimally useless elements.

By definition see that

U = [

u∈U

{x ∈ Z ^ℓ _>0 : x > _unif u}.

For x ∈ U min , let I 1 = {i 6 ℓ : x i > 1} and I 2 = {j 6 ℓ : y j > 1} (with y as above).

Now I 1 ∩ I 2 = ∅ else if i ∈ I 1 ∩ I 2 then x − e i is also useless (via y − e i ) contradicting minimality. We may assume that both I 1 and I 2 are non-empty, since otherwise we would have x = y = 0. Evidently min I 1 < min I 2 as y < lex x.

By the Mann-Dickson lemma we know that U min is finite, but now we will be able to get an explicit bound on max(kuk _∞ : u ∈ U min ):

Fix a pair of disjoint non-empty subsets I 1 ∪ I 2 ⊂ {1, . . . , ℓ} with min I 1 < min I 2 , and let n 1 = |I ₁ |, n 2 = |I ₂ |, with n = n 1 + n 2 6 ℓ. We define a (d + 1)-by-n matrix M where the columns are indexed by the elements of I 1 ∪ I 2 , and the row numbers run from 0 to d. If j ∈ I ₁ then M _0,j = 1 and M _i,j = (a _j ) _i for 1 6 i 6 d; if j ∈ I ₂ then M _0,j = −1 and M _i,j = −(a _j ) _i for 1 6 i 6 d. Then the top row of the equation M ( ^x y ) = 0 with x ∈ Z ⁿ _>0

and y ∈ Z ⁿ _>0

gives that kyk 1 = kxk 1 and the i ^th row yields that P

j 6 ℓ x j (a j ) i = P

j 6 ℓ y j (a j ) i for 1 6 i 6 d, so together they yield that P

j6ℓ x _j a _j = P

j6ℓ y _j a _j .

By the minimality of x there cannot exist (x ∗ , y ∗ ) ∈ Z ⁿ _>0

× Z ⁿ _>0

such that M ( ^x y

) = 0 and x ∗ < unif x. Indeed, by construction of I 1 and I 2 we would have (after extending by zeros) that y ∗ < lex x ∗ , thus implying that x ∗ is useless – contradicting the fact that x is minimally useless.

Using Lemma 6.4, as applied to the matrix M with K ₁ = max a∈A kak ∞ := K and K ₂ = 1, we conclude that

kxk ∞ 6 2 ^2ℓ (d + 1) ^ℓ(d+1) ℓ ^d+2 K ^(d+1)(ℓ+3) + 2 ^ℓ (d + 1) ^ℓ(d+1) K ^ℓ(d+1) 6 2 ^2ℓ+1 (d + 1) ^ℓ(d+1) ℓ ^d+2 K ^(d+1)(ℓ+3)

In [12, Lemma 1], Nathanson and Ruzsa proved that for all U ^′ ⊂ U _min B(N, U ^′ ) := |{x ∈ Z ^s _>0 : kxk ₁ = N, x > _unif u for all u ∈ U ^′ }|

is equal to a fixed polynomial in N , once N > ℓ max u∈U

kuk ∞ . Indeed, let U ^′ = {u ₁ , . . . , u _m }, where each u j = (u 1,j , u 2,j , . . . , u s,j ) ∈ Z ^ℓ _>0 . Letting u ^∗ _i = max j6m u i,j , and

u ^∗ = (u ^∗ ₁ , u ^∗ ₂ , . . . , u ^∗ _ℓ ), we have that

B (N, U ^′ ) = |{x ∈ Z ^ℓ _>0 : kxk 1 = N, x > _unif u ^∗ }|

= |{x ∈ Z ^ℓ > 0 : kxk ₁ = N − ku ^∗ k ₁ }|

=

N − ku ^∗ k ₁ + ℓ − 1 ℓ − 1

provided N > ku ^∗ k ₁ , which is a polynomial in N . Since ku ^∗ k ₁ 6 ℓ max u∈U

kuk ∞ , our claim

follows.

Then by inclusion-exclusion we have

|NA| = |{x ∈ Z ^ℓ _>0 : kxk 1 = N, x is not useless}|

= X

U

⊂U

(−1) ^|U

^| B(N, U ^′ ) which is a polynomial in N once N > N _Kh (A) where

N Kh (A) 6 2 ^2ℓ+1 (d + 1) ^ℓ(d+1) ℓ ^d+3 K ^(d+1)(ℓ+3) 6 (2ℓw(A)) ^(d+4)ℓ ,

as K := max a∈A kak _∞ 6 w(A). To obtain the last displayed inequality we assumed that d > 2 (as we use Lemma 2.1 for d = 1 which gives N Kh (A) 6 w(A) − 1), and ℓ > d + 1.

7. Structure bounds in the general case: proof of Theorem 1.3

We start by introducing the central structural result of this section. As a reminder, we say that p ∈ ex(H(A)) if there is a vector v ∈ span(A − A) \ {0} and a constant c such that hv, pi = c and hv, xi > c for all x ∈ H(A) \ {p}.

Lemma 7.1 (Decomposing P (A)) . Let B ∪ {0} ⊂ A ⊂ Z ^d with |A| = ℓ and B a non- empty linearly independent set. Suppose that 0 ∈ ex(H(A)). Let A ⁺ denote the set of x ∈ P (A) ∩ C B with the property that, for all b ∈ B, x − b / ∈ P(A) ∩ C B . Then A ⁺ has the following two properties:

C _B ∩ P (A) = A ⁺ + P (B ∪ {0}) (7.1)

and

A ⁺ ⊂ NA for some N 6 N 0 (A) := 2 ^11d

d ^12d

ℓ ^3d

w(A) ^8d

.

Lemma 7.1 is straightforward for large N ((7.1) was already given in [5, Proposition 4]) but our focus is on getting an effective bound on such N.

The only other ingredient in the proof of Theorem 1.3 is the following classical lemma:

Lemma 7.2 (Carath´eodory). Let A ⊂ R ^d be a finite set, and let V := span(A − A). If dim V = r, then

H(A) = [

B⊂ex(H(A))

|B|=r+1 span(B−B)=V

H(B).

Proof. After an affine transformation one may assume that V = R ^d . Then see [5, Lemma 4]

for the proof, in which the union is taken over all B ⊂ A with |B| = d +1 and span(B −B) = R ^d . The equality as claimed, where B ⊂ ex(H(A)), then follows from general fact that

H(A) = H(ex(H(A)) (see Lemma A.1).

Proof of Theorem 1.3. Let

v ∈ (NH(A) ∩ (a 0 N + Λ A−A )) \ ( [

b∈ex(H (A))

(bN − E (b − A))). (7.2) We will show that v ∈ NA for all N > (d + 1)N 0 (A).

Let V = span(A − A) and r = dim V as above. By Lemma 7.2 there exists a set B ⊂ ex(H(A)) with |B| = r + 1 such that v ∈ NH (B ^∗ ) and span(B − B ) = V . Write B = {b ₀ , b 1 , . . . , b r }. Since v ∈ NH (B ) we can write v = P r

i=0 c i b i for some real c i > 0 such that P r

i=0 c _i = N . Since N > (d + 1)N ₀ (A) there must be some c _i > N ₀ (A). After permuting coordinates, we will assume that c r > N 0 (A). Thus

b r N − v =

r−1

X

i=0

c i (b r − b i ) ∈ (N − N 0 (A)) · H(b r − B),

so that b r N − v ∈ C b

−B ⊂ C b

−A . By the assumption (7.2) we also have b r N − v / ∈ E (b r − A) and b r N − v ∈ Λ b

−A . Hence b r N − v ∈ P (b r − A). We may now apply Lemma 7.1 to the sets b r − A and (b r − B) \ {0}; the hypotheses are satisfied since b r ∈ ex(H(A)) implies 0 ∈ ex(H(b r − A)). Furthermore, w(b r − A) = w(A). We thus obtain

b r N − v ∈ C b

−B ∩ P (b r − A) = A ⁺ + P (b r − B ^∗ ) for some set A ⁺ ⊂ C b

−B ∩ P (b r − A) with A ⁺ ⊂ N 0 (A)(b r − A).

Now let us write

b r N − v = u + w, with u ∈ A ⁺ and w ∈ P (b r − B). Thus u + w = P r−1

i=0 c i (b r − b i ), with c i ∈ R >0 for all i and P r−1

i=0 c i 6 N − N 0 (A). Expressing u and w with respect to the basis (b r − B) \ {0}, and noting that u ∈ A ⁺ ⊂ C b

−B ∩ N 0 (A)(b r − A), we infer that w = P r−1

i=0 γ i (b r − b i ) with

γ i 6 c i and γ i ∈ Z >0 for all i. Hence w ∈ (N − N 0 (A))(b r − B).

Putting everything together we have

b r N − v = u + w ∈ N 0 (A)(b r − A) + (N − N 0 (A))(b r − B )

⊂ N 0 (A)(b r − A) + (N − N 0 (A))(b r − A) = N(b r − A).

Hence v ∈ NA as required.

The proof shows that N Str (A) 6 (d+1)N 0 (A) = (d+1)2 ^11d

d ^12d

ℓ ^3d

w(A) ^8d

6 (dℓ w(A)) ^13d

as we may take d > 2 (after Lemma 2.1) .

It remains to prove Lemma 7.1. The condition x ∈ P(A) ∩ C B but x − b / ∈ P(A) ∩ C B

in the definition of A ⁺ is a minimality-type condition on x. As our argument for analysing the set A ⁺ will not stay within C B , it turns out to be convenient to separate the P (A) part and the C B part of this condition; this motivates the following definition.

Definition 7.3 (Absolutely B-minimal). Let B ∪ {0} ⊂ A ⊂ Z ^d , with A finite. We say that u ∈ P (A) is absolutely B-minimal with respect to A if u − b / ∈ P (A) for all b ∈ B. Let S _abs (A, B) denote the set of absolutely B-minimal elements.

Let S _abs (A, ∅) = P (A) and use the convention that C ∅ = {0}. By this definition S _abs (A, B) ⊂ S(A, B), though these sets needn’t be equal, so being a B-minimal element is a weaker condition than being an absolutely B -minimal element.

For a subset U ⊂ R ^d and x ∈ R ^d , we define dist(x, U ) := inf

u∈U kx − uk ∞ .

Lemma 7.4 (Controlling the absolutely B-minimal elements). Let B ∪ {0} ⊂ A ⊂ Z ^d with |A| = ℓ > 2, and assume that B is a (possibly empty) linearly independent set. Let r := dim span(A) and suppose that 0 ∈ ex(H(A)). If x ∈ S abs (A, B ) and dist(x, C B ) 6 X then x ∈ NA for some N ∈ Z >0 with

N 6 (X + 1)2 ^10dr d ^11d

^r ℓ ^3dr w(A) ^7d

^r .

Lemma 7.4 is the main technical result of this section. The hypotheses allow r to be less than d, even though we will only apply the lemma when r = d, since our proof involves induction on r. Similarly, we do not assume that Λ A = Z ^d ∩ span(A), as this property would not necessarily be preserved by the induction step. Deducing Lemma 7.1 is straightforward:

Proof of Lemma 7.1. If x ∈ A ⁺ then we can partition B = B ^′ ∪ B ^′′ so that b ^′ ∈ B ^′ implies x − b ^′ ∈ / C B and b ^′′ ∈ B ^′′ implies x − b ^′′ ∈ C B \ P (A).

Writing x with respect to the basis B, we get x = ℓ + X

b

∈B

c b

b ^′′ where ℓ = X

b

∈B

ℓ b

b ^′

with c b

> 1 for all b ^′′ ∈ B ^′′ and ℓ b

∈ [0, 1) for all b ^′ ∈ B ^′ .

Since kℓk _∞ 6 dw(A), this implies that dist(x, C B

) 6 dw(A). Furthermore, for all b ^′′ ∈ B ^′′

we have x − b ^′′ ∈ P(A). Hence / x ∈ S _abs (A, B ^′′ ). By Lemma 7.4 as applied to B ^′′ and X = dw(A), we may conclude that x ∈ NA for N > N 0 (A) as in Lemma 7.1.

To establish (7.1), note that A ⁺ + P (B ∪ {0}) ⊂ P(A) ∩ C B by definition. On the other hand if y ∈ P(A) ∩ C _B and there exists some b ₁ ∈ B with y − b ₁ ∈ P(A) ∩ C _B then we replace y by y − b 1 . We repeat this with b 2 , . . . until the process terminates, which it must do since the sum of the coefficients of y with respect to the basis B decreases by 1 at each step. We are left with y − b 1 − · · · − b k ∈ A ⁺ so that y ∈ A ⁺ + P (B ∪ {0}).

It remains is to prove Lemma 7.4. Following the proofs in [6, 3] we now show that in certain favourable circumstances, S abs (A, B ) may be controlled in terms of the Davenport constant of Z ^d /Λ B . However this is not used in our proof of Lemma 7.4 (except when d = 1) but, for reasons of motivation, it is helpful to understand why this type of argument fails.

Lemma 7.5. Let B ∪ {0} ⊂ A ⊂ Z ^d , with A finite and B a basis of R ^d . Suppose that C A = C B . Let Z ^d /Λ B := G. Then S _abs (A, B) ⊂ NA, where N = max(1, D(G) − 1) and D(G) is the Davenport constant of G.

Proof. Let x ∈ S _abs (A, B), and assume that x 6= 0. Then write x = a 1 + a 2 + · · · + a N

(x)

for some a i ∈ A. If there were a subsum P

i∈I a i ≡ 0 mod Λ B , then since C A = C B

we would have P

i∈I a i ∈ C A ∩ Λ B ⊂ C B ∩ Λ B . But since B is a basis of R ^d we have C B ∩ Λ B = P (B) ∪ {0}, so P

i∈I a i ∈ P (B) ∪ {0}. By minimality of N A (x) we also have P

i∈I a i 6= 0. Therefore x ∈ P(A) + y for some non-zero y ∈ P(B ), contrary to the assumption that x ∈ S abs (A, B ). Hence N A (x) 6 max(1, D(G) − 1), which also takes care

of the x = 0 case.

If C B is a strict subset of C A then the above argument doesn’t necessarily work, as P

i∈I a i ≡ 0 mod Λ B does not automatically imply that P

i∈I a i ∈ P (B) ∪ {0}: Indeed the key issue is how an element a 1 + · · · + a N = x ∈ P (A) ∩ C B can have partial sums P

i∈I a i ∈ / C B . Sketch of our proof of Lemma 7.4. The easy cases are r = 1 (which follows from any of the existing literature [11, 14, 5, 6], or from Lemma 7.5) and B = ∅ (which is dealt with in Lemma 7.11 below). From these base cases, we will construct a proof by induction on r. We may assume, therefore, that r > 2 and B is non-empty. For this sketch, we will also assume that r = d. There are three main phases to the induction step.

• We provide an extra restriction on the region of R ^d where S _abs (A, B) can lie, by showing that if x ∈ S _abs (A, B) then dist(x, ∂(C _A )) 6 Y , where ∂(C _A ) is the topological boundary of C A and Y is some explicit bound. ⁶ The bound dist(x, ∂(C A )) 6 Y is a generalisation of a basic result from the one dimensional case – the classical ‘Frobenius postage stamp’ problem – in which the boundary of C A is just {0} and one shows that the exceptional set E (A) is finite. Since ∂(C A ) is a union of d − 1 dimensional facets, there is some non-zero linear map α : R ^d −→ R for which dist(x, ker α) 6 Y .

• We combine the distance condition from above with the hypotheses of Lemma 7.4, giving dist(x, C B ) 6 X and dist(x, ker α) 6 Y . In turn, we show that this implies dist(x, C B ∩ ker α) 6 f(X, Y, A) (for some explicit function f (X, Y, A)), by a quantitative linear algebra argument. For this part, one should have in mind the situation of two rays, both starting from the origin. If x is in a neighbourhood of both rays separately, then x will be in some neighbourhood of the origin. The size of this neighbourhood will be determined by the

6 If r 6 d − 1 then one cannot use the topological boundary here, since ∂(C A ) = C A in this case, but this

issue may be circumvented.