Bounds on the lattice point enumerator via slices and projections

Gardner, Gronchi and Zong posed the problem to find a discrete analogue of M. Meyer's inequality bounding the volume of a convex body from below by the geometric mean of the volumes of its slices with the coordinate hyperplanes. Motivated by this problem, for which we provide a first general bound, we study in a more general context the question to bound the number of lattice points of a convex body in terms of slices as well as projections.


Introduction
One of the central questions in Geometric Tomography is to determine or to reconstruct a set K in the n-dimensional Euclidean space R n by some of its lower dimensional "structures" (see [15]). Usually, these are projections on and sections with lower dimensional subspaces of R n . A classical and very well-known example in this context is the famous Loomis-Whitney inequality [29], which compares the volume of a non-empty compact set K to the geometric mean of its projections onto the coordinate hyperplanes: Here vol(K) denotes the volume, i.e., the n-dimensional Lebesgue measure of the set K, and vol n−1 (K|e ⊥ i ) the (n − 1)-dimensional volume of the orthogonal projection of K onto the coordinate hyperplane orthogonal to the ith unit vector e i . Equality is attained, e.g., if K = [a 1 , b 1 ] × ... × [a n , b n ], a i , b i ∈ R, is a rectangular box. For various generalizations and extensions of this inequality we refer to [8] and the references within.
Loomis and Whitney proved (1.1) by observing that it suffices to prove it when K is the pairwise non-overlapping disjoint union of equal cubes which is then a purely combinatorial problem. In particular, this combinatorial version implies (and is actually equivalent to) the following discrete variant of (1.1) where #(M ) = |M ∩ Z n | is the lattice point enumerator (with respect to Z n ). So (1.1) and (1.2) are equivalent statements for compact sets. The discrete version (1.2) was also independently proven by Schwenk and Munro [36]. Due to the comparison of n-and (n − 1)-dimensional volumes in (1.1), it is easy to see that there is no lower bound on the volume in terms of the geometric mean of vol n−1 (K|e ⊥ i ). However, if we further assume that K ∈ K n , i.e., it belongs to the family of convex and compact sets, and if we replace projections by sections, then it was shown by M. Meyer [33] (1. 3) vol(K) where equality is attained, if and only if K is a generalized crosspolytope, i.e., K = conv{a 1 e 1 , −b 1 e 1 , ..., a n e n , −b n e n }, for some a i , b i ≥ 0. Observe that n! 1/n /n is asymptotically 1/e. In [14], Gardner, Gronchi and Zong posed the question to find a discrete analogue of M. Meyer's inequality (1.3); more precisely, they asked Question. Let n ∈ N. Is there a constant c n > 0 such that for all K ∈ K n (1.4) #(K) As in the case of the Loomis-Whitney inequality, a discrete version (1.4) would imply the analogous inequality for the volume, and hence, by (1.3) we certainly have c n ≤ n! 1/n /n (cf. (2.9)). In the plane, Gardner et al. [14] proved for any K ∈ K 2 . The elongated cross-polytope K = conv{±e 1 , ±he 2 } shows that 1 √ 3 is asymptotically best possible, i.e., for h → ∞. Hence, in contrast to the Loomis-Whitney inequality, (1.3) has no equivalent discrete version, since in the plane the constant in Meyer's inequality is 1/ √ 2. The short answer to the question above by Gardner et al. is "No!" for arbitrary convex bodies and n ≥ 3 (cf. Proposition 3.1). Restricted to the set K n o of origin-symmetric convex bodies, however, we have the following result.
We do not believe that this inequality is best possible. Instead, we propose 3 1−n n to be the right constant (cf. Example 3.2). While it is not possible to bound the volume of a symmetric convex set K from above in terms of n i=1 vol n−1 (K ∩ e ⊥ i ), Feng, Huang and Li proved in [12] that there is a constantc n ≤ (n − 1)! such that for any K ∈ K n o there exists an orthogonal basis u 1 , ..., u n ∈ R n such that: In a recent preprint, Alonso-Gutiérrez and Brazitikos [2] improved this result considerably: they showed that up to a universal constant the best possiblec n is equal to the isotropic constant L K which is bounded from above by n 1/4 (see [24]). Moreover, they proved that this is valid for any centered convex body, i.e., for a convex body whose centroid is at the origin. Inspired by this, we prove the following inequaltities: There exists a basis b 1 , ..., b n of the lattice Z n such that and there exists t i ∈ Z n , 1 ≤ i ≤ n, such that Observe that due to Brunn's concavity principle (see, e.g., [3, Theorem 1.2.1]) the volume maximal slice vol n−1 (K ∩ (t + u ⊥ )), t ∈ R n , of an originsymmetric convex body K ∈ K n o is always the central slice, i.e., t = 0. This is no longer true regarding lattice points which explains the difference bewtween (1.6) and (1.7). Here we have the following kind of a discrete Brunn's concavity principle.
and the inequality is best possible.
From (1.7) we get for K ∈ K n o immediately an inequality of the type with c n = O(n 2 ). This may be regarded as a lattice version of the wellknown slicing problem for volumes asking for the correct order of a constant c such that for all centered convex bodies K ∈ K n there exists a u ∈ R n \{0} such that To this day, the best known bound c ≤ n 1/4 is due to Klartag [24]. With respect to the discrete slicing inequality (1.8) we prove If K ∈ K n o , the constant can be replaced by O(n). Finally, we will give an example that shows that all the constants in (1.6), (1.7), (1.8) and (1.10) must be at least of order √ n.
Theorem 1.5. For n ∈ N there exists a sequence of symmetric convex bodies where H ranges over all affine hyperplanes in R n and c > 0 is a universal constant.
We want to remark that the slicing problem (1.9) has been extensively studied also for other measures. For instance, Koldobsky [26] proved for measures µ that admit a continuous density (see also [25] and the references within). Chasapis, Giannopoulos and Liakopoulos [11] extended this result to lower dimensional sections of not necessarily symmetric convex bodies where F ranges over all k-dimensional subspaces of R n and µ is a measure with a locally integrable density function. In [1] the authors obtained an inequality similar to (1.11) for K ∈ K n o and the lattice point enumerator where F ranges over all k-dimensional linear subspaces with dim(F ∩ Z n ) = k. In the case k = n − 1 and convex bodies of "small" volume, Regev [34] proved via a probablistic approach such an inequality with the constant O(n) instead of O(n) n−1 . Finally, we discuss a reverse Loomis-Whitney inequality in the sprit of (1.5), Campi, Gritzmann, and Gronchi [9] showed that there exists a constantd n ≥ c/n, where c is an absolute constant, such that (1.12) vol(K) where again u 1 , . . . , u n form an suitable orthonormal basis. In [27], Koldobsky, Saroglou and Zvavitch showed that the optimal order of the constant d n is of size n −1/2 . In order to get a meaningful discrete version of (1.12) we have to project so that Z n |u ⊥ i is again a lattice, i.e., u i ∈ Z n , and we have to count the lattice points of K|u ⊥ i with respect to this lattice. Theorem 1.6. Let K ∈ K n o with dim(K ∩ Z n ) = n. There exist linearly independent vectors v 1 , ..., v n ∈ Z n such that The paper is organized as follows. In the next section we recall briefly some basic definitions and tools from Convex Geometry and Geometry of Numbers needed for the proofs. Section 3 is devoted to the slicing inequalities, in particular, we provide the proofs of the Theorems 1.1 to 1.5 as well as of Lemma 1.3. The proof of Theorem 1.6 is given in Section 4, and in the final section we discuss improvements for the special class of unconditional bodies.

Preliminaries
For a non-zero vector x ∈ R n , we denote its orthogonal complement by x ⊥ and its Euclidean norm by |x|. We write B n = {x ∈ R n : |x| ≤ 1} for the Euclidean unit ball and C n = [−1, 1] n for the symmetric cube. The interval between x, y ∈ R n is defined as [x, y] = {λx + (1 − λ)y : λ ∈ [0, 1]}. For two sets A, B ⊆ R n the Minkowski addition is defined elementwise, i.e., Similarly, for a scalar λ ∈ R, one defines λA = {λ a : a ∈ A} and we write −A = (−1)A.
A convex body is a compact convex set K ⊆ R n . We say that K is origin symmetric, if K = −K. The set of all convex bodies is denoted by K n and the set of all origin symmetric convex bodies is denoted by K n o . The support function of a convex body K is defined for x ∈ R n as h K (x) = sup y∈K x, y . If the origin is an interior point of K, the polar body of K is defined as Moreover, for such K, the gauge function | · | K : R n → R ≥0 is defined by |x| K = min{µ ≥ 0 : x ∈ µK}; it is | · | K = h K (·) [35, Lemma 1.7.13]. The volume vol(K) of a convex body K is its n-dimensional Lebesgue measure. If K is contained in an k-dimensional space F , we denote by vol k (K) its kdimensional Lebesgue measure in F . It is a famous open problem in Convex Geometry to find the best possible lower bound on the volume product vol(K)vol(K ), where K ∈ K n o . Mahler conjectured that it is 4 n /n! and it is known to be true with π n /n! [28]. Here we will just use For X ⊆ R n , we denote the convex hull of X by conv(X). If X is finite, conv(X) is called a polytope and if, in addition, X ⊆ Z n , we call conv(X) a lattice polytope.
In general, a lattice Λ ⊆ R n is a discrete subgroup of R n of the form Zv is a primitive sublattice of Λ. The polar lattice of Λ is defined as There are several duality relations between Λ and Λ of which we recall a few here (cf. e.g. [32,Proposition 1.3.4]). First of all the determinants of Λ and Λ are linked by the simple formula det(Λ) det(Λ ) = 1. Further, a k-dimensional subspace L ⊆ R n is a lattice subspace of Λ, if and only if L ⊥ is an (n − k)-dimensional lattice subspace of Λ . In particular, every lattice hyperplane H of Λ possesses a primitive normal vector v ∈ Λ and the determinant of Λ ∩ H is given by |v | det Λ. Moreover, the orthogonal projection Λ |L is a k-dimensional lattice and we have the following relation: where on the left-hand side, the polarity operation is taken within the space L.
For a set A ⊂ R n we denote by |A| its cardinality and for the lattice point enumerator of a set A ⊆ R n with respect to a lattice Λ ⊆ R n we write Minkowski established via his successive minima various fundamental relations between the volume and lattice point properties of a symmetric convex body. For K ∈ K n o and a lattice Λ ⊆ R n , both full-dimensional, the ith successive minimum is defined as Here we also need a discrete variant of the upper bound going back to Betke et al. [6]. For K ∈ K n o they proved which was later improved in [20]. The currently best known upper bound is due to Malikiosis [31], which, in particular, implies for In general, linearly independent lattice points a i ∈ Λ, 1 ≤ i ≤ n, corresponding to the successive minima, i.e., a i ∈ λ i (K, Λ) K, do not form a basis of Λ. It was shown by Mahler (cf. [18, Sec. 2.10]), however, that there exists a lattice basis b 1 , ..., b n ∈ Λ such that Next we also need a lower bound on the product of the successive minima of a convex body K with the ones of the polar body K (cf. [17, Theorem 23.2]): Regarding upper and lower bounds on the volume in terms of the lattice point enumerator we mention here two results. First, van der Corput [18, and Blichfeldt [7] showed for K ∈ K n with dim(K ∩ Λ) = n Finally, the volume and the lattice point enumerator are equivalent "on a large scale", i.e., for any n-dimensional convex body K ⊆ R n , n-dimensional lattice Λ ⊆ R n and t ∈ R n one has (cf. e.g. [37, Lemma 3.22]) det Λ In this paper we will mostly deal with the standard lattice Z n since all the results can easily be generalized to arbitrary lattices.
For more information on Geometry of Numbers and/or Convex Geometry we refer to the books [15,17,18,35]. Proposition 3.1. Let n ≥ 3 be fixed. There exists no positive number c > 0 such that for all K ∈ K n .

Slicing inequalities for the lattice point enumerator
Proof. We first prove it for n = 3. For an integer r ∈ N, let T k be the simplex with vertices {0, e 1 , e 1 + k e 2 , k e 3 } (see Figure 1). Then, #(T k ) = 2(k + 1) and also #( and so for k → ∞ the left hand sides tends to 0. For n ≥ 4 we just can consider, e.g., the simplices conv(T k ∪{e 4 , . . . , e n }).
Roughly speaking, the simplex T k from above falsifies (3.1) because the two skew segments [0, ke 3 ] and [e 1 , e 1 + k e 2 ] are both "long", but do not generate any additional points in T k . Such a construction is not possible in the symmetric case. In fact, if K ∈ K n o possesses 2h + 1 points on the coordinate axis Re n , any interior point v ∈ K ∩ e ⊥ n will contribute O v,n (h) points to K. Here, O v,n hides a constant that only depends on v and n. However, unlike the simplex above, a symmetric convex body always contains at least #(K)/3 n -many interior lattice points (see [16]). Motivated by this heuristic, we conjecture the following polytopes to be extremal in which is why we conjecture the optimal constant in (3.1) to be 3 − n−1 n ≈ 1/3.
In order to prove the lower bound in Theorem 1.1, we would first like to understand the behaviour of #(·) with respect to affine transformations. An important tool for this is the index of a sublattice Λ ⊆ Λ, i.e., the number of different cosets a + Λ , a ∈ Λ. If dim Λ = dim Λ, this number is known to be finite and is given by det(Λ )/ det(Λ). The following lemma will be used extensively throughout this paper.
A ∈ Z n,n be a regular matrix and t ∈ R n . Then, In particular, we have which is best possible, and for m ∈ N, which is best possible up to a factor (1 + 1 2m−1 ) n . Proof. Let Λ = 2AZ n ⊆ Z n and let Γ i ⊆ Z n , 1 ≤ i ≤ 2 n | det A|, be the cosets of Λ in Z n . Consider two points y j = A x j + t ∈ A K + t, where x j ∈ K, j = 1, 2, that belong to a common Γ i , say. For such points, we have y 1 − y 2 ∈ Λ. Thus, by the symmetry of K, we have That means Since for any two finite sets A, B ⊆ R n (cf. In fact, Wills [38] showed that for any lattice polygon P ⊆ R 2 and t ∈ R 2 one has #(P + t) ≤ #(P ). We conjecture that for any lattice polytope P ∈ K n o #(P + t) ≤ 2 n−2 #(P ).
ii) If K is not necessarily symmetric, (3.2) and (3.3) fail. In that case, counterexamples are given by the simplices T k in the proof of Proposition 3.1. Basically, the reason for this is that T k contains O(k) lattice points, while in a translation or dilation of T k one may find a rectangular triangle spanned by two orthogonal segments of length O(k) each, lying in a hyperplane of the form e ⊥ i + t, t ∈ Z 3 . Such a triangle contributes O(k 2 ) points. Again, letting k → ∞ shows that the inequalities cannot be generalized to the non-symmetric case. This has also been observed independently by Lovett and Regev in [30].
Next we come to the proof of Theorem 1.1.
We note that by the symmetry and convexity of K as well as the definition of 2 Λ i we readily have φ (a 1 , . . . , a n ) ∈ (K ∩ Z n ) n−1 . Moreover, since a i ∈ e ⊥ i , 1 ≤ i ≤ n, the map is injective and so Together with (3.6) we obtain Next, we want to reverse the inequality in Theorem 1.1. Apparently, one cannot get an upper bound on #(K) in terms of the geometric mean of the sections K ∩ e ⊥ i . Here we have to replace the e i by a lattice basis b 1 , ..., b n of Z n that "suits" the body K. Our strategy will then be to decompose Z n = ∪ j∈Z {x ∈ Z n : x, b i = j} and estimate the sections parallel to b ⊥ i against the central one. As for the volume Brunn's concavity principle states that vol K ∩ (t + L) ≤ vol(K ∩ L), for any k-dimensional linear subspace L ⊆ R n , t ∈ R n and K ∈ K n o . So the volume-maximal section of K parallel to L is indeed always the one containing the origin. Unfortunately, this is false in the discrete setting as the following example shows: Let K = conv ± ([0, 1] n−1 × {1}) and for 1 ≤ k ≤ n − 1, let L k = span{e 1 , ..., e k }. Then #(K ∩ L k ) = 1, but #(K ∩ (e n + L k )) = 2 k .
Indeed, the deviation between the central section and the maximal section is extremal for K as above, as claimed in Lemma 1.3 which we prove next.
Proof of Lemma 1.3. We may assume t ∈ Z n . In that case, Λ = L ∩ Z n and Λ = Λ + t are k-dimensional (affine) sublattices of Z n and both of them intersect exactly 2 k cosets of Z n /2Z n : Let Γ ∈ Z n /2Z n such that Λ ∩ Γ = ∅. Then Λ ∩ Γ ∈ Λ/2Λ and since two different cosest do not have a common point, we see that Λ intersects at most |Λ/2Λ| = 2 k cosets of Z n /2Z n . Conversely, since Λ arises as a section of Z n with a linear subspace, there exists a complementary lattice Λ ⊆ Z n such that Λ ⊕ Λ = Z n . Then, every cosetΓ ∈ Λ/2Λ defines a unique cosetΓ ⊕ 2Λ of Z n /2Z n and so Λ meets at least 2 k cosets of Z n /2Z n . Regarding the affine lattice Λ , we note that the translation by t serves as a bijection between cosets in Λ and the cosets in Λ . Now consider two points x, y ∈ K ∩ Λ belonging to a common coset Γ of Z n /2Z n . By the symmetry of K, we have 1 2 (x − y) ∈ K ∩ Λ and so Thus, by (3.5), every coset of Z n /2Z n that is present in Λ ∩ K contains at most 1 2 (#(K ∩ Λ) + 1) points of K ∩ Λ . Hence, |K ∩ Λ | ≤ 2 k 1 2 (|K ∩ Λ| + 1).
Now we are ready for the proof of the reverse Meyer Theorem 1.2.
Proof of Theorem 1.2. By induction on the dimension, we will show that for any n-dimensional convex body K ∈ K n o and any n-dimensional lattice Λ, there exists a basis b 1 , ..., b n of Λ and vectors t 1 , ..., t n ∈ Λ such that From this, (1.7) follows by considering Λ = Z n and taking the nth root. Moreover, (1.6) follows immediately from (1.7) and Lemma 1.3.
First, we assume dim(K ∩ Λ) = n. For any b ∈ Λ we may write where t b ∈ Λ is chosen to be the translation that maximizes the number of lattice points in a section parallel to b ⊥ . Now let b 1 , . . . , b n ∈ Λ be a basis of Λ obtained from (2.5) with respect to the polar body K , i.e., we have |b i | K ≤ iλ i (K , Λ ), 1 ≤ i ≤ n. For the vectors b i we denote the above translation vectors t b i by t i . Then, on account of h(K, b i ) = |b i | K we conclude from (3.8) where for the last inequality we used λ i (K , Λ ) ≥ 1 which follows from the assumption dim(K ∩ Λ) = n via (2.6).
Using the upper bound of Minkowski's theorem (2.3), the lower bound on the volume product (2.1) and van der Corput's inequality (2.7), we estimate Substituting this into (3.9) yields the desired inequality (3.7) for this case. It remains to consider the case dim(K ∩ Λ) < n, so let K ∩ Λ ⊆ H for some (n − 1)-dimensional lattice subspace H ⊆ R n . Let Γ = Λ ∩ H and next we apply our induction hypothesis to Γ and K ∩ H. Hence, we find a basis y 1 , ..., y n−1 of Γ and vectors t 1 , ..., t n−1 ∈ Γ such that which is equivalent to where b n ∈ Λ is a primitive normal vector of H. Unfortunately, the independent system {y 1 , ..., y n−1 , b n } is in general not a basis of Λ . In fact, the y i 's are not elements of Λ in the first place. In view of (2.2), we have Γ = Λ |H = Λ |b ⊥ n . So there are vectors b i ∈ (y i + Rb n ) ∩ Λ . For these vectors, one has b ⊥ i ∩ H = y ⊥ i ∩ H. By our assumption on K, this means In view of (3.11) and (3.12), {b 1 , ..., b n } is the desired basis and our proof is complete.
Remark 3.5. We remark that if we just want to find linearly independent lattice points a i ∈ Z n , 1 ≤ i ≤ n, for the slices in Theorem 1.2 instead of a basis, then one can save one factor of n in the bounds of Theorem 1.2. For if, we replace in the proof above the basis vectors b i by linearly independent lattice points a i ∈ λ i (K )K ∩ Z n , 1 ≤ i ≤ n. In this way we do not need the estimate (2.5). In particular this leads to In the remainder of this section, we want to generalize (3.13) to the nonsymmetric case.
If we consider (3.8), we see that we can also estimate #K, if K is not symmetric; in that case one replaces the factor (2 h(K, b) + 1) on the right hand side by the number of hyperplanes parallel to b ⊥ that intersect K. As it will turn out, the challenge then is to compare the number of lattice points in K − K to the number of points in K.
As for the volume, this is known as the Rogers-Shephard inequality, which asserts that The simplices T k given in the proof of Proposition 3.1 show that there is no similar inequality for the lattice point enumerator: While we have #(T k ) = O(k), in T k − T k we find the triangle conv{0, k/2e 2 , ke 3 } which contains O(k 2 ) lattice points. Since k can be arbitrarily large, there is no constant c n > 0 depending only on the dimension such that #(K − K) ≤ c n #(K). However, the simplices T k are extremely flat and therefore easily admit a large hyperplane section. Our strategy in order to prove Theorem 1.4 will be to argue that convex bodies K whose difference body K − K contains disproportionately many lattice points are automatically flat. This reasoning is inspired by the proof of Theorem 4 in [4].
First we assume that 1 2 (K +c) = c+ 1 2 (−c+K) does not contain any integral lattice point. Then the well-known flatness theorem implies that (cf. [5]) In view of (3.14) we are done. So we can assume that there is a lattice point a ∈ 1 2 (c + K) ∩ Z n . By the choice of c (cf. (3.15)), we get Now in order to bound λ * in this case we use (3.10) applied to (K − K) and Z n , which implies Together with (3.14) we obtain (3.17) #(K) n ≤ n!4 n #(K − K)# K ∩ (t + y ⊥ ) n .
After taking the n-th root, we have as desired.
We finish the section by proving Theorem 1.5 providing a lower bound on the constants presented in Theorem 1.2 and 1.4.
Proof of Theorem 1.5. We consider a lattice Λ ⊆ R n such that Λ is selfpolar, i.e., Λ = Λ , and λ 1 (B n , Λ) = c √ n, where c is an absolute constant. Such lattices have been detected by Conway and Thompson [22,Theorem 9.5]. We will use a volume approximation argument for the Euclidean ball rB n = {x ∈ R n : |x| ≤ r}, where r → ∞.
For x ∈ R n \ {0} and α ∈ R, let H(x, α) = {y ∈ R n : x, y = α} be the corresponding hyperplane. For r > 0 let a r ∈ R n and α r ∈ R such that # Λ (rB n ∩ H(a r , α r )) is maximal. Since Λ is self-polar, we may assume that a r ∈ Λ and α ∈ Z. In order to control the limit r → ∞ we want to find a sequence of radii (r j ) j∈N ⊆ N such that r j → ∞ and H(a r j , α r j ) is constant. To this end, fix a primitive vector a 0 ∈ Λ. Van der Corput's inequality (2.7) yields where ω i denotes the volume of the i-dimensional Euclidean unit ball and we used that the determinant of Λ ∩ a ⊥ 0 is given by |a 0 | det Λ = |a 0 |, since the determinant of any self-polar lattice is 1.
On the other hand, if r is large enough, rB n contains n linearly independent points of Λ. Thus, the maximal section rB n ∩H(a r , α r ) contains (n−1) affinely independent points of Λ; Otherwise, we might choose another point x ∈ rB n ∩ Λ and replace H(a r , α r ) by the affine hull of rB n ∩ H(a r , α r ) ∩ Λ and x. This yields a hyperplane that contains more lattice points of rB n than H(a r , α r ), contradicting the maximality. Hence, Blichfeldt's inequality (2.8) yields Combining with (3.18), we obtain |a r | ≤ 2 n−1 n! |a 0 |, for almost all r ∈ N.
Since this bound is independent of r, we find a sequence (r j ) j∈N ⊆ N that tends to infinity such that a r j = a, for all j and some primitive a ∈ Λ independent of j.
Hence, we may assume that α r j ≤ |a| 2 . Since α r is integral, we even find a sequence of radii (r j ) j∈N ⊆ N such that H(a r j , α r j ) = H(a, α) =: H for all j and a fixed α ∈ N. We choose K j = r j B n . In order to estimate the limit, we want to apply (2.9) to r j B n and r j B n ∩ H. The latter body may be viewed as a ball of radius r j − o(r j ) that is embedded in an (n − 1)-space together with a translation of Λ ∩ a ⊥ . Thus, by (2.9), where c > 0 is an absolute constant. In the last step we used the assumption that λ 1 (B n , Λ) = c √ n and Stirling's formula to estimate the volumes, which are known to be π n 2 /Γ( n 2 + 1), where Γ(·) is the Eulerian Gamma-function. Taking the n-th root yields the claim.

Discrete version of the reverse Loomis-Whitney inequality
The goal of this section is to prove Theorem 1.6. First, for a given v ∈ Z n we have to estimate the number of points in the projection of K|v ⊥ with respect to the projected lattice Z n |v ⊥ against the number of points in (K ∩ Z n )|v ⊥ .
Proof. For short we write K = K|v ⊥ and Λ = Z n |v ⊥ . Consider a line ⊆ v ⊥ that contains at least 5 points from K ∩ Λ, i.e., for some y ∈ Λ \ {0} and m ≥ 2. Since v ∈ K, the length of Rv ∩ K is at least 2|v|. Therefore, the length of the segment (i · y + Rv) ∩ K is at least |v|, for any |i| ≤ m 2 , as can be seen by considering the triangle given by v, −v and a point w ∈ K with w|v ⊥ = my (cf. Fig. 3). So, as long as i ≤ m/2 Figure 3. The lattice points in the projection close to the origin admit a preimage in K that is long enough to ensure a lattice point. the section (i · y + Rv) contains a lattice point. Since m ≥ 2, at least 1/3 of the points in ∩ Λ ∩ K \ {0} have a preimage in K ∩ Z n . Hence, with we know that |A| ≤ 3 (K ∩ Z n )|v ⊥ and so it suffices to prove To this end, let R 1 , ..., R 4 n−1 ⊆ Λ be the cosets of 4Λ in Λ and consider two distinct points x, y ∈ K ∩ R k . Then x − y ∈ 4Λ and thus 1 2 If A = ∅, we immediately get and the claim of the lemma follows. So let A = ∅. Then |K ∩ R k | ≤ 2|A| which leads to |K ∩ Λ| ≤ 4 n−1 · 2|A| and thus (4.1).
Remark 4.2. The inequality of Lemma 4.1 is essentially best-possible, in the sense that in any dimension, there is a convex body K ∈ K n o with e n ∈ K, such that To see this, let u = (1, 2, 4, ..., 2 n−1 ) T ∈ R n . We have C n ∩ u ⊥ ∩ Z n = {0}; Suppose there was a non-zero point x ∈ C n ∩ u ⊥ ∩ Z n . Let i be the largest index such x i = 0. By symmetry, we may assume that x i > 0. It follows from On the other hand, we have (C n ∩ u ⊥ )|e ⊥ n = C n−1 ; Let x ∈ {±1} n−1 be a vertex of C n−1 . Then, by (4.2), Thus, the convex body K = conv (C n ∩ u ⊥ ) ∪ {±e n } has the desired properties. Now we are ready for the proof of Theorem 1.6.
Proof of Theorem 1.6. We abbreviate λ i = λ i (K) and let v i ∈ λ i K ∩ Z n , 1 ≤ i ≤ n, be linearly independent. Due to our assumption we have λ n ≤ 1 and so v i ∈ K. So we may apply Lemma 4.1 to obtain To this end we set S i = Z ∩ Rv i , 1 ≤ i ≤ n. Then |S i | = 2 1/λ i + 1. Now we choose a subset Z i ⊆ Z such that the projection Z i → Z|v ⊥ i is bijective. Clearly, Z i + S i ⊆ Z + Z and so we have So we obtain In view of Lemma 3.3, (3.3) we have |Z + Z| ≤ #(2K) ≤ 4 n #K = 4 n |Z| and thus Finally we use (2.4), i.e., n i=1 2/λ i + 1 ≥ 3 −n |Z| in order to get (4.3). Remark 4.3.
i) The above proof does not depend on the particular properties of the lattice Z n . So one obtains the same statement for an arbitrary n-dimensional lattice Λ ⊆ R n . More precisely, if K ∈ K n o fullfills dim(K ∩ Λ) = n, we have where v i ∈ λ i (K, Λ)K ∩ Λ are linearly independent. ii) Also, the above approach yields a reverse Loomis-Whitney-type inequality, if one aims for a lattice basis, instead of merely independent lattice vectors. However, in order to apply Lemma 4.1 one has to ensure that the basis is contained in K. For the basis {b 1 , ..., b n } from equation (2.5), this means that one has to enlarge K by a factor n. So in this case, we obtain

Unconditional Bodies
A convex body K is called unconditional, if it is symmetric to all the coordinate hyperplanes, i.e., (±x 1 , ..., ±x n ) ∈ K, for any x ∈ K. For such bodies we can improve some of our inequalities. We start with Lemma 3.3, (3.3) Lemma 5.1. Let K ∈ K n be unconditional and m ∈ N. Then #mK ≤ (2m − 1) n #K and the inequality is sharp.
Next, let K ⊆ R n be an arbitrary unconditional convex body. Consider the unconditional body K that we obtain by multiplying the first coordinates in K by m. The lattice points in K and K can be partitioned into intervals parallel to e 1 . The intervals that we see in K are exactly the intervals of K, multiplied by m. So by the 1-dimensional case we have #K ≤ (2m − 1)#K. If we repeat this argument for every coordinate, we end up with the desired inequality. The cubes K = − (1 − 1 2m ), 1 − 1 2m n show that the inequality is sharp.
We conjecture (2m − 1) n to be the right constant also for arbitrary symmetric convex bodies. Lemma 5.1 yields a slightly improved version of Theorem 1.1 for the class of unconditional bodies. Proof. Again, we write K i = K ∩ e ⊥ i and h i = h(K, e i ). Note that h i is attained by a multiple of e i , since K is unconditional. This implies and we obtain Note that for an unconditional body K ⊆ R n one has K ∩ e ⊥ i = K|e ⊥ i , 1 ≤ i ≤ n. Therefore, Proposition 5.2 is also a sharpening of Theorem 1.6. In fact, following the lines of the proof of the discrete reverse Loomis-Whitney inequality in Section 4, the above proof is a simplification of the proof in Section 4.
Moreover, the inequalities of Theorem 1.2 and 1.4 hold with constant 1 for unconditional bodies, by the Loomis-Whitney inequality.
As for the discrete Brunn inequality, a constant 1 is obtained when intersecting an unconditional body K with a coordinate subspace L, since every slice K ∩(L+t) is mapped into the central slice injectively by the orthogonal projection onto L. Moreover, for any hyperplane H there is a coordinate i such that the projection H → e ⊥ i is bijective and maps lattice points in H to lattice points in e ⊥ i (The index i can be chosen to be an index for which the normal vector v of H is non-zero). Therefore, the maximal hyperplane section with respect to #(·) can always be chosen to be a coordinate section.
However, for general subspaces L we cannot hope for a constant 1 in the discrete Brunn inequality, as the next example illustrates. On the other hand, we have #(C n ∩ {x ∈ R n : x, u = 1}) = n, because, in view of (4.2), the points x k = e k − k−1 j=0 e j , 1 ≤ k ≤ n are contained in this section and as in Remark 4.2, by considering the maximal non-zero coordinate of a lattice point in this section, there are no further points.

Acknowledgements
We thank Eduardo Lucas Marín and Matthias Schymura for helpful comments and suggestions.