A Blichfeldt-type inequality for the surface area

Monatsh Math 154, 135–144 (2008) DOI 10.1007/s00605-008-0530-8 Printed in The Netherlands

A Blichfeldt-type inequality for the surface area By

Martin Henk1 and Jo¨rg M. Wills2 1

Universit€at Magdeburg, Magdeburg, Germany 2 Universit€at Siegen, Siegen, Germany

Dedicated to Uli Betke on the occasion of his 60th birthday Received May 15, 2007; accepted in revised form September 18, 2007 Published online March 25, 2008 # Springer-Verlag 2008 Abstract. In 1921, Blichfeldt gave an upper bound on the number of integral points contained in a convex body in terms of the volume of the body. More precisely, he showed that #ðK \ Zn Þ 4 n! volðKÞ þ n, whenever K
Rn is a convex body containing n þ 1 affinely independent integral points. Here we prove an pffiffiffianalogous inequality with respect to the surface area FðKÞ, namely #ðK \ Zn Þ < volðKÞ þ ðð n þ 1Þ=2Þ ðn  1Þ! FðKÞ. The proof is based on a slight improvement of Blichfeldt’s bound in the case when K is a non-lattice translate of a lattice polytope, i.e., K ¼ t þ P, where t 2 Rn n Zn and P is an n-dimensional polytope with integral vertices. Then we have #ððt þ PÞ \ Zn Þ 4 n! volðPÞ. Moreover, in the 3-dimensional case we prove a stronger inequality, namely #ðK \ Zn Þ < volðKÞ þ 2 FðKÞ. 2000 Mathematics Subject Classification: 52C07, 11H06 Key words: Lattice polytopes, volume, surface area

1. Introduction Let Kn be the set of all convex bodies in the n-dimensional Euclidean space R . For a subset S
Rn and the integral lattice Zn let GðSÞ be the lattice point enumerator of S, i.e., GðSÞ ¼ #ðS \ Zn Þ. By volðSÞ we denote, as usual, the volume, i.e., the n-dimensional Lebesgue measure, of S. The problem to bound GðKÞ, K 2 Kn , in terms of continuous functionals, as e.g. the intrinsic volumes, has a long history in convexity (cf. e.g., [2, 7, 12]), and the first general upper bound with respect to the volume is due to Blichfeldt [4] n

GðKÞ 4 n! volðKÞ þ n;

ð1:1Þ

provided dimðK \ Zn Þ ¼ n, i.e., K contains n þ 1 affinely independent lattice points of Zn . This bound is best possible for any number of lattice points, as, for instance, the simplex Sk ¼ convf0; k e1 ; e2 ; . . . ; en g, k 2 N, shows. Here ei denotes the i-th canonical unit vector and so we have GðSk Þ ¼ k þ n and volðSk Þ ¼ k=n!. Our main result is an inequality analogous to (1.1), but now with respect to the volume and the surface area FðKÞ of the body.

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Theorem 1.1. Let K 2 Kn with dimðK \ Zn Þ ¼ n. Then pffiffiffi nþ1 ðn  1Þ! FðKÞ: GðKÞ < volðKÞ þ 2 In contrast to Blichfeldt’s inequality, the volume is now weighted by the factor 1pffiffiinstead of n!, which is apparently best possible. We conjecture that the factor nþ1 can be omitted in this inequality. In dimension 2, this follows easily from 2 Pick’s identity [2, pp. 38], and the 3-dimensional case is an immediate consequence of the next theorem Theorem 1.2. Let K 2 K3 . Then 4 GðKÞ < volðKÞ þ FðKÞ þ 3  pffiffiffi : 3
Since the surface area of a 3-dimensional convex body containing a simplex with integral p vertices is not less than the surface area of the simplex S1 , which is equal ffiffiffi to ð3 þ 3Þ=2 > 3  3 p4 ffiffi
, we obtain Corollary 1.3. Let K 2 K3 with dimðK \ Z3 Þ ¼ 3. Then GðKÞ < volðKÞ þ 2FðKÞ: We remark that an inequality of the form GðKÞ < volðKÞ þ ðn  1Þ! FðKÞ would be tight in the sense that ðn  1Þ! in front of the surface area can not be replaced by cðn  1Þ! for a constant c < 1. To see this pffiffiffi we note that for the simplex S1 with n þ 1 lattice points we have FðS1 Þ ¼ ðn þ nÞ=ðn  1Þ!. In fact, the family of simplices Sk , k 2 N, shows that even for large GðKÞ the factor of FðKÞ in Theorem 1.1 cannot be lowered substantially, i.e., it has to be of size
ððn  2Þ!Þ. For the classcial problem to estimate the asymptotic behavior of GðKÞ  volðKÞ in large regions we refer to the survey [11]. By construction it is clear that the constant 2 in Corollary 1.3 in front of FðKÞ is not optimal. By the argument given before the Corollary it is easy to see that it can be lowered to 1.96. On the other hand the simplex S1 shows that it has to be at least 1.62. The inequality in Theorem 1.1 may also be regarded as a counterpart to a wellknown lower bound on GðKÞ due to Bokowski, Hadwiger and Wills [6]. They proved that 1 volðKÞ  FðKÞ < GðKÞ; ð1:2Þ 2 and this inequality is best possible. The proof of Theorem 1.1 is based on a lemma on lattice points in a translate of a lattice polytope. To this end we denote by Pn
Kn the set of all lattice polytopes, i.e., polytopes having integral vertices. Lemma 1.4. Let P 2 Pn with dimðP \ Zn Þ ¼ n, and let t 2 Rn n Zn . Then Gðt þ PÞ 4 n!volðPÞ; and the inequality is best possible for any number of lattice points.

A Blichfeldt-type inequality

137

In other words, if we have a non-lattice translate of P then we can slightly improve Blichfeldt’s bound (1.1) by n. This does not mean, however, that t þ P has less lattice points than P. For instance, for n > 2 and m 2 N let Tm be the so called Reeve simplex Tm ¼ convf0; e1 ; . . . ; en1 ; mvg, where v ¼ e1 þ e2 þ    þ en . Then the vertices are the only lattice points in Tm , but ð1=2Þv þ Tm contains the m lattice points v; 2v; . . . ; mv. In the 2-dimensional case the situation is different and for a detailed discussion of lattice points in translates of lattice polygons we refer to [10]. Since (1.1) and the inequality in Lemma 1.4 depend only on the volume, it is easy to generalize them to an arbitrary lattice 
Rn with determinant det  > 0. Then, with the setting as before, we have volðKÞ volðPÞ iÞ #ðK \ Þ 4 n! þ n; and iiÞ #ððt þ PÞ \ Þ 4 n! : ð1:3Þ det  det  In the case of Theorem 1.1 we conjecture that the right statement for general lattices is Conjecture 1.1. Let 
Rn be a lattice and let K 2 Kn with dimðK \ Þ ¼ n. Then volðKÞ FðKÞ #ðK \ Þ < þ ðn  1Þ! ; det  det n1 where det n1 is the minimal determinant of an ðn  1Þ-dimensional sublattice of . In the 2-dimensional case the correctness of the inequality is again an easy consequence of Pick’s identity. It is also not hard to verify such an inequality with an additional factor of order n in front of the surface area, and we will give an outline of a proof of this result in the last section (see Corollary 4.2). For a corresponding conjecture regarding the lower bound (1.2) we refer to [17, 18]. In Section 1 we will prove Lemma 1.4 and Theorem 1.1. The proof of Theorem 1.2 is based on results on the inner=outer parallel body of a convex body and is given in the second section. 2. Proof of Lemma 1.4 and Theorem 1.1 The proof of Lemma 1.4 will be an immediate consequence of the fact that for n linearly independent points a1 ; . . . ; an 2 Zn and the associated half-open Plattice n parallelepiped C ¼ f i¼1 i ai : 0 4 i < 1g one has GðCÞ ¼ j detða1 ; . . . ; an Þj:

ð2:1Þ

Observe, both sides just describe the index of the sublattice generated by a1 ; . . . ; an with respect to Zn (see e.g., [8, p. 22]). Proof of Lemma 1.4. Let P
Rn be a lattice polytope and let t 2 Rn n Zn . Let S1 ; . . . ; Sm  P be n-dimensional lattice simplices such that P ¼ [m i¼1 Si and dimðSi \ Sj Þ 4 n  1 P for i 6¼ j. For instance, we can take any lattice triangulation of P. Then volðPÞ ¼ m i¼1 volðSi Þ and m X Gðt þ Si Þ: Gðt þ PÞ 4 i¼1

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Hence it suffices to prove the statement for an n-dimensional lattice simplex S, say. Without loss of generality let 0; a1 ; . . . ; an be the vertices of S, ai 2 Zn , and let C be the half-open parallelepiped generated by a1 ; . . . ; an. Then by (2.1) we have GðCÞ ¼ j detða1 ; a2 ; . . . ; an Þj ¼ n! volðSÞ:

ð2:2Þ

Next we observe that for any vector t 2 Rn Gðt þ CÞ ¼ GðCÞ:

ð2:3Þ

This is aP well-known fact, but forPsake of completeness we give a short argument: n n n     Let t ¼ P i¼1 i ai . For b ¼ t þ i¼1 i ai 2 ðt þ CÞ \ Z the vector f ðbÞ defined n n  by f ðbÞ ¼ i¼1 ði þ i  bi þ i cÞai is contained in C \ Z . Here bxc denotes the largest integer not bigger then x. It is easy to see that f is a bijection between ðt þ CÞ \ Zn and C \ Zn , and hence we have verified (2.3). Finally, since t þ S
ðt þ CÞ [ ft þ a1 ; t þ a2 ; . . . ; t þ an g;

ð2:4Þ

we get Gðt þ SÞ 4 Gðt þ CÞ for t 2 Rn n Zn. Together with (2.3) and (2.2) we obtain the desired inequality for S, and thus for the lattice polytope P. In order to show that it is best possible let Sk be the simplex defined in the introduction. Then k ¼ n! volðSk Þ and if we translate Sk by 12 e1, for instance, then e1 ; . . . ; ke1 are the only lattice points in 12 e1 þ Sk . & Remark 2.1. If t ¼ 0 then (2.4) gives GðSÞ 4 GðCÞ þ n because a1 ; . . . ; an are the only points in S not contained in C. Thus we get by the same argument Blichfeldt’s inequality GðKÞ 4 n! volðKÞ þ n. For the proof of Theorem 1.1 we also need some facts about lattice polytopes. Let P
Rn be a lattice polytope. Then we can describe it as P ¼ fx 2 Rn : ai  x 4 bi ; 1 4 i 4 mg;

ð2:5Þ

for some ai 2 Zn , bi 2 Z. Here x  y denotes the inner product, and by kk we denote the associated Euclidean norm. Without loss of generality let Fi ¼ P \ fx 2 Rn : ai  x ¼ bi g be the facets of P, 1 4 i 4 m. We may also assume that the vectors ai are primitive vectors, i.e., convf0; ai g \ Zn ¼ f0; ai g. In this case we have (cf. e.g. [13, Proposition 1.2.9]) detðaff Fi \ Zn Þ ¼ kai k;

ð2:6Þ

n

where detðaff Fi \ Z Þ is the determinant of the ðn  1Þ-dimensional sublattice of Zn contained in the affine hull of Fi . Proof of Theorem 1.1. Let K 2 Kn with dimðK \ Zn Þ ¼ n. By the monotonicity of volðÞ and FðÞ it suffices to prove the conjecture for the n-dimensional lattice polytope P ¼ convfK \ Zn g. Let Cn be the cube of edge length 1 centered at the origin. Let L1 ¼ fz 2 P \ Zn : z þ Cn
Pg and L2 ¼ ðP \ Zn Þ n L1 . Obviously, we have #L1 4 volðPÞ;

ð2:7Þ

A Blichfeldt-type inequality

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and it remains to bound the size of the set L2 . To this end let P be given as in (2.5) with facets F1 ; . . . ; Fm . For each lattice point z 2 L2 there exists a facet Fi such that z þ Cn intersects Fi , i.e., there exists an x 2 Cn with ai  z þ ai  x > bi . Hence we have 1 ai  z > bi  ai  x 5 bi  jai j; 2 where j  j denotes the l1 -norm. Since the left hand side is an integer we obtain   1 jai j  1: ai  z 5 bi  i with i ¼ ð2:8Þ 2 Thus L2


m [ ðQi \ Zn Þ;

ð2:9Þ

i¼1

where Qi ¼ convfFi ; Fi  ði =kai k2 Þai g is the prism with basis Fi and height i =kai k in the direction ai . Next we claim that for 1 4 i 4 m pffiffiffi nþ1 ðn  1Þ!voln1 ðFi Þ þ ðn  1Þ; ð2:10Þ GðQi Þ < 2 where voln1 ðÞ denotes the ðn  1Þ-dimensional volume. Each lattice point in such a prism Qi is contained in one of the layers Hi ðjÞ ¼ Qi \ fx 2 Rn : ai  x ¼ bi  jg;

j ¼ 0; 1; . . . ; i :

Of course, Hi ð0Þ ¼ Fi is an ðn  1Þ-dimensional lattice polytope with respect to the lattice Fi ¼ affðFi Þ \ Zn . On account of (2.6) we get from Blichfeldt’s inequality (see (1.3) i)) GðHi ð0ÞÞ ¼ #ðFi \ Fi Þ 4 ðn  1Þ!

voln1 ðFi Þ þ ðn  1Þ: kai k

ð2:11Þ

Now let vi 2 Zn be any lattice vector in the lattice hyperplane fx 2 Rn : ai  x ¼ bi  1g and let wi be a lattice vector in Fi . Since Hi ðjÞ ¼ Hi ð0Þ  j we have n



Hi ðjÞ \ Z ¼

1 kai k2

ai

 Hi ð0Þ  j ai þ jðwi  vi Þ \ Zn ¼ ðj ti þ Fi Þ \ Zn ; kai k2 1

with ti ¼ wi  vi  1=kai k2 ai 2 fx 2 Rn : ai  x ¼ 0g. Since ai is primitive and j 4 i < kai k2 we find that jti 2 Rn n Zn for 1 4 j 4 i. Thus we may apply in these cases Lemma 1.4, or more precisely (1.3) ii), and obtain GðHi ðjÞÞ 4 ðn  1Þ!

voln1 ðFi Þ ; kai k

j ¼ 1; . . . ; i :

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Together with (2.11) we get GðQi Þ ¼ GðHi ð0ÞÞ þ

i X

GðHi ðjÞÞ

j¼1

d12 jai je ðn  1Þ! voln1 ðFi Þ þ ðn  1Þ kai k pffiffiffi nþ1 ðn  1Þ! voln1 ðFi Þ þ ðn  1Þ; < 2 and so we have verified (2.10). Finally, in order to prove the inequality of the theorem we have to consider the lattice points which we count more than once in the right hand side of (2.9), and we claim m X #L2 4 GðQi Þ  mðn  1Þ: ð2:12Þ 4

i¼1

To this end we consider the vertices v1 ; . . . ; vk of P. Let gn1 ðvj Þ be the number of facets containing vj and let f0 ðFi Þ be the number of vertices of the facet Fi . Obviously, we have m X i¼1

f0 ðFi Þ ¼

k X

gn1 ðvj Þ:

ð2:13Þ

j¼1

Since each facet has at least n vertices P and each vertex is contained in at least n facets we conclude from (2.13) that m i¼1 f0 ðFi Þ 5 maxfm; kgn 5 k þ mðn  1Þ. This shows (2.12) and so, in view of (2.10) we get pffiffiffi nþ1 ðn  1Þ! FðPÞ: #L2 < 2 Together with (2.7) we obtain the desired inequality. & 3. Proof of Theorem 1.2 For the proof of Theorem 1.2 we need a bit of the theory of intrinsic volumes for which we refer to [16]. Let K 2 Kn and let Bn be the n-dimensional unit ball of volume n . The outer parallel body K þ Bn of K at distance  is the Minkowski sum of K and Bn , i.e., K þ Bn ¼ fx þ y : x 2 K; y 2 Bn g. Its volume can be described by the so called Steiner polynomial n X volðK þ Bn Þ ¼ Vi ðKÞni ni ; ð3:1Þ i¼0

where Vi ðKÞ is called the i-th intrinsic volume of K. In particular, we have Vn ðKÞ ¼ volðKÞ, Vn1 ðKÞ ¼ ð1=2ÞFðKÞ, and V0 ðKÞ ¼ 1. It was conjectured by Wills that n X GðKÞ 4 Vi ðKÞ; ð3:2Þ i¼0

A Blichfeldt-type inequality

141

but, in general, this inequality does not hold (see [3, 9]). In dimension three, however, it is true [15] and so we have GðKÞ 4 V3 ðKÞ þ V2 ðKÞ þ V1 ðKÞ þ 1:

ð3:3Þ

The inner parallel body of K at distance  is given by the set K  Bn ¼ fx 2 K : x þ Bn  Kg: If K  Bn is non-empty then we trivially have ðK  Bn Þ þ Bn  K. Proof of Theorem 1.2. According to (3.3) and (3.1) we obtain GðKÞ 4 V3 ðKÞ þ V2 ðKÞ þ V1 ðKÞ þ 1

  2 3 1 1 1 þ 3 pffiffiffi < V3 ðKÞ þ V1 ðKÞ1 pffiffiffi þ V2 ðKÞ2 pffiffiffi


  3  1 þ 1  3 pffiffiffi
  4 1=2 B3 Þ þ 1  pffiffiffi : ¼ volðK þ
3


Hence, if K 
1=2 B3 6¼ ; we get with  ¼ 1  3p4 ffiffi
GðK 
1=2 B3 Þ < volððK 
1=2 B3 Þ þ
1=2 B3 Þ þ  4 volðKÞ þ : On the other hand it was shown in [14, Korollar 1] that GðKÞ  GðK  31=2 B3 Þ 4 FðKÞ þ 2: Combining the last two inequalities yields GðKÞ 4 GðK  31=2 B3 Þ þ FðKÞ þ 2 4 GðK 
1=2 B3 Þ þ FðKÞ þ 2 < volðKÞ þ FðKÞ þ ð2 þ Þ: & In the context with the conjectured inequality (3.2) it was shown by Bokowski [5] that for n 4 5 GðKÞ 4 volðK þ n1=n Bn Þ: With n ¼ 1=n this leads, as in the proof above, to GðP  n Bn Þ 4 volðPÞ n where P ¼ convfK \ Zn g. In order to estimate the remaining lattice points GðPÞ  GðP  n Bn Þ, which are close to the boundary of P, we can proceed as in the proof of Theorem 1.1 where we bound the size of the set L2 . This leads, roughly speaking, for n 4 5 to an inequality of the form   1 GðKÞ < volðKÞ þ n þ ðn  1Þ! FðKÞ; 2 which is stronger than the one of Theorem 1.1. Since the improvement, however, is marginal we omit a detailed proof.

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4. The inequality for arbitrary lattices In order to present an inequality as in Theorem 1.1 for arbitrary lattices we need some basic facts and notions from geometry of numbers for which we refer to [8]. For a lattice 
Rn let  ¼ fy 2 Rn : y  b 2 Z for all b 2 g be its polar lattice. The length (norm) of a shortest non-zero lattice vector in a lattice  is denoted by 1 ðÞ, and an ðn  1Þ-dimensional sublattice of  with minimal determinant is denoted by n1. Then we have (cf. e.g. [13, Proposition 1.2.9]) det    ð Þ ¼ det  : ð4:1Þ 1

n1

Moreover, we need the so called Dirichlet-Voronoi cell DVðÞ of a lattice  consisting of all points whose nearest lattice point in  is the origin, i.e., DVðÞ ¼ fx 2 Rn : kxk 4 kx  bk for all b 2 g: Then volðDVðÞÞ ¼ det  and the smallest radius of a ball containing DVðÞ is called the inhomogeneous minimum of  and will be denoted by ðÞ. So in the case of the integral lattice Zn the Dirichlet-Voronoi cell is just the cube of edge pffiffiffi length 1 centered at the origin and ðZn Þ ¼ n=2, 1 ðZn Þ ¼ 1, det Znn1 ¼ 1 and ðZn Þ ¼ Zn . With these notations we can generalize Theorem 1.1 as follows. Theorem 4.1. Let 
Rn be a lattice and let K 2 Kn with dimðK \ Þ ¼ n. Then GðKÞ 4

volðKÞ FðKÞ þ ððÞ1 ð Þ þ 1Þðn  1Þ! : det  det n1

Observe, in the case  ¼ Zn we get essentially the inequality of Theorem 1.1. By fundamental results of Banaszczyk [1] it is known that ðÞ ð Þ 4 cn; 1

for some universal constant c and so we have Corollary 4.2. Let 
Rn be a lattice and let K 2 Kn with dimðK \ Þ ¼ n. Then GðKÞ