on finite lattice coverings

J. geom. 72 (2001) 132 – 1500047–2468/01/020132 – 19 $ 1.50 + 0.20/0© Birkhauser Verlag, Basel, 2001

On finite lattice coverings

Martin Meyer and Uwe Schnell

Abstract. We consider finite lattice coverings of strictly convex bodies K. For planar centrally symmetric K wecharacterize the finite arrangements Cn such that conv Cn ⊂ Cn +K, where Cn is a subset of a covering lattice forK (which satisfies some natural conditions). We prove that for a fixed lattice the optimal arrangement (measuredwith the parametric density) is either a sausage, a so-called double sausage or tends to a Wulff-shape, dependingon the parameter. This shows that the Wulff-shape plays an important role for packings as well as for coverings.Further we give a version of this result for variable lattices. For the Euclidean d-ball we characterize the lattices,for which the optimal arrangement is a sausage, for large parameter.

Mathematics Subject Classification (2000): 52C15, 52C17 (11H31).Key words: Covering, Lattice Covering, Convex Body, Wulff-shape.

1. Introduction and results

In the following let K be a centrally symmetric strictly convex body in the Euclidean 2-spaceE2. By hK and fK we denote its support function (or tac-function) and distance function,respectively (see e.g. [10], p. 15 and p. 107). A finite point set Cn with n elements is calleda covering arrangement if conv Cn ⊂ Cn + K. The density of a covering arrangement ismeasured by the parametric density

ϑ(Cn, K, �) = nV(K)

V(conv Cn + �K), (1)

where V denotes the volume and � is a parameter. The definition also applies for negative �.In this case conv Cn +�K is the inner parallel body of conv Cn with respect to K and we letϑ(Cn, K, �) = ∞ if the denominator is 0. If not explicitly mentioned we assume that � ≥ 0.Finite coverings in E2 without the parameter have already been studied by Bambah, Rogers,Woods, Zassenhaus and G. Fejes Toth (see [1], [2], [3] and [8]). Under various aspects theyshowed relations between classical infinite coverings and finite coverings. Results for finitecoverings, which are optimal with respect to the parametric density, have been proved in[5]. These authors investigate non-lattice coverings as well as lattice coverings. Any finitepoint set can be approximated by a sequence of lattice subsets and so there is no differenceto the non-lattice case. Of course the corresponding lattices have small determinants andmany lattice points can be omitted. Therefore we restrict ourselves to a class of lattices,which contains all thin covering lattices and for which we can give a complete description.

132

Vol. 72, 2001 On finite lattice coverings 133

If µ(K, L) is the covering radius of K with respect to L and λ1(K, L) is the first successiveminimum of K with respect to L (see e.g. [10], p. 123), then the following restrictions tothe lattices L are natural

(i) µ(K, L) = 1,(ii) λ1(K, L) > 1.

We denote the set of lattices with (i) and (ii) by L(K). For µ = µ(K, L) and λ1 =λ1(K, L) holds the inequality µ2 − (µ − λ1/2)2 ≥ det L/V(K) ([10], p. 125). As aconsequence we have for the density of a covering lattice with λ1 ≤ 1 that V(K)/ det L ≥(λ1 −λ2

1/4)−1 ≥ 4/3. In particular the thinnest covering lattice is contained in L(K), sinceϑ(K) = maxµ(K,L)≤1 V(K)/ det L ≤ 2π3−3/2 ≤ 1.21.

From condition (ii) it follows that only ‘complete‘ arrangements as used in [5] are con-sidered, i.e. Cn ⊂ L with conv Cn ⊂ Cn + K ⇒ (L ∩ conv Cn) = Cn. So instead ofCn we study Pn = conv Cn, i.e. lattice polygons with Pn ⊂ (L ∩ Pn) + K, and call themcovering polygons. In the following we are looking for covering polygons, which attain theminimum in

min

{nV(K)

V(Pn + �K): Pn covering polygon, G(Pn, L) ≤ n

}, (2)

where G(Pn, L) = card (Pn ∩ L) is the lattice point enumerator. Note that Pn may haveless than n lattice points. This slight modification guaranties that max V(Pn + �K) ismonotonously increasing in n and so irregularities are avoided, which come from numbertheoretic problems.

A bone is a linear arrangement with two additional points at both ends. For the non-latticeproblem the bone is optimal in some cases (see [5]). But it cannot be represented as acovering polygon for a lattice L ∈ L(K) because of the completeness condition.

The restriction to L(K) has the following consequence, which will be proved in Section 2.

LEMMA 1. Let K be strictly convex and L ∈ L(K). If there is a point contained in threetranslates of K, then it is a boundary point of these translates.

REMARK. The lemma does not hold for general K. Let K be the unit square and L be thelattice with basis {(1, 0), (1/2, 3/4)}. Then µ(K, L) = 1, λ1(K, L) = 3/2 and (1/2, 1/2)

is contained in the boundary of two translates and in the interior of one translate.

To characterize the set of covering polygons we introduce the following notations.

134 Martin Meyer and Uwe Schnell J. Geom.

DEFINITION 1. Let K be a strictly convex centrally symmetric convex body and L ∈L(K). Then let Fs(K, L) be the set directions

Fs(K, L) = {u ∈ L : K ∩ (K + u) �= ∅}.Further let

Fd(K, L) = {u ∈ L : u = u1 + u2, u1, u2 ∈ Fs(K, L),

conv {0, u1, u1 + u2} is a covering polygon}and F(K, L) = Fs(K, L) ∪ Fd(K, L).

REMARKS. (1) Fs(K, L) �= F(K, L) is possible (e.g. if K is the circle and L is thehexagonal lattice).

(2) The translates corresponding to Fs(K, L) can also be called neighbours. The Dirichlet-Voronoi cell D = {x ∈ E2 : fK(x) ≤ fK(x − u), for all u ∈ L} of L with respect to K isconvex, L + D is a lattice tiling and D is a parallelogram or a hexagon (see [10], p. 169).Since all neighbours with respect to D are neighbours (with respect to K) it follows thatthere are at least six neighbours.

(3) For u ∈ Fs(K, L) obviously Sn(u) = conv {i · u : i = 0, . . . , n − 1} is a coveringpolygon, which is called sausage.

(4) Elements of F(K, L)\Fs(K, L) do not allow a sausage. But there are closelyrelated covering polygons: Let u ∈ Fd(K, L) and u = u1 + u2 with u1, u2 ∈ Fs(K, L).By symmetry, the quadrangles Qk = conv {0, u1, u1 +ku, ku} and Q′

k = conv {0, u1, u1 +(k − 1)u, ku} for k = 1, 2, . . . are covering polygons (see Figure 1).

Because of λ1(K, L) > 1 there are no lattice points in the interior of Qk and Q′k and so

G(Qk, L) = 2(k + 1) and G(Q′k, L) = 2k + 1. We call these lattice quadrangles double

sausage.

(5) F(K, L) is finite.

THEOREM 1. Let K be a planar centrally symmetric strictly convex body and L ∈ L(K).Then an L-polygon P is a covering polygon if and only if

(i) int P = ∅ and P is a sausage, i.e. a segment of a lattice line parallel to u ∈ Fs(K, L).or

(ii) int P �= ∅ and all edges of P are parallel to vectors in F(K, L).

For fixed n and fixed lattice L we are looking for covering polygons with at most n latticepoints with maximal V(Pn + �K). We prove that the optimal Pn is either a sausage, or adouble sausage or tends to a Wulff-shape, depending on the parameter. This shows thatthe Wulff-shape plays an important role not only for packings ([4], [13], [12]) but also forcoverings.


Figure 1 Q4 and Q′4 for K = B2 and the hexagonal lattice

To formulate the theorem we need some more notation. For a lattice L let L∗ be the polarlattice (see e.g. [10], p. 23). For a primitive u ∈ L there is a corresponding vector u⊥ ∈ L∗,which is orthogonal to u and has length ‖u⊥‖ = ‖u‖/ det L. We define

(1) hs := hs(K, L) = max{hK(u⊥) : u ∈ Fs(K, L)}.(2) hm := hm(K, L) = max{hK(u⊥) : u ∈ F(K, L)}.

LEMMA 2. 2hs(K, L) ≥ hm(K, L).

DEFINITION 2. For L ∈ L(K) let F(K, L) = {u1, . . . , ur} and

ci = 1

2− �hK(u⊥

i ), i = 1, . . . , r.

Then the Wulff-shape for the parameter � is

W(�) = W(L, K, �) = {x : u⊥i x ≤ ci, i = 1, . . . , r}

REMARK. For � < 12hm

clearly W(�) �= ∅.

THEOREM 2. For fixed L ∈ L(K) with 2hs > hm and large n the optimal coveringpolygon Pn

(a) is a sausage in direction u, where u ∈ Fs(K, L) corresponds to hs, for � > 12(2hs−hm)

.(b) is a double sausage in direction u, where u ∈ F(K, L)\Fs(K, L) corresponds to

hm, for 12hm

< � < 12(2hs−hm)

.

(c) tends to W(�), when it is normalized by the factor R(W(�))/R(Pn), for � < 12hm

.

Further, part (c) is also true for negative �.


REMARKS. If 2hs = hm, then case (a) is omitted. If Fs(K, L) = F(K, L) or hs = hm,then case (b) is omitted.

Now we consider covering polygons for K with respect to variable lattices in L(K).Theorem 2 can be transferred to the following statement for variable lattices.

THEOREM 3. There are parameters 0 < �1 < �2 such that for large n the optimalcovering polygon Pn with respect to K and any lattice L ∈ L(K)

(a) is a sausage, for � > �2.(b) is a double sausage, for �1 < � < �2.(c) tends to W(L0, K, �), when it is normalized by the factor R(W(L0, K, �))/R(Pn),

for � < �1, where L0 is the thinnest covering lattice for K.

Further, part (c) is also true for negative �.

In the last section we consider lattice coverings in arbitrary dimension d ≥ 2. In this case, itis not easy to show that there are lattices satisfying the conditions µ(K, L) = 1 < λ1(K, L).If there is such a lattice L, then 2L + K is a packing and it follows for the optimal latticepacking density δ(K)

δ(K) ≥ V(K)

det(2L)≥ 2−dϑ(K) ≥ 2−d.

Up to a constant this is the theorem of Minkowski-Hlawka and so it seems that L(K) �= ∅ isdifficult to prove. The problem is to find lattices, which are simultaneously good coveringlattices and good packing lattices and has been investigated by Butler [7]. Even the existenceof lattice coverings with L\{0} + K �= Ed is an open problem. For the Euclidean d-ballBd we give a characterization of lattices, for which the optimal arrangement is a sausage,for large � and n.

Let L be a covering lattice for Bd . We can define the sausage directions Fs(Bd, L) analogous

to the planar case. They can be divided into two subsets.

Fo(Bd, L) = {u ∈ Fs(B

d, L) : ‖u‖ ≥ ‖v‖ for all v ∈ Fs(Bd, L)},

Fk(Bd, L) = Fs(B

d, L) \ Fo(Bd, L).

If Sn(u) is a sausage for u ∈ Fs(Bd, L), then V(Sn(u) + �Bd) only depends on the length

of u. For u ∈ Fo(Bd, L) let wl = ‖u‖. Further let wk = max{‖u‖ : u ∈ Fk(B

d, L)}.If u ∈ Fs(B

d, L), then Sn(u) + Bd covers a cylinder with radius r = (1 − wl2/4)1/2. On

the other hand let

η = minu∈Fo(Bd,L)

min{dist (v, lin (u)) : v ∈ L\lin (u)}.


THEOREM 4. Let L be a covering lattice for Bd . The optimal arrangement is a sausagefor large n and large � if and only if r < η.

Since the condition in Theorem 4 is satisfied for covering lattices with λ1(Bd, L) > 1, we

obtain the following consequence.

COROLLARY 1. LetLbe a covering lattice withλ1(Bd, L) > 1. Then the optimal arrange-

ment is a sausage for large � and n.

2. Characterization of the covering polygons

Proof of Lemma 1. Let w1, w2 ∈ L and x ∈ K ∩ (w1 + K) ∩ (w2 + K).

(1) Let x be contained in the interior of three translates. Then there is an r < 1 with0, w1, w2 ∈ x + rK. If w1 and w2 are collinear, then without restriction w2 = 2w1.From x = 2w1 + y, with y ∈ K, it follows w1 ∈ K, which contradicts L ∈ L(K).Hence, by a well-known covering criterion (see [10], p. 281), L is a covering lattice for rK.Consequently µ(K, L) = rµ(rK, L) ≤ r < 1, which is a contradiction to L ∈ L(K).

(2) Let x be contained in the interior of two translates and in the boundary of one translate.Then there is an x′ close to x, which is as in case (1).

(3) Let x be contained in int K and in the boundary of w1 +K and w2 +K. We distinguishtwo cases:

(a) (w1 + int K) ∩ (w2 + int K) �= ∅. Let y be contained in the intersection. Then thepoints in conv {x, y} close to x are as in case (1).

(b) (w1 + int K)∩ (w2 + int K) = ∅. Then there is a line g separating w1 +K and w2 +K.Since K is strictly convex it follows (w1 + K) ∩ (w2 + K) = {x}. A consequence of thesymmetry of K is x = (w1+w2)/2. So 2x = w1+w2 is a lattice point and x ∈ (2x+ int K)

and we are in case (2).

For the proof of Theorem 1 we introduce the following set, which is related to theDirichlet-Voronoi cell.

For the convex body K and a covering lattice L let

S(u) = S(K, L, u) = {x ∈ E2 : x ∈ u + K, x /∈ v + int (K), for all v ∈ L\{u}}.

LEMMA 3. If L ∈ L(K), then S(0) is a connected body.

Proof. A consequence of Lemma 1 is that the boundary of S(0) consists of finitely manyarcs, which belong to the boundary of translates of K and do not have common points intheir interior. Since λ1(K, L) > 1 we have int S(0) �= ∅ and the assertion follows. �


Proof of Theorem 1. (1) Let P be a covering polygon.

(i) If int P = ∅, then P is a line segment in a direction u ∈ L. Since necessarily K ∩ (K +u) �= ∅ it follows u ∈ Fs(K, L).

(ii) Now let int P �= ∅. Assume that P has an edge parallel to u ∈ L\F(K, L). Lete = conv {v1, v2} be contained in the edge, with v1, v2 ∈ L, v2 − v1 primitive and letg = aff (e). Then f = e\((v1 + K) ∪ (v2 + K)) has nonempty intersection with certaintranslates of K. The centers of these translates are contained in different lattice linesparallel to g because of λ1(K, L) > 1 and since K is strictly convex, the intersectionswith e have different lengths. There are translates of K from each side of g covering f .The intersections cannot overlap because of Lemma 1. Hence f is partitioned twice byintervals of different lengths, where these two partitions are symmetric with respect to themidpoint of f (or e). If the partitions consist of more than one interval, then a partitionpoint is contained in the interior of an interval of the other partition, which contradictsLemma 1. It follows that e is covered by v1 + K and v2 + K and at most one furthertranslate v3 + K. In the first case we have u ∈ Fs(K, L). In the second case there arepoints x1, x2 ∈ e with x1 ∈ (v1 + K) ∩ (v3 + K) and x2 ∈ (v2 + K) ∩ (v3 + K).Further (v3 − v1), (v3 − v2) ∈ Fs(K, L) and so x3 = (v1 + v3)/2 ∈ (v1 + K) ∩ (v3 + K)

and x4 = (v2 + v3)/2 ∈ (v2 + K) ∩ (v3 + K). From conv {v1, x1, x3} ⊂ v1 + K,conv {v2, x2, x4} ⊂ v2 +K and conv {v3, x3, x1, x2, x4} ⊂ v3 +K it follows u ∈ Fd(K, L).

(2) (i) The sausage in (i) is obviously a covering polygon.

(ii) In case (ii) a lattice point w outside P is separated from P by the affine hull g of anedge. Since g is covered by translates from the other side of g and S(w) is connected itfollows that S(w) ∩ int P = ∅ for all w ∈ L\P . If w1, w2 ∈ L\P , then the boundary of(w1 + K) ∩ (w2 + K) is contained in bd (S(w1)) ∪ bd (S(w2)) and so (w1 + K) ∩ (w2 +K)∩ int (P) = ∅. By Lemma 1 the plane can be decomposed in regions, which are coveredby exactly one or exactly two translates. It follows that int P and by the compactness of K

also P is covered by the translates w + K with w ∈ P ∩ L.

3. Optimal covering polygons for a fixed lattice

Proof of Lemma 2. Let u ∈ F(K, L) with hK(u⊥) = hm. If u ∈ Fs(K, L) then evenhs = hm. So we assume that u ∈ F(K, L)\Fs(K, L). Hence u = u1 + u2 with u1, u2 ∈Fs(K, L). Since (·)⊥ is a rotation of π/2 and a dilatation with factor (det L)−1 it followsu⊥ = (u1+u2)

⊥ = u⊥1 +u⊥

2 and consequently hm = hK(u⊥) ≤ hK(u⊥1 )+hK(u⊥

2 ) ≤ 2hs.For the proof of Theorem 2 we make use of some properties of mixed volumes (for detailssee [11], p. 270 ff). If K and L are convex bodies, then

V(L + �K) = V(L) + 2�V(L, K) + �2V(K), (3)


where V(L, K) is a mixed volume. If L = P is a polygon with edges of length f1, . . . , fr

and unit normal vector vi, then

V(P, K) = 1/2r∑

i=1

fihK(vi). (4)

If P is a lattice polygon and k = G( bd P, L), then Pick’s identity (see e.g. [9], p. 778)states

V(P) = (G(P, L) − 1) det L − det L/2 · k. (5)

In the following we consider covering polygons P with edges in directions u1, . . . , ur

∈ F(K, L). Let fi = ki‖ui‖, ki ∈ N, be the edgelengths. Pick’s formula then says that

V(P) = (G(P, L) − 1) det L − det L/2r∑

i=1

ki.

From (3), (4) and ‖u⊥i ‖ = ‖ui‖/ det L it follows

V(P + �K) = (G(P, L) − 1) det L(6)

+ det Lr∑

i=1

fi/‖ui‖(�hK(u⊥i ) − 1/2) + �2V(K).

LEMMA 4. Let K be a centrally symmetric strictly convex body and L ∈ L(K). For� < 1/(2hm(K, L)) let W = W(L, K, �). Then there is a sequence {Wn} of coveringpolygons with G(Wn, L) ≤ n and

V(Wn + �K) ≥ (n − 1) det L − 2√

det LV(W)√

n + o(√

n).

Proof. Let αn be minimal with G(αnW, L) ≥ n. Then obviously α = O(√

n). We canchoose lattice lines parallel to the edges of αnW in a distance bounded by a constant suchthat the intersection Qn of the corresponding halfplanes is a centrally symmetric latticepolygon with G(Qn, L) ≤ n + c

√n, for a constant c.

Assume that W is not a parallelogram. Then there is an edge e of Qn such that Qn iscontained in the cone, which is generated by the affine hulls of the adjacent edges. Withoutrestriction let 0 be their intersection point. This cone is the positive hull of a lattice triangleT with one edge e′ parallel to e. Erhart’s formula (see e.g. [9], p. 780) yields that G(kT, L) =ak2 + bk + 1, for suitable integers a, b and k = 0, 1, . . . . If e = ke′ and p = G(e′, L) − 1,then G(e, L) = pk + 1. If we exchange the supporting line aff (e) by aff ((k − 1)e′), thenthe number of lattice points is increased by

d(k) = G(kT, L) − G((k − 1)T, L) − (pk + 1) + (p(k − 1) + 1)

= (2k − 1)a + b − p.


Qn is centrally symmetric and we can do the reverse process for −e and delete d(k + 1)

lattice points. Repeating this operation q times we obtain a lattice polygon with

G(Qn, L) + (d(k) + · · · + d(k − q + 1)) − (d(k + 1) + · · · + d(k + q))

lattice points. With d(k − r) − d(k + r + 1) = −2a(2r + 1) it follows that the number oflattice points is decreased by 2a

∑q−1r=0 (2r + 1) = 2aq2. Let Wn be the covering polygon

obtained by this operation, where q is chosen such that

G(Qn, L) − 2aq2 ≤ n < G(Qn, L) − 2a(q − 1)2.

From 2a(q − 1)2 ≤ c√

n it follows

G(Wn, L) ≥ n − 2a(q2 − (q − 1)2) = n − O(n1/4). (7)

If W is a parallelogram, then there is a sausage direction u, which does not appear as anedge. So we can cut off lattice points from one vertex parallel to u and obtain new coveringpolygons. As above we delete d(q) = aq2 +bq+1 lattice points, where G(Qn, L)−d(q) ≤n < G(Qn, L)−d(q− 1). It follows q = O(n1/4) and again we obtain a covering polygonWn with (7).

There is a constant β such that (det L)−1V((αn − β)W) ≤ G(Qn, L) ≤ n + c√

n and soαn ≤ √

n det L/V(W)+o(√

n). By definition, (αn)−1Wn → W and therefore V(Wn, W) =

αnV(W, W) + o(√

n). Consequently

V(Wn, W) ≤ √n det L · V(W) + o(

√n). (8)

Let u1, . . . , ur be the directions of edges of Wn and let ci be as in Definition 2. Then for� < 1/(2hm) we have ci > 0 and hW(u⊥

i /‖u⊥i ‖) ≤ ci/‖u⊥

i ‖. Equality holds, if there is anedge of W in direction ui. Since this is the case for the edges of Wn up to at most one edgeof length o(

√n) it follows with (4)

V(Wn, W) = det L/2r∑

i=1

fici/‖ui‖ + o(√

n).

Finally the assertion follows with (6) for Wn, (7) and (8). �

Proof of Theorem 2. Let P be a covering polygon with edges in directions u1, . . . , ur ∈F(K, L). Then let fi = ki‖ui‖, ki ∈ N, be the edgelengths. If P has m ≤ n lattice points,then by (6) we have

V(P + �K) = (m − 1) det L

+ det Lr∑

i=1

fi/‖ui‖(�hK(u⊥i ) − 1/2) + �2V(K).


For fixed n we have to maximize

f(P) = f(P, K, �) = (m − 1) +r∑

i=1

ki(�hK(ui) − 1/2).

Let Son be a sausage with n lattice points in the direction corresponding to hs. Obviously

f(Son) = 2�hs(n − 1). (9)

We can realize an optimal double sausage D0n with n lattice points for any n such that

f(Don) = (n − 1) + (n − 2)(�hm − 1/2) + c, (10)

with a constant c ≥ −1. For fixed � > 1/(2hm) consider a covering polygon P , which isnot a sausage or a double sausage. If P is contained in the convex hull of two adjacent latticelines, then f(P) ≤ f(Do

n), for large m. Otherwise P contains a number of i lattice points inthe interior with i → ∞ for m → ∞. Consequently f(P) ≤ (n−1) + (n−i)(�hm−1/2) <

f(Don), if m is large. It follows that for � > 1/(2hm) and large n the double sausage is

optimal among all covering polygons with dimension 2. So part (a) and (b) follow fromcomparing (9) and (10).

For fixed 0 ≤ � < 12hm

and W = W(L, K, �) we have, as in the proof of Lemma 4, thatV(P, W) ≤ det L/2

∑ri=1 fici/‖ui‖, for any covering polygon P , and we obtain

V(P + �K) ≤ det L(m − 1) − 2V(P, W) + �2V(K). (11)

If Pn is the optimal covering polygon for n, then it follows from Lemma 4 that

G(Pn, L) det L − 2V(Pn, W) ≥ n det L − 2√

det L · V(W)√

n + o(√

n).

Immediate consequences areG(Pn, L) = n−o(n) andV(Pn, W) ≤√n det LV(W) +o(

√n).

Hence we obtain for the perimeter F(Pn) = O(√

n) and∑r

i=1 ki = O(√

n). From Pick’sidentity we conclude that

V(Pn) = n det L + o(n). (12)

From Minkowski’s inequality (see e.g. [11], p. 317) it follows V(Pn, W) ≥√V(Pn) · V(W) = √

n det L · V(W) + o(√

n) and so we have

V(Pn, W) = √n det L · V(W) + o(

√n). (13)

Now let us consider the sequence Pn/R(Pn). Let Pnν/R(Pnν) be a converging sub-sequence with limit P0 (by the Theorem of Blaschke there are such sequences). FromV(Pnν)(R(Pnν))

−2 → V(P0) and (12) we obtain

R(Pnν) = √V(Pnν)/V(P0) + o(

√nν) = √

nν det L/V(P0) + o(√

nν).


Together with (13) and the homogeneity and continuity of the mixed volumes it follows

V(W, P0) = limV(W, Pnν)

R(Pnν)= √

V(W) · V(P0),

i.e. it holds equality in Minkowski’s inequality. This is only possible if P0 is homothetic toW and from R(P0) = 1 it follows P0 = W/R(W). Since Pnν/R(Pnν) was chosen arbitrarilywe obtain Pn/R(Pn) → W/R(W).

Finally we consider negative�. Let P be a covering polygon with respect to K and L ∈ L(K)

with edges e1, . . . , er and outer unit normals v1, . . . , vr. Then the inner parallel body P+�K

is the intersection of the halfspaces corresponding to the lines parallel to the edges withdistance |�|hK(v1), . . . , |�|hK(vr). The strips ei + |�|hK(vi)conv {0, −vi} may overlap ormay have points outside P if there are acute angles. If α is the angle between two adjacentedges ei, ej , then this area can be estimated by

2(|�|hK(vi))(|�|hK(vj))/sin(α) ≤ 2�2D(K)2/ sin(α).

α is the angle between two vectors u1, u2 ∈ F(K, L) and consequently ‖u1‖‖u2‖sin α ≥ det L. From λ1(K, L) > 1 it follows that L is admissible for K and so det L ≥(K), where (K) is the critical determinant of K. Together with ‖u1‖, ‖u2‖ ≤ 2D(K) weobtain sin(α) ≥ (K)/(4D(K)2). Since the total volume of the strips is |�| ∑r

i=1 V1(ei)hK

(vi) = 2|�|V(P, K), we conclude that for any covering polygon P

V(P + �K) = V(P) + 2�V(P, K) + �2c, (14)

where 0 ≤ c ≤ A and A is a constant depending only on K. Hence the formula (6) alsoholds for � < 0 up to a constant, which can be neglected in the proof of Lemma 4 andpart (c).

4. Optimal covering polygons for variable lattices

The optimal sausage direction u is determined by

sm := sm(K) = max{V1(K/ lin (u)⊥)‖u‖ : u ∈ bd (2K)}. (15)

(Note that for packings the direction minimizing this expression is optimal.) Obviouslysm is 2V(Q), where Q denotes the maximal centrally symmetric quadrangle, which canbe inscribed K. If L is the lattice with fundamental domain Q, then L ∈ L(K) and sm isattained for L, i.e. sm = 2 det Lhs(K, L).

While the direction of the optimal sausage is independent of �, the optimal double sausage(maximized over all lattices in L(K)) depends on �.


Of course, we want to compare the maximal coefficient of n in (6), which is sm for thesausage. So we define, motivated by (6) and (10),

ds (�) = supL∈L(K)(�hm(K, L) + 1/2) det L. (16)

Since it is difficult to determine ds (�) for given K, we give the following estimates.

LEMMA 5. Let K ⊂ E2 be a centrally symmetric strictly convex body and � ≥ 0. Then

(a) ds (�) ≥ 3/4sm� + 3/8V(K)/δ(K), where δ(K) denotes the density of the densestlattice packing for K.

(b) There is an α < 1 such that hm(K, L) det L ≤ αsm(K) holds for any L ∈ L(K).

Proof. (a) Let x be the direction corresponding to sm and ‖x‖ = 1. Further let 2l be theintersection of K and the line g = lin (x) and h = 1/2V1(K/g⊥) such that sm = 4hl.There is a point a ∈ bd K such that a + lx ∈ bd K. Let u = 3lx and u′ = a + 2lx

and let L be the lattice with basis {u, u′}. If h′ denotes the distance between a and g,then det L = 3lh′. The hexagon H = conv {±a, ±(a + lx), ±lx} is affinely regular andinscribed in K. Since the critical determinant (K) is one third of the area of the minimalinscribed affinely regular hexagon (see [10], p. 243, there is a misprint in Theorem 1), itfollows det L = V(H) ≥ 3(K) = 3/4V(K)/δ(K). L is a covering lattice for H and soµ(K, L) ≤ 1 and by Lemma 1 µ(K, L) = 1. By the choice of a we have λ1(K, L) ≥ 1and equality only if K is a parallelogram, which is not strictly convex. Hence L ∈ L(K).Further we have hm(K, L) ≥ hK(u⊥) = h‖u‖/ det L = h/h′ and so

ds (�) ≥ 3�lh + 3/2lh′ = 3/4sm� + 1/2 · V(H) ≥ 3/4sm� + 3/8V(K)/δ(K).

(b) Let h0 = (K)/(2D(K)) > 0, where D(K) denotes the diameter of K. For any line g

through the center of K and the line g′ parallel to g with distance h0 we consider the valueof (V1(g ∩ K) + V1(g

′ ∩ K))/(2V1(g ∩ K)). There is a line such that the maximal value α

is attained and since K is strictly convex it follows α < 1.

Now let L ∈ L(K) and u ∈ F(K, L). Further let {u, u′} be a basis of L and h′ the distanceof u′ to lin (u) such that det L = ‖u‖h′. From λ1(K, L) > 1 it follows det L ≥ (K) andtogether with ‖u‖ ≤ 2D(K) we obtain h′ ≥ h0. The line segment conv {0, u} is coveredby K, u + K and a third translate with distance ≥ h0 to lin (u). By the definition of α wehave ‖u‖ ≤ 2αV1(K ∩ lin (u)). It follows hK(u⊥) det L = 1/2 · V1(K/ lin (u)⊥)‖u‖ ≤V1(K/ lin (u)⊥)αV1(K ∩ lin (u)) ≤ αsm and part (b) is proved. �

Proof of Theorem 3. Let L0 be the thinnest covering lattice for K. Then we define �1 =inf{� : ds (�) ≥ det L0} and �2 = sup{� : ds (�) ≥ sm�}. From Lemma 5 we haveds (�) ≤ αsm� + det L0/2. Consequently �2 ≤ det L0/(2sm(1 − α)) < ∞.


Further �1 > 0 is clear. For � = det L0/sm we have �sm = det L0. From Lemma 5 itfollows with V(K) ≥ det L that ds (det L0/sm) ≥ 3/4 det L0 +3/8 det L0 > det L0. Fromthe continuity of ds we conclude �1 < det L0/sm < �2.

Since ds is monotonously increasing and ds (λ�) < λ ds (�), for λ > 1, it follows that themaximum value of {det L0, sm�, ds (�)} is det L0, for � < �1 and ds (�), for �1 < � < �2

and sm�, for � > �2.

For a covering polygon Pn with respect to a fixed lattice L ∈ L(K) we conclude from (11),(9) and (10) that V(Pn +�K) ≤ max{1, 2�hs(K, L), (�hm(K, L)+1/2)} ·det L ·n+O(1),for any �. Consequently for covering polygons Pn with respect to any L ∈ L(K) and forany parameter � we obtain

V(Pn + �K) ≤ max{det L0, sm�, ds (�)} · n + O(1). (17)

Now, for fixed � and n, let Pn be an optimal covering polygon for K with respect to thelattice Ln ∈ L(K).

For � > �2 the maximum in (17) is sm�. Hence Pn has to be a sausage, for large n, andpart (a) follows. For �1 < � < �2 the maximum in (17) is ds (�). Obviously, Pn cannot be asausage. If �hm(K, Ln)−1/2 ≤ 0, then V(Pn+�K) ≤ n det Ln+O(1) ≤ n det L0+O(1),which is worse than n ds (�)+o(n). Hence �hm(K, Ln)−1/2 > 0. In this case, the doublesausage is optimal among all Pn ⊂ Ln, which are not sausages, as shown in the proof ofTheorem 2. So part (b) is proved.

Now let � < �1. If � ≥ 1/(2hm(K, Ln)), then V(Pn + �K) ≤ n ds (�) + O(1), which isobviously not optimal. So � < 1/(2hm(K, Ln)), for large n. As in (11) we have

V(Pn + �K) ≤ n det Ln − 2V(W(Ln, K, �), Pn) + o(√

n).

On the other hand, by the optimality of Pn and Lemma 4 for L0 it follows

V(Pn + �K) ≥ n det L0 − 2√

det L0V(W(L0, K, �))√

n + o(√

n).

From det Ln ≤ det L0 we obtain

V(W(Ln, K, �), Pn) ≤ √det L0V(W(L0, K, �))

√n + o(

√n). (18)

We can assume that Ln is convergent and has limit L0. Any accumulation point of the set⋃n F(K, Ln) is contained in F(K, Ln). So, for large n, we have AnF(K, Ln) ⊂ F(K, L0),

where An maps Ln into L0 and An tends to the identity. By definition of the Wulff-shapesW(L0, K, �) ⊂ (1 + εn)W(Ln, K, �) with εn → 0 and from (18) we obtain

V(W(L0, K, �), Pn) ≤ √det L0V(W(L0, K, �))

√n + o(

√n).


We already showed in the proof of Theorem 2 that this is sufficient to conclude Pn/R(Pn) →W(L0, K, �)/R(W(L0, K, �)).

For negative � the Wulff-shapes are nonempty. Again we use (14) and the proof is the sameas for positive � because the constant A in (14) does not depend on L.

5. A criterion for lattice coverings of the d-ball

In this section we investigate arbitrary covering lattices for the d-ball Bd with d ≥ 2.For the covering lattice L we consider finite covering arrangements C ⊂ L. An optimalcovering arrangement with respect to � and L consisting of at most n elements is denotedby C�,n. Further let D�,n = D(conv C�,n). Since the diameter is attained, there is adirection u�,n with ‖u�,n‖ = 1 and an x�,n such that x�,n, x�,n + D�,n u�,n ∈ C�,n. LetR�,n = R(conv C�,n/ lin (u�,n)

⊥). In a first step we show that optimal arrangements arenearly one dimensional, i.e. R�,n is bounded, and that the diameter of an optimal coveringarrangement is not much shorter than the diameter of an optimal sausage.

LEMMA 6. There are constants c0, c1, c2, depending only on d and L and a parameter�0 = �0(d, L) such that

(i) R�,n ≤ c0,(ii) D�,n ≥ n wl − c1 n

�− c2,

for all � ≥ �0.

REMARK. A similar result has already been shown for the non-lattice case in [6].

Proof. From Steiner’s formula for the outer parallel body (see [11], p. 197) we have

V(conv C�,n + �Bd) =d∑

i=0

Vi(conv C�,n)κd−i�d−i,

where V0, . . . , Vd are the intrinsic volumes. On the other hand, if v ∈ Fo(Bd, L), then for

the sausage Sn(v) holds

V(Sn(v) + �Bd) = κd�d + (n − 1) wl κd−1�

d−1.

Since C�,n is optimal, it follows

(n − 1) wl ≤ V1(conv C�,n) +d∑

i=2

Vi(conv C�,n) · κd−i

κd−1�1−i.


From Cauchy’s integral representation of the intrinsic volumes (see [11], p. 295) we obtainVi(conv C�,n) ≤ n · Vi(B

d) and consequently there is a constant c such that

(n − 1) wl ≤ V1(conv C�,n) + n c

�, for � ≥ 1. (19)

conv C�,n is contained in a cylinder Z�,n over a (d − 1)-ball with radius R�,n and lengthD�,n. Therefore we have V1(conv C�,n) ≤ V1(Z�,n) = D�,n+ R�,n · V1(B

d) and so

(n − 1) wl ≤ D�,n + R�,n · V1(Bd) + n c

�, � ≥ 1. (20)

Further conv C�,n contains a triangle with length D�,n and height R�,n and with Cauchy’sformula and the monotony of the intrinsic volumes we conclude

D�,n · R�,n/2 ≤ V2(conv C�,n) ≤ n V2(Bd). (21)

Since V1 is up to a factor the mean width it follows from (19) that D�,n ≥ c′ V1(conv C�,n) ≥c′′ n for � ≥ �(d, L) ≥ 1. Together with (21) this shows that R�,n is bounded and (i) isproved. Finally, (ii) follows from (i) and (20). �

Next we give an upper bound for the diameter of any covering arrangement. For this aimlet su = max{u · v : v ∈ Fs(B

d, L)} for directions u with ‖u‖ = 1.

LEMMA 7. Let Cn ⊂ L be a covering arrangement with card Cn ≤ n and let u be adirection in which D(conv Cn) is attained. Then

D(conv Cn) ≤ (n − 1) su.

Proof. We can assume that 0 ∈ Cn and that there is an y ∈ Cn such that ‖y‖ = D(conv Cn).The line segment conv ({0, y}) is covered by a sequence of translates t1 + Bd, . . . , tm + Bd

with ti ∈ Cn, t1 = 0 and tm = y, such that (ti+1 + Bd) ∩ (ti + Bd) �= ∅, i.e. ti+1 − ti ∈Fs(B

d, L). For u = y/‖y‖ it follows

D(conv Cn) = ‖y‖ = u ·m−1∑i=1

(ti+1 − ti) ≤ (m − 1) su ≤ (n − 1) su.

�

LEMMA 8. Let L be a covering lattice for Bd with η > r. For large � and n the diameterof conv C�,n is attained in a direction v

‖v‖ with v ∈ Fo(Bd, L).


Proof. We can assume that 0 ∈ C�,n and that there is an y�,n ∈ C�,n such that ‖y�,n‖ =D(conv C�,n). Further let u�,n = y�,n

‖y�,n‖ . From Lemma 6 and Lemma 7 it follows

n su�,n ≥ D�,n ≥ n wl − c1 n

�− c2. (22)

For sufficiently large � we have limn→∞ su�,n > wk and so for large � and n su�,n = u�,n ·v,with v ∈ Fo(B

d, L).

Let P�,n be the length of the projection of conv {0, y�,n} into lin (v). Then

P�,n = D�,n · cos( � (u�,n, v)) = D�,n · u�,n v

‖v‖ = D�,n · su�,n

wl

and from (22) we obtain

P�,n ≥(

1 − c1

� wl

− c2

wl n

)2

· n wl ≥(

1 − 2 c1

� wl

− 2 c2

wl n

)· n wl

(23)= n wl − 2 c1 n

�− 2 c2.

As in the proof of Lemma 7 let t1, . . . , tm ∈ C�,n with m = m�,n ≤ n, conv {0, y�,n} ⊂⋃mi=1(ti + Bd), t1 = 0, tm = y�,n and ti+1 − ti ∈ Fs(B

d, L). We can divide t1, . . . , tm intosubsets

{0, t1, . . . , ts1}, {ts1+1, . . . , ts2}, . . . , {tsq+1, . . . , tsq+1 = y�,n}such that Sj = conv {tsj+1, . . . , tsj+1} is a line segment parallel to v, for j = 0, . . . , q = q�,n

(with s0 = −1) and tsj+1 − tsj is not parallel to v, j = 1, . . . , q. Further let lj be the lengthof Sj , j = 0, . . . , q. The lemma is proved if the assumption q ≥ 1 leads to a contradiction.

For this aim we first give an upper bound for q.

If w = max{ v‖v‖ · z : z ∈ Fs(B

d, L), z �= v}, then w < wl and we have

P�,n =(

m−1∑i=1

(ti+1 − ti)

)· v

‖v‖ ≤ (m − q) wl + q w ≤ n wl − q(wl − w).

From (23) we conclude that there are constants c3, c4 such that

q ≤ c3 n

�+ c4. (24)

Now let j be the triangle conv (Sj ∪ {tsj+1+1}). By the definition of η it contains a

quadrangle Qj with one side S′j of length l

′j = r

ηlj − w contained in Sj and height r.

Let l be the number of segments conv {ti, ti+1} in S′j with length wl and k the number of

these segments with length ≤ wk. Then (l + 2) wl + k wk ≥ l′j .


If ‖ti+1 − ti‖ = wl, then ti+1 + Bd and ti + Bd have a common boundary point in the sideS

′′j parallel to S

′j . These are contained in the interior of j and so a further point of C�,n

is needed. Since wl > 1 a translate x + Bd contains at most two of these points. So weneed at least nj = 3

2 l + k translates to cover Qj . Let w = min{wl

3 , wl − wk} > 0. Then itfollows

l′j − 2wl ≤ l wl + k wk =

(3

2l + k

)wl − 3

2l(wl

3

)− k (wl − wk)

(25)

≤(

3

2l + k

)· (wl − w) = nj (wl − w).

Further, to cover Sj\S ′j , we need at least n

′j = lj−l

′j

wltranslates. Then n ≥ ∑q

j=0

(nj + n′j − c), for a constant c. With c′ = c + (2wl + w)/(wl − w) − w/wl and (25)

we conclude

n ≥q∑

j=0

(l′j − 2wl

wl − w+ lj − l

′j

wl

− c

)

≥(

r

η· 1

wl − w+

(1 − r

η

)· 1

wl

)·

q∑j=0

lj − c′ (q + 1) ≥ 1

w′ P�,n − c′ (q + 1),

with w′ < wl. It follows that P�,n ≤ w′ n + c′ w′ (q + 1). Together with (23), (24) anddivision by n we obtain

wl − 2 c1

�− 2 c2

n≤ w′ + c′ w′

(c3

�+ c4 + 1

n

).

Taking the limit n → ∞ it follows wl ≤ w′ + (c′ w′ c3 + 2 c1)/�, which is a contradictionfor large �. �

Proof of Theorem 4. First let η > r. Without restriction let 0 ∈ C�,n and y�,n ∈ C�,n with‖y�,n‖ = D�,n. By Lemma 8 for sufficiently large � and n there exists an u ∈ Fo(B

d, L)

with lin (u) = lin (y�,n). We assume that for arbitrary large � and n C�,n is not a sausage.For these � and n there is a q�,n ∈ C�,n with q�,n /∈ lin (u). Then dist(q�,n, lin (u)) ≥ η.Let = conv {0, q�,n, y�,n}. contains a quadrangle with length r/η D�,n and height r.As in the proof of Lemma 8 with and D�,n instead of j and lj we obtain

n ≥ r

η· 1

wl − w· D�,n +

(1 − r

η

)· 1

wl

· D�,n − c

=(

r

η· 1

wl − w+

(1 − r

η

)· 1

wl

)· D�,n − c ≥ 1

w′ · D�,n − c


with w′ < wl. From Lemma 6 we get wl − c1�

− c2n

≤ w′ + cn

, a contradiction for large �

and n.

For η ≤ r let u ∈ Fo(Bd, L) and v ∈ L ∩ Bd with dist (v, lin (u)) = η. Further let

Cn = {x1, . . . , xn−4, xa, xb, xc, xd} with xi = (i − 1) · u, i = 1, . . . , n − 4, and xa = v,xb = −v, xc = (n − 5) u + v, xd = (n − 5) u − v. Then Cn is a covering arrangement withcard Cn ≤ n and conv Cn is a parallelogram with length (n − 5) wl and height η. It follows

V(conv Cn + �Bd) = �dκd + �d−1V1(conv Cn)κd−1 + �d−2V2(conv Cn)κd−2

≥ �dκd + �d−1(n − 5) wl κd−1 + �d−2(n − 5) wl η κd−2

> �dκd + �d−1(n − 1) wl κd−1 = V(Sn(u) + �Bd),

for sufficiently large n.

REMARK. The configurations in the proof of Theorem 4 have the shape of bones.This shows that bones play an important role for lattice coverings as well as for arbitrarycoverings [5].

Proof of Corollary 1. Let u ∈ Fo(Bd, L). L ∩ lin (u) + Bd covers an infinite cylinder

with radius r = (1 − w2l /4)1/2. Because of λ1(B

d, L) > 1 this cylinder contains no latticepoints in L\ lin (u). Therefore we have η > r and with Theorem 4 the assertion follows.

References

[1] Bambah, R. P. and Rogers, C. A., Covering the plane with convex sets, J. London Math. Soc. 27 (1952),304–314.

[2] Bambah, R. P., Rogers, C. A. and Zassenhaus, H., On coverings with convex domains, Acta Arith. 9 (1964),191–207.

[3] Bambah, R. P. and Woods, A. C., On plane coverings with convex domains, Mathematika 18 (1971), 91–97.[4] Betke, U. and Boroczky, Jr., K., Large lattice packings and the Wulff-shape, Mathematika, to appear.[5] Betke, U., Henk, M. and Wills, J. M., A new approach to coverings, Mathematika 42 (1995), 251–263.[6] Betke, U. and Meyer, M., Finite parametric covering densities for large parameter, in preparation.[7] Butler, G. J., Simultaneous packing and covering in Euclidean space, Proc. London Math. Soc. 25 (1972)

no. 3, 721–735.[8] Fejes Toth, G., Finite coverings by translates of centrally symmetric convex domains Discrete Com-

put. Geom. 2 (1987), 353–364.[9] Gritzmann, P. and Wills, J. M., Lattice points, Ch. 3.2 in: Handbook of Convex Geometry, P. M. Gruber,

J. M. Wills eds., North Holland, Amsterdam 1993.[10] Gruber, P. M. and Lekkerkerker, C. G., Geometry of Numbers, North Holland, Amsterdam, 1987.[11] Schneider, R., Convex geometry: The Brunn-Minkowski theory, Cambridge University Press, Cambridge,

1993.


[12] Schnell, U., Periodic sphere packings and the Wulff-shape, Beitrage Algebra Geom. 40 (1999) no. 1,125–140.

[13] Wills, J. M., Lattice packings of spheres and the Wulff-shape, Mathematika 86 (1996), 229–236.

M. Meyer and U. SchnellMathematisches InstitutUniversitat Siegen57068 SiegenGermanye-mail: [email protected]

[email protected]

Received 19 May 1999.

on finite lattice coverings

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