ON FINITE LATTICE COVERINGS

ON FINITE LATTICE COVERINGS M. MEYER AND U. SCHNELL Abstract. We consider nite lattice coverings of strictly convex bod-

ies K . For planar centrally symmetric K we characterize the nite arrangements Cn such that conv Cn Cn + K , where Cn is a subset of a covering lattice for K (which satis es some natural conditions). We prove that for a xed lattice the optimal arrangement (measured with the parametric density) is either a sausage, a socalled double sausage or tends to a Wul{shape, depending on the parameter. This shows that the Wul{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.

1. Introduction and Results In the following let K be a centrally symmetric strictly convex body in the Euclidean 2{space E 2. By hK and fK we denote its support function and distance function, respectively (see e. g. [GL]). A nite point set Cn with n elements is called a covering arrangement if conv Cn Cn + K . The density of a covering arrangement is measured by the parametric density nV (K ) #(Cn; K; %) = V ( conv (1) Cn + %K ) ; where V denotes the volume and % is a parameter. The de nition also applies for negative %. In this case conv Cn + %K is the inner parallel body of conv Cn with respect to K and we let #(Cn ; K; %) = 1 if the denominator is 0. If not explicitely mentioned we assume that % 0. Finite coverings in E 2 without the parameter have already been studied by Bambah, Rogers, Woods, Zassenhaus and G. Fejes Toth (see [BR], [BRZ], [BW] and [F]). Under various aspects they showed relations between classical in nite coverings and nite coverings. Results for nite coverings, which are optimal with respect to the parametric density, have been proved in [BHW]. These authors investigate non{lattice coverings as well as lattice coverings. Any nite point set can be approximated by a sequence of lattice subsets and so there is no dierence to the non{lattice case. Of course the corresponding lattices have small determinants and many 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. If (K; L) is the covering radius of K with respect to L and 1(K; L) is the 1

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M. MEYER AND U. SCHNELL

rst succesive minimum of K with respect to L (see e. g. [GL]), then the following restrictions to the lattices L are natural (i) (K; L) = 1, (ii) 1(K; L) > 1. We denote the set of lattices with (i) and (ii) by L(K ). For = (K; L) and 1 = 1(K; L) holds the inequality 2 ? ( ? 1 =2)2 det L=V (K ) ([GL], p. 125). As a consequence we have for the density of a covering lattice with 1 1 that V (K )= det L (1 ? 21=4)?1 4=3. In particular the thinnest covering lattice is contained in L(K ), since #(K ) = max(K;L)1 V (K )= det L 2 3?3=2 1:21. From condition (ii) it follows that only `complete` arrangements as used in [BHW] are considered, i. e. Cn L with conv Cn Cn + K ) (L \ conv Cn ) = Cn . So instead of Cn we study Pn = conv Cn , i. e. lattice polygons with Pn (L \ Pn ) + K , and call them covering polygons. In the following we are looking for covering polygons, which attain the minimum in nV ( K ) (2) min V (P + %K ) : Pn covering polygon; G(Pn ; L) n ; n

where G(Pn ; L) = card (Pn \ L) is the lattice point enumerator. Note that Pn may have less than n lattice points. This slight modi cation guaranties that max V (Pn + %K ) is monotonously increasing in n and so irregularities

are avoided, which come from number theoretic problems. A bone is a linear arrangement with two additional points at both ends. For the non{lattice problem the bone is optimal in some cases (see [BHW]). But it cannot be represented as a covering polygon for a lattice L 2 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 2 L(K ). If there is a point contained in three translates 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 the lattice with basis f(1; 0); (1=2; 3=4)g. Then (K; L) = 1; 1(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. De nition 1. Let K be a strictly convex centrally symmetric convex body and L 2 L(K ). Then let Fs (K; L) be the set directions Fs(K; L) = fu 2 L : K \ (K + u) 6= ;g: Further let Fd(K; L) = fu 2 L : u = u1 + u2; u1; u2 2 Fs(K; L); conv f0; u1; u1 + u2 g is a covering polygon g

ON FINITE LATTICE COVERINGS

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Figure 1. Q4 and Q04 for K = B 2 and the hexagonal lattice.

and F (K; L) = Fs (K; L) [ Fd (K; L). Remarks: (1) Fs(K; L) 6= F (K; L) is possible (e. g. if K is the circle and L is the hexagonal lattice). (2) The translates corresponding to Fs (K; L) can also be called neighbours. The Dirichlet{Voronoi cell D = fx 2 E 2 : fK (x) fK (x ? u), for all u 2 Lg of L with respect to K is convex, L + D is a lattice tiling and D is a parallelogram or a hexagon (see [Sch]). Since all neighbours with respect to D are neighbours (with respect to K ) it follows that there are at least six neighbours. (3) For u 2 Fs (K; L) obviously Sn (u) = conv fi u : i = 0; : : : ; n ? 1g is a covering polygon, which is called sausage. (4) Elements of F (K; L)nFs(K; L) do not allow a sausage. But there are closely related covering polygons: Let u 2 Fd(K; L) and u = u1 + u2 with u1 ; u2 2 Fs(K; L). By symmetry, the quadrangles Qk = conv f0; u1; u1 + ku; kug and Q0k = conv f0; u1; u1 +(k ? 1)u; kug for k = 1; 2; : : : are covering polygons (see Figure 1). Because of 1 (K; L) > 1 there are no lattice points in the interior of Qk and Q0k and so G(Qk ; L) = 2(k + 1) and G(Q0k ; L) = 2k + 1. We call these lattice quadrangles double sausage. (5) F (K; L) is nite. Theorem 1. Let K be a planar centrally symmetric strictly convex body and L 2 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 2 Fs (K; L). or (ii) int P 6= ; and all edges of P are parallel to vectors in F (K; L).

For xed n and xed lattice L we are looking for covering polygons with at most n lattice points with maximal V (Pn + %K ). We prove that the

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optimal Pn is either a sausage, or a double sausage or tends to a Wul{ shape, depending on the parameter. This shows that the Wul{shape plays an important role not only for packings ([BB], [W], [S]) but also for coverings. To formulate the Theorem we need some more notation. For a lattice L let L be the polar lattice (see e. g. [GL]). For a primitive u 2 L there is a corresponding vector u? 2 L , which is orthogonal to u and has length ku?k = kuk= det L. We de ne (1) hs := hs (K; L) = maxfhK (u? ) : u 2 Fs (K; L)g. (2) hm := hm (K; L) = maxfhK (u? ) : u 2 F (K; L)g. Lemma 2. 2hs(K; L) hm (K; L). De nition 2. For L 2 L(K ) let F (K; L) = fu1; : : : ; urg and ci = 21 ? %hK u?i ; i = 1; : : : ; r: Then the Wul{shape for the parameter % is n o W (%) = W (L; K; %) = x : u?i x ci ; i = 1; : : : ; r

Remark: For % < 2h1m clearly W (%) =6 ;. Theorem 2. For xed L 2 L(K ) with 2hs > hm and large n the optimal

covering polygon Pn (a) is a sausage in direction u, where u 2 Fs (K; L) corresponds to hs , for % > 2(2hs1?hm ) . (b) is a double sausage in direction u, where u 2 F (K; L)nFs(K; L) corresponds to hm , for 2h1m < % < 2(2hs1?hm ) . (c) tends to W (%), when it is normalized by the factor R(W (%))=R(Pn), for % < 2h1m . 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 optimal covering polygon Pn with respect to K and any lattice L 2 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 %.


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In the last section we consider lattice coverings in arbitrary dimension

d 2. In this case, it is not easy to show that there are lattices satisfying the conditions (K; L) = 1 < 1 (K; L). If there is such a lattice L, then 2L + K is a packing and it follows for the optimal lattice packing density (K ) V (K ) 2?d #(K ) 2?d : (K ) det(2 L)

Up to a constant this is the Theorem of Minkowski{Hlawka and so it seems that L(K ) 6= ; is dicult to prove. The problem is to nd lattices, which are simultaneously good covering lattices and good packing lattices and has been investigated by Butler [B]. Even the existence of lattice coverings with Lnf0g + K 6= E d is an open problem. For the Euclidean d{ball B d 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 B d . We can de ne the sausage directions Fs (B d ; L) analogous to the planar case. They can be divided into two subsets. Fo(Bd ; L) = fu 2 Fs(Bd; L) : kuk kvk for all v 2 Fs(Bd ; L)g; Fk (Bd ; L) = Fs(Bd ; L) n Fo(Bd ; L): If Sn (u) is a sausage for u 2 Fs (B d ; L), then V (Sn (u) + %B d ) only depends on the length of u. For u 2 Fo (B d ; L) let wl = kuk. Further let wk = maxfkuk : u 2 Fk (B d ; L)g. If u 2 Fs (B d ; L), then Sn (u) + B d covers a cylinder with radius r = (1 ? wl2 =4)1=2. On the other hand let = mind minfdist (v; lin (u)) : v 2 Ln lin (u)g: u2Fo (B ;L)

Theorem 4. Let L be a covering lattice for Bd . The optimal arrangement is a sausage for large n and large % if and only if r < . Since the condition in Theorem 4 is satis ed for covering lattices with 1(B d ; L) > 1, we obtain the following consequence. Corollary 1. Let L be a covering lattice with 1(Bd; L) > 1. Then the optimal arrangement is a sausage for large % and n.

2. Characterization of the covering polygons Proof of Lemma 1: Let w1; w2 2 L and x 2 K \ (w1 + K ) \ (w2 + K ). (1) Let x be contained in the interior of three translates. Then there is an r < 1 with 0; w1; w2 2 x + rK . If w1 and w2 are collinear, then without restriction w2 = 2w1. From x = 2w1 + y , with y 2 K , it follows w1 2 K , which contradicts L 2 L(K ). Hence, by a well{known covering criterion (see [GL], p. 281), L is a covering lattice for rK . Consequently (K; L) = r(rK; L) r < 1, which is a contradiction to L 2 L(K ).

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(2) Let x be contained in the interior of two translates and in the boundary of one translate. Then there is an x0 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 distinguish two cases: (a) (w1 + int K ) \ (w2 + int K ) 6= ;. Let y be contained in the intersection. Then the points in conv fx; y g close to x are as in case (1). (b) (w1+ int K )\(w2 + int K ) = ;. Then there is a line g seperating w1 +K and w2 + K . Since K is strictly convex it follows (w1 + K ) \ (w2 + K ) = fxg. A consequence of the symmetry of K is x = (w1 + w2)=2. So 2x = w1 + w2 is a lattice point and x 2 (2x + int K ) and we are in case (2). For the proof of Theorem 1 we introduce the following set, which is related to the Dirichlet{Voronoi cell. For the convex body K and a covering lattice L let S (u) = S (K; L; u) = fx 2 E 2 : x 2 u + K; x 2= v + int (K ); for all v 2 Lnfugg: Lemma 3. If L 2 L(K ), then S (0) is a connected body.

Proof: A consequence of Lemma 1 is that the boundary of S (0) consists of nitely many arcs, which belong to the boundary of translates of K and do not have common points in their interior. Since 1(K; L) > 1 we have int S (0) 6= ; 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 2 L. Since necessarily K \ (K + u) 6= ; it follows u 2 Fs (K; L). (ii) Now let int P 6= ;. Assume that P has an edge parallel to u 2 LnF (K; L). Let e = conv fv1 ; v2g be contained in the edge, with v1; v2 2 L, v2 ? v1 primitive and let g = a (e). Then f = en((v1 + K ) [ (v2 + K )) has nonempty intersection with certain translates of K . The centers of these translates are contained in dierent lattice lines parallel to g because of 1(K; L) > 1 and since K is strictly convex, the intersections with e have dierent 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 by intervals of dierent lengths, where these two partitions are symmetric with respect to the midpoint of f (or e). If the partitions consist of more than one interval, then a partition point is contained in the interior of an interval of the other partition, which contradicts Lemma 1. It follows that e is covered by v1 + K and v2 + K and at most one further translate v3 + K . In the rst case we have u 2 Fs (K; L). In the second case there are points x1; x2 2 e with x1 2 (v1 + K ) \ (v3 + K ) and x2 2 (v2 + K ) \ (v3 + K ). Further (v3 ? v1); (v3 ? v2 ) 2 Fs (K; L) and so x3 = (v1 + v3 )=2 2 (v1 + K ) \ (v3 + K ) and x4 = (v2 + v3 )=2 2 (v2 + K ) \ (v3 + K ). From conv fv1; x1; x3g v1 + K , conv fv2; x2; x4g v2 + K and conv fv3 ; x3; x1; x2; x4g v3 + K it follows u 2 Fd (K; L).


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(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 ane hull g of an edge. Since g is covered by translates from the other side of g and S (w) is connected it follows that S (w) \ int P = ; for all w 2 LnP . If w1; w2 2 LnP , 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 covered by 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 2 P \ L. 3. Optimal covering polygons for a fixed lattice Proof of Lemma 2: Let u 2 F (K; L) with hK (u?) = hm. If u 2 Fs(K; L) then even hs = hm . So we assume that u 2 F (K; L)nFs(K; L). Hence u = u1 + u2 with u1 ; u2 2 Fs (K; L). Since ()? is a rotation of =2 and a dilatation with factor (det L)?1 it follows u? = (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 details see [Sch]). If K and L are convex bodies, then (3) V (L + %K ) = V (L) + 2%V (L; K ) + %2V (K ); 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=2

(4)

r X i=1

fi hK (vi):

If P is a lattice polygon and k^ = G( bd P; L), then Pick's identity (see e. g. [GL]) states (5) V (P ) = (G(P; L) ? 1) det L ? det L=2 k^: In the following we consider covering polygons P with edges in directions u1 ; : : : ; ur 2 F (K; L). Let fi = ki kui k; ki 2 N, be the edgelengths. Pick's

formula then says that

V (P ) = (G(P; L) ? 1) det L ? det L=2 From (3), (4) and ku?i k = kui k= det L it follows V (P + %K ) = (G(P; L) ? 1) det L (6)

+ det L

r X i=1

r X i=1

ki :

fi =kuik(%hK (u?i ) ? 1=2) + %2V (K ):

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Lemma 4. Let K be a centrally symmetric strictly convex body and L 2 L(K ). For % < 1=(2hm(K; L)) let W = W (L; K; %). Then there is a sequence fWn g of covering polygons with G(Wn ; L) n and p p p V (Wn + %K ) (n ? 1) det L ? 2 det LV (W ) n + o( n): Proof: Let n be minimal with G(n W; L) n. Then obviously = p O( n). We can choose lattice lines parallel to the edges of n W in a distance bounded by a constant such that the intersection Qn of the corresponding p halfplanes is a centrally symmetric lattice polygon 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 is contained in the cone, which is generated by the ane hulls

of the adjacent edges. Without restriction let 0 be their intersection point. This cone is the positive hull of a lattice triangle T with one edge e0 parallel to e. Erhart's formula (see e. g. [GL]) yields that G(kT; L) = ak2 + bk +1, for suitable integers a; b and k = 0; 1; : : : . If e = ke0 and p = G(e0 ; L) ? 1, then G(e; L) = pk +1. If we exchange the supporting line a (e) by a ((k ? 1)e0), then the 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) = ?P2a(2r + 1) it follows that ?1 (2r + 1) = 2aq 2. Let the number of lattice points is decreased by 2a rq=0 Wn be the covering polygon obtained by this operation, where q is chosen such that G(Qn; L) ? 2aq 2 n < G(Qn; L) ? 2a(q ? 1)2: p From 2a(q ? 1)2 c n it follows (7) G(Wn; L) n ? 2a(q 2 ? (q ? 1)2) = n ? O(n1=4): If W is a parallelogram, then there is a sausage direction u, which does not appear as an edge. So we can cut o lattice points from one vertex parallel to u and obtain new covering polygons. As above we delete d(q) = aq 2 + 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 polygon Wn with (7). There is a constant )?1 V ((n ? )W ) G(Qn ; L) n + p such that (det Lp p c n and so n n det L=V (W ) + o( n). pBy de nition, (n)?1 Wn ! W and therefore V (Wn ; W ) = n V (W; W ) + o( n). Consequently p p (8) V (Wn ; W ) n det L V (W ) + o( n):


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Let u1 ; : : : ; ur be the directions of edges of Wn and let ci be as in De nition 2. Then for % < 1=(2hm) we have ci > 0 and hW (u?i =ku?i k) ci =ku?i k. Equality holds, if there is an edge of W in direction ui . Since p this is the case for the edges of Wn up to at most one edge of length o( n) it follows with (4)

V (Wn ; W ) = det L=2

r X i=1

p fici =kuik + 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 2 F (K; L). Then let fi = ki kui k; ki 2 N, be the edgelengths. If P has m n lattice points, then by (6) we have V (P + %K ) = (m ? 1) det L + det L

r X i=1

fi =kuik(%hK (u?i ) ? 1=2) + %2V (K ):

For xed n we have to maximize

f (P ) = f (P; K; %) = (m ? 1) +

r X i=1

ki(%hK (ui) ? 1=2):

Let Sno be a sausage with n lattice points in the direction corresponding to hs. Obviously (9) f (Sno ) = 2%hs(n ? 1): We can realize an optimal double sausage Dn0 with n lattice points for any n such that (10) f (Dno ) = (n ? 1) + (n ? 2)(%hm ? 1=2) + c; with a constant c ?1. For xed % > 1=(2hm ) consider a covering polygon P , which is not a sausage or a double sausage. If P is contained in the convex hull of two adjacent lattice lines, then f (P ) f (Dno ), for large m. Otherwise P contains a number of i lattice points in the interior with i ! 1 for m ! 1. Consequently f (P ) (n ? 1) + (n ? i)(%hm ? 1=2) < f (Dno ), 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 from comparing (9) and (10). For xed 0 % < 2h1m and W =PW (L; K; %) we have, as in the proof of Lemma 4, that V (P; W ) det L=2 ri=1 fi ci =kuik, for any covering polygon P , and we obtain (11) V (P + %K ) det L(m ? 1) ? 2V (P; W ) + %2V (K ): If Pn is the optimal covering polygon for n, then it follows from Lemma 4 that p p p G(Pn; L) det L ? 2V (Pn; W ) n det L ? 2 det L V (W ) n + o( n):

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? p p and i=1 ki = O( n). From Pick's identity we conclude that (12) V (Pn ) = n det L + o(n):

pn)

Immediate consequences are G(Pn ; L) = n o(n) and V (Pn ; W ) p n det PrLV (W ) +o( n). Hence we obtain for the perimeter F (Pn ) = O(

From Minkowski's inequality (seepe. g. [Sch]) it followsp p V (Pn ; W ) V (Pn) V (W ) = n det L V (W ) + o( n) and so we have

p p V (Pn; W ) = n det L V (W ) + o( n): Now let us consider the sequence Pn =R(Pn ). Let Pn =R(Pn ) be a converging subsequence with limit P0 (by the Theorem of Blaschke there are such sequences). From V (Pn )(R(Pn ))?2 ! V (P0 ) and (12) we obtain p p R(P ) = V (P )=V (P ) + o(pn ) = n det L=V (P ) + o(pn ):

(13)

n

n

0

0

Together with (13) and the homogenity and continuity of the mixed volumes it follows Pn ) = pV (W ) V (P ); V (W; P0) = lim V R(W; 0 (Pn ) i. e. it holds equality in Minkowski's inequality. This is only possible if P0 is homothetic to W and from R(P0) = 1 it follows P0 = W=R(W ). Since Pn =R(Pn ) was chosen arbitrarily we obtain Pn=R(Pn) ! W=R(W ). Finally we consider negative %. Let P be a covering polygon with respect to K and L 2 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 with distance j%jhK (v1); : : : ; j%jhK (vr). The strips ei + j%jhK (vi) conv f0; ?vig may overlap or may have points outside P if there are acute angles. If is the angle between two adjacent edges ei ; ej , then this area can be estimated by 2(j%jhK (vi ))(j%jhK (vj ))= sin() 2%2D(K )2= sin(): is the angle between two vectors u1; u2 2 F (K; L) and consequently ku1kku2k sin det L. From 1(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 ku1k; ku2k 2D(K ) we obtain sin() (K )=(4D(K )2). Pr Since the total volume of the strips is j%j i=1 V1 (ei )hK (vi) = 2j%jV (P; K ), we conclude that for any covering polygon P (14) V (P + %K ) = V (P ) + 2%V (P; K ) + %2c; where 0 c A and A is a constant depending only on K . Hence the formula (6) also holds for % < 0 up to a constant, which can be neglected in the proof of Lemma 4 and part (c).


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4. Optimal covering polygons for variable lattices The optimal sausage direction u is determined by (15) sm := sm (K ) = maxfV1(K= lin (u)?)kuk : u 2 bd (2K )g: (Note that for packings the direction minimizing this expression is optimal.) Obviously sm is 2V (Q), where Q denotes the maximal centrally symmetric quadrangle, which can be inscribed K . If L is the lattice with fundamental domain Q, then L 2 L(K ) and sm is attained 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 coecient of n in (6), which is sm for the sausage. So we de ne, motivated by (6) and (10), (16) ds (%) = supL2L(K ) (%hm(K; L) + 1=2) det L: Since it is dicult to determine ds (%) for given K , we give the following estimates. Lemma 5. Let K E 2 be a centrally symmetric strictly convex body and % 0. Then (a) ds (%) 3=4sm% + 3=8V (K )= (K ), where (K ) denotes the density of the densest lattice packing for K . (b) There is an < 1 such that hm (K; L) det L sm (K ) holds for any L 2 L(K ). Proof: (a) Let x be the direction corresponding to sm and kxk = 1. Further let 2l be the intersection of K and the line g = lin (x) and h = 1=2V1(K=g ?) such that sm = 4hl. There is a point a 2 bd K such that a + lx 2 bd K . Let u = 3lx and u0 = a + 2lx and let L be the lattice with basis fu; u0g. If h0 denotes the distance between a and g , then det L = 3lh0. The hexagon H = conv fa; (a + lx); lxg is anely regular and inscribed in K . Since the critical determinant (K ) is one third of the area of the minimal inscribed anely regular hexagon (see [GL], p. 243, there is a misprint in Theorem 1), it follows det L = V (H ) 3(K ) = 3=4V (K )= (K ). L is a covering lattice for H and so (K; L) 1 and by Lemma 1 (K; L) = 1. By the choice of a we have 1 (K; L) 1 and equality only if K is a parallelogram, which is not strictly convex. Hence L 2 L(K ). Further we have hm (K; L) hK (u? ) = hkuk= det L = h=h0 and so ds (%) 3%lh + 3=2lh0 = 3=4sm % + 1=2 V (H ) 3=4sm % + 3=8V (K )= (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 0 parallel to g with distance h0 we consider the value of (V1(g \ K ) + V1(g 0 \ K ))=(2V1(g \ K )). There is a line such that the maximal value is attained and since K is strictly convex it follows < 1.

12


Now let L 2 L(K ) and u 2 F (K; L). Further let fu; u0g be a basis of L and h0 the distance of u0 to lin (u) such that det L = kukh0 . From 1(K; L) > 1 it follows det L (K ) and together with kuk 2D(K ) we obtain h0 h0. The line segment conv f0; ug is covered by K; u + K and a third translate with distance h0 to lin (u). By the de nition of we have kuk 2V1(K \ lin (u)). It follows hK (u? ) det L = 1=2 V1(K= lin (u)? )kuk V1(K= lin (u)?)V1(K \ lin (u)) sm and part (b) is proved.

Proof of Theorem 3: Let L0 be the thinnest covering lattice for K . Then we de ne %1 = inf f% : ds (%) det L0 g and %2 = supf% : ds (%) sm%g. From Lemma 5 we have ds (%) sm % + det L0=2. Consequently %2 det L0=(2sm(1 ? )) < 1. Further %1 > 0 is clear. For % = det L0=sm we have %sm = det L0. From Lemma 5 it follows with V (K ) det L that ds (det L0 =sm ) 3=4 det L0 + 3=8 det L0 > det L0. From the continuity of ds we conclude %1 < det L0 =sm < %2. Since ds is monotonously increasing and ds (%) < ds (%), for > 1, it follows that the maximum value of fdet L0 ; sm%; ds (%)g is det L0 , for % < %1 and ds (%), for %1 < % < %2 and sm %, for % > %2. For a covering polygon Pn with respect to a xed lattice L 2 L(K ) we conclude from (11), (9) and (10) that V (Pn + %K ) maxf1; 2%hs(K; L); (%hm(K; L)+ 1=2)g det L n + O(1), for any %. Consequently for covering polygons Pn with respect to any L 2 L(K ) and for any parameter % we obtain (17) V (Pn + %K ) maxfdet L0 ; sm%; ds (%)g n + O(1): Now, for xed % and n, let Pn be an optimal covering polygon for K with respect to the lattice Ln 2 L(K ). For % > %2 the maximum in (17) is sm %. Hence Pn has to be a sausage, for large n, and part (a) follows. For %1 < % < %2 the maximum in (17) is ds (%). Obviously, Pn cannot be a sausage. 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 double sausage is optimal among all Pn Ln , which are not sausages, as shown in the proof of Theorem 2. So part (b) is proved. Now let % < %1. If % 1=(2hm (K; Ln)), then V (Pn + %K ) n ds (%) + O(1), which is obviously not optimal. So % < 1=(2hm(K; Ln)), for large n. As in (11) we have p 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 p p p V (Pn + %K ) n det L0 ? 2 det L0 V (W (L0; K; %)) n + o( n): From det Ln det L0 we obtain p p p V (W (Ln; K; %); Pn) det L0V (W (L0; K; %)) n + o( n): (18)


13

We can assume that SLn 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 de nition of the Wul{shapes W (L0; K; %) (1 + n )W (Ln ; K; %) with n ! 0 and from (18) we obtain p p p 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 sucient to conclude Pn =R(Pn ) ! W (L0 ; K; %)=R(W (L0; K; %)). For negative % the Wul{shapes are nonempty. Again we use (14) and the proof is the same as 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 nite covering arrangements C L. An optimal covering arrangement with respect to % and L consisting of at most n elements is denoted by C%;n . Further let D%;n = D( conv C%;n ). Since the diameter is attained, there is a direction u%;n with ku%;n? k = 1 and an x%;n such that x%;n ; x%;n + D%;n u%;n 2 C%;n . ? Let R%;n = R conv C%;n = lin (u%;n ) . In a rst step we show that optimal arrangements are nearly one dimensional, i.e. R%;n is bounded, and that the diameter of an optimal covering arrangement 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 [BM]. Proof: From Steiner's formula for the outer parallel body (see [Sch]) we have

V conv C%;n + %B d =

d X i=0

Vi ( conv C%;n)d?i %d?i ;

where V0; : : : ; Vd are the intrinsic volumes. On the other hand, if v 2 Fo(Bd; L), then for the sausage Sn(v) holds V Sn (v) + %B d = d %d + (n ? 1) wl d?1 %d?1 : Since C%;n is optimal, it follows (n ? 1) wl V1( conv C%;n) +

d X i=2

Vi( conv C%;n) d?i %i?1 : d?1

14


From Cauchy's integral representation of the instrinsic volumes (see [Sch]) we obtain Vi( conv C%;n) n Vi(B d ) and consequently there is a constant c such that (19) (n ? 1) wl V1( conv C%;n ) + n%c ; for % 1: conv C%;n is contained in a cylinder Z%;n over a (d ? 1){ball with radius R%;n and length D%;n . Therefore we have V1( conv C%;n) V1(Z%;n) = D%;n+ R%;n V1 (Bd ) and so (20) (n ? 1) wl D%;n + R%;n V1(B d ) + n c ; % 1: % Further conv C%;n contains a triangle with length D%;n and height R%;n and

with Cauchy's formula and the monotony of the instrinsic volumes we conclude (21) D%;n R%;n=2 V2 ( conv C%;n) n V2(Bd ): Since V1 is up to a factor the mean width it follows from (19) that D%;n c0 V1( conv C%;n ) c00 n for % %(d; L) 1. Together with (21) this shows that R%;n is bounded and (i) is proved. Finally, (ii) follows from (i) and (20). Next we give an upper bound for the diameter of any covering arrangement. For this aim let su = maxfu v : v 2 Fs (B d ; L)g for directions u with kuk = 1. Lemma 7. Let Cn L be a covering arrangemnet with card Cn n and let u be a direction in which D( conv Cn ) is attained. Then D( conv Cn ) (n ? 1) su : Proof: We can assume that 0 2 Cn and that there is an y 2 Cn such that kyk = D( conv Cn). The line segment conv (f0; yg) is covered by a sequence of translates t1 + B d ; : : : ; tm + B d with ti 2 Cn , t1 = 0 and tm = y , such that (ti+1 + B d ) \ (ti + B d ) 6= ;, i.e. ti+1 ? ti 2 Fs (B d ; L). For u = y=ky k it follows

D( conv Cn) = kyk = u

m ?1 X i=1

(ti+1 ? ti ) (m ? 1) su (n ? 1) su :

Lemma 8. For large % and n the diameter of conv C%;n is attained in a direction kvvk with v 2 Fo (B d ; L). Proof: We can assume that 0 2 C%;n and that there is an y%;n 2 C%;n %;n . From Lemma 6 such that ky%;n k = D( conv C%;n ). Further let u%;n = kyy%;n k and Lemma 7 it follows (22) n su%;n D%;n n wl ? c1%n ? c2:


15

For suciently large % we have limn!1 su%;n > wk and so for large % and n su%;n = u%;n v, with v 2 Fo(B d ; L). Let P%;n be the length of the projection of conv f0; y%;n g into lin (v ). Then P%;n = D%;n cos (\(u%;n; v)) = D%;n uk%;nvkv = D%;n suw%;n l and from (22) we obtain

P%;n 1 ? %cw1 ? wc2n l l

2

(23)

n wl 1 ? %2 wc1 ? w2 cn2 n wl l l 2 c n 1 = n wl ? % ? 2 c2:

As in the proof of Lemma 7 let t1 ; : : : ; tm 2 C%;n with m = m%;n n, conv f0; y%;ng [mi=1 (ti + B d ), t1 = 0, tm = y%;n and ti+1 ? ti 2 Fs (B d ; L). We can divide t1 ; : : : ; tm into subsets

f0; t1; : : : ; ts1 g; fts1+1 ; : : : ; ts2 g; : : : ; ftsq +1; : : : ; tsq+1 = y%;n g such that Sj = conv ftsj +1 ; : : : ; tsj+1 g 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 length of Sj , j = 0; : : : ; q. The Lemma is proved if the assumption q 1 leads to a contradiction. For this aim n we rst give an upper bound o for q . v d If w = max kvk z : z 2 Fs (B ; L); z = 6 v , then w < wl and we have P%;n =

m ?1 X

!

(ti+1 ? ti ) kvv k (m ? q ) wl + q w n wl ? q (wl ? w): i=1

From (23) we conclude that there are constants c3, c4 such that (24)

q c3%n + c4: ?

Now let j be the triangle conv Sj [ ftsj+1 +1 g . By the de nition of it contains a quadrangle Qj with one side Sj0 of length lj0 = r lj ? w contained in Sj and height r. Let l be the number of segments conv fti ; ti+1 g in Sj0 with length wl and k the number of these segments with length wk . Then (l + 1) wl + k wk lj0 l wl + k wk . If kti+1 ? ti k = wl, then ti+1 + B d and ti + B d have a common boundary point in the side Sj00 parallel to Sj0 . These are contained in the interior of j and so a further point of C%;n is needed. Since wl > 1 a translate x + B d contains at most two of these points. So we need at least nj = 32 l + k

16


translates to cover Qj . Let we = min w3l ; wl ? wk > 0. Then it follows 0l ? wl l wl + k wk = 3 l + k wl ? 3 l wl ? k (wl ? wk ) j 2 2 3 (25) 23 l + k (wl ? we) = nj (wl ? we): 0 , we need at least n0 = lj ?lj translates. Then n Further, to cover S n S j j j wl Pq 0 0 j =0 (nj + nj ? c), for a constant c. With c = c + w=(wl ? we) and (25) we conclude 0

q X

lj0 lj ? lj0 n + w ?c l j =0 wl ? we X q 1 r 1 r w ? we + 1 ? w lj ? c0 (q + 1) w10 P%;n ? c0 (q + 1); l l j =0

with w0 < wl . It follows that P%;n w0 n + c0 w0 (q + 1). Together with (23), (24) and division by n we obtain w ? 2 c1 ? 2 c2 w0 + c0 w0 c3 + c4 + 1 : l

% n % n Taking the limit n ! 1 it follows wl w0 + (c0 w0 c3 + 2 c1)=%, which is a contradiction for large %.

Proof of Theorem 4: First let > r. Without restriction let 0 2 C%;n and y%;n 2 C%;n with ky%;n k = D%;n . By Lemma 8 for suciently large % and n there exists an u 2 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 2 C%;n with q%;n 2= lin (u). Then dist (q%;n ; lin (u)) . Let = conv f0; q%;n; y%;n g. contains a quadrangle with length r= D%;n and height r. As in the proof of Lemma 8 with and D%;n instead of j and lj we obtain 1 r r D + 1? 1 D ?c n wl ? we %;n wl %;n = r 1 + 1 ? r 1 D%;n ? c 10 D%;n ? c w ? we w w

l

l

with w0 < wl . From Lemma 6 we get wl ? c%1 ? cn2 w0 + nc , a contradiction for large % and n. For r let u 2 Fo (B d ; L) and v 2 L \ B d with dist (v; lin (u)) = . Further let Cn = fx1; : : : ; xn?4; xa; xb ; xc; xd g 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 2 U (B d ; L) and conv Cn is a parallelogram with length (n ? 5) wl


17

and height . It follows V conv Cn + %B d = %dd + %d?1 V1( conv Cn)d?1 + %d?2 V2( conv Cn)d?2 %dd + %d?1(n ? 5) wl d?1 + %d?2(n ? 5) wl d?2 > %dd + %d?1 (n ? 1) wl d?1 = V (Sn (u) + %B d ); for suciently large n. Remark: The con gurations 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 arbitrary coverings [BHW]. Proof of Corollary 1: Let u 2 Fo(Bd ; L). L \ lin (u) + Bd covers an in nite cylinder with radius r = (1 ? wl2=4)1=2. Because of 1(B d ; L) > 1 this cylinder contains no lattice points in Ln lin (u). Therefore we have > r and with Theorem 4 the assertion follows. References [BR] R. P. Bambah, C. A. Rogers, `Covering the plane with convex sets', J. London Math. Soc., 27 (1952), 304{314. [BRZ] R. P. Bambah, C. A. Rogers, H. Zassenhaus, Òn coverings with convex domains', Acta Arith., 9 (1964), 191{207. [BW] R. P. Bambah, A. C. Woods, Òn plane coverings with convex domains', Mathematika, 18 (1971), 91{97. [BB] U. Betke, K. Boroczky, Jr., `Large lattice packings and the Wul{shape', Mathematika, to appear. [BHW] U. Betke, M. Henk, J. M. Wills,. À new approach to coverings', Mathematika, 42 (1995), 251{263. J. Reine Angew. Math., (1994), 165{191. [BM] U. Betke, M. Meyer, `Finite parametric covering densities for large parameter', in preparation. [B] G. J. Butler, `Simultaneous packing and covering in Euclidean space', Proc. London Mat. Soc., (3) 25 (1972), 721{735. [F] G. Fejes Toth, `Finite coverings by translates of centrally symmetric convex domains', Discrete Comput. Geom., 2 (1987), 353{364. [GL] P.M. Gruber, C.G. Lekkerkerker, `Geometry of Numbers', North Holland, Amsterdam, 1987. [Sch] R. Schneider, `Convex geometry: The Brunn-Minkowski theory', Cambridge University Press, Cambridge, 1993. [S] U. Schnell, `Periodic sphere packings and the Wul{shape', Beitr. z. Algebra u. Geometrie, 40 (1999), No. 1, 125{140. [W] J.M. Wills, `Lattice packings of spheres and the Wul{shape', Mathematika 86 (1996), 229{236. Mathematisches Institut, Universitat Siegen, D-57068 Siegen, Germany

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