SUCCESSIVE-MINIMA-TYPE INEQUALITIES. U. Betke, M. Henk and J.M. Wills

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1 SUCCESSIVE-MINIMA-TYPE INEQUALITIES U. Betke, M. Henk and J.M. Wills Abstract. We show analogues of Minkowski s theorem on successive minima, where the volume is replaced by the lattice point enumerator. We further give analogous results to some recent theorems by Kannan and Lovász on covering minima. 1. Introduction Throughout this paper E d denotes the d-dimensional euclidean space and the set of all convex bodies compact convex sets in E d is denoted by K d. Further K d 0 denotes the 0-symmetric convex bodies, i.e. K K d with K = K, and a convex body K K d is called strictly convex if the segment xy intersects the interior of K for all x,y K, x y. The set of lattices L E d with lattice determinant detl) > 0 is denoted by L d, and the lattice of all points with integral coordinates in E d is denoted by Z d. The k-th coordinate of a point x E d is denoted by x k, and α α ) denotes for α R the largest smallest) integer α α). The i-th successive minimum λ i K, L), 1 i d, for K K d 0, dimk) = d, with respect to a lattice L L d is defined by λ i K, L) = min{λ R λ > 0, dimλk L) i}. Between the volume V and the successive minima Minkowski established for K K d 0, dimk) = d, L L d the following relations cf. [EGH], pp. 8, [GL], pp. 13, [M]) λ 1 K, L)) d V K) d detl), 1.1) λ 1 K, L)... λ d K, L)V K) d detl), 1.) λ 1 K, L)... λ d K, L)V K) d detl). 1.3) d! All these inequalities are tight. The theorem on successive minima of Minkowski 1.) is a deep result in geometry of numbers with many applications and is an improvement of 1.1) since λ 1 K, L) λ d K, L). There are several analogues of these results, e.g. by Mahler, Weyl, Hlawka cf. [EGH], [GL], [H]). In the main part of our paper we give some analogues of 1.1), 1.) and 1.3) where V is replaced by the lattice point enumerator GK, L) = cardk L). Key words and phrases. Lattice Points, Successive Minima, Covering Minima. 1 Typeset by AMS-TEX

2 U. BETKE, M. HENK AND J.M. WILLS The results yield in particular a generalization of the following inequalities by Minkowski [M], pp. 79) which are closely related to 1.1). For K K d 0, L L d, dimk) = d, with λ 1 K, L) = 1 holds if, in addition, K is strictly convex then 3 d, 1.4) d ) The covering minima introduced by Kannan&Lovász [KL] form another sequence of numbers associated with a convex body and a lattice. For K K d, dimk) = d, and L L d the i-th covering minimum µ i K, L), 1 i d, is defined by µ i K, L) = min{t R tk + L meets every d i)-dimensional affine subspace}. E.g. the last covering minimum µ d K, L) is the classical inhomogeneous minimum cf. [GL], pp. 98) and µ 1 K, L)) 1 is called the L-width of K. Kannan&Lovász [KL] showed several analogies and relations between the λ i and the µ i. In particular they proved that there are constants α d,β d > 0 only depending on d, such that for K K d and L L d holds: ) d α d GK, L) 1, 1.6) µ 1 K, L) and if K K d 0 ) d β d GK, L) µ 1 K, L) d. 1.7) Here we obtain an analogous result where the lattice point enumerator is replaced by the volume. In analogy to 1.1) we have. Lattice Points and Successive Minima Theorem.1. Let K K0, d dimk) = d, and L L d. Then ) d λ 1 K, L) + 1,.1) if K, in addition, is strictly convex None of these inequalities can be improved. ) d 1..) λ 1 K, L)

3 SUCCESSIVE-MINIMA-TYPE INEQUALITIES 3 Remark. Obviously inequality.) is an improvement of.1) only if /λ 1 K, L) is an integer. Proof. It suffices to prove the theorem for the standard lattice Z d, since GK, Z d ) = GAK,AZ d ) and λ i K, Z d ) = λ i AK,AZ d ) for every linear map A with deta) 0. Let p = /λ 1 K, Z d ) + 1. First suppose that there are two lattice points g = g 1,...,g d ) T, h = h 1,...,h d ) T, g h, in K with g i h i mod p, i = 1,...,d..3) By the convexity of K and from p > /λ 1 K, Z d ) follows that the lattice point g1 h 1 p,..., g ) T d h d = 1 ) p p g + 1 p ) h.4) belongs to Z d \{0}) intλ 1 K, Z d )K). This is a contradiction to the definition of λ 1 K, Z d ) and so there exist no lattice points g,h K, g h, satisfying.3). Hence each lattice point g K corresponds uniquely to a representation g 1,...,g d ) where g i denotes the residue class with respect to p of the i-th coordinate of g. There are at most p d of such representations, so we get.1). For the cube C d q = {x E d x i q, 1 i d}, q N, follows GC d q, Z d ) = q +1) d = /λ 1 C d q, Z d )+1 ) d and this shows that.1) cannot be improved. For the proof of.) let p = /λ 1 K, Z d ) and g,h K Z d, g h, such that.3) holds. From /p λ 1 K, Z d ) follows that the lattice point.4) lies in the boundary of λ 1 K, Z d )K. By the strict convexity of K this implies g = h. So as above) each pair g, g with g K Z d \{0}) corresponds uniquely to a residue class vector g 1,...,g d ) T which shows.). To show that.) is tight, we construct a standard example. Let C d = {x E d 0 x i 1, 1 i d} and P the 0-symmetric polytope P = conv{c d, C d }. We have GP, Z d ) = d+1 1 and λ 1 P, Z d ) = 1. With elementary considerations the existence of a strictly 0-symmetric convex body K in fact of infinitely many) follows with P K, GK) = GP) and λ i K, Z d ) = 1, i = 1,...,d. This shows that.) cannot be improved. Let us remark that for λ 1 K, Z d ) = 1 the inequalities.1) and.) become Minkowski s inequalities 1.4) and 1.5). In the case d = we can improve.1) and.) in the following way Theorem.. Let K K0, dimk) =, and L L. Then λ 1 K, L) + 1 λ K, L) + 1,.5) if K, in addition, is strictly convex λ 1 K, L) None of these inequalities can be improved. λ K, L) 1..6)

4 4 U. BETKE, M. HENK AND J.M. WILLS Remark. Again, in general inequality.6) is not an improvement of.5). Proof. It obviously suffices to prove the theorem for the lattice Z. Let z 1,z be linearly independent lattice points with z i λ i K, Z )K, i = 1,, and such that the segment z 1 z is free of other lattice points. Then the triangle conv{0,z 1,z } contains no other lattice points expect 0,z 1,z, and so z 1,z are a basis of Z cf. [GL], p. 0). Hence we may assume cf. [GL], p. ) Now we have for each x = x 1,x ) T K z 1 = 1,0) T and z = 0,1) T. λ K, Z ) x 1 1 or λ K, Z ) x 1;.7) otherwise the lattice point x 1 / x 1,x / x ) T would belong to the interior of conv{±z 1, ±z,λ K, Z )x}) λ K, Z )K which contradicts the definition of the second successive minimum. Now let p i = /λ i K, Z ) + 1, i = 1,, and let f : E E the linear map with f x 1,x ) T) = x 1, ) T x. p 1 p With /p i < λ i K, Z ) we get from.7) fk) Z = {0}..8) Let g = g 1,g ) T, h = h 1,h ) T, g h, two lattice points of K with g i h i mod p i, i = 1,..9) By the convexity of fk) it follows, that the lattice point g1 h 1, g ) T h = 1 p 1 p fg) + 1 f h).10) belongs to fk) Z \{0}) which contradicts.8). Hence there are no two lattice points of K with property.9) and so each lattice point g K corresponds uniquely to a representation g 1,g ) where g i denotes the residue class with respect to p i of the i-th coordinate of g. There are at most p 1 p of such representations, so we get.5). For the proof of.6) let p i = /λ i K, Z ), i = 1,, g, h and fk) be as above. From /p i λ i K, Z ) and.7) follows intfk)) Z = {0}. Hence the lattice point.10) belongs to the boundary of K. With the strict convexity of fk) this implies g = h and as in the proof of.) we get.6). The examples in the proof of Theorem.1. shows that both inequalities are tight. On account of Minkowski s theorem on successive minima 1.) we conjecture that Theorem.. can be generalized to

5 SUCCESSIVE-MINIMA-TYPE INEQUALITIES 5 Conjecture.1. Let K K0, d dimk) = d, and L L d. Then λ i K, L) + 1, if K, in addition, is strictly convex ) 1 λ i K, L) The following proposition shows that inequalities of this type exist Proposition.1. Let K K0, d dimk)= d, and L L d. Then ) i λ i K, L) + 1. Proof. Let λ 1 K, L),...,λ j K, L) 1, λ j+1 K, L),...,λ d K, L) > 1 and let z 1,...,z j be j linearly independent lattice points with z i λ i K, L)K) L, 1 i j. Further let L be the linear subspace spanned by z 1,...,z j and K = K L, L = L L. We clearly have λ i K, L) = λ i K, L), 1 i j, and with Blichfeldt s theorem [GL], p. 6) and 1.) for K, L it follows: GK, L) = j! VjK) detl + j with equality only for j = 1. j ) i λ i K, L) + 1, On the other hand Conjecture.1. is in the sense stronger than Minkowski s second theorem that it is easy to derive the latter from the former: Proposition.. If an inequality holds for all K K d 0 and all L L d, then λ i K, L) + c i V K) detl) λ i K, L). ), c i R, Proof. Let K K0. d Then we have for µ R, µ 0, λ i K,µL) = µλ i K, L). Further we have by elementary properties of the Riemann integral V K) ) detl) = lim µ 0 µd GK,µL) lim µ µ 0 λ i K,µL) + c i = λ i K, L). The lower bound 1.3) is much easier to prove than Minkowski s theorem 1.). The same seems to hold for the case of lattice points as we have as satisfactory general lower bound:

6 6 U. BETKE, M. HENK AND J.M. WILLS Theorem.3. Let K K d 0, dimk) = d, L L d, and λ 1 K, L) 1. Then GK, L) d 1 λ ) d 1K, L) 1 d! λ i K, L)..11) In general the constants cannot be improved. Proof. Again, it suffices to prove the theorem for the lattice Z d. For convenience we write ν i = 1/λ i K, Z d ), 1 i d. Let z 1,...,z d be d linearly independent points in Z d with ν i z i K, 1 i d, and let Q K be the crosspolytope with vertices ±ν i z i, 1 i d. Denoting by e i the i-th coordinate unit vector and by P the crosspolytope with vertices ±ν i e i, 1 i d, we have GK, Z d ) GQ, Z d ) GP, Z d ) and V P) = d /d!ν 1... ν d ). Hence it is only necessary to prove GP, Z d ) 1 1 ) d V P)..1) ν 1 Let ρ = 1 1 ν 1 and E j denote the plane spanned by e 1,...,e j, 1 j d. The total orthogonal complement of the plane E j is denoted by Ej. We show by induction that for each z j ρp Z d Ej GP z j + E j ), Z d ) V j ρp z j + E j )),.13) where V j denotes the j-dimensional volume. For j = 1 we have V 1 ρp z 1 +E 1 )) = V 1 P z 1 + E 1 )) 1. As for any segment S we have GS, Z d ) V 1 S) 1 the assertion is proved. Now let z j+1 ρp Z d Ej+1 and let η R be the maximal number such that ηe j+1 + z j+1 ρp. Then we have GP z j+1 + E j+1 ), Z d ) i= i= GP ie j+1 + z j+1 + E j ), Z d ) V j ρp ie j+1 + z j+1 + E j ) ) = V j ρp z j+1 + E j )) i= It follows GP z j+1 + E j+1 ), Z d ) V j ρp z j+1 + E j )) where f is the function given by fx) = { 1 x η) j, for x,η] 0, for x [η, ). 1 i ) j. η fi) + fi + 1),.14)

7 SUCCESSIVE-MINIMA-TYPE INEQUALITIES 7 Now f is convex on [0, ) and from this we obtain by elementary properties of the integral Along with.14) it follows fi) + fi + 1) GP z j+1 + E j+1 ), Z d ) η 0 fx)dx = η j + 1. η j + 1 V j ρp z j+1 + E j )) = V j+1 ρp z j+1 + E j+1 )), which proves.13). In particular for j = d.13) is equivalent to.1). It remains to show that.11) cannot be improved in general. To this end we consider the regular crosspolytope P d = conv{±e 1,...,±e d }). From Ehrhart s theorems cf. [GL], pp. 135) follows that for k N d d GkP d, Z d ) = k i G i P d ), GkP d, Z d ) = k i 1) d i G i P d ),.15) where GP d, Z d ) denotes the number of lattice points in the interior of P d and G i P d ) are constants. Especially we have G d P d ) = V P d ) = d /d!. Next we observe GkP d, Z d ) = GkP d, Z d ) + GkP d 1, Z d ) + GkP d 1, Z d ) and thus we obtain from.15) G d 1 P d ) = d 1 d 1)!. On the other hand the formula.11) yields GkP d, Z d ) 1 1 ) d V kp d ) = d k d! kd d d i ) 1 k Comparing the coefficients of k d and k d 1 with the coefficients of GkP d, Z d ) in.15) we see, that we can do no better than in the theorem. Minkowski s inequalities 1.) and 1.3) appear to be much more symmetric than Theorem.. and Theorem.3. By a slight weakening of Theorem.3. we obtain a corollary, which is up to the Gauss-brackets completely symmetric to Theorem. in the same way as 1.) to 1.3): Corollary.1. Let K K0, d dimk) = d, L L d, and λ d K, L). Then GK, L) 1 ) d! λ i K, L) 1. In general the constants cannot be improved. Proof. On account of λ 1 K, L) λ d K, L) the assertion follows from.11). In the proof of Theorem.3..1) appears to have some interest of its own, as it relates volume, lattice number and successive minima for crosspolytopes. Thus there is the natural question for a formula of this kind, which holds for all 0- symmetric convex bodies, and for a corresponding upper bound. Certainly.1) is not true for all K K d 0 as the class of open boxes with edges parallel to the coordinate axes shows. But this class suggests: ) i.

8 8 U. BETKE, M. HENK AND J.M. WILLS Conjecture.. Let K K d 0, dimk) = d, L L d, and λ d K, L). Then V K) detl) 1 λ ) ik, L) GK, L). 3. Covering Minima For the volume of a convex body we have the following lower bound with respect to the covering minima Theorem 3.1. Let K K d and L L d. Then there is a constant τ d, only depending on d, with 0 < τ d d!) 1 and µ 1 K, L)) d V K) τ d detl). 3.1) Proof. Since V K) = V A 1 K) deta) and µ i K, L) = µ i A 1 K,A 1 L) for every linear map A with deta) 0 it suffices to prove the theorem for the lattice Z d. For a convex body K K d Kannan&Lovász [KL] proved µ 1 K, Z d ) = λ 1 K K), Z d )) 1, where K K) denotes the polar body of the difference body K K of K. So µ 1 K, Z d ) d V K) = λ 1 K K), Z d ) ) d V K). 3.) From Roger s and Shephard s theorem on the difference body cf. [GL], p. 3) and Bourgain s and Milman s theorems on the polar body cf. [EGH], p. 31, [KL]) we have with a constant c 1 d d ) V K) V K K) ) d c1 V K K) ) 1 3.3) d From 3.) and 3.3) we obtain with 1.1) and d d ) 1 c1d ) d = d τ d µ 1 K, Z d ) d V K) d τ d λ1 K K), Z d ) ) d V K K) ) 1 τ d. The regular crosspolytope shows that τ d d!) 1. The constants α d,β d in 1.6), 1.7) and τ d in 3.1) are not best possible. We conjecture Conjecture 3.1. Let K K d and L L d. Then µ 1 K, L)) d V K) detl). d! From µ 1 K, L) µ d K, L) and.6) follows µ 1 K, L)... µ d K, L)V K) τ d detl), i.e. an analogue of Minkowski s theorem on successive minima 1.), although not with best constant. As a direct consequence of a result by Nosarzewska, Hadwiger and Wills one obtains for the surface area F and the lattice Z d

9 SUCCESSIVE-MINIMA-TYPE INEQUALITIES 9 Proposition 3.1. Let K K d, dimk) = d, and 1 i d. Then None of this inequalities can be improved. µ i K, Z d )V K) < 1 FK). 3.4) Proof. For lattice-point-free K K d with respect to the standard lattice Z d Nosarzewska d = ), Wills d = 3,4) and Hadwiger general d) cf. [GL], p. 8, [EGH], p. ) proved V K) < 1 FK). From this follows for general K Kd µ d K, Z d )V K) < 1 FK). On account of µ i K, Z d ) µ d K, Z d ), 1 i d, shows this 3.4). Now let q N, q 3, and Q q = { x E d x i q, 1 i d 1, x d 1 1 }. q Then µ 1 Q q, Z d ) > 1, V Q q ) < 1 FQ q) and lim µ 1Q q, Z d )V Q q )FQ q ) 1 = 1 q, hence none of the inequalities can be improved. References [EGH] P. Erds, P.M. Gruber, J. Hammer, Lattice points, Longman Scientific & Technical, New York, [GL] P.M. Gruber, C.G. Lekkerkerker, Geometry of numbers, North Holland, Amsterdam, [H] M. Henk, Inequalities between successive minima and intrinsic volumes of a convex body, Mh. Math ), [KL] R. Kannan, L. Lovász, Covering minima and lattice-points free convex bodies, Ann. of Math ), [M] H. Minkowski, Geometrie der Zahlen, Teubner, Leipzig, Mathematisches Institut, Universitt Siegen, Hlderlinstrasse 3, D-W-5900 Siegen, Federal Republic of Germany. address: Wills@hrz.uni-siegen.dbp.de