A discrete analogue for Minkowski's second theorem on successive minima

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by  Romanos Malikiosis

A Discrete Analogue for Minkowski’s Second
0
Theorem on Successive Minima
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0
2

n
a Romanos-Diogenes Malikiosis
J

1
2
July 5th, 2009

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T
N
. Abstract
h
t
a
m Themainresultofthispaperisaninequalityrelatingthelatticepointenu-
merator of a 3-dimensional, 0-symmetric convex body and its successive
[
minima. Thisisanexampleofgeneralization ofMinkowski’stheoremson

1 successiveminima,wherethevolumeisreplacedbythediscreteanalogue,
v the lattice point enumerator. This problem is still open in higher dimen-
9
sions,however,weintroduceastrongerconjecturethatshowsapossibility
2
of proof by induction on thedimension.
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3
.
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1 Introduction
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1
:
v
Xi AsubsetK ofRdiscalledaconvexbodyifitisconvex,compact,withnonempty
interior (sometimes the technical condition K int(K) is required,but we will
ar notneedit here). The setofallconvexbodies⊂ofRd willbe denotedas d, and
K
the subset of the 0-symmetric elements will be denoted as d. Furthermore,
d will denote the set of all lattices in Rd. We will denoteKth0e ith successive
L
minimum of K d with respect to Λ d by λ (K,Λ), i.e.,
∈K0 ∈L i
λ =inf λ>0 dim(λK Λ)>i ,
i
{ | ∩ }
where dim(A) for A Rd is the dimension of the vector space spanned by all
⊂
vectors of the set A.
Minkowski proved the following two inequalities relating the volume of K
∈
d, i.e., its d-dimensional Lebesgue measure, with its successive minima, with
K0
1
respect to a lattice Λ:
d
vol(K) 2
6 (1)
det(Λ) λ (K,Λ)
(cid:18) 1 (cid:19)
and
d d
1 2 vol(K) 2
6 6 (2)
d! λ (K,Λ) det(Λ) λ (K,Λ)
i i
i=1 i=1
Y Y
which is an improvement of (1). The above are known as the first and second
theorem on successive minima, respectively. About a century later, in 1993,
Betke, Henk, and Wills [BHW] stated discrete analogues of these theorems,
wherethe volumeisreplacedbythe lattice pointpointenumerator,G(K,Λ):=
#(K Λ). They proved the analogue for the first theorem, which predicts the
∩
following inequality:
d
2
G(K,Λ)6 +1 (3)
λ (K,Λ)
(cid:20) 1 (cid:21)
As for the second theorem, they stated a conjecture, which they verified in the
planar case. The conjectural inequality is:
d
2
G(K,Λ)6 +1 (4)
λ (K,Λ)
i=1(cid:20) i (cid:21)
Y
In section 3.2 we shall see a proof for the 3-dimensional case of the above con-
jecture. There is a notion of induction in the proof; we need statements about
intersections of a given convex body by hyperplanes passing through lattice
points. Theresultingconvexbodies, whosedimensionis d 1,arenotcentrally
−
symmetric, in general. Therefore, it is necessary to extend the definition of the
successive minima, as well as the results referred to them, to the class of all
convex bodies, not necessarily 0-symmetric, namely
λ (K,Λ):=λ (1DK,Λ),
i i 2
where DK stands for the difference body of K, i.e.,
DK =K K = x y x,y K .
− { − | ∈ }
Under this definition, Minkowski’s theorems still hold; this is a simple conse-
quence of the Brunn-Minkowski theorem ([GL],pp. 12 and 32), which predicts
that
vol(K)6vol(1DK).
2
Insection3.1,wewillprovideaproofforinequality(3)forthenon0-symmetric
case; we conjecture that inequality (4) still holds when K is not 0-symmetric,
and we will prove some weaker estimates, exactly as in [M].
2
2 Some auxiliary lemmata
We will first need the following standard convention; remember that the suc-
cessive minima of K d with respect to a lattice Λ are those of 1DK. By
∈ K 2
definition of the successive minima λ (K,Λ), there are d linearly independent
i
lattice vectors ai =ai(K,Λ) such that
ai(K,Λ) λi(K,Λ)DK Λ.
∈ 2 ∩
Then, we constructa basis of Λ, denoted by ei =ei(K,Λ), 16i6d, such that
lin(a1,...,ai)=lin(e1,...,ei)
for all i, 16i6d. Furthermore, we define the following subgroups of Λ:
Λi(K):=Ze1 Zei
⊕···⊕
We will usually abbreviate the notation to Λi. It should be noted that there is
an abuse of notation here; it is evident that the choice of the ai’s and the ei’s,
as well as the Λi’s, is not always unique. However, by this notation we shall
always mean a choice of vectors or subgroups with the above properties. The
main property that will be used later is
int(λiDK) Λ Λi−1 (5)
2 ∩ ⊂
Lemma 2.1 Let K d, Λ d. For each real n , satisfying n > 2/λ , we
i i i
∈ K ∈ L
have
DK n (Λ Λi−1)=∅
i
∩ \
.
Proof: Assume otherwise; then the intersection
1 DK (Λ Λi−1)
ni ∩ \
would be nonempty. The left part of this intersection is a subset of
int(λiDK),
2
since n >2/λ . Therefore, the intersection
i i
int(λiDK) (Λ Λi−1)
2 ∩ \
would be nonempty; but this is absurd, since it contradicts (5) above.(cid:3)
Thefollowingisanadaptationoflemma2.1in[H],forthecaseofnon-symmetric
convex bodies. The proof is identical, so we will not provide it here.
3
Lemma 2.2 Let K d, Λ d, and Λ a sublattice of Λ. Then
∈K ∈L
G(K,Λ)6 det(eΛ)G(DK,Λ).
det(Λ)
e
e
The following two lemmata will be used for the proof of inequality (4) in the
3-dimensional case. Notice that they are statements in d dimensions.
Lemma 2.3 Let K Rd be a convex body, Λ d, such that K Λ=∅. For
⊂ ∈L ∩
a real t>1, there is some v Λ such that
∈
K (v+tΛ)=∅.
∩
Proof: Take v Λ such that #(conv(v,K) Λ) is minimal. If this number is
∈ ∩
greater than 1, then there is some w Λ, w = v, such that w conv(v,K).
∈ 6 ∈
Hence, conv(w,K) conv(v,K), and v / conv(w,K), contradicting the min-
⊂ ∈
imality of #(conv(v,K) Λ). Thus, conv(v,K) Λ = v . We claim that
K (v+tΛ)=∅. Assum∩e not; then there is some∩u Λ s{uc}h that v+tu K.
∩ ∈ ∈
By convexity,and the fact that t>1, we get v+u conv(v,K), which implies
u=0, or v K, a contradiction, since K Λ=∅.∈This concludes the proof.(cid:3)
∈ ∩
The next lemma generalizes the above:
Lemma 2.4 Let K Rd be a convex body, and Λ d. Let S Λ be finite,
⊂ ∈ L ⊂
and r be a positive integer, such that
(1) (K S) rΛ=∅.
− ∩
(2) DS r(Λ 0 )=∅.
∩ \{ }
Now let t>r be an integer. There is a set S′ Λ, obtained by translating each
⊂
v S by some vector r w(v), where w(v) Λ, such that
∈ · ∈
(1)′ (K S′) tΛ=∅.
− ∩
(2)′ DS′ t(Λ 0 )=∅.
∩ \{ }
4
Proof: We will prove it by induction on #(S). If #(S) = 1, i.e. S = v ,
{ }
we use Lemma 2.3 for K v and the lattice rΛ. Since t > r, there is some
w(v) Λ, suchthat (K v−) (r w(v)+tΛ)=∅. Put S′ = v+r w(v) , and
we ca∈n easily see that (1−)′ is∩satis·fied. It should be noted tha{t when· #(S})=1,
conditions (2) and (2)′ hold vacuously.
Now,assumethat#(S)>1. Takev S+rΛ,suchthat#(conv(v,K) (S+rΛ))
∈ ∩
is minimal. Again, as in the proof of Lemma 2.3, we must have conv(v,K)
∩
(S+rΛ)= v . ApplyinductionforK =conv(v,K)andS =S (S (v+rΛ));
{ } \ ∩
wehave#(S)=#(S) 1. Let’sseewhy(1)and(2)aresatisfiedforK,S (same
−
r,Λ); (2) is obviously satisfied, as S eS. If (1) were not esatisfied, there would
⊂
be some we S and u Λ such that w +ru conv(v,K). By meineimality
∈ ∈ ∈
assumption, w+ru = v. But v /eS +rΛ, a contradiction. Thus, (1) and (2)
∈
hold for K, S,eand by induction there is some S′ Λ, obtained from S by
⊂
translating each u S by r w(u), we(u) Λ, such that
∈ · ∈
e e e e
e (K S′) tΛ=∅,
− ∩
and e e
DS′ t(Λ 0 )=∅.
∩ \{ }
Now, simply take S′ = S′ v . (2)′ is satisfied for S′; if x,y S′, then
∪ {e } ∈
x y / t(Λ 0 ) from the above. If x S′ and y = v, then again from the
ab−ove,∈v x\/{tΛ}, since ve K′. If x=y∈=v, we have nothing to provee, so
− ∈ ∈
e
DS′ t(Λ 0 )=∅.
∩ \{ }
(1)′ is also satisfied for K,S′; assume not. Then, there would be some w S′,
∈
u Λ such that w+tu K. If w S′, then w+tu K, which contradicts
(K∈ S′) tΛ = ∅. If w∈= v, then v∈+tu K, and by∈convexity, v+ru K,
− ∩ ∈ ∈
hence u = 0, by minimality assumptieon, and v K, aecontradiction. This
∈
coencluedes the proof.(cid:3)
3 Inequalities for G(K,Λ)
Throughout the rest of the paper we will use the notation
2
q (K,Λ)= +1 .
i
λ (K,Λ)
(cid:20) i (cid:21)
Also, when Λ is the standard lattice Zd, we write G(K) instead of G(K,Zd).
5
3.1 The general case
The method of the following proof is similar to the proof of theorem 1.5 in [H],
from a slightly different viewpoint.
Theorem 3.1.1 Let K d, Λ d, q = q (K,Λ). Let also n ,...,n be a
i i 1 d
∈ K ∈ L
sequence of integers satisfying
n divides n , 16i6d 1.
i+1 i
• −
q 6n , 16i6d.
i i
•
Then,
d
G(K,Λ)6 n .
i
i=1
Y
Proof: Let ei =ei(K,Λ) and define
Λ=Zn e1 Zn ed.
1 d
⊕···⊕
By lemma 2.2, e
d
det(Λ)
G(K,Λ)6 G(DK,Λ)=G(DK,Λ) n .
i
det(Λ)
i=1
e Y
e e
It suffices to prove that G(DK,Λ)=1, or equivalently
DKe (Λ 0 )=∅.
∩ \{ }
This follows from lemma 2.1 and theefact that
d
Λ 0 n (Λ Λi−1)
i
\{ }⊂ \
i=1
[
e
(we remind that n >q >2/λ ). Indeed, let g Λ 0 be arbitrary,and let k
i i i
∈ \{ }
be minimal such that
g Zn1e1 Znkeek.
∈ ⊕···⊕
Sincenk dividesalln1,...,nk−1byhypothesis,wehaveg nkΛ. Byminimality
of k, we also have g / Λk−1, hence g n (Λ Λk−1), as d∈esired.(cid:3)
k
∈ ∈ \
6
Weobtaininequality(3)forthenon0-symmetriccaseaswellbysimplyputting
n =q , for all i, 16i6d. We remind the following definition, given in [M]:
i 1
Definition 3.1.1 Let C denote the least positive constant, such that for any
d
sequence of d positive integers, x 6 x ,6 6 x , there exists a sequence of
1 2 d
···
integers y ,y ,...,y satisfying:
1 2 d
a. x 6y , for all i, 16i6d.
i i
b. y divides y , for all i, 16i6d 1.
i i+1
−
y y y
c. 1 2··· d 6C .
d
x x x
1 2 d
···
We can always choose n ,...,n with the properties given in theorem 3.1.1,
1 d
such that
d d
n 6C q .
i d i
i=1 i=1
Y Y
The estimates for C in [M], yield the following:
d
Theorem 3.1.2 Let K d, Λ d. Then
∈K ∈L
d
4
G(K,Λ)6 (√3)d−1 q (K,Λ).
i
e
i=1
Y
If K d, then
∈K0
d−1 d
4 40
G(K,Λ)6 3 q (K,Λ).
i
e 9
r ! i=1
Y
Notice that when K is non 0-symmetric, we cannot disregard all q that are
i
less than or equal to 2, as it was done in [M]; the reason is that we may have
q = = q = 2, and K Λ have full affine dimension (while if K was 0-
1 d
··· ∩
symmetric, this would mean that K Λ = 0 ). Consider, for example, the
∩ { }
d-dimensional cube [0,1]d. So, in the non 0-symmetric case we cannot reduce
the base to (3 40/9)d−1, using this argument.
p
7
3.2 The case d = 3
We will introduce an inductive method in order to prove inequality (4); this
method works up to dimension 3, and it is inadequate for higher dimensions.
As we will see in the next section, stronger versions of lemmata 2.3 and 2.4
might be needed for this method to work in all dimensions.
LetK d,Λ d. Fixabasisei =ei(K,Λ)ofΛ,thatsatisfiestheproperties
given i∈n Kse0ction∈2L. We will write each vector x of Rd with coordinates with
respect to this basis:
x = (x ,...,x )
1 d
= x e1+ +x ed.
1 d
···
Define
K[t]:= x K x =t ,
d
{ ∈ | }
i.e., the subset of K whose elements have fixed height, or the intersection of K
by a hyperplane parallel to the vector subspace spanned by e1,...,ed−1. We
canwrite G(K,Λ)intermsoflattice pointenumeratorsofconvexbodies whose
dimension is d 1; this is the point where induction could be used. Namely,
−
G(K,Λ)= G(K[t] ted,Λd−1).
−
t∈Z
X
ThebodiesK[t] ted areprojectionsoftheintersectionsK[t]onthevectorsub-
spacespannedby−e1,...,ed−1 alongthelatticevectored. Asbefore,Λd−1 isthe
Z-span of e1,...,ed−1. Apart from K[0] which is 0-symmetric, the other pro-
jections are not necessarily 0-symmetric. This is the main reason for extending
inequalities (3) and (4) for the non symmetric case.
Next, we observe that
1D(K[t] ted) 1DK,
2 − ⊂ 2
therefore, for 16i6d 1
−
λ (K[t] ted,Λd−1)>λ (K,Λ),
i i
−
which implies
q (K[t] ted,Λd−1)6q (K,Λ),
i i
−
for 16i6d 1. Assuming that inequality (4) holds for d 1, we would have
− −
d−1
G(K[t] ted,Λd−1)6 q (K,Λ),
i
−
i=1
Y
8
forall t Z. Only the factor q is missing;normally,it wouldbe obtainedfrom
d
∈
the number of the nonempty intersections, K[t]. But it is not always the case
that this number is less than q (K,Λ).
d
The next technique is to group all intersections whose heights are congruent
modulo q . Then, the above sum becomes
d
qd−1
G(K,Λ)= G(K[t] ted,Λd−1).
−
Xr=0 t≡r(Xmodqd)
It suffices to prove that for each fixed r, we have
d−1
G(K[t] ted,Λd−1)6 q .
i
−
t≡r(Xmodqd) Yi=1
Ofcourse,wecouldhavemorethanoneconvexbodyintheabovesum,however,
the above collection of convex bodies K[t] ted, t r(modq ) satisfies some
d
− ≡
restricting conditions, namely
(1) D(K[t] ted) q (Λd−1 Λi−1)=∅forall t r(modq )and16i6d 1.
i d
− ∩ \ ≡ −
(2) ((K[t] ted) (K[t′] t′ed)) q Λd−1 =∅ for all t,t′ r(modq ), t=t′.
d d
− − − ∩ ≡ 6
The above are consequences of lemma 2.1. Indeed, for the first one we observe
that
D(K[t] ted) q (Λd−1 Λi−1) DK q (Λ Λi−1),
i i
− ∩ \ ⊂ ∩ \
and the latter is empty since q >2/λ . As for the second, if
i i
((K[t] ted) (K[t′] t′ed)) q Λd−1 =∅,
d
− − − ∩ 6
therewouldexistsomev Λd−1 suchthatq v+(t t′)ed K[t] K[t′] DK.
d
Since q t t′, and t=t′,∈the intersection − ∈ − ⊂
d
| − 6
DK q (Λ Λd−1)
d
∩ \
would be nonempty, again a contradiction by lemma 2.1.
It is natural to state the following conjecture:
Conjecture 3.2.1 Let K ,...,K Rd be convex bodies and Λ d. Also,
1 n
⊂ ∈ L
let e1,...,ed be a basis of Λ, and denote by Λi the Z-span of 0,e1,...,ei, and
let q >q > >q be positive integers satisfying
1 2 d+1
···
9
(1) DK q (Λ Λi−1)=∅ for all 16j 6n and 16i6d.
j i
∩ \
(2) (K K ) q Λ=∅ for all 16j,l 6n, j =l.
j l d+1
− ∩ 6
Then
n d
G(K ,Λ)6 q .
j i
j=1 i=1
X Y
Fromthe aboveanalysis,it is clearthatthe aboveconjecture implies inequality
(4) for one dimension higher. We will verify this conjecture for d = 1,2, thus
proving inequality (4) in all dimensions up to three. A statement in support of
this conjecture is that condition (2) is too restricting for the convex bodies K ,
j
given the fact that q is smaller than the rest of the q ’s. It simply says that
d+1 i
notwotranslatesofK andK , j =l,byvectorsofq Λintersect. Inthenext
j l d+1
6
section, we will see a reduction of this conjecture, which is more convincing.
Proof of Conjecture 3.2.1, d=1: Without loss of generality, we may assume
that Λ=Z. Let K =[a ,b ], 16j 6n. Conditions (1) and (2) read
j j j
(1) b a <q for all 16j 6n.
j j 1
−
(2) (K K ) q Z=∅ for all 16j,l 6n, j =l.
j l 2
− ∩ 6
If b a > q , then the union of K with all its translates by multiples of q
1 1 2 1 2
cover−all of R, so by condition (2) we must have n=1, therefore
n
G(K ,Λ)=G(K )6q
j 1 1
j=1
X
by (1). If b a < q , there is a translate of each K by some multiple of q ,
1 1 2 j 2
−
2 6 j 6 n, that lies in (b ,a +q ), again by (2). Since they do not intersect
1 1 2
each other by (2), we will have
n
G(K )6G([a ,a +q ))=q 6q .(cid:3)
j 1 1 2 2 1
j=1
X
Proof of Conjecture 3.2.1, d=2: Let
n
D =dim DK q Λ .
j 3
∩
(cid:18)(cid:18)j=1 (cid:19) (cid:19)
[
10