Math. Ann. 296, 625~535 (1993)
SpringerVerlag i 993
New bounds in some transference theorems
in the geometry of numbers
W. Banaszczyk
Institute of Mathematics, L6d~, University, Banacha 22, PL90238 L6d~, Poland
Received July 4, 1992
Mathematics Subject Classification (1991): 11H06, 11H60, 52C07
Introduction
The aim of this paper is to give new bounds in certain inequalities concerning
mutually reciprocal lattices in R". To formulate the problem, we have to introduce
some notation and terminology.
We shall treat IR" as an ndimensional euclidean space with the norm II II and
metric d. The inner product of vectors u, v will be denoted by uv; we shall usually
write u 2 instead of uu. The closed and open unit balls in F," will be denoted by
B, and B'., respectively. If A c IR ", then span A and A z denote the linear subspace
spanned over A and the orthogonal complement of A in P~".
A lattice in IR" is an additive subgroup of IR" generated by n linearly indepen
dent vectors. The family of all lattices in IR" will be denoted by A..
Given a lattice L e A., we define the polar (dual, reciprocal) lattice L* in the
usual way:
L* = {u ~ ~":uv~Z for each v~L} .
One has L**= L. By d(L) we shall denote the determinant of L, i.e. the n
dimensional volume of a fundamental domain of L. Naturally, d(L*) = d(L)a.
A convex body in/R" is a compact convex subset of IR" containing interior
points. The family of all osymmetric convex bodies in ~" will be denoted by cg,.
Given a convex body U ~ cg., we define the polar body U ~ in the usual way:
U ~ = {u ~lRn:uv_< 1 for each v~ U} .
By 1[ [Iu we shall denote the norm on 1R" induced by U (the Minkowski functional of
U). By dv we shall denote the metric induced by [I [Jr. Thus l[ 11 = 1[ IJn. and d = de..
Given a lattice L e A. and a convex body U ~ cg., we denote
p(L, U) = max{dr(u, L): u ~ R"}
= min{r > 0:L + rU = 1R"},
2t(L, U) = min{r > 0:di m span(Lc~rU) ~_ i} (i = 1 ..... n) .
626 W. Banaszczyk
The quantity #(L, U) is called the covering radius of L with respect to U. The
quantities hi(L, U) are called the successive minima of L with respect to U. To
simplify the notation, we shall write #(L) and 2i(L) instead of #(L, B,) and 2~ (L, B,),
respectively.
Given a convex body U ~ cg,, we define
~(U) = sup max 2~(L, U) 2. ~+I( L*, U ~ ,
LeAn ].~i~n
r/(U) = sup #(L, U)21(L*, U ~ ,
L~An
~(U)= sup sup inf d(uv, 7Z) 1 dv(u,L)llvl[vo.
LeAn u~Rn\L v~L*
uvCZ
It is clear that the quantities ~(U), q(U) and ~(U) are affine invariants of U. The
obvious inequality/~(L, U) < n ,~,(L, U) implies that q(U) < n~(U). It is also
clear that ~/(U) < ~(U).
For each n = 1, 2 ..... we define
4, = sup r r/, sup r/(U), (, = sup ((U).
Ue~n Ue~dn Ue~gn
Then q. < and q, < Mahler [11] proved that 4, < (n!) 2 for n > 1. See
also [6], VIII, Sect. 5, Theorem VI, where (n!) 2 is replaced by n!. Recently,
Lagarias et al. [10] proved that ~(B,)< ~ n 2 for n > 7. Similar bounds were
obtained independently in [4], Theorem (2.1).
The inequalities ~/(B.) < ~2 n s/z for n > 5 follow from Lemma 1.4 of [3] (see also
[4], Lemma (1.4) and the subsequent remarks). Lagarias et al. [10] proved that
11n <= n a/2 for every n.
That (, < oo was proved by Khinchin [9]. Cassels [6], XI, Sect. 3, Theorem VI
gave the bound (, < 21 "(n!) 2. Babai [1] proved that (. < C" for some universal
constant C. The inequalities ((B.) < 12n(n + 1) can be derived from Lemma 7 of
[2]; see also [4], Lemma (1.1) and the subsequent remarks. Hastad [7] proved that
((B,) < 6n 2 + 1.
In the present paper we show that there exists an universal constant C such that
~(B.), q(B,), ~(B.) < Cn for every n. Examples show that ~(B,), ~7(B,), ~(B,) > cn
where c is some other universal constant. For the numerical values of C and c, see
the final remarks in Sect. 2. The corresponding inequalities for convex bodies other
than B, are discussed in Sect, 3.
The determination of the quantities ~(U), r/(U) and ~(U) is a classical problem
in the geometry of numbers. The corresponding results belong to the socalled
transference theorems; a detailed description is given in Cassels' book [6].
Recently, these questions became a subject of intensive investigations in integer
programming; we refer the reader to the introductions in [1, 7, 10]. Another source
of motivation is commutative harmonic analysis, more precisely, the theory of
characters of additive subgroups and quotient groups of topological vector spaces.
This point of view is presented exhaustively in monograph [5], especially, in
Section 3. See also the survey article [4].
Papers [24, 7, 10] all found on the idea of KorkinZolotarev bases. The
method used in the present paper is entirely different. Taken from commutative
New bounds in the geometry of numbers 627
harmonic analysis, it consists in investigating certain probability measures on
lattices and Fourier transforms of such measures. The proofs are nonconstructive.
1 Gaussianlike measures on lattices
Let # be a finite Borel measure on IR". For the purpose of the present paper, it is
convenient to define the Fourier transform p ^ of # by the formula
#^ (x) = S e2~ixr dll(y) (x ~ IR").
Nn
Similarly, given an integrable complexvalued function f on N", by the Fourier
transform f ^ of f we shall mean the function
/^( x) = f e:~xYf(y)dy ( xeR").
Rn
The symbol dy denotes integration with respect to the ndimensional Lebesgue
measure on ~". Under such a definition, the function e ~x: is equal to its Fourier
transform.
Let L be a lattice in IR". By aL we shall denote the probability measure on
L given by the formula
aL(A)= E e"X2 / Z e'~2 (A = L).
x~A / xEL
By gOL we shall denote the function on IR" defined by the formula
0L(u) = Z / Y (u JR").
xcL+u / x~L
To simplify the notation, we shall write
p(A) = Z e"X2 (A c IR").
x~A
Then aL(A) = p(A)/p(L) for A c L, and tpL(u) = p(L + u)/p(L) for u ~ IR".
In the remaining part of this section, n is a fixed positive integer, and L is an
arbitrary, but fixed lattice in N".
(1.1) Lemma, Let a, b be positive numbers such that ab = 7r. Let p be the measure on
L given by the formula p({x}) = e "~2. Then
(i) ~, e2~i~reaxe=b"/ed(L)i ~ ee"iuze "b(y+z)2 (u,y~]R");
xeL+u z~L*
(ii) p^ (y) = b n/2 d(L)I ~, e,btr+,)~ (y ~ IR").
zEL*
Proof Point (i) follows from the Poisson summation formula; see e.g. [8, (31.46)
(c)]. Point (ii) is a direct consequence of (i). []
(1.2) Corollary. One has try, = tpL* and of. = ~OL.
628 W. Banaszczyk
Proof It is enough to prove that ~ = ~ot:, Let # be the measure on L given by the
formul a #({x}) = e  ~. Then o'L = ll/#(L), so that ~2 = #~/#^ (0), and the result
follows from (1.1) (ii).
Let k = 1 ..... n and let x ~ IR". By Xk we shall denote the kth coordinate ofx. If
82
f is a function on IR", we shall write fkk = ~ f
(1.3) Lemma. Given arbitrary u ~ 1R", a > 0 and k = I ..... n, one has
1
I f u = O, then ~c ~  .
 2a
E e~
x~L+u J xsL ~ a
Proof Let # be the measure on L + u given by the formula p((x}) = e "~, Then
#^( y) = ~ e z'~ixyeax 2 (y ~"),
x~L+u
x~L+u
Setting here y = O, we get
x 2 e 2~xy e": (y e IR ~) .
E x~ e": =  (d~) ' ~(o).
xeL+u
(1)
Let us denote b = ~. It follows from (1.1) (ii) that
r
I~k(y ) = b ":2 d(L) 1 ~ e 2"i"" [  2rob + 4rceb2(yk + z~)23e ~b('+z?
for y s IR". Setting y = O, we get
It~(O) = b"/2 d(L) 1 ~ e z"i"~ [  2rob + 4rt 2b2z~]e "b~2 .
z~L*
Let v be the measure on L* given by the formul a v({z}) = e  ~:. Then
(2)
~_, eZ"~U" e ~bz~ = v^(  u),
zeL*
(3)
z~ e 2~i"~ e '~b: =  (4rc2) 1 Vk~( U).
z~L*
Setting u = y = 0 in (1.1) (i), we see that
(4)
v ~ (0):= ~. e~b'Z = d(L) b"/2 ~ e"x~ (5)
:t~L* xeL
New bounds in the geometry of numbers 629
Now, from (1)(5) we derive
1 v ^(  u) l Vk~k(u)
X  2a v ^(0~ +4a z v ^( 0~ (6)
Take an arbitrary v e IR". By (1.1) (i), we have
d(L) 1 b.J2 v ^ (v) = d(L) 1 b./2 ~, e2~i~ enbz2
zEL*
: Z eaX2 1 [e_a(x+v)2
xeL+v = ~ xvL ~ + e a( x v) 2]
= eaV= y' eaX2 cosh 2axv > eav2 2 eaX2.
x~L xcL
Hence, by (5),
v ^ (v)/v ^ (0) >__ ea~2 (v e F,"). (7)
The functions v ^ and  Vk~ are positivedefinite, being the Fourier transforms
of the positive measures e '~b~ and 4rc2z2ke '~b~, respectively. Therefore
v^(  u) < v^(0) and vA(  u) < V~k(O ). It follows directly from (7) that
1 1 1
^ ^ =. If u = 0, then (6)
 Vkk(O)/v (0) < 2a. Now, (6) implies that ~c < 2a + 2a a
1
yields ~c < 2aa' because Vk~k(0) < 0. []
(1.4) Lemma. For each a > 1, one has
(i) ~, e~taI x2 ~_ a hI2 ~,, e=~=;
xeL xeL
(ii) ~ e '~" ~ < 2a "/z ~_, e '~= (u e IR") .
x~L+u xeL
Proof. Let us consider the function
f(a) = ~, e "~'x2 (a >= l ).
x~L
From (1.3) we obtain
7r 7[ n
if(a) = ~ ..LE x2 e"~'~'= = ~ ,=,E ,,eLE x~ e '~'x=
mz a x7 =aix2 n
<~ ~_ ~ e =2af ( a) ( a>l ).
Ct ~Vt x~L
T/
Hence [logf(a)]' < 2a etc., which yieldsf(a) < a"/2f(1) for a > 1. This proves (i).
To prove (ii), take an arbitrary u e IR" and consider the function
g( a) = ~ e =~'*= ( a> 1).
xeL+u
630 W. Banaszczyk
Applying (t.3) and then (i), we may write
7~ n 1
?ITC a ~ haIx2
< 2~ 2 e <=nal+'/2f(1) ( a>l ).
a ~ xeL
Hence
a a
g(a)   g(1) = f g'(t)dt < nf(1)j" t ' +'z dt
! 1
=2f ( 1) [ a "/2 1] ( a>l ). (8)
By (1.2), the function q~L is positivedefinite, being the Fourier transform of the
positive measure aL.. Thus
g(l) p(L + u)
   = ~OL(U) < q~L(O) = 1 ,
f(l ) p(L)
i.e. g(1) _<f(1). Now (8) yields
9(a) = 2f(l) [a "/2  1] + 0(1) < 2a"/~f(1)  f ( 1) < 2a"/2f(1). []
(1.5) Lenlma. For each c > (27z)t/2, one has
(i) p( r\cx/n B',) < [c~e e"r p(r),
(ii) p((L + u)\cx/n B') < 2[c 2x/~~ e'~]" p(L) (u e IR") .
Proof. For each t e (0, 1), we have
Z e~tx2= Z e~t(1ox2 enX2
xeL xeL
> ~ e,.1t)x: e~X: > e~.t)c2. ~ e,,x2.
xeL xeL
x2~czn xZ>=c:n
On the other hand, (1.4) (i) says that
Thus
E e r ex2< t~/2 E e~X2"
x~L xeL
E e~X2 < tn/2 e~tlt)czn Z enX2
x~L xeL
x2~_c2n
which can be written as
p(L\c /n B') < It e"C2(1  0]" p(L).
Setting here t = (2~ce)  t we obtain (i). To prove (ii), it is enough to apply (1.4) (ii)
instead of (1.4) (i). []
New bounds in the geometry of numbers 631
2 Transference theorems
(2.1) Theorem. Let L be an arbitrary lattice in IR', n > 1. Then
2~(L)2,_~+1(L*)<n ( i = 1,...,n).
Proof. If n = 1, then )~l (L) 21(L* ) = 1. If n = 2, then it is not hard to verify that
2t(L)22(L*) < 2"3 1/:. So, we may assume that n > 3.
Let us suppose the contrary, that there is an index i = 1 ..... n with
2i(L) 4,_ i+ ~ (L*) > n. Replacing L by sL for a suitably chosen coefficient s, we may
assume that
3
2~(L) > ~ x/n, (9)
4
2._,+1(L*) >
(10)
Let K be the subgroup of ]R" generated by L n ~ B',, and let H be the
subgroup generated by L* c~{.v/nB',. Denot e M = span K and N = span H.
Fr om (9) and (10) we get dim M<i  1 and dim N<n i. Hence
di mM + di mN > n + 1, and we can find a vector ueM J with [lull
= 3 1/2.
As u e M we have uv = 0 for each v s K. Hence
a~ (u)= Y. ~rL({v))cos 2~zuv = Z + Z ~r,({v})cos 2auv
v~L vEK veL\K
>=
Now, (1.5) (i) implies
( rt ( L\K) <=
so that
Y'. ~L({V}) Z ~L( {V}) =I   2r
v~K v~L\K
that
~L(t ~N//~B'n)<I~N~e9ml 613 <0.15,
r163 (u) > 0.7. (11)
Since Hull = 31J2 andn > 3,we have B, u c 3 B,, which means that
Hence, by (1.5) (ii),
< 2[ 2~e ~] 3 p( L*). (12)
Next, as u e N we have
p( n + u) = e ""2 a( n) =< e "/3 p( L*). (13)
632 W. Banaszczyk
From (12) and (13) we derive
p(L* + u) = p( ( L*\n) + u) + p(H + u) < 0.4p(L*),
i.e. ~oL,(u) < 0.4. In view of (1.2), this means that o~ (u) < 0.4, which contradicts
(11). []
(2.2) Theorem. Let L be an arbitrary lattice in IR", n > 1. Then 21(L)#(L*) < n.
Proof If n = 1, then 21(L)#(L*) = If n = 2, then it is not hard to verify that
21(L)/~(L*) < 2 1:2. Therefore we may assume that n > 3.
Suppose the contrary, that :~t(L)#(L*) > n. Replacing L by sL for a suitably
chosen coefficient s, we may assume that
21 (L) > 21/2 x/~, (14)
)~I(L*) > 21/z x/n (15)
Condition (15) means that there is some u elR" such that (L* +u)c~
2 1:2/~ B, = ~. Then, by (1.5)(ii),
p(L* + u) = p(iL* + u)\2 l/z x/n B.) < 2[2 1/2 2x/~e"/233 p(L*),
whence
1
cpL.(u ) < ~. (16)
On the other hand, (14) means that L n 21/2 ~ B. = {0}, and then
e~( u) = Z + Z aL({V})COS2~UV
VEL v~L
v 2 < n/2 v 2 >= hi2
> Y E
v~L v~L
v2<n/2 v2>n[2
Now, (1.5) (i) implies that
aL({v}) = 1  2 eL (L\21/2x/~ B').
1
eL(L\21/2 v/~ B.) < [21/2 2x/~ e~/213 < 4'
and therefore e~ (u) > In view of (1.2), this contradicts (16). []
(2,3) Theorem. Let L be an arbitrary lattice in R", n > 1, and let u ~ IRn\ L. Then
there exists some v ~ L* with cos 21ruv <= 0.1 and Ilvll d(u, L) <= n.
Proof. Let us suppose that n ~ 2, the case n = 1 being trivial. Replacing L by sL
and u by su for a suitably chosen coefficient s, we may assume that d(u, L) = x/~.
Then (1.5) (ii) yields
p(L + u) = p((L + u)\x/n B~,) < 2 [ 2x/'f~ e~] 2 p(L).
Hence, by (1.2),
ab(u) = ~oL(u) < 2[V/2~ e~] 2 . (17)
New bounds in the geometry of numbers 633
On the other hand, (1.5) (i) implies that
Let us denote
Then we may write
x:= aL.(L*\x/nB',) < [ 2x/~e"] 2 .
9=
~" e ~v2 cos 2nuv / ~ e ~vz .
vEL* l v~L*
v2<n v2<n
(18)
obt u) = Y + Z ~L.({v))cos2=u~
vsL* vEL*
u2<n u2>=n
> OaK. (L* n ~ B'n)  ~L.(L*\v/n B'.)
 ~9(1  x)  K. (19)
Now, (17)(19) imply that ~9 < 0.1. Consequently, there must exist a vector v
L* c~ x/n B', with cos 2nuv < 0.1. []
Theorem (2,1) says that r < n for every n. Its proof, based on (1.5), allows
one to deduce that
~( B.) <~( l +O( n 1;2) ) asn~oo.
On the other hand, there is a result of Conway and Thompson (see [12], Ch. II,
Theorem 9.5) which asserts that one can construct a sequence of lattices L, e A,
such that L* = L, for every n, and
as n * oo. (20)
n
2~(L.) > ~e (1 + o(1))
Consequently, one has
/,/
~(B.) __> ~ne (1 + o(1)) as n ~ oo.
As L*=L,, we have d( L,) = 1 for every n, which implies
#(L,)" vol(B,) > 1. Thus
( n~ 1/2
:(L.) > \F~eJ (1 + o(1)) as n ~ oo.
Theorem (2.2) says that q(B.) < n for every n. The proof shows that
asn~.
that
(21)
q(B.) ~ ~(1 + 0(n1/2))
On the other hand, (20) and (21) imply that
> n
q(B.) = ~ (1 + o(1)) as n, oo.
634 w. Banaszczyk
Let L be an arbitrary lattice in IR", n > 2. Choose any u ~ L with [lull = 21(L). It
is not hard to see that there exists a vector v eL* with uv = 1 and
v2<l +3~[ r/( B,_l ) ] 2 (cf. the proof of Theorem (3.3) in [4]). As
r/{B,_a) <  1), it follows that I[vll < 3 1/2 n.
From (2.3) we get ((B,) < 5n for every n. The constant 5 may be replaced by
a smaller one, perhaps even just by 1. From the proof of (2.3) one can deduce that
~(Bn) < 2n (1 + 0(n1/2)) as n ~ oo
On the other hand, from (20) and (21) we obtain
( ( B.) > n (1 + o(1)) as n > ~.
7~e
3 Other convex bodies
Let U be a symmetric convex body in ~". According to the John theorem, there
exists an ndimensional ellipsoid D such that D c U c v/~D. Thus, from
(2.1)(2.3) it follows that ~(U), ~(U), ((U) < Cn 3/2 for some universal constant C.
The technique used in Sections 2 and 3 allows one to show that if U is symmetric
through each of the coordinate hyperplanes, then ~(U), q(U), ((U) < C1 n log n for
some other constant Ca. Moreover, if U is the unit ball in l~, 1 < p < 0% then ~(U),
q(U), ((U) < C2 n 1~ n. The proofs will be given in a separate paper. On the
other hand, the following fact is true:
(3.1) Theorem. There exists an universal constant c such that, given an arbitrary
convex body U ~ ~,, one can find a lattice L ~ An with 21(L, U)21(L*, U ~ > cn.
This implies that ~(U), q(U), ((U) > cl n for each U e r Theorem (3.1) follows
easily from Siegel's mean value theorem; the details of the proof will be given
elsewhere.
Let us close with the following remark: it follows easily from the Bourgain
Milman theorem on the product of volumes of polar bodies and from the
Minkowski convex body theorem that
;q( L,U) ),I ( L*,U~ ( L~An, U~Cg,,n= I, 2 .... )
for some universal constant C.
References
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Ann. 273, 653664 (1986)
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Complutense Madr. 2 (special issue), 3546 (1989)
New bounds in the geometry of numbers 635
5. Banaszczyk, W.: Additive subgroups of topological vector spaces. (Lect. Notes Math., vol
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