THE (αj2β)-INEQUALITY ON A TORUS
YURI BILU
A
A theorem of Macbeath asserts thatµ(AjB) min(1, µ(A)jµ(B)) for any subsets A and B of a finite-dimensional torus. We conjecture that, when the obvious exceptions are excluded, a stronger inequality
µ(AjB) min(1, µ(A)jµ(B)jmin(µ(A), µ(B))) holds, and we prove this conjecture under some technical restrictions.
1. Introduction
Let A and B be subsets of the torusrl r\r and let AjB l oajb:a ? A, b ? Bq. Throughout the paper we use the notation
α l µ(A), β l µ(B), γ l µ(AjB), (1)
where µ is the normalized Haar measure on r and µ the corresponding inner measure. (Recall that by definitionµ(A) l sup µ(F) over all closed F 9 A.) Macbeath [15] proved that
γ min(1, αjβ) (2)
(for the one-dimensional torus, (2) was established earlier by Raikov [21]). The result of Macbeath is sometimes called the (αjβ)-inequality, by analogy with the classical (αjβ)-theorems of Mann and Kneser on the addition of integer sequences [16, 12, 17, 9, 19].
The (αjβ)-inequality was extended to second countable connected compact abelian groups by Shields [26], to connected locally compact abelian groups by Kneser [13], and to unimodular connected locally compact groups by Kemperman [11]. Recently an new and elegant proof of Kemperman’s result was found by Ruzsa [24].
In the present paper we restrict ourselves to the case of a torus, asking a different question : can (2) be strengthened ? Simple examples show that, in general, the answer is ‘ no ’.
E 1.1. Let χ:r be a non-zero character and I, J intervals on of lengthα and β, respectively. Putting A l χ−"(I) and B l χ−"(J), we obtain equality in
(2). (An interal of length λ 1 on the one-dimensional torus l \ is the projection of an interval in of length λ.)
Thus, one may hope to improve on (2) only after having excluded certain ‘ extremal ’ cases. We suggest the following conjecture (the (αj2β)-inequality).
Received 24 November 1995 ; revised 24 April 1996 and 31 October 1997. 1991 Mathematics Subject Classification 11B99.
C 1.2. Let A and B be subsets of r and α, β, γ defined as in (1). Suppose that
α β and γ 1. (3)
Then either
γ αj2β (4)
or there exists a non-zero characterχ:r and closed intervals I, J 7 such that χ(A) 7 I, χ(B) 7 J,
length(I ) γkβ, length(J) γkα. (5)
As a particular case, we quote a conjecture of Macbeath [15] : if α, β 0 and γ l αjβ 1, then Al χ−"(I)BM and B l χ−"(J)BM, where χ, I, J are as in Example
1.1 and M is a set of measure 0.
It is easy to see that the inequalities in (5) cannot be improved, and closed intervals cannot be replaced by open.
N. Here and below the subscript indicates the projection from to . E 1.3. Put
A"l([0,α]Doαjβkεq), B"l([0,β]Do2βkεq)
with 0 ε β α 1\3. Further, let χ be an arbitrary non-zero character, Al χ−"(A") and Blχ−"(B"). Then γlαj2βkε αj2β, but for any non-zero
characterχh the sets χh(A) and χh(B) are not contained in open intervals of length γkβ andγkα, respectively; this is obvious for χh l χ, and an easy exercise for χh χ.
In this paper we confirm Conjecture 1.2, and, in particular, the conjecture of Macbeath in the case whenα is small enough and the ratio α\β is not too large. The precise formulation of our result is as follows.
T 1.4. For any τ 1 there exists a constant c(τ) 0 such that the
conjecture isalid with (3) replaced by
τ−"α β α c(τ). (6)
In the important particular case when Al B we obtain the following.
C 1.5. There exists an absolute constant c 0 with the following
property. Let A9 r and suppose that α l µ(A) c and γ l µ(AjA) 3α. Then
there exists a non-zero characterχ:r such that χ(A) is a subset of a closed interal
of lengthγkα.
The one-dimensional case of Corollary 1.5 was obtained by Moskvin, Freiman and Yudin [18, Lemma 2]. Our argument can be regarded as a development of their method. In particular, like them we make an essential use of Freiman’s fundamental theorem on the addition of finite sets, quoted here as Lemma 2.2.4 (see the proof of Proposition 3.3). Some new ideas were needed for extending the argument to arbitrary dimension and distinct summands ; for the latter purpose we used a result of Ruzsa [25], based on the ideas of Plu$ nnecke [20].
In Section 2 we collect miscellaneous auxiliary facts to be used in the argument. In Section 3 we prove Lemma 3.1, which can be considered as a crude version of Theorem 1.4. The proof of Theorem 1.4 occupies Section 4.
2. Auxiliary material 2.1. Conex bodies
In this subsection S9 sis a symmetric conex body, that is, a convex bounded set, symmetric with respect to the origin, and containing a neighbourhood of the origin.
The S-norm ons is defined by RxR
Sl infoλ:λ−"x ? Sq. The S-norm of a linear
functionalφ:s is RφR
Sl supoQφ(x)Q:RxRS 1q. The kth successive minimum λk is the smallest λ with the following property: there exist linearly independent
e",…,ek? ssuch thatReiRS λ. Recall the second inequality of Minkowski:
2s\s! λ"(λ
sVol S 2s.
LetΓ be a lattice in s. We say that S isΓ-thick if the set SEΓ generates a finite index subgroup ofΓ. We shall say simply thick instead of s-thick.
L 2.1.1 (Mahler). Let λ",…,λs be the successie minima of S. Then there
exists a basis e",…,esof ssuch that
λi ReiRS max(1, i\2) λi for 1 i s. (7)
Proof. See [3, Chapter 8, Corollary of Theorem 7]. Actually, it is proved there
that there exists a basis with ReiRS max(1, i\2) λi. However, we may assume that Re"RS… ResRS, rearranging e",…,esif necessary. ThenReiRS λiby the definition of successive minima.
Any basis with this property will be referred to as a Mahler’s basis for S. Similarly one defines Mahler’s bases for S of an arbitrary latticeΓ.
R 2.1.2. It is natural to make the following remark, though it is irrelevant to the topic of this paper. We do not know whether max(1, i\2) in Mahler’s lemma can be improved, but it certainly cannot be replaced by 1, because, starting from dimension 3, there exist thick symmetric convex bodies containing no basis of the integral lattice. Here is a simple example in dimension 3 : put
a"l(0,1,1), a#l(1,0,1), a$l(1,1,0),
and let S be the convex hull ofopa",pa#,pa$q. Then S contains no integral points except the origin andpa",pa#,pa$, and the lattice generated by a",a#,a$ has index
2 in$. Similar examples can be constructed in the higher dimensions.
A. All implicit constants in this subsection depend only on the dimension s.
P 2.1.3. Let e",…,esbe a Mahler basis for S and let xl x"e"j(jxses.
Then
xi λ−"
i RxRS for1 i s. (8)
Proof. Assuming that the basis e",…,esis orthonormal, we write (x, ei) instead
of xi. We have to prove thatβiλi 1, where βil maxo(x, ei)\RxRS: x? q. Fix i and find ai? ssatisfyingRa
iRSl 1 and (ai, ei)l βi. Denote by Sithe convex hull of 2s points, two of them beingpaiand the remaining 2(sk1) are pej\RejRS, where j i. Clearly, Si7 S. Consequently,
Vol S Vol Sil 2sβiReiRS s!Re"RS(ResRS βiλi λ"(λs βiλiVol S, whenceβiλi 1, as desired.
P 2.1.4. Let S be thick. Then
Vol S QSEsQ Vol S. (9)
Proof. If x"e"j(jxses? S then QxiQ βi, where the βi were defined in the
previous proof. Therefore
QSEsQ (2β"j1)((2β sj1). Since S is thick,λi 1. Therefore βi λ−"
i 1, whence 2βij1 βi. We obtain QSEsQ β"(β
s (λ"(λs)−" Vol S. Further, if maxiQxQReiRS (2s)−" then x l x"e"j(jx
ses? S, because RxRS 1\2. Therefore QSEsQ s i=" [2(2sReiRS)−"j1] (Re"R S(ResRS)−" (λ"(λs)−" Vol S. The proposition is proved.
R 2.1.5. Note that the assumption ‘S is thick’ is needed only for the second inequality in (9).
Actually, much more precise estimates for the number of lattice points are available. See [8, Section 3.1] and references there.
2.2. Addition of finite sets
We quote here some results on the addition of finite sets of integers, to be used in our argument.
A. In this subsection A and B are finite sets of integers.
We denote by min A and max A the minimal and the maximal element and put l(A)l max Akmin A, m(A) l maxoQaQ:a ? Aq;
gcd(A) denotes the greatest common divisor of the elements of A. (10) L 2.2.1 (Ruzsa). If QAjBQ σQBQ then QAjAQ σ$QBQ.
Proof. Ruzsa [23, Lemma 3.3] proves that, ifQAjBQ σQBQ, then for any positive integers k and l we have
)
(Aj(jA)k(Aj(jA))
σk+lQBQ.JMKML
k
JMKML
l
In particular,QAjAkAQ σ$QBQ, which yields QAjAQ σ$QBQ.
Ruzsa utilizes the graph-theoretic method of Plu$ nnecke [20] (see also [22, 19]). L 2.2.2 (Freiman). Suppose that 0 ? AEB and gcd(ADB) l 1. Then (a) if l(B) l(A) QAQjQBQk3, then QAjBQ l(A)jQBQ;
(b) if max(l(A), l(B)) QAQjQBQk2, then QAjBQ QAQjQBQjmin(QAQ, QBQ)k3.
Proof. See Freiman [4]. Simpler proofs were recently suggested by Steinig [28],
Lev and Smeliansky [14, Theorem 2] and Hamidoune [10]. See also Stanchescu [27]. P 2.2.3. Suppose that 0 ? AEB. Then
(a) if gcd(A)l gcd(B) l 1 and max(l(A), l(B)) QAQjQBQk3, then QAjBQ l(A)jQBQ (and QAjBQ l(B)jQAQ by symmetry);
(b) if gcd(ADB) l 1, but gcd(B) 1, then QAjBQ QAQj2QBQk2.
Proof. (a) Without restricting generality we may assume that min Bl 0. Put
al max A, Bh l BE[0, l(A)], Bd l BBBh.
Then the sets A and Bh meet the condition of Lemma 2.2.2(a), whence QAjBhQ l(A)jQBhQ.
Further, the sets AjBh and ajBd are disjoint: any element of the former is smaller than any element of the latter. Therefore
QAjBQ QAjBhQjQajBdQ l(A)jQBhQjQBdQ l l(A)jQBQ. (b) See [14, Lemma 2].
L 2.2.4 (Freiman). Suppose that 0 ? A and QAjAQ σQAQ, where σ is a
positie real number. Then there exist an integer s c"(σ), a thick symmetric conex
body S9 s and a homomorphism φ:s such that Vol S c#(σ)QAQ and
φ(SEs)8 A.
Proof. This is a fundamental result of Freiman [5, 6]. A different (and simpler)
proof was recently suggested by Ruzsa [25]. See also [19] for an exposition of Ruzsa’s proof, and [1] for a proof close to Freiman’s original.
The following proposition is a useful complement to Lemma 2.2.4.
P 2.2.5. In Lemma 2.2.4 the conex body S can be chosen so that RφRSl m(A).
Proof. Put Sh l SEox ? s:φ(x) m(A)q. Obviously, φ(ShEs)8 A but we
Thus, let be the subspace of s spanned by ShEs. Put Sd l ShE and Γ l sE . Then by Proposition 2.1.4
Vol (Sd)
detΓ QSdEΓQ QSEsQ Vol S QAQ,
where all implicit constants depend only on σ. Obviously, Sd is Γ-thick and RφdRSdl m(A), where φd l φQ . Identifying with dim and Γ with dim , we
obtain the result.
3. The main lemma
It is well known that the following three properties of a set A7 rare equivalent : (J1) µ(cA) l 0, where cA is the boundary of A;
(J2) the indicator function of A (which is 1 on A and 0 outside A) is Riemann integrable ;
(J3) for any infinite sequenceoakqk?uniformly distributed onrwe have lim
N _
Qok ? :ak? A and QkQ NqQ
2N l µ(A).
Sets with any of these properties are usually called Jordan-measurable. For brevity, we refer to them as Jordan sets.
The goal of this section is the following lemma.
L 3.1. Let A 9 rbe a non-empty open Jordan set with µ(A) l α. Assume
that AjA is also a Jordan set, and that µ(AjA) σµ(A), where σ is a positie real
number. Then there exists a non-zero characterχ such that χ(A) lies in an interal of
length O(αc"
), where c"lc"(σ)0 and the constant implied by the O(…) depends only
onσ.
For anyθ ? rand A9 r put
(θ, A) l on ? :θn ? Aq. For any N 0 we also put
(θ, A, N) l (θ, A)E(kN, N). Forη ? and ε 0 we write
(η, ε) l (η, [kε, ε]), (η, ε, N) l (η, [kε, ε], N ).
R 3.2. Sets (θ, A) with an open A are called Bohr sets; they generate Bohr’s topology on. An efficient application of Bohr sets to additive problems was recently given by Ruzsa [25]. See also [7, 2].
Call an elementθ of rgenericif it does not belong to a proper closed subgroup of r. In these terms the theorems of Kronecker and Weyl can be expressed as follows :
(Kronecker) if θ is generic in r and A is an open subset of r, then θ(θ, A) is
dense in A,
(Recall that the asymptotic density dX of a set X9 is limN _(2N )−" QXE(kN, N)Q,
provided that the limit exists.)
For the proof of Lemma 3.1 we may assume, preserving generality, that 0r? A.
A. Until the end of this section, A is an open Jordan subset of r, such that AjA is also a Jordan set, and
µ(A) l α, µ(AjA) σα, 0r? A.
In this section constants implied by, and O(…) depend only on σ.
P 3.3. Let θ ? rbe generic. For any N N! (where N!0 depends on
A andθ) there exist η l η(N) ? and ε l ε(N) 0 such that
(θ, A, N) 7 (η, ε), (11)
α ε αc", (12)
where c"lc"(σ)0. Moreoer, for any X,NN! we hae
Q(η(N), 2ε(N), X)Q αc"X. (13)
Proof. For any N 0 put N* l max (θ, A, N). Since A is open,
gcd((θ, A, N)) l 1 and N 2N*, (14)
when N is large enough. By the theorem of Weyl, for all sufficiently large N we have Q(θ, A, N)Q "#αN, Q(θ,AjA,2N)Q3µ(AjA)N. (15) Now define N! so that (14) and (15) hold for all NN!, and assume that NN! in the sequel. By (15)
Q(θ, A, N)j(θ, A, N)Q Q(θ, AjA, 2N)Q 6σQ(θ, A, N)Q.
Since 0r? A, we have 0 ? (θ, A, N). By Lemma 2.2.4 together with Proposition 2.2.5,
there exist an integer s 1, a thick symmetric convex body S 9 s and a homomorphismφ:s such that
Vol S Q(θ, A, N)Q αN, φ(SEs)8 (θ, A, N), RφR
S N. (16)
Since gcd((θ, A, N)) l 1, the homomorphism φ is surjective.
If sl 1 then φ: is either the identity map or the negation. In both the cases S8 [kN*, N*], whence
N 2N* Vol S αN.
Thus,α 1, and the assertion is trivial with ε l 1\2 and any η ? .
Now suppose that s 2. Since φ:s is surjective, φ(e!)l1 for some e!?s. Prolong φ by linearity to a linear functional on s and put
l ker φ, Γ l Es,
S!lSE .
Let e",…,es−" be a Mahler basis for S! with respect to the lattice Γ. Obviously,
e!,e",…,es−"is a basis of s.
We haveφ(x!e!j…xs−"es−")l x!. Define ψ:s by ψ(x!e!jx"e"j…xs−"es−")l x" (recall that s 2). Since e!,e",…,es−
Letψ! be the restriction of ψ to and let ?SEsbe such thatφ() l N*. Put ε l 3Rψ!RS!andηh l ψ()\N*. Then
RηhφkψRS ε. (17)
Indeed, fix x? sand put yl (φ(x)\N*) . Then ykx ? and RyR
S 2RxRSbecause Qφ(x)Q RφRSRxRS NRxRS 2N*RxRS.
(recall thatRφRS N by (16)). We obtain
Q(ηhφkψ) (x)Q l Qψ(ykx)Q l Qψ!(ykx)QRψ!RS!RykxRS! "$εRykxRS εQxQS, which proves (17).
Putη l η. We have to establish (11)–(13).
Proof of (11). Fix n? (θ, A, N). There exists x ? SEs such that φ(x) l n.
By (17)
Qηhnkψ(x)Q l Q(ηhφkψ) (x)Q εRxRS ε. Sinceψ(x) ? , we obtain n ? (η, ε).
Proof of(12). Redefine the inner product ons
to make the basis e!,e",…,es−" orthonormal. The restriction of this inner product to induces a Lebesgue measure
on , which will be denoted by Vol . Since e",…,es−
"is an orthonormal basis of , we have detΓ l 1.
The volume of the convex hull of S! and p is 2s−"N* Vol
S!. Since this convex
hull is contained in S, we haveαN Vol S N Vol S!. Hence Vol S!α. Now by Proposition 2.1.3, for any x? we have
Qψ!(x)Q\RxRS! λ−" " (Vol S!)"/(s−") α"/(s−") αc",
where λ" is the first successive minimum of S! with respect to Γ. This shows that ε Qψ!QS! αc".
We now prove thatε α. For any T 0 let ΣTbe the domain in # defined by the inequalitiesQxQ T and QηhxkyQ ε. By (16) and (17), for any x ? SEsthe point (φ(x), ψ(x)) belongs to ΣNE#. Since S is thick, so is ΣN. By Proposition 2.1.4,
αN Q(θ, A, N)Q Q(η, ε, N)Q QΣNE#Q Vol ΣNl 2εN, which proves thatε α.
Proof of (13). Now let ΣT be the domain in # defined by the inequalities
QxQ T and QηhxkyQ 2ε. If ΣXis thick, then by Proposition 2.1.4 Q(η, 2ε, X)Q QΣXE#Q Vol ΣXl 8εX αc"X, as wanted.
Now suppose thatΣXis not thick. Put Yl minoT:ΣTis thickq. Then there is a line Λ in # such that ΣTE# 9 Λ for any positive T Y; in particular, ΣXE# 9 Λ. SinceΣNis thick, Y N.
For any n? (η, ε, Y), the vertical line x l n intersects Λ inside the strip QηhxkyQ ε (the intersection point is (n, m), where m is the nearest integer to ηhn). In particular, this is the case for nl Y*, because
Since Y X N!, we have Y2Y*, whence the vertical line xlY intersects Λ inside the strip QηhxkyQ 2ε. It follows that for any positive T Y we have (η, 2ε, T) l HE(kT, T), where H is the projection of ΛE# on the first coordinate. Let a be the positive generator of H. If X athen(η, 2ε, X) l o0q and there is nothing to prove. If X a then
Q(η, 2ε, X)Q X l QHE(kX, X)Q X QHE(kY, Y)Q Y QΣYE#Q Y VolΣY Y l 8ε αc". This completes the proof of (13) and of Proposition 3.3.
P 3.4. As in Proposition 3.3, let θ be generic. Then there exist η ? and
a positie ε αc"such that(θ, A) 7 (η, ε) and d(η, ε) αc".
Proof. For all N N!, let η(N) and ε(N) be the quantities defined in Proposition
3.3. There is a sequence Nj _ such that the sequences oηjq and oεjq converge (where we writeηjl η(Nj) andεjl ε(Nj)). Denote byη and ε the corresponding limits. Then α ε αc", in particular ε 0.
Further, fix n? (θ, A). Then nηj? [kεj,εj] for all sufficiently large j. Therefore nη ? [kε, ε], which proves that(θ, A) 7 (η, ε).
To estimate the asymptotic density of(η, ε), fix X N!. For sufficiently large j we have Nj X and εj ε\N2. Also, for any n ? (η, ε, X) we have nηj nη ? [kε, ε], whence
nηj? [kεN2, εN2]7 [k2εj, 2εj]
when j is large enough. Thus,(η, ε, X) 9 (ηj, 2εj, X ) for all sufficiently large j. By (13),
Q(η, ε, X)Q
X
Q(ηj, 2εj, X )Q X αc".
Sending X to infinity, we obtain d(η, ε) αc". The proposition is proved.
Proof of Lemma3.1. Fix a genericθ ? rand letη and ε be from Proposition 3.4.
C 1. If η is not generic in then α 1.
Proof. Ifη is not generic, then it belongs to the torsion of , say, mη l 0 for
some non-zero m? . In this case (η, ε) is a union of several complete residue classes mod m. Since A is open, (θ, A) intersects all residue classes mod m. Therefore
(η, ε) l . This yields α 1 since d(η, ε) αc".
To simplify notation, put Il [kε, ε].
C 2. If (θ, η) is generic in ri, then α 1.
Proof. Suppose that (θ, η) is generic. Then d((θ, η), AiI) l 2εµ(A) by the
theorem of Weyl. On the other hand, since(θ, A) 7 (η, ε), we have ((θ, η), AiI) l (θ, A),
The assertion of Lemma 3.1 is trivial when α 1. Hence we may assume that η is generic but (θ, η) is not. Since (θ, η) is not generic, there exist a character
χ:r and an integer m, not both zero, such that χ(θ) l mη. Moreover, both χ and
mare non-zero, because bothη and θ are generic. We define the pair (χ, m) in a unique way requiring that m is positive and as small as possible.
C 3. If m 1 then α 1.
Proof. Let∆ be the subgroup of ri consisting of (x, y) satisfying χ(x) l my.
Then (θ, η) ? ∆, and by the minimality of m, the element (θ, η) is generic in ∆. Denote by π":∆ r and π#:∆ the projections to the first and second coordinate, respectively. Put Ul π−"
" (A). Then(θ, A) l ((η, θ), U). By the theorem of Kronecker, (η, θ) (θ, A) is dense in U. It follows that η(θ, A) is dense in π#(U). On the other hand,η(θ, A) 7 η(η, ε) 9 I. Therefore π#(U)9I.
For any x? π#(U) we have xjZm9 π#(U), where Zm is the cyclic subgroup of order m. If m 1 then the set xjZmis not contained in an interval shorter than 1km−" 1\2. Consequently, 2ε 1\2, whence α 1.
Thus, we may assume that ml 1, whence χ(θ) l η. It follows that χ(θ(θ, A)) 7 η(η, ε) 9 I. Again by the theorem of Kronecker, θ(θ, A) is dense in A, and we obtain finally χ(A) 7 I. The lemma is proved.
4. Proof of Theorem 1.4
To begin, we fix some conventions. In this section A, B9 r. We define α, β and γ as in (1) and fix τ 1. We put C l ADB. We assume that
τ−"α β α, γ αj2β.
We can assume that 0r? AEB, translating A and B if necessary. Further, if β l 0
then triviallyγ αj2β. Therefore γ α β 0.
All constants implied by the symbols, and O(…) depend only on τ. 4.1. Open Jordan sets
In this subsection we assume that the sets A, B, AjA, AjB and BjB are open Jordan sets. Then C and CjC are open Jordan sets as well. As in Section 3, fix a genericθ ? r.
P 4.1.1. For sufficiently large N we hae the inequality
Q(θ, C, N)j(θ, C, N)Q "&µ(CjC)N. (18)
Proof. Since CjC is Jordan and µ(CjC) 0, there exists a closed Jordan set
F9 CjC such that µ(F) "#µ(CjC). Since θ(θ,C) is dense in C, we have
F9 CjC 7
n? (θ,C)
(θnjC). Since F is compact, for some N!0 we have
F9
n? (θ,C,N!)
Now pick n? (θ, F). By (19), there exists n"?(θ,C,N!) such that nθkn"θ?C. Then nkn"?(θ,C) and Qnkn"QQnQjN!. Therefore for any NN! we have
(θ, F, NkN!)7(θ,C,NkN!)j(θ,C,N)7(θ,C,N)j(θ,C,N). (20) On the other hand, since F is Jordan, for sufficiently large N we have
Q(θ, F, NkN!)Q"#µ(F)(NkN!)"&µ(CjC)N, (21) and the result follows from (20) and (21).
P 4.1.2. There exists a non-zero character χ:r such that χ(C) 7 [kε, ε], where 0 ε αc#
. Here c# is a positie constant, depending on τ.
Proof. For sufficiently large N we have
Q(θ, A, N)Q "#αN, Q(θ,B,N)"#βN,
Q(θ, AjB, 2N)Q 3γN, Q(θ, C, N)Q 2µ(C) N. (22)
Now fix N such that (18) and (22) hold. Then
Q(θ, A, N)j(θ, B, N)Q Q(θ, AjB, 2N)Q 3(αj2β) N 1 2 3 4 (6j3τ) βN 9αN Q(θ, B, N)Q, Q(θ, A, N)Q. By Lemma 2.2.1 Q(θ, A, N)j(θ, A, N)Q Q(θ, B, N)Q, Q(θ, B, N)j(θ, B, N)Q Q(θ, A, N)Q. Hence µ(CjC) N Q(θ, C, N)j(θ, C, N)Q Q(θ, A, N)j(θ, A, N)QjQ(θ, A, N)j(θ, B, N)Q jQ(θ, B, N)j(θ, B, N)Q Q(θ, A, N)QjQ(θ, B, N)Q Q(θ, C, N)Q µ(C) N.
Thus,µ(CjC) µ(C). By Lemma 3.1, there exists a non-zero character χ such thatχ(C) lies in an interval of length O(µ(C)c#
), where c#lc#(τ)0. Since 0r? C, we
may assume this interval to be of the form [kε, ε], where ε µ(C)c# αc#. This proves the proposition.
A characterχ is primitie if ker χ is connected. Any non-zero character χ can be uniquely presented as qχ!, where χ! is primitive and qlq(χ) a positive integer (it is equal to the number of components of kerχ).
By Proposition 4.1.2, there exists a positiveε αc#such thatχ(C) 7 [kε, ε]
for
some non-zero character χ. However, there can be several characters with this property ; choose one with the minimal alue of q(χ).
By a coordinate system on r we mean a system of closed one-dimensional subgroups",…,r, such thatrl "&(&
r, where for anyian isomorphism i% is fixed. Given a coordinate system, we write an element of r
as (x",…,xr), where xi? .
Write our character asχ l q(χ) χ!, and fix a coordinate system such that none of
",…,ris a subgroup of kerχ!. Then χ!(x",…,xr)l ν"x"j(νrxr, whereν",…,νr
are non-zero integers with gcd(ν",…,νr)l 1.
As in [15], let P",…,Prbe distinct odd primes, all greater than ql q(χ), and let ZPi
be the cyclic subgroup ofiof order Pi. Then Gl ZP
i&(&ZPris a cyclic subgroup
ofrof order Pl P"…P
r. By the construction, GEker χ!l0, and even GEkerχl0, since gcd(P, q)l 1. Therefore χ maps G isomorphically onto the cyclic group ZP .
For any X9 rdefine a set of integers X4 as follows : X4l ok ? :QkQ P\2 and (k\P)? χ(XEG)q.
Obviously,QX4Q l QXEGQ. It is important to notice that QxQ εP for any x ? C4. P 4.1.3. If P",…,Prare sufficiently large thengcd(C4 )l 1.
Proof. Since C is open and contains the origin, it also contains the r points
(0, … , 0, (1\Pi), 0, … , 0) for 1 i r,
provided that the Pi are large enough. Therefore χ(CEG) 8 oqν"\P",…,qνr\Prq. When the Pi are large enough we haveQqνiP\PiQ P\2, whence
C48 oqν"P\P",…,qνrP\Prq.
It follows that dl gcd(C4) divides q, because the νi are distinct from zero and gcd(ν",…,νr)l 1.
Now write C4l odn",…,dnkq. Then Qdni\PQ ε, whence
(q\d) χ!(CEG)lon"\P,…,nk\Pq9 [kε\d, ε\d].
We may assume P",…,Pr to be so large that χ!(C) is contained in the (ε\q)-neighbourhood of χ!(CEG). It follows that (q\d)χ!(C) is contained in the (ε\d)-neighbourhood of (q\d)χ!(CEG), that is, (q\d)χ!(C)7[k2ε\d, 2ε\d]. From the minimality of q we conclude that dl 1. The proposition is proved.
P 4.1.4. If ε 1\4 then A4jB4 7 A fjB.
Proof. If a? A4 and b ? B4 then obviously ((ajb)\P)? χ((AjB)EG). Since
QaQ εP and QbQ εP, we have QajbQ 2εP P\2, whence ajb ? A fjB. The proposition is proved.
Sinceε αc#, we haveε "% when αc$(τ). We shall assume this in the sequel. It is worth noting that this is the single point in the whole argument where we need the extra conditionα c$(τ).
Letδ 0 be so small that γj5δ αj2β; if α β then we require in addition that αkδ βjδ. Assume that Piare so large that
δP 3,
)
QAEGQ P kα)
δ,)
QBEGQ P kβ)
δ,)
Q(AjB)EGQ P kγ)
δ, (23) and(*) for any x? A (respectively, B) there exists xh ? AEG (respectively, BEG) such thatχ(xkxh) ? [kδ, δ].
We can also assume thatQB4Q QA4Q. Indeed, if β α then βjδ αkδ, which yields Bg A4by (23) (recall thatQA4Q l QAEGQ and QB4Q l QBEGQ). If α l β then the assertion is symmetric in A and B, and we can interchange them if it happens thatQB4Q QA4Q.
By Proposition 4.1.3 we have gcd(A4DB4) l 1. Since 0r? AEB, we have 0 ? A4EB4.
Since A4jB4 7 A fjB, we have
QA4jB4Q QA fjBQ (γjδ) P ((αkδ)j2(βkδ)kδ) P QA4Qj2QB4Qk3. Hence gcd(A4)l gcd(B4) l 1 by Proposition 2.2.3(b), and
max(l(A4), l(B4)) QA4QjQB4Qk3
by Lemma 2.2.2(b). It follows now from Proposition 2.2.3(a) that
l(A4) QA4jB4QkQB4Q (γkβj2δ) P, l(B4) QA4jB4QkQA4Q (γkαj2δ) P. Therefore χ(AEG) is contained in an interval of length γkβj2δ. By (*), χ(A) is a subset of an interval of lengthγkβj4δ. Sending δ to zero, we conclude that χ(A) is contained in an interval of lengthγkβ. Similarly χ(B) is a subset of an interval of lengthγkα. This completes the proof of Theorem 1.4 in the case when A and B are open Jordan sets.
4.2. Closed sets
We begin with a very simple lemma.
L 4.2.1. Let X be a subset of rwithµ(X) 0. Then for any fixed λ 1
there exist only finitely many charactersχ:r such that µ(χ(X)) λ.
Proof. We use induction in r. Thus, put rl 1 and let X 7 have positive
measure. Fix a density point x of X and findε 0 such that any open interval I which contains x and is of length at mostε satisfies µ(IEX) λµ(I).
Any character χ: is the multiplication by an integer ν l ν(χ). If QνQ ε−"
then we have an interval I of lengthν−" such that µ(IEX) λµ(I). The character χ
maps I faithfully onto, whence µ(χ(X)) µ(χ(IEX)) λ. Thus, µ(χ(X)) λ yields Qν(χ)Q ε−", which proves the lemma in the case r l 1.
Now consider arbitrary r 1 and present r as r−"i. By the theorem of Fubini, there exists x? r such that µ(XE(xjr−")) 0 and µ(XE(xj)) 0. (Here we denote by µ the normalized Lebesgue measures on xjr−" and xj, respectively.) Translating X, we may assume that xl 0r. By induction, there are
finitely many possibilities for the restrictionsχQr−"andχQ. Hence there are finitely
many possibilities forχ. The lemma is proved.
In this subsection A and B are closed sets. Fix an epimorphismr rand denote by Oε9 rthe image of the open ball inrhaving centre in the origin and radiusε. Obviously, Oε is an open Jordan set.
Since A and B are closed, so is Fl AjB. Therefore
Fl
ε!
(FjOε),
and similarly for A and B. Pick δ 0 such that γjδ αj2β and α βjδ in the caseα β. Let ε l ε(δ) be such that
Since A is compact, its open covering 5a?A(ajOε) has a finite subcovering. Arguing similarly with B, we obtain finite subsets A#9 A and B# 9 B such that A9 Ah B A#jOεand B9 Bh B B#jOε. Both Ah and Bh are open Jordan sets (each of them is a union of finitely many translates of Oε), and so are the sets AhjAh, AhjBh and BhjBh (which are unions of finitely many translates of O#ε). We can assume that µ(Ah) µ(Bh): if α β, this follows from α βjδ; if α l β, then interchange A and B if necessary.
Now
µ(AhjBh) µ(FjO#ε) γjδ αj2β µ(Ah)j2µ(Bh).
Therefore there exists a non-zero characterχ mapping Ah and Bh into intervals of lengthγkβjδ and γkαjδ, respectively.
Formally, we cannot now sendδ to 0, because the character χ depends on δ, so we have to write it asχδ. However, by Lemma 4.2.1, there are at most finitely many possibilities forχδ. Therefore we have a sequenceδn 0 such that all theχδ
nare equal
to the same characterχ. This completes the proof of the theorem in the case when A and B are closed.
4.3. Arbitrary sets
Now we make no additional assumptions about A and B. Since the closure A` is compact, for anyδ 0 and any character χ there exists a finite set A(χ, δ) 9 A with the following property : for any x? A there is xh ? A(χ, δ) such that χ(xkxh) ? [kδ, δ]. Similarly we define B(χ, δ).
For every smallδ 0 we want to find a non-zero character χ l χδmapping the sets A(χ, δ) and B(χ, δ) into intervals of length γkβjδ and γkαjδ, respectively. Such a χ would map A and B into intervals of length γkβj3δ and γkαj3δ, respectively, and we would be able to complete the proof using Lemma 4.2.1 in the same manner, as at the end of the previous subsection.
Thus, fixδ 0 so small that γ αj2βk3δ and αkδ β in the case α β. Let Ah 7 A and Bh 7 B be closed sets such that µ(Ah) αkδ and µ(Bh) βkδ. Again, we may assume that µ(Ah) µ(Bh).
By Lemma 4.2.1, there exist only finitely many charactersχ mapping Ah into an interval of lengthγkαjδ. Put
Ad l AhD
0
χA(χ, δ)
1
,the union being over the characters quoted in the previous sentence. Since we added to the set Ah only finitely many new elements, the set Ad is closed and µ(Ad) l µ(Ah). In the same manner define the set Bd.
Since AdjBd is closed, we have
µ(AdjBd) µ(AjB) l γ (αkδ)j2(βkδ) µ(Ad)j2µ(Bd).
Therefore there exists a non-zero characterχ mapping Ad and Bd into intervals of lengthγkβjδ and γkαjδ, respectively. This χ maps Ah into an interval of length γkβjδ, whence A(χ, δ) 9 Ad. Thus, χ maps A(χ, δ) into an interval of length
γkβjδ; by the similar reason it maps B(χ, δ) into an interval of length γkαjδ. This completes the proof of Theorem 1.4.
Acknowledgements. I am pleased to thank Gregory Freiman for having initiated
this research. Some ideas of this paper came up in our conversation. I also thank Daniel Berend, Seva Lev, and the referee for valuable suggestions.
This work was done during the author’s post-doctoral stage at the Max-Planck-Institut fu$ r Mathematik. I am grateful to Prof. F. Hirzebruch and MPI fu$r Mathematik for their kind invitation and excellent working conditions. A considerable part of the paper was written when I was visiting Macquarie University, New South Wales, Australia. I am pleased to thank the Mathematical Department of that university, and especially William W. L. Chen and Alfred J. van der Poorten for their generous hospitality.
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