Systems of quadratic diophantine inequalities
par WOLFGANG
MULLER
RÉSUMÉ. Soient
Q1,..., Qr
des formesquadratiques
avec descoefficients réels. Nous prouvons que pour
chaque 03B5
> 0 lesystème
|Q1(x)| 03B5,..., |Qr (x)| 03B5
desinégalités
a une solution entière non-triviale si lesystème Q1 (x) = 0,...,Qr(x) =
0 a unesolution réelle
non-singulière
et toutes les formes03B1 =
(03B11, ... , 03B1r)
~ Rs,03B1~
0 sont irrationnelles avec rang > 8r.ABSTRACT. Let
Q1, ... , Qr
bequadratic
forms with real coeffi- cients. We prove that for any ~ > 0 the system ofinequalities
|Q1 (x)| ~,... , |Qr(x) |
~ has a nonzero integer solution, pro- vided that the systemQ1 (x) = 0,...,Qr(x) =
0 has a nonsin-gular
real solution and all forms in the real pencilgenerated by Q1,
... ,Qr
are irrational and have rank > 8r.1. Introduction
Let
Q1, ... , Qr
bequadratic
forms in s variables with real coefficients.We ask whether the
system
ofquadratic inequalities
has a nonzero
integer
solution for every E > 0. If someQi
is rational andE is small
enough
then for x E Zs theinequality IQi(X)1 I
E isequivalent
tothe
equation
= 0. Hence if all forms are rational then forsufficiently
small E the
system (1.1)
reduces to asystem
ofequations.
In this caseW. SCHMIDT
[10] proved
thefollowing
result. Recall that the realpencil generated by
the formsQ~, ... , Qr
is defined as the set of all formswhere a =
(a 1,
... ,ar)
EJRr, a #
0. The rational andcomplex pencil
aredefined
similarly. Suppose
thatQ1, ... , Qr
are rationalquadratic
forms.Then the system =
0, ... , Q, (x) -
0 has a nonzerointeger
solutionprovided
that1A real quadratic form is called rational if its coefficients are up to a common real factor rational. It is called irrational if it is not rational.
(i)
thegiven
forms have a commonnonsingular
realsolution,
and either(iia)
each form in thecomplex pencil
has rank >4r2
+4r,
or(iib)
each form in the rationalpencil
has rank >4r3
+4r2.
Recently,
R. DIETMANN[7]
relaxed the conditions(iia)
and(iib).
Hereplaced
themby
the weaker conditions(iia’)
each form in thecomplex pencil
has rank >2r2
~-3r,
or(iib’)
each form in the rationalpencil
has rank >2r3
if r is even and rank>
2r 3+
2r if r is odd.If r = 2 the existence of a
nonsingular
real solution of = 0 andQ2 (x)
= 0 follows if one assumes that every form in the realpencil
is indefinite
(cf.
SWINNERTON-DYER[11]
and COOK[6]).
As notedby
W. SCHMIDT
[10]
this is false for r > 2.We want to consider
systems
ofinequalities (1.1)
without hidden equa-lities. A natural condition is to assume that all forms in the real
pencil
areirrational. Note that if
Qa
is rational and c is smallenough,
then(1.1)
andx E Z~
imply Qa(x)
= 0. We proveTheorem 1.1. Let
Q1, ... , Qr
bequadratic forms
with realcoefficient.
Then
for
every E > 0 thesystem (1.1)
has a nonzerointeger
solution pro- vided that(i)
thesystem Q 1 (x)
=0, ... ,
= 0 has anonsingular
realsolution, (ii)
eachform
in the realpencil
is irrational and has rank > 8r.In the case r = 1 much more is known. G.A. MARGULIS
[9] proved
thatfor an irrational
nondegenerate
formQ
in s > 3 variables the setI
x E is dense in R
(Oppenheirra conjecture)
In the case r > 1 all known results assume that the formsQi
arediagonal2.
For more information onthese results see E.D. FREEMAN
[8]
and J.BRUDERN,
R.J. COOK[4].
In 1999 V. BENTKUS and F. GÖTZE
[2]
gave acompletely
differentproof
of the
Oppenheim conjecture
for s > 8. We use a multidimensional variant of their method to countweighted
solutions of thesystem (1.1).
To do thiswe introduce for an
integer parameter
N > 1 theweighted exponential
sumHere
Qa
is definedby (1.2), e(:c)
=exp(2Jriz)
asusual ,
and2Note added in proof: Recently, A. GORODNIK studied systems of nondiagonal forms. In
his paper On an Oppenheim-type conjecture for systems of quadratic forms, Israel J. Math.
149 (2004), 125-144, he gives conditions (different from ours) that guarantee the existence of a nonzero integer solution of (1.1). His Conjecture 13 is partially answered by our Theorem l.l.
denotes the fourfold convolution
of pN,
thedensity
of the discrete uniformprobability
distribution on[- N,
Since WN is aprobability density
on Z~ one
trivially
obtains 1. Thekey point
in theanalysis
ofBENTKUS and G6TZE is an estimate of
SN(a
+E)Snr(a - E)
in terms ofE alone. Lemma 2.2
gives
ageneralization
of their estimate to the case r > 1. It isproved
via the doublelarge
sieveinequality.
It shows thatfor
N-2 (e(
1 theexponential
sumsSN(a - e)
andSN(a
+E)
cannotbe
simultaneously large.
This information is almost sufficient tointegrate
18N(a)1 ]
within therequired precision.
As a secondingredient
we use for0
To
1Tl
the uniform boundNote that
(1.5)
is false if the realpencil
contains a rational form. Theproof
of
(1.5)
followsclosely
BENTKUS and G6TZE[2]
and uses methods from thegeometry
of numbers.2. The double
large
sieve boundThe
following
formulation of the doublelarge
sieveinequality
is due toBENTKUS and G6TZE
[2].
For a vector T =(Tl, ... , Ts)
withpositive
realcoordinates write
T-1
=(Tll,
... , and setLemma 2.1
(Double large sieve).
Let p, v denote measures on and letS,
T be s-dimensional vectors withpositive
coordinates. Writewhere
(., .)
denotes the standard scalarproduct
in I~s and g, h : JRs -+ C aremeasurable
functions.
Thenwhere
The
implicit
constant is an absolute one. Inparticular, if I
1 and1 and 1L, v are
probability
measures, thenRemark. This is Lemma 5.2 in
[1].
For discrete measures the lemma isdue to E. BOMBIERI and H. IWANIEC
[3].
Thegeneral
case follows from the discrete oneby
anapproximation argument.
Lemma 2.2. Assume that each
form
in the realpencil of Ql, - - ., Qr
hasrank > p. Then the
exponential
sum(1.3) satisfies
Proof.
Set S =SN(a - E)SN(a + E).
We start withTo
separate
the variables m and n in theweight
function writewhere B =
(-1/2,1/2]8
and h denotes the(finite)
Fourier seriesSince w = PN * pN * pN we find
h(O)
=hN(O)2,
whereNow set
Using (2.5)
we findNote that
a(m)
andb(n)
areindependent
of E.Furthermore, by
Bessel’sinequality
Hence
t . -- I 1---
We are now in the
position
toapply
Lemma 2.1. Denoteby Ai,..., As
the
eigenvalues
ofQE
ordered in such a way that~~1 ~ > ’ ’ ’ > I.
ThenQE
= where U isorthogonal
and A = SetAl/2 ==
diag(IÀlll/2,..., IÀsI1/2),
E =diag(sgn(Ai),..., sgn(As))
andFurthermore, let p
denote the uniformprobability
distribution on and v the uniformprobability
distribution onN.
ChooseSj =
1 + Then x C .M
implies x
EB(T) and yEN implies
y E
B(S).
If followsby (2.3)
thatwhere
Hence
with = +
tl/2 N) min(N,
1 +t-1N-1~.
To prove(2.4)
we haveto consider the case
N-2 ~E~
1only.
Otherwise the trivial bound 1 is sufficient. SinceAj = Àj(E)
variescontinuously
onRr B fol
and
Aj (cc)
=cAj (E)
for c > 0 there exist constants 0Cj
oo suchthat
If
N-2
1 thenIÀjl I
« 1 andA (I Aj 1)
K 1 forall j
s. Further-more,
for j
p we findIÀjl I
x1,E I
and GAltogether
thisyields
3. The uniform bound
Lemma 3.1
(H.
DAVENPORT[5]).
LetAi,xi
+ ... +hjszs be s
linear
forms
with real andsymmetric coe,f,jicient
matrix Denoteby ~~.~~
the distance to the nearestinteger. Suppose
that P > 1. Then the numberof
x E 7GS such thatis «
(Mi ...
HereMl, ... , Ms
denotes thefirst
sof
the 2s succes-sive minima
of
the convexbody defined by F(x, y) 1,
wherefor
x, y E IEBSLemma 3.2. Assume that each
form
in the realpencil of Ql, ... , Qr is
irrational. Then
for
anyfixed
0To Tl
ocProof.
We start with oneWeyl
step.Using
the definition of u~N we findHere the first sum is over all m2, m3, nl, n2, n3, z E ZS with
N,
It is an s-dimensional box with sides
parallel
to the coordinate axes and sidelength
« N.By Cauchy’s inequality
it follows thatHere we used the well known bound
where
Ii
are intervals oflength
» 1 and ei denotes the i-th unit vector.Set
We claim that
To see this set
where denotes the fractional
part
of x. ThenN(a)
for allm =
(ml, ... , ms)
with 1 16N. Note that if zl and Z2 are countedin
D",(a)
then zl - Z2 is counted inN(a).
It follows thatTo estimate
N(a)
we use Lemma 3.1 with P = 16N andLj (z)
=2(ei, Qax)
for 1 i s. This
yields
where
Ml,a ~ ~ ~ Ms~a
are the first s from the 2s successive minima of the convexbody
defined in Lemma 3.1.Now suppose that there exists an E >
0,
a sequence of real numbersNun -
oo anda(n)
withTo Tl
such thatBy (3.1)
and(3.2)
thisimplies
Since
(
) we obtain andthis proves
By
the definition of the successive minima there existx~n~ , (n) E ZS such
that ... , 1 are
linearly independent
andMi,
=Hence for 1 ~
i, j s
Since
I
Cyi,t Tl
thisinequalities imply «Tl
1. This proves that theintegral
vectorsare contained in a bounded box. Thus there exists an infinite sequence with = for k > 1. The
compactness
of{o;
E JRsI To
Tl~ implies
that there is asubsequence
of with=
0;(0)
and10;(0)1 ~ Tl. xJnk) yjnk)
for
1 ~ j ~
s.Then Xj and Yj
are well defined andWe claim that Xs are
linearly independent. Indeed,
suppose thatthere are qj such that = 0. Then qjyj = 0
by (3.4).
Thisimplies ~~=1 yj)
= 0 and the linearindependence
of(xj, yj) yields
qj = 0 for
all j.
The matrixequation 2(~(o) (~i,... ,~5) = (~/i,..., ys) implies
thatQ,,,(o)
is rational.By
ourassumptions
this isonly possible
ifa(o)
=0, contradicting la(o) I
>To
> 0. Thiscompletes
theproof
of theLemma. D
Lemma 3.3. Assume that each
form
in the realpencil
irrational and has rank > 1. Then there exists a
function Tl(N)
such thatTl(N)
tends toinfinity
as N tends toinfinity
andfor
every 6 > 0Proof.
We first prove that there exist functionsTo(N) Tl(N)
such thatTo(N) J. 0
andI
oo for N - oo andFrom Lemma 3.2 we know that for each m E N there exist an with
Without loss of
generality
we assume that isincreasing.
ForN"z
N defineTo(N) = ~, Tl(N)
= m and for NNl
setTo(N)
=Tl(N)
= 1.Obviously
this choice satisfies(3.5). Replacing To(N) by max(To(N), N-1)
we can assume thatN-1 To(N)
1.Finally,
Lemma 2.2 with p > 1
yields
4. The
integration procedure
In this section we use Lemma 2.2 to
integrate
It is here wherewe need the
assumption
p > 8r.Lemma 4.1. For 0 U T
set B(U, T) = faER’ I
Ulal T~
anddefine
Furthermore,
let h be a measurablefunction
with 0h(a)
S(1
~-k > r.
If
eachform
in the realpencil generated by Ql,
... ,Qr
has rank > p with p > 8r andif "((U, T) > 4p/(8r)N -p/4
thenProof.
Set B =B(U,T) and q
=-y (U, T).
For 1 > 0 defineIf L denotes the least non
negative integer
suchthat 7
>2-L-1
thenISN(a)l ~y 2-L
and for any M > Lwhere
DM
=~a
E B ~IISN(a)1 ] 2-M-1}. By
Lemma 2.2with some constant C
depending
onQ1,... ,Qr. By considering C-’12SN(a)
instead of
SN (a)
we may assume C = 1. If aE BL
and a + EE Bl
it followsthat
If
N-1
thisimplies N-224(l+1)/P - 6,
say, and if1,El
>N-1
thisimplies 1,E I
>2-~~+~/P == p,
say. Note that p if2g~1+1)/P N2,
andthis is true for all 1 M if
HT
We choose M as the
largest integer
less orequal
toThen the
assumption y > 4p/(8r)N -p/4 implies
LM, (4.1 )
andTo estimate the
integral
overBl
wesplit J3/
in a finite number of subsets.If
0
choose any(31 E Bl
and setBL(~l~ _ {a Ila - 8}.
Ifa E then
la - {3ll ~
p.0 choose 02
EBl B
and set
{o; 1 1 a - ,(32~
GThen I a - I
> p andla - (3212
p for all a UB (,(32) }. Especially 1,81 - fl2 ]
>p. In this way we construct a sequence
{3l,..., 13m
ofpoints
inBi
withl,8i
- p for i~ j.
This construction terminates afterfinitely
manysteps.
To see this note that the balls with center(3i
and radiusp/2
aredisjoint
and contained in a ball with center 0 and radius T +p/2.
Thus
vol(KT+P/2)
and thisimplies
m «(1 +T/p)T.
SinceI we obtain
Note that for a E if
1,3il
> 1. If U > 1 the first sum isempty
and the second sum is(8/prU-(k-r).
If U 1 then the first sum contains G
p-" summands;
Thus both sumsare bounded
by (~/p)r.
Thisyields
Altogether
we obtainby (4.2)
and the definition of6,
p, L5. Proof of Theorem 1.1
We
apply
a variant of theDavenport-Heilbronn
circle method to countweighted
solutions of(1.1).
Without loss ofgenerality
we may assume E == 1. Otherwiseapply
Theorem 1.1 to the formsE-lQi.
We choosean even
probability density X
withsupport
in~-1, 1~
andx(x) > 1/2
for1/2. By choosing X sufficiently
smooth we may assume that its Fourier transform satisfiesi(t) = f x(x)e(tx)
dx «(1
-+-Itl)-r-3.
SetThen
K(a) _ ~i=1 x(ai). By
Fourier inversion we obtain for aninteger
parameter N > 1Our aim is to prove for N >
No,
say,with some constant c > 0. This
certainly implies
the existence of a non-trivial solution of
(1.1),
since the contribution of the trivial solution x = 0 toA(N)
is N-S and s > p > 8r. To prove(5.1)
we divide R’ in amajor
arc, a minor arc and a trivial arc. For 6 > 0 set
where
Tl (N)
denotes the function of Lemma 3.3.Using
the boundK(a)
«(1
+ Lemma 4.1(with
the choice U =Tl(N)
and the trivialestimate
1) implies
Furthermore,
Lemma 4.1 with U =Na-2
and T =Tl(N), together
withLemma 3.3
yield
Thus
(5.1)
follows if we can prove that the contribution of themajor
arc isLemma 6.1. Assume that each
form
in the realpencil of Q1, ... , Qr
hasrank > p. Let g, h : be measurable
functions
withlgl
1 andlhl
1. Thenfor
N > 1Proof.
Note that the bound is trivial forN-2.
Hence we assumeN-2.
Denoteby Ai, ... , As
theeigenvalues
ofQa
ordered in sucha way that
~1 ~ > ’ ’ ’ ~ ~ I A, 1.
ThenQa
=UT AU,
where U isorthogonal
and A =
diag(Ai , ... , A,).
Write x =(1, 7),
where x =(xl, ... ,
and~ _
(~p+i?... ?
Thenwhere
and h is defined
similarly.
IfN-2
thenby (2.6) lail
^lal
WN 2
fori p. Now we
apply
the doublelarge
sieve bound(2.3).
For1 j
p be the continuous uniformprobability
distribution on and set
9(;f)
=IÀpl-l/2xp)
and = Then
Together
with(6.1)
this proves the lemma.For a E 9Jt we want to
approximate by
where 7r =
IB * IB * IB * IB
is the fourfold convolution of the continuous uniform distribution on B =(-1/2,1/2~5.
Setg(u)
=e( Q a (u)).
Denoteby
9Ul the directional derivative
of g
in direction ul, and set 9UIU2 ==(gul)U2.
We use the
Taylor
seriesexpansions
Applying
the third of these relations tof(T)
= + the second to=
gul (X
+-FU2)
and the first toj(T)
= + we find forTogether
we obtain theexpansion
Multiplying
withWN(X), summing
over x E andintegrating
with respect to the
probability
measure 7ryields
where
Go(a)
is definedby (6.2),
and
An
elementary
calculationyields
Since gu and guv are sums of odd functions
(in
at least one of the compo-nents of
u)
we infer = 0 andG3(a)
= 0.Furthermore,
the trivialbound
guvz"(x)
Wlal3N3
+lal2N
foryields
This is
sharp enough
to prover -
To deal with
Go
andG2
we need a bound forwhere
L(x) _ (x, Qau)
and 0j
2.Using
the definition of WN and 7r we find thatu)
isequal
toExpanding L(xl
+ x2 + X3 +X4)
andQa(xl
+ X2 + X3 +X4)
this can bebounded
by
I I
Here
Applying
Lemma 6.1 to the doubleintegral
over xl and x2 andestimating
the
integral
over X3 and X4trivially
we obtainuniformly
inlul
W 1Setting
we conclude for
sufficiently
smallSimilarly,
theexplicit expression
of and the definition ofyield
Hence
Altogether
we haveproved
that for p > 8r7.
Analysis
of the termsHo(N)
andH2(N)
Lemma 7.l. Denote
by
thefourfold
convolutionof
the continuous uni-form probability
distribution onBN
=(-N- 1/2,
N +1/2]8
andby fN
thedensity of
7rN. Thenand
Proof. By
Fourier inversion and the definition of wN and 7r = 7ro we findThis proves the first assertion of the Lemma.
Similarly,
This
implies
With the abbreviations
Lm
=2 (x, Qmu~
andLm
=2(u, Qmv)
the inner-most
integral
can be calculated as rHere we used the relations
Since
we find
w,j-.
Altogether
we concludeSince 7rN has
compact support
andf N
is two timescontinuously
differen-tiable, partial integration yields
- -11 - _0 -
This
completes
theproof
of the second assertion of theLemma,
sinceFinally,
we proveNote that
Hence, by
Fourier inversionThis
completes
theproof
of Lemma 7.1. We remark that we used the fourfold convolution in the definitionof wN,
7rN,IN
for the above treatmentof
H2(N) only.
At all otherplaces
of theargument
a twofold convolutionwould be sufficient for our purpose. 0
Lemma 7.2. Assume that the
system Ql (x)
=0, ... , Qr (x)
= 0 has anonsingular
realsolutions,
thenwhere A denotes the s-dimensional
Lebesgue
measure.Proof.
This isproved
in Lemma 2 of[10].
Note that if asystem
ofhomoge-
neous
equations
=0, ... ,
= 0 has anonsingular
realsolution,
then it has a
nonsingular
real solution with1/2.
Now we
complete
theproof
of Theorem 1.1 as follows. For c > 0 and N > 0 setThen
By
Lemma 7.1and
With Lemma 7.2 this
yields
for
N > No,
say.Together
with(6.3)
thiscompletes
theproof
of Theo-rem 1.1. D
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507-516.
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Wolfgang MULLER
Institut fiir Statistik Technische Universitat Graz 8010 Graz, Austria
E-mail: w.muellerOtugraz.at