A further omega result for the ellipsoid problem in algebraic number fields

LXII.2 (1992)

A further omega result for the ellipsoid problem in algebraic number fields

by

Ulrich Rausch (Marburg)

The classical ellipsoid problem, i.e., the problem of evaluating asymp- totically the number of lattice points in a multidimensional ellipsoid, was generalized in [2] to algebraic number fields, and upper and lower estimates for the pertinent lattice rest were proved (

).

While the O-result obtained there holds generally, the Ω-theorems im- pose several restrictions on the parameters involved. For instance, the ellip- soids under consideration are supposed to be centered at lattice points. Also, the arithmetic nature of the underlying Hermitian forms plays a significant role.

In the present paper I give a somewhat weaker Ω-estimate valid in the general case, which fits well into what is known for the rational field.

We begin by formulating the problem. Let K be an algebraic number field of degree [K : Q] = n = r

+ 2r

(in the standard notation), d its discriminant, and r = r

+ r

− 1. Let e

= 1 for p = 1, . . . , r

and e

= 2 for p = r

+ 1, . . . , n.

For a fixed rational integer k ≥ 2, consider the set T of all column vectors

, . . . , ν

, ν

, . . . , ν

, . . . , ν

, . . . , ν

)

∈ C

such that

ν

∈ R (j = 1, . . . , k; p = 1, . . . , r

) ,

ν

= ν

(j = 1, . . . , k; p = r

+ 1, . . . , r + 1) ; let 0 ∈ T denote the zero vector, and

ν^(p)

= (ν

, . . . , ν

)

for ν ∈ T .

(¹) There is a misprint on p. 330, line −10, of [2]: For “endlichen Grad ¨uberQ” read

“endlichen Grad ¨uber

Q

We call ν, α ∈ T congruent with respect to a given system a = (a

, . . . , a

) of non-zero ideals a

⊆ K, in symbols

ν ≡ α (a) ,

if there exist numbers µ

∈ a

(j = 1, . . . , k) such that

ν

− α

= µ

(p = 1, . . . , n; j = 1, . . . , k) , where µ

denotes the pth conjugate of µ

.

For p = 1, . . . , r + 1, let Q

∈ C

be a positive definite Hermitian matrix, real for p = 1, . . . , r

, and let Q

be the Hermitian form in k variables arising from it; we write briefly Q for the system of these forms.

Further, let x = (x

, . . . , x

) ∈ R

, R

denoting the set of positive real numbers.

For convenience we supplement x

, Q

(and thus Q

) for r+1 < p ≤ n by setting

x

= x

,

= Q

for p = r

+ 1, . . . , r + 1 . The counting function

A

(x) = A

(x; Q, a, α)

is now defined as the number of vectors ν ∈ T satisfying ν ≡ α (a) and Q

(ν

) ≤ x

for p = 1, . . . , r + 1

(or, equivalently, for p = 1, . . . , n). The lattice rest in question is P

(x) = P

(x; Q, a, α) = A

(x) − C

X

with

X = x

. . . x

=

r+1

Y

p=1

x

, C

= 2

ω

ω

|d|

N (a

. . . a

) √ D , where

D =

n

Y

p=1

det Q

, ω

= π

/Γ (

l + 1) ,

and N denotes the ideal norm in K. Then we have the following result.

Theorem. Always (i.e., for all choices of k, Q, a, α) P

(x) = Ω

(X

) as X → ∞ .

P r o o f. We shall use extensively the contents of [2]. By e Q

we denote the Hermitian form associated with the matrix (Q

)

, and c stands for the system of ideals c

= (a

d)

(j = 1, . . . , k), where d is the different of K. c

, . . . , c

are positive constants which, as well as all O-, -, and

-constants, may depend on K, k, Q, a, and α.

For λ = (λ

, . . . , λ

) ∈ R

, λ

= λ

(p = r

+ 1, . . . , r + 1), let

(1) a(λ) := X

≡0(c),^Q(

e

e

,

where e Q(ν) = λ means e Q

(ν

) = λ

for p = 1, . . . , n, and α · ν denotes the real number

k

X

j=1 n

X

p=1

α

ν

. Since ν and −ν occur together in (1), a(λ) is real.

If a(λ) were always zero, the series Θ  πe

x

u

, . . . , πe

x

u

; e Q, c, 0, −α



in the proof of [2, Hilfssatz 6.1] would reduce to its constant term 1, and it would follow that

J

(P

(x)) := (4πε)

R

R^r+1

P

(x

e

, . . . , x

e

) exp



− 1 4ε

r+1

X

p=1

v

 dv

= 0 for all x ∈ R

and all ε > 0 .

From this we would easily conclude, either with the aid of [2, Hilfssatz 4.3]

(= [1, Theorem 3.1]) or directly (cf. the remark at the beginning of [1, Sect. 3]), that P

(x) = 0 for all x ∈ R

. But this is impossible since A

(x) is certainly not continuous. Thus

M := {λ : a(λ) 6= 0} 6= ∅ .

By [2, Hilfssatz 6.1] (with δ = 1/(4n)) we have for x ∈ R

, 0 < ε ≤ 1/2 (2) J

(P

(x)) = S

(x, ε) + O(ε

X

) ,

where

S

(x, ε) = c

X

X

λ∈M

a(λ)

×

r+1

Y

p=1

 e

(π

e

λ

x

)

cos(2π q

e

λ

x

−

(e

k + 1)π)

 . We proceed to select one dominant term from S

(x, ε).

Let ν ≡ 0 (c), λ = e Q(ν), such that λ

. . . λ

=: Λ ≤ Λ

. The vector ν is made up of the conjugates of numbers ν

∈ c

(j = 1, . . . , k). Multiplying all of these by the same suitably chosen unit, we can obtain a µ ≡ 0 (c) such that λ

:= e Q(µ) satisfies

λ

. . . λ

= Λ and λ

≤ c

Λ

(p = 1, . . . , n) ,

hence

k

X

j=1

|µ

|

≤ c

Λ

(p = 1, . . . , n)

by [2, Hilfssatz 5.1]. Since these inequalities have only finitely many solu- tions µ ≡ 0 (c), it follows that Λ belongs to some discrete subset of R

. Consequently, we may choose from M a λ

= (λ

, . . . , λ

) (c-constants!) such that

λ

. . . λ

= min{λ

. . . λ

: λ ∈ M } .

The inequality between arithmetic and geometric means yields for λ ∈ M

r+1

X

p=1

e

λ

λ

=

n

X

p=1

λ

λ

≥ n with equality only for λ = λ

, hence

(3)

r+1

X

p=1

e

λ

λ

≥ n + 2c

for λ ∈ M, λ 6= λ

, since the left-hand member attains only discrete values.

Now, for t > 1, let x

= 1

e

λ

(t + ϑ

√

t), where 0 ≤ ϑ

≤ 3 (p = 1, . . . , r + 1) , and

ε = T /t, where T ≥ 1 ; the ϑ

’s and T will be chosen later. Then

(4) c

t ≤ x

≤ c

t (p = 1, . . . , r + 1) and, if t ≥ (3n/c

)

,

(5) επ

r+1

X

p=1

e

λ

x

≤ T π

(n + c

) =: c

T , while for λ 6= λ

we have by (3)

Y := επ

r+1

X

p=1

e

λ

x

≥ T π

(n + 2c

) =: c

+ c

1 − c

T , say, so that

Y ≥ (c

+ c

)T + c

Y .

Hence the terms with λ 6= λ

contribute to S

(x, ε) at most c

e

X

X

λ∈M

|a(λ)|

r+1

Y

p=1

e

(π

e

λ

x

)

, which, for t ≥ 2c

T , is

 e

ε

 e

X

by [2, Hilfssatz 5.6] and (4).

Regarding λ = λ

, (4) and (5) yield X

|a(λ

)|

r+1

Y

p=1

e

(π

e

λ

x

)

 e

X

. As ϑ

, . . . , ϑ

run independently through [0, 3], the values of each term

2π q

e

λ

x

= 2π q

e

(t + ϑ

√ t) cover an interval of length > 2π; hence the product

r+1

Y

p=1

cos

 2π

q

e

λ

x

−

(e

k + 1)π



takes every value between −1 and +1. Thus, if ξ denotes either of the numbers +1 and −1, we can always find ϑ

’s such that

ξS

(x, ε) ≥ e

{c

− c

e

}X

,

provided t ≥ c

T . Choosing here T = c

large enough to make the term in curly brackets positive, we obtain from (2) the following result:

There exist x ∈ R

and ε > 0 such that X is arbitrarily large, (6) c

X

≤ ε ≤ c

X

,

and

(7) ξJ

(P

(x)) ≥ c

X

. Now assume

ξP

(x) ≤ %X

for X ≥ X

. Then, as in [2, Hilfssatz 8.3], it follows that

ξJ

(P

(x)) ≤ c

%X

+ O(1)

for X ≥ e

X

and every ε according to (6). Comparing this to (7) yields

% ≥ c

, hence the assertion.

References

[1] U. R a u s c h, A summation formula in algebraic number fields and applications, I , J.

Number Theory 36 (1990), 46–79.

[2] —, Zum Ellipsoidproblem in algebraischen Zahlk¨orpern, Acta Arith. 58 (1991), 309–

333.

KERSCHENSTEINERSTRASSE 18 D-W-3550 MARBURG

GERMANY

Received on 7.5.1991

and in revised form on 24.3.1992 (2136)