DISTRIBUTION OF LATTICE POINTS ON HYPERBOLIC SURFACES

VSEVOLOD F. LEV

Let Λ′, Λ′′ ⊂ ℝ^{s} have the same number of points on each hyperbolic surface |x_{1} ⋯ x_{s}| = C. We investigate the case when Λ′, Λ′′ are sublattices of ℤ^{s} of the same prime index and show that then Λ′ and Λ′′ must coincide up to re-enumerating the coordinate axes and changing their directions.

1. Introduction

If the opposite is not stated explicitly, we denote vectors by lower-case italic and Greek letters and their coordinates by the same letters with appropriate lower indices. This paper is aimed to prove the following result.

Theorem 1. Let Λ′, Λ′′ be two sublattices of ℤ^{s} of the same prime index, having the same number of points on each surface |x_{1} ⋯ x_{s}| = C. Then Λ′′ may be obtained from Λ′ (and vice versa) by re-enumerating the coordinate axes and changing their directions.

It would be interesting to extend this theorem onto the case of ℤ^{s}-sublattices of nonprime indices, and also to investigate the case of general ℝ^{s}-sublattices, when rotations around the bisectrices ±x_{1} = ⋯ = ±x_{s} should also be taken into account. However, this may present some unexpected difficulties, as shows already for s = 2 the example of

Λ′ = {m ∈ ℤ^{2} | m_{1} ≡ 0 (mod 4)},
Λ′′ = {m ∈ ℤ^{2} | m_{1}, m_{2} ≡ 0 (mod 2)}.

Both these lattices are of the index 4, and they evidently have the same number of points on each hyperbola |x_{1}x_{2}| = C.

One may also investigate surfaces of other types, say, spheres. A special class of hyperbolas-like surfaces arises in the theory of uniform distribution, where one often faces functions of the form

Z(t, Λ) = ∑_{m∈Λ} \frac{1}{(\overline{m}_{1} ⋯ \overline{m}_{s})^{t}} ,

1991 Mathematics Subject Classification. Primary: 11P21; Secondary: 52Cxx, 11Hxx.

2 VSEVOLOD F. LEV

here \overline{m}_{i} = max\{1, |m_{i}|\}. It is natural to consider those Λ′, Λ′′ for which Z(t, Λ′) = Z(t, Λ′′) and evidently this means that on each surface of the form \overline{x}_{1} ⋯ \overline{x}_{s} = C there is the same number of points of Λ′ and Λ′′. Using the method of this paper, one can prove that Theorem 1 remains valid also for these surfaces instead of pure hyperbolas (though the proof would require substantially greater amount of calculations).

The author has no information about other results concerning this kind of problems.

2. Sublattices of prime index

In this section we establish the structure of ℤ^{s}-sublattices of prime index, and thus reduce the considered problem to an algebraic one.

In what follows p will stand for a fixed prime number. For λ ∈ ℤ (or λ ∈ 𝔽_{p}) denote

δ_{p}(λ) = \begin{cases}1, & if λ ≡ 0 (mod p), \\ 0, & if λ ≢ 0 (mod p). \end{cases}

Lemma 1. Let a ∈ 𝔽_{p}^{s} and assume a ≠ 0. Then the set

Λ_{a} = {m ∈ ℤ^{s} | δ_{p}(a_{1}m_{1} + ⋯ + a_{s}m_{s}) = 1}

is a ℤ^{s}-sublattice of index p and moreover, every ℤ^{s}-sublattice of index p is of this form for a properly chosen a.

Proof. Suppose a_{s} ≠ 0, denote by a′_{s} the inverse to a_{s} in 𝔽_{p} and let a_{i}^{∗} = −a_{i}a′_{s} (i = 1, …, s − 1). Consider the matrix

E = ( \begin{matrix}1 & 0 & ⋯ & 0 & 0 \\ 0 & 1 & ⋯ & 0 & 0 \\ . & . & & . & . \\ . & . & & . & . \\ . & . & & . & . \\ 0 & 0 & ⋯ & 1 & 0 \\ a_{1}^{∗} & a_{2}^{∗} & ⋯ & a_{s−1}^{∗} & p \end{matrix} )

Its columns form a basis for some lattice Λ of the index | det E| = p, while each column obviously belongs to Λ_{a}. Hence Λ ⊆ Λ_{a}, and therefore either Λ_{a} = Λ, or Λ_{a} = ℤ^{s}. Obviously it is the first possibility that really holds, thus Λ_{a} = Λ is a lattice of index p.

Conversely, let Λ ⊂ ℤ^{s} be a lattice of index p generated by some matrix E. Then det E = ±p, and, therefore, there exists a vector a ∈ 𝔽_{p}^{s}, a ≠ 0 such that E^{T} a ≡ 0 (mod p) (more precisely, all the coordinates of E^{T} a divide by p). Hence, each column e of E satisfies a_{1}e_{1} + ⋯ + a_{s}e_{s} ≡ 0 (mod p), thus Λ ⊆ Λ_{a}. But (by the already proved part of the lemma) both these lattices are of the same index p, so Λ = Λ_{a}. □

DISTRIBUTION OF LATTICE POINTS 3

Obviously, Λ_{a} does not change if we multiply a by a non-zero residue λ ∈ 𝔽_{p}. If we permute the coordinates of a, the new lattice may be obtained from the original one by re-enumerating the coordinate axes. And if we change signs of some of the coordinates a_{i}, this maps the lattice symmetrically relative to the hyperplanes x_{i} = 0. This shows that if

(1)  T_{a}(C) = ∑_{|m_{1}···m_{s}|=C} δ_{p}(a_{1}m_{1} + ⋯ + a_{s}m_{s})

is the number of points of Λ_{a} on the surface |x_{1} ⋯ x_{s}| = C, then Theorem 1 will follow from (and is actually equivalent to) the following theorem:

Theorem 1′. Assume T_{a}(C) = T_{b}(C) for some fixed non-zero a, b ∈ 𝔽_{p}^{s} and every integer C. Then b may be obtained from a (and vice versa) by means of multiplying by a non-zero residue λ ∈ 𝔽_{p}, reordering the coordinates and changing the signs of some of them.

We write a ∼ b if a and b are related as described in this theorem.

Show now that without loss of generality we may restrict ourselves with the case when all the coordinates of a, b differ from 0. This will include two steps. As the first step, we suppose T_{a}(C) = T_{b}(C) for every C and prove that a and b have the same number of non-zero coefficients. To this end, assume

a = (a_{1}, …, a_{s′}, 0, …, 0),
b = (b_{1}, …, b_{s′′}, 0, …, 0),

where a_{1}, …, a_{s′}, b_{1}, …, b_{s′′} ≠ 0, and define

α = (a_{1}, …, a_{s′}),
β = (b_{1}, …, b_{s′′}).

If (m_{1}, …, m_{s}) contributes a non-zero term to the sum T_{a}(1), then (m_{1}, …, m_{s′}) contributes a non-zero term to T_{α}(1), and it is easily seen that every such (m_{1}, …, m_{s′}) will be induced in this way by exactly 2^{s−s′} vectors (as the coordinates m_{s′+1}, …, m_{s} can be independently picked out to be ±1). The same applies to T_{b}(1) and T_{β}(1), and therefore

(2)  T_{a}(1) = 2^{s−s′}T_{α}(1), T_{b}(1) = 2^{s−s′′}T_{β}(1),

and for brevity we denote this common value by T. Furthermore, if q is prime then

(3)  T_{a}(q) = 2^{s−s′}T_{α}(q) + 2^{s−s′}(s − s′)T_{α}(1) = 2^{s−s′}T_{α}(q) + (s − s′)T,
(4)  T_{b}(q) = 2^{s−s′′}T_{β}(q) + 2^{s−s′′}(s − s′′)T_{β}(1) = 2^{s−s′′}T_{β}(q) + (s − s′′)T.

4 VSEVOLOD F. LEV

By the Dirichlet theorem, for each z = 1, …, p − 1 there exists a prime q_{z} under q_{z} ≡ z (mod p), and for z = 0 choose q_{0} = p. Substituting this q_{z} into (3), (4) and summing up over all z one obtains

(5)  2^{s−s′} ∑_{z=0}^{p−1} T_{α}(q_{z}) + p(s − s′)T = 2^{s−s′′} ∑_{z=0}^{p−1} T_{β}(q_{z}) + p(s − s′′)T.

But the sums in the latter identity may be explicitly evaluated:

∑_{z=0}^{p−1} T_{α}(q_{z}) =

= ∑_{z=0}^{p−1} ∑_{ε_{1},...,ε_{s′}=0}^{1} ∑_{i=1}^{s′} δ_{p}((−1)^{ε_{1}}a_{1} + ⋯ + (−1)^{ε_{i}}q_{z}a_{i} + ⋯ + (−1)^{ε_{s′}}a_{s′}) =

= ∑_{ε_{1},...,ε_{s′}=0}^{1} ∑_{i=1}^{s′} ∑_{z=0}^{p−1} δ_{p}((−1)^{ε_{1}}a_{1} + ⋯ + (−1)^{ε_{i}}za_{i} + ⋯ + (−1)^{ε_{s′}}a_{s′}) =
= 2^{s′}s′,

and similarly,

∑_{z=0}^{p−1} T_{β}(q_{z}) = 2^{s′′}s′′.

Hence (5) yields

2^{s}s′ + p(s − s′)T = 2^{s}s′′ + p(s − s′′)T,
s′(2^{s} − pT) = s′′(2^{s} − pT),

and so s′ = s′′, unless p = 2. But if p = 2, then

T = T_{a}(1) = \begin{cases}2^{s}, & if s′ ≡ 0 (mod 2), \\ 0, & if s′ ≢ 0 (mod 2), \end{cases}

therefore 2^{s} ≠ pT and s′ = s′′ also in this case. Observe, that this completely settles Theorem 1′ for p = 2, and in what follows we will assume p to be odd.

Next (and this is our second step), show that T_{α}(C) = T_{β}(C) provided T_{a}(C) = T_{b}(C) (for each C). For C = 1 this follows from (2) (in view of s′ = s′′); and for C > 1 we have

T_{a}(C) = 2^{s−s′}T_{α}(C) + 2^{s−s′} ∑_{d|C, d<C} k_{C/d}T_{α}(d),

where k_{C/d} are some combinatorial coefficients which do not depend on a, but only on s − s′ and on the system of exponents in the prime decomposition of C/d. A similar equality holds for T_{b}(C), and thus the required conclusion follows by induction on C.

DISTRIBUTION OF LATTICE POINTS 5

Now, if we prove Theorem 1′ for vectors with non-zero coordinates, then T_{α}(C) = T_{β}(C) implies α ∼ β, and hence a ∼ b. Thus we arrive to the final form of the theorem to be proved:

Theorem 1′′. Let p be an odd prime, and let a, b ∈ (𝔽_{p}^{×})^{s}. Assume T_{a}(C) = T_{b}(C) for all integer C. Then a ∼ b.

3. Even Dirichlet characters

The initial point of our proof is as follows: along with T_{a}(C), we consider all the values T_{c}(C) for c ∈ (𝔽_{p}^{×})^{s}, sum them up and then select T_{a}(C) using characters of the group (𝔽_{p}^{×})^{s}. But characters of (𝔽_{p}^{×})^{s} are closely related to the Dirichlet characters of 𝔽_{p}^{×}. And of all the characters of 𝔽_{p}^{×} of particular interest for us will be even ones (those for which χ(−1) = 1). Even characters form a subgroup of index 2 in the group of all characters, and we denote this subgroup by X. Also, denote by G the multiplicative group of all squares of 𝔽_{p}^{×}, and by Ĝ the dual group of characters of G. Most of the properties of X we use will be derived from the following simple lemma.

Lemma 2. The mapping Ĝ → X defined by

χ ↦ (n ↦ χ(n^{2})) (n ∈ 𝔽_{p}^{×})

establishes an isomorphism Ĝ ≅ X.

Proof. The above defined mapping is obviously an injective homomorphism, and |Ĝ| = |G| = (p − 1)/2 = |X|. □

Corollary 1.

i) X is cyclic; ii) For λ ∈ 𝔽_{p}^{×},

∑_{χ∈X} χ(λ) = \begin{cases}(p − 1)/2, & if λ = ±1, \\ 0, & if λ ≠ ±1; \end{cases}

iii) Each complex multiplicative function on X (that is each character of X) is of

the form

χ ↦ χ(λ)

for some fixed λ ∈ 𝔽_{p}^{×} (and even λ ∈ G).

Proof.

i) X ≅ Ĝ ≅ G, and G is cyclic.

ii) Follows from the orthogonality relation for group G. iii) Follows from the canonical homomorphism between G and ˆ Ĝ (the group of char-

acters of Ĝ).

□

By χ_{0} we denote the principal character of 𝔽_{p}^{×}. By \overline{χ} we denote the character conjugated to χ.

Lemma 3. Let X_{0} ⊆ X be a set of even characters, such that χ_{0} ∈ X_{0}, and if χ ∈ X_{0} then necessarily \overline{χ} ∈ X_{0}. Let also f : X_{0} → ℂ be a complex function on X_{0} such that from χ_{1} ⋯ χ_{r} = χ_{0} follows f(χ_{1}) ⋯ f(χ_{r}) = 1 for every r ≥ 1 and χ_{1}, …, χ_{r} ∈ X_{0}. Then f can be extended to a multiplicative function on all X.

Proof. First, observe that f can be extended to the subgroup X̃_{0} ⊆ X generated by X_{0}: if

χ = χ_{1} ⋯ χ_{r} (χ_{1}, …, χ_{r} ∈ X_{0}),

we put

f(χ) = f(χ_{1}) ⋯ f(χ_{r}),

and it is easily verified that this correctly defines f as a multiplicative function on X̃_{0}.

Next, if χ is a generating element of X, then all elements of X̃_{0} are powers of χ, and we choose ν to be the smallest positive integer with χ^{ν} in X̃_{0}. Now, we choose the value of f(χ) to satisfy f^{ν}(χ) = f(χ^{ν}), and then we define f(χ^{n}) = f^{n}(χ) for every integer n.

□

4. Proof of Theorem 1′′

We start with fulfilling the plan outlined in the previous section: by (1), we have:

T_{a}(C) =

= (p − 1)^{−s} ∑_{c} ∑_{m_{1}···m_{s}=±C} δ_{p}(c_{1}m_{1} + ⋯ + c_{s}m_{s}) ∑_{χ} χ_{1}(c′_{1}a_{1}) ⋯ χ_{s}(c′_{s}a_{s}),

where c = (c_{1}, …, c_{s}) runs over all elements of (𝔽_{p}^{×})^{s}; next, χ runs over all collections (χ_{1}, …, χ_{s}) of Dirichlet characters; finally, c′_{i} (i = 1, …, s) stands for the inverse of c_{i} in 𝔽_{p}^{×}. For our purposes, it is sufficient to consider only those C for which C ≢ 0 (mod p).

DISTRIBUTION OF LATTICE POINTS 7

Then c_{i}m_{i} runs over 𝔽_{p}^{×} together with c_{i}, and using multiplicativity of characters we get

T_{a}(C) =

= (p − 1)^{−s} ∑_{χ} χ_{1}(a_{1}) ⋯ χ_{s}(a_{s}) ∑_{m_{1}···m_{s}=±C} χ_{1}(m_{1}) ⋯ χ_{s}(m_{s}) ·

· ∑_{c} \overline{χ}_{1}(c_{1}m_{1}) ⋯ \overline{χ}_{s}(c_{s}m_{s}) δ_{p}(c_{1}m_{1} + ⋯ + c_{s}m_{s}) =

= (p − 1)^{−s} ∑_{χ} σ(χ) χ_{1}(a_{1}) ⋯ χ_{s}(a_{s}) ∑_{m_{1}···m_{s}=±C} χ_{1}(m_{1}) ⋯ χ_{s}(m_{s}),

where

σ(χ) = ∑_{c} \overline{χ}_{1}(c_{1}) ⋯ \overline{χ}_{s}(c_{s}) δ_{p}(c_{1} + ⋯ + c_{s}).

Consider the latter sum. If λ ∈ 𝔽_{p}^{×} is fixed, then λc runs over (𝔽_{p}^{×})^{s} together with c, hence

σ(χ) = ∑_{c} \overline{χ}_{1}(λc_{1}) ⋯ \overline{χ}_{s}(λc_{s}) δ_{p}(λ(c_{1} + ⋯ + c_{s})) =
= \overline{χ_{1} ⋯ χ_{s}}(λ) σ(χ),

and so σ(χ) = 0 if the product χ_{1} ⋯ χ_{s} is not the principal character χ_{0}. On the other hand, for χ_{1} ⋯ χ_{s} = χ_{0} and at least one of the characters χ_{i} distinct from χ_{0}, we use a well-known representation of δ_{p} with complex exponents to obtain

σ(χ) =
= \frac{1}{p} ∑_{z=0}^{p−1} ∑_{c_{1},...,c_{s}=1}^{p−1} \overline{χ}_{1}(c_{1}) ⋯ \overline{χ}_{s}(c_{s}) e^{2πi \frac{c_{1}+···+cs}{p} z} =

= \frac{1}{p} ∑_{z=1}^{p−1}(χ_{1} ⋯ χ_{s})(z) ∑_{c_{1},...,c_{s}=1}^{p−1} \overline{χ}_{1}(zc_{1}) ⋯ \overline{χ}_{s}(zc_{s}) e^{2πi \frac{zc_{1}+···+zcs}{p}} =

= \frac{1}{p} ∑_{z=1}^{p−1} (∑_{c_{1}=1}^{p−1} \overline{χ}_{1}(c_{1})e^{2πi \frac{c_{1}}{p}} ) ⋯ (∑_{c_{s}=1}^{p−1} \overline{χ}_{s}(c_{s})e^{2πi \frac{cs}{p}} ) =
= \frac{p − 1}{p} τ(\overline{χ}_{1}) ⋯ τ(\overline{χ}_{s}),

where the Gaussian sums τ(\overline{χ}_{i}) are known to be distinct from 0. Also, if χ_{1} = ⋯ = χ_{s} = χ_{0} then obviously σ(χ) ≠ 0. We see that σ(χ) differs from 0 if and only if χ_{1} ⋯ χ_{s} = χ_{0}, and thus

T_{a}(C) = (p − 1)^{−s}∑_{χ}^{∗}σ(χ) χ_{1}(a_{1}) ⋯ χ_{s}(a_{s}) ∑_{m_{1}···m_{s}=±C} χ_{1}(m_{1}) ⋯ χ_{s}(m_{s}),

8 VSEVOLOD F. LEV

where the sum marked by asterisk extends to all collections of Dirichlet characters with the product χ_{0}. As to the internal sum, it may be written as

∑_{\begin{matrix}m_{1}···m_{s}=C \\ m_{1},...,m_{s}≥1 \end{matrix}}(χ_{1}(m_{1}) + χ_{1}(−m_{1})) ⋯ (χ_{s}(m_{s}) + χ_{s}(−m_{s})),

and this vanishes if at least one of the characters χ_{i} is odd. But otherwise χ_{i}(m_{i}) + χ_{i}(−m_{i}) = 2χ_{i}(m_{i}) and therefore

T_{a}(C) = 2^{s}(p − 1)^{−s}∑_{χ}^{∗∗}σ(χ) χ_{1}(a_{1}) ⋯ χ_{s}(a_{s}) ∑_{m_{1}···m_{s}=C} χ_{1}(m_{1}) ⋯ χ_{s}(m_{s}),

where the sum marked by two asterisks extends to all collections of even Dirichlet characters with the product χ_{0}, and m_{i} here and in all subsequent sums attain only positive values. Group now together those collections χ which differ only by a permutation. If χ′ and χ′′ are two such collections then obviously σ(χ′) = σ(χ′′), and also

∑_{m_{1}···m_{s}=C} χ′_{1}(m_{1}) ⋯ χ′_{s}(m_{s}) = ∑_{m_{1}···m_{s}=C} χ′′_{1}(m_{1}) ⋯ χ′′_{s}(m_{s})

for each C, hence

T_{a}(C) = 2^{s}(p − 1)^{−s}∑_{φ}^{∗∗∗}σ(φ) ( ∑_{χ∈P (φ)} χ_{1}(a_{1}) ⋯ χ_{s}(a_{s})) ∑_{m_{1}···m_{s}=C} φ_{1}(m_{1}) ⋯ φ_{s}(m_{s}),

here φ runs over all ordered collections of even characters with the product χ_{0}, and χ runs over the set P (φ) of all collections which may be obtained by a permutation of φ.

A similar equality holds, of course, for T_{b}(C), and thus

(6)  ∑_{φ}^{∗∗∗}u(φ) ∑_{m_{1}···m_{s}=C} φ_{1}(m_{1}) ⋯ φ_{s}(m_{s}) = 0,

where we set

u(φ) = σ(φ) ( ∑_{χ∈P (φ)} χ_{1}(a_{1}) ⋯ χ_{s}(a_{s}) − ∑_{χ∈P (φ)} χ_{1}(b_{1}) ⋯ χ_{s}(b_{s})) .

The next step is to show that the internal sums in (6), being considered as functions of C, are linearly independent, and so u(φ) = 0.

We first derive from (6) that for every r ≥ 1 and every system of residues n_{1}, …, n_{r} ∈ 𝔽_{p}^{×} holds

(7)  ∑_{φ}^{∗∗∗}u(φ) \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{r}=1 \end{matrix} φ_{i_{1}}(n_{1}) ⋯ φ_{i_{r}}(n_{r}) = 0,

here and below indices of dashed sums (∑′) run over pairwise distinct values; for example, in the latter sum (i_{1}, …, i_{r}) runs over all r-element subsets of the set {1, …, s}.

To obtain (7) we use induction on r. For r = 1 the result follows immediately if in (6) we choose C = q_{1}, where q_{1} ≡ n_{1} (mod p) is prime. For r ≥ 2 we choose pairwise distinct primes q_{i} under q_{i} ≡ n_{i} (mod p) (i = 1, …, r) and let C = q_{1} ⋯ q_{r}. Next, let Ω run over all partitions of the set {1, …, r}. Denote by ∑^{(Ω)} the sum over all those i_{1}, …, i_{r} for which i_{ν} = i_{μ} if and only if ν and μ fall into the same class under partition Ω. We have:

∑_{φ}^{∗∗∗}u(φ) ∑_{m_{1}···m_{s}=C} φ_{1}(m_{1}) ⋯ φ_{s}(m_{s}) =

(8)  = ∑_{φ}^{∗∗∗}u(φ) ∑_{i_{1},...,i_{r}=1}^{s} φ_{i_{1}}(q_{1}) ⋯ φ_{i_{r}}(q_{r}) =

= ∑_{Ω} ∑_{φ}^{∗∗∗}u(φ) \begin{matrix}s \\ ∑ & (Ω) \\ i_{1},...,i_{r}=1 \end{matrix} φ_{i_{1}}(n_{1}) ⋯ φ_{i_{r}}(n_{r}).

Now, if Ω is the partition to r sets of one element in each set, then the internal sum in the latter equality equals

\begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{r}=1 \end{matrix} φ_{i_{1}}(n_{1}) ⋯ φ_{i_{r}}(n_{r});

and if Ω = {S_{1}, …, S_{t}}, where S_{1} ∪ … ∪ S_{t} = {1, …, r}, t < r then this internal sum equals

\begin{matrix}s \\ ∑ & ′ \\ j_{1},...,j_{t}=1 \end{matrix} φ_{j_{1}}(N_{1}) … φ_{j_{s}}(N_{t}),

here N_{ν} = ∏_{i∈S_{ν}} n_{i}. Since by the induction hypothesis

∑_{φ}^{∗∗∗}u(φ) \begin{matrix}s \\ ∑ & ′ \\ j_{1},...,j_{t}=1 \end{matrix} φ_{j_{1}}(N_{1}) ⋯ φ_{j_{t}}(N_{t}) = 0,

then (7) follows from (8) and (6).

Specifically, (7) for r = s yields

(9)  ∑_{φ}^{∗∗∗}u(φ) \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{s}=1 \end{matrix} φ_{i_{1}}(n_{1}) ⋯ φ_{i_{s}}(n_{s}) = 0,

here (i_{1}, …, i_{s}) runs over all permutations of the set {1, …, s}.

Let χ be a fixed collection of even characters with χ_{1} ⋯ χ_{s} = χ_{0}. Multiplying (9) by \overline{χ}_{1}(n_{1}) ⋯ \overline{χ}_{s}(n_{s}) and summing up over all n ∈ (𝔽_{p}^{×})^{s} we obtain

∑_{φ}^{∗∗∗}u(φ) \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{s}=1 \end{matrix} (∑_{n_{1}=1}^{p−1}(\overline{χ}_{1}φ_{i_{1}})(n_{1})) ⋯ (∑_{n_{s}=1}^{p−1}(\overline{χ}_{s}φ_{i_{s}})(n_{s})) = 0,

thus u(χ) = 0 by the orthogonality relation, since there exists precisely one φ and precisely one permutation (i_{1}, …, i_{s}) with

χ_{1} = φ_{i_{1}}, …, χ_{s} = φ_{i_{s}}.

This shows that

(10)  ∑_{χ∈P (φ)} χ_{1}(a_{1}) ⋯ χ_{s}(a_{s}) = ∑_{χ∈P (φ)} χ_{1}(b_{1}) ⋯ χ_{s}(b_{s})

provided that φ_{i} are even and φ_{1} ⋯ φ_{s} = χ_{0}.

Furthermore, it follows from (10) that

(11)  \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{s}=1 \end{matrix} φ_{i_{1}}(a_{1}) ⋯ φ_{i_{s}}(a_{s}) = \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{s}=1 \end{matrix} φ_{i_{1}}(b_{1}) ⋯ φ_{i_{s}}(b_{s}).

Indeed, if {φ_{1}, …, φ_{s}} break into t groups with l_{1}, …, l_{t} coinciding characters in each group and distinct characters in distinct groups, then (11) is obtained from (10) by multiplying by l_{1}! ⋯ l_{t}!. Next, (11) may be rewritten as

(12)  \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{s}=1 \end{matrix} φ_{1}(a_{i_{1}}) ⋯ φ_{s}(a_{i_{s}}) = \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{s}=1 \end{matrix} φ_{1}(b_{i_{1}}) ⋯ φ_{s}(b_{i_{s}}).

If now 1 ≤ r ≤ s and φ_{i} are even with φ_{1} ⋯ φ_{r} = χ_{0}, we define φ_{r+1} = ⋯ = φ_{s} = χ_{0} and get

(13)  \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{r}=1 \end{matrix} φ_{1}(a_{i_{1}}) ⋯ φ_{r}(a_{i_{r}}) = \begin{matrix}s \\ ∑ & ′ \\ i_{1},...,i_{r}=1 \end{matrix} φ_{1}(b_{i_{1}}) ⋯ φ_{r}(b_{i_{r}}).

Prove that the latter equality remains valid if we extend the summation over all collections i_{1}, …, i_{r}, probably with coinciding i_{ν} and even with r > s. Using the above introduced notation, write

∑_{i_{1},...,i_{r}=1}^{s} φ_{1}(a_{i_{1}}) ⋯ φ_{r}(a_{i_{r}}) = ∑_{Ω} \begin{matrix}s \\ ∑ & (Ω) \\ i_{1},...,i_{r}=1 \end{matrix} φ_{1}(a_{i_{1}}) ⋯ φ_{r}(a_{i_{r}}).

For a fixed Ω = {S_{1}, …, S_{t}} the internal sum in the right-hand side equals

\begin{matrix}s \\ ∑ & ′ \\ j_{1},...,j_{t}=1 \end{matrix} Φ_{1}(a_{j_{1}}) ⋯ Φ_{t}(a_{j_{t}}),

DISTRIBUTION OF LATTICE POINTS 11

where Φ_{ν} = ∏_{i∈S_{ν}} φ_{i}, and by (13) this does not change if we replace a by b (if t > s we can not use (13) but then this last sum obviously equals 0).

Our next observation is that

∑_{i_{1},...,i_{r}=1}^{s} φ_{1}(a_{i_{1}}) ⋯ φ_{r}(a_{i_{r}}) = ∏_{j=1}^{r}(φ_{j}(a_{1}) + ⋯ + φ_{j}(a_{s}))

and hence we have proved that

(14)  ∏_{j=1}^{r}(φ_{j}(a_{1}) + ⋯ + φ_{j}(a_{s})) = ∏_{j=1}^{r}(φ_{j}(b_{1}) + ⋯ + φ_{j}(b_{s}))

as soon as φ_{1}, …, φ_{r} are even characters with φ_{1} ⋯ φ_{r} = χ_{0}.

We now use Lemma 3. For X_{0} we choose the set of all those characters φ under φ(b_{1}) + ⋯ + φ(b_{s}) ≠ 0, and define f : X_{0} → ℂ by

f(φ) = \frac{φ(a_{1}) + ⋯ + φ(a_{s})}{φ(b_{1}) + ⋯ + φ(b_{s})}.

Then (14) guarantees fulfilling of the conditions of Lemma 3, and using also Corollary 1(iii) we conclude that there exists λ ∈ 𝔽_{p}^{×} such that

(15)  φ(a_{1}) + ⋯ + φ(a_{s}) = (φ(b_{1}) + ⋯ + φ(b_{s})) φ(λ)

for every φ ∈ X_{0}. But if in (14) we choose r = 2 and φ_{1} = φ, φ_{2} = \overline{φ} where φ is even but φ(b_{1}) + ⋯ + φ(b_{s}) = 0, we see that also φ(a_{1}) + ⋯ + φ(a_{s}) = 0 and thus (15) holds for all even φ, regardless of whether φ ∈ X_{0} or not. By multiplicativity of φ, we can write (15) in the form

(16)  ∑_{i=1}^{s} φ(a_{i}) = ∑_{i=1}^{s} φ(λb_{i}).

Multiplying (16) by \overline{φ}(n) for n ∈ 𝔽_{p}^{×} and summing up over all even characters φ, we obtain

∑_{i=1}^{s} (∑_{φ} φ(a_{i}n′)) = ∑_{i=1}^{s} (∑_{φ} φ(λb_{i}n′)) ,

thus using Corollary 1(ii) we conclude that a and λb have the same number of coordinates equal ±n. And this shows that a may be obtained from b by multiplication by λ, reordering the coordinates and changing signs of some of them.

This completes the proof.

12 VSEVOLOD F. LEV

5. Acknowledgment

I am grateful to Prof. N.M. Korobov who drew my attention to this problem and suggested the use of the Dirichlet theorem for solving it.

School of Math. Sciences, Tel-Aviv University, 69978 Tel-Aviv, Israel

E-mail address: seva@@math.tau.ac.il