A NOTE ON CHARACTER SUMS
PÉTER CSIKVÁRI
Abstract. We will investigate certain character sums. We will prove
some discrepancy-type inequalities for incomplete sums.
1. Introduction.
We will investigate incomplete sums, in particular we will estimate incomplete character sums. The main result in this direction is Vinogradov’s
theorem. Throughout this paper we will write e(α) = e2πiα .
Theorem (Vinogradov [3]): Let q, x, y be positive integers, 0 < x < y ≤ q.
Let a1 , a2 , . . . , aq ∈ C and
µ ¶
q
X
jt
aj e
F (t) =
,
q
j=1
P
Further let A = qj=1 aj = F (0). Then
¯ y
¯
q−1
¯X
y − x + 1 ¯¯
1 X |F (l)|
¯
A¯ ≤
an −
¯
¯
¯ 2q
q
|| ql ||
n=x
l=1
where ||x|| = minn∈Z |x − n|.
A consequence of this theorem is the famous Pólya-Vinogradov inequality,
which states that if χ is a non-principal character mod q then for any positive
integer n
¯ n
¯
¯X
¯
√
¯
¯
χ(k)¯ ≪ q log q.
¯
¯
¯
k=1
In this paper we will prove a lower bound for incomplete sums in terms
of the |F (l)|’s and we will apply it for some character sums. In
Pnwhat follows
let χ be a primitive
P character modulo q. Let Sn = Sn (χ) = k=1 χ(k) and
Lm = Lm (χ) = m
n=1 Sn (χ). We will prove the following theorems.
P
Theorem 1.1. Let a1 , a2 , . . . , aq be complex numbers, Ak = kj=1 aj , A =
³ ´
P
Aq . Let F (l) = qj=1 aj e jlq . Then for 1 ≤ l ≤ q − 1 we have
¶ µ ¶
µ
µ ¶¶ X
q µ
j
jl
l
F (l) = 1 − e
Aj − A e
q
q
q
j=1
2000 Mathematics Subject Classification. Primary: 11L07. Secondary: 11L40.
Key words and phrases. Character sums, exponential sums.
1
2
PÉTER CSIKVÁRI
and
1
max
2π 1≤l≤q−1
Ã
|F (l)|
|| ql ||
!
¯
q ¯
X
¯
¯
k
¯Ak − A¯ .
≤
¯
q ¯
k=1
We will also prove a variant of this theorem concerning intervals of given
length.
Theorem 1.2. Let 1 ≤ k ≤ (q − 1)/2 be a fixed positive integer, let
P
a1 , . . . , aq , C be complex numbers and Bi = i+k
j=i+1 aj . Then we have
¯
¯ q
µ ¶¯
q
j ¯ 1X
2 ¯¯X
aj e
|Bi − C| .
¯≤
¯
π¯
q ¯ k
j=1
i=1
We will prove a variant of the Pólya-Vinogradov inequality concerning
Lm (χ).
Theorem 1.3. For every primitive character χ¯ mod q there exist
a constant
¯
¯
Cq m ¯
c and a complex number Cq = Cq (χ̄) for which ¯Lm (χ) + τ (χ̄) ¯ ≤ cq 3/2 for all
m where τ (χ) is the Gaussian sum
µ ¶
q
X
n
.
χ(n)e
τ (χ) =
q
n=1
As an application of Theorem 1.1 we will show
Theorem 1.4. For every primitive character χ mod q there exist an n and
m
1 √
q and |Lm + Cτ (qχ̄)
an m such that 1 ≤ n, m ≤ q and |Sn | ≥ 2π
| ≥ 4π1 2 q 3/2 .
Remark: The first statement of Theorem 1.4 is known [1],[2].
2. Proofs of the theorems
Proof of Theorem 1.1. By partial summation we get:
µ ¶
µ ¶ X
q
q
X
jl
jl
(Aj − Aj−1 )e
aj e
=
=
q
q
j=1
j=1
q
X
µ
¶¶
µ ¶
µ µ ¶
(j + 1)l
l
jl
−e
+ Aq e
=
Aj e
=
q
q
q
j=1
µ
µ ¶¶ X
µ ¶
µ ¶
q
l
l
jl
= 1−e
+ Aq e
=
Aj e
q
q
q
j=1
µ
µ ¶¶ X
¶ µ ¶
q µ
j
jl
l
= 1−e
Aj − A e
q
q
q
j=1
A NOTE ON CHARACTER SUMS
3
µ ¶ µ
µ ¶¶ X
µ ¶
q
l
l
j
jl
+Aq e
+ 1−e
.
Ae
q
q
q
q
j=1
Now we will show that for 1 ≤ l ≤ q − 1 we have
µ ¶¶ X
µ ¶
µ ¶ µ
q
l
jl
j
l
+ 1−e
Ae
= 0.
Aq e
q
q
q
q
j=1
³ ´ 1−e nl ³ ´
³ ´
( )
Indeed, let k0 = 0 and kn = e ql + · · · + e nlq = 1−e ql e ql ; then kq = 0
(q)
since 1 ≤ l ≤ q − 1. Then we have
µ
µ ¶¶ X
µ ¶
q
l
j
jl
1−e
Ae
q
q
q
j=1
Thus
µ
µ ¶¶ X
µ ¶
q
l
jl
A
1−e
je
=
q
q
q
j=1
µ
µ ¶¶ X
µ
µ ¶¶ X
q
q
A
A
l
l
=
1−e
1−e
j(kj − kj−1 ) = −
kj =
q
q
q
q
j=1
j=1
µ ¶
µ ¶¶
µ ¶X
q µ
A
l
jl
l
= −Ae
.
=− e
1−e
q
q j=1
q
q
F (l) =
µ
µ ¶¶ X
¶ µ ¶
q µ
j
l
jl
1−e
.
Aj − A e
q
q
q
j=1
¯
³ ´¯
¯
¯
Since ¯1 − e ql ¯ ≤ 2π|| ql || it follows that
¯
q ¯
1 |F (l)| X ¯¯
j ¯¯
≤
¯Aj − q A¯ .
2π || ql ||
j=1
Proof of Theorem 1.2. The proof is very similar to the previous one:
µ ¶
µ ¶ X
q
q
X
i
i
Bi e
=
=
(Bi − C)e
q
q
i=1
i=1
µ
¶¶
µ µ ¶
q
X
i−k+1
i
+ ··· + e
ai e
=
q
q
i=1
³ ´
k
µ
¶
q
X
i−k+1 e q −1
³ ´
ai e
=
=
1
q
e
−
1
i=1
q
³ ´
k
¶
µ ¶
µ
q
i
−k + 1 1 − e q X
³ ´
.
ai e
=e
q
q
1 − e 1 i=1
q
¤
4
PÉTER CSIKVÁRI
¯
³ ´¯
¯
¯
Since ¯1 − e 1q ¯ ≤
¯
³ ´¯
¯
¯
and ¯1 − e kq ¯ ≥ 4k
we have
q
¯ q
µ ¶¯¯
q
j ¯ 1X
2 ¯¯X
aj e
|Bi − C| .
¯
¯≤
π ¯ j=1
q ¯ k i=1
2π
q
¤
Proof of Theorem 1.3. First we start from the identity
µ ¶
q
1 X
kl
χ(k) =
.
χ̄(l)e
τ (χ̄) l=1
q
Then we have
µ ¶
µ ¶
q
q
n
n
n
X
X
X
1 X
1 X
kl
kl
χ(k) =
=
=
χ̄(l)e
χ̄(l)
e
τ (χ̄) l=1
q
τ (χ̄) l=1
q
k=1
k=1
k=1
³ ´
q
e nlq − 1 µ l ¶
1 X
=
.
χ̄(l) ³ ´
e
τ (χ̄) l=1
q
e l −1
q
It follows that
³ ´
l
¶
¶
m µ µ
e
X
q
nl
1 X
χ̄(l) ³ ´
−1 =
e
χ(k) =
τ
(
χ̄)
q
e ql − 1 n=1
n=1 k=1
l=1
³ ´  ³ ´

ml
l
µ
¶
q
e
−
1
e
q
l
1 X
 ³q´
e
=
− m =
χ̄(l) ³ ´
l
l
τ (χ̄) l=1
q
e q −1 e q −1
³ ´
¶
µ µ ¶
q
e 2lq
1 X
Cq m
ml
=
−1 −
χ̄(l) ³ ³ ´
´2 e
τ (χ̄) l=1
q
τ (χ̄)
e ql − 1
q
m X
n
X
where
q
Cq =
X
l=1
Hence
³ ´
e ql
.
χ̄(l) ³ ´
e ql − 1
¯
¯
q−1
m X
n
¯X
¯
1 3/2
2
C
m
1 X
¯
¯
q
≤
q .
χ(k)
+
≤
¯
¯ √
l
2
¯ n=1
¯
τ
(
χ̄)
q
2
||
16||
q
l=1
k=1
Proposition:
q
X
k=1
kχ(k) =
q
τ (χ̄)
q
X
l=1
³ ´
e ql
.
χ̄(l) ³ ´
e ql − 1
¤
A NOTE ON CHARACTER SUMS
In other words
5
q
τ (χ̄) X
Cq =
kχ(k).
q k=1
Proof. Let us use
P the previous theorem with m = tq where t is a large positive
integer. Since qk=1 χ(k) = 0 we have
¯ ¯ q
¯
¯ tq
n
n
¯ ¯ XX
¯
¯X X
C
tq
C
tq
¯
¯
q ¯
q ¯
χ(k)
+
=
t
χ(k)
+
¯ ¯
¯=
¯
¯ ¯
¯
¯
τ
(
χ̄)
τ
(
χ̄)
n=1 k=1
n=1 k=1
¯ Ã q
!¯ ¯ Ã q
!¯
¯ X
X
Cq q ¯¯
Cq q ¯¯ ¯¯
¯
kχ(k) −
= ¯t
(q − k + 1)χ(k) +
¯ = ¯−t
¯ ≤ cq 3/2 .
¯
τ (χ̄) ¯ ¯
τ (χ̄) ¯
k=1
k=1
Since t can be arbitrarily large we get
q
X
k=1
kχ(k) =
Cq q
.
τ (χ̄)
¤
Remark: We also could have proved this result directly.
If χ(−1) = 1 then one can easily see that Cq = 0.
Proof of Theorem 1.4. We start out from the Gaussian sum τ (χ).
Let us apply Theorem 1.1. Note that Sq = 0, thus
µ
µ ¶¶ X
µ ¶
q
1
n
τ (χ) = 1 − e
Sn e
q
q
n=1
Now we have
¯
µ ¶¯ X
q
X
¯ q
¯
2π
1
¯
|Sn |.
|Sn | ≤
q = |τ (χ)| ≤ ¯¯1 − e
q ¯ n=1
q n=1
1 √
Thus there exists an n for which |Sn | ≥ 2π
q, which proves the first statement. To prove the second statement we apply Theorem 1.1 to the sequence
an = Sn ; in this case Am = Lm and
√
q
q
qX
q Cq q
qX
Cq n
n
(q − k + 1)χ(k) = −
kχ(k) = −
A=
=
q
n k=1
n k=1
n τ (χ̄)
τ (χ̄)
by the Proposititon. Thus we can apply Theorem 1.1 to obtain
µ
µ ¶¶ X
µ ¶
q
1
n
τ (χ) = 1 − e
=
Sn e
q
q
n=1
=
µ
¶ µ ¶
µ ¶¶2 X
q µ
Cq n
n
1
e
.
Ln +
1−e
q
τ
(
χ̄)
q
n=1
6
PÉTER CSIKVÁRI
Hence
√
¯
¯
¯
µ ¶¯2 X
¯
¯ q ¯
¯
C
n
1
q
¯ Ln +
¯
¯≤
q = |τ (χ)| ≤ ¯¯1 − e
¯
¯
¯
q
τ
(
χ̄)
n=1
¯
µ ¶2 X
q ¯
¯
2π
Cq n ¯¯
¯
≤
¯Ln + τ (χ̄) ¯ .
q
n=1
Thus there exists an n for which |Ln +
Cq n
|
τ (χ̄)
≥
1 3/2
q .
4π 2
¤
As a consequence of it we show that for some 1 ≤ m ≤ q we have
|Lm | > cq 3/2 with some positive absolute constant c, which with the PólyaVinogradov theorem shows that at least c logq q n’s of the interval [1, q] we
have |Sn | ≫ q 1/2 .
Proposition: For some 1 ≤ m ≤ q we have |Lm | > cq 3/2 with some positive
absolute constant c.
Proof. We will prove the proposition with c = 8π1 2 .
If χ(−1) = 1 then this is a trivialPconsequence of Theorem 1.4 since in
this case Cq = 0. It is also trivial if | qn=1 Sn (χ)| ≥ 8π1 2 q 3/2 since we choose
P
m = q. Finally if | qn=1 Sn (χ)| ≤ 8π1 2 q 3/2 then by Theorem 1.4 for some
1 ≤ m ≤ q we have
¯ ¯
¯
¯ m
¯ ¯ m
q
q
¯
¯
¯
¯X
¯ ¯X
X
X
m
¯ ¯m
¯
¯
¯ ¯
Sn (χ)¯ − ¯
Sn (χ)¯
Sn (χ)¯ ≥ ¯
Sn (χ) −
¯
¯ ¯ q n=1
¯
¯ n=1
¯ ¯ n=1
q n=1
≥
1 3/2
1
1
q − 2 q 3/2 = 2 q 3/2 .
2
4π
8π
8π
¤
Acknowledgement. The author thanks András Sárközy for his numerous
advice and comments.
References
[1] A. Sárközy:Some remarks concerning irreguralities of distribution of sequences of
integers in arithmetic progressions, IV, Acta Math. Acad. Sci. Hung. 30(1-2) (1977),
pp. 155-162
[2] A. V. Sokolovskii: On a theorem of A. Sárközy, Acta Arithmetica 41 (1982), pp.
27-31 (in Russian).
[3] I. M. Vinogradov: Elements of Number Theory, Problem 12. of Chapter V.
Eötvös Loránd University, Department of Algebra and Number Theory,
H-1117 Budapest, Pázmány Péter sétány 1/C, Hungary
E-mail address: csiki@cs.elte.hu