p-adic Gauss integrals from the Poison summarizing formula

0.12 MB·English

by  D. V. Prokhorenko

p-adic Gauss integrals from the Poison summarizing
formula
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1 D.V. Prokhorenko
0 ∗
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n January 5, 2011
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J

4

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Abstract
p
-
h
t In the present paper we show how to obtain the well-known formula for Gauss
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m
sums and the Gauss reciprocity low from the Poison summarizing formula by using
[

some ideas of renormalization and ergodic theories. We also apply our method to
1
v
obtain new simple derivation of the standard formula for p-adic Gauss integrals.
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∗Institute of Spectroscopy, Russian Academy of Sciences, 142190 Moskow Region, Troitsk,
[email protected]
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1 Notations
Let S(R) be a Schwartz space on real line (the space of all smooth functions decaying at
infinity faster than each inverse polynomial with all their derivatives). Let f(x) S(R) be
∈
some function from the Schwartz space. The Fourier transform of f(x) is defined by the
following formula:
fˆ(k) = (2π)−12 f(x)eikxdx (1)
Z
The inverse Fourier transform is defined as follows
fˇ(x) = (2π)−21 f(k)e−ikxdk (2)
Z
It is easy to prove that the Fourier transform ant its inverse map the Schwartz space into
the Schwartz space. The Fourier theorem states that f(x) S(R)
∀ ∈
fˇˆ(x) = fˆˇ(x) = f(x). (3)
Let f(x) S(R). We have
∈
2π(m+1/2)
fˆ(k) = (2π)−21 f(x)eikxdx = (2π)−21 f(x)eikxdx
Xk∈Z Xk∈ZZ Xk∈Z mX∈Z2π(mZ−1/2)
2π(m+1/2)
= (2π)−21 f(x)eikxdx = (2π)21 f(2πm). (4)
mX∈Z Xk∈Z2π(mZ−1/2) mX∈Z
Therefore we have the following Poison summarizing formula
fˆ(k) = (2π)21 f(2πm). (5)
k∈Z m∈Z
X X
The rigorous proof of this formula see for example in [1, 2].
Let p be an integer primer number. The Gauss sums are defined as follows:
p−1
q q
G.S.( ) := exp(2πi k2). (6)
p p
k=0
X
Let us introduce the Legendre symbols (p). By definition
q
q
( ) = 1 if integer x : q = x(mod p) (7)
p ∃
2
and
q
( ) = 1 if integer x : q = x(mod p) (8)
p − ∀ 6
The integer numbers q such that (q) = 1 are called quadratic residues, and the integer
p
numbers such that (q) = 1 are called quadratic nonresidues. Note that
p −
p−1
exp(2πik) = 0 (9)
k=0
X
Note also that the number of quadratic residues is equal to the number of quadratic non-
residues. Therefore, we have
q q 1
G.S.( ) = ( )G.S.( ). (10)
p p p
Let α C and Rα > 0. The Gauss integral, by definition, is the following integral
∈
π
e−αx2dx = , (11)
α
Z r
where we take such branch of square root that it > 0 if α real and α > 0.
One can proof (by using (11) that
e[−αx2(k) = 1 e−14kα2. (12)
√2α
2 The Gauss sums and the Gauss reciprocity low
Instead of calculation
p−1
1 1
G.S.( ) := exp(2πi k2) (13)
p p
k=0
X
we will calculate
1 2πik2
G.S.ε( ) := exp( εk2),
p p −
k∈Z
X
ε > 0 (14)
Let us clear the question how G.S.ε(1) connects with G.S.(1).
p p
For small ε > 0 we can regard the function e−εx2 as constant function at each interval
[mp,(m + 1)p] for each integer m. So, the part of G.S.ε(1) calculated for k = mp, mp +
p
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1,...(m+1)p 1canbeapproximatedbye−ε(mp)2G.S.(1). Obtainedsumoverpointsmp, m
− p ∈
Z we can approximate by integral. The mistake will be of order O(1), ε 0. Therefore
→
1 1
G.S.ε( ) = G.S.( ) exp( εx2)+O(1)
p p −
x∈pZ
X
+∞
1 1
= G.S.( ) exp( εx)dx+O()
p p −
−Z∞
1 1 π
= G.S.( ) +O(1), ε 0. (15)
p p ε →
r
Therefore we have the following representation
1 1 1 π
G.S.ε( ) = G.S.( ) +O(1), ε 0. (16)
p p p ε →
r
From other hand, let us denote by f(x) the following function:
x
f(x) := exp(πi εx). (17)
p −
Let us calculate the Fourier transform of f(x). We have
1 k
fˆ(k) = exp( )
2(ε 2πi) −(ε πi)
− p − p
q 1 ki pε
exp( (+ )) (18)
≈ 4πi −(π) πi
p
− p
q
In result, we have
p ki pε
fˆ(k) eiπ4exp( (+ )). (19)
≈ 4π −(π) πi
r p
Using Poison summarizing formula we find:
1
G.S.ε( ) = (2π)21 fˆ(2πk)
p
k∈Z
X
+∞
p 4 πik  kpε
1 iπ
= (2π)2 e4 exp( p) exp( )+O()
4π −  ×  − 
r Xk=1 −Z∞
1ei4π p 4π1 4 πik
= (2π)2 exp( p)+O(), ε . (20)
8 π ε p −  →
r r k=1
X
In result we have:
1 ei4π 1 π 1 4 πik
G.S.ε( ) = ( )2 exp( p)+O(), ε . (21)
p 2√2√p ε −  →
k=1
X
4
So we need to calculate the sum in right hand side of last equality. Squares of numbers
1, 2, 3, 4 are 0, 1, 0, 1 modulo 4 respectively. So
4 πik
−iπ
exp( p) = √e  , if p = (mod),
− 
k=1
X
4 πik
iπ
exp( p) = √e , if p = (mod), (22)
− 
k=1
X
Therefore from (16) and (21) we have
q q
G.S.( ) = √p( ), if p = 1(mod),
p p
q q
G.S.( ) = i√p( ), if p = 3(mod), (23)
p p
that has to be proven.
Let us now use this method to prove the Gauss reciprocity low. Put by definition
q πiqx
G.S.ε( ) := exp( εx) (24)
p p −
x∈Z
X
and
πiq
fq(x) := exp( x εx) (25)
p p −
By the same method as we prove (16) it is easy to prove that
q q 1 π
G.S.ε( ) = G.S.( ) +O(1), ε 0. (26)
p p p ε →
r
It is easy also to calculate that
p ikp pε
fˆq(k) ei4πexp( (+ )) (27)
p ≈ 4πq −(πq) iπq
r
Let us now apply to (24) the method used for derivation (21). We obtain
q
G.S.ε( ) = (2π)12 fˆq(2πk)
p p
k∈Z
X
+∞
= (2π)21 p eiπ4 1 exp( kpε)dp
4πq 4 − 
r −Z∞
4q
πi p
exp( k )+O(), ε . (28)
× −  q →
k=1
X
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In result
q eiπ4 1 π 1 4q πi p
G.S.ε( ) = ( )2 exp( k )+O(), ε . (29)
p 2√2√pq ε −  q →
k=1
X
So we need to calculate:
4q
πi p
Σ := exp( k ). (30)
−  q
k=1
X
Let a and b be integer reciprocity primer numbers. There exists a ring homomorphism
Z Z Z (31)
ab a b
→ ×
which assign to each element x(modab) Z the element (x(moda),x(modb)) Z Z
ab a b
∈ ∈ ×
x(modab) (x(moda),x(modb)) (32)
7→
Well known theorem (Chinese remainder theorem) states that the homomorphism just de-
scribed is an isomorphism.
a and b are reciprocity primer numbers. So there exists such integers r and s that
ra+sb = 1. Therefore if (x,y) Z Z then its inverse image under the homomorphism
a b
∈ ×
just defined is
sbx+ray. (33)
If the element α Z is a square (i.e. α = β2 for some integer β) then its image (x,y) is
ab
∈
also a square i.e. there exists integers (z,w) such that x = z2,y = w2. It follows from this
remark that if q = 2 then
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4q
πi p
Σ = exp( k )
−  q
k=1
X
4 q
πi πi
= exp( spn) ( rpm), (34)
−  − q
n=1 m=
X X
where s and r are integer satisfying to
4r +qs = 1. (35)
Let us calculate r and s. Consider the case q = 1(mod), i.e. q = 4f +1 for some integer f.
In this case s = 1 and r = f = −1+q(modq) = (modq). Consider the case q = 4f 1
− − 4  −
i.e. q = 1(mod). In this case s = 1 and r = f = 1+q(modq) = (modq).
− − 4 
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Let us calculate Σ for the case q = 4f +1 for some integer f. In this case
Σ = 2√2e−iπ4( 1)q−21G.S.(p) if p = (mod)
− q
Σ = 2√2eiπ4( 1)q−21G.S.(p) if p = (mod). (36)
− q
Consider now the case q = 4f 1 for some integer f. In this case we have
−
Σ = 2√2eiπ4( 1)q−21G.S.(p) if p = (mod)
− q
Σ = 2√2e−iπ4( 1)q−21G.S.(p) if p = (mod). (37)
− q
In result (26) and (29) implies
q 1 p 1
G.S.( ) = G.S.( ) α(p,q). (38)
p √p q √q
q−1p−1
Here by definition α(p,q) = ( 1) 2 2 β(p,q), where
−
β(p,q) = 1 if p = (mod), and q =  (mod)
β(p,q) = i if p = (mod), and q =  (mod)
−
β(p,q) = i if p = (mod), and q =  (mod)
β(p,q) = 1 if p = (mod), and q =  (mod) (39)
And by using the expression of Gauss sums trough the Legendre symbols we obtain
q p p−1q−1
( )( ) = ( 1) 2 2 , (40)
p q −
what has to be proven. The equality (40) is named the Gauss reciprocity low.
3 p-adic Gauss integrals
Let p — be an integer primer number. Let Q be a field of rational number. Each r Q :
∈
r = 0 can be represented as follows:
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a
r = pγ, (41)
b
where γ Z, a,b Z, a = 0(mod p), a = 0(mod p). The field Q of p-adic numbers by
p
∈ ∈ 6 6
definition is a completion of Q with respect to the following norm:
a
a = pγ = p−γ. (42)
| |p |b |p
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It is easy to prove that satisfy to the following ultrametric property:
p
|·|
a+b max( a , b ). (43)
p p p
| | ≤ | | | |
Each a Q can be represented as follows:
p
∈
a = p−γ(a +a p+a p2 +...), (44)
0 1 2
where γ Z, a ,a ,a 0,1,...,p 1 and the series converges with respect to the p-adic
0 1 2
∈ ∈ { − }
norm .
p
|·|
Let a = p−γ(a +a p+a p2+...) Q , γ Z, a ,a ,a 0,1,...,p 1 . By definition
0 1 2 p 0 1 2
∈ ∈ ∈ { − }
the fractional part a of a is defined as follows:
{ }
a = p−γ(a +a p+...+a pγ−1) if γ > ,
0 1 γ−1
{ }
a = 0 if γ . (45)
{ } ≤
For each x Q put by definition:
p
∈
χ (x) = exp(πi x ). (46)
p
{ }
Let B = x Q x 1 be an unit ball in Q with the center at zero. Note that
0 p p p
{ ∈ || | ≤ }
the ultrametric condition (43) implies that each point of unit ball B is its center. B is a
0 0
compact grope with respect to ”+” as a group operation. So there exists a Haar measure
dµ at Q which we will normalize such that µ(B ) = 1.
p 0
More about p-adic analysis see in [3].
LetR+ = x R x 0 . LetΩ(x) beafunctionΩ : R+ Rsuch thatΩ(x) = 1 ifx 
{ ∈ | ≥ } → ≤
and Ω(x) = 0 otherwise.
Theorem. Let p be a primer integer number such that p = 2. Then
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I := χ (ax2 +bx)dµ(x) = Ω( b ), if a 
p p p
| | | | ≤
Z
B0
−1 b2 b
I = χ (ax2 +bx)dµ(x) = λ (a) a 2χ ( )Ω( ) if a > , (47)
p p | |p p −4a |a|p | |p
Z
B0
where
λ (a) = 1 if a = pN, N even,
p p
| | −
a
λ (a) = ( 0) if N odd and p =  (mod ),
p p −
a
λ (a) = i( 0) if N odd and p =  (mod ) (48)
p
p −
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Proof.
a) Let a 1. In this case
p
| | ≤
I = χ (ax2 +bx)dµ(x) = χ (bx)dµ(x) = Ω( b ). (49)
p p p
| |
Z Z
B0 B0
b) Now let a > 1. If b > 1 then
p
| | a p
(cid:12) (cid:12)
(cid:12) (cid:12)
I = χp(ax2 +bx)d(cid:12)µ(x)(cid:12) = χp b2 χp a x2 + bx+ b2 dµ(x)
−4a a 4a
BZ0 (cid:16) (cid:17)BZ0 (cid:16) (cid:16) (cid:17) (cid:17)
2
χp b2 χp a x+ b dµ(x) (50)
−2a 4a
(cid:16) (cid:17)BZ0 (cid:16) (cid:16) (cid:17) (cid:17)
Put
b
c := = p−N(c +c p+...), (51)
2a 0 1
and
a = p−γ(a +a p+...), (52)
0 1
and
x = x +x +... . (53)
0 1
Let us find the coefficient at x in bx. It is evident, that this coefficient is equal to
N+γ−1
2a c
0 0
(54)
p
and
2a c
0 0 = 0. (55)
{ p } 6
But the term ax2 could be represented as follows
ax2 = integer part+terms which do not depend of x (56)
N+γ−
So the integral (50) is equal to zero.
b) So consider the case a > 1, b 1. But each point of unit ball is its center. So,
p
| | a p ≤
using substitution x x b we ob(cid:12)tain(cid:12):
7→ − 4a (cid:12)(cid:12) (cid:12)(cid:12)
I = χ b2 χ (ax2)dµ(x) (57)
p p
−4a
(cid:16) (cid:17)BZ0
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Let us represent a as follows a = 1 (a + a p + ...). At first consider the case when a is
pN 0 1 0
odd. We can represent a as follows
q
a = d2, (58)
pN
where q is a primer number p = q and d is some p-adic number such that d = 1. We have
p
6 | |
I = χ b2 χ (ax2)dµ(x)
p p
−4a
(cid:16) (cid:17)BZ0
q
= χ b2 χ ( x2)dµ(x) (59)
p −4a p pN
(cid:16) (cid:17)BZ0
Here we use the substitution x d−1x. We have
→
b2 1
I = χ ( ) J, (60)
p −4a pN
where, by definition
pN
πiqk
J = exp( ). (61)
pN
k=1
X
Let us define
πiq
J := exp x εk , (62)
ε
{ pN − }
k∈Z
X
and
πiq
f := exp x εk , ε >  (63)
ε { pN − }
It is easy to prove
+∞
1
J = J exp( εx)dx+O(), ε . (64)
ε pN − →
−Z∞
Therefore
1
J = J√πε+O(1), ε 0. (65)
ε pN →
From other hand
pN ikpN pNε
fˆ eiπ4exp( (+ )). (66)
ε ≈ s4πq − πq πiq
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