SUR LES FORMULES EXPLICITES DE LA THÉORIE DES
NOMBRES PREMIERS
ANDRE WEIL (1952)
In analytic number theory we come upon several formulas that rely on both the sum
over all the non-trivial zeros of the zeta function and a sum over prime numbers.
It is not dicult to extend these formulas, often called explicit formulas, to much
more general cases and it is truly a simple exercise to do so.
I would like to recognize Marcel Riesz...
Let us recall the denition of Hecke's L-functions, mit Grossencharakteren. Let
k
be a number eld of degree
k,
then
kv
d
over the rational numbers. If
shall designate the completion of
k
kp , kp∗ in place of
units in the
p
of
k
eld
kp = kv .
Let
p
and we may use
kv , kv∗ ; and in this case we write
p-adic
Up
vr (1 ≤ r ≤ r1 )
eld and
ki
An idele of
k
with respect to
vλ
for (1
ηr = 1
k is an element a = (av ) of the group
and
kc∗
v
is discrete then
v;
and similarly
vi (r1 + 1 ≤ i ≤ r1 + r2 )
k.
Let
where
kr
kλ designate
the
will denote a real
ηi = 2.
and
Y
If
for the (compact) group of
≤ λ ≤ r1 + r2 )
a complex completion. We set
kv .
instead of
designate real and imaginary archimedian valuations of
completion of
is a valuation of
with respect to the absolute value
∗
given by v . Let kv be the multiplicative group of units in
it corresponds to a prime ideal
v
such that
ap ∈ Up
for almost
c
all
p
(that is to say, for all but nitely many
in the natural way shall be designated by
ideles (the image of
k∗
in
Ik ).
equivalently, a character of
let
m(p)
Ik
Let
Ik ;
The group of ideles, topologized
and let
Ck = Ik /Pk .
be the smallest whole number such that
nite and is in fact
0
Pk
Let
which is trivial on
χ(1, 1, ...a, ...) = 1)and ap 6≡ 1 mod pm .
Date
p).
Pk .
χ
If
be the group of principal
Ck ,
or
is a prime ideal of
k,
be a character of
p
χ(a) = 1
for all
: November 2012.
1
p.
Therefore,
(i.e.
χ , m(p) is always
Q m(p)
f= p
is an ideal
By of the continuity of
for all but nitely many
a ∈ Up
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
of
k
f
and
conductor
is called the
χ.
of
Y
Furthermore, on the subgroup
2
kλ∗ ⊂ Ik
,
λ
χ
i.e. the product of the archimedian completions,
r1Y
+r2
χ(a1 , ...., ar1 +r2 ) =
(0.1)
λ=1
aλ
|aλ |
takes on the form:
−fλ
|aλ |iηλ ϕλ
where each fr is either 0 or 1, fi is a whole number and
Now, let
of
kp
a = (av )
n(p) = vp (ap )) and
Q
p. Therefore pn(p) is an
(I suppose
almost all
be designated by
λ
be an idele. For each
and
ap = 1
av = 1
(a).
for every prime divisor
p
of
f
ap
determines a principal ideal
n(p)
integral or a fractional ideal in
k
p
for every prime divisor
,
f, χ(a) = 1
f
of
then
if
(a) = 1
. Therefore, if
χ(a)
and
aλ = 1
on ideals
a
of
k
is 0 for
which will
aλ = 1
for every
depends only on the ideal
χ(a) = χ(a)
pn(p)
for all
λ
and
a = (a);
that are prime to
We set,
X χ(a)
Y
χ(p)
L(s) =
=
1−
N (a)s
N (p)s
a
p
(0.2)
where the sum is taken over all integral ideals of
f,
is a real number.
by the denition of the ideles,
By the denition of
we have therefore dened a function
f.
p
ϕλ
k
that are prime to the conductor
and the product is taken over prime ideals that don't divide f. If
character which takes on the value
ζk (s)
of the eld
1
everywhere, then
L(s)
χ
is the trivial
is the zeta function
k.
From equation 0.2 we get:
0
L /L(s) = −
(0.3)
∞
XX
log(N p)χ(p)n (N p)−ns
p n=1
If we let
s = σ+it , the series and products above converge absolutely and uniformly
in the half plane
and
L0 /L(s)
For
a ∈ Ik
σ ≥ 1 + a for a > 0; therefore in the same half-plane, L(s), L(s)−1 ,
are bounded.
and
a = (a),
χ1 (a) = ||a||iτ χ(a)
is
let
||a|| =
Y
|aλ |ηλ N (a)−1 .
λ
is also a character of
L1 (s) = L(s + iτ ).
We say that
Ck
χ, χ1
Because
||a|| = 1
for
a ∈ Pk ,
and the L-function associated to
are
associates.
χ1
Among all characters
associated to a given character, there is one for which the exponents
ϕλ
given in
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
0.1 satisfy
P
ηλ ϕλ = 0.
3
Without loss of generality, we may therefore assume that
this condition is satised.
Let
∆
be the discriminant of
k
and let
A = (2π)−d |∆|N f.
SR
(1 ≤ λ ≤ r1 + r2 )
Y ηλ sλ + |fλ | G(s) = (2r1 A)s/2 Γ
2
λ
Y ηλ (1 − sλ ) + |fλ | 1−2
G1 (s) = G(1 − s̄) = (2r1 A) 2
Γ
2
sλ = s + iϕλ
(0.4)
λ
where
·
designates complex conjugation. Then,
−1
η(1−s)+|f|
Γ
Gη,f (s) = (2/η)s Γ ηs+|f|
2
2
X 1
−
+ iϕr
Y
2
1
G(s) = G(s)/G1 (s) = 2 r
Gηλ ,fλ (sλ )
As− 2
(0.5)
λ
By the well-know property of the
strip
σ0 ≤ σ ≤ σ1
N = d(σ1 −
Now let
for
Γ-function, |G(s)|−1
c > πd/4,
is
O(ec|t| )
and, in the same strip,
1
2 ) if we exclude a neighborhood of the poles of
Λ(s) = G(s)L(s).
χ
is
O(|t|N )
where
G(s).
A fundamental theorem of Hecke says that
meromorphic function on the whole plane with poles at
has no poles if
|G|
uniformly in the
s = 0, 1
if
χ
Λ(s)
is a
is trivial and
is not trivial and (in both cases) it satises a functional equation:
Λ(1 − s̄) = κΛ(s)
(0.6)
κ
|κ| = 1.
Λ(s)
as a
denite integral which shows simultaneously the functional equation and that
Λ(s)
where
is a constant such that
is bounded in the strip
is trivial.
Let
σ0 ≤ σ ≤ σ1
σ0 ≤ σ ≤ σ1
L1 (s) = s(s − 1)L(s)
we have that
of equation 0.2,
This is shown by expressing
excluding neighborhoods of the poles if
and let
|L1 (s)| ≤ Cec|t|
L(s) is bounded on σ = σ1
σ0 ≤ 0 < 1 ≤ σ1 .
where
and
c
and
C
χ
Then inside of
are constants. Because
L(1 − s̄) is bounded on σ = σ0 .
We
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
4
have that
|L1 (s)| = |s(s − 1)L(s)| = |s(s − 1)G(1 − s̄)L(1 − s)|
and it follows that
|L1 (s)|
is
O(|t|N )
on
σ = σ0
and on
large. From the classical theorems we conclude that
σ0 ≤ σ ≤ σ1 , |t| > 1,
where
D
σ = σ1
if
where
a>0
the smaller circles is
the strip
|L1 (s)| ≤ D|t|
|m| ≥ 2,
a
Tm
is
is a constant. We arrive at this by applying, for
O(log|t|).
Then, by 0.2 and 0.6, all the zeros of
L(s)
1+
O(e|s|
since
and there is therefore a constant
in the interval
m < Tm < m + 1
)
such that
α
with center
L1 (s)
Λ(s)
in
are in
that satisfy
and for each integer
Λ(s)
has no zero in the
is
1+
O(e|s|
)
for all
σ ≥ 1 + a,
is bounded in
−a ≤ σ ≤ 1 + a
for
in the half-plane
a>0
σ ≥ 1,
it follows that
σ ≤ −a
and
|Λ(s)|
is
. Finally,
(away from the poles) if follows that
is an entire function of order 1, where we set
δχ = 1
if
χ
is trivial
otherwise. We have therefore,
Λ(s) = aebs [s(s − 1)]−δχ
where
>0
in this half-plane and due to 0.6, also in the half-plane
Λ(s)
0
|G(s)|
is bounded in the half-plane
[s(s − 1)]δχ Λ(s)
a, b
Y
s
(1 − )es/ω
ω
ω
are constant and the product is taken over the zeros
with their multiplicities.
Λ0 /Λ(s) − Λ0 /Λ(s0 ) =
ω
of
Λ(s)
counted
2
Now we have therefore for some
X
ω
1
Λ(s)
so it follows that the number of zeros of
O(logT )
On the other hand,
and
r, R
|t − Tm | ≤ α/ log |m|.
band
since
on the strip
is xed and one sees that the number of zeros of
0 ≤ σ ≤ 11and
T ≤ |t| ≤ T + 1
is suciently
N
example theorem D of DP, p. 49 at two circles of constant radius
1 + a + it
N
s, s0 :
1
1
−
s − ω s0 − ω
− δχ
1
1
1
1
−
−
−
s s − 1 s0
s0 − 1
The classical reasoning of Hadamard applied to the work of Landau (Vorlesungen uber Zahlenthevol. II, p.14-15) also shows that Λ(s) has no zeroes to the right of σ = 1, and consequentially
not on σ = 0. This reasoning doesn't happen automatically if χ is a real, nontrivial, character
but in this case L(s) is the quotient of the zeta function of a quadratic extension of k by the zeta
function of k and therefore the result is still valid.
2
If s = 0 is a zero of Λ, the corresponding factor should be replaced by s and this won't have any
eect of what follows. This circumstance cannot come up in any case see footnote 1.
orie,
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
where, in particular, for
s = 1 − s¯0 ,
0
Re [Λ /Λ(s0 )] =
X
5
taking 0.6 into account,
Re
ω
1
s0 − ω
− δχ Re
1
1
+
s0
s0 − 1
s0 = 1 + a + it, 0 ≤ a ≤ 1, this gives
X
1
1
a
−
δ
|Λ0 /Λ(s0 )| ≥ Re |Λ0 /Λ(s0 )| ≥
+
χ
(a + 1)2 + (t − γ)2
a+1 a
ω
where
Re
designates the real part. For
where we've set
this line,
ω = β + iγ .
G0 /G(s)
is
But
O(log|t|),
L0 /L(s)
is bounded to the right of
and therefore
σ = 1 + a;
|Λ0 /Λ(s0 )| ≤ A0 log |t| + A1
on
and this
gives us
X
ω
where
1
(a + 1)2 X
a
≤
≤ A00 log |t| + A01
2 + (t − γ)2
a2 + (t − γ)2
a2
(a
+
1)
ω
A0 , A1, A00 , A01
a
depend on
s = σ + it, −a ≤ σ ≤ 1 + a
t = Tm ,
and
t.
but not on
On the other hand, if we take
|t − γ| ≥ α log |m|
where
, elementary
reasoning (DP th. 26, p 71-72) shows that,
√
|Λ0 /Λ(s) − Λ0 /Λ(s0 )| ≤
X
2
a
2δχ
+ 2
(a + 1 − σ) log |m|
2 + (t − γ)2
α
(a
+
1)
a
+ t2
ω
and because of the preceding discussion, the rst term is
B0 , B1 do
not depend on
an integer
a.
Since
|Λ0 /Λ(s0 )|
m, |m| ≥ 2, s = σ + iTm
and
is
≤ B0 (log |m|)2 + B1
O(log |m|),
where
we have nally that for
−a ≤ σ ≤ 1, 0 ≤ a ≤ 1
2
|Λ0 /Λ(s)| ≤ B (log |m|)
(0.7)
This discussion leads to the following result. Let
F (x) is a complex-valued function
dened on the real line which possesses a Mellin Transform
ˆ
∞
Φ(s) =
3
F (x)e(s− 2 )x dx
1
−∞
which is holomorphic in a band
that there exists an
assures that
3
Φ(s)
a0 > 0
−a ≤ σ ≤ 1 + a
such that
F (x)e
. More precisely we can suppose
( 21 +a0 )|x|
exists and is holomorphic in
is an element of
−a0 ≤ σ ≤ 1 + a0
L1 .
This
and we suppose
To conform with standard usage we should say that Φ(s) is actually the transformation of
v −1/2 F (log v) dened for v > 0.
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
a > 0 such that a ≤ 1 and a < a0
further that there exists an
uniformly in the strip
−a ≤ σ ≤ a + 1.
and
T > 2, T 0 > 2,
Taking
6
Φ(s) is o(log |t|)−2
there will be integers
l, m such that |T − Tm | < 1,| − T 0 − Tl | < 1.The number of zeros ω = β + iγ
of
Λ(s)
whose imaginary parts γ falls between T and Tm is O(log T ) and therefore the sum
X
Φ(ω) taken over these zeros tends to zero when T is increased indenitely and the
ω
same goes for
−T 0
and
Tl .
We will therefore consider the integral of
on the contour of the rectangle formed from the lines
Φ,
By virtue of 0.7 and the hypothesis on
tends to zero with
1/T
and the side
Φ(s)d log Λ(s)
σ = −a, σ = 1, t = Tm , t = Tl .
the integral taken on the side
t = Tl
tends to zero with
1/T 0 .
t = Tm
Taking into
account 0.6 we get
X
(0.8)
Φ(ω) = δχ [Φ(0) − Φ(1)] + I(Tm , Tl )
mod o(1)
−T 0 ≤γ≤T
where
o(1) designates the additive group of functions on T, T 0
1/T, 1/T 0
and
1
I(t, t ) =
2πi
ˆ
1+a+it
0
Since
L0 /L(s)
Φ(s)d log Λ(s) − Φ(1 − s̄)d log Λ(s)
1+a+it0
σ = 1 + a,
bounded on the line
tends toward 0 with
1/T
I(−T , Tl )
with
0
1/T ,
replacing
pears as a sum of an integral analogue
0
with
the function
(0.9)
I(T, −T 0 ).
and
0
and
I(T, −T 0 )
I(Tm, Tl )
which go to zero with
G(s).
In
G0 /G(s)
O(log|t|), I(T, Tm )
whereby in 0.8 we can replace
Λ(s)
I0 (T, −T )
is
by
G(s)L(s); I(T, −T 0 )
taken on
L(s)
and
0
ap-
I1 (T, −T )
of
Applying 0.3 we obtain the equation
−1
I0 (T, −T ) =
2π
ˆ
T
0
dt
−T 0
Xˆ
p,n
∞
Hp,n (u)eitu du
−∞
where
Hp,n (u) =
and
i
log N p h
n
n
( 21 +a)u
−n
n −( 12 +a)u
χ(p)
F
(u
+
log
N
p
)en
+
χ(p)
F
(u
−
log
N
p
)e
N pn/2
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
ˆ
ˆ
1+1+iT
2πiI1 (T, −T 0 ) =
G0 /G(s)
Φ(s)d log G1 (s).
−a−iT 0
has no pole in the half-plane
these functions are also
poles for some
σ0 , σ1 .
−a+iT
Φ(s)d log G(s) −
1+a−iT 0
But
7
O(log |t|)
σ>0
σ0 ≤ σ ≤ σ1
in
and therefore not in
σ<1
and
outside of a neighborhood of the
It follows that
I1 (T, −T 0 ) ≡
1
2πi
ˆ
1
2 +iT
Φ(s)d log G(s)
mod o(1)
1
0
2 −iT
By applying the same reasoning again we get
I1 (T, −T 0 ) ≡ J0 (T, −T 0 ) +
rX
1 +r2
Jλ (T, −T 0 )
mod o(1)
λ=1
where
J0 (T, −T 0 ) =
and
1
Jλ (T, −T ) =
2πi
ˆ
0
log A
2πi
ˆ
1
2 +iT
Φ(s)ds
1
0
2 −iT
1
2 +iT
1
0
2 −iT
Φ(s − iϕλ )d log Gηλ ,fλ (s)
We have therefore
(0.10)
X
Φ(ω) ≡ δχ [Φ(0) − Φ(1)] + I0 (T, −T 0 ) + J0 (T, −T 0 ) +
−T 0 <γ<T
rX
1 +r2
Jλ (T, −T 0 )
λ=1
F (x)e−iϕλ x is Φ( 21 + it − iϕλ ) so we
ˆ ∞
Jλ (T, −T 0 ) =
F (x)eiϕλ x Hηλ, fλ (x; T, −T 0 )dx
Now, the Fourier transform of
−∞
where
Hη, f (x; T, −T 0 ) =
1
2πi
ˆ
T
−T 0
1
eixt d log Gη,f ( + it)
2
can write
mod o(1)
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
8
Now we let
1
Hη,f (x) =
2πi
ˆ
∞
e
ixt
−∞
2
d log Gη,f
η
1
1
+ it − G2,0
+ it
2
2
and because of the known identities and values, we see that the dierence between
the integral in the second term and the same integral taken with the limits
T
≤ 2(e
has absolute value
for
η=2
and
Cf
−πT
depends on
+e
f
−πT
0
) for η = 1,f = 0 or 1 and ≤ Cf (T
but not on
T
T 0.
or
−T 0
−1
+T
and
0−1
)
Furthermore because of known
formulas we have
H1,f (x) =
(−1)1−f
,
+ e−x/2
ex/2
H2,f (x) =
1 − e−|f x|/2
|ex/2 + e−x/2 |
It follows that
Jλ (T, −T 0 ) ≡
ηλ
2
ˆ
∞
F (x)e−iϕλ x Hηλ fλ (x)dx+
−∞
ηλ
2
ˆ
∞
F (x)e−iϕλ x H2,0 (x; T ; −T 0 )dx
−∞
In order to go further we will place more precise hypotheses on
we will assume that
(1)
F (x)
F (x)
F (x).
Henceforth
satises the following conditions
is continuous and continuously dierentiable everywhere except at a
nite number of points
ai ,
at which
F (x)
and its derivative
most a discontinuity of the rst kind and also that
F 0 (x)
have at
F (ai ) = 21 [F (ai + 0) +
F (ai ) − 0)].
(2) There exists a
These imply that for
uniformly in
takes
1
b > 0 such that F (x) and F 0 (x) are O e−( 2 +b)|x| as x → ∞.
0
1
0 < a0 < b, F (x)e( 2 +a )
−a ≤ σ ≤ 1 + a0 .
0 < a < a0 < b, a ≤ 1.
is
L1
and that
Φ(s)
is
O(|t|−1 )
The preceding results are valid provided that one
For a convenient choice of constant
|F (x)| ≤ Ce−( 2 +b)|x|
1
C,
we have
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
If we let
9
δ =b−a:
|Hp,n (u)|
e−δ|u|
≤
+ N pnb eδ|u| inf(N p−n(1+2b) , e−(1+2b)|u| ) ≤ 2N p−n(1+a)
C log N p
N pn(1+b)
from which we get that
ˆ
∞
|Hp,n (u)| du ≤ 2C
∞
It follows that the series
and denes an
1
L
H(u) =
P
log N p
N pn(1+a)
Hp,n (u) is absolutely and uniformly convergent
function and0.9 can be written
I0 (T, −T 0 ) =
−1
2π
ˆ
ˆ
T
∞
H(u)eitu du
dt
−T 0
∞
H(u)
is continuous and continuously dierentiable away from the
ai ± log N pn
where it and its derivative have at most discontinuities of the
Furthermore,
points
1
1
+
δ
1 + 2a + δ
rst kind and its value is the average of the left and right limits.
conditions, the Fourier inversion formula applies, that is to say that
−H(0)
as
Under these
I0
tends to
T → ∞.
In order to obtain the common limit of the two terms in 0.10 for
have to evaluate the limit of
Jλ (T, −T 0 )
in the case where
T = T 0 → ∞ we'll
ηλ = 0, fλ = 0.
That is,
we'll have to evaluate the limit of
1
2πi
as
ˆ
T
Ψ(t)d log G2,0
−T
1
+ it
2
=
1
π
ˆ
T
−T
1
+ it dt
Ψ(t)< Γ0 /Γ
2
T → ∞, where Ψ(t) is the Fourier transform of some function F1 (x) = F (x)e−iϕδ x
which satises conditions 1 and 2 earlier.
But the function
as
t → ∞.
< Γ0 /Γ
1
2
+ it
is an even function as has the form
log |t|+O(t−2 )
log |t| is a distribution which was determined by
+ it is therefore a distribution which diers from
The Fourier transform of
0
4
L. Schwartz That of < Γ /Γ
1
2
the previous one only by a continuous function. It turns out to be the distribution
−πP F (|ex/2 − e−x/2 |−1 ), where the symbol P F
4
will be dened as follows. Let
α(x)
L. Schwartz, Theorie des Distributions, vol. II (Act. Sc et Ind. no 1122, Paris, Hermann et cie,
1951, (formule (VII, 7; 18)
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
be an
L1
function on
(−∞, 1)
(1, ∞)
and on
β(x) = |x|α(x)
such that
10
satises
condition 1 above. We let,
ˆ
∞
ˆ
λ→+∞
−∞
But since
∞
α(x)dx = lim
PF
α(x)
is such that
1 − e−λ|x| α(x)dx − 2β(0) log λ .
−∞
|x|α(x)
satises condition 1, and since if
ϕ(x)
is con-
tinuous and continuously dierentiable on a compact support the formula
ˆ
∞
P F α(ϕ) = P F
α(x)ϕ(x)dx
−∞
denes a distribution
0
P F αequal
as is the same as Schwartz's
α
to the function
Pf
on every interval not containing
distribution except for a couple of Dirac's
tributions. Thus stated, it follows from the theorem's of Schwartz that if
Ψ(t)
δ
dis-
F1 (x)and
are as above, we have that
ˆ
T
lim
T →∞
−T
ˆ ∞
1
F1 (x)dx
Ψ(t)< Γ0 /Γ
+ it dt = −πP F
;
x/2 − e−x/2 |
2
|e
−∞
and we can also verify this directly, as follows. Since
1
0
< Γ /Γ
+ it
2
diers from
log |t|
by an
L2
function, then if we use Plancherel's theorem then it
suces to check the formula for
´T
−T
Ψ(t) log |t|dt.
Without changing anything, we could replace
F1 (−x)
Ψ(t), F1 (x)
by
Ψ(t) + Ψ(−t), F1 (x) +
or we could just assume that those functions are even; verifying the result
directly for
F1 (0) = 0
F1
equal to 1 for
|x| < 1
and 0 otherwise, we reduce to the case where
and therefore, taking account property 1 with
F2 (x) = |x|F1 (x), F2
is
an even function with a nite number of discontinuities (of the rst kind) and is
O(e−|x|/2 )
at innity. We have demonstrated the following:
ˆ
∞
F2 (x)dx = −
0
2
lim
π T →∞
ˆ
T
∞
ˆ
log t
0
F2 (x) cos(tx)dx dt
0
or, after integrating by parts,
ˆ
0
∞
2
lim
F2 (x)dx =
π T →∞
ˆ
0
T
ˆ
0
∞
F2 (x) sin(tx)
dtdx
t
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
11
The double integral is absolutely convergent, so we may write
ˆ
ˆ
∞
xT
F2 (x)
0
0
!
ˆ ∞
ˆ
ˆ ∞
sin t
sin t
π ∞
F2 (x)dx −
F2 (x)
dt dx =
dt dx
t
2 0
t
xT
0
and it is immediate that the last term tends toward 0 as
T →∞
which proves the
result. We have shown the following:
If F (x) satises conditions 1, 2, the sum
Φ(ω) taken over the zeros ω = β + iγ
X
ω
of L(s) that satisfy 0 ≤ β ≤ 1, |γ| < T tends toward a limit as T → ∞ and this
limit has the value
ˆ
lim
T →∞
X
F (x) ex/2 − e−x/2 dx + F (0) log A
−∞
|γ|<T
−
(0.11)
∞
Φ(ω) = δχ
−
∞
XX
log N p χ(p)n F (log N pn ) + χ(p)−n F (log N p−n )
n/2
Np
p n=1
rX
1 +r2
ˆ
∞
F (x)e−iϕλ x Kηλ ,fλ (x)dx
PF
−∞
λ=1
Where we set
K1,f (x) =
e( 21 + f)
e−f|x|/2
,
K
=
2,f
ex/2 − e−x/2 |ex − e−x |
This is the most general form of the explicit formula. Naturally, we can loosen
the restrictions somewhat on
F.
We will apply these results to a version of the Riemann Hypothesis. For this, we'll
need the following lemma:
In order for L(s) to satisfy the Riemann Hypothesis, it is necessary and sucient
that the common values of the two sides in 0.11are positive for every function F of
the form
ˆ
∞
F (x) = F0 (x) ∗ F0 (−x) =
F0 (x + t)F0 (t)dt
−∞
where F0 is a function satisfying 1&2 above.
F
is then continuous and is the anti-derivative of a function
F 0 which
is continuous
everywhere except for a nite number of points where it can have discontinuities of
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
the rst kind (these are the points
of
F0 ).
F
Now,
satises condition
F
the Mellin Transform of
ω
L(s)
of
2
and if
ω0 = β 0 + γ0
F
and
with
Φ.
β0 6=
ai
if the
Φ0
are the points of discontinuity
is the Mellin Transform of
Φ(s) = Φ0 (s)Φ0 (1 − s̄).
is
in the critical strip are on
this choice of
Therefore, if all the zeros
σ =
L(s)
has a zero in the strip
1
− iγ0 ), Φ0 (s) = Ψ0 (z), Φ(s) = Ψ(s)
2
Ψ(z) = Ψ0 (z)Ψ(z̄) and that F0 (x)eiγ0 x , F (x)eiγ0 x are the Fourier transforms
Ψ0 (z), Ψ(z)
transform of all functions of the form
P (x)e
on the real axis (?). Since the Fourier-
−Ax2
where
P
is a polynomial and
Ψ0 (z) = P (z)e−Az
is a function of the same form, it follows that if we take
will satisfy conditions 1 and 2. For one such choice of
the same form, and therefore
P
then
1
2 , the rst term in 0.11 is positive for
Suppose to the contrary that
of the functions induced from
F0 (x)
F0
1
2 . Let
z = i(s −
so that
ai − aj
12
Φ(s)
will be
Φ(ω) will be absolutely convergent.
0 2
O(e−A t )
for all
Ψ0 (z), Ψ(z)
A0 < A
A>0
2
then
will be of
and therefore
In order to prove the lemma it will suce to
P (z) and A. Indeed, for
− iγ0 , and, in particular,
show that this sum is negative for a convenient choice of
every zero
ω
L(s) in the critical band, let η = i ω − 21
− iγ0 = i(β0 − 12 ). Let Q(z) be the polynomial
of
η 0 = i ω0 −
1
2
zeros all the distinct
the
η
take
η
other than
η0
and
η¯0
that satisfy
are real or come in conjugate pairs, so
P (z) = zQ(z)Q(−z).
that for
X
Now, if
m
Q
having as simple
|<(η)| ≤ 2.
Now due to 0.6
actually has real coecients. We
is the order of
ω0
as a zero of
L(s),
we get
A > 1:
2
X
2
1 2
1
4
Φ(ω) = − 2m β0 −
|Q(η0 )| e2A(β0 − 2 ) +
P (η)2 e−2Aη
2
|<η|>2
≤ −2m β0 −
1
2
2
2
|Q(η0 )| e2A(β0 − 2 ) + e−3A
1
4
X
2
|P (η)2 e−η |
|<η|>2
and it's clear that the last term is negative for
A
suciently large.
We will now use the lemma to give a necessary and sucient condition for
functions constructed over any eld
to do this let
of
k,
let
kv∗
Ck = Ik/ Pk
k
L-
to satisfy the Riemann Hypothesis. In order
be the group of idele classes of
k.
For every valuation
v
be identied with the groups of ideles whose components are all equal
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
1 except for at at the place
an isomorphism on
the function
Ck
kv∗
v.
we identify
||a|| on Ik ,
Ik
Since the canonical homomorphism from
kv∗
Ck .
also with its image in
and since it is equal to
1 on Pk
n(x)
x ∈ kp∗
is dened for
More generally, if
|| · ||
ϕ
induces
|·|
on
kρ∗
and
| · |2
Ck
to
is
Earlier, we dened
it is dened on the quotient
and it is a homomorphism to the positive reals. The kernel of this map
compact. The function
13
on
ki∗
and
N p−n(x)
Ck0
is
kp∗
if
G
to
on
(x) = pn(x) .
by
is a homomorphism with a compact kernel from a group
G
the positive reals then we can normalize the Haar measure on
that the measure of a compact set of
G
determined by
by the condition
1 ≤ ϕ(ξ) ≤ M
is
log M. Ifϕ
maps to a discrete subgroup of the reals then we can normalize the Haar measure
1 ≤ ϕ(ξ) ≤ M
by the condition that on a compact set (open or closed) dened by
is
log M + O(1)
M → ∞.
as
In the last case if the image is generated by
measure of
kerϕ
G
normalized by ϕ.
by
is called
|| · ||
and
will be
d× xas
log α.
α>1
the
In either case the Haar measure thus determined on
We write
dξ
the Haar measure on
Ck
as the Haar measure on
kv∗
normalized by||
· ||
normalized
also. This notation
is to remind you that the measure is invariant under (multiplicative) translation
by
kv∗ and
+
d x/|x|
that
Up
in fact, up to a constant it is equal to
and on an imaginary eld it is
d× x = drdθ/πr.
of units is
aλ = a1 > 0
this group in
If we let
x = reiθ
Ck .
Ik
Then
γk
Let
γk
zero on almost all
x)
Γ.
for all
γk .
Furthermore, the characters
χ
Let
F0 (χ, x)
p
be the image of
is isomorphic to the group of positive reals and
ηλ ϕλ = 0 are those that take the value 1 on γk
the discrete topology on
ap = 1
formed by the ideles such that
Ck
of
Ck
is
that satisfy
whereby the group
Γof characters
Ck0 ,
and so we put
is isomorphic to the group of characters on the compact group
all
then we have
we have that the measure of the compact group
for all the others (archimedian ones).
0
the direct product of Ck and
P
d x/2|x|.
On a real eld it is
log N p.
Let's consider the subgroup of
and
kp∗
On a eld
+
d+ x/||x||.
be a function dened on
Γ×R
which is
χ (that is to say that for all but nitely many χ, F0 (χ, x) = 0 for
and such that for each
χ,F0 (χ, x)
satises conditions 1. and 2. For
we let
Ω(ξ) =
X
χ∈Γ
F0 (χ, − log ||ξ||)χ(ξ)
ξ ∈ Ck
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
and for each
14
χ∈Γ
ˆ
∞
F (χ, x) =
F0 (χ, x + t)F0 (χ, t)dt
−∞
Under these conditions we have that
X
P (ξ) = Ω(ξ) ∗ Ω(ξ −1 ) =
F (χ, − log ||ξ||)χ(ξ)
χ∈Γ
where the
∗
function on
designates the product in
Ck .
Upon using
F (χ, x)
Ck
dξ , P (ξ)
, with the measure
as
F (x)
is a positive
we see the second term of0.11 is
positive. After some easy calculation we arrive at
ˆ
X P (x) − P (1)θv (||x||)
1
1
P (ξ)(||ξ|| 2 + ||ξ||− 2 )dξ −
D(P ) = ρP (1) +
1
Ck
v
with
ρ = log |∆| + log(2π)
−d
−
rX
1 +r2
ˆ
θλ (e−x )Kηλ,0 (x)dx
−∞
v, θp (t) = 0
for all
d× x ≥ 0
∞
PF
λ=1
θv (1) = 1
where we take
||x − 1|| · ||x||− 2
for
t 6= 1
for all
p
θλ
and
are taken
so that the above equation is meaningful. For example they could be the constant
function
θλ (t) = 1
for all t. The value of
The earlier lemma implies that
D(P )
D(P ) ≥ 0
is independent of this choice.
for all functions
before is a necessary and sucient condition for the
Riemann Hypothesis. Now
D
condition can be expressed as
P
as a function of
D
P
obtained as stated
L function over k
to satisfy the
is a distribution on
Ck
and this
is of positive type. It hardly need be said that
the Riemann Hypothesis is no more easily solved in this form than in the classical
form. On the other hand, these ideas make clear the analogy between number elds
and function elds. If
eld of
q
k
is an algebraic function eld of dimension 1, over a nite
elements we can still dene the group
Ck
the idele classes of
k
(which here
is totally disconnected and is the projective limit of discrete groups) and similarly
kv∗ relative
the sub-groups
For an idele
to the idele
respect to
a,
a.
||a||
we let
to the valuations
||a|| = q − deg(a)
v
where
of
a
k
(which here is totally discrete).
is the divisor naturally associated
As in the earlier discussion, we normalize the Haar measure with
on
Ck
and on
manner as above, this time
kv∗ .
We can dene a distribution
ρ = (2g − 2) log q
where
g
D
on
Ck
in the same
is the genus and we take
SUR LES FORMULES EXPLICITES DE LA THÉORIE DES NOMBRES PREMIERS
15
θv (1) = 1 and θv (t) = 0 otherwise for all v. These calculations are much easier than
earlier, and they show the accuracy of the Riemann Hypothesis for all
constructed on
k
is equivalent to the inequality
D(P ) ≥ 0
L-functions
for all functions
P (ξ) = Ω(ξ) ∗ Ω(ξ −1 )
on
Ck
where
Ω
is a function on
Ck
cosets (?) of an open subgroup of
which is compactly supported and constant on
Ck .
Of course, in the case of function elds, the
Riemann Hypothesis is already solved and therefore the distribution
D
attached to
such a eld is of positive type.
Translated by NAVA BALSAM (WITHOUT PERMISSION)