Will a physicist prove the
Riemann Hypothesis?
Marek Wolf
Cardinal Stefan Wyszynski University,
Faculty of Mathematics and Natural Sciences,
Warsaw
Riemann: the greatest
mathematician ever.
1900: Hilbert’s list of problems for the XX
centaury:
the first part of the eighth problem: The
Riemann Hypothesis
2000: Clay Institute: problems for
millennium.
15 years old Gauss conjectured:
π(x) := ]{p − prime, p < x}
Z
π(x) ∼
2
x
du
:= Li(x)
ln(u)
Remark:
Li(x) = γ + ln(ln(x)) +
∞
X
(ln(x))n
n=1
nn!
γ ≈ 0.577216P
is the Euler–Mascheroni
constant
n
1
γ = limn→∞
−
ln(n)
.
k=1 k
Euler’s Gold Formula
−1
∞
X
Y
1
1
=
ζ(s) :=
1− s
s
n
p
p
n=1
This formula is valid only for <[s] > 1.
There are no zeros of ζ(s) for <[s] > 1.
1 is neither prime nor composite.
Euler has derived expressions giving values of
zeta for even arguments (Bn are Bernoulli
numbers):
|B2m |π 2m
ζ(2m) =
2(2m)!
Bm+1
ζ(−m) = −
m+1
R.Apery (1998): ζ(3) is irrational.
W.I. Zudilin (2001): One of the numbers:
ζ(5), ζ(7) ζ(9), ζ(11)
is irrational.
The path leading to the proof of the Gauss
conjecture was outlined by Riemann in the
paper: “Ueber die Anzal der Primzahlen unter
einer gegebenen Grosse”
(Monatsberichte der
Berliner Akademie, November 1859). First
Riemann has continued analytically ζ(s) to
the whole complex plane with exception of
s = 1: at s = 1 ζ(s) has a pole.
Riemann has shown that the integral (s 6= 1):
Z
Γ(−s) +∞ (−x)s dx
ζ(s) =
2πi +∞ e x − 1 x
where the contour:
6
is equal to
<[s] = 1.
P∞
n=1 1/n
s
--
on the right of the line
Trivial zeros: s = −2, −4, −6, . . . i.e.
ζ(−2n) = 0. Besides that there exist
infinity of zeros ρ = σ + it in the
critical strip 0 ≤ <[ρ] = σ ≤ 1. If ρ
is zero, then also ρ and 1 − ρ are
zeros. Zeros are located symmetrically
around the critical line <[s] = 12 .
The Riemann hypothesis
All non-trivial zeros are lying on the
critical line
1
s = + it
2
i.e. are of the form ρ = 21 + iγ
von Mangoldt(1905):
7
T
T
N(T ) = 2π
ln 2πe
+ 8 + O(ln(T ))
Hardy (1914) There is infinitely many zeros of ζ(s)
on the critical line.
ζ(ρi ) = 0
1
2
3
4
5
6
7
8
9
10
11
12
13
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
+ i14.134725142
+ i21.022039639
+ i25.010857580
+ i30.424876126
+ i32.935061588
+ i37.586178159
+ i40.918719012
+ i43.327073281
+ i48.005150881
+ i49.773832478
+ i52.970321478
+ i56.446247697
+ i59.347044003
i = 1, 2, . . . , 26
14
15
16
17
18
19
20
21
22
23
24
25
26
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
+ i60.831778525
+ i65.112544048
+ i67.079810529
+ i69.546401711
+ i72.067157674
+ i75.704690699
+ i77.144840069
+ i79.337375020
+ i82.910380854
+ i84.735492981
+ i87.425274613
+ i88.809111208
+ i92.491899271
Next Riemann obtained the exact formula
for π(x). Let
R(x) =
∞
X
µ(m)
m=1
m
Li(x 1/m )
where µ(n) is the Möbius function:

for n = 1
1
µ(n) =
0
when p2|n

(−1)r when n = p1p2 . . . pr
Then
π(x) = R(x) −
X
R(x ρ)
(∗)
ρ
where the sum runs over zeros of ζ(s), i.e.
ζ(ρ) = 0.
π(x) ≈ Li(x)
Littlewood has proved that Li(x) − π(x)
changes infinitely many times sign. Here is
the computer illustration of the formula (*):
π(x) = R(x) −
X
R(x ρ)
(∗)
ρ
gdzie suma przebiega po wszystkich zerach
ζ(s), tzn. ζ(ρ) = 0. Pochodna tego wzoru
to suma δ zlokalizowanych na liczbach
pierwszych.
Wząr (*) moňna sprawdziŁ na komputerze
π(x) ≈ Li(x)
√
π(x) = Li(x)+O(ln(x) x)
Z grubsza n-ta liczba pierwsza i Li−1(n)
maję 12 tych samych cyfr.
x
102
103
104
105
106
107
108
109
1010
1011
1012
1013
1014
π(x)
Li(x)
Li(x) − π(x)
25
30
5
168
178
10
1229
1246
17
9592
9630
38
78498
78628
130
664579
664918
339
5761455
5762209
754
50847534
50849235
1701
455052511
455055615
3104
4118054813
4118066401
11588
37607912018
37607950281
38263
346065536839 346065645810
108971
3204941750802 3204942065692
314890
x
1015
1016
1017
1018
1019
1020
1021
1022
1023
1024
1025
π(x)
Li(x) − π(x)
29844570422669
1052619
279238341033925
3214632
2623557157654233
7956589
24739954287740860
21949555
234057667276344607
99877775
2220819602560918840
222744647
21127269486018731928
597394250
201467286689315906290
1932355520
1925320391606803968923
7250186752
18435599767349200867866 17146872832
176846309399143769411680 55160980939
Jacques Salomon Hadamard (1865 – 1963)
and Charles-Jean -tienne Gustave Nicolas,
Baron de la VallÚe Poussin (1866 – 1962)
proved in 1896 the PNT: there are no zeros
of ζ(s) on the line 1 + it.
J.P. Gram(1903): 15 zeros are on the critical line
..
.
A.Turing (1953): 1104 zeros are on the critical line
(“in an optimistic hope that a zero would be found off critical line”)
..
.
D.H. Lehmer (1956): 25000 zeros are on the critical
line
..
.
S. Wedeniwski: 2.5 × 1011 zeros are on the critical
line: s = 12 + it |t| < 29, 538, 618, 432.236
X. Gourdon(2004): 1013 zeros are on the critical line.
How to prove the
Riemann Hypothesis?
Mertens hypothesis:
M(x) =
√
X
µ(n)
n<x
If |M(x)| < x then Riemann hypothesis is
fulfilled.
Littlewood: RH is equivalent to the:
M(x) = O(x 1/2)
A.M. Odłyżko and H.J.J. te Riele in 1985
have disproved Mertens hypothesis. They
have calculated first 2000 zeros of ζ(s) with
accuracy 100-105 digits – It took 40 hours
on the CDC CYBER 750 + 10 hours on the
Cray-1.
|!
RH is equivalent to
1√
x ln(x) for all x ≥ 2657.
8π
Criteria for RH involving integrals
|π(x)−Li(x)| ≤
RH is true ⇔
Z ∞
Z ∞
1 − 12t 2
3−γ
ln(|ζ(σ
+
it)|dσdt
=
π
.
2 )3 1
(1
+
4t
32
0
2
Z
RH is true ⇔
<(s)= 12
ln(|ζ(s)|)
|ds| = 0.
|s|2
Lagarias (2000): An Elementary
Problem Equivalent to the Riemann
Hypothesis: RH is equivalent to the
inequalities:
X
σ(n) ≡
d ≤ Hn + exp(Hn ) log(Hn )
d|n
for each n = 1, 2, . . .. Here HnPare
n-th harmonic numbers Hn = nj=1 1j .
The de Bruijn-Newmana constant (1950,
1976):
1 2 1
z 1
1
− z2 − 41
+
ξ(iz) =
z −
π
Γ
ζ z+
2
4
2 4
2
RH ⇔ all zeros of ξ(iz) are real.
Φ(t) =
∞
X
(2π 2n4e 9t − 3πn2e 5t )e −πn
2 e 4t
n=1
t ≥ 0. Then ξ is a Fourier transform of Φ:
Z ∞
1 z ξ
=
Φ(t) cos(zt)dt
8 2
0
Larger class: H(z, λ) is the Fourier
λt 2
transform of Φ(t)e and
H(z, 0) = 18 ξ( 12 z) N. G. De Bruijn
proved that (1950):
1. H(z, λ) has only real zeros for
λ ≥ 12
2. If H(z, λ) has only real zeros for
some λ0, then H(z, λ) has only real
zeros for each λ0 > λ.
Ch. Newman (1976) has proved that there
exists such parameter λ1, that H(z, λ1) has
at least one non-real zero. Thus there exists
such constant Λ in the interval
−∞ < Λ < 21 , that H(z, λ) has real zeros
⇔ λ > Λ. Riemann Hypothesis is equivalent
to Λ ≤ 0.
Newman believes Λ ≥ 0.
Csordas et al (360 digits) (1988) − 50 < Λ
...
te Riele (250 digitsr) (1991)
...
−5 < Λ
Odyzko (2000) − 2.7 · 10−9 < Λ
k = 1020 + 71810732, γk+1 − γk < 0.000145
“ if RH is true, it is barely true”
Maybe RH is
undecidable
(P. Cohen, Fields medallist 1966)
|!
B. van der Pol
Bulletin of the AMS 53 (1947), pp. 976-981
ζ( 12 + it)
=
1
+
it
2
Z
∞
e
−x/2
x
be c − e
x/2
e −ixt dx
−∞
paper, scissors, rotor, source of light,
photocell, sinusoidal current of variable
frequency.
The Poly’a– Hilbert
Conjecture: “ζ( 12 + i Ĥ) = 0”
RH is true, because complex parts of
the non-trivial zeros correspond to
eigenvalues of the positive self-adjoint
operator. Was proposed around 1910,
first published in 1973.
letter of G. Polya (1887-1985) to
Odlyzko from January 1982:
Montgomery (1973):Assume RH: ρ = 21 + iγ.
2!
Z β
X
sinπu
du
1=
1−
πu
α
0
0<γ,γ ≤T
2πα ≤γ−γ 0 ≤ 2πβ
ln T
ln T
F. Dyson recognized in the above formula
the same dependence as in the behavior of
the differences between pairs of eigenvalues
of random Hermitian matrices.
A. Odlyzko performed computer experiments
The 1020-th zero of the Riemann zeta
function and 70 million of its neighbors,
thousands of hours on Cray-1 i Cray X-MP
The 1020-th zero of the Riemann zeta
function and 175 million of its neighbors,
1992 revision of 1989
The 1021-st zero of the Riemann zeta
function,
The 1022-nd zero of the Riemann zeta
function, (∼ 109 zeros)
/(2π))
Odlyzko (1987): δn = (γn+1 − γn ) ln(γn2π
1
N
X
1≤n≤N,
k≥0
δn +δn+1 +...+δn+k ∈[α,β]
Z
1∼
β
1−
α
sinπu
πu
2!
du
The results confirmed the GUE (Gaussian
unitary ensemble) distribution: the gaps
between imaginary parts of consecutive
nontrivial zeros of ζ(s) display the same
behavior as the differences between pairs of
eigenvalues of random Hermitian matrices.
P. Sarnak wrote: “At the phenomenological
level this is perhaps the most striking
discovery about the zeta function since
Riemann.”
1
Ĥ = (xp + px)
2
I
I
I
Ĥ has the classical counterpart
describing a chaotic, unstable and
bounded dynamics.
The dynamics of the Riemann does
not have reversal time symmetry.
And the Riemann dynamics is
one-dimensional
The number of levels of Ĥ < E :
7
E
E
ln
− 1 + +. . .
N(E ) =
2π
2π
8
T
T
7
N(T ) =
ln
+ +O(ln(T ))
2π
2πe
8
ψE (x) ∼
const
ζ
1/2−iE
|x|
1
− iE
2
M. Shlesinger has investigated a very special
one-dimensional random walk which can be
linked with the RH. The probability of
jumping to other sites with steps having a
displacement of ±l sites involves directly the
Möbius function:
1
1
µ(l)
p(±l) = C
±
, β > 0,
1+β
1+β−
2
l
l
where C =
factor.
1
1
ζ(1+β)+ ζ(1+β)
is a normalization
Some general properties of the "structure function"
λ(k) being
Fourier of the probabilities p(l):
P the
ikl
λ(k) = l e p(l), enabled Shlesinger to locate the
complex zeros inside the critical strip, however the
result of J. Hadamard and Ch. J. de la
Vallée–Poussin that ζ(1 + it) 6= 0 can not be
recovered by this method. What is interesting the
existence of off critical line zeros is not in
contradiction with behavior of λ(k) following from
the laws of probability.
Dynamics of point particle bouncing
inside the circular billiard
T (β, ψ) = (β + π − 2ψ, ψ)
Survival probability P(t) that the particle will
not escape till time t:
∞
1 X
n(φ(n) − µ(n))
tP(t) =
8π n=1
2π
× g
− θ0 − + g (θ0 − )
n
φ(n) Euler’s totient function: the number of
positive integers m ≤ n that are relatively
prime to n: gcd(m, n) = 1
2
x x >0
g (x) =
0 x ≤0
Then they proved that RH is
equivalent to
lim lim δ (tP1(t) − 2/) = 0
→0 t→∞
be true for every δ > −1/2.
The functional equation can be written in
non–symmetrical form:
π
2Γ(s) cos s ζ(s) = (2π)s ζ(1 − s).
2
The Kramers–Wannier duality relation for the
partition function Z (J) of the two dimensional Ising
model with parameter J expressed in units of kT
Z (J) = 2N (cosh(J))2N (tanh(J))N Z (e
J),
where N denotes the number of spins and e
J is related
−2e
J
to J via e
= tanh(J),
The Lee–Yang circle theorem on the zeros of the
partition function. Let Z (β, z) denote the grand –
canonical partition function. Phase transitions are
connected with the singularities of the derivatives of
Z (β, z), and they appear when Z (β, z) = 0. The
Lee–Yang theorem asserts that in the
thermodynamical limit, when the sum for partition
function involves infinite number of terms, all zeros
of Z (β, z) for a class of spin models are pure
imaginary and lie in the complex plane of the
magnetic field z on the unit circle: |z| = 1.
s = 21 + it can be mapped into the unit circle
s → u = s/(1 − s) = ( 12 + it)/( 12 − it) : |u| = 1.
In 1975 S.M. Voronin proved theorem on the
universality of the Riemann ζ(s) function:
Let 0 < r < 1/4 and f (s) be a complex
function continuous for |s| ≤ r and
analytical in the interior of the disk. If
f (s) 6= 0, then for every > 0 there exists
real number T = T (, f ) such that:
3
< .
+i T
max f (s) − ζ s +
4
|s|≤r
J. Derbyshire “Prime Obsession” (2001) pp. 357-358:
JD: Andrew, you have gazed on more non-trivial
zeros of the Riemann zeta function than any person
alive. What do you think about this darn
Hypothesis? Is it true, or not?
AO: Either it’s true, or else it isn’t.
JD: Oh, come on, Andrew. You must have some
feeling for an answer. Give me a probability. Eighty
percent it’s true, twenty percent it’s false? Or what?
AO: Either it’s true, or else it isn’t.