1.
Theorem.
Define
P
n
(T
)
=
∑
u
= 1
n
(
n
- 1
u
- 1
)
T
u
u
!
,
Φ
0
1
= e
3
2
t
1
,
Φ
n
α
=
∑
j
1
+...
+j
α
= n
P
j
1
(2t
1
)
...
P
j
α
(2t
α
)
e
3
2
(t
1
+...
+t
α
)
for
n
≥
1
,
Ψ
α
=
1
π
d
log
λ
1
q
1
- q
1
λ
1
∧
...
∧
1
π
d
log
λ
α
q
α
- q
α
λ
α
,
Riemann
's
hypothesis
is
equivalent
to
the
upper
bound
lim
sup
n
|
∑
α
= 1
n
<
Φ
n
α
,
Ψ
α
>
<
Φ
0
1
,
Ψ
1
>
α
+1
|
1
n
≤
1
where
λ
is
the
modular
lambda
function,
q
j
= e
i
π
τ
j
τ
j
=
ie
t
j
λ
j
=
λ
(
τ
j
),
and
triangular
brackets
denote
integration
over
the
product
of
the
arcs
where
each
λ
j
ranges
between
the
cusp
at
1
and
the
cusp
at
0.
Proof.
It
is
the
radius
of
convergence
at
s
=
3
2
+
1
+z
1
- z
of
the
Maclaurin
series
of
π
s
8
ζ
(s
)
ζ
(s
- 1
)
(1
- 4
1
- s
)
Γ
(s
)
.
The
unit
disk
in
the
z
plane
maps
to
the
half
- plane
to
the
right
of
3
2
.
Note
that
α
+1
in
the
denominator
can
be
replaced
by
just
α.
The
exponent
is
3
2
instead
of
5
2
since
θ
(ie
t
)
4
dt
=
e
- t
π
d
log
1
-
λ
λ
.
Remarks.
1.
In
this
context,
one
of
Rieman
's
obserations
is
that
if
we
write
λ
'
= 1
-
λ
τ
'
=
- 1
τ
q
'
= e
i
π
τ
'
t
'
=
- t
then
for
any
two
values
a, b
of
λ,
say
with
1
>
a
>
b
>
0
,
1
π
∫
a
b
e
(s
- 1
)t
d
log
λ
q
λ
'
=
1
π
∫
a
'
b
'
e
(s
-
1
)t
'
d
log
λ
'
q
'
λ
.
Since
a
', b
'
are
in
the
reversed
order
it
is
convenient
to
write
it
=
1
π
∫
b
'
a
'
e
(s
- 1
)t
'
d
log
q
'
λ
λ
'
.
We
like
to
have
the
lower
limit
be
the
larger
one.
In
order
to
make
a
well
- defined
one
- form
which
descends
through
the
branching
at
the
point
where
λ
= 1
/2
,
which
maps
to
j
= 1728
,
we
must
choose
one
of
the
two
forms,
we
choose
1
π
d
log
λ
q
λ
'
.
By
symmetry,
if
we
integrate
e
st
- t
times
this
form
along
the
path
where
λ
passes
from
1
to
0
,
then
j
passes
from
infinity
to
1728,
the
j
invariant
of
the
Gaussian
integers,
and
back
again.
We
use
the
same
form
for
both
halves
of
the
arc,
but
reverse
the
sign
of
the
form
on
the
return
journey
to
cancel
the
reversal
of
orientation
only.
This
defines
an
entire
function
h
(s
)
which
satisfies
the
functional
equation
h
(s
)
= h
(2
- s
)
,
and
our
L
series
satisfies
8
ζ
(s
)
ζ
(s
- 1
)
(1
- 4
1
- s
)
=
π
s
Γ
(s
)
(
h
(s
)
+
1
s
- 2
-
1
s
).
Since
also
1
s
- 2
-
1
s
is
symmetric,
the
whole
of
8
π
- s
Γ
(s
)
ζ
(s
)
ζ
(s
- 1
)
(1
- 4
1
- s
)
is
invariant
under
s
↦
2
- s
.
2.
Actually
ζ
(s
)
ζ
(s
- 1
)
satisfies
two
different
functional
equations.
The
one
we
've
seen
gives
invariance
of
ζ
(s
)
ζ
(s
- 1
)
(1
- 4
1
- s
)
Γ
(s
)
π
- s
under
s
↦
2
- s
.
But
also
the
invariance
under
s
↦
1
- s
of
the
of
first
factor
in
ζ
(s
)
Γ
(s
/2
)
π
-
s
2
ζ
(s
- 1
)
Γ
(
s
- 1
2
)
π
s
- 1
2
combined
with
the
invariance
under
interchanging
the
factors,
gives
invariance
of
the
product
also
under
s
↦
2
- s
.
From
one
function
satisfying
two
functional
equations,
we
deduce
that
2
1
- s
Γ
(s
)
Γ
(1
- s
/2
)
Γ
(1
/2
- s
/2
)
must
satisfy
a
functional
equation
of
its
own.
That
is,
the
Gamma
function
satisfies
the
functional
equation
that
2
1
- s
Γ
(s
)
Γ
(1
- s
/2
)
Γ
(1
/2
- s
/2
)
must
be
an
odd
function
of
s
.
This
fact
is
equivalent
to
the
Legendre
duplication
formula.
3.
While
Ψ
α
is
the
wedge
product
of
all
1
i
d
τ
j
+
1
π
d
log
λ
j
1
-
λ
j
,
once
multiplied
out,
all
but
one
of
the
resulting
integrals
is
non
- convergent.
One
can
multiply
by
a
sequence
of
functions
tending
to
1
to
which
a
convergence
theorem
applies,
and
take
the
limit
of
convergent
integrals.
For
example,
multiplying
by
e
-
μ
(t
1
2
+...
+t
α
2
)
,
which
is
a
product
of
functions
of
the
separate
coordinates,
<
Φ
n
α
,
Ψ
α
>
=
lim
μ
→
0
∑
u
+v
=
α
p
+q
= n
(
α
u
)C
q
v
(
μ
)
∑
j
1
+...
+j
u
= p
∏
s
= 1
u
1
π
∫
1
0
P
j
s
(2t
)
e
3
2
t
-
μ
t
2
d
log
λ
1
-
λ
where
C
q
v
(
μ
)
=
∑
j
1
+..
+j
v
= q
∏
s
= 1
v
∫
-
∞
∞
P
j
s
(2t
)
e
5
2
t
-
μ
t
2
dt
.
The
limits
in
the
first
integral
denote
that
the
integral
is
taken
as
λ
ranges
from
1
to
0
.
Thus
<
Φ
n
α
,
Ψ
α
>
is
a
limit
of
polynomials
of
degree
α
in
period
integrals
from
1
to
0
with
coefficients
which
are
known
quadratic
algebraic
functions
of
μ.
4.
Instead
of
looking
at
the
L
series
for
sums
of
4
squares,
which
is
8
ζ
(s
)
ζ
(s
- 1
)
(1
- 4
1
- s
)
we
can
look
at
8
ζ
(s
)
ζ
(s
- 1
)
4
- s
(4
- 2
s
)
(2
- 2
s
)
This
is
the
same
series
but
with
even
terms
given
a
negative
sign.
Amazingly
convenient
that
a
sum
of
four
squares
is
odd
if
and
only
if
the
number
of
odd
summands
is
odd
(meaning
1
or
3
).
This
means
that
if
we
were
to
think
of
θ
(0,
τ
)
4
and
θ
(1
/2,
τ
)
4
as
basic
homogeneous
coordinates
(when
we
mod
the
upper
half
plane
by
τ
↦
τ
+2
and
its
conjugate
under
τ
↦
- 1
/
τ
),
replacing
one
by
the
other,
which
in
terms
of
λ
is
the
linear
fractional
transformation
λ
↦
λ
(
λ
- 1
)
has
the
nice
effect
that
it
sends
our
complicated
d
log
λ
/
(1
-
λ
)
directly
to
d
log
λ
In
other
words,
8
ζ
(s
)
ζ
(s
- 1
)
4
- s
(4
- 2
s
)
(2
- 2
s
)
Γ
(s
)
=
π
s
- 1
∫
0
∞
y
s
- 1
d
log
(
λ
/q
)
In
the
right
side,
that
λ
/q
is
the
familiar
q
series
(with
the
common
divisor
of
q
removed
)
16
-
128q
+
704
q
2
-
3072
q
3
.....
and
here
y
is
just
the
imaginary
part
of
τ,
so
τ
is
going
from
0
to
i
∞
and
y
= i
τ
.
Here
then
we
have
removed
a
little
convergence
problem,
as
λ
passes
nicely
from
1
to
0.
We
can
even
let
Re
(s
)
= 3
/2
.
A
zero
of
Riemann
's
zeta
function
gives
by
translating
one
unit
to
the
right,
a
zero
of
ζ
(s
- 1
)
.
Now
fixing
this
value
of
s
gives
that
the
as
lambda
passes
from
1
to
0
the
integral
of
y
s
- 1
d
log
(16
-
128q
+
704
q
2
-
3072
q
3
.....
)
is
equal
to
0.
That
is
to
say,
this
is
one
of
those
one
- forms
which
corresponds
to
an
un
- covering
now.
An
identification.
5.
I
would
like
to
thank
Robert
McKay
for
suggesting
this
problem.
By
turning
back
to
what
I
think
may
have
been
Riemann
's
original
point
of
view,
techniques
reminiscent
of
wave
equations
might
simplify
the
Merten
approach,
but
if
the
relation
with
cycles
and
moduli
of
abelian
varieties
is
too
distant,
a
new
analytic
expression
may
be
what
first
simplifies
Mertens
'
calculation.
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