Multiplicative functions in function fields
Adam J Harper
21st January 2016
Plan of the talk:
I
The talk reports on joint work with A. Granville and K.
Soundararajan.
I
Review of multiplicative functions over the integers, and in
function fields.
I
Halász-type theorems on mean values, and ideas about the
proofs.
I
(If time permits) Comment on a difference between the
integers and function fields (variation of mean values).
Multiplicative functions, I
Over the integers, a multiplicative function f : N → C is one that
satisfies
f (mn) = f (m)f (n) ∀ (m, n) = 1.
The classical generating function associated with f is the Dirichlet
series
∞
X
f (n)
.
F (s) :=
ns
n=1
If we have sufficient rough control on the size of f (n) (e.g. the
pointwise bound |f (n)| ≤ 1), then F (s) will converge absolutely in
the half plane <(s) > 1, and there will satisfy
!
∞
Y
X
f (p k )
.
F (s) =
1+
p ks
p
k=1
Multiplicative functions, II
Over the polynomial ring Fq [x], with q a prime power, a
multiplicative function f : M → C is one that satisfies
∀ coprime F , G ∈ M,
f (FG ) = f (F )f (G )
where M denotes the set of monic polynomials in Fq [x].
The natural generating function is the power series
X
F(z) :=
f (F )z deg(F ) .
F ∈M
If we have the pointwise bound |f (F )| ≤ 1, then F(z) will
converge absolutely in the disc |z| < 1/q (since there are q n monic
polynomials of degree n), and there will satisfy
!
∞
Y
X
k kdeg(P)
F(z) =
1+
f (P )z
,
P∈P
k=1
where P denotes the set of irreducible monic polynomials in Fq [x].
Mean values
Returning to the integer setting, we would like to use information
about the Dirichlet
series F (s) to deduce things about the mean
P
value (1/x) n≤x f (n). (And similarly in the function field
setting.)
Key point: If we are lucky, the values f (n) might be so well
behaved that F (s) enjoys very good properties, such as analytic
continuation to a large part of the complex plane. Lots of classical
analytic number theory works like this (think of Dirichlet
characters χ(n), for example). But what can we do if we don’t
know so much about F (s)?
To simply state some results, let Λf (n) denote the Dirichlet series
P
Λf (n)
coefficients of −F 0 (s)/F(s) = ∞
n=1 ns , where <(s) > 1, say.
Q
P
f (p k )
Also let Fx (s) := p≤x 1 + ∞
denote the finite Euler
k=1 p ks
product, truncated to primes p ≤ x.
Theorem (Halász’s theorem over the integers)
Suppose f : N → C is a multiplicative function, and suppose
|Λf (n)| ≤ Λ(n) for all n. Then for all large x we have
1 X
log log x
f (n) (1 + M)e −M +
,
x
log x
n≤x
where the quantity M = M(f , x) is defined by
Fx (1 + it) −M
.
e
log x := max 1 + it |t|≤log x
Because of the condition |Λf (n)| ≤ Λ(n), we have
!
∞
X
Y
1
1+
|Fx (1 + it)| ≤
log x,
pk
p≤x
k=1
so we certainly always have M 1.
In my opinion, the crucial feature of Halász’s theorem is that it is a
general lossless bound.
In other words, if we use the trivial lower bound M 1 in the
theorem, we recover the trivial bound
1 X
f (n) 1.
x
n≤x
If we have any better bound for M, then we will get a non-trivial
mean value bound.
This feature is quite rare in most analytic arguments (typically one
“gives away” a few logarithms at the beginning).
In the function field setting,
the power series
P let Λf (F ) denote
F0
deg(F
)
coefficients of z F (z) = F ∈M Λf (F )z
. Note that here, the
analogue of the von Mangoldt function is given by Λ(F ) = deg(P)
if F is a power of the irreducible P, and Λ(F ) = 0 otherwise.
Also let Fn (z) be the power series corresponding to the
multiplicative function with Λf (F ) = 0 if deg(F ) ≥ n.
Theorem (Granville, H., Soundararajan, 2015)
Suppose f : M → C is a multiplicative function, and that
|Λf (F )| ≤ Λ(F ) for all F . Then for all large n we have
1
n
q
X
f (M) (1 + M)e −M ,
M∈M,
deg(M)=n
where now M = M(f , q, n) is defined by
e −M n := max |Fn (z)|.
|z|=1/q
Idea of the proof (in function fields), I
We can write
Z
X
1
dz
1 1
Fn (z) n+1 + small error.
f (M) = n
n
q
q 2πi |z|=1/q
z
M∈M,
deg(M)=n
Bounding the integral trivially using the triangle inequality, we get
Z
q
|Fn (z)||dz| ≤ max |Fn (z)|,
≤
2π |z|=1/q
|z|=1/q
so we are off from what we want by roughly a factor of n.
Key point: It is often quite difficult to bound an L1 norm in an
efficient way, and even bounding an L2 norm may not give a good
result. Motivated by additive number theory (the circle method),
we want to have three factors around, so we can pull one out and
apply Parseval to the other two factors.
Idea of the proof (in function
P fields), II
n
So we want to write (1/q )
M∈M, f (M) as an integral of a
deg(M)=n
product of three power series, up to a small error.
To get two factors, a natural thing is to write
Z
X
1
dz
1 1
zFn0 (z) n+1
f (M) ≈
n
n
q
nq 2πi |z|=1/q
z
M∈M,
deg(M)=n
=
1 1
nq n 2πi
Z
|z|=1/q
zFn0 (z)
Fn (z)
Fn (z)
dz
z n+1
.
Note here that we have gained a factor 1/n on the outside of the
0
P
n
integral, and z F
Λf (F )z deg(F ) is an
F ∈M,
Fn (z) =
deg(F )≤n−1
arithmetically meaningful power series (whose average behaviour
we could hope to control).
Idea of the proof (in function fields), III
To get three factors there are various options, for example we can
write
0
0
Z 1
zFn (z)
zFn (z)
0
Fn (z) =
1+
zFn (tz)dt
Fn (z)
Fn (z)
0
0
Z 1
zFn (z)
tzFn0 (tz)
dt
=
1+
Fn (tz)
.
Fn (z)
Fn (tz)
t
0
The contribution from the 1 in the bracket can be absorbed into
our small error, so it remains to bound
0
Z 1Z
1 1
zFn (z) tzFn0 (tz)
dt
Fn (tz)dz .
nq n 2πi 0 |z|=1/q Fn (z)
Fn (tz)
t
One can pull out a factor max |Fn (tz)| from the inner integral, and
use Cauchy–Schwarz and Parseval to efficiently bound everything
else there.