Main terms in moments Brian Conrey AIM and Bristol Cetraro July - - PowerPoint PPT Presentation

main terms in moments
SMART_READER_LITE
LIVE PREVIEW

Main terms in moments Brian Conrey AIM and Bristol Cetraro July - - PowerPoint PPT Presentation

Main terms in moments Brian Conrey AIM and Bristol Cetraro July 8, 2019 At the Amalfi conference in 1989: Theorem : (Conrey and Ghosh) log 9 T Z T 1 | (1 / 2 + it ) | 6 dt 10 . 13 a 3 T 9! 0 2 ( k 1) 2 k 1 k


slide-1
SLIDE 1

Main terms in moments

Brian Conrey AIM and Bristol

July 8, 2019

Cetraro

slide-2
SLIDE 2
slide-3
SLIDE 3

At the Amalfi conference in 1989:

Theorem : (Conrey and Ghosh)

1 T Z T |ζ(1/2 + it)|6 dt ≥ 10.13 a3 log9 T 9!

ak = Y

p

✓ 1 − 1 p ◆(k−1)2 k−1 X

j=0

k−1

j

2 pj

slide-4
SLIDE 4

Conrey and Ghosh conjecture: 1992 Conrey and Gonek conjecture: 1998

slide-5
SLIDE 5

Keating and Snaith formula:

g1 = 1 g2 = 2 g3 = 42 g4 = 24024

g5 = 701149020

1 T Z T |ζ(1/2 + it)|2k dt ∼ gkak logk2 T k2!

Conjecture (KS):

Z

U(N)

| det(I − U)|2k dU = (N + 1)(N + 2)2 . . . (N + k)k(N + k + 1)k−1 . . . (N + 2k − 1) 1 · 22 · · · · · kk · (k + 1)k−1 · · · · · (2k − 1) ∼ gk N k2 k2!

gk = k2! 1 · 22 · · · · · kk · (k + 1)k−1 · · · · · (2k − 1)

slide-6
SLIDE 6

where Theorem:

slide-7
SLIDE 7

BA,B(s) =

X

n=1

τA(m)τB(m) ms

Y

α∈A

ζ(s + α) =

X

n=1

τA(n) ns

Let A and B be sets of small complex numbers and Then, with s=1/2+it

Conjecture (CFKRS) The Recipe

where

Z ψ ✓ t T ◆ Y

α∈A

ζ(s + α) Y

β∈B

ζ(1 − s + β) dt

= Z ψ ✓ t T ◆ X

U⊂A,V ⊂B |U|=|V |

✓ t 2π ◆−(U+V ) BU∪V −,V ∪U −(1) dt + O(T 1−δ)

and ψ ∈ C∞[1, 2]

Note that each term has a total of |A| |B| singularities; but the sum is analytic .

slide-8
SLIDE 8

Conjecture (C, Farmer, Keating, Rubinstein, Snaith) where

slide-9
SLIDE 9

RMT analogue

This matches perfectly with the recipe!

Theorem (CFKRS). Let Z(A, B) = Y

α∈A β∈B

z(α + β) where z(x) = (1 − e−x)−1. Then Z

U(N)

Y

α∈A

ΛX(e−α) Y

β∈B

ΛX∗(e−β) dX = X

S⊂A T ⊂B |S|=|T |

e−N(S+T )Z(S ∪ T −, T ∪ S−)

slide-10
SLIDE 10

Conrey, Iwaniec, Soundararajan proved a sixth moment estimate for Dirichlet L-functions which found the full 9th degree polynomial above. Nathan Ng proved that the sixth moment of zeta with all lower order terms can be obtained from precise information about the shifted divisor problem. Chandee and Li obtained the leading order term (the 24024) assuming RH for the 8th moment of this family.

slide-11
SLIDE 11

There is interest in how new main terms enter the picture in moment problems as the order of the moment grows. Classically we can do the second moment of zeta by diagonal analysis; but the fourth moment requires shifted convolution sums. For averages of quadratic L-functions one uses diagonal analysis for the first and second moment and then Soundararajan’s Poisson formula for quadratic characters for the third moment. New main terms arise from the “square” values of k after using Poisson. For averaging cusp form L-functions at the center via the Peterson formula one initially uses only the diagonal terms at the start; then Kowalski, Michel and Vanderkaam show how to use parts of the Kloosterman sum to obtain some off-diagonal contributions; further investigation leads to off-off-diagonal contributions to the main term. In the asymptotic large sieve applications of Conrey, Iwaniec and Soundararajan (eg for the sixth moment of Dirichlet L-functions) the final main terms seem to be located in a remote part of the complex plane, far from other contributing singularities. In Zhang and Diaconu using multiple Dirichlet series found a polar term at 3/4!

Combinatorics of main terms

slide-12
SLIDE 12

Moments of long Dirichlet polynomials

slide-13
SLIDE 13

+T X

m6=n

ambn √mn ˆ ψ ✓ T 2π log n m ◆ = T ˆ ψ(0) X

n≤X

anbn n Z ∞ ψ ✓ t T ◆ X

m≤X

am m1/2+it X

n≤X

bn n1/2−it dt

slide-14
SLIDE 14

We can often rewrite this as The off-diagonal piece is

X

h6=0 0<m+hX

ambm+h p m(m + h) ˆ ψ ✓ T 2π log ✓ 1 + h m ◆◆ ∼ X

h6=0 0<mX

ambm+h m ˆ ψ ✓ Th 2πm ◆

slide-15
SLIDE 15

The case of t-aspect for Ramanujan tau L-function

slide-16
SLIDE 16

S(X, h) := X

n≤X

τ ∗(n)τ ∗(n + h)

τ ∗(n) = τ(n)n−11/2 S(X, h) ⌧h X2/3

1 p X X

h≤ √ X

|S(X, h)| ⌧ X1/2+✏

Good (1983) How does one average the moments of a cusp form L-function in t-aspect? The shifted convolutions play a role.

slide-17
SLIDE 17

X

n≤X

τ ∗(n)τ ∗(n + h) ⌧ X1−δ

uniformly for h ⌧ X2−η

Blomer (2005)

slide-18
SLIDE 18

S(X,1) for 1< X < 10000

slide-19
SLIDE 19

uniformly for H ⌧ X1/2

T(X; H) := X

n≤X

  • X

h≤H

τ ∗(n + h)

  • 2

= CXH + O(X2/3+✏H5/3)

Good:

slide-20
SLIDE 20

T(X;H) for X = 5000; 1< H < 100

slide-21
SLIDE 21

Z ψ ✓ t T ◆

  • X

n≤X

τ ∗(n) n1/2+it

  • 2

dt

⇠ ( T ˆ ψ(0) P

n≤X τ ∗(n)2 n

X ⌧ T 2 R ψ t

T

  • |Lτ ∗(1/2 + it)|2 dt

X T 2

Z ψ ✓ t T ◆ |Lτ ∗(1/2 + it)|2 dt ∼ T ˆ ψ(0) X

n≤T 2

τ ∗(n)2 n

But Averaging a long tau polynomial

= T ˆ ψ(0) X

n≤X

τ ∗(n) n + 2T X

h6=0 T mX

τ ∗(m)τ ∗(m + h) m ˆ ψ ✓ Th 2πm ◆ + o(T)

slide-22
SLIDE 22

So, it must be the case that

for X T 2

2 X

h6=0 T mX2

τ ∗(m)τ ∗(m + h) m ˆ ψ ✓ Th 2πm ◆ ∼ C ˆ ψ(0) log T 2 X

X

n≤X

τ ∗(n)2 n ∼ C log X

where C is such that

slide-23
SLIDE 23

The case of zeta

slide-24
SLIDE 24

M2(α) =    α4 if 0 < α < 1 −α4 + 8α3 − 24α2 + 32α − 14 if 1 < α < 2 2 if α > 2

Zeta-polynomials

1 T Z T

  • X

n≤X

dk(n) ns − Resw=1−s ζ(s + w)kXw w

  • 2

dt ∼ Mk(α) ak logk2 T k2!

where s = 1/2 + it and X = T α. For example

slide-25
SLIDE 25
slide-26
SLIDE 26

Let and

ΛU(s) =

N

X

j=0

Scj(U)sj ζ(s)k =

X

n=1

dk(n) ns

To what extent do the coefficients

  • f ΛU(s)k behave like dk(n)nit?
slide-27
SLIDE 27

Diaconis - Gamburd formula

Z

U(N)

Scj1(U) . . . Scjk(U)Sch1(U) . . . Sch`(U) du

If j1 + · · · + jk = h1 + · · · + h` ≤ N, then

is the number of k × ` matrices with non-negative integer entries and row sums j1, . . . , jk and column sums h1, . . . h`.

For example, Z

U(N)

Scj(U)Sck(U) du = 0 Z

U(N)

|Scj(U)|2 du = 1 and Z

U(N)

|Scj(U)Sck(U)|2 dU = 1 + min{j, k} if j 6= k and j + k  N.

slide-28
SLIDE 28

∼ ak N k2 Z

U(N)

  • X

j1+···+jk≤αN ji≤N

Scj1 . . . Scjk

  • 2

dU

Conjecture

Sandro Bettin assumes the recipe and proves this.

1 T(log T)k2 Z T

  • X

m≤T α

dk(m) m1/2+it − Ress=1 ζ(s)kT αs s

  • 2

dt

slide-29
SLIDE 29

Z

U(N)

  • X

j1+···+jk≤αN

Scj1 . . . Scjk

  • 2

du = X

m≤αN

Ik(m, N)

By orthogonality

Ik(m; N) := Z

U(N)

  • X

j1+···+jk=m ji≤N

Scj1(U) . . . Scjk(U)

  • 2

dU

where

slide-30
SLIDE 30

Keating, Rodgers, Roddity-Gershon, Rudnick formula

If m < N then By the functional equation this also works for (k-1)N , m < kN. It’s not so clear what the formula looks like when N < m < 2n.

Ik(m; N) = ✓k2 − 1 + m m ◆

slide-31
SLIDE 31

Keating, Rodgers, Roddity-Gershon, Rudnick formula

Ik(m; N) is equal to the number of k × k matrices with non-negative integer entries at most N in size whose rows are weakly increasing, columns are weakly decreasing and whose anti-diagonal sums to kN − m

(Gelfond-Tsetlin patterns)

slide-32
SLIDE 32

Zeros of R

U(25) ΛU(x)3ΛU ∗(x)3 dU = P25 m=0 Ik(m, 25)x2m

slide-33
SLIDE 33

Ik(m; N) = γk(c)N k2−1 + O(N k2−2) c = m/N

γk(c) = 1 k!G(1 + k2) Z

[0,1]k δc(w1 + · · · + wk)∆(w1, . . . , wk)2dw

Keating, Rodgers, Roddity-Gershon, Rudnick where G is the Barnes function.

γk(c) = M 0

k(c)

k2!

slide-34
SLIDE 34

1 X Z 2X

X

  • X

x≤n≤x+H

dk(n)

  • 2

dx − @ 1 X Z 2X

X

X

x≤n≤x+H

dk(n) dx 1 A

2

Conjecture: Keating, Rodgers, Roddity-Gershon, Rudnick; Rodgers, Soundararajan Lester has made progress on the divisor problem in short intervals. Keating, Rodgers, Roddity-Gershon, Rudnick prove the function field analogue of this.

log X log X

H

→ c ∈ (0, k)

with

∼ ak(q)γk(c)H ✓ log X H ◆k2−1

slide-35
SLIDE 35

X

1≤a≤q (a,q)=1

  • X

n≡a mod q n≤X

dk(n) − 1 φ(q) X

(n,q)=1 n≤X

dk(n)

  • 2

Conjecture: Keating, Rodgers, Roddity-Gershon, Rudnick; Rodgers, Soundararajan Rodgers and Soundararajan prove this for delta<c<2-delta (assuming GRH). Keating, Rodgers, Roddity-Gershon, Rudnick prove a function field analogue of this.

log X log q → c ∈ (0, k)

with

∼ ak(q)γk(c)H(log X)k2−1

slide-36
SLIDE 36

γk(c) = 1 k!G(1 + k2) Z

[0,1]k δc(w1 + · · · + wk)∆(w1, . . . , wk)2dw

g(u) = Z 1 exp(−tx) dx

γk(c) = 1 G(1 + k)2 Z ∞

−∞

exp(2πiuc)Dk(2πiu) du

Dk(t) = det

k×k

⇣ g(i+j−2)(u) ⌘

Basor, Ge, Rubinstein

slide-37
SLIDE 37

Dk(t) = Dk(0) exp ∞ X

m=1

cm(k) m tm !

cM(k) = 1 (M − 1)(M − 2)

M−3

X

m=0

(m + 2)cm+2(k) (cM−m−2(k − 1) + cM−m−2(k + 1) − 2cM−m−2(k)) c1(k) = −k 2

c2(k) = k2 4(4k2 − 1)

D satisfies a Painleve equation which leads to a recursion formula:

slide-38
SLIDE 38

Back to moments

  • f zeta
slide-39
SLIDE 39

X

n≤X

τA(n)τB(n + h)

We need information about The delta method of Duke, Friedlander and Iwaniec (1993) can provide the needed conjecture. Bill Duke John Friedlander Henryk Iwaniec

Divisor correlations

slide-40
SLIDE 40

Delta method conjecture

The poles of this Dirichlet series can be determined by replacing the exponential by Dirichlet characters and finding the coefficient of the trivial character (i.e. zeta).

hτA(m)τB(m + h)im=u ⇠

X

q=1

rq(h)hτA(m)e(m/q)im=uhτB(n)e(n/q)in=u

hτA(m)e(m/q)im=u = 1 2πi Z

|w−1|=✏

DA(w, e(1/q))uw−1 dw where DA(w, e(1/q)) =

X

n=1

τA(n)e(n/q) nw rq(h) is the Ramanujan sum and DA is the Estermann function

slide-41
SLIDE 41

Z ∞ ψ( t T ) X

m≤T r n≤T r

τA(m)τB(n) √mn ⇣m n ⌘it dt

Assuming delta-conjecture

+O(T 1−δ) where 1 ≤ r < 2 and

= Z 1 ψ( t T ) ✓ BA,B(1; T r) + X

α2A β2B

✓ t 2π ◆αβ BA0,B0(1; T r) ◆ A0 = A − {α} ∪ {−β}, B0 = B − {β} ∪ {−α}

slide-42
SLIDE 42

This relies on the identity

where and

slide-43
SLIDE 43

X

j=0

τA0∪{− ˆ

β}(pj)τB0∪{−ˆ α}(pj)

pj Equating Euler products, the identity is:

Suppose ˆ α ∈ A and ˆ β ∈ B. Let A0 = A − {ˆ α} and B0 = B − {ˆ β}. Then

Identifying `geometric’ factors and the location of the poles of both sides of the identity is a matter of equating the `linear’ term of the Euler products; i.e. the value at p of the arithmetic function. So, in some sense the identification of the whole p-factor is more complex.

=

X

h=0

1 ph(1−ˆ

α− ˆ β) ∞

X

j=0

τA0(pj) pj(1−ˆ

α) ∞

X

k=0

τB0(pk) pk(1− ˆ

β) ∞

X

d,q=0

µ(pq) pd+q(2−ˆ

α− ˆ β) ∞

X

m=0

τA0(pm+d+q) pm(1−ˆ

α) ∞

X

n=0

τB0(pn+d+q) pn(1− ˆ

β)

slide-44
SLIDE 44

Thus, we see the `one-swap’ terms arise from the standard shifted divisor problem.

What is the RMT analogue of the input from the (averaged over h) delta method?

Where are the rest of the terms from the recipe???

slide-45
SLIDE 45

What if r > 2? Say ` < r < ` + 1

To find the higher-swap terms we will need convolutions of shifted divisors …

slide-46
SLIDE 46

Conrey - Keating approach

is related to

X

m,n<T r

τA(m)τB(n) √mn ˆ ψ T 2π log m n

  • X

A=A!∪···∪A` B=B1∪···∪B`

X

M1...M`=N1...N` (Mi,Ni)=1

`

Y

j=1

@ X

mj,nj

τAj(mj)τBj(nj) √mn 1 A ˆ ψ T 2π log m1 . . . m` n1 . . . n`

   M1m1 = N1n1 + h1 . . . M`m` = N`n` + h`   

Note that

ˆ ψ T 2π log m1 . . . m` n1 . . . n`

  • ∼ ˆ

ψ( T 2π X hi niNi )

Split A and B up into A = A1 ∪ · · · ∪ A` and B = B1 ∪ · · · ∪ B`. Then

subject to Q mi, Q ni ≤ T r and

which controls the ranges of the sums.

slide-47
SLIDE 47

hτA(m)τB(n)i(∗)

m=u

⇠ 1 M

X

q=1

rq(h)hτA(m)e(mN/q)im=uhτB(n)e(nM/q)in= uN

M

Delta method conjecture with general linear constraint

hτA(m)e(mN/q)im=u = 1 2πi Z

|w−1|=✏

DA(w, e N q

  • )uw−1 dw

where DA(w, e N q

  • ) =

X

n=1

τA(n)e(nN/q) nw .

(∗) : mM − nN = h

where The poles of this Dirichlet series can be determined by replacing the exponential by Dirichlet characters and finding the coefficient of the trivial character (i.e. zeta).

slide-48
SLIDE 48

1 2⇡i Z

(2)

Xs T 2⇡ −`s X

(M1,N1)=···=(M`,N`)=1 N1...N`=M1...M` ✏j ∈{−1,+1}

Z

0<v1,...,v`<∞

ˆ (✏1v1 + · · · + ✏`v`)

`

Y

j=1

 1 (2⇡i)2 ZZ

|wj −1|=✏ |zj −1|=✏

M −zj

j

N s+1−wj

j

X

hj,qj

rqj(hj) hs+2−wj−zj

j

vs+1−wj−zj

j

DAj(wj, e Nj qj

  • )DBj(zj, e

Mj qj

  • )

T 2⇡ wj+zj−2dwjdzj dvj ds s

1 2⇡i Z

(2)

Xs Z ∞ (t) X

U(`)⊂A V (`)⊂B

✓Tt 2⇡ ◆− P

ˆ ↵∈U(`) ˆ ∈V (`)

(ˆ α+ ˆ β+s)

×B(As − U(`)s + V (`)−, B − V (`) + U(`)s

−, 1) dt ds

s

= Connecting divisor correlations and the recipe where U(`) denotes a set of cardinality ` with precisely one element from each

  • f A1, . . . , A` and similarly V (`) denotes a set of cardinality ` with precisely one

element from each of B1, . . . , B` .

slide-49
SLIDE 49

If we sum this over all the ways to split up A and B we get what the recipe predicts times a factor But this is the number of automorphisms of the *-system.

Automorphisms

`!2`2k−2`

slide-50
SLIDE 50

50

We can use this approach to discover a formula for γk(c), c > 1.

Z

z,w

(η + X (zj + wj))k2 Q

j

⇥ (1 − zj)k(1 − wj)k⇤ ∆(z)2∆(w)2 Q

j

⇥ zk

j wk j

⇤ Q

i,j [(1 − wi − zj)(1 + wi + zj)] dz dw

It is a linear combination of the functions

where η is the fractional part of c and the products are for 1 ≤ i, j ≤ c

slide-51
SLIDE 51

Wooley has pointed out the connection with counting points on varieties and Manin’s idea of counting points on certain varieties by counting points on a stratified set of subvarieties; this idea may be relevant here. Trevor Wooley

  • Y. I.Manin
slide-52
SLIDE 52

Manin stratification

slide-53
SLIDE 53

We can see an example already from the case k=2. We expect that is a good approximation to zeta(s)^2 when X >>T^2.

X

m≤X |h|≥1

d(m)d(m + h) ˆ ψ(hT/m)

But the analysis of averaged over h fails to reveal a large main term of size X/T^2 as well as a secondary main term that reflects the change in behavior as X passes T^2.

X

n≤X

d(n) ns − Resw=1−s ζ(s + w)2Xw w

slide-54
SLIDE 54

There is a closer analogy between moments of zeta and averages

  • f characteristic polynomials than just that the main terms agree once

we insert the arithmetic factor. In a shifted moment of zeta we let the arithmetic factors go to 1 and we replace zeta(1+x) by z(x). The shift alpha for zeta becomes exp(alpha) in RMT. Finally N becomes t/(2 pi). At that point ALL of the main terms agree. It stands to reason that we can learn something by carefully analyzing all of the pieces from both points of view. And, if we regard the matrix size N in RMT as the infinite prime it stands to reason that we should investigate carefully what happens with the finite primes as well.

slide-55
SLIDE 55

The same circle of ideas works for averages of ratios

  • f zeta-functions and characteristic polynomials.

In particular, using ratios, we can revisit the Bogomolny-Keating papers on `Hardy-Littlewood implies GUE’

  • n a similar footing as this work.

GUE

slide-56
SLIDE 56

RMT problems

  • 1. Find (simple) exact formulas for for m>N?
  • 2. Do the satisfy a Painleve?
  • 3. What is the RMT version of CK V?
  • 4. Does the answer to 3 lead to a recursion formula for
  • 5. Can we find the analogue of for O(N) and USp(2N)?
  • 6. Does 5 lead to a formulation of CK V for moments in other families?

Ik(m; N) Ik(m; N)

Ik(m; N) ? Ik(m; N)

slide-57
SLIDE 57

Function field problems

  • 1. Can we do pair correlation over function fields?
  • 2. Can we reproduce any of KRRR for fixed q?
  • 3. Analogue of CK I-V for function fields?
  • 4. Analogue of divisors in short intervals and in arithmetic progressions

for other families?

slide-58
SLIDE 58

Number field problems

  • 1. Do Rodgers-Soundararajan with shifts and with power savings.
  • 2. What is the precise connection between the answer to 1 and

the moment polynomials? Work out the sizes (asymptotics?) and the relationship between these averages for various lambda and ranges of T and X. When X > T^2 is it best to average

  • ver h first, possibly with Voronoi (see Jutila, Ivic)?

Connections with Manin stratification.)

Mλ(X, T; W) := X

m≤X h≥1

λ(m)λ(m + h) m W ✓hT m ◆

  • 3. Let
  • 4. Extend Nathan Ng’s work to rigorously obtain averages
  • f long Dirichlet polynomials with general divisor coefficients.
  • 5. Asymptotics of 10th moment of L-functions in cusp form families
  • 6. Rigorously derive n-correlation conjecture in a range [0,2]

from H-L conjectures

Sλ(X; T) := X

m≤X

  • H

X

h=1

λ(m + h)

  • 2

and

slide-59
SLIDE 59

FRG Project

A newly funded NSF project with PIs Conrey, Iwaniec, Keating, Soundararajan, Wooley and senior scientists Brad Rodgers and Caroline Turnage-Butterbaugh will be devoted to understanding these questions. If you are interested in this project send me email at conrey@aimath.org with subject line FRG

slide-60
SLIDE 60

The End