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 | ζ (1 / 2 + it ) | 6 dt ≥ 10 . 13 a 3 T 9! 0 � 2 ◆ ( k − 1) 2 k − 1 � k − 1 ✓ 1 − 1 j Y X a k = p j p p j =0

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

Keating and Snaith formula: ( N + 1)( N + 2) 2 . . . ( N + k ) k ( N + k + 1) k − 1 . . . ( N + 2 k − 1) Z | det( I − U ) | 2 k dU = 1 · 2 2 · · · · · k k · ( k + 1) k − 1 · · · · · (2 k − 1) U ( N ) N k 2 g k ∼ k 2 ! k 2 ! g k = 1 · 2 2 · · · · · k k · ( k + 1) k − 1 · · · · · (2 k − 1) g 1 = 1 g 2 = 2 g 3 = 42 g 4 = 24024 g 5 = 701149020 Conjecture (KS): log k 2 T Z T 1 | ζ (1 / 2 + it ) | 2 k dt ∼ g k a k T k 2 ! 0

Theorem: where

Conjecture (CFKRS) The Recipe ∞ τ A ( n ) Y X ζ ( s + α ) = Let A and B be sets of small complex numbers and n s α ∈ A n =1 and ψ ∈ C ∞ [1 , 2] Then, with s=1/2+it ✓ t ◆ Y Z Y ζ ( s + α ) ζ (1 − s + β ) dt ψ T α ∈ A β ∈ B ✓ t ✓ t ◆ − ( U + V ) ◆ Z X B U ∪ V − ,V ∪ U − (1) dt + O ( T 1 − δ ) = ψ 2 π T U ⊂ A,V ⊂ B | U | = | V | where ∞ τ A ( m ) τ B ( m ) X B A,B ( s ) = m s n =1 Note that each term has a total of |A| |B| singularities; but the sum is analytic .

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

RMT analogue Theorem (CFKRS). Let Y Z ( A, B ) = z ( α + β ) α ∈ A β ∈ B where z ( x ) = (1 − e − x ) − 1 . Then Z Y Y Λ X ( e − α ) Λ X ∗ ( e − β ) dX U ( N ) α ∈ A β ∈ B X e − N ( S + T ) Z ( S ∪ T − , T ∪ S − ) = S ⊂ A T ⊂ B | S | = | T | This matches perfectly with the recipe!

Conrey, Iwaniec, Soundararajan proved a sixth moment estimate for Dirichlet L-functions which found the full 9th degree polynomial above. Chandee and Li obtained the leading order term (the 24024) assuming RH for the 8th moment of this family. 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.

Combinatorics of main terms 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!

Moments of long Dirichlet polynomials

✓ t Z ∞ ◆ X a m b n X ψ n 1 / 2 − it dt m 1 / 2+ it T 0 m ≤ X n ≤ X a n b n = T ˆ X ψ (0) n n ≤ X ✓ T ◆ a m b n 2 π log n X ˆ + T ψ √ mn m m 6 = n

The off-diagonal piece is ✓ T ✓ ◆◆ a m b m + h 1 + h ˆ X 2 π log ψ p m m ( m + h ) h 6 =0 0 <m + h  X We can often rewrite this as ✓ Th ◆ a m b m + h X ˆ ψ ∼ m 2 π m h 6 =0 0 <m  X

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

How does one average the moments of a cusp form L-function in t-aspect? The shifted convolutions play a role. τ ∗ ( n ) = τ ( n ) n − 11 / 2 Good (1983) X τ ∗ ( n ) τ ∗ ( n + h ) S ( X, h ) := n ≤ X S ( X, h ) ⌧ h X 2 / 3 1 X | S ( X, h ) | ⌧ X 1 / 2+ ✏ p X √ h ≤ X

Blomer (2005) X τ ∗ ( n ) τ ∗ ( n + h ) ⌧ X 1 − δ n ≤ X uniformly for h ⌧ X 2 − η

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

Good: 2 � � � � X X = CXH + O ( X 2 / 3+ ✏ H 5 / 3 ) τ ∗ ( n + h ) � � T ( X ; H ) := � � � � n ≤ X h ≤ H � � uniformly for H ⌧ X 1 / 2

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

Averaging a long tau polynomial 2 ✓ t ◆ � � � τ ∗ ( n ) � Z X � � ψ dt � � n 1 / 2+ it T � � n ≤ X � � ✓ Th τ ∗ ( n ) τ ∗ ( m ) τ ∗ ( m + h ) ◆ = T ˆ X X ˆ ψ (0) + 2 T + o ( T ) ψ 2 π m n m n ≤ X h 6 =0 T  m  X ( τ ∗ ( n ) 2 T ˆ X ⌧ T 2 ψ (0) P n ≤ X n ⇠ � t | L τ ∗ (1 / 2 + it ) | 2 dt X � T 2 R � ψ T But ✓ t τ ∗ ( n ) 2 ◆ Z | L τ ∗ (1 / 2 + it ) | 2 dt ∼ T ˆ X ψ (0) ψ T n n ≤ T 2

So, it must be the case that ✓ Th ψ (0) log T 2 ◆ τ ∗ ( m ) τ ∗ ( m + h ) ˆ ∼ C ˆ X 2 ψ 2 π m m X h 6 =0 T  m  X 2 for X � T 2 where C is such that τ ∗ ( n ) 2 X ∼ C log X n n ≤ X

The case of zeta

Zeta-polynomials 2 log k 2 T � � Z T ζ ( s + w ) k X w � � 1 d k ( n ) X � � − Res w =1 − s dt ∼ M k ( α ) a k � � n s k 2 ! T w � � 0 n ≤ X � � where s = 1 / 2 + it and X = T α . For example  α 4 if 0 < α < 1  − α 4 + 8 α 3 − 24 α 2 + 32 α − 14 M 2 ( α ) = if 1 < α < 2 2 if α > 2 

Let ∞ d k ( n ) ζ ( s ) k = X n s n =1 and N X Sc j ( U ) s j Λ U ( s ) = j =0 To what extent do the coe ffi cients of Λ U ( s ) k behave like d k ( n ) n it ?

Diaconis - Gamburd formula If j 1 + · · · + j k = h 1 + · · · + h ` ≤ N , then Z Sc j 1 ( U ) . . . Sc j k ( U ) Sc h 1 ( U ) . . . Sc h ` ( U ) du U ( N ) is the number of k × ` matrices with non-negative integer entries and row sums j 1 , . . . , j k and column sums h 1 , . . . h ` . For example, Z Z | Sc j ( U ) | 2 du = 1 Sc j ( U ) Sc k ( U ) du = 0 U ( N ) U ( N ) and Z | Sc j ( U ) Sc k ( U ) | 2 dU = 1 + min { j, k } U ( N ) if j 6 = k and j + k  N .

Conjecture 2 � � Z T ζ ( s ) k T α s � � 1 d k ( m ) X � � m 1 / 2+ it − Res s =1 dt � � T (log T ) k 2 s � � 0 m ≤ T α � � 2 � � � � ∼ a k Z � � X Sc j 1 . . . Sc j k dU � � N k 2 � � U ( N ) � � j 1+ ··· + jk ≤ α N � � ji ≤ N Sandro Bettin assumes the recipe and proves this.

By orthogonality 2 � � � � Z X X � � du = I k ( m, N ) Sc j 1 . . . Sc j k � � � � U ( N ) j 1 + ··· + j k ≤ α N m ≤ α N � � where 2 � � � � Z � � X I k ( m ; N ) := Sc j 1 ( U ) . . . Sc j k ( U ) dU � � � � U ( N ) � � j 1+ ··· + jk = m � � ji ≤ N

Keating, Rodgers, Roddity-Gershon, Rudnick formula If m < N then ✓ k 2 − 1 + m ◆ I k ( m ; N ) = m 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.

Keating, Rodgers, Roddity-Gershon, Rudnick formula I k ( 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)

U (25) Λ U ( x ) 3 Λ U ∗ ( x ) 3 dU = P 25 m =0 I k ( m, 25) x 2 m Zeros of R

Keating, Rodgers, Roddity-Gershon, Rudnick I k ( m ; N ) = γ k ( c ) N k 2 − 1 + O ( N k 2 − 2 ) c = m/N where 1 Z [0 , 1] k δ c ( w 1 + · · · + w k ) ∆ ( w 1 , . . . , w k ) 2 dw γ k ( c ) = k ! G (1 + k 2 ) G is the Barnes function. γ k ( c ) = M 0 k ( c ) k 2 !

Conjecture: Keating, Rodgers, Roddity-Gershon, Rudnick; Rodgers, Soundararajan 2 2 � � 0 1 Z 2 X Z 2 X � � 1 @ 1 X X � � d k ( n ) d k ( n ) dx dx − � � A X X � � X X x ≤ n ≤ x + H x ≤ n ≤ x + H � � ◆ k 2 − 1 ✓ log X ∼ a k ( q ) γ k ( c ) H H log X → c ∈ (0 , k ) with log X H Lester has made progress on the divisor problem in short intervals. Keating, Rodgers, Roddity-Gershon, Rudnick prove the function field analogue of this.

Conjecture: Keating, Rodgers, Roddity-Gershon, Rudnick; Rodgers, Soundararajan 2 � � � � 1 � � X X X d k ( n ) − d k ( n ) � � φ ( q ) � � � � 1 ≤ a ≤ q n ≡ a mod q ( n,q )=1 � � ( a,q )=1 n ≤ X n ≤ X ∼ a k ( q ) γ k ( c ) H (log X ) k 2 − 1 log X with log q → c ∈ (0 , k ) Rodgers and Soundararajan prove this for delta<c<2-delta (assuming GRH). Keating, Rodgers, Roddity-Gershon, Rudnick prove a function field analogue of this.

Basor, Ge, Rubinstein 1 Z [0 , 1] k δ c ( w 1 + · · · + w k ) ∆ ( w 1 , . . . , w k ) 2 dw γ k ( c ) = k ! G (1 + k 2 ) Z ∞ 1 γ k ( c ) = exp(2 π iuc ) D k (2 π iu ) du G (1 + k ) 2 −∞ ⇣ ⌘ g ( i + j − 2) ( u ) D k ( t ) = det k × k Z 1 g ( u ) = exp( − tx ) dx 0