The derivative of the Riemann-zeta function, Toeplitz determinants and the The derivative of the Riemann-zeta function, Riemann- Hilbert Toeplitz determinants and the problem

Francesco Riemann-Hilbert problem Mezzadri

The problem

Where we are Francesco Mezzadri

Question to answer

Motivations

The Riemann- Hilbert approach Random Matrices, L-functions and primes Comments 30 October 2008 ETH, Z¨urich

Work in collaboration with Man Yue Mo The derivative of the Riemann-zeta function, Outline Toeplitz determinants and the Riemann- Hilbert problem 1 The problem

Francesco Mezzadri 2 Where we are The problem

Where we are 3 Question to answer Question to answer

Motivations 4 Motivations The Riemann- Hilbert approach

Comments 5 The Riemann-Hilbert approach

6 Comments The derivative of the Riemann-zeta function, The problem Toeplitz determinants and the • The j.p.d.f. for the eigenvalues for matrices in the CUE: Riemann- Hilbert problem 1 Y 2 P (θ , . . . , θ ) = eiθj − eiθk . Francesco CUE 1 N (2π)N N! Mezzadri 1≤j

The problem Where we are • Characteristic polynomial of a matrix in the CUE: Question to answer N Motivations X N−k Λ(z) = det(Iz − U) = ak z . The Riemann- Hilbert k=0 approach

Comments What can we say about the distributions of the roots of

dΛ(z) Λ0(z) = ? dz The derivative of the Riemann-zeta function, The problem Toeplitz determinants and the Riemann- Hilbert problem 1 1

Francesco 0.8 0.8

Mezzadri 0.6 0.6

0.4 0.4 The problem

0.2 0.2 Where we are 0 0 Question to Im(z) Im(z) answer −0.2 −0.2

Motivations −0.4 −0.4

The Riemann- −0.6 −0.6

Hilbert −0.8 −0.8 approach −1 −1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 Comments Re(z) Re(z) N = 20 N = 50 The derivative of the Riemann-zeta function, The problem Toeplitz determinants and the 0 Riemann- Define λ := (1 − |z |)N. Hilbert problem λ Francesco N Mezzadri z0 The problem

Where we are

Question to answer

Motivations

The Riemann- Hilbert approach

Comments

1 Does the limit distribution Q(λ) = limN→∞ Q(λ; N) exist? 2 What does Q(λ) look like? The derivative of the Riemann-zeta function, Where we are Toeplitz determinants and the Riemann- Theorem (FM 2003) Hilbert problem 1 Francesco Q(λ) ∼ , λ → ∞. Mezzadri λ2

The problem Where we are Conjecture (FM 2003) Question to answer 4 1/2 Motivations Q(λ) ∼ λ , λ → 0. The Riemann- 3π Hilbert approach Theorem (Due˜nez,Farmer, Froehlich, Hughes, FM, and Comments Phan 2008)

4 14 Q(λ) = λ1/2 − λ3/2 + O(λ5/2). 3π 15π The derivative of the Riemann-zeta function, Question to answer Toeplitz determinants and the Riemann- What happens in between? Hilbert problem

Francesco 0.3 Mezzadri

The problem 0.25 Where we are 0.2 Question to answer 0.15 Motivations 0.1 The Riemann- Hilbert approach 0.05 Comments 1 2 3 4 5 6 λ Zeros of Λ0(z), N = 100 The derivative of the Riemann-zeta function, Motivations Toeplitz determinants and the Riemann- • The Riemann zeta function Hilbert problem ∞ X 1 Francesco ζ(s) = , Re(s) > 1. Mezzadri ns n=1 The problem

Where we are • The local correlations the zeros of ζ(1/2 + it) for large t Question to answer are the same as those of eigenvalues of matrices in the Motivations CUE. The Riemann- Hilbert • The local statistical properties of ζ(1/2 + it) as t → ∞ approach are accurately modelled by Λ(z). Comments Theorem (Speiser 1934) The Riemann hypothesis is equivalent to the statement that 0 1 ζ (s) has no zeros to the left of the critical line Re(s) = 2 . The derivative of the Riemann-zeta function, Motivations Toeplitz determinants and the Riemann- Hilbert ζ0(s) Λ0(z) problem 1 log t N/(2π) Francesco 2π 2π Mezzadri Right of the line Re(s) = 1/2 Interior of the unit circle

The problem

Where we are 0.3 0.3

Question to 0.25 0.25 answer 0.2 0.2

Motivations 0.15 0.15

The Riemann- 0.1 0.1 Hilbert 0.05 0.05 approach 1 2 3 4 5 6 1 2 3 4 5 6 Comments Re(s) λ Re. part of zeros of ζ0(s). Zeros of Λ0(z), N = 100. 105 zeros, t ∈ [106, 106 + 60, 000] The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants Q(λ; N) can be written as and the Riemann- 2 Hilbert N problem Q(λ; N) = − 4π(N − 1) Francesco Mezzadri 2 ZZ ∂ 1 × DN−1[exp(iϕ )](w, α, z) The problem ∂α2 Where we are C  Question to 2
2 answer + DN−1[exp(iϕ )](w, α, z) d w , Motivations α=0 The Riemann- 1 2 Hilbert where ϕ (θ, w, α, z) and ϕ (θ, w, α, z) are approach   Comments w α ϕ1(θ, w, α, z) =Re − N(z − eiθ) (N(z − eiθ))2  w   α  ϕ2(θ, w, α, z) =Re − Im N(z − eiθ) (N(z − eiθ))2

with |z| = 1 − λ/N. The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the Riemann- Hilbert • Why the Riemann-Hilbert Problem (RHP)? problem • There is a close connection between the RHP and Francesco Mezzadri integrable operators.

The problem • This is the kernel of an integrable operator: Where we are
Question to sin π(x − y) answer K(x, y) = . π(x − y) Motivations

The Riemann- Hilbert • The Christoffel-Darboux formula is another one: approach

Comments N−1 X pj (x)pj (y) pN (x)pN−1(y) − pN−1(x)pN (y) = CN . (pj , pj ) x − y j=0 The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the • Toeplitz operators are integrable. Riemann- Hilbert • Let ϕ(z) be analytic in some annulus problem

Francesco  −1 Mezzadri Γρ = z : ρ < |z| < ρ , 0 < ρ < 1

The problem and consider Where we are

Question to answer TN−1 = {ϕj−k }0≤j,k≤N−1 . Motivations

The Riemann- Hilbert approach • TN−1 induces a map on the space of trigonometric Comments polynomials τN−1 : PN−1 → PN−1 defined by

N−1 k X j 1 τN−1z = ϕj−k z , z ∈ S , 0 ≤ k ≤ N − 1. j=0 The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants PN−1 j and the • If p(z) = j=0 aj z then Riemann- Hilbert problem     τN−1p (z) = (1 − KN−1) p (z) Francesco Z Mezzadri 0 0 0 = p(z) − KN−1(z, z )p(z )dz 1 The problem S Where we are • Question to KN−1 is an integrable operator: answer

Motivations N 0 −N
0
0 1 z (z ) − 1 1 − ϕ(z ) The Riemann- KN−1(z, z ) = 0 Hilbert 2πi z − z approach

Comments • Then

DN−1 = det (TN−1) = det (1 − KN−1) . The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants • and the Let Σ be an oriented contour in C. An operator on Riemann- L2(Σ, |dz|) is called integrable if its kernel is of the form Hilbert problem Pk 0 Francesco j=1 fj (z)gj (z ) Mezzadri K(z, z0) = . z − z0 The problem

Where we are 2 Question to • The action of K on L (Σ, |dz|) is answer Motivations k X The Riemann- Kh(z) = πi f (z)Hhg (z), z ∈ Σ, Hilbert j j approach j=1 Comments where H is the Cauchy Principal Value Operator

Z 0   1 f (z ) 0 Hf (z) = lim 0 dz . →0 πi {z0∈Σ, |z−z0|>} z − z The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the Riemann- • The resolvent R is an integrable operator too: Hilbert problem Pk 0 Fj (z)Gj (z ) Francesco R(z, z0) = j=1 , Mezzadri z − z0 The problem where Where we are Question to −1 answer Fj = (1 − K) fj = (1 + R) fj Motivations −1 Gj = (1 − K) gj = (1 + R) gj The Riemann- Hilbert approach

Comments

It is a remarkable fact that the functions FJ and Gj can be computed in terms of the solution of a Riemann-Hilbert Matrix Factorization Problem. The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the • An k × k matrix function M(z) is the unique solution (if it Riemann- Hilbert exists) of the RHP (Σ, v) if problem • M(z) is analytic in C \ Σ, Francesco • M+(z) = M−(z)v(z), z ∈ Σ, Mezzadri • M(z) → I as z → ∞. The problem • v(z) is called jump matrix. Where we are

Question to answer Recipe to compute the resolvent R:

Motivations 1 Let f = (f ,..., f )t and g = (g ,..., g )t . Construct the The Riemann- 1 k 1 k Hilbert following jump matrix: approach Comments  2πi  v(z) = I − fg t . 1 + iπ hg, f i

2 Find the solution (if it exists) of the RHP (Σ, v). The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the 3 We have Riemann- Hilbert problem t −1 F = (F1,..., Fk ) = (1 ∓ hg, f i) M±f . Francesco Mezzadri t −1 t
−1 G = (G1,..., Gk ) = (1 ∓ hg, f i) M± g. The problem Where we are 4 Insert F and G into the formula Question to answer Pk 0 0 j=1 Fj (z)Gj (z ) Motivations R(z, z ) = , z − z0 The Riemann- Hilbert approach

Comments The Riemann-Hilbert approach is particularly powerful when the integrable operator K depends on one or more asymptotic parameters and singularities need to be han- dled. The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants Theorem (Deift 1999) and the Riemann- Hilbert Let ϕt be the symbol of a Toeplitz determinant DN−1, then problem Francesco Z 2 Mezzadri d log DN−1 −1 dϕt X 0 = (1 − ϕt ) Ft,j (ζ)Gt,j (ζ)dζ, dt 1 dt The problem S j=1 Where we are T t N T (Ft,1, Ft,2) = M+(ζ)(ζ , 1) Question to answer 1 − ϕ  −1  T (G , G )T = t (Mt )T (ζ) ζ−N , −1 , Motivations t,1 t,2 2πi + The Riemann- Hilbert approach where Mt,+(ζ) is the boundary value from the left of the

Comments solution of the RHP

 −1 N  t t ϕt −(ϕt − 1)ζ 1 M+(ζ) = M−(ζ) −N , ζ ∈ S ζ (ϕt − 1) 2 − ϕt Mt (ζ) = I + O(ζ−1), ζ → ∞ The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the Riemann- Γ4 Hilbert problem Γ λ 2 z = 1 − N Francesco Γ1 Mezzadri Γ3

The problem z· 1· 1·/z Where we are

Question to answer

Motivations

The Riemann- Hilbert approach

Comments n 1 o − 2 −1 Γ1 = |ζ| = 1 − N , Γ2 = Γ1 −1 Γ3 = {|ζ| = ρ, ρ < 1, ρ = O(1)} , Γ4 = Γ3 The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the The previous RHP becomes Riemann- Hilbert σ problem 3 (1,t) t − 2 M (ζ) = M (ζ)ϕt |ζ| < ρ Francesco Mezzadri  N  σ3 (1,t) t 1 ζ − 2 M (ζ) = M (ζ) ϕt between Γ3 and Γ1 The problem 0 1

Where we are  −1 N  σ3 (1,t) t 1 (1 − ϕt )ζ − 2 1 Question to M (ζ) = M (ζ) ϕt between Γ1 and S answer 0 1

Motivations   σ3 (1,t) t 1 0 2 1 The Riemann- M (ζ) = M (ζ) −N −1 ϕt between S and Γ2 Hilbert ζ (1 − ϕt ) 1 approach  1 0 σ3 Comments M(1,t)(ζ) = Mt (ζ) ϕ 2 between Γ and Γ ζ−N 1 t 2 4 σ3 (1,t) t 2 −1 M (ζ) = M (ζ)ϕt |ζ| > ρ The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz (1,t) determinants Then M (ζ) satisfies the jump conditions and the Riemann- 1 −ϕ ζN  Hilbert ν(1)(ζ) = t |ζ| = ρ problem 0 1 Francesco  N  Mezzadri (1) 1 ζ ν (ζ) = ζ ∈ Γ1 The problem 0 1

Where we are   (1) 1 0 Question to ν (ζ) = −N ζ ∈ Γ2 answer −ζ 1 Motivations  1 0 ν(1)(ζ) = |ζ| = ρ−1 The Riemann- ϕ−1ζ−N 1 Hilbert t approach (1,t) λ
±1 Comments M has singularities at z = 1 − N :

σ3 (1,t) (1,t) − 2 M (ζ) = (Mλ + O(ζ − z))ϕt ζ → z σ3 (1,t) (1,t) −1 2 −1 M (ζ) = (M−λ + O(ζ − z ))ϕt ζ → z The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the Riemann- Hilbert problem Γ4 Francesco Γ λ Mezzadri 2 z = 1 − N Γ The problem 1 Γ3 D1 Where we are · ·−1 Question to z z answer

Motivations

The Riemann- Hilbert approach

Comments

n − 1 o D1 = ζ : |ζ − 1| ≤ N 2 The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the Riemann- Hilbert problem

Francesco (1) (1,t) Mezzadri • Outside D1, ν → I as N → ∞. Thus, M (ζ) → I .

The problem • We need to solve the RHP exactly in D1, then match it Where we are with the approximate solution outside. Question to answer • The substitution

Motivations ξ = N log ζ

The Riemann- Hilbert maps D1 to C. approach • The RHP inside D is mapped into Comments 1 The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the Riemann-  ξ Hilbert (2,t) (2,t) 1 e problem M (ξ) = M (ξ) ξ ∈ Γ + − 0 1 1 Francesco Mezzadri   (2,t) (2,t) 1 0 M+ (ξ) = M− (ξ) −ξ ξ ∈ Γ2 The problem −e 1 Where we are „ « β u −i 2 − (ξ−γ (λ)) σ3 Question to (2,t) (ξ−γN (λ)) N answer M (ξ) = (C+ + O(ξ − γN (λ)))e Motivations ξ → γN (λ) The Riemann- „ « β u Hilbert i + σ3 (2,t) (ξ+γ (λ))2 (ξ+γN (λ)) approach M (ξ) = (C− + O(ξ + γN (λ)))e N Comments ξ → −γN (λ) M(2,t)(ξ) = I + O(ξ−1)
ξ → ∞,

λ
where γN (λ) = N log 1 − N and β, u are related to the parameters in ϕ1 and ϕ2. The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the • The solution of the original RHP is Riemann- Hilbert ( − 1 problem I + O(N 2 ) ζ ∈ \ D ; (1,t) C 1 Francesco M (ζ) =  − 1  (2,t) Mezzadri I + O(N 2 ) M (ζ) ζ ∈ D1.

The problem Where we are (2,t) • The Toeplitz determinant DN−1 in terms of M Question to answer becomes Motivations ∞ Z 1 Z X d log ϕt The Riemann- log D = Nηˆ + kηˆ ηˆ + (ξ) Hilbert N−1 0 k −k dt approach k=1 0 iR Comments  −1  t  1 −eξ × tr M(2,t)(ξ) M(2,t)(ξ) dξdt e−ξ 1 − 1 +O(N 2 ) The derivative of the Riemann-zeta function, The Riemann-Hilbert approach Toeplitz determinants and the Riemann- Hilbert problem ξ (2,t) σ3 Francesco • If we define Y (ξ) = M e 2 , then Mezzadri

The problem ∂ξY (ξ) = A(ξ)Y (ξ) Where we are 3 + 3 − X Ai X Ai σ3 Question to A(ξ) = i + i + , answer (ξ − γN (x)) (ξ + γN (x)) 2 i=1 i=1 Motivations

The Riemann- + − Hilbert where the Ai s and Ai s are complicated functions of the approach parameters in the symbols ϕ1 and ϕ2. Comments • This equation has three multiple poles at ±γN (x) and ∞. The derivative of the Riemann-zeta function, Comments Toeplitz determinants and the Riemann- Hilbert problem

Francesco • Szeg˝o’stheorem does not apply because of essential Mezzadri singularities in the symbols. The problem • The RHP problem needs to be solved exactly in a Where we are neighbourhood of these points and matched with the Question to answer approximate solution outside. Motivations • Similar techniques as in the study of the double-scaling The Riemann- Hilbert limit of RMT but with a more complicated singularity approach structure (three essential singularities) Comments • Isomonodromic problem with three multiple poles