Moments of Traces for Circular -ensembles Tiefeng Jiang University - - PowerPoint PPT Presentation

Moments of Traces for Circular β-ensembles

Tiefeng Jiang

University of Minnesota

This is a joint work with Sho Matsumoto

September 19, 2014

Outline

Moments for Haar Unitary Matrices (D.E. Thm) Background for Circular β-Ensembles Moments for Circular β-Ensembles Proof by Jack Polynomials CLT

• 1. Moments for Haar Unitary Matrices

◮ What is Haar-invariant unitary matrix Γn? Mathematically, Γn : Haar probability measure on U(n) : set of n by n unitary matrices.

• 1. Moments for Haar Unitary Matrices

◮ What is Haar-invariant unitary matrix Γn? Mathematically, Γn : Haar probability measure on U(n) : set of n by n unitary matrices. Statistically, Assume the entries of Y = Yn×n are i.i.d. CN(0, 1). Two ways to generate such matrices

• 1. Moments for Haar Unitary Matrices

◮ What is Haar-invariant unitary matrix Γn? Mathematically, Γn : Haar probability measure on U(n) : set of n by n unitary matrices. Statistically, Assume the entries of Y = Yn×n are i.i.d. CN(0, 1). Two ways to generate such matrices 1) The matrix Q in QR (Gram-Schmidt) decomposition of Y

• 1. Moments for Haar Unitary Matrices

◮ What is Haar-invariant unitary matrix Γn? Mathematically, Γn : Haar probability measure on U(n) : set of n by n unitary matrices. Statistically, Assume the entries of Y = Yn×n are i.i.d. CN(0, 1). Two ways to generate such matrices 1) The matrix Q in QR (Gram-Schmidt) decomposition of Y 2) Γn

d

= Y(Y∗Y)−1/2

◮ Theorem (Diaconis and Evans: 2001) (a) a = (a1, · · · , ak), b = (b1, · · · , bk) with aj, bj ∈ {0, 1, 2, · · · }. For n ≥ k

j=1 jaj ∨ k j=1 jbj,

◮ Theorem (Diaconis and Evans: 2001) (a) a = (a1, · · · , ak), b = (b1, · · · , bk) with aj, bj ∈ {0, 1, 2, · · · }. For n ≥ k

j=1 jaj ∨ k j=1 jbj,

E  

k

• j=1

(Tr(Uj

n))aj(Tr(Uj n))bj

  = δab

k

• j=1

jajaj!

◮ Theorem (Diaconis and Evans: 2001) (a) a = (a1, · · · , ak), b = (b1, · · · , bk) with aj, bj ∈ {0, 1, 2, · · · }. For n ≥ k

j=1 jaj ∨ k j=1 jbj,

E  

k

• j=1

(Tr(Uj

n))aj(Tr(Uj n))bj

  = δab

k

• j=1

jajaj! (b) For j and k, E

• Tr(Uj

n)Tr(Uk n)

• = δjk · j ∧ n.

COE U(n)=CUE CSE O(n) Sp(n) Compact Groups Classical Circular Ensembles Circular Ensembles and Haar-invariant Matrices from Classical Compact Groups

COE U(n)=CUE CSE O(n) Sp(n) Compact Groups Classical Circular Ensembles Circular Ensembles and Haar-invariant Matrices from Classical Compact Groups Diaconis (2004) believes there is a good formula for COE and CSE

• 2. Background for Circular β-Ensembles

◮ Probability density function eiθ1, · · · , eiθn : eigenvalues of Haar-invariant unitary matrix. pdf: f(θ1, · · · , θn|β = 2)

• 2. Background for Circular β-Ensembles

◮ Probability density function eiθ1, · · · , eiθn : eigenvalues of Haar-invariant unitary matrix. pdf: f(θ1, · · · , θn|β = 2) f(θ1, · · · , θn|β) = Const ·

• 1≤j<k≤n

|eiθj − eiθk|β β > 0, θi ∈ [0, 2π)

• 2. Background for Circular β-Ensembles

◮ Probability density function eiθ1, · · · , eiθn : eigenvalues of Haar-invariant unitary matrix. pdf: f(θ1, · · · , θn|β = 2) f(θ1, · · · , θn|β) = Const ·

• 1≤j<k≤n

|eiθj − eiθk|β β > 0, θi ∈ [0, 2π) This model: circular β-ensemble by physicist Dyson for study of nuclear scattering data. Matrix model by Killip & Nenciu (04)

◮ Three Important Circular Ensembles COE (β = 1), CUE (β = 2), CSE (β = 4)

◮ Three Important Circular Ensembles COE (β = 1), CUE (β = 2), CSE (β = 4) Construction of COE and CUE U = Un×n : Haar unitary

◮ Three Important Circular Ensembles COE (β = 1), CUE (β = 2), CSE (β = 4) Construction of COE and CUE U = Un×n : Haar unitary U follows CUE

◮ Three Important Circular Ensembles COE (β = 1), CUE (β = 2), CSE (β = 4) Construction of COE and CUE U = Un×n : Haar unitary U follows CUE UTU follows COE

◮ Three Important Circular Ensembles COE (β = 1), CUE (β = 2), CSE (β = 4) Construction of COE and CUE U = Un×n : Haar unitary U follows CUE UTU follows COE CSE is similar but a bit involved (see Mehta)

◮ Three Important Circular Ensembles COE (β = 1), CUE (β = 2), CSE (β = 4) Construction of COE and CUE U = Un×n : Haar unitary U follows CUE UTU follows COE CSE is similar but a bit involved (see Mehta) Entries of CUE : roughly independent CN(0, 1) (Jiang, AP06)

◮ Three Important Circular Ensembles COE (β = 1), CUE (β = 2), CSE (β = 4) Construction of COE and CUE U = Un×n : Haar unitary U follows CUE UTU follows COE CSE is similar but a bit involved (see Mehta) Entries of CUE : roughly independent CN(0, 1) (Jiang, AP06) Entries of COE : roughly CN(0, 1) (but dependent) (Jiang, JMP09)

Moments for Circular β-Ensembles

Moments for Circular β-Ensembles

◮ Bad news from COE: Let Mn be COE. By elementary check E

• |Tr(Mn)|2

= 2n n + 1

Moments for Circular β-Ensembles

◮ Bad news from COE: Let Mn be COE. By elementary check E

• |Tr(Mn)|2

= 2n n + 1 Moments depend on n

Moments for Circular β-Ensembles

◮ Bad news from COE: Let Mn be COE. By elementary check E

• |Tr(Mn)|2

= 2n n + 1 Moments depend on n Later results: E

• |Tr(Mn)|2

not depend on n only at β = 2

Moments for Circular β-Ensembles

◮ Bad news from COE: Let Mn be COE. By elementary check E

• |Tr(Mn)|2

= 2n n + 1 Moments depend on n Later results: E

• |Tr(Mn)|2

not depend on n only at β = 2 This suggest: moments for general β-ensemble depend on n

◮ Notation λ = (λ1, λ2, · · · ) : partition

◮ Notation λ = (λ1, λ2, · · · ) : partition |λ| = λ1 + λ2 + · · · : weight

◮ Notation λ = (λ1, λ2, · · · ) : partition |λ| = λ1 + λ2 + · · · : weight mi(λ) : multi of i in (λ1, λ2, · · · )

◮ Notation λ = (λ1, λ2, · · · ) : partition |λ| = λ1 + λ2 + · · · : weight mi(λ) : multi of i in (λ1, λ2, · · · ) l(λ)= # of positive λi in λ : length

◮ Notation λ = (λ1, λ2, · · · ) : partition |λ| = λ1 + λ2 + · · · : weight mi(λ) : multi of i in (λ1, λ2, · · · ) l(λ)= # of positive λi in λ : length zλ =

• i≥1

imi(λ)mi(λ)!

◮ Notation λ = (λ1, λ2, · · · ) : partition |λ| = λ1 + λ2 + · · · : weight mi(λ) : multi of i in (λ1, λ2, · · · ) l(λ)= # of positive λi in λ : length zλ =

• i≥1

imi(λ)mi(λ)! pλ = l(λ)

i=1 pλi, where pk(x1, x2, · · · ) = xk 1 + xk 2 + · · ·

◮ Notation λ = (λ1, λ2, · · · ) : partition |λ| = λ1 + λ2 + · · · : weight mi(λ) : multi of i in (λ1, λ2, · · · ) l(λ)= # of positive λi in λ : length zλ =

• i≥1

imi(λ)mi(λ)! pλ = l(λ)

i=1 pλi, where pk(x1, x2, · · · ) = xk 1 + xk 2 + · · ·

λ = (3, 2, 2) : |λ| = 7, m2(λ) = 2, m3(λ) = 1, l(λ) = 3, pλ = (

i λ3 i ) · ( i λ2 i )2

α > 0, K ≥ 1, n ≥ 1, define A =

• 1 −

|α − 1| n − K + α δ(α ≥ 1) K B =

• 1 +

|α − 1| n − K + α δ(α < 1) K

α > 0, K ≥ 1, n ≥ 1, define A =

• 1 −

|α − 1| n − K + α δ(α ≥ 1) K B =

• 1 +

|α − 1| n − K + α δ(α < 1) K Let θ1, · · · , θn ∼ f(θ1, · · · , θn|β), α = 2/β. Zn = (eiθ1, · · · , eiθn), pµ(Zn) = pµ(eiθ1, · · · , eiθn)

Theorem (a) If n ≥ K = |µ|, then A ≤ E

• |pµ(Zn)|2

αl(µ)zµ ≤ B

Theorem (a) If n ≥ K = |µ|, then A ≤ E

• |pµ(Zn)|2

αl(µ)zµ ≤ B (b) If |µ| = |ν|, then E

• pµ(Zn)pν(Zn)
• = 0.

Theorem (a) If n ≥ K = |µ|, then A ≤ E

• |pµ(Zn)|2

αl(µ)zµ ≤ B (b) If |µ| = |ν|, then E

• pµ(Zn)pν(Zn)
• = 0.

If µ = ν and n ≥ K = |µ| ∨ |ν|, then

• E
• pµ(Zn)pν(Zn)
• ≤ max{|A−1|, |B−1|}·α(l(µ)+l(ν))/2(zµzν)1/2

Theorem (a) If n ≥ K = |µ|, then A ≤ E

• |pµ(Zn)|2

αl(µ)zµ ≤ B (b) If |µ| = |ν|, then E

• pµ(Zn)pν(Zn)
• = 0.

If µ = ν and n ≥ K = |µ| ∨ |ν|, then

• E
• pµ(Zn)pν(Zn)
• ≤ max{|A−1|, |B−1|}·α(l(µ)+l(ν))/2(zµzν)1/2

(c) ∃ C = C(β) s.t. ∀m ≥ 1, n ≥ 2

• E
• |pm(Zn)|2

− n

• ≤ Cn32nβ

m1∧β

Take β = 2, then A = B = 1. We recover ◮ Theroem (Diaconis and Evans: 2001) a = (a1, · · · , ak), b = (b1, · · · , bk) with aj, bj ∈ {0, 1, 2, · · · }. For n ≥ k

j=1 jaj ∨ k j=1 jbj,

E  

k

• j=1

(Tr(Uj

n))aj(Tr(Uj n))bj

  = δab

k

• j=1

jajaj!

Corollary ∀ β > 0, (a) lim

n→∞ E

• pµ(Zn)pν(Zn)
• = δµν

2 β l(µ) zµ;

Corollary ∀ β > 0, (a) lim

n→∞ E

• pµ(Zn)pν(Zn)
• = δµν

2 β l(µ) zµ; (b) lim

m→∞ E

• |pm(Zn)|2

= n for any n ≥ 2.

Corollary µ = ν : K = |µ| ∨ |ν|. If n ≥ 2K, then (a)

• E[|pµ(Zn)|2]

αl(µ)zµ − 1

• ≤ 6|1 − α|K

n ;

Corollary µ = ν : K = |µ| ∨ |ν|. If n ≥ 2K, then (a)

• E[|pµ(Zn)|2]

αl(µ)zµ − 1

• ≤ 6|1 − α|K

n ; (b)

• E
• pµ(Zn)pν(Zn)
• ≤ 6|1 − α|K

n · α(l(µ)+l(ν))/2(zµzν)1/2.

◮ Exact formula The exact formula gives E[|p1(Zn)|2] = 2 β n n − 1 + 2β−1

◮ Exact formula The exact formula gives E[|p1(Zn)|2] = 2 β n n − 1 + 2β−1 =     

2n n+1,

if β = 1 1, if β = 2

n 2n−1,

if β = 4

◮ Exact formula The exact formula gives E[|p1(Zn)|2] = 2 β n n − 1 + 2β−1 =     

2n n+1,

if β = 1 1, if β = 2

n 2n−1,

if β = 4 Exact formula is given next

Proofs by Jack Polynomial

◮ Jack Polynomial Jack polynomial J(α)

λ

= J(α)

λ (x1, · · · , xn) is symmetric in x1, · · · , xn

Proofs by Jack Polynomial

◮ Jack Polynomial Jack polynomial J(α)

λ

= J(α)

λ (x1, · · · , xn) is symmetric in x1, · · · , xn

α = 1, it is Schur polynomial

Proofs by Jack Polynomial

◮ Jack Polynomial Jack polynomial J(α)

λ

= J(α)

λ (x1, · · · , xn) is symmetric in x1, · · · , xn

α = 1, it is Schur polynomial α = 2, it is Zonal polynomial

Proofs by Jack Polynomial

◮ Jack Polynomial Jack polynomial J(α)

λ

= J(α)

λ (x1, · · · , xn) is symmetric in x1, · · · , xn

α = 1, it is Schur polynomial α = 2, it is Zonal polynomial α = 1/2, it is Zonal spherical function

Proofs by Jack Polynomial

◮ Jack Polynomial Jack polynomial J(α)

λ

= J(α)

λ (x1, · · · , xn) is symmetric in x1, · · · , xn

α = 1, it is Schur polynomial α = 2, it is Zonal polynomial α = 1/2, it is Zonal spherical function Orthogonal property: Zn = (eiθ1, . . . , eiθn)

Proofs by Jack Polynomial

◮ Jack Polynomial Jack polynomial J(α)

λ

= J(α)

λ (x1, · · · , xn) is symmetric in x1, · · · , xn

α = 1, it is Schur polynomial α = 2, it is Zonal polynomial α = 1/2, it is Zonal spherical function Orthogonal property: Zn = (eiθ1, . . . , eiθn)

• [0,2π)n J(α)

λ (Zn)J(α) µ (¯

Zn)

• 1≤p<q≤n

|eiθp − eiθq|2/α dθ1 · · · dθn = δλµ · δ(l(λ) ≤ n) · explicit const

Write J(α)

λ

=

• ρ:|ρ|=|λ|

θλ

ρ(α)pρ

pρ =

• λ:|λ|=|ρ|

Θλ

ρ(α)J(α) λ

Write J(α)

λ

=

• ρ:|ρ|=|λ|

θλ

ρ(α)pρ

pρ =

• λ:|λ|=|ρ|

Θλ

ρ(α)J(α) λ

For |µ| = |ν| = K, E

• pµ(Zn)pν(Zn)
• =
• λ⊢K: l(λ)≤n

Θλ

µ(α)Θλ ν(α)E(J(α) λ J(α) λ )

Use explicit form of E(J(α)

λ J(α) λ )

relationship between θλ

ρ(α) and Θλ ρ(α)

Use explicit form of E(J(α)

λ J(α) λ )

relationship between θλ

ρ(α) and Θλ ρ(α)

we have E

• pµ(Zn)pν(Zn)
• =

αl(µ)+l(ν)zµzν

• λ⊢K: l(λ)≤n

θλ

µ(α)θλ ν(α)

Cλ(α) N α

λ (n)

Use explicit form of E(J(α)

λ J(α) λ )

relationship between θλ

ρ(α) and Θλ ρ(α)

we have E

• pµ(Zn)pν(Zn)
• =

αl(µ)+l(ν)zµzν

• λ⊢K: l(λ)≤n

θλ

µ(α)θλ ν(α)

Cλ(α) N α

λ (n)

Cλ(α) =

• (i,j)∈λ
• (α(λi − j) + λ′

j − i + 1)(α(λi − j) + λ′ j − i + α)

Use explicit form of E(J(α)

λ J(α) λ )

relationship between θλ

ρ(α) and Θλ ρ(α)

we have E

• pµ(Zn)pν(Zn)
• =

αl(µ)+l(ν)zµzν

• λ⊢K: l(λ)≤n

θλ

µ(α)θλ ν(α)

Cλ(α) N α

λ (n)

Cλ(α) =

• (i,j)∈λ
• (α(λi − j) + λ′

j − i + 1)(α(λi − j) + λ′ j − i + α)

• N α

λ (n) =

• (i,j)∈λ

n + (j − 1)α − (i − 1) n + jα − i Young diagram

Main proof: play Cλ(α) play N α

λ (n)

use orthogonal relations of θλ

µ(α)

◮ Examples E[|p1(Zn)|4] = 2nα2(n2 + 2(α − 1)n − α) (n + α − 1)(n + α − 2)(n + 2α − 1)

◮ Examples E[|p1(Zn)|4] = 2nα2(n2 + 2(α − 1)n − α) (n + α − 1)(n + α − 2)(n + 2α − 1) =       

8(n2+2n−2) (n+1)(n+3) ,

if β = 1 2, if β = 2

2n2−2n−1 (2n−1)(2n−3),

if β = 4

E

• p2(Zn)p1(Zn)2

E

• p2(Zn)p1(Zn)2
• =

2α2(α − 1)n (n + α − 1)(n + 2α − 1)(n + α − 2)

E

• p2(Zn)p1(Zn)2
• =

2α2(α − 1)n (n + α − 1)(n + 2α − 1)(n + α − 2) =       

8 (n+1)(n+3),

if β = 1 0, if β = 2

−1 (2n−1)(2n−3),

if β = 4

Similar results also hold for Macdonald polynomials

Central Limit Theorem

Theorem (eiθ1, · · · , eiθn): any β-circular ensemble. g(z) = m

k=0 ckzk

Xn := n

j=1 g(eiθj).

Then Xn − µn → CN(0, σ2) where µn = nc0 and σ2 = 2 β

m

• k=1

k|ck|2.

Central Limit Theorem

Theorem (eiθ1, · · · , eiθn): any β-circular ensemble. g(z) = m

k=0 ckzk

Xn := n

j=1 g(eiθj).

Then Xn − µn → CN(0, σ2) where µn = nc0 and σ2 = 2 β

m

• k=1

k|ck|2. Next consider g(z) = ∞

k=0 ckzk for β = 1, 4;

β = 2 by D. E.

Theorem Zn := (eiθ1, · · · , eiθn): CSE (β = 4) {aj, bj}: ∞

j=1(j log j)(|aj|2 + |bj|2) < ∞

σ2 := ∞

j=1 j(|aj|2 + |bj|2)

Theorem Zn := (eiθ1, · · · , eiθn): CSE (β = 4) {aj, bj}: ∞

j=1(j log j)(|aj|2 + |bj|2) < ∞

σ2 := ∞

j=1 j(|aj|2 + |bj|2)

Then, ∞

j=1

• aj tr(Zj

n) + bj tr(Zj n)

• → U + iV

where (U, V) ∼ N2(0, Σ), Σ = 1 4 ∞

j=1 j|aj + ¯

bj|2 2 · Im(∞

j=1 jajbj)

2 · Im(∞

j=1 jajbj)

∞

j=1 j|aj − ¯

bj|2

• .

Idea to prove CLT: estimate 2nd moment to truncate infinite series to finite sum. Then use moment ineq.

Idea to prove CLT: estimate 2nd moment to truncate infinite series to finite sum. Then use moment ineq. Proposition β = 4. ∃ K > 0 s.t. E[| n

j=1 eimθj|2] ≤ Km log(m + 1) for m ≥ 1,

n ≥ 2.

E[|

n

• j=1

eimθj|2] = 1 4m2

• λ⊢m:l(λ)≤n

(θλ

(m))2

Cλ Nλ(n) where

E[|

n

• j=1

eimθj|2] = 1 4m2

• λ⊢m:l(λ)≤n

(θλ

(m))2

Cλ Nλ(n) where Nλ(n) =

• (i,j)∈λ
• 1 +

1 2n − 2i + j

• .

E[|

n

• j=1

eimθj|2] = 1 4m2

• λ⊢m:l(λ)≤n

(θλ

(m))2

Cλ Nλ(n) where Nλ(n) =

• (i,j)∈λ
• 1 +

1 2n − 2i + j

• .

θλ

(m)(α) =

• (i,j)∈λ

(i,j)=(1,1)

(1 2(j − 1) − (i − 1)),

E[|

n

• j=1

eimθj|2] = 1 4m2

• λ⊢m:l(λ)≤n

(θλ

(m))2

Cλ Nλ(n) where Nλ(n) =

• (i,j)∈λ
• 1 +

1 2n − 2i + j

• .

θλ

(m)(α) =

• (i,j)∈λ

(i,j)=(1,1)

(1 2(j − 1) − (i − 1)), Cλ(α) =

• (i,j)∈λ
• (1

2(λi − j) + λ′

j − i + 1)

×(1 2(λi − j) + λ′

j − i + 1

2)

• ,

Main contribution: λ = (r, s, 1m−r−s) (θλ

(m))2

Cλ = (m − r − s + 1) (m + r − s + 1)(m + r − s)(m − r + s)(m − r + s − 1) ×22r−2s[(r − 1)!]2(2r − 2s + 1)·(2s − 2)! [(s − 1)!]2(2r − 1)! .

Main contribution: λ = (r, s, 1m−r−s) (θλ

(m))2

Cλ = (m − r − s + 1) (m + r − s + 1)(m + r − s)(m − r + s)(m − r + s − 1) ×22r−2s[(r − 1)!]2(2r − 2s + 1)·(2s − 2)! [(s − 1)!]2(2r − 1)! . 1 K · m

• (n − r + 1)(n − s + 1)

≤ Nλ(n) ≤ K · m

• (n − r + 1)(n − s + 1)

. if m ≥ n. Done!

Thanks for your patience!