Random Matrices And Random Partitions Normal Convergence, Volume 1.pdf

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Random Matrices and Random Partitions Normal Convergence

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World Scientific Series on Probability Theory and Its Applications Series Editors: Zhenghu Li (Beijing Normal University, China) Yimin Xiao (Michigan State University, USA)

Vol. 1 Random Matrices and Random Partitions: Normal Convergence by Zhonggen Su (Zhejiang University, China)

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World Scientific Series on Probability Theory and Its Applications Volume 1

Random Matrices and Random Partitions Normal Convergence

Zhonggen Su Zhejiang University, China

World Scientific NEW JERSEY

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LONDON

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SINGAPORE

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BEIJING

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SHANGHAI

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HONG KONG

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TA I P E I

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CHENNAI

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Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Su, Zhonggen. Random matrices and random partitions normal convergence / by Zhonggen Su (Zhejiang University, China). pages cm. -- (World scientific series on probability theory and its applications ; volume 1) Includes bibliographical references and index. ISBN 978-9814612227 (hardcover : alk. paper) 1. Random matrices. 2. Probabilities. I. Title. QA273.43.S89 2015 519.2'3--dc23 2015004842 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

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Preface

This book is intended to provide an introduction to remarkable probability limit theorems in random matrices and random partitions, which look rather different at a glance but have many surprising similarities from a probabilistic viewpoint. Both random matrices and random partitions play a ubiquitous role in mathematics and its applications. There have been a great deal of research activities around them, and an enormous exciting advancement had been seen in the last three decades. A couple of excellent and big books have come out in recent years. However, the work on these two objects are so rich and colourful in theoretic results, practical applications and research techniques. No one book is able to cover all existing materials. Needless to say, these are rapidly developing and ever-green research fields. Only recently, a number of new interesting works emerged in literature. For instance, based on Johansson’s work on deformed Gaussian unitary ensembles, two groups led respectively by Erd¨ os-Yau and Tao-Vu successfully solved, around 2010, the long-standing conjecture of Dyson-Gaudin-Mehta-Wigner’s bulk universality in random matrices by developing new techniques like the comparison principles and rigidity properties. Another example is that with the help of concepts of determinantal point processes coined by Borodin and Olshanski, around 2000, in the study of symmetric groups and random partitions, a big breakthrough has been made in understanding universality properties of random growth processes. Each of them is worthy of a new book. This book is mainly concerned with normal convergence, namely central limit theorems, of various statistics from random matrices and random partitions as the model size tends to infinity. For the sake of writing and learning, we shall only focus on the simplest models among which are circular

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unitary ensemble, Gaussian unitary ensemble, random uniform partitions and random Plancherel partitions. As a matter of fact, many of the results addressed in this book are found valid for more general models. This book consists of three parts as follows. We shall first give a brief survey on normal convergence in Chapter 1. It includes the well-known laws of large numbers and central limit theorems for independent identically distributed random variables and a few methods widely used in dealing with normal convergence. In fact, the central limit theorems are arguably regarded as one of the most important universality principles in describing laws of random phenomena. Most of the materials can be found in any standard probability theory at graduate level. Because neither the eigenvalues of a random matrix with all entries independent nor the parts of random partitions are independent of each other, we need new tools to treat statistics of dependent random variables. Taking this into account, we shall simply review the central limit theorems for martingale difference sequences and Markov chains. Besides, we shall review some basic concepts and properties of convergence of random processes. The statistic of interest is sometimes a functional of certain random process in the study of random matrices and random partitions. We will be able to make use of functional central limit theorems if the random processes under consideration is weakly convergent. Even under the stochastic equicontinuity condition, a slightly weaker condition than uniform tightness, the Gikhmann-Skorohod theorem can be used to guarantee convergence in distribution for a wide class of integral functionals. In Chapters 2 and 3 we shall treat circular unitary ensemble and Gaussian unitary ensemble respectively. A common feature is that there exists an explicit joint probability density function for eigenvalues of each matrix model. This is a classic result due to Weyl as early as the 1930s. Such an explicit formula is our starting point and this makes delicate analysis possible. Our focus is upon the second-order fluctuation, namely asymptotic distribution of a certain class of linear functional statistics of eigenvalues. Under some smooth conditions, a linear eigenvalue statistic satisfies the √ central limit theorem without normalizing constant n, which appears in classic L´evy-Feller central limit theorem for independent identically distributed random variables. On the other hand, either indicator function or logarithm function does not satisfy the so-called smooth condition. It turns out that the number of eigenvalues in an interval and the logarithm of characteristic polynomials do still satisfy the central limit theorem after √ suitably normalized by log n. The log n-phenomena is worthy of more

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ix

attention since it will also appear in the study of other similar models. In addition to circular and Gaussian unitary ensembles, we shall consider their extensions like circular β matrices and Hermite β matrices where β > 0 is a model parameter. These models were introduced and studied at length by Dyson in the early 1960s to investigate energy level behaviors in complex dynamic systems. A remarkable contribution at this direction is that there is a five (resp. three) diagonal sparse matrix model representing circular β ensemble (resp. Hermite β ensemble). In Chapters 4 and 5 we shall deal with random uniform partitions and random Plancherel partitions. The study of integer partitions dates back to Euler as early as the 1750s, who laid the foundation of partition theory by determining the number of all distinct partitions of a natural number. We will naturally produce a probability space by assigning a probability to each partition of a natural number. Uniform measure and Plancherel measure are two best-studied objects. Young diagram and Young tableau are effective geometric representation in analyzing algebraic, combinatorial and probabilistic properties of a partition. Particularly interesting, there exists a nonrandom limit shape (curve) for suitably scaled Young diagrams under both uniform and Plancherel measure. This is a kind of weak law of large numbers from the probabilistic viewpoint. To proceed, we shall further investigate the second-order fluctuation of a random Young diagram around its limit shape. We need to treat separately three different cases: at the edge, in the bulk and integrated. It is remarkable that Gumbel law, normal law and Tracy-Widom law can be simultaneously found in the study of random integer partitions. A basic strategy of analysis is to construct a larger probability space (grand ensemble) and to use the conditioning argument. Through enlarging probability space, we luckily produce a family of independent geometric random variables and a family of determinantal point processes respectively. Then a lot of well-known techniques and results are applicable. Random matrices and random partitions are at the interface of many science branches and they are fast-growing research fields. It is a formidable and confusing task for a new learner to access the research literature, to acquaint with terminologies, to understand theorems and techniques. Throughout the book, I try to state and prove each theorem using language and ways of reasoning from standard probability theory. I hope it will be found suitable for graduate students in mathematics or related sciences who master probability theory at graduate level and those with interest in these fields. The choice of results and references is to a large

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extent subjective and determined by my personal point of view and taste of research. The references at the end of the book are far from exhaustive and in fact are rather limited. There is no claim for completeness. This book started as a lecture note used in seminars on random matrices and random partitions for graduate students in the Zhejiang University over these years. I would like to thank all participants for their attendance and comments. This book is a by-product of my research project. I am grateful to the National Science Foundation of China and Zhejiang Province for their generous support in the past ten years. I also take this opportunity to express a particular gratitude to my teachers, past and present, for introducing me to the joy of mathematics. Last, but not least, I wish to thank deeply my family for their kindness and love which is indispensable in completing this project. I apologize for all the omissions and errors, and invite the readers to report any remarks, mistakes and misprints. Zhonggen Su Hangzhou December 2014

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Contents

Preface 1.

Normal Convergence 1.1 1.2 1.3 1.4

2.

Classical central limit theorems . . The Stein method . . . . . . . . . The Stieltjes transform method . Convergence of stochastic processes

. . . .

. . . .

. . . .

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. . . .

Introduction . . . . . . . . . . . . Symmetric groups and symmetric Linear functionals of eigenvalues Five diagonal matrix models . . . Circular β ensembles . . . . . . .

. . . . . . . . polynomials . . . . . . . . . . . . . . . . . . . . . . . .

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33 38 47 57 78 89

Introduction . . . . . . . . . . . . . Fluctuations of Stieltjes transforms . Number of eigenvalues in an interval Logarithmic law . . . . . . . . . . . Hermite β ensembles . . . . . . . . .

Random Uniform Partitions 4.1 4.2 4.3

1 19 25 29 33

Gaussian Unitary Ensemble 3.1 3.2 3.3 3.4 3.5

4.

1

Circular Unitary Ensemble 2.1 2.2 2.3 2.4 2.5

3.

vii

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. 89 . 98 . 112 . 125 . 136 153

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 153 Grand ensembles . . . . . . . . . . . . . . . . . . . . . . . 160 Small ensembles . . . . . . . . . . . . . . . . . . . . . . . 169 xi

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4.4 4.5 5.

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A functional central limit theorem . . . . . . . . . . . . . 180 Random multiplicative partitions . . . . . . . . . . . . . . 200

Random Plancherel Partitions 5.1 5.2 5.3 5.4 5.5

Introduction . . . . . . . Global fluctuations . . . Fluctuations in the bulk Berry-Esseen bounds for Determinantal structure

207 . . . . . . . . . . . . . . . . . . . . . . . . . . . character ratios . . . . . . . . .

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207 220 237 244 253

Bibliography

261

Index

267

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Chapter 1

Normal Convergence

1.1

Classical central limit theorems

Throughout the book, unless otherwise specified, we assume that (Ω, A, P ) is a large enough probability space to support all random variables of study. E will denote mathematical expectation with respect to P . Let us begin with Bernoulli’s law, which is widely recognized as the first mathematical theorem in the history of probability theory. In modern terminology, the Bernoulli law reads as follows. Assume that ξn , n ≥ 1 is a sequence of independent and identically distributed (i.i.d.) random variables, P (ξn = 1) = p and P (ξn = 0) = 1 − p, where 0 < p < 1. Denote Pn Sn = k=1 ξk . Then we have Sn P −→ p, n

n → ∞.

In other words, for any ε > 0,   S n − p > ε → 0, P n

(1.1)

n → ∞.

It is this law that first provide a mathematically rigorous interpretation about the meaning of probability p that an event A occurs in a random experiment. To get a feeling of the true value p (unknown), what we need to do is to repeat independently a trial n times (n large enough) and to count the number of A occurring. According to the law, the larger n is, the higher the precision is. Having the Bernoulli law, it is natural to ask how accurate the frequency Sn /n can approximate the probability p, how many times one should repeat the trial to attain the specified precision, that is, how big n should be. With this problem in mind, De Moivre considered the case p = 1/2 and 1

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proved the following statement: Z b   Sn − n 2 1 P a ≤ 1√ 2 ≤ b ≈ √ e−x /2 dx. n 2π a 2

(1.2)

Later on, Laplace further extended the work of De Moivre to the case p 6= 1/2 to obtain Z b   2 1 Sn − np p √ ≤b ≈ P a≤ e−x /2 dx. (1.3) 2π a np(1 − p) Formulas (1.2) and (1.3) are now known as De Moivre-Laplace central limit theorem (CLT). p Note ESn = np, V ar(Sn ) = np(1 − p). So (Sn − np)/ np(1 − p) is a normalized random variable with mean zero and variance one. Denote √ −x2 /2 φ(x) = e / 2π, x ∈ R. This is a very nice function from the viewpoint of function analysis. It is sometimes called bell curve since its graph looks like a bell, as shown in Figure 1.1.

Fig. 1.1

Bell curve

The Bernoulli law and De Moivre-Laplace CLT have become an indispensable part of our modern daily life. See Billingsley (1999a, b), Chow (2003), Chung (2000), Durrett (2010) and Fischer (2011) for a history of the central limit theorem and the link to modern probability theory. But what is the proof? Any trick? Let us turn to De Moivre’s original proof of (1.2). To control the left hand side of (1.2), De Moivre used the binomial formula   n 1 P (Sn = k) = k 2n

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and invented together with Stirling the well-known Stirling formula (it actually should be called De Moivre-Stirling formula) √ n! = nn e−n 2πn(1 + o(1)). √ Setting k = n/2 + nxk /2, where a ≤ xk ≤ b, we have √ S − n  nn e−n 2πn(1 + o(1)) 1 n 2 p √ P 1√ = xk = n · 2 k k e−k (n − k)n−k e−(n−k) 2πk 2π(n − k) 2 n 2 1 e−xk /2 (1 + o(1)). = √ 2πn Taking sum over k yields the integral of the right hand side of (1.2). Given a random variable X, denote its distribution function FX (x) under P . Let X, Xn , n ≥ 1 be a sequence of random variables. If for each continuity point x of FX , FXn (x) → FX (x),

n → ∞, d

then we say Xn converges in distribution to X, and simply write Xn −→ X. In this terminology, (1.3) is written as S − np d p n −→ N (0, 1), n → ∞, np(1 − p) where N (0, 1) stands for a standard normal random variable. As the reader may notice, the Bernoulli law only deals with frequency and probability, i.e., Bernoulli random variables. However, in practice people are faced with a lot of general random variables. For instance, measure length of a metal rod. Its length, µ, is intrinsic and unknown. How do we get to know the value of µ? Each measurement is only a realization of µ. Suppose that we measure repeatedly the metal rod n times and record Pn the observed values ξ1 , ξ2 , · · · , ξn . It is believed that k=1 ξk /n give us a good feeling of how long the rod is. It turns out that a claim similar to the Bernoulli law is also valid for general cases. Precisely speaking, assume that ξ is a random variable with mean µ. ξn , n ≥ 1 is a sequence of i.i.d. Pn copy of ξ. Let Sn = k=1 ξk . Then Sn P −→ µ, n → ∞. (1.4) n This is called the Khinchine law of large numbers. It is as important as the Bernoulli law. As a matter of fact, it provides a solid theoretic support for a great deal of activity in daily life and scientific research. The proof of (1.4) is completely different from that of (1.1) since we do not know the exact distribution of ξk . To prove (1.4), we need to invoke

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the following Chebyshev inequality. If X is a random variable with finite mean µ and variance σ 2 , then for any positive x > 0 P (|X − µ| > x) ≤

σ2 . x2

In general, we have Ef (X) , f (x) where f : R 7→ R is a nonnegative nondecreasing function. We remark that the Chebyshev inequalities have played a fundamental role in proving limit theorems like the law of large numbers. Having (1.4), we next naturally wonder what the second order fluctuation is of Sn /n around µ? In other words, is there a normalizing constant an → ∞ such that an (Sn − nµ)/n converges in distribution to a certain random variable? What is the distribution of the limit variable? To attack these problems, we need to develop new tools and techniques since the De Moivre argument using binomial distribution is no longer applicable. Given a random variable X with distribution function FX , define for every t ∈ R, P (X > x) ≤

ψX (t) = EeitX Z = eitx dFX (x). R

Call ψX (t) the characteristic function of X. This is a Fourier transform of FX (x). In particular, if X has a probability density function pX (x), then Z ψX (t) = eitx pX (x)dx; R

while if X takes only finitely or countably many values, P (X = xk ) = pk , k ≥ 1, then ∞ X ψX (t) = eitxk pk . k=1

Note the characteristic function of any random variable is always well defined no matter whether its expectation exists. Example 1.1. (i) If X is a normal random variable with mean µ and 2 2 variance σ 2 , then ψX (t) = eiµt−σ t /2 ; (ii) If X is a Poisson random variable with parameter λ, then ψX (t) = it eλ(e −1) ; (iii) If X is a standard Cauchy random variable, then ψX (t) = e−|t| .

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Some basic properties are listed below. (i) ψX (0) = 1, |ψX (t)| ≤ 1 for any t ∈ R. (ii) ψX (t) is uniformly continuous in any finite closed interval on R. (iii) ψX (t) is nonnegative definite. According to Bochner’s theorem, if any function satisfying (i), (ii) and (iii) above must be a characteristic function of a random variable. (iv) ψX (t) = ψX (−t) for any t ∈ R. (v) If E|X|k < ∞, then ψX (t) is k times differentiable, and (k)

ψX (0) = ik EX k .

(1.5)

Hence we have the Taylor expansion at zero: for any k ≥ 1 ψX (t) =

k X il ml l=0

l!

tl + o(|t|k )

(1.6)

and ψX (t) =

k−1 X l l=0

|t|k i ml l t + βk θk , l! k!

where ml := ml (X) = EX l , βk = E|X|k , |θk | ≤ 1. (vi) If X and Y are independent random variables, then ψX+Y (t) = ψX (t)ψY (t). Obviously, this product formula can be extended to any finitely many independent random variables. (vii) The distribution function of a random variable is uniquely determined by its characteristic function. Specifically speaking, we have the following inversion formula: for any FX -continuity points x1 and x2 Z T −itx1 1 e − e−itx2 FX (x2 ) − FX (x1 ) = lim ψX (t)dt. T →∞ 2π −T it In particular, if ψX (t) is absolutely integrable, then X has density function Z ∞ 1 pX (x) = e−itx ψX (t)dt. 2π −∞ P∞ P∞ On the other hand, if ψX (t) = k=1 ak eitxk with ak > 0 and k=1 ak = 1, then X is a discrete random variable, and P (X = xk ) = ak ,

k = 1, 2, · · · .

In addition to basic properties above, we have the following L´evy continuity theorem, which will play an important role in the study of convergence in distribution.

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d

Theorem 1.1. (i) Xn −→ X if and only if ψXn (t) → ψX (t) for any t ∈ R. (ii) If ψXn converges pointwise to a function ψ, and ψ is continuous at t = 0, then ψ(t) is a characteristic function of some random variable, say d

X, and so Xn −→ X. Having the preceding preparation, we can easily obtain the following CLT for sums of independent random variables. Theorem 1.2. Let ξ, ξn , n ≥ 1 be a sequence of i.i.d. random variables Pn with mean µ and variance σ 2 . Let Sn = k=1 ξk . Then Sn − nµ d √ −→ N (0, 1), σ n

n → ∞.

(1.7)

This is often referred to as Feller-L´evy CLT. Its proof is purely analytic. For sake of comparison, we quickly review the proof. Proof. Without loss of generality, we may and do assume µ = 0, σ 2 = 1. By hypothesis it follows   t n . ψSn /√n (t) = ψξ √ n Also, using (1.6) yields  t  1 t2 =1− ψξ √ +O n 2n n for each t. Hence we have  1 n  t2 +O ψSn /√n (t) = 1 − 2n n 2 → e−t /2 . By Theorem 1.1 and (i) of Example 1.1, we conclude the desired (1.7).  Remark 1.1. Under the assumption that ξn , n ≥ 1 are i.i.d. random variables, the condition σ 2 < ∞ is also necessary for (1.7) to hold. See Chapter 10 of Ledoux and Talagrand (2011) for a proof. In many applications, it is restrictive to require that ξn , n ≥ 1 are i.i.d. random variables. Therefore we need to extend the Feller-L´evy CLT to the non-i.i.d. cases. In fact, there have been a great deal of work toward this direction. For the sake of reference, we will review below some most commonly used CLT, including the Lindeberg-Feller CLT for independent not necessarily identically distributed random variables, the martingale CLT, the CLT for ergodic stationary Markov chains.

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Assume that ξn , n ≥ 1 is a sequence of independent random variables, Pn Pn 2 Eξn = µn , V ar(ξn ) = σn2 < ∞. Let Sn = k=1 ξk , Bn = k=1 σk . Assume further that Bn → ∞. Introduce the following two conditions. Feller condition: 1 max σ 2 → 0. Bn 1≤k≤n k Lindeberg condition: for any ε > 0 n 1 X E(ξk − µk )2 1(|ξk −µk |≥ε√Bn ) → 0. Bn k=1

Obviously, Feller condition is a consequence of Lindeberg condition. Moreover, we have the following Lindeberg-Feller CLT. Theorem 1.3. Under Feller condition, the ξn satisfies the CLT, that is Pn Sn − k=1 µk d √ −→ N (0, 1), n → ∞ Bn if and only if Lindeberg condition holds. It is easy to see that if there is a δ > 0 such that 1

n X

1+δ/2

Bn

E|ξk |2+δ → 0,

(1.8)

k=1

then Lindeberg condition is satisfied. The condition (1.8) is sometimes called Lyapunov condition. Corollary 1.1. Assume that ξn , n ≥ 1 is a sequence of independent Bernoulli random variables, P (ξn = 1) = pn , P (ξn = 0) = 1 − pn . If P∞ n=1 pn (1 − pn ) = ∞, then Pn (ξk − pk ) d pPk=1 −→ N (0, 1), n → ∞. (1.9) n p (1 − p ) k k=1 k The Lindeberg-Feller theorem has a wide range of applications. In particular, it implies that the normal law exists universally in nature and human society. For instance, the error in measurement might be caused by a large number of independent factors. Each factor contributes only a very small part, but none plays a significant role. Then the total error obeys approximately a normal law.

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Next turn to the martingale CLT. First, recall some notions and basic properties of martingales. Assume that An , n ≥ 0 is a sequence of nondecreasing sub-sigma fields of A. Let Xn , n ≥ 0 be a sequence of random variable with Xn ∈ An and E|Xn | < ∞. If for each n ≥ 1 E(Xn |An−1 ) = Xn−1

a.e.

we call {Xn , An , n ≥ 0} a martingale. If An = σ{X0 , X1 , · · · , Xn }, we simply write {Xn , n ≥ 0} a martingale. If {Xn , An , n ≥ 0} is a martingale, setting dn := Xn −Xn−1 , then {dn , An , n ≥ 0} forms a martingale difference sequence, namely E(dn |An−1 ) = 0

a.e.

Conversely, given a martingale difference sequence {dn , An , n ≥ 0}, we can form a martingale {Xn , An , n ≥ 0} by n X Xn = dk . k=1

Example 1.2. (i) Assume that ξn , n ≥ 1 is a sequence of independent Pn random variables with mean zero. Let S0 = 0, Sn = k=1 ξk , A0 = {∅, Ω}, An = σ{ξ1 , · · · , ξn }, n ≥ 1. Then {Sn , An , n ≥ 0} is a martingale. (ii) Assume that X is a random variable with finite expectation. Let An , n ≥ 0 is a sequence of non-decreasing sub-sigma fields of A. Let Xn = E(X|An )

a.e.

Then {Xn , An , n ≥ 0} is a martingale. We now state a martingale CLT due to Brown (1971). Theorem 1.4. Assume that {dn , An , n ≥ 0} is a martingale difference Pn Pn sequence, set Sn = k=1 dk , Bn = k=1 Ed2k . If the following three conditions (i)
P 1 max E d2k Ak−1 −→ 0, n → ∞ Bn 1≤k≤n (ii) n
P 1 X E d2k Ak−1 −→ 1, n → ∞ Bn k=1

(iii) for any ε > 0 n 1 X Bn

k=1


P E d2k 1(|dk |≥ε√Bn ) Ak−1 −→ 0,

n→∞

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are satisfied, then S d √ n −→ N (0, 1), Bn

n → ∞.

(1974) presented an improved version under the following slightly weaker conditions: (i0 ) there is a constant M > 0 such that 1 E max d2 ≤ M, Bn 1≤k≤n k

∀n ≥ 1;

1 P max d2 −→ 0, Bn 1≤k≤n k

n → ∞;

(ii0 )

(iii0 ) n 1 X 2 P dk −→ 1, Bn

n → ∞.

k=1

The interested reader is referred to Hall and Heyde (1980) for many other limit theorems related to martingales. Let E be a set of at most countable points. Assume that Xn , n ≥ 0 is a random sequence with state space E. If for any states i and j, any time n ≥ 0, P (Xn+1 = j|Xn = i, Xn−1 = in−1 , · · · , X0 = i0 ) = P (Xn+1 = j|Xn = i) = P (X1 = j|X0 = i),

(1.10)

then we call Xn , n ≥ 0 a time-homogenous Markov chain. Condition (1.10), called the Markov property, implies that the future is independent of the past given its present state. Denote pij the transition probability: pij = P (Xn+1 = j|Xn = i). The matrix P := (pij ) is called the transition matrix. It turns out that both the X0 and the transition matrix P will completely determine the (n)
law of a Markov chain. Denote the n-step transition matrix P(n) = pij , (n)

where pij = P (Xn = j|X0 = i). Then a simple chain rule manipulation shows P(n+m) = P(n) P(m) .

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This is the well-known Chapman-Kolmogorov equation. Moreover, P(n) = Pn . (n) State j is accessible from state i, denoted by i → j, if pij > 0 for some n ≥ 1. State i and j communicate with each other, denoted by i ↔ j, if i → j and j → i. A Markov chain is irreducible if any two states communicate each other. The period di of a state i is the greatest common divisor of all n that (n) satisfy pii > 0. State i is aperiodic if di = 1, and otherwise it is periodic. Denote by τi the hitting time τi = min{n ≥ 1; Xn = i}. A state i is transient if P (τi = ∞|X0 = i) > 0, and i is recurrent if P (τi < ∞|X0 = i) = 1. A recurrent state i is positive recurrent if E(τi |X0 = i) < ∞. An irreducible aperiodic positive recurrent Markov chain is called ergodic. If a probability distribution π on E satisfies the following equation: π = πP, then we call π a stationary distribution. If we choose π to be an initial distribution, then the Xn is a stationary Markov chain. In addition, if for any i, j πi pij = πj pji , then the Xn is reversible. In particular, d

(X0 , X1 ) = (X1 , X0 ). An irreducible aperiodic Markov chain is ergodic if and only if it has a stationary distribution. Theorem 1.5. Assume that Xn , n ≥ 0 is an ergodic Markov chain with P stationary distribution π. Assume that f : E → 7 R is such that i∈E f (i)πi is absolutely convergent. Then n−1 X 1X f (Xi ) = f (i)πi n→∞ n i=0

lim

i∈E

a.e.

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11

This is a type of law of large numbers for Markov chains. See Serfozo (2009) for a proof. Let L02 be the subspace of L2 (π) consisting of functions f : E 7→ R with P Eπ f := i∈E f (i)πi = 0. We shall give a sufficient condition under which Pn−1 the linear sum Sn (f ) := i=0 f (Xi ) satisfies the CLT. To this end, we introduce the transition operator TX T g(i) = g(j)pij . j∈E
Trivially, T g(i) = E g(X1 ) X0 = i . Assume that there exists a function g such that f = g − T g. (1.11) Then it easily follows from the martingale CLT
Sn (f ) d √ −→ N 0, σf2 n with limit variance σf2 = kgk22 − kT gk22 , where k · k2 denotes the L2 -norm. If f is such that ∞ X T nf (1.12) n=0

is convergent in L2 (π), then the solution of the equation (1.11) do exist. P∞ Indeed, g = n=0 T n f just solves the equation. It is too restrictive to require that the series in (1.12) is L2 -convergent. An improved version is Theorem 1.6. Let Xn , n ≥ 0 is an ergodic Markov chain with stationary distribution π. Assume f ∈ L02 satisfies the following two conditions: (i) L

2 T n f −→ 0,

n → ∞;

(ii) ∞ X

kT n f k22 − kT n+1 f k22


1/2

< ∞.

n=0

Then we have
Sn (f ) d √ −→ N 0, σf2 n with limit variance σf2

n  X X

n+1

2  k 2

= lim T f 2− T k f 2 . n→∞

k=0

k=0

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In the preceding paragraphs, we have seen that the characteristic functions is a powerful tool in proving convergence in distribution and identifying the limit distribution. It is particularly successful in the study of partial sums of independent or asymptotically independent random variables. However, it is sometimes not an easy task to compute the characteristic function of a random variable of interest. In the rest of this section and next sections we will briefly introduce other methods and techniques, among which are the moment method, the replacement trick, the Stein method and the Stieltjes transform method. The moment method is closely related to an interesting old problem. Is the distribution of X uniquely determined by its moments? If not, what extra conditions do we require? Suppose X has finite moments of all or(k) ders. Then according to (1.5), ψX (0) = ik mk where mk = EX k , k ≥ 0. However, it does not necessarily follow ψX (t) =

∞ k X i mk k=0

k!

tk .

(1.13)

Example 1.3. Consider two random variables X and Y , whose probability density functions are as follows ( 2 √ 1 e−(log x) /2 , x > 0 2π x pX (x) = 0, x≤0 and ( pY (x) =

2 √ 1 e−(log x) /2 (1 2π x

0,

+ sin(2π log x)), x > 0 x ≤ 0.

Then it is easy to check that X and Y have all moments finite and EX k = EY k for any k ≥ 1. If the following Carleman condition ∞ X

−1/2k

m2k

=∞

(1.14)

k=1

is satisfied, then (1.13) holds, and so the distribution of X is uniquely determined by its moments. A slightly stronger condition for (1.14) to hold is lim inf k→∞

1 1/2k m < ∞. k 2k

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Example 1.4. (i) Assume X ∼ N (0, σ 2 ), then for k ≥ 1  m2k (X) = σ 2k (2k − 1)!!, m2k+1 (X) = 0. (ii) Assume X is a Poisson random variable with parameter λ, then for k≥1 EX(X − 1) · · · (X − k + 1) = λk . (iii) Assume X is a random variable with density function given by  1√ 4 − x2 , |x| ≤ 2, ρsc (x) = 2π (1.15) 0, otherwise, then for k ≥ 1 

1 m2k (X) = k+1 m2k+1 (X) = 0.

2k k


,

(iv) Assume X is a random variable with density function given by ( q 1 4−x 0 < x ≤ 4, x , ρM P (x) = 2π (1.16) 0, otherwise, then for k ≥ 1   1 2k mk (X) = . k+1 k

Fig. 1.2

Wigner semicircle law

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Fig. 1.3

Marchenko-Pastur law

We remark that ρsc and ρM P (see Figures 1.2 and 1.3) are often called Wigner semicircle law and Marchenko-Pastur law in random matrix literature. They are respectively the expected spectrum distributions of Wigner random matrices and sample covariance matrices in the large dimensions. It is now easy to verify that these moments satisfy the Carleman condition (1.14). Therefore normal distribution, Poisson distribution, Wigner semicircle law and Marchenko-Pastur law are all uniquely determined by their moments. Theorem 1.7. Let Xn , n ≥ 1 be a sequence of random variables with all moments finite. Let X be a random variable whose law is uniquely determined by its moments. If for each k ≥ 1 mk (Xn ) → mk (X),

n→∞

d

then Xn −→ X. When applying Theorem 1.7 in practice, it is often easier to work with cumulants rather than moments. Let X be a random variable with all moments finite. Expand log ψX (t) at t = 0 as follows log fX (t) =

∞ k X i τk k=1

k!

tk .

We call τk the kth cumulant of X. Example 1.5. (i) If X ∼ N (µ, σ 2 ), then τ1 = µ,

τ2 = σ 2 ,

τk = 0,

∀k ≥ 3.

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15

(ii) If X is a Poisson random variable with parameter λ, then τk = λ,

∀k ≥ 1.

The cumulants possess the following nice properties. Fix a a constant. (i) shift equivariance: τ1 (X + a) = τ1 (X) + a; (ii) shift invariance: τk (X + a) = τk (X),

∀k ≥ 2;

τk (aX) = ak τk (X),

∀k ≥ 2;

(iii) homogeneity:

(vi) additivity: if X and Y are independent random variables, then τk (X + Y ) = τk (X) + τk (Y ),

∀k ≥ 2;

(v) relations between cumulants and moments: k−1 X k − 1  τk = mk − τl mk−l l−1 l=1

X

=

(−1)α1 +···+αk −1 (α1 + · · · + αk − 1)!

α1 ,··· ,αk

k Y l=1

k! mαl , (1.17) αl !lαl l

where the summation is extended over all nonnegative integer solutions of the equation α1 + 2α2 + · · · + kαk = k, and XY mk = τ|B| λ B∈λ

where λ runs through the set of all partitions of {1, 2, · · · , n}, B ∈ λ means one of the blocks into which the set if partitioned and |B| is the size of the block. Since knowledge of the moments of a random variable is interchangeable with knowledge of its cumulants, Theorem 1.7 can be reformulated as Theorem 1.8. Let Xn , n ≥ 1 be a sequence of random variables with all moments finite. Let X be a random variable whose law is uniquely determined by its moments. If for each k ≥ 1 τk (Xn ) → τk (X), d

then Xn −→ X.

n→∞

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This theorem is of particular value when proving asymptotic normality. Namely, if τ1 (Xn ) → 0,

τ2 (Xn ) → 1

and for k ≥ 3 τk (Xn ) → 0, d

then Xn −→ N (0, 1). To conclude this section, we will make a brief review about Lindeberg replacement strategy by reproving the Feller-L´evy CLT. To this end, we need an equivalent version of convergence in distribution, see Section 1.4 below. Lemma 1.1. Let X, Xn , n ≥ 1 be a sequence of random variables. Then d Xn −→ X if and only if for each bounded thrice continuously differentiable function f with kf (3) k∞ < ∞, Ef (Xn ) −→ Ef (X),

n → ∞.

Theorem 1.9. Let ξn , n ≥ 1 be a sequence of i.i.d. random variables with Pn mean zero and variance 1 and E|ξn |3 < ∞. Let Sn = k=1 ξk . Then it follows Sn d √ −→ N (0, 1), n

n → ∞.

Proof. Let ηn , n ≥ 1 be a sequence of i.i.d. normal random variables Pn with mean zero and variance 1, and let Tn = k=1 ηk . Trivially, T √n ∼ N (0, 1). n According to Lemma 1.1, it suffices to show that for any bounded thrice continuously differentiable function f with kf (3) k∞ < ∞, S  T  n n Ef √ − Ef √ −→ 0 n → ∞. n n To do this, set Rn,k =

k−1 X l=1

ξl +

n X l=k+1

ηl ,

1 ≤ k ≤ n.

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Then S  T  n n Ef √ − Ef √ n n n h   1 i  1 X = Ef √ (Rn,k + ξk ) − Ef √ (Rn,k + ηk ) n n =

k=1 n h X

 1   1 i Ef √ (Rn,k + ξk ) − Ef √ Rn,k n n

k=1 n h X

  1 i  1 (1.18) Ef √ (Rn,k + ηk ) − Ef √ Rn,k . n n k=1 √ Applying the Taylor expansion of f at Rn,k / n, we have by hypothesis   1   1 Ef √ (Rn,k + ξk ) − Ef √ Rn,k n n  1 ξ  ξ2  1 1 k = Ef 0 √ Rn,k √ + Ef 00 √ Rn,k k n n n 2 n 3 ξ 1 k + Ef (3) (R∗ ) 3/2 6 n   1 1 1 = Ef 00 √ Rn,k + 3/2 Ef (3) (R∗ )ξk3 , (1.19) 2n n 6n √ √ where R∗ is between (Rn,k + ξk )/ n and Rn,k / n. Similarly, we also have  1   1  Ef √ (Rn,k + ξk ) − Ef √ Rn,k n n   1 1 1 Ef 00 √ Rn,k + 3/2 Ef (3) (R∗∗ )ηk3 , (1.20) = 2n n 6n √ √ where R∗∗ is between (Rn,k + ξk )/ n and Rn,k / n. Putting (1.19) and (1.20) back into (1.18) yields S  T  n n Ef √ − Ef √ n n n X 1 1 Ef (3) (R∗ )ξk3 + 3/2 Ef (3) (R∗∗ )ηk3 . = 6n3/2 6n −

k=1

Noting kf (3) k∞ < ∞ and E|ξk |3 < ∞ and E|ηk |3 < ∞, we obtain S  T 
n n Ef √ − Ef √ = O n−1/2 . n n The assertion is now concluded.



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Proof of Theorem 1.2 To apply Theorem 1.9, we need to use the truncation technique. For any constant a > 0, define ξ¯k (a) = ξk 1(|ξk |≤a) ,

1 ≤ k ≤ n.

Obviously, ξ¯k (a), 1 ≤ k ≤ n are i.i.d. bounded random variables. Let µk (a) = E ξ¯k (a) and σ ¯k2 (a) = V ar(ξ¯k (a)) for 1 ≤ k ≤ n. So according to Theorem 1.9, it follows Pn

¯

− µk (a)) d √ −→ N (0, 1), σ ¯1 (a) n

k=1 (ξk (a)

n → ∞.

Since a > 0 is arbitrary, then by selection principle, there is a sequence of constants an > 0 such that an → ∞ and Pn ¯ k=1 (ξk (an ) − µk (an )) d √ −→ N (0, 1). (1.21) σ¯1 (an ) n In addition, it is easy to see µk (an ) → 0,

σ¯n (an ) → 1.

Hence by (1.21), Pn

¯

k=1 (ξk (an )

√ n

− µk (an ))

d

−→ N (0, 1).

(1.22)

Finally, it follows from the Chebyshev inequality n  X 1  P √ Sn − (ξ¯k (an ) − µk (an )) −→ 0. n

(1.23)

k=1

Combining (1.22) and (1.23) together, we conclude the assertion (1.7).  Remark 1.2. The Lindeberg replacement strategy makes clear the fact that the CLT is a local phenomenon. By this we mean that the structure of the CLT does not depend on the behavior of any fixed number of the increments. Only recently was it successfully used to establish the Four Moment Comparison theorem for eigenvalues of random matrices, which in turn solves certain long-standing conjectures related to universality of eigenvalues of random matrices. See Tao and Vu (2010).

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The Stein method

The Stein method was initially conceived by Stein (1970, 1986) to provide errors in the approximation by the normal distribution of the distribution of the sum of dependent random variables of a certain structure. However, the ideas presented are sufficiently abstract and powerful to be able to work well beyond that purpose, applying to approximation of more general random variables by distributions other than normal. Besides, the Stein method is a highly original technique and useful in quantifying the error in the approximation of one distribution by another in a variety of metrics. This subsection serves as a basic introduction of the fundamentals of the Stein method. The interested reader is referred to nice books and surveys, say, Chen, Goldstein and Shao (2010), Ross (2011). A basic starting point is the following Stein equation. Lemma 1.2. Assume that ξ is a random variable with mean zero and variance σ 2 . Then ξ is normal if and only if for every bounded continuously differentiable function f (kf k∞ , kf 0 k∞ < ∞), Eξf (ξ) = σ 2 Ef 0 (ξ).

(1.24)

Proof. Without loss of generality, assume σ 2 = 1. If ξ ∼ N (0, 1), then by the integration by part formula, we easily obtain (1.24). Conversely, assume (1.24) holds. Fix z ∈ R and consider the following first order ordinary differential equation f 0 (x) − xf (x) = 1(−∞,z] (x) − Φ(z),

(1.25)

where Φ(·) denotes the standard normal distribution function. Then a simple argument shows that there exists a unique solution: ( 2 R 2 ∞ ex /2 x e−t /2 (Φ(z) − 1(−∞,z] (t))dt, R fz (x) = 2 2 ∞ ex /2 x e−t /2 (Φ(z) − 1(−∞,z] (t))dt. Note 1 −z2 /2 e , z → ∞. z p < π/2 and kfz0 k∞ ≤ 2. By hypothesis, it

1 − Φ(z) ∼ It is not hard to see kfz k∞ follows

P (ξ ≤ z) − Φ(z) = Efz0 (ξ) − Eξfz (ξ) = 0. We now conclude the proof.



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As an immediate corollary to the proof, we can derive the following Stein continuity theorem. Theorem 1.10. Assume that ξn , n ≥ 1 is a sequence of random variables with mean zero and variance 1. If for every bounded continuously differentiable function f , Eξn f (ξn ) − Ef 0 (ξn ) → 0,

n→∞

d

then ξn −→ N (0, 1). Remark 1.3. The above Stein equation can be extended to a non-normal random variable. Assume that ξ has the (q +2)th moment finite, f is (q +1) times bounded continuously differentiable, then Eξf (ξ) =

q X τk+1 k=0

k!

Ef (k) (ξ) + εq ,

where τk is the kth culumant of ξ, the remainder term admits the bound εq ≤ cq kf (q+1) kE|ξ|q+2 ,

cq ≤

1 + (3 + 2q)q+2 . (q + 1)!

As the reader may notice, if we replace the indicator function 1(−∞,z] by a smooth function in the preceding differential equation (1.25), then its solution will have a nicer regularity property. Let H be a family of 1Lipschitz functions, namely  H = h : R 7→ R, |h(x) − h(y)| ≤ |x − y| . Consider the following differential equation: f 0 (x) − xf (x) = h(x) − Eh(ξ),

(1.26)

where ξ ∼ N (0, 1). Lemma 1.3. Assume h ∈ H. There exists a unique solution of (1.26): ( 2 R 2 ∞ ex /2 x e−t /2 (Φ(z) − h(t))dt, R fh (x) = 2 2 ∞ ex /2 x e−t /2 (Φ(z) − h(t))dt. Moreover, fh satisfies the following properties r π 0 , kfh00 k∞ ≤ 2. kfh k∞ ≤ 2, kfh k∞ ≤ 2

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We omit the proof, which can be found in Chen, Goldstein and Shao (2010). It is easily seen that for any random variable W of interest, Eh(W ) − Eh(ξ) = Efh0 (W ) − EW fh0 (W ). Given two random variables X and Y , the Wasserstein distance is defined by dW (X, Y ) = sup |Eh(X) − Eh(Y )|. h∈H

Note the Wasserstein distance is widely used in describing the distributional approximations. In particular, r πp sup |P (X ≤ x) − P (Y ≤ x)| ≤ dW (X, Y ). 2 x∈R Let G be the family of bounded continuously differentiable functions with bounded first and second derivatives, namely r n o π 0 G = f : R 7→ R, kf k∞ ≤ 2, kf k∞ ≤ , kf 00 k∞ ≤ 2 . 2 Taking Lemma 1.3 into account, we immediately get Theorem 1.11. Let ξ ∼ N (0, 1), W a random variable. Then we have dW (W, ξ) ≤ sup |EW f (W ) − Ef 0 (W )|. f ∈G

To illustrate the use of the preceding Stein method, let us take a look at the normal approximation of sums of independent random variables below. Example 1.6. Suppose that ξn , n ≥ 1 is a sequence of independent random Pn variables with mean zero, variance 1 and E|ξn |3 < ∞. Let Sn = i=1 ξi , then n S  4 X n E|ξi |3 . (1.27) dW √ , ξ ≤ 3/2 n n i=1 √ Proof. Writing Wn = Sn / n, we need only to control the supremum √ of EWn f (Wn ) − Ef 0 (Wn ) over G. Set Wn,i = (Sn − ξi )/ n. Then by independence and noting Eξi = 0 n

1 X Eξi f (Wn ) EWn f (Wn ) = √ n i=1 n


1 X = √ Eξi f (Wn ) − f (Wn,i ) . n i=1

(1.28)

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Using the Taylor expansion of f at Wn,i , we have 2 ξi 1 ∗ ξi f (Wn ) − f (Wn,i ) = f 0 (Wn,i ) √ + f 00 (Wn,i ) , n 2 n

(1.29)

∗ where Wn,i is between Wn and Wn,i . Inserting (1.29) into (1.28) and noting Eξi2 = 1 yields

EWn f (Wn ) =

n n 1 X 1X ∗ Ef 0 (Wn,i ) + 3/2 Ef 00 (Wn,i )ξi3 . n i=1 2n i=1

Subtracting Ef 0 (Wn ) in both sides gives n
1X E f 0 (Wn,i ) − f 0 (Wn ) n i=1

EWn f (Wn ) − Ef 0 (Wn ) =

+

n 1 X ∗ Ef 00 (Wn,i )ξi3 . 2n3/2 i=1

(1.30)

It follows from the mean value theorem ξi ‡ f 0 (Wn,i ) − f 0 (Wn ) = f 00 (Wn,i )√ , n

(1.31)

‡ where Wn,i is between Wn and Wn,i . Thus combining (1.30) and (1.31), and noting kf 00 k∞ ≤ 2 and E|ξi | ≤ 1 ≤ E|ξi |3 , we have (1.27) as desired. 

Recall the ordered pair (W, W 0 ) of random variables is exchangeable if d

(W, W 0 ) = (W 0 , W ). d

Trivially, if (W, W 0 ) is an exchangeable pair, then W = W 0 . Also, assuming g(x, y) is antisymmetric, namely g(x, y) = −g(y, x), then Eg(W, W 0 ) = 0 if the expectation exists. Theorem 1.12. Assume that (W, W 0 ) is an exchangeable pair, and assume there exists a constant 0 < τ ≤ 1 such that
E W 0 |W = (1 − τ )W. If EW 2 = 1, then 1 dW (W, ξ) ≤ √ 2π

r

 3 1 0 E(|W 0 − W |2 |W ) 2 E W − W . + E 1− 2τ 3τ

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Proof.

Given f ∈ G, define F (x) =

Rx 0

23

f (t)dt. Then it obviously follows

EF (W 0 ) − EF (W ) = 0.

(1.32)

On the other hand, using the Taylor expansion for F at W , we obtain 1 F (W 0 ) − F (W ) = F 0 (W )(W 0 − W ) + F 00 (W )(W 0 − W )2 2 1 + F 000 (W ∗ )(W 0 − W )3 6 1 = f (W )(W 0 − W ) + f 0 (W )(W 0 − W )2 2 1 00 + f (W ∗ )(W 0 − W )3 , 6 where W ∗ is between W and W 0 . This together with (1.32) in turn implies 1 1 Ef (W )(W 0 − W ) + Ef 0 (W )(W 0 − W )2 + Ef 00 (W ∗ )(W 0 − W )3 = 0. 2 6 Note by hypothesis E(W 0 |W ) = (1 − τ )W , Ef (W )(W 0 − W ) = E f (W )(W 0 − W )|W




= Ef (W )E W 0 − W |W




= −τ EW f (W ). Hence we have EW f (W ) =

1 1 Ef 0 (W )(W 0 − W )2 + Ef 00 (W ∗ )(W 0 − W )3 . 2τ 6τ

Subtracting Ef 0 (W ) yields 
 1 EW f (W ) − Ef 0 (W ) = −Ef 0 (W ) 1 − E (W 0 − W )2 W 2τ 1 00 ∗ + Ef (W )(W 0 − W )3 . 6τ Thus it follows from the Cauchy-Schwarz inequality and the fact kf 0 k∞ ≤ p 2/π and kf 00 k∞ ≤ 2, r r 
2 EW f (W ) − Ef 0 (W ) ≤ 2 E 1 − 1 E (W 0 − W )2 W π 2τ 3 1 0 + E W − W . 3τ The proof is complete.



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As an application, we shall revisit below the normal approximation of sums of independent random variables. Use notations in Example 1.6. A key ingredient is to construct Wn0 in such a way that (Wn , Wn0 ) is an exchangeable pair and E(Wn0 |Wn ) = (1 − τ )Wn for some 0 < τ ≤ 1. Let {ξn0 , n ≥ 1} be an independent copy of {ξn , n ≥ 1}. Let I be a uniform random variable taking values 1, 2, · · · , n, independent of all other random variables. Define
1 Wn0 = √ Sn − ξI + ξI0 . n Let An = σ{ξ1 , · · · , ξn }. Trivially, An is a sequence of increasing σ-fields and Wn ∈ An . Some simple manipulation shows

1 E Wn0 − Wn An = √ E − ξI + ξI0 An n 1 = − Wn , n which implies τ = 1/n. In addition,

1 E (Wn0 − Wn )2 An = E (ξI − ξI0 )2 |An n n 1 X 2 1 ξ = + 2 n n i=1 i and n 
1 1 X 2 E (Wn0 − Wn )2 Wn = + 2 E ξi |Wn . n n i=1

Hence we have 
2 1 E 1− E (Wn0 − Wn )2 Wn 2τ n  1  X 2 2 1 1 ξi Wn + 2E =E 1− 2τ n n i=1 n

=

2 1 X 2 2 (ξ − Eξ )|W E n i i 4n2 i=1

≤


2 1 X 2 E ξi − Eξi2 2 4n i=1

n

n n 1 X 1 X 4 2 2 2 ≤ 2 E(ξi − Eξi ) ≤ 2 Eξ . 4n i=1 n i=1 i

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Finally, note 3 E Wn0 − Wn =

1 n3/2

3 E ξI − ξI0 n

≤

8 1X E|ξi |3 . 3/2 n n i=1

Applying Theorem 1.12, we immediately obtain v n n  r 1 1u S uX 8 1X n t dW √ , ξ ≤ Eξi4 + 3/2 E|ξi |3 . n 2π n i=1 3n n i=1 In particular, when ξn , n ≥ 1 are i.i.d. random variables with Eξn4 < ∞, then  S A n dW √ , ξ ≤ √ n n for some numerical constant A.

1.3

The Stieltjes transform method

Stieltjes transforms, also called Cauchy transforms, of functions of bounded variation are other important tools in the study of convergence of probability measures. It actually plays a particularly significant role in the asymptotic spectrum theory of random matrices. Given a probability measure µ on the real line R, its Stieltjes transform is defined by Z 1 dµ(x) sµ (z) = R x−z for any z outside the support of µ. In particular, it is well defined for all complex numbers z in C \ R. Some elementary properties about sµ (z) are listed below. Set z = a + iη. (i) sµ (z) = sµ (z). So we may and do focus on the upper half plane, namely η > 0 in the sequel. (ii) |sµ (z)| ≤

1 . |η|

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(iii) Z

1 dµ(x). |x − z|2 R So Im(sµ (z)) has the same sign as Im(z). (iv) 1 sµ (z) = − (1 + o(1)), z → ∞. z R (v) If mk (µ) := R xk dµ(x) exists and are finite for every k ≥ 0, then it follows ∞ 1 X mk (µ) . sµ (z) = − z zk Im(sµ (z)) = Im(z)

k=0

So sµ (z) is closely related to the moment generating function of µ. (vi) sµ (z) is holomorphic outside the support of µ. Example 1.7. (i) Let µ := µsc be a probability measure on R with density function ρsc given by (1.15). Then its Stieltjes transform is √ z z2 − 4 ssc (z) = − + . (1.33) 2 2 (ii) Let µ := µM P be a probability measure on R with density function given by (1.16). Then its Stieltjes transform is 1p 1 sM P (z) = − + z(z − 4). 2 2z Theorem 1.13. Let µ be a probability measure with Stieltjes transform sµ (z). Then for any µ-continuity points a, b (a < b) Z 1 b sµ (λ + iη) − sµ (λ − iη) µ(a, b) = lim dλ η→0 π a 2i Z 1 b = lim Im(sµ (λ + iη))dλ. η→0 π a Proof. Let X be a random variable whose law is µ. Let Y be a standard Cauchy random variable, namely Y has probability density function 1 pY (y) = , y ∈ R. π(1 + y 2 ) Then the random variable Zη := X + ηY has probability density function Z 1 λ − y  pY dµ(y) pη (λ) = η R η Z η 1 dµ(y) = π R (λ − y)2 + η 2 1 = Im(sµ (λ + iη)). π

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27

d

Note that Zη −→ X as η → 0. This implies lim P (Zη ∈ (a, b)) = P (X ∈ (a, b)).

η→0

We now conclude the proof.



Compared with the Fourier transform, an important advantage of Stieltjes transform is that one can easily find the density function of a probability measure via the Stieltjes transform. In fact, let F (x) be a distribution function induced by µ and x0 ∈ R. Suppose that limz→x0 Im(sµ (z)) exists, denoted by Im(sµ (x0 )). Then F is differentiable at x0 , and F 0 (x0 ) = Im(sµ (x0 ))/π. The Stieltjes continuity theorem reads as follows. Theorem 1.14. (i) If µ, µn , n ≥ 1 is a sequence of probability measures on R such that µn ⇒ µ,

n→∞

then for each z ∈ C \ R, sµn (z) → sµ (z),

n → ∞.

(ii) Assume that µn , n ≥ 1 is a sequence of probability measures on R such that as n → ∞, sµn (z) → s(z),

∀z ∈ C \ R

for some s(z). Then there exists a sub-probability measure µ (µ(R) ≤ 1) such that Z 1 s(z) = dµ(x) R x−z and for any continuous function f decaying to 0 at infinity, Z Z f (x)dµn (x) → f (x)dµ(x), n → ∞. R

R

(iii) Assume µ is a deterministic probability measure, and µn , n ≥ 1 is a sequence of random probability measures on R. If for any z ∈ C \ R, P

sµn (z) −→ sµ (z),

n→∞

then µn weakly converges in probability to µ. Namely, for any bounded continuous function f , Z Z P f (x)dµn (x) −→ f (x)dµ(x), n → ∞. R

R

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The reader is referred to Anderson, Guionnet and Zeitouni (2010), Bai and Silverstein (2010), Tao (2012) for its proof and more details. To conclude, let us quickly review the Riesz transform for a continual diagram. Let √
2 u arcsin u2 + 4 − u2 , |u| ≤ 2, Ω(u) = π (1.34) |u|, |u| > 2 its Riesz transform is defined by RΩ (z) =

 1 exp z

Z

2

−2

(Ω(u) − |u|)0  du u−z

for each z ∈ / [−2, 2]. It is easy to compute 1p 2 z z − 4, RΩ (z) = − + 2 2 which implies by (1.33) RΩ (z) = ssc (z).

(1.35)

This is not a coincidence! As we will see in Chapter 5, Ω(u) (see Figure 1.4 below) turns out to be the limit shape of a typical Plancherel Young diagram. The equation (1.35) provides another evidence that there is a close link between random matrices and random Plancherel partitions.

Fig. 1.4

Ω curve

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1.4

29

Convergence of stochastic processes

Let (S, ρ) be a metric space, S a σ-field generated by its topology. A mapping X : Ω 7→ S is said to be measurable if for any A ∈ S X −1 (A) = {ω : X(ω) ∈ A} ∈ A. We also call X a S-valued random element. The most commonly studied random elements include real (complex) random variable, random vector, random processes, Banach-valued random variable. Denote by PX the law of X under P : PX (A) := P ◦ X −1 (A) = P (ω ∈ Ω : X(ω) ∈ A),

A ∈ S.

By definition, a sequence of random variables Xn weakly converges to a random variable X if PXn ⇒ PX , and write simply Xn ⇒ X. The following five statements are equivalent: (i) for any bounded continuous function f , Ef (Xn ) → Ef (X),

n → ∞;

(ii) for any bounded uniformly continuous function f , Ef (Xn ) → Ef (X),

n → ∞;

(iii) for any closed set F , lim sup P (Xn ∈ F ) ≤ P (X ∈ F ); n→∞

(iv) for any open set G, lim inf P (Xn ∈ G) ≥ P (X ∈ G); n→∞

(v) for any measurable X-continuity set A, lim P (Xn ∈ A) = P (X ∈ A).

n→∞

The reader is referred to Billingsley (1999a) for the proof and more details. In addition, (ii) can be replaced by (ii0 ) for any bounded infinitely differentiable function f , Ef (Xn ) → Ef (X),

n → ∞.

It can even be replaced by (ii00 ) for any continuous function f with compact support, Ef (Xn ) → Ef (X),

n → ∞.

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d

In the special cases S = R and Rk , Xn ⇒ X is equivalent to Xn −→ X. In the case S = R∞ , Xn ⇒ X if and only if for each k ≥ 1 Xn ◦ πk−1 ⇒ X ◦ πk−1 ,

n→∞

where πk denotes the projection from R∞ to Rk . The case of C[0, 1] is more interesting and challenging. Assume Xn ⇒ X. Then for any k ≥ 1 and any k points t1 , t2 , · · · , tk ∈ [0, 1] Xn ◦ πt−1 ⇒ X ◦ πt−1 , 1 ,··· ,tk 1 ,··· ,tk

n→∞

(1.36)

where πt1 ,··· ,tk is a projection from C[0, 1] to Rk , i.e., πt1 ,··· ,tk (x) = (x(t1 ), · · · , x(tk )). However, the condition (1.36) is not a sufficient condition for Xn to weakly converge to X. We shall require additional conditions. The Xn is said to be weakly relatively compact if every subsequence has a further convergent subsequence in the sense of weak convergence. According to the subsequence convergence theorem, Xn is weakly convergent if all the limit variables are identical in law. Another closely related concept is uniform tightness. The Xn is uniformly tight if for any ε > 0, there is a compact subset Kε in (S, ρ) such that P (Xn ∈ / Kε ) < ε,

for all n ≥ 1.

The celebrated Prohorov’s theorem tells that the Xn must be weakly relatively compact if Xn is uniformly tight. The converse is also true in a separable complete metric space. A major point of this theorem is that the weak convergence of probability measures rely on how they concentrate in a compact subset in a metric space. In C[0, 1], the so-called Ascoli-Arzel`a lemma completely characterizes a relatively compact subset: K ⊆ C[0, 1] is relatively compact if and only if (i) uniform boundedness: sup sup |x(t)| < ∞, t∈R x∈K

(ii) equi-continuity: for any ε > 0 there is a δ > 0 such that sup sup |x(s) − x(t)| < ε. |s−t|≤δ x∈K

Note that under condition (ii), (i) can be replaced by the condition (i0 ) sup |x(0)| < ∞. x∈K

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Combining the Ascoli-Arzel` a lemma and Prohorov’s theorem, we can readily give a criterion for Xn to weakly converge to X in C[0, 1]. Assume that we are given a sequence of continuous random processes X and Xn , n ≥ 1 in C[0, 1]. Then Xn ⇒ X if and only if (i) finite dimensional distributions converge, namely (1.36) holds; (ii) for any ε > 0, there is a finite positive constant M such that P (|Xn (0)| > M ) < ε,

for all n ≥ 1;

(iii) for any ε > 0 and η > 0, there is a δ > 0 such that   P sup |Xn (s) − Xn (t)| > η < ε, for all n ≥ 1.

(1.37)

(1.38)

|s−t|<δ

To illustrate how to use the above general framework, we shall state and prove the Donsker invariance principle. Theorem 1.15. Let ξn , n ≥ 1 be a sequence of i.i.d. real random variables defined in a common probability space (Ω, A, P ), and Eξ1 = 0 and Pn V ar(ξ1 ) = 1. Define Sn = i=1 ξi , n ≥ 1, and define 1 nt − [nt] Xn (t) = √ S[nt] + √ ξ[nt]+1 , n n

0 ≤ t ≤ 1.

(1.39)

Then Xn ⇒ B,

n→∞

where B = (B(t), 0 ≤ t ≤ 1) is a standard Brownian motion. Proof. We need to verify the conditions (1.36), (1.37) and (1.38). Indeed, (1.36) directly follows from the Feller-L´evy CLT. (1.37) is trivial since Xn (0) = 0, and (1.38) follows from the L´evy maximal inequality for sums of independent random variables. The detail is left to the reader.  We remark that the random process constructed in (1.39) is a polygon √ going through points (k/n, Sk / n). It is often referred to as a partial sum process. The Donsker invariance principle has found a large number of applications in a wide range of fields. To apply it, we usually need the following mapping theorem. Let (S1 , ρ1 ) and (S2 , ρ2 ) be two metric spaces, h : S1 7→ S2 a measurable mapping. Assume that X, Xn , n ≥ 1 is a sequence of S1 -valued random elements and Xn ⇒ X. It is natural to ask under what hypothesis the h(Xn ) still weakly converges to h(X). Obviously, if h is continuous, then we have h(Xn ) ⇒ h(X). Moreover, the same still holds if h is a measurable mapping and PX (Dh ) = 0 where Dh is the set of all discontinuity points of h. As a simple example, we can compute the

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√ limiting distribution of max1≤k≤n Sk / n. Indeed, let h(x) = sup0≤t≤1 x(t) where x ∈ C[0, 1]. Then h is continuous and 1 h(Xn ) = √ max Sk , h(B) = sup B(t). n 1≤k≤n 0≤t≤1 Hence it follows 1 d √ max Sk −→ sup B(t), n → ∞. n 1≤k≤n 0≤t≤1 Another example is to compute the R 1 limiting distribution of the weighted Pn sum k=1 kξk /n3/2 . Let h(x) = 0 x(t)dt where x ∈ C[0, 1]. Then h is continuous and Z 1 n 1 X h(Xn ) = 3/2 kξk + op (1), h(B) = B(t)dt, n 0 k=1 where op (1) is negligible. Hence it follows Z 1 n 1 X d kξ −→ B(t)dt. k n3/2 k=1 0 More interesting examples can be found in Billingsley (1999a). In addition to R∞ and C[0, 1], one can also consider weak convergence of random processes in D[0, 1], C(0, ∞) and D(0, ∞). As the reader might notice, in proving the weak convergence of Xn in C[0, 1], the most difficult part is to verify the uniform tightness condition (1.38). A weaker version than (1.38) is stochastic equicontinuity: for every ε > 0 and η > 0, there is a δ > 0 such that sup P (|Xn (s) − Xn (t)| > η) < ε, for all n ≥ 1. (1.40) |s−t|<δ

Although (1.40) does not guarantee that the process Xn converges weakly, we can formulate a limit theorem for comparatively narrow class of functionals of integral form. Theorem 1.16. Suppose that X = (X(t), 0 ≤ t ≤ 1) and Xn = (Xn (t), 0 ≤ t ≤ 1) is a sequence of random processes satisfying (1.36) and (1.40). Suppose that g(t, x) is a continuous function and there is a nonnegative function h(x) such that h(x) ↑ ∞ as |x| → ∞ and |g(t, x)| lim sup sup = 0. a→∞ 0≤t≤1 |x|>a h(x) If supn≥1 sup0≤t≤1 E|h(Xn (t))| < ∞, then as n → ∞ Z 1 Z 1

d g t, X(t) dt. g t, Xn (t) dt −→ 0

0

This theorem is sometimes referred to as Gikhman-Skorohod theorem. Its proof and applications can be found in Chapter 9 of Gikhman and Skorohod (1996).

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Chapter 2

Circular Unitary Ensemble

2.1

Introduction

For n ∈ N, a complex n × n matrix Un is said to be unitary if Un∗ Un = Un Un∗ = In . This is equivalent to saying Un is nonsingular and Un∗ = Un−1 . The set Un of unitary matrices forms a remarkable and important set, a compact Lie group, which is generally referred to as the unitary group. This group has a unique regular probability measure µn that is invariant under both left and right multiplication by unitary matrices. Such a measure is called Haar measure. Thus we have induced a probability space (Un , µn ), which is now known as Circular Unitary Ensemble (CUE). By definition the columns of an n×n random unitary matrix are orthogonal vectors in the n dimensional complex space Cn . This implies that the matrix elements are not independent and thus are statistically correlated. Before discussing statistical correlation properties, we shall have a quick look at how to generate a random unitary matrix. Form an n × n random matrix Zn = (zij )n×n with i.i.d. complex standard normal random variables. Recall Z = X + iY is a complex standard normal random variable if X and Y are i.i.d. real normal random variable with mean 0 and variance 1/2. The Zn is almost sure of full rank, so apply Gram-Schmidt orthonormalization to its columns: normalize the first column to have norm one, take the first column out of the second column and normalize to have norm one, and so on. Let Tn be the map induced by Gram-Schmidt algorithm, then the resulting matrix Tn (Zn ) is unitary and even is distributed with Haar measure. This is easy to prove and understand. Indeed it holds Tn (Un Zn ) = Un Tn (Zn ) for any unitary matrix d

d

Un ∈ Un . Since Un Zn = Zn , so Un Tn (Zn ) = Tn (Zn ) as required. 33

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Given a unitary matrix Un , consider the equation Un v = λv, where λ is a scalar and v is a vector in Cn . If a scalar λ and a nonzero vector v happen to satisfy this equation, then λ is called an eigenvalue of Un and v is called an eigenvector associated with λ. The eigenvalues of Un are zeros of the characteristic polynomial det(zIn − Un ). It turns out that all eigenvalues are on the unit circle T := {z ∈ C; |z| = 1} and are almost surely distinct with respect to product Lebesgue measure. Call these {eiθ1 , · · · , eiθn } with 0 ≤ θ1 , · · · , θn < 2π. Note that for any sequence of n points on T there are matrices in Un with these points as eigenvalues. The collection of all matrices with the same set of eigenvalues constitutes a conjugacy class in Un . The main question of interest in this chapter is: pick a Un ∈ Un according to Haar measure, how are {eiθ1 , · · · , eiθn } distributed? The most celebrated result is the following Weyl formula. Theorem 2.1. The joint probability density for the unordered eigenvalues of a Haar distributed random matrix in Un is Y
1 eiθj − eiθk 2 , pn eiθ1 , · · · , eiθn = (2.1) n (2π) n! 1≤j
where the product is by convention 1 when n = 1. The proof is omitted, the reader is referred to Chapter 11 of Mehta (2004). See also Chapter 2 of Forrester (2010). Weyl’s formula is the starting point of the following study of the CUE. In particular, it gives a simple way to perform averages on Un . For a class function f that are constant on conjugacy classes, Z Z

f (Un )dµn = f eiθ1 , · · · , eiθn pn eiθ1 , · · · , eiθn dθ1 · · · dθn . Un

[0,2π]n

Obviously, U1 has only one eigenvalue and is uniformly distributed on T. U2 has two eigenvalues whose joint probability density is 2
1 p2 eiθ1 , eiθ2 = 2 eiθ1 − eiθ2 . 8π It is easy to compute the marginal density for each eigenvalue by integrating out the other argument. In particular, each eigenvalue is also uniformly distributed on T. But these
two eigenvalues are not independent of each other; indeed p2 eiθ1 , eiθ2 tends to zero as θ1 and θ2 approach each other.

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35

Interestingly, these properties hold for general n. To see this, let for any set of n-tuple complex numbers (x1 , · · · , xn )
. ∆(x1 , · · · , xn ) = det xj−1 k 1≤j,k≤n Then the Vandermonde identity shows Y ∆(x1 , · · · , xn ) =

(xk − xj ).

1≤j
Define Sn (θ) = e−i(n−1)θ/2

n X k=1

eikθ =

sin nθ/2 , sin θ/2

(2.2)

where by convention Sn (0) = n. Thus we have by using the fact that the determinant of a matrix and its transpose are the same Y eiθj − eiθk 2 = ∆(eiθ1 , · · · , eiθn ) 2 1≤j

= det Sn (θk − θj ) 1≤j,k≤n . This formula is very useful to computing some eigenvalue statistics. In order to compute the m-dimensional marginal density, we also need a formula of Gaudin (see Conrey (2005)), which states Z 2π Sn (θj − θ)Sn (θ − θk )dθ = 2πSn (θj − θk ). 0

As a consequence, we have Z 2π

det Sn (θj − θk ) 1≤j,k≤n dθn = 2π det Sn (θj − θk ) 1≤j,k≤n−1 . 0

Repeating this yields easily Z

1 pn,m eiθ1 , · · · , eiθm := pn eiθ1 , · · · , eiθn dθm+1 · · · dθn n n!(2π) [0,2π]n−m Z
1 det Sn (θj − θk ) n×n dθm+1 · · · dθn = n n!(2π) [0,2π]n−m
(n − m)! 1 det Sn (θj − θk ) 1≤j,k≤m . = m n! (2π) In particular, the first two marginal densities are
1 pn,1 eiθ = , 0 ≤ θ ≤ 2π (2.3) 2π and 
1  1 1 − (Sn (θ1 − θ2 ))2 . (2.4) pn,2 eiθ1 , eiθ2 = 2 n(n − 1) (2π)

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Hence each eigenvalue is still uniformly distributed on T. The CUE and its eigenvalue distributions naturally appear in a variety of problems from particle physics to analytic number theory. It is indeed a very special example of three ensembles introduced and studied by Dyson in 1962 with a view to simplifying the study of energy level behavior in complex quantum systems, see Dyson (1962). More generally, consider n identically charged particles confined to move on T in the complex plane. Each interacts with the others through the usual Coulomb potential, − log eiθj − eiθk , which gives rise to the Hamiltonian X Hn (θ1 , · · · , θn ) = − log eiθj − eiθk . 1≤j
This induces the Gibbs measure with parameter n and β > 0 Y
1 eiθj − eiθk β , pn,β eiθ1 , · · · , eiθn = (2π)n Zn,β

(2.5)

1≤j
where n is the number of particles, β stands for the inverse temperature, and Zn,β is given by Zn,β =

Γ( 21 βn + 1) . [Γ( 12 β + 1)]n

The family of probability measures defined by (2.5) is called Circular β Ensemble (CβE). The CUE corresponds to β = 2. Viewed from the opposite perspective, one may say that the CUE provides a matrix model for the Coulomb gas at the inverse temperature β = 2. In Section 2.5 we shall see a matrix model for general β. In this chapter we will be particularly interested in the asymptotic behaviours of various eigenvalue statistics as n tends to infinity. Start with the average spectral measures. Let eiθ1 , · · · , eiθn be eigenvalues of a Haar distributed unitary matrix Un . Put them together as a probability measure on T: n

νn =

1X δeiθk . n k=1

Theorem 2.2. As n → ∞, νn ⇒ µ where µ is a uniform measure on T.

in P,

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37

The proof is basically along the line of Diaconis and Shahshahani (1994). We need a second-order moment estimate as follows. Lemma 2.1. For any integer l 6= 0, n X 2 E eilθk = min(|l|, n).

(2.6)

k=1

Proof.

It is easy to see n X 2 E eilθk = n + n(n − 1)Eeil(θ1 −θ2 ) .

(2.7)

k=1

In turn, in virtue of (2.2) and (2.4) it follows n(n − 1)Eeil(θ1 −θ2 ) Z 2π Z 2π   1 il(θ1 −θ2 ) 2 = e 1 − (S (θ − θ )) dθ1 dθ2 n 1 2 (2π)2 0 0 Z 2π Z 2π n X 2 1 il(θ1 −θ2 ) ik(θ1 −θ2 ) =− e e dθ1 dθ2 (2π)2 0 0 k=1 Z 2π Z 2π   X 1 =− eil(θ1 −θ2 ) n + ei(m−k)(θ1 −θ2 ) dθ1 dθ2 2 (2π) 0 0 1≤m6=k≤n

= −]{(m, k) : m − k = l,  |l| − n, |l| ≤ n, = 0, |l| > n.

1 ≤ m 6= k ≤ n} (2.8)

Substituting (2.8) into (2.7) immediately yields (2.6).



Proof of Theorem 2.2. It is enough to show that the Fourier transform of νn converges in probability to that of µ. Let for each integer l 6= 0 Z 2π Z 2π 1 1 e−ilθ dνn (θ), µ ˆ(l) = e−ilθ dµ(θ). νˆn (l) = 2π 0 2π 0 Trivially µ ˆ(l) = 0. We shall need only prove E νˆn (l) = 0,

E|ˆ νn (l)|2 → 0.

Since each eigenvalue is uniform over T, then n

E νˆn (l) = E

1 X −ilθk e n k=1 −ilθ1

= Ee

= 0.

(2.9)

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Also, it follows from (2.6) E|ˆ νn (l)|2 =

|l| n2

whenever n ≥ |l|. Thus (2.9) holds as n → ∞. The proof is concluded.  Theorem 2.2 means that for every bounded continuous function f Z 2π n
P 1
1X f eiθk −→ f eiθ dθ. n 2π 0 k=1

This is a kind of law of large numbers, and is very similar to the Khinchine law of large numbers for sums of i.i.d. random variables in standard probability theory, see (1.4). One cannot see from such a first-order average the difference between eigenvalues and sample points chosen at random from the unit circle T. However, a significant feature will appear in the secondorder fluctuation, which is the main content of the following sections. 2.2

Symmetric groups and symmetric polynomials

We shall first introduce the irreducible characters of symmetric groups and state without proofs character relations of two kinds. Then we shall define four classes of symmetric polynomials and establish a Schur orthonormality formula of Schur polynomials and power polynomials with respect to Haar measure. Most of the materials can be found in Macdonald (1995) and Sagan (2000). Throughout the section, n is a fixed natural number. Consider the symmetric group , Sn , consisting of all permutations of {1, 2, · · · , n} using composition as the multiplication. Assume σ ∈ Sn , denote by rk the number of cycles of length k in σ. The cycle type, or simply the type, of σ is an expression of the form
1r1 , 2r2 , · · · , nrn . For example, if σ ∈ S5 is given by σ(1) = 2, σ(2) = 3, σ(3) = 1, σ(4) = 4, σ(5) = 5
then it has cycle type 12 , 20 , 31 , 40 , 50 . The cycle type of the identity permutation is (1n ). In Sn , permutations σ and τ are conjugates if σ = ϑτ ϑ−1

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39

for some ϑ ∈ Sn . The set of all permutations conjugate to a given σ is called the conjugacy class of σ and is denoted by Kσ . Conjugacy is an equivalent relation, so the distinct conjugacy classes partition Sn . It is not hard to see that two permutations are in the same conjugacy class if and only if they have the same cycle type, and the size of a conjugacy class with Qn cycle type (1r1 , 2r2 , · · · , nrn ) is given by n!/ i=1 iri ri !. Let Md be the set of all d × d matrices with complex entries, and Ld the group of all invertible d × d matrices with multiplication. A matrix representation of Sn is a group homomorphism X : Sn −→ Ld . Equivalently, to each σ ∈ Sn is assigned X(σ) ∈ Ld such that (i) X(1n ) = Id the identity matrix, and (ii) X(στ ) = X(σ)X(τ ) for all σ, τ ∈ Sn . The parameter d is called the degree of the representation. Given a matrix representation X of degree d, let V be the vector space of all column vectors of length d. Then we can multiply v ∈ V by σ ∈ Sn using σv = X(σ)v. This makes V into a Sn -module of dimension d. If a subspace W ⊆ V is closed under the action of Sn , that is, w ∈ W ⇒ σw ∈ W

for all σ ∈ Sn ,

then we say W is a Sn -submodule of V . A non-zero matrix representation X of degree d is reducible if the Sn module V contains a nontrivial submodule W . Otherwise, X is said to be irreducible. Equivalently, X is reducible if V has a basis B in which every σ ∈ Sn is assigned a block matrix of the form   A(σ) 0 X(σ) = , 0 B(σ) where the A(σ) are square matrices, all of the same size, and 0 is a nonempty matrix of zeros. Two representations X and Y are equivalent if and only if there exists a fixed matrix T such that Y (σ) = T X(σ)T −1 ,

for all σ ∈ Sn .

The number of inequivalent irreducible representations is equal to the number of conjugacy classes of Sn .

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A classical theorem of Maschke implies that every matrix representation of Sn having positive dimension is completely reducible. In particular, let X be a matrix representation of Sn of degree d > 0, then there is a fixed matrix T such that every matrix X(σ), σ ∈ Sn has the form  (1)  X (σ) 0 ··· 0   0 X (2) (σ) · · · 0   −1 T X(σ)T =  , .. .. . . . .   . . . . (k) 0 0 · · · X (σ) where each X (i) is an irreducible matrix representation of Sn . To every matrix representation X assigns one simple statistic, the character defined by χX (σ) = trX(σ), where tr denotes the trace of a matrix. Otherwise put, χX is the map trX

Sn −→ C. It turns out that the character contains much of the information about the matrix representation. Here are some elementary properties of characters. Lemma 2.2. Let X be a matrix representation of Sn of degree d with character χX , then (i) χX (1n ) = d; (ii) χX is a class function, that is, χX (σ) = χX (τ ) if σ and τ are in the same conjugacy class; (iii) if Y is a matrix representation equivalent to X, then their characters are identical: χX ≡ χY . Let χ and ψ be any two functions from Sn to C. The inner product of χ and ψ is defined by 1 X hχ, ψi = χ(σ)ψ(σ). n! σ∈Sn

In particular, if χ and ψ are characters, then 1 X hχ, ψi = χ(σ)ψ(σ). n!

(2.10)

σ∈Sn

Theorem 2.3. If χ and ψ are irreducible characters, then we have character relation of the first kind hχ, ψi = δχ,ψ , where δχ,ψ stands for Kronecker delta.

(2.11)

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Since character is a class function, then we can rewrite (2.10) and (2.11) as 1 X hχ, ψi = |K|χ(K)ψ(K) = δχ,ψ , n! K

where we mean by χ(K) and ψ(K) that χ and ψ act in K respectively, |K| denotes the number of K and the sum is over all conjugacy classes of Sn . This implies that the modified character table r  |K|  U= χ(K) n! χ,K has orthonormal rows. Hence U , being a square, is a unitary matrix and has orthonormal columns. Thus we have proven the character relation of the second kind as follows. Theorem 2.4. X

χ(K)χ(L) =

χ

n! δK,L , |K|

(2.12)

where the sum is take over all irreducible characters. A partition of n is a sequence (λ1 , λ2 , · · · , λl ) of non-increasing natural numbers such that l X

λi = n,

i=1

where the λi ’s are called parts, l is called the length. If λ = (λ1 , λ2 , · · · , λl ) is a partition of n, then we write λ 7→ n. We Pl also use the notation |λ| = i=1 λi . The cycle type of a permutation in Sn naturally gives a partition of n. Conversely, given a
λ 7→ n, let rk = ]{i; λi = k}, then we have a cycle type 1r1 , 2r2 , · · · , nrn . Thus there is a natural one-to-one correspondence between partitions of n and conjugacy classes of Sn . As a consequence, the number of irreducible characters is equal to the number of partitions of n. Let Pn be the set of all partitions of n. We need to find an ordering on Pn . Since each partition is a sequence of integer numbers, then a natural ordering is the ordinary lexicographic order. Let λ = (λ1 , λ2 , · · · , λl ) and µ = (µ1 , µ2 , · · · , µm ) be partitions of n. Then λ < µ in lexicographic order if, for some index i, λj = µj

for j < i

and

λi < µi .

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This is a total ordering on Pn . For instance, on P6 we have





16 < 2, 14 < 22 , 12 < 23 < 3, 13 < 3, 2, 1




< 32 < 4, 12 < 4, 2 < 5, 1 < 6 . Another ordering is the following dominance order. If λ1 + λ2 + · · · + λi ≥ µ1 + µ2 + · · · + µi for all i ≥ 1, then λ is said to dominate µ, written as λ D µ. The lexicographic order is a refinement of the dominance order in the sense that λ ≥ µ if λ, µ ∈ Pn satisfy λ D µ. Next we shall describe a graphic representation of a partition. Suppose λ = (λ1 , λ2 , · · · , λl ) 7→ n. The Young diagram (shape) of λ is an array of n boxes into l left-justified rows with row i containing λi boxes for 1 ≤ i ≤ l. The box in row i and column j has coordinates (i, j), as in a matrix, see Figure 2.1.

Fig. 2.1

Young diagram

A Young tableau of shape λ, tλ , is an array obtained by putting the numbers 1, 2, · · · , n into the boxes bijectively. A Young tableau tλ is standard if the rows are increasing from left to right and the columns are increasing from top to bottom. Let ti,j stand for the entry of t in position (i, j). Clearly there are n! Young tableau for any shape λ 7→ n. Two tableaux tλ1 and tλ2 are row equivalent, tλ1 ∼ tλ2 , if corresponding rows of the two tableaux contain the same elements. A tabloid of shape λ is  {tλ } = tλ1 , tλ1 ∼ tλ .

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The number of tableaux in any given equivalence class is λ! := λ1 !λ2 ! · · · λl !. Thus the number of tabloids of shape λ is just n!/λ!. Given a σ ∈ Sn , define σtλ = (σti,j ). λ

To illustrate, if σ = (1, 2, 3) ∈ S3 , λ = (2, 1) 7→ 3, t = 

λ

σt = (σti,j ) =

23 1



 12 , then 3

 .

This induces an action on tabloids by letting   σ tλ = σtλ . Suppose that the tableau tλ has columns C1 , C2 , · · · , Cλ1 . Let X κCj = sgn(σj )σj , σj ∈SCj

where SCj is a symmetric group of permutations of numbers from Cj . Let κtλ = κC1 κC2 · · · κCλ1 . This is a linear combinations of elements of Sn , so κtλ ∈ C[Sn ]. Now we can pass from tabloid tλ to polytabloid  etλ = κtλ tλ . Some basic properties are summarized in the following lemma.  Lemma 2.3. (i) For any λ 7→ n, etλ , tλ is a standard Young tableau is independent; (ii) For any λ 7→ n,  S λ =: span etλ , tλ is a standard Young tableau  = span etλ , tλ is a Young tableau ;  (iii) S λ , λ 7→ n form a complete list of irreducible Sn -modules. Let χλ be the character of matrix representation associated with S λ , and dλ the number of standard Young tableaux of shape λ, then we have Theorem 2.5. χλ (1n ) = dimS λ = dλ and X λ7→n

χ2λ (1n ) =

X λ7→n

d2λ = n!.

(2.13)

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Formula (2.13) is often referred to as the Burnside identity. Some more information about partitions will be found in Chapters 4 and 5. Consider the ring Z[x1 , · · · , xn ] of polynomials in n independent variables x1 , · · · , xn with rational integer coefficients. The symmetric group Sn acts on this ring by permuting the variables, and a polynomial is symmetric if it is invariant under this action. Let Λn := Λn [x1 , · · · , xn ] be the subring formed by the symmetric polynomials. We will list four classes of widely used symmetric polynomials, all indexed by partitions. • Elementary symmetric polynomials. For each integer r ≥ 0 the rth elementary symmetric polynomial er is the sum of all products of r distinct variables xi , so that e0 = 0 and for r ≥ 1 X er = xi1 xi2 · · · xir . 1≤i1
For each partition λ = (λ1 , λ2 , · · · , λl ) define eλ = eλ1 eλ2 · · · eλl . er , r ≥ 0 are algebraically independent over Z, and every element of Λn is uniquely expressible as a polynomial in the er . • Complete symmetric polynomials. For each integer r ≥ 0 the rth complete symmetric polynomial hr is the sum of all monomials of total degree r in the variables x1 , x2 , · · · , xn . In particular, h0 = 1 and h1 = e1 . By convention, hr and er are defined to be zero for r < 0. Define hλ = hλ1 hλ2 · · · hλl for any partition λ = (λ1 , λ2 , · · · , λl ). The hr , r ≥ 0 are algebraically independent over Z, and Λn = Z[h1 , h2 , · · · , hn ] • Schur symmetric polynomials. For each partition λ = (λ1 , λ2 , · · · , λl ) with length l ≤ n, consider the determinant λ +n−j
. det xi j 1≤i,j≤n This is divisible in Z[x1 , x2 , · · · , xn ] by each of the differences xj − xi (1 ≤ Q i < j ≤ n), and hence by their product 1≤i<j≤n (xj − xi ), which is the
. Define Vandermonde determinant det xn−j i 1≤i,j≤n λ +n−j

sλ := sλ (x1 , · · · , xn ) =

det(xi j

)1≤i,j≤n , n−j det(xi )1≤i,j≤n

(2.14)

where sλ = 0 if the numbers λj + n − j (1 ≤ j ≤ n) are not all distinct.

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The quotient (2.14) is a symmetric and homogeneous polynomial of degree |λ|, that is, in Λn . It is called the Schur polynomial in the variable x1 , x2 , · · · , xn , corresponding to the partition λ. The Schur polynomial sλ (x1 , · · · , xn ) with l ≤ n form a Z-basis of Λn . Each Schur polynomial can be expressed as a polynomial in the elementary symmetric polynomials er , and as a polynomial in the complete symmetric polynomial hr . The formulas are: sλ = det(hλi −i+j )1≤i,j≤n where l(λ) ≤ n, and sλ = det(eλ0i −i+j )1≤i,j≤n where λ0 is a conjugate partition with l(λ0 ) ≤ n. In particular, we have s(n) = hn ,

s(1n ) = en .

• Power sum polynomials. For each r ≥ 1 the rth is pr := pr (x1 , · · · , xn ) =

n X

xri .

i=1

We define pλ = pλ1 pλ2 · · · pλl for each partition λ = (λ1 , λ2 , · · · , λl ). Note that pλ , λ 7→ n do not form a Z-basis of Λn . For instance, 1 1 h2 = p2 + p21 2 2 does not have integral coefficients when expressed in terms of the pλ . In
general, for any partition λ = 1r1 , 2r2 , · · · of n, define zλ =

∞ Y

iri ri !.

(2.15)

i=1

Then we can express hn , en as linear combinations of the pλ as follows X hn = zλ−1 pλ λ7→n

and en =

X

(−1)n−l(λ) zλ−1 pλ .

λ7→n

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The Schur polynomial sλ can also be expressed as linear combinations of the pλ X sλ = zρ−1 χλ (ρ)pρ , ρ7→|λ|

where χλ (ρ) is the value of irreducible character χλ at permutations of cycle-type ρ. Conversely, we have the following inverse formula, which is called Schur-Weyl duality. Theorem 2.6. pλ =

X

χρ (λ)sρ .

(2.16)

ρ7→|λ|

We now define an inner product on Λn . Suppose f ∈ Λn , let
f (Un ) = f eiθ1 , · · · , eiθn , where Un is an n × n unitary matrix with eigenvalues eiθ1 , · · · , eiθn . Thus f : Un 7→ C is invariant under unitary transforms. Suppose we are given two symmetric polynomials f, g ∈ Λn , their inner product is defined by Z hf, gi = f (Un )g(Un )dµn . Un

It turns out that Schur polynomials are orthonormal with respect to this inner product, which is referred to as Schur orthonormality. In particular, we have Theorem 2.7. hsλ , sτ i = δλ,τ 1(l(λ)≤n) . Proof.

According to (2.1), we have Z hf, gi = f (Un )g(Un )dµn Un Z

1 = f eiθ1 , · · · , eiθn g e−iθ1 , · · · , e−iθn n (2π) n! [0,2π]n Y eiθj − eiθk 2 dθ1 · · · dθn . · 1≤j
(2.17)

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If λ and τ are two partitions of lengths ≤ n, then by (2.14) Z

1 hsλ , sτ i = sλ eiθ1 , · · · , eiθn sτ e−iθ1 , · · · , e−iθn (2π)n n! [0,2π]n Y eiθj − eiθk 2 dθ1 · · · dθn · 1≤j
Z

1 det ei(λk +n−k)θj det e−i(τk +n−k)θj dθ1 · · · dθn = n (2π) n! [0,2π]n  1 = (2.18) det(ei(λk +n−k)θj ) det(e−i(τk +n−k)θj ) 1 , n! where [f ]1 denotes the constant term of f . A simple algebra shows   det(ei(λk +n−k)θj ) det(e−i(τk +n−k)θj ) 1 = n!δλ,τ , which together with (2.18) implies hsλ , sτ i = δλ,τ . We conclude the proof.



Having Schur orthonormality (2.17), we can further compute the inner product of power sum polynomials. For any partitions µ and ν, applying the Schur-Weyl duality (2.16) immediately yields X X χρ (µ)χσ (ν)hsρ , sσ i hpµ , pν i = ρ7→|µ| σ7→|ν|

=

X X

χρ (µ)χσ (ν)δρ,σ 1(l(ρ)≤n)

ρ7→|µ| σ7→|ν|

X

= δ|µ|,|ν|

χρ (µ)χρ (ν)1(l(ρ)≤n) .

ρ7→|µ|

When |µ| = |ν| ≤ n, the sum is taken over all partitions of |µ|, and so the character relation (2.12) of the second kind shows hpµ , pν i = zµ δµ,ν ,

(2.19)

where zµ is defined as in (2.15).

2.3

Linear functionals of eigenvalues

Let f : T 7→ R be a square integrable function, that is, f ∈ L2 (T, dµ). Define the Fourier coefficients by Z 2π
1 ˆ fl = f eiθ e−ilθ dθ, −∞ < l < ∞, 2π 0

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so that fˆ0 is the average of f over T. Since f is real, then fˆ−l = fˆl . In this section we
shall focus on the fluctuation of linear eigenvalue Pn statistic k=1 f eiθk around the average nfˆ0 . P∞ Theorem 2.8. If f ∈ L2 (T, dµ) is such that l=1 l|fˆl |2 < ∞, then n X

d f eiθk − nfˆ0 −→ N 0, σf2 , n → ∞ (2.20) k=1

where

σf2

=2

P∞

l=1

l|fˆl |2 .

This theorem goes back to Szeg¨o as early as in 1950s, and is now known as Szeg¨ o’s strong limit theorem. There exist at least six different proofs with slight different assumptions on f in literature, and the most classical one uses the orthogonal polynomials on the unit circle T. Here we prefer to prove the theorem using the moment method of Diaconis and Evans (2001), Diaconis (2003). The interested reader is referred to Simon (2004) for other five proofs. See also a recent survey of Deift, Its and Krasovsky (2012) for extensions and applications. Lemma 2.4. Suppose that Z = X + iY is a complex standard normal random variable, then for any non-negative integers a and b EZ a Z¯ b = a!δa,b . Proof.

Z can clearly be written in polar coordinates as follows: Z = γeiθ ,

where γ and θ are independent, θ is uniform over [0, 2π], and γ has density 2 function 2re−r , r ≥ 0. It easily follows EZ a Z¯ b = Eγ a+b eiθ(a−b) = Eγ a+b Eeiθ(a−b) = Eγ 2a δa,b = a!δa,b , as desired.



Lemma 2.5. (i) Suppose that Zl , l ≥ 1 is a sequence of i.i.d. complex standard normal random variables. Then for any m ≥ 1 and any nonnegative integers a1 , a2 , · · · , am and b1 , b2 , · · · , bm m X n m X n a l Y bl Y E eilθk e−ilθk l=1 k=1 m √ Y

l=1 k=1

m √
al Y
b lZl lZ l l

=E

l=1

l=1

(2.21)

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 Pm Pm whenever n ≥ max l=1 lal , l=1 lbl . (ii) For any integer j, l ≥ 1, n n X X E eilθk e−ijθk = δj,l min(l, n). k=1

Proof.

(2.22)

k=1

Recall the lth power sum polynomial n
X pl eiθ1 , · · · , eiθn = eilθk . k=1

Then it follows m X n Y

eilθk

al

= pλ eiθ1 , · · · , eiθn




l=1 k=1

and m X n Y

e−ilθk

b l

= pµ (eiθ1 , · · · , eiθn )

l=1 k=1



where λ = 1a1 , 2a2 , · · · , mam and µ = 1b1 , 2b2 , · · · , mbm . According to the orthogonality relation in (2.19), we have m X n m X n a l Y bl Y E eilθk e−ilθk l=1 k=1

l=1 k=1

= hpλ , pµ i m Y = δλ,µ lal al !

(2.23)

l=1

whenever |λ|, |µ| ≤ n. Now we can get the identity (2.21) using Lemma 2.4. Turn to (2.22). We immediately know from (2.23) that the expectation is zero if j 6= l, while the case j = l has been proven in (2.6).  As an immediate consequence, we have the following Theorem 2.9. For each integer m ≥ 1 n X  √ d l Zl , eilθk , 1 ≤ l ≤ m −→


1≤l≤m ,

n → ∞.

k=1

In particular, it holds m n X X
d 2 ˆ fl e−ilθk − nfˆ0 −→ N 0, σm,f , l=−m 2 where σm,f =2

k=1

Pm

l=1

l|fˆl |2 .

n→∞

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P∞ Proof of Theorem 2.8. Since l=−∞ |fˆl |2 < ∞, then we can express f in terms of Fourier series ∞ X
f eiθ = fˆl e−ilθ , l=−∞

from which it follows n ∞ n X X X
f eiθk = fˆl e−ilθk k=1

l=−∞

= nfˆ0 +

k=1 ∞ X

fˆl

l=1

n X

e−ilθk +

k=1

∞ X

fˆl

l=1

n X

e−ilθk .

k=1

It is sufficient for us to establish the following statement: there exists a sequence of numbers mn with mn → ∞ and mn /n → 0 such that (i) ∞ n X X P fˆl e−ilθk −→ 0; (2.24) l=mn +1

k=1

(ii) mn X

fˆl

l=1

n X

d

e−ilθk −→

k=1

∞ X

l|fˆl |2

1/2

NC (0, 1),

(2.25)

l=1

where NC (0, 1) denotes a complex standard normal random variable. Indeed, for any sequence of numbers mn with mn → ∞, we have by (2.22) ∞ n ∞ X 2 X X E fˆl e−ilθk = min(l, n)|fˆl |2 l=mn +1

k=1

≤

l=mn +1 ∞ X

l|fˆl |2

l=mn +1

→ 0, which directly implies (2.24) using the Markov inequality. For (2.25), note the moment identity (2.21) is applicable to yield mn mn n n X a X  X ¯ X ilθk b E fˆl e−ilθk fˆl e l=1 k=1 mn X

l=1

=E

l=1

l=1

=

k=1

mn b √ a X ¯√ fˆl l Z l fˆl l Zl

mn X l=1

l|fˆl |2

a

a!δa,b

(2.26)

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whenever n ≥ mn (a + b). If mn /n → 0, then for any non-negative integer numbers a and b, the assumption n ≥ mn (a+b) can be guaranteed for sufficiently large n. Now we can conclude the claim (2.25) by letting n → ∞ in (2.26). The proof is now complete.  Remark 2.1. A remarkable feature in Theorem 2.8 is that there is no normalizing constant in left hand side of (2.20). Recall that there is a normal√ izing constant 1/ n in the central limit theorem for sums of i.i.d. random variables with finite variance. This further manifests that the eigenvalues of the CUE spread out more regularly on the unit circle T than independent uniform points. This phenomena also appears in the central limit theorem for linear functional of eigenvalues of the Gaussian Unitary Ensemble (GUE), see Chapter 3 below. The following result shows that even when ∞ X l|fˆl |2 = ∞, l=1


Pn the central limit theorem for k=1 f eiθk after properly scaled still holds under a weak additional assumption. Recall that a positive sequence {ck } is said to be slowly varying if for any α > 0 cbαkc = 1. lim k→∞ ck Theorem 2.10. Suppose that f ∈ L2 (T, dµ) is such that Bn :=

n X

l|fˆl |2 ,

n≥1

l=1

is slowly varying. Then n 
1 X d √ f eiθk − nfˆ0 −→ N (0, 1). 2Bn k=1 Proof.

As in the proof of Theorem 2.8, it follows n ∞ n n ∞ X X X X
¯ X ilθk f eiθk − nfˆ0 = fˆl e−ilθk + fˆl e . k=1

l=1

k=1

l=1

It is enough to prove ∞ n 1 X ˆ X −ilθk d √ fl e −→ NC (0, 1). Bn l=1 k=1

k=1

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Because Bn , n ≥ 1 is slowly varying, there must be a sequence of integers mn such that as n → ∞ mn mn → ∞, →0 n and Bmn → 1. (2.27) Bn We shall establish the following statements: (i) ∞ X

1 √ Bn

fˆl

l=mn +1

n X

P

e−ilθk −→ 0;

(2.28)

k=1

(ii) mn n X 1 X d √ fˆl e−ilθk −→ NC (0, 1). Bn l=1 k=1

According to (2.22), ∞ n X 2 X E fˆl e−ilθk = l=mn +1

k=1

=

∞ X l=mn +1 n X

(2.29)

|fˆl |2 min(l, n) ∞ X

l|fˆl |2 + n

l=mn +1

l=n+1

Summing by parts, 2

∞ X

|fˆl |2 =

l=n+1

∞ X

(Bl+1 − Bl )

l=n

=

∞ X l=n

1 l+1

Bn Bl − . l(l + 1) n + 1

Since Bn , n ≥ 1 is slowly varying, ∞ n X Bl → 1. Bn l(l + 1) l=n

Putting these together implies ∞ n 1 X ˆ X −ilθk 2 E fl e → 0, Bn l=mn +1

which in turn implies (2.28).

k=1

|fˆl |2 .

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Turn to (2.29). Fix non-negative integers a and b. Since mn /n → 0, then n ≥ mn (a + b) for sufficiently large n. So (2.21) is applicable to yield mn mn n n X a X  X ¯ X ilθk b E fˆl e−ilθk fˆl e l=1

=

mn X

k=1

l|fˆl |2

a

l=1

k=1

a!δa,b .

(2.30)

l=1

Thus (2.29) is valid from (2.27) and (2.30). The proof is now complete.  To conclude this section, we shall look at two interesting examples. The first one is the distribution of values taken by the logarithm of characteristic polynomial of a random unitary matrix. Recall the characteristic polynomial of a matrix Un is defined by the determinant det(zIn − Un ). Fix z = eiθ0 and assume Un is from the CUE. Since eiθ0 is almost surely not an eigenvalue of Un , then n
Y
det eiθ0 In − Un = eiθ0 − eiθk 6= 0. k=1


It is fascinating that the logarithm of det eiθ0 In −Un after properly scaled weakly converges to a normal distribution, analogous to Selberg’s result on the normal distribution of values of the logarithm of the Riemann zeta function. This was first observed by Keating and Snaith (2000), which argued that the Riemann zeta function on the critical line could be modelled by the characteristic polynomial of a random unitary matrix. Theorem 2.11. As n → ∞,
d 1 √ log det(eiθ0 In − Un ) − inθ0 −→ NC (0, 1), log n where log denotes the usual branch of the logarithm defined on C \ {z : Re(z) ≤ 0}. Proof.

First observe n X

log det eiθ0 In − Un − inθ0 = log 1 − ei(θk −θ0 ) . k=1

According to Weyl’s formula, ei(θk −θ0 ) ,


d 1 ≤ k ≤ n = eiθk ,


1≤k≤n .

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Hence it suffices to prove n
d 1 X √ log 1 − eiθk −→ NC (0, 1). log n k=1

Note that it follows for any n ≥ 1 n X k=1

n ∞ X
1 X ilθk e , log 1 − eiθk = − l l=1

a.s.

(2.31)

k=1

Indeed, for any real r > 1 ∞  X eiθk  1 ilθk log 1 − e , =− r lrl l=1

and so n X k=1

n ∞  X 1 X ilθk eiθk  =− e . log 1 − r lrl l=1

k=1

Thus we have by virtue of (2.22) ∞ n ∞ X 2 X X 2 1 1 11 ilθk E min(l, n) − 1 − 1 e = l rl l2 r l l=1

k=1

l=1

=

∞ 2 2 X 11 + . − 1 − 1 l rl l2 r l

n X 1 1 l=1

l=n+1

Letting r → 1+ easily yields n ∞ 2 X X 1 1 ilθk − 1 e E → 0, l rl k=1

l=1

which in turn implies (2.31). Now we need only prove n ∞ 1 X 1 X ilθk d √ e −→ NC (0, 1). log n l=1 l k=1

The proof is very similar to that of Theorem 2.8. Let mn = n/log n so that mn → ∞ and mn /n → 0. We shall establish the following statements: (i) n ∞ X 1 1 X ilθk P √ e −→ 0; log n l=m +1 l k=1 n

(2.32)

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(ii) mn n 1 X 1 X ilθk d √ e −→ NC (0, 1). log n l=1 l k=1

According to (2.22), it holds n ∞ X 1 X ilθk 2 e = E l k=1

l=mn +1

n X l=mn +1

(2.33)

∞ X 1 1 +n l l2 l=n+1

= O(log log n), which together with the Markov inequality directly implies (2.32). To prove (2.33), note for any non-negative integers a and b, mn mn n n X 1 X −ilθk b 1 X ilθk a X e e E l l k=1

l=1

=

mn X 1 a l=1

l

l=1

k=1

a!δa,b

= (1 + o(1))(log n)a a!δa,b , as desired.



The second example of interest is the numbers of eigenvalues lying in an arc. For 0 ≤ a < b < 2π, write Nn (a, b) for the number of eigenvalues eiθk with θk ∈ [a, b]. Particularly speaking, Nn (a, b) =

n X

1(a, b) (θk ).

k=1

Since each eigenvalue eiθk is uniform over T, then n(b − a) . 2π The following theorem, due to Wieand (1998) (see also Diaconis and Evans (2001)), shows that the fluctuation of Nn (a, b) around the mean is asymptotically normal. It is worth mentioning that the asymptotic variance log n (up to a constant) is very typical in the study of numbers of points like eigenvalues in an interval. The reader will again see it in the study of GUE and random Plancherel partitions. ENn (a, b) =

Theorem 2.12. For 0 ≤ a < b < 2π, as n → ∞ Nn (a, b) − n(b−a) d 2π √ −→ N (0, 1). 1 log n π

(2.34)

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(2.34) is actually a direct consequence of Theorem 2.10. Indeed,


f eiθ = 1(a, b) (θ). Then a simple calculation shows
b−a 1 fˆ0 = , fˆl = e−ila − e−ilb , 2π l2πi and so n X Bn : = l|fˆl |2

l 6= 0

l=1

=

n 2 1 X 1 −ila e − e−ilb 4π 2 l l=1

n
1 X1 = 2 + 2 cos l(b − a) . 2 4π l l=1

On the other hand, an elementary calculus shows n 1 X cos l(b − a) → 0, n → ∞. log n l l=1

Hence Bn is a slowly varying and 1 Bn → , log n 2π 2 The proof is complete.

n → ∞. 

The above theorem deals only with the number of eigenvalues in a single arc. In a very similar way, employing the Cram´er-Wald device, one may prove a finite dimensional normal convergence for multiple arcs. Theorem 2.13. As n → ∞, the finite dimensional distribution of the processes Nn (a, b) − n(b−a) 2π √ , 0 ≤ a < b < 2π 1 log n π converges to those of a centered Gaussian process {Z(a, b) : 0 ≤ a < b < 2π} with the covariance structure  1, if a = a0 and b = b0 ,     1  , if a = a0 and b 6= b0 ,    2 1 0 0 EZ(a, b)Z(a0 , b0 ) = 2 , if a 6= a and b = b ,     − 12 , if b = a0 ,     0, otherwise. Proof.

See Theorem 6.1 of Diaconis (2001).



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Five diagonal matrix models

This section is aimed to establish a five diagonal sparse matrix model for the CUE and to provide an alternate approach to asymptotic normality of the characteristic polynomials and the number of eigenvalues lying in an arc. We first introduce basic notions of orthogonal polynomials and Verblunsky coefficients associated to a finitely supported measure on the unit circle, and quickly review some well-known facts, including the Szeg¨o recurrence equations and Verblunsky’s theorem. The measure we will be concerned with is the spectral measure induced by a unitary matrix and a cyclic vector. Two matrices of interest to us are upper triangular Hessenberg matrix and CMV five diagonal matrix, whose Verblunsky coefficients can be expressed in a simple way. Then we turn to a random unitary matrix distributed with Haar measure. Particularly interesting, the associated Verblunsky coefficients are independent Θv -distributed complex random variables. Thus as a consequence of Verblunsky’s theorem, we naturally get a five diagonal matrix model for the CUE. Lastly, we rederive Theorems 2.11 and 2.12 via a purely probabilistic approach: use only the classical central limit theorems for sums of independent random variables and martingale difference sequences. Assume we are given a finitely supported probability measure dν on exactly n points eiθ1 , eiθ2 , · · · , eiθn with masses ν1 , ν2 , · · · , νn , where νi > 0 Pn and i=1 νi = 1. Let L2 (T, dν) be the of square integrable functions on T with respective to dν with the inner product given by Z

hf, gi = f eiθ g eiθ dν. T  Applying the Gram-Schmidt algorithm to the ordered set 1, z, · · · , z n−1 , we can get a sequence of orthogonal polynomials Φ0 , Φ1 , · · · , Φn−1 , where Φ0 (z) = 1,

Φk (z) = z k + lower order.

Define the Szeg¨ o dual by Φ∗k (z) = z k Φk (¯ z −1 ). Namely, Φk (z) =

k X l=0

cl z l

⇒

Φ∗k (z) =

k X

c¯k−l z l .

l=0

As Szeg¨ o discovered, there exist complex constants α0 , α1 , · · · , αn−2 ∈ D, where D := {z ∈ C, |z| < 1}, such that for 0 ≤ k ≤ n − 2 Φk+1 (z) = zΦk (z) − α ¯ k Φ∗k (z)

(2.35)

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and Φ∗k+1 (z) = Φ∗k (z) − αk zΦk (z).

(2.36)

Expanding z n in this basis shows that there exists an αn−1 , say αn−1 = eiη ∈ T, 0 ≤ η < 2π, such that if letting Φn (z) = zΦn−1 (z) − α ¯ n−1 Φ∗n−1 (z),

(2.37)

then Φn (z) = 0

in L2 (T, dν).

Define ρk =

p 1 − |αk |2 ,

0 ≤ k ≤ n − 1,

then it follows from recurrence relations (2.35) and (2.36) kΦ0 k = 1,

kΦk k =

k−1 Y

ρl ,

k ≥ 1.

(2.38)

l=0

The orthonormal polynomial φk is defined by φk (z) =

Φk (z) . kΦk k

We call αk , 0 ≤ k ≤ n − 1 the Verblunsky coefficients associated to the measure dν, which play an important role in the study of unitary matrices. We sometimes write αk (dν) for αk to emphasize the dependence on the underlying measure dν. A basic fact we need below is Theorem 2.14. There is a one-to-one correspondence between the finitely supported probability measure dν on T and complex numbers α0 ,α1 ,· · · , αn−1 with α0 ,α1 , · · · , αn−2 ∈ D and αn−1 ∈ T. This theorem is now called Verblunsky’s theorem (also called Favard’s theorem for the circle). The reader is referred to Simon (2004) for the proof (at least four proofs are presented). It is very expedient to encode the Szeg¨o recurrence relation (2.35) and (2.36). Let Bk (z) = z

Φk (z) . Φ∗k (z)

It easily follows B0 (z) = z,

Bk+1 (z) = zBk (z)

¯k (z) 1−α ¯k B , 1 − αk Bk (z)

z ∈ T,

(2.39)

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which shows that the Bk (z) can be completely expressed in terms of the Verblunsky coefficients αk . Note in view of (2.39), Bk is a finite Blaschke product of degree k + 1. Define a continuous function ψk (θ) : [0, 2π) 7→ R via
Bk eiθ = eiψk (θ) . (2.40) ψk (θ) is the absolute Pr¨ afer phase of Bk , so the set of points  iθ  iθ e : Bn−1 (e ) = α ¯ n−1 = eiθ : ψn−1 (θ) ∈ 2πZ + η is the support of dν. Also, ψk (θ) is a strictly increasing function of θ. To avoid ambiguity, we may choose a branch of the logarithm in (2.39) so that ψ0 (θ) = θ,

ψk+1 (θ) = ψk (θ) + θ + Υ(ψk , αk ),

(2.41)

where Υ(ψ, α) = −2Im log(1 − αeiψ ). Let C[z] be the vector space of complex polynomials in the variable z. Consider the multiplication operator Π : f (z) = zf (z) in C(z). We easily obtain an explicit expression of Π in the basis of orthonormal polynomials φk , 0 ≤ k ≤ n − 1. In particular,       0 φ0 φ0  φ1   φ1   0          .  ..    . L  . Π  .  = Hn  ..  +  .  ,       φn−2   φn−2   0   Φn φn−1 φn−1 kΦn−1 k
L L where Hn = Hij 0≤i,j≤n−1 is an lower triangular Hessenberg matrix given by  Qi  −αj−1 α ¯ i l=j+1 ρl , j ≤ i − 1,    −αi−1 α ¯i, j = i, L Hij =  ρi , j = i + 1,    0, j > i + 1. A simple algebra further shows that the characteristic polynomial of HnL is equal to the nth polynomial Φn (z) defined in (2.37). Namely, det(zIn − HnL ) = Φn (z), iθ1

iθn

HnL .

(2.42)

which implies e , · · · , e are the spectrum of So, the spectral analL ysis of Hn can give relations between the zeros of orthogonal polynomials and the Verblunsky coefficients. However, HnL is a far from sparse matrix

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L and the entries Hij depend on the Verblunsky coefficients αk and ρk in a complicated way. This makes difficult this task. Moreover, the numerical computations of zeros of high degree orthogonal polynomials becomes a nontrivial problem due to the Hessenberg structure of HnL . To overcome this difficulty, Cantero, Moral, and Vel´azquez (2003) used a simple and ingenious idea. Applying the Gram-Schmidt proce −1 2 −2 dure to the first n of the ordered set 1, z, z , z , z , · · · rather than  1, z, · · · , z n−1 , we can get a sequence of orthogonal Laurent polynomials, denoted by χk (z), 0 ≤ k ≤ n−1. We will refer to the χk as the standard right orthonormal L-polynomial with respect to the measure dν. Interestingly, the χk can be expressed in terms of the orthonormal polynomial φk and its Szeg¨ o dual φ∗k as follows:

χ2k+1 (z) = z −k φ2k+1 (z), χ2k (z) = z −k φ∗2k (z). Similarly, applying the Gram-Schmidt procedure to the first n of the or dered set of 1, z −1 , z, z −2 , z 2 , · · · , we can get another sequence of orthogonal Laurent polynomial, denoted by χk∗ . We call the χk∗ the standard left orthogonal L-polynomial. It turns out that the χk and χk∗ are closely related to each other through the equation:
χk∗ (z) = χ ¯k z −1 . Define  Ξk =

α ¯ k ρk ρk −αk



for 0 ≤ k ≤ n − 2, while Ξ−1 = (1) and Ξn−1 = (α ¯ n−1 ) are 1 × 1 matrices. Then it readily follows from the Szeg¨o recurrence relation that      1 χ2k (z) χ2k−1 (z) −α2k−1 1 = , χ2k∗ (z) χ2k−1∗ (z) 1 −¯ α2k−1 ρ2k−1     χ2k−1 (z) χ2k−1∗ (z) = Ξ2k−1 , χ2k (z) χ2k∗ (z)     χ2k∗ (z) χ2k (z) z = Ξ2k . χ2k+1∗ (z) χ2k+1 (z) It can be further written as a five term recurrence equation: zχ0 (z) = α ¯ 0 χ0 (z) + ρ0 χ1 (z),

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  ρ2k−2 α ¯ 2k−1 −α2k−2 α ¯ 2k−1 χ2k−2 (z) z = χ2k−1 (z) ρ2k−2 ρ2k−1 −α2k−2 ρ2k−1    ρ2k−1 α ¯ 2k ρ2k−1 ρ2k χ2k (z) + . χ2k+1 (z) −α2k−1 α ¯ 2k −α2k−1 ρ2k Construct now the n × n block diagonal matrices χ2k−1 (z) χ2k (z)

L = diag(Ξ0 , Ξ2 , Ξ4 , · · · ),

M = diag(Ξ−1 , Ξ1 , Ξ3 , · · · )

and define Cn = ML,

Cnτ = LM.

(2.43)

It is easy to check that L and M are symmetric unitary matrices, and so both Cn and Cnτ are unitary. A direct manipulation of matrix product shows that Cn is a five diagonal sparse matrix. Specifically speaking, if n = 2k, then Cn is equal to   α ¯0 ρ0 0 0 0 ··· 0 0   ρ0 α ¯ 1 ρ1 α ¯ 2 ρ1 ρ2 0 ··· 0 0   ¯ 1 −α0 α   ρ ρ −α ρ −α α 0 ··· 0 0 0 1 1 ¯ 2 −α1 ρ2   0 1   0 ρ2 α ¯ 3 −α2 α ¯ 3 ρ3 α ¯4 · · · 0 0   0 ,    0 0 ρ2 ρ3 −α2 ρ3 −α3 α4 · · · 0 0   .. .. .. .. .. ..   .. ..   . . . . . . . .    0 0 0 0 0 · · · −αn−3 α ¯ n−2 −αn−3 ρn−2  0 0 0 0 0 · · · ρn−2 α ¯ n−1 −αn−2 α ¯ n−1 while if n = 2k + 1, then Cn is equal to   α ¯0 ρ0 0 0 ··· 0 0 0   ρ0 α ¯ 1 ρ1 α ¯ 2 ρ1 ρ2 · · · 0 0 0   ¯ 1 −α0 α   ρ ρ −α ρ −α α 0 0 0 0 1 1 ¯ 2 −α1 ρ2 · · ·   0 1   0 ρ2 α ¯ 3 −α2 α ¯3 · · · 0 0 0   0 .    0 0 ρ2 ρ3 −α2 ρ3 · · · 0 0 0   .. .. .. .. .. ..   .. ..   . . . . . . . .    0 0 0 0 · · · ρn−3 α ¯ n−2 −αn−3 α ¯ n−2 ρn−2 α ¯ n−1  0 0 0 0 · · · ρn−3 ρn−2 −αn−3 ρn−2 −αn−2 α ¯ n−1 The multiplication operator Π : f (z) 7→ zf (z) can be explicitly expressed in the basis of χk , 0 ≤ k ≤ n − 1 as follows. If n = 2k, then       0 χ0 χ0   χ1   χ1   0         ..     , Π  ...  = Cn  ...  +  (2.44) .          χn−2   χn−2   0  Φn χn−1 χn−1 z k−1 kΦn−1 k

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while if n = 2k + 1, then 

χ0 χ1 .. .





χ0 χ1 .. .







0 0 .. .

                        .  = Cn  + Π       0  χn−3    χn−3         n  χn−2   χn−2   ρn−2 zk−1 Φ kΦn−1 k  n χn−1 χn−1 −αn−2 zk−1 Φ kΦn−1 k

(2.45)

The analog in the basis of χk∗ , 0 ≤ k ≤ n − 1 holds with Cn replaced by Cnτ . Call Cn and Cnτ the CMV matrices associated to α0 , α1 , · · · , αn−1 . Similarly to the equation (2.42), we have Lemma 2.6. In the above notations, det(zIn − Cn ) = Φn (z). Proof.

If n = 2k, then by (2.44) 

χ0 χ1 .. .





         (zIn − Cn )  =     χn−2    χn−1

0 0 .. . 0 Φn (z) z k−1 kΦn−1 k

    .   

Denote by Cn,k the k × k subminor matrix of Cn . Applying to solve the above system with respect to χn−1 (z), we get    zI n−1 − Cn,n−1 det  χn−1 (z) =  det(zIn − Cn )  ··· 1

=

Φn (z) det(zIn−1 − Cn,n−1 ) , z k−1 kΦn−1 k det(zIn − Cn )

which implies det(zIn − Cn ) Φn (z) = . det(zIn−1 − Cn,n−1 ) Φn−1 (z)

0 .. . 0 Φn (z) z k kΦn−1 k

     

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Similarly, if n = 2k + 1, applying Cram´er’s rule to solve the initial system (2.45) with respect to χn−2 (z) gives 1 χn−2 (z) = det(zIn − Cn )   0 0   .. ..   zI . .   n−2 − Cn,n−2   0 0 · det     n (z)   ··· ρn−2 zk−1ΦkΦ −ρ α ¯ n−2 n−1 n−1 k   Φn (z) ¯ n−1 ··· −αn−2 zk−1 kΦn−1 k z + αn−2 α = zρn−2

Φn (z) k−1 z kΦn−1 k

det(zIn−2 − Cn,n−2 ) det(zIn − Cn )

and so det(zIn − Cn ) Φn (z) = . det(zIn−2 − Cn,n−2 ) Φn−2 (z) Thus we find by induction that for n ≥ 1 det(zIn − Cn ) Φn (z) = . det(z − Cn,1 ) Φ1 (z) Since Φ1 (z) = z − α ¯ 0 = det(z − Cn,1 ), then the claim follows.



In what follows we will be concerned with the spectral measure of a unitary matrix. Let Un be a unitary matrix from Un , e1 = (1, 0, · · · , 0)τ a cyclic vector. Construct a probability measure dν on T such that Z z m dν = hUnm e1 , e1 i, m ≥ 0. T

Note that dν is of finite support. Indeed, let eiθ1 , eiθ2 , · · · , eiθn be the eigenvalues of Un , then there must exist a unitary matrix Vn such that   iθ1 e 0 ··· 0  0 eiθ2 · · · 0   ∗  Un = Vn  . .. . . ..  Vn . .  . . .  . 0

0 · · · eiθn

Furthermore, Vn may be chosen to consist of eigenvectors v1 , v2 , · · · , vn . If we further require that v11 := q1 > 0, v12 := q2 > 0, · · · , v1n := qn > 0, then Vn is uniquely determined. In addition, it easily follows q12 + q22 + · · · + qn2 = 1

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because of orthonormality of Vn . Thus it follows n X qj2 eimθj , m ≥ 0. hUnm e1 , e1 i = j=1

 So dν is supported by eiθ1 , eiθ2 , · · · , eiθn and
ν eiθj = qj2 , j = 1, 2, · · · , n. Having the measure dν on T, we can produce the Verblunsky coefficient. s αk (dν) We shall below write αk (Un , e1 ) for αk (dν) to indicate the underlying matrix and cyclic vector. The following lemmas provide us with two nice examples of unitary matrices, whose proofs can be found in Simon (2004). Lemma 2.7. Given a sequence of complex numbers α0 , α1 , · · · , αn−2 ∈ D and
αn−1 ∈ T, construct an upper triangular Hessenberg matrix HnU = U by letting Hij 0≤i,j≤n−1 Qj−1  −αi−1 α ¯ j l=i ρl , i < j,    −αi−1 α ¯i, i = j, U (2.46) Hij =  ρ , i = j + 1,   j 0, i > j + 1.
Then αk HnU , e1 = αk , 0 ≤ k ≤ n − 1. Lemma 2.8. Given a sequence of complex numbers α0 , α1 , · · · , αn−2 ∈ D and αn−1 ∈ T, construct a CMV matrix Cn as in (2.43). Then αk (Cn , e1 ) = αk , 0 ≤ k ≤ n − 1. What is the distribution of the αk (Cn , e1 ) if Un is chosen at random from the CUE? To answer this question, we need to introduce a notion of Θv -distributed random variable. A complex random variable Z is said to be Θv -distributed (v > 1) if for any f Z Z
(v−3)/2 v−1 Ef (Z) = f (z) 1 − |z|2 dz. 2π D For v ≥ 2 an integer, there is an intuitive geometric interpretation for Z. Lemma 2.9. If X = (X1 , · · · , Xn , Xn+1 ) ∈ Rn+1 is uniform over the ndimensional unit sphere Sn , then for any 1 ≤ k ≤ n, Z Γ( n+1 2 ) f (x1 , · · · , xk ) Ef (X1 , · · · , Xk ) = ) Bk 2π k/2 Γ( n−k+1 2
(n−k−1)/2 · 1 − x21 − · · · − x2k dx1 · · · dxk , (2.47)

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 where Bk = (x1 , · · · , xk ) : x21 + · · · + x2k < 1 . In particular, X1 is Beta(1/2, n/2)-distributed and X1 + iX2 is Θn distributed. Proof. (2.47) is actually a direct consequence of the following change-ofvariables formula using matrix volume: Z Z f (v)dv = (f ◦ φ)(u) Jφ (u) du (2.48) V

U

where V ⊆ Rn and U ⊆ Rm with n ≥ m, f is integrable on V, φ : U 7→ V is sufficiently well-behaved function, dv and du denote respectively the volume element in V ⊆ Rn , and Jφ (u) is the volume of Jacobian matrix Jφ (u). To apply (2.48) in our setting, let φk (x1 , · · · , xn ) = xk ,

1≤k≤n

and φn+1 (x1 , · · · , xn ) = 1 − x21 − · · · − x2n n

n

So the S is the graph of B under the Jacobian matrix of φ is  1 0 0  0 1 0   Jφ =  0 0  0  0 0  0

∂φn+1 ∂φn+1 ∂φn+1 ∂x1 ∂x2 ∂x3


1/2

.

mapping φ = (φ1 , · · · , φn+1 ). The ··· ··· .. . ··· ···

0 0

0 0

1 0

0 1

∂φn+1 ∂φn+1 ∂xn−1 ∂xn

    .   

This is an n + 1 × n rectangular matrix, whose volume is computed by q Jφ = det(J τ Jφ ) φ

= 1 − x21 − · · · − x2n


−1/2

.

Hence according to (2.48), we have Ef (X1 , · · · , Xk ) Z = f (x1 , · · · , xk )ds Sn

Z
−1/2 Γ( n+1 2 ) = f (x1 , · · · , xk ) 1 − x21 − · · · − x2n dx1 · · · dxn (n+1)/2 2π Bn Z Γ( n+1 2 ) f (x1 , · · · , xk )dx1 · · · dxk = 2π (n+1)/2 Bk Z
−1/2 · 1 − x21 − · · · − x2n dxk+1 · · · dxn (2.49) x2k+1 +···+x2n <1−x21 −···−x2k

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where ds denotes the uniform measure on Sn . On the other hand, it is easy to compute Z
−1/2 1 − x21 − · · · − x2n dxk+1 · · · dxn x2k+1 +···+x2n <1−x21 −···−x2k


(n−k−1)/2 = 1 − x21 − · · · − x2k Z
−1/2 1 − x21 − · · · − x2n−k dx1 · · · dxn−k Bn−k (n−k+1)/2

=


(n−k−1)/2 π 1 − x21 − · · · − x2k . n−k+1 Γ( 2 )

Substituting (2.50) into (2.49) immediately get (2.47).

(2.50) 

Remark 2.2. Lemma 2.9 can also be proved by using the following wellknown fact: let g1 , · · · , gn+1 be a sequence of i.i.d. standard normal random variables, then 1 d (g1 , · · · , gn+1 ). (X1 , · · · , Xn+1 ) = 2 2 (g1 + · · · + gn+1 )1/2 To keep notation consistent, Z is said to be Θ1 -distributed if Z is uniform on the unit circle. Theorem 2.15. Assume that Un is a unitary matrix chosen from Un at random according to the Haar measure dµn . Then the Verblunsky coefficients αk (Un , e1 ) are independent Θ2(n−k−1)+1 -distributed complex random variables. The key to proving Theorem 2.15 is the Householder transform, which will transfer unitary matrix into an upper triangular Hessenberg form. Write Un = (uij )n×n . Let w = (w1 , w2 , · · · , wn )τ where u21  1 |u21 | 1/2 w1 = 0, w2 = − − , |u21 | 2 2α wl = −

(2α2

ul1 , − 2α|u21 |)1/2

l≥3

where α > 0 and α2 : = |u21 |2 + |u31 |2 + · · · + |un1 |2 = 1 − |u11 |2 . Trivially, it follows w∗ w = 1.

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Define 

1 0  Rn = In − 2ww∗ =  .  ..

0

··· Vn−1

0

   . 

0 This is a reflection through the plane perpendicular to w. It is easy to check Rn−1 = Rn∗ = Rn and  u11 (u12 , u13 , · · · , u1n )Vn−1     u21      u31  Rn Un Rn =  ,   Vn−1 Un,n−1 Vn−1   Vn−1  .     ..  

un1 where Un,n−1 is the n − 1 × n − 1 submatrix of Un by deleting the first row and the first column. Take a closer look at the first column. The first element of Rn Un Rn is unchanged, u11 ; while the second is   u21 
 u21  u32  1 − 2w2 w2∗ , −2w2 w3∗ , · · · , −2w2 wn∗  .  = α , |u21 |  ..  un1 and the third and below are zeros. So far we have described one step of the usual Householder algorithm. To make the second entry nonnegative, we need to add one further conjugation. Let Dn differ from the identity matrix by having (2, 2)-entry e−iφ with φ chosen appropriately and form Dn Rn Un Rn Dn∗ . Then we get the desired matrix   u (u , u , · · · , u )V 11 12 13 1n n−1  p1 − |u |2    11   0  .   Vn−1 Un,n−1 Vn−1 ..     . 0 Proof of Theorem 2.15. We shall apply the above refined Householder algorithm to a random unitary matrix Un . To do this, we need the following

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realization of Haar measure: choose the first column at random from the unit sphere Sn−1 ; then choose the second column from the unit sphere of vectors orthogonal to the first; then the third column and so forth. In this way we get a Haar matrix because it is invariant under left multiplication by any unitary matrix. Now the first column of Un is a random vector from the unit sphere Sn−1 . After applying the above refined Householder algorithm, the new first column take the form (¯ α0 , ρ0 , 0, · · · , 0)τ where α ¯ 0 = u11 the original (1, 1) entry of U and so is by Lemma 2.9 Θ -distributed, while ρ0 = n 2n−1 p 2 1 − |α0 | as desired. The other columns are still orthogonal to the first column and form a random orthogonal basis for the orthogonal complement of the first column. Remember Haar measure is invariant under both right and left multiplication by a unitary. For the subsequent columns the procedure is similar. Assume the (k − 1)th column is   α ¯ k−2 ρ0 ρ1 · · · ρk−3  −¯   αk−2 α0 ρ1 · · · ρk−3    .   ..     −¯ αk−2 αk−3    .   ρk−2     0     ..   . 0 Let   ρ0 ρ1 · · · ρk−2  −α0 ρ1 · · · ρk−2      ..   .      −αk−3 ρk−2  X= ,   −αk−2     0     ..   . 0 then X is a unit vector orthogonal to the first k − 1 columns. Namely X is an element of the linear vector space spanned by the last n−k+1 columns in Un . Its inner product with the kth column, denoted by α ¯ k−1 , is distributed as the entry of a random vector from 2(n−k + 1)-sphere and is independent of α0 , α1 , · · · , αk−2 . This implies that αk−1 is Θ2(n−k)+1 -distributed.

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We now multiply the matrix at hand from left by the appropriate reflection and rotation to bring the kth columns into the desired form. Note neither these operations alter the top k rows and so the inner product of the kth column with X is unchanged. But now the kth column is uniquely determined; it must be α ¯ k−1 X +ρk−1 ek+1 , where ek+1 = (0, · · · , 0, 1, 0 · · · 0)τ . We then multiply on the right by RD∗ , but this leaves the first k column unchanged, while orthogonally intermixing the other columns. In this way, we obtain a matrix whose first k column confirm to the structure HnU . While the remaining columns form a random basis for the orthogonal complements of the span of those k columns. In this way, we can proceed inductively until we reach the last column. It is obliged to be a random orthonormal basis for the one-dimensional space orthogonal to the preceding n − 1 columns and hence a random unimodular multiple, say α ¯ n−1 , of X. This is why the last Verblunsky coefficient is Θ1 -distributed. We have now conjugated Un to a matrix in the form of Hessenberg as in Lemma 2.7. Note the vector e1 is unchanged under the action of each of the conjugating matrices, then
αk (Un , e1 ) = αk HnU , e1 = αk . We conclude the proof.  Combining Lemma 2.8 and Theorem 2.15 together, we immediately have Theorem 2.16. Let α0 , α1 , · · · , αn−1 be a sequence of independent complex random variables and αk is Θ2(n−k−1)+1 -distributed. Define the CMV matrix Cn as in (2.43), then its eigenvalues are distributed according to (2.1). Cn is called a five diagonal matrix model of the CUE. It first appeared in the work of Killip and Nenciu (2004). The rest of this section will be used to rederive Theorems 2.11 and 2.12 with help of the Verblunsky coefficients αk and the Pr¨afer phase ψk introduced above. Start by an identity in law due to Bourgade, Hughes, Nikeghbali and Yor (2008). Lemma 2.10. Let Vn ∈ Un be a random matrix with the first column v1 uniformly distributed on the n-dimensional unit complex sphere. If Un−1 ∈ Un−1 is distributed with Haar measure dµn−1 and is independent of Vn ,

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then  Un := Vn

1 0



0

(2.51)

Un−1

is distributed with Haar measure dµn on Un . Proof.

We shall prove for a fixed matrix M ∈ Un d

M Un = Un . Namely,  M Vn

1 0



0



d

= Vn

Un−1

1 0

0

 .

Un−1

Write Vn = (v1 , v2 , · · · , vn ). Since v1 is uniform, then so is M v1 . By conditioning on v1 = v and M v1 = v, it suffices to show     1 0 1 0 d . (2.52) = (v, v2 , · · · , vn ) (v, M v2 , · · · , M vn ) 0 Un−1 0 Un−1 Choose a unitary matrix A such that Av = e1 . Since A(v, M v2 , · · · , M vn ) is unitary, then it must be equal to   1 0 0 Xn−1 for some Xn−1 ∈ Un−1 . Similarly,  A(v, v2 , · · · , vn ) = for some Yn−1 ∈ Un−1 . It is now easy to see   1 1 0 0 0 Xn−1

0 Un−1



d

=



1 0



0 Yn−1

1 0

0



Yn−1

d

1 0

0



Un−1

d

since Xn−1 Un−1 = Un−1 and Yn−1 Un−1 = Un−1 by rotation invariance of Haar measure. Thus by virtue of invertibility of A, (2.52) immediately follows, which concludes the proof.  Lemma 2.11. Let Un be a random matrix from the CUE with the Verblunsky coefficients α0 , α1 , · · · , αn−1 . Then d

det(In − Un ) =

n−1 Y

(1 − αk ).

k=0

(2.53)

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Proof. To apply Lemma 2.10, we particularly choose a Vn as follows. Let v1 be a random vector uniformly distributed on the n-dimensional unit complex sphere. Define
Vn = v1 , e2 + a2 (v1 − e1 ), · · · , en + an (v1 − e1 ) , where e1 , · · · , en are classic base in Rn , a2 , · · · , an are such that Vn is unitary, that is, hv1 , ek i ak = , k = 2, 3, · · · , n. hv1 − e1 , e1 i According to Lemma 2.10, it follows    1 0 d , (2.54) det(In − Un ) = det In − Vn 0 Un−1 where Un−1 is distributed with Haar measure dµn−1 independently of v1 . It remains to computing the determinant on the right hand side of (2.54). Note    1 0    1 0 = det det In − Vn − Vn det Un−1 . (2.55) ∗ 0 Un−1 0 Un−1 ∗ Set Un−1 = (u2 , u3 , · · · , un ) and w1 = v1 − e1 . Then   1 0 − Vn ∗ 0 Un−1       0 0 − (en + an w1 ) . − (e2 + a2 w1 ), · · · , = −w1 , un u2 So by the multi-linearity property,   −w11 0    −w21   1 0    − V det = det   n . ∗ ∗ . U − I 0 Un−1  . n−1  n−1 −wn1 ∗ = −w11 det(Un−1 − In−1 ).

(2.56)

Substituting (2.56) in (2.55), we get    1 0 = −w11 det(In−1 − Un−1 ). det In − Vn 0 Un−1 Observe w11 = v11 − 1 and v11 ∼ Θ2n−1 -distributed. Thus it follows d

det((In − Un ) = (1 − α0 ) det(In−1 − Un−1 ). Proceeding in this manner, we have d

det(In − Un ) =

n−1 Y

(1 − αk ),

k=0

as required.



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Remark 2.3. The identity (2.53) can also be deduced using the recurrence relation of orthogonal polynomials Φk (z). Indeed, according to Theorem 2.16, d

det(In − Un ) = det(In − Cn ).

(2.57)

On the other hand, by Lemma 2.6 and (2.37) det(In − Cn ) = Φn (1) = Φn−1 (1) − α ¯ n−1 Φ∗n−1 (1). Note by (2.40) Φ∗n−1 (1) = e−iψn−1 (0) . Φn−1 (1) Hence we have   det(In − Cn ) = Φn−1 (1) 1 − α ¯ n−1 e−iψn−1 (0) . Inductively, using (2.35) we get det(In − Cn ) =

n−1 Y

 1−α ¯ k e−iψk (0) .

k=0

Observe that ψ0 (0) = 0 and ψk (0) depends only on α0 , α1 , · · · , αk−1 . Using the conditioning argument and the rotation invariance of αk , we can get d

det(In − Cn ) =

n−1 Y

(1 − αk ),

(2.58)

k=0

which together with (2.57) implies (2.53). It is worth mentioning that the identity (2.58) is still valid for the CβE discussed in next section. We also need some basic estimates about the moments of Θv -distributed random variables. Lemma 2.12. Assume Z is Θv -distributed for some v ≥ 1. (i) |Z| and argZ are independent real random variables. Moreover, argZ is uniform over (0, 2π) and |Z| is distributed with density function p|Z| (r) = (v − 1)r(1 − r2 )(v−3)/2 ,

0 < r < 1.

(ii) EZ = EZ 2 = 0,

E|Z|2 =

2 , v+1

E|Z|4 =

8 . (v + 1)(v + 3)

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Proof. (i) directly follows from (2.47), while a simple computation easily yields (ii).  Proof of Theorem 2.11. Without loss of generality, we may and do assume θ0 = 0. According to Lemma 2.11, it suffices to prove the following asymptotic normality n−1 X 1 d √ log(1 − αk ) −→ NC (0, 1). log n k=0

(2.59)

Since |αk | < 1 almost surely for k = 0, 1, · · · , n − 2, then log(1 − αk ) = −

∞ X 1 l=1

l

αkl .

Taking summation over k, we have the following n−1 X

log(1 − αk ) = −

k=0

∞ n−2 X1 X l=1 k=0

= −

n−2 X

l

αk −

αkl + log(1 − αn−1 ) n−2 ∞ n−2 1 X 2 XX 1 l α + log(1 − αn−1 ) αk − 2 l k k=0

k=0

l=3 k=0

=: Zn,1 + Zn,2 + Zn,3 + Zn,4 . Firstly, we shall prove Z d √ n,1 −→ NC (0, 1). log n

(2.60)

n−2  1 X 1 d √ |αk | cos ηk −→ N 0, 2 log n k=0

(2.61)

n−2  1 X 1 d √ |αk | sin ηk −→ N 0, 2 log n k=0

(2.62)

It is equivalent to proving

and

where ηk = argαk . We only prove (2.61) since (2.62) is similar. In view of Lemma 2.12, E|αk |2 =

1 , n−k

E|αk |4 =

2 (n − k)(n − k + 1)

and E cos2 ηk =

1 , 2

E cos4 ηk ≤ 1.

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Since ηk is uniform, it is easy to check n−2 1 X 1 E|αk |2 E cos2 ηk → log n 2 k=0

and n−2 X

E|αk |4 ≤

k=0

π2 . 3

Hence (2.61) is now a direct consequence of the Lyapunov CLT. Secondly, to deal with Zn,2 , note by Lemma 2.12 n−2 n−2 π2 X 2 2 X E αk = E|αk |4 ≤ . 3 k=0

k=0

This, together with the Markov inequality, easily implies 1 P √ Zn,2 −→ 0. log n

(2.63)

Thirdly, it is easy to check EZn,3 = 0 and E|Zn,3 |2 ≤

∞ n−2 X1 X l=3 k=0

l

E|αk |2l ≤

2π 2 . 3

So, it follows 1 P √ Zn,3 −→ 0. log n

(2.64)

Finally, for Zn,4 , note αn−1 = eiη is uniformly distributed on T, then almost surely αn−1 is not equal to 1, and so log(1 − αn−1 ) < ∞. Hence it follows 1 P √ Zn,4 −→ 0. log n

(2.65)

Gathering (2.60), (2.63), (2.64) and (2.65) together implies (2.59).  Turn to Theorem 2.12. Let αk = αk (Un , e1 ), the Verblunsky coefficients associated to (Un , e1 ), and construct the Bk and ψk as in (2.39) and (2.40), then  iθ  e : Bn−1 (eiθ ) = e−iη = eiθ : ψn−1 (θ) ∈ 2πZ + η is the eigenvalues of Un . In particular, the number of angles lying in the arc (a, b) ⊆ [0, 2π) is approximately (ψn−1 (b) − ψn−1 (a))/2π; indeed, it follows ψn−1 (b) − ψn−1 (a) Nn (a, b) − ≤ 1. 2π

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In this way, it suffices to show that asymptotically, ψn−1 (b) and ψn−1 (a) follow a joint normal law. This is what Killip and Nenciu (2008) employed in the study of general CβE. Lemma 2.13. Assume a, b ∈ R and α ∼ Θv . Define

˜ Υ(a, α) = −2Im log 1 − αeia , Υ(a, α) = 2Im αeia . Then we have ˜ EΥ(a, α) = E Υ(a, α) = 0, 4 ˜ ˜ α) = E Υ(a, α)Υ(b, cos(b − a), v+1 48 ˜ , E Υ(a, α)4 = (v + 1)(v + 3) 16 ˜ , E|Υ(a, α) − Υ(a, α)|2 ≤ (v + 1)(v + 3) 8 E|Υ(a, α)|2 ≤ . v+1 Proof. implies

(2.66) (2.67) (2.68) (2.69) (2.70)

The fact that α follows a rotationally invariant law immediately EΥ(a, α) = 0.

Specifically, for any 0 ≤ r < 1, Z 2π
1 log 1 − rei(θ+a) dθ = 0 2π 0 ˜ by the mean value principle for harmonic functions. E Υ(a, α) = 0 is similar and simpler. For (2.67), note Im(αeia ) = |α| sin(a + argα) and so it follows by Lemma 2.12 ˜ ˜ α) E Υ(a, α)Υ(b, = 4E|α|2 sin(a + argα) sin(b + argα) Z Z
(v−3)/2 4(v − 1) 1 2π 3 = r 1 − r2 sin(θ + a) sin(θ + b)drdθ 2π 0 0 4 cos(b − a). = v+1

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Similarly, we have Z Z
(v−3)/2 4 16(v − 1) 1 2π 5 r 1 − r2 sin (θ + a)drdθ 2π 0 0 48 . = (v + 1)(v + 3)

˜ E Υ(a, α)4 =

Applying Plancherel’s theorem to the power series formula for Υ gives ∞ 2 X 2 ˜ α) = E Υ(a, α) − Υ(a, E|α|2l . l2 l=2   π2 − 1 E|α|4 ≤2 6

(2.67) easily follows from Lemma 2.12. Lastly, combining (2.67) and (2.69) implies (2.70).



Lemma 2.14. Assume that ak , Θk , γk , k ≥ 0 are real valued sequences satisfying Θk+1 = Θk + δ + γk with 0 < δ < 2π. Then we have n X 1 − eiδ ak eiΘk k=1

≤ 2 max |ak | + 1≤k≤n

Proof.

n X

|ak − ak+1 | +

k=1

n X

|ak γk |.

(2.71)

k=1

Note n
X eiδ − 1 ak eiΘk k=1

= =

n X k=1 n X k=1

ak ei(Θk +δ) − eiΘk ak e

iΘk+1

−e

iΘk







−

n X


ak eiΘk+1 − ei(Θk +δ) .

k=1

Then (2.71) easily follows using summation by parts and the fact 1−eiγk ≤ |γk |. 

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Proof of Theorem 2.12. Start with 1-dimensional convergence. Note for any 0 < a < 2π ψn−1 (a) − na =

n−1 X


ψk (a) − ψk−1 (a) − a

k=1

=

n−2 X


Υ ψk (a), αk ,

k=0




where Υ ψk (a), αk is defined as in (2.41). Since ψk (a) depends only on α0 , · · · , αk−1 , then it follows from Lemma 2.13
E Υ(ψk (a), αk ) α0 , · · · , αk−1 = 0.
Namely, Υ ψk (a), αk , 0 ≤ k ≤ n − 2 is a martingale difference sequence. Define

˜ ψk (a), αk = −2Im αk eiψk (a) . Υ
˜ ψk (a), αk , 0 ≤ k ≤ n − 2 is also a martingale difference Similarly, Υ sequence. Moreover, by Lemma 2.13 again n−2 X

˜ k (a), αk ) 2 ≤ 3π 2 . E Υ(ψk (a), αk ) − Υ(ψ

k=0

Thus it suffices to prove n−2 X
d 1 ˜ ψk (a), αk −→ √ Υ N (0, 1). 2 log n k=0

It is in turn sufficient to verify n−2
P 1 X ˜ k (a), αk )2 α0 , · · · , αk−1 −→ E Υ(ψ 1 2 log n k=0

and n−2
4 1 X ˜ E Υ ψk (a), αk ) → 0. 2 log n k=0

These directly follow from Lemma 2.12. Next turn to 2-dimensional convergence. We need to verify n−2
P 1 X ˜ k (a), αk )Υ(ψ ˜ k (b), αk ) α0 , · · · , αk−1 −→ E Υ(ψ 0. log n k=0

(2.72)

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For distinct numbers a and b, we have by Lemma 2.12
˜ k (a), αk )Υ(ψ ˜ k (b), αk ) α0 , · · · , αk−1 E Υ(ψ
2 = cos ψk (b) − ψk (a) . n−k Define ak =

1 , n−k

Θk = ψk (b) − ψk (a),

δ =b−a

and

γk = Υ ψk (b), αk − Υ ψk (a), αk , then by (2.41) Θk+1 = Θk + δ + γk . Applying Lemma 2.14 and noting E|γk | ≤

8 , (n − k)1/2

we have X n−2 1 − ei(b−a) E k=0

1 ei(ψk (b)−ψk (a)) ≤ 9. n−k

(2.72) is now valid. Thus by the martingale CLT
d 1 √ ψn−1 (a) − na, ψn−1 (b) − nb −→ (Z1 , Z2 ), 2 log n where Z1 , Z2 are independent standard normal random variables. 2.5



Circular β ensembles

The goal of this section is to extend the five diagonal matrix representation to the CβE. Recall that the CβE represents a family of probability measures on n points of T with density function pn,β (eiθ1 , · · · , eiθn ) defined by (2.5). As observed in Section 2.1, pn,2 describes the joint probability density of eigenvalues of a unitary matrix chosen from Un according to Haar measure. Similarly, pn,1 (pn,4 ) describes the joint probability density of eigenvalues of an orthogonal (symplectic) matrix chosen from On (Sn ) according to Haar measure. However, no analog holds for general β > 0.

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The following five diagonal matrix model discovered by Killip and Nenciu (2004) plays an important role in the study of CβE. Theorem 2.17. Assume that α0 , α1 , · · · , αn−1 are independent complex random variables and αk ∼ Θβ(n−k−1)+1 -distributed for β > 0. Construct the CMV matrix Cn as in (2.43), then the eigenvalues of Cn obey the same law as pn,β . The rest of this section is to prove the theorem. The proof is actually an ordinary use of the change of variables in standard probability theory. Let HnU be as in (2.46), then we have by Lemmas 2.7 and 2.8
αk (Cn , e1 ) = αk HnU , e1 = αk . So it suffices to prove the claim for HnU . Denote the ordered eigenvalues of HnU by eiθ1 , · · · , eiθn . Then there must be a unitary matrix Vn = (vij )n×n such that   iθ1 0 ··· 0 e  0 eiθ2 · · · 0   ∗  (2.73) HnU = Vn  . .. . . ..  Vn .  .. . . .  0

0 · · · eiθn

Vn may be chosen to consist of eigenvectors v1 , v2 , · · · , vn . We also further require that v11 := q1 > 0, v12 := q2 > 0, · · · , v1n := qn > 0, thus Vn is uniquely determined. It easily follows q12 + q22 + · · · + qn2 = 1

(2.74)

because of orthonormality of Vn . The following lemma gives an elegant identity between the eigenvalues and eigenvectors and the Verblunsky coefficients. Lemma 2.15. n Y l=1

Proof.

ql2

n−2 Y iθ
n−l−1 e j − eiθk 2 = 1 − |αl |2 .

Y 1≤j
Define A and Q by  1  eiθ1  A= ..  .

l=0

1 eiθ2 .. .

··· ··· .. .

1



eiθn .. .

   

ei(n−1)θ1 ei(n−1)θ2 · · · ei(n−1)θn

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and q12 0  Q= .  .. 

0 q22 .. .

··· ··· .. .



0 0 .. .

   

0 0 · · · qn2 then it follows n Y l=1

ql2

iθ e j − eiθk 2 = det(AQA∗ ).

Y 1≤j
On the other hand, define  Φ0 (eiθ1 )  Φ1 (eiθ1 )  B= ..  .

 Φ0 (eiθ2 ) · · · Φ0 (eiθn ) Φ1 (eiθ2 ) · · · Φ1 (eiθn )   , .. .. ..  . . . iθ1 iθ2 iθn Φn−1 (e ) Φn−1 (e ) · · · Φn−1 (e ) where Φ0 , Φ1 , · · · , Φn−1 are monic orthogonal polynomials associated to the Verblunsky coefficients α0 , α1 , · · · , αn−1 . Then it is trivial to see det(A) = det(B). In addition, from the orthogonality property of the Φl , it follows   kΦ0 k2 0 · · · 0   0 kΦ1 k2 · · · 0   BQB ∗ =  . , . .. ..   .. . 0 0 0 · · · kΦn−1 k2 Pn where kΦl k2 = j=1 qj2 |Φl (eiθj )|2 . Hence according to (2.38), n−1 Y det(AQA∗ ) = det(BQB ∗ ) = kΦl k2 l=0

=

n−2 Y

1 − |αl |


2 n−l−1

l=0

just as required.



A key ingredient to the proof of Theorem 2.17 is to look for a proper change of variables and to compute explicitly the corresponding determinant of the Jacobian. For any |t| < 1, it follows from (2.73)   (1 − teiθ1 )−1 · · · 0
−1   ∗ .. .. .. In − tHnU = Vn  (2.75)  Vn . . . . iθn −1 0 · · · (1 − te )

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Applying the Taylor expansion of (1 − x)−1 , and equating powers of t of the (1, 1) entries on both sides of (2.75), we can get the following system of equations

α ¯0 =

n X

qj2 eiθj

j=1

∗ + ρ20 α ¯1 = ∗ + ρ20 ρ21 α ¯2 =

n X j=1 n X

qj2 ei2θj qj2 ei3θj

(2.76)

j=1

.. . ∗ + ρ20 ρ21 · · · ρ2n−2 α ¯ n−1 =

.. . n X

qj2 einθj

j=1

where the ∗ denotes terms involving only variables already having appeared on the left hand side of the preceding equations. In this way, we can naturally get a one-to-one mapping from (α0 , α1 , · · · , αn−1 ) to (eiθ1 , · · · , eiθn , q1 , · · · , qn−1 ). Recall that αk , 0 ≤ k ≤ n − 2 has an independent real and imaginary part, while αn−1 , eiθj have unit modulus. We see that the number of variables is equal to 2n − 1. In particular, let αk = ak + ibk and define J to be the determinant of the Jacobian matrix for the change of variables, namely Vn−2 J=

V

k=0 dak ∧ dbk Vn−1 Vn l=1 dql j=1

d¯ αn−1 dθj

where ∧ stands for the wedge product. We shall compute explicitly the J following Forrester and Rains (2006) below. First, taking differentials on both sides of (2.74) immediately yields

qn dqn = −

n−1 X j=1

qj dqj .

(2.77)

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Similarly, taking differentials on both sides of (2.76) gives n n X X qj2 eiθj dθj eiθj qj dqj + i d¯ α0 = 2 j=1 n X

j=1

ρ20 d¯ α1 = 2 ρ20 ρ21 d¯ α2 = 2

n X j=1 n X

ei3θj qj dqj + i3

j=1

j=1 n X

qj2 ei3θj dθj

(2.78)

j=1

.. . ρ20 ρ21

qj2 ei2θj dθj

ei2θj qj dqj + i2

.. .

· · · ρ2n−2 d¯ αn−1

=2

n X

e

inθj

qj dqj + in

n X

qj2 einθj dθj .

j=1

j=1

Forming the complex conjugates of all these equations but last, we get n n X X qj2 e−iθj dθj e−iθj qj dqj − i dα0 = 2 j=1 n X

j=1

ρ20 dα1 = 2 ρ20 ρ21 dα2 = 2

n X j=1 n X

e−i2θj qj dqj − i2

e−i3θj qj dqj − i3

qj2 e−i3θj dθj

(2.79)

j=1

j=1

.. . ρ20 ρ21 · · · ρ2n−3 dαn−2 = 2

j=1 n X

qj2 e−i2θj dθj

.. . n X

e−i(n−1)θj qj dqj − i(n − 1)

j=1

n X

qj2 e−i(n−1)θj dθj .

j=1

Now taking the wedge products of both sides of these 2n − 1 equations in (2.78) and (2.79), and using (2.77) shows n−2 Y 4(n−l−2) n−2 ^ ^ ρ20 ρ21 · · · ρ2n−2 ρl αn−1 d¯ αl ∧ dαl d¯ l=0

= (2i)n−1 qn2

n−1 Y

l=0

ql3 D eiθ1 , · · · , eiθn

l=1


where D eiθ1 , · · · , eiθn is defined by # " j xk − xjn  −j D(x1 , · · · , xn ) = det  xk − x−j n

^
n−1 l=1

" j,k=1,··· ,n−1

[xnk − xnn ]k=1,··· ,n−1

dql

n ^

dθj ,

(2.80)

#



j=1

jxjk −jx−j k

j=1,··· ,n−1

k=1,··· ,n

 .

[nxnk ]k=1,··· ,n (2.81)

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Lemma 2.16. We have 4 1≤j
Q (n−1)(n−2)/2

D(x1 , · · · , xn ) = (−1)

Proof. By inspection, the determinant D(x1 , · · · , xn ) is a symmetric function of x1 , · · · , xn which is homogeneous of degree n. Upon multiplying columns 1 and n by x2n−3 , we see that D(x1 , · · · , xn ) becomes a 1 polynomial in x1 , so it must be of the form p(x1 , · · · , xn ) Qn 2n−3 , j=1 xj where p(x1 , · · · , xn ) is a symmetric polynomial of x1 , · · · , xn of degree 2n(n − 1). We see immediately from (2.81) that D(x1 , · · · , xn ) = 0 when x1 = x2 . Furthermore, it is straightforward to verify that  ∂ j x1 D(x1 , · · · , xn ) = 0, j = 1, 2, 3 ∂x1 when x1 = x2 . This is equivalent to saying ∂j ∂xj1

D(x1 , · · · , xn ) = 0,

j = 1, 2, 3

when x1 = x2 . The polynomial p(x1 , · · · , xn ) must thus contain as a factor Q (x1 − x2 )4 , and so 1≤j
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In the cofactor, we add n−3 times the first column to the (n−1)th column. −(2n−7) Then we see the coefficient of x2 is given by a cofactor of the following 2 × 2 matrix ! −(n−4) −(n−4) −(n−4) −(n−4) x2 − xn x2 − (n − 1)xn . −(n−3) −(n−3) −(n−3) x2 − xn −(n − 3)xn Proceeding in this manner, we see the coefficient of x−5 k−1 is given by the determinant of the n + 1 × n + 1 matrix   xk − xn · · · xn−1 − xn xk ··· xn  x−1 − x−1 · · · x−1 − x−1  −x−1 ··· −x−1  k  n n n n−1 k   2 2 2 2 2 2 2xk ··· 2xn  xk − xn · · · xn−1 − xn   . .. .. .. .. .. ..     . . . . . .  n−1  n−1 n−1 n−1 n−1 n−1  xk − xn · · · xn−1 − xn (n − 1)xk · · · (n − 1)xn  nxnk ··· nxnn xnk − xnn · · · xnn−1 − xnn Interchange rows to get  −1 the top two −1  xk − x−1 · · · xn−1 − x−1 −x−1 ··· −x−1 n n n k  x −x  · · · xn−1 − xn xk ··· xn k n     2 2 2 2 2 2 ··· 2xn 2xk  xk − xn · · · xn−1 − xn   . (2.82) .. .. .. .. .. ..     . . . . . .  n−1 n−1 n−1 n−1  n−1  xk − xn−1  (n − 1)x · · · (n − 1)x · · · x − x n n n n−1 k nxnk ··· nxnn xnk − xnn · · · xnn−1 − xnn 2n−1 We postpone deciding the coefficient of x−1 . k , but we turn to the term xn In the determinant of (2.82), we first subtract the kth column from columns 1, 2, ·· · , k − 1 to get  −1 −1 −1 −x−1 ··· −x−1 x−1 n k k − xn−1 · · · xn−1 − xn  x −x  xk ··· xn n−1 · · · xn−1 − xn  k   2  2 2 2 2 2 2xk ··· 2xn  xk − xn−1 · · · xn−1 − xn   . .. .. .. .. .. ..     . . . . . .  n−1 n−1 n−1 n−1 n−1   xk − xn−1  · · · x − x (n − 1)x · · · (n − 1)x n n n−1 n−1 k xnk − xnn−1 · · · xnn−1 − xnn nxnk ··· nxnn Then we add n times the kth column to (n + 1)th column to see the coefficient x2n−1 is given by the determinant of the n − 1 × n − 1 matrix n   of−1 −1 −1 xk − x−1 −x−1 ··· −x−1 n n−1 · · · xn−2 − xn−1 k   x −x xk ··· xn n−1 · · · xn−2 − xn−1   k   2 2x2k ··· 2x2n   xk − x2n−1 · · · x2n−2 − x2n−1 .  .. .. .. .. .. ..     . . . . . .  n−3 n−3 n−3 n−3 n−3 n−3   xk − xn−1 · · · xn−2 − xn−1 (n − 3)xk · · · (n − 3)xn−1  n−2 n−2 n−2 xn−2 − xn−2 · · · (n − 2)xn−2 n−1 · · · xn−2 − xn−1 (n − 2)xk n−1 k

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Repeating this operation, we get the coefficient of x3k+1 is −x−1 k . In sumQn 4(j−1) −(2n−3) Qn is (−1)k−1 . mary, the coefficient of the j=1 xj j=1 xj Assume n = 2k + 1. Then we can almost completely repeat the proceQn 4(j−1) −(2n−3) Qn is dure above to see that the coefficient of the j=1 xj j=1 xj k (−1) . Qn Q 4(j−1) Finally, note the coefficient of j=1 xj in 1≤j


Proceed to computing the determinant J. We have Lemma 2.17. Qn−2 |J| = Proof.


1 − |αl |2 Qn . qn l=1 ql

l=0

Note dαk = dak + idbk , d¯ αk = dak − idbk .

It easily follows  d¯ αk ∧ dαk = det

1 −i 1 i

 dak ∧ dbk

= 2idak ∧ dbk .

(2.83)

Inserting (2.83) into (2.80) gives Qn−1 iθn
qn2 l=1 ql3 J = Qn−2 Qn−2 4(n−l−2) D eiθ1 ,··· ,e . 2 l=0 ρl l=0 ρl According to Lemmas 2.15 and 2.16, it immediately follows Qn−1 Y q 2 l=1 ql3 eiθj − eiθk 4 |J| = Qn−2 nQn−2 4(n−l−2) 2 1≤j


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Lemma 2.18. Let  2 ∆n = (q1 , q2 , · · · , qn−1 ) : qi > 0, q12 + · · · + qn−1 <1 . Then Z ∆n

n Γ( β2 )n 1 Y β−1 qj dq1 · · · dqn−1 = , qn j=1 2n−1 Γ( βn 2 )

q 2 where qn = 1 − q12 − · · · − qn−1 . Proof. We only consider the case of n ≥ 2 since the other case is trivial. Start with n = 2. Using the change of variable, we have Z 1 Z 2
β/2−1 β−1 1 Y β−1 qj dq1 = 1 − q12 q1 dq1 0 ∆2 q2 j=1 Z 1 1 β/2−1 = (1 − q1 )β/2−1 q1 dq1 2 0 =

Γ( β2 )2 . 2Γ(β)

Assume by induction that the claim is valid for some n ≥ 2. It easily follows Z n+1 1 Y β−1 qi dq1 · · · dqn ∆n+1 qn+1 i=1 Z n
β/2−1 Y 2 qiβ−1 dq1 · · · dqn = 1 − qn2 − q12 − · · · − qn−1 2 2 ≤1−qn q12 +···+qn−1

Z =

1

1 − qn2


β/2−1

i=1

qnβ−1 dqn

0

 · 1−

n−1 Y

Z

2 2 q12 +···+qn−1 ≤1−qn i=1

qiβ−1

β/2−1 2 qn−1 q12 dq1 · · · dqn−1 . − · · · − 1 − qn2 1 − qn2

(2.84)

Making a change of variable, the inner integral becomes Z Y β−1
β/2−1 n−1 2 (1 − qn2 )(n−1)β/2 1 − q12 − · · · − qn−1 qi dq1 · · · dqn−1 , ∆n

i=1

which is in turn equal to (1 − qn2 )(n−1)β/2

Γ( β2 )n 2n−1 Γ( nβ 2 )

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using the induction hypothesis. Substituting in (2.84) gives Z Z 1 n+1 Γ( β2 )n 1 Y β−1 qi dq1 · · · dqn = (1 − qn2 )nβ/2−1 qnβ−1 dqn 2n−1 Γ( nβ ) ∆n+1 qn+1 i=1 0 2 =

Γ( β2 )n+1 2n Γ( (n+1)β ) 2

.

We conclude by induction the proof.



Proof of Theorem 2.17. As remarked above, (2.73)
naturally induces a one-to-one mapping from eiθ1 , · · · , eiθn , q1 , · · · , qn−1 to (a0 , b0 , · · · , an−2 , bn−2 , αn−1 ). Let fn,β and hn,β be their respective joint probability density functions. Then it follows by Lemmas 2.17 and 2.15
fn,β eiθ1 , · · · , eiθn , q1 , · · · , qn−1 = hn,β (a0 , b0 , · · · , an−2 , bn−2 , αn−1 )|J| n−2

Y
β(n−l−1)/2 1 β n−1 Qn (n − 1)! 1 − |αl |2 = (2π)n qn l=1 ql l=0

β n−1 = (n − 1)! (2π)n

Y 1≤j
n Y iθ e j − eiθk β 1 qlβ−1 . qn l=1




This trivially implies that eiθ1 , · · · , eiθn is independent of (q1 , · · · , qn−1 ). Integrating out the qj over ∆n , we get by Lemma 2.18 Z

gn,β eiθ1 , · · · , eiθn := fn,β eiθ1 , · · · , eiθn , q1 , · · · , qn−1 dq1 · · · dqn−1 ∆n

=

n! (2π)n Zn,β

Y

iθ e j − eiθk β .

1≤j
Dividing by n! to eliminate the ordering of eigenvalues, we conclude the proof as desired.  Having a CMV matrix model, we can establish the following asymptotic normal fluctuations for the CβE. Theorem 2.18. Let eiθ1 , · · · , eiθn be chosen on the unit circle according to the CβE. Then as n → ∞ (i) for any θ0 with 0 ≤ θ0 < 2π, q

2 β

1

n X

log n

j=1


d log 1 − ei(θj −θ0 ) −→ NC (0, 1);

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(ii) for any 0 < a < b < 2π, Nn (a, b) − n(b−a) d 2π q −→ N (0, 1), 1 2 π β log n where Nn (a, b) denotes the number of the angles θj lying in the arc between a and b. Proof.

Left to the reader.



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Chapter 3

Gaussian Unitary Ensemble

3.1

Introduction

Let Hn be the set of all n × n Hermitian matrices. To each matrix H ∈ Hn assign a probability measure as follows 2n(n−1)/2 − 1 trH 2 e 2 Pn (H)dH = dH, (3.1) (2π)n2 /2 where dH is the Lebesgue measure on the algebraically independent entries of H. Pn is clearly invariant under unitary transform, namely Pn (U HU ∗ ) = Pn (H) for every unitary matrix U , see Chapter 2 of Deift and Gioev (2009) for a proof. The probability space (Hn , Pn ) is called Gaussian Unitary Ensemble (GUE). It is the most studied object in random matrix theory. As a matter of fact, the GUE is a prototype of a large number of matrix models and related problems. Note that the GUE can be realized in the following way. Let zii , 1 ≤ i ≤ n be a sequence of i.i.d real standard normal random variables, zij , 1 ≤ i < j ≤ n an array of i.i.d. complex standard normal random variables ∗ independent of the zii ’s. Then An := (zij )n×n where zji = zij , i < j will induce a probability measure as given by (3.1) in Hn . A remarkable feature of the GUE is that the eigenvalues have an explicit nice probability density function. Let λ1 , · · · , λn be n real unordered eigenvalues of An , then they are almost surely distinct to each other and are absolutely continuous with respect to Lebesgue measure on Rn . In particular, we have Theorem 3.1. Let pn (x) denote the joint probability density function of λ = (λ1 , · · · , λn ), then n Y Y 2 1 2 Q |x − x | e−xk /2 , (3.2) pn (x) = j i n n/2 (2π) k=1 k! 1≤i<j≤n k=1 89

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where x = (x1 , · · · , xn ) ∈ Rn . This theorem, due to Weyl, plays an important role in the study of GUE. Its proof can be found in the textbooks Anderson, Guionnet and Zeitouni (2010), Deift and Gioev (2009). (3.2) should be interpreted as follows. Let f : Hn 7→ R be an invariant function, i.e., f (H) = f (U HU ∗ ) for each H ∈ Hn and unitary matrix U . Then Z Ef (H) =

f (x)pn (x)dx. Rn

It is worthy to remark that there are two factors in the righthand side of (3.2). One is the product of n standard normal density functions, while the other is the square of Vandermonde determinant. The probability that two eigenvalues neighbor each other very closely is very small. Hence intuitively speaking, eigenvalues should locate more neatly than i.i.d. normal random points in the real line. It is the objective of this chapter that we shall take a closer look at the arrangement of eigenvalues from global behaviours. In order to analyze the precise asymptotics of pn (x), we need to introduce Hermite orthogonal polynomials and the associated wave functions. Let hl (x), l ≥ 0 be a sequence of monic orthogonal polynomials with respect 2 to the weight function e−x /2 with h0 (x) = 1. Then 2 2 dl hl (x) = (−1)l ex /2 l e−x /2 dx [l/2] X xl−2i , l ≥ 1. (3.3) = l! (−1)i i 2 i!(l − 2i)! i=0 Define ϕl (x) = (2π)−1/4 (l!)−1/2 hl (x)e−x so that we have Z

2

/4

(3.4)

∞

ϕl (x)ϕm (x)dx = δl,m ,

∀l, m ≥ 0.

−∞

Now a simple matrix manipulation directly yields  1 1 ··· 1  x1 x2 · · · xn Y  (xj − xi ) = det  . .. . . .  .. . .. . 1≤i<j≤n

    

xn−1 xn−1 · · · xn−1 n 1 2  h0 (x2 ) · · · h0 (xn )  h1 (x2 ) · · · h1 (xn )    = det   . (3.5) .. .. ..   . . . hn−1 (x1 ) hn−1 (x2 ) · · · hn−1 (xn ) 

h0 (x1 ) h1 (x1 ) .. .

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Furthermore, substituting (3.5) into (3.2) and noting (3.4) immediately leads to the following determinantal expression for pn (x). Proposition 3.1. pn (x) =


1 det Kn (xi , xj ) n×n , n!

(3.6)

where Kn is defined by Kn (x, y) =

n−1 X

ϕl (x)ϕl (y).

(3.7)

l=0

Such a expression like (3.6) turns out to be very useful in the study of asymptotics of eigenvalues. In fact, the GUE is one of the first examples of so-called determinantal point processes (see Section 3.3 below for more details). A nice observation about the kernel Kn is the following: Z ∞ Kn (x, z)Kn (z, y)dz = Kn (x, y). −∞

As an immediate consequence, we can easily obtain any k-dimensional marginal density. Let pn,k (x1 , · · · , xk ) be the probability density function of (λ1 , · · · , λk ), then it follows pn,k (x1 , · · · , xk ) =


(n − k)! det Kn (xi , xj ) k×k . n!

(3.8)

In particular, we have pn,1 (x) =

1 Kn (x, x). n

We collect some basic properties of Hermite wave functions ϕl (x) below. See Szeg¨ o (1975) for more information. Lemma 3.1. For l ≥ 1, it follows (i) recurrence equation √ √ xϕl (x) = l + 1ϕl+1 (x) + lϕl−1 (x); (ii) differential relations √ x ϕ0l (x) = − ϕl (x) + lϕl−1 (x), 2  1 x2  ϕ00l (x) = − l + − ϕl (x); 2 4

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(iii) Christoffel-Darboux identities l−1 X

√ ϕm (x)ϕm (y) =

m=0 l−1 X

ϕ2m (x) =

l·

ϕl (x)ϕl−1 (y) − ϕl (y)ϕl−1 (x) , x−y

x 6= y

√
l · ϕ0l (x)ϕl−1 (x) − ϕl (x)ϕ0l−1 (x) ;

(3.9)

(3.10)

m=0

(iv) boundedness κ := sup kϕl k∞ < ∞.

(3.11)

l≥0

The next lemma, known as Plancherel-Rotach formulae, provides asymptotic behavior formulae for the Hermite orthogonal polynomials. Lemma 3.2. We have as n → ∞ (i) for |x| < 2 − δ with δ > 0, r  √ 2 π 1 k+1 1/4 θ − + O(n−1 ), cos nα(θ) + n ϕn+k ( nx) = π (4 − x2 )1/4 2 4 where x = 2 cos θ, α(θ) = θ − sin 2θ/2, k = −1, 0, 1 and the convergence is uniform in x. The asymptotics is also valid for |x| < 2 − δn with δn−1 = o(n2/3 ); (ii) for x = ±2 + ζn−2/3 with ζ ∈ R,
√ n1/12 ϕn ( nx) = 21/4 Ai(ζ) + O(n−3/4 ) , where Ai(ζ) stands for the standard Airy function; (iii) for |x| > 2 + δ,
√ 1 e−(2n+1)β(θ)/2 1 + O(n−1 ) , n1/4 ϕn ( nx) = √ π sinh θ where x = 2 cosh θ with θ > 0 and β(θ) = sinh 2θ/2 − θ. As a direct application, we obtain a limit of the first marginal probability density after suitably scaled. Proposition 3.2. Define p¯n (x) =

√

√ npn,1 ( nx).

(i) We have as n → ∞ p¯n (x) → ρsc (x) uniformly on any closed interval of (−2, 2), where ρsc was defined by (1.15).

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(ii) Given s > 0, there exist positive constants c1 and c2 such that for each n≥1  3/2 c1 s  (3.12) p¯n 2 + 3/2 ≤ 1/3 e−c2 s , n n s  3/2 s  c1 p¯n −2 − 3/2 ≤ 1/3 e−c2 s . n n s (iii) Given |x| > 2 + δ with δ > 0, there exist positive constants c3 and c4 such that for each n ≥ 1 3/2

p¯n (x) ≤ c3 e−c4 n|x|

.

Proof. To start with the proof of (i). Note it follows from (ii) and (iii) of Lemma 3.1 √
√ 1 p¯n (x) = √ Kn nx, nx n √ √ √ √ = ϕ0n ( nx)ϕn−1 ( nx) − ϕn ( nx)ϕ0n−1 ( nx) √ √ √ √ √ = nϕn−1 ( nx)2 − n − 1ϕn ( nx)ϕn−2 ( nx). (3.13) Fix a δ > 0. For each x such that |x| ≤ 2 + δ, we have by Lemma 3.1 (i)  √ √ π 2 2 + O(n−1 ) cos nα(θ) − nϕn−1 ( nx)2 = 4 π(4 − x2 )1/2 and √

√ √ nϕn ( nx)ϕn−2 ( nx)   2 θ π θ π = cos nα(θ) − − + O(n−1 ), cos nα(θ) + − 2 1/2 2 4 2 4 π(4 − x )

where x = 2 cos θ, α(θ) = θ − sin 2θ/2 and the convergence is uniform in x. Now a simple algebra yields 1 p p¯n (x) = 4 − x2 + O(n−1 ), 2π as desired. Proceed to prove (ii). Note that it follows by Lemma 3.1
√ d Kn (x, x) = n ϕ00n (x)ϕn−1 (x) − ϕn (x)ϕ00n−1 (x) dx √ = − nϕn (x)ϕn−1 (x). Since Kn (x, x) vanishes at ∞, we have Z ∞ √ ϕn (u)ϕn−1 (u)du, Kn (x, x) = n x

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from which it follows p¯n (x) =

√

Z

∞

n

√ √ ϕn ( nu)ϕn−1 ( nu)du.

x

By the Cauchy-Schwarz inequality, Z 1/2 Z √ √  ∞ ϕn ( nu)2 du p¯n (x) ≤ n x

∞

1/2 √ . ϕn−1 ( nu)du

x

2/3

Let x = 2 + s/n Z ∞ √ n

. We will below control Z ∞  √ √ 1 u 2 2 ϕn n(2 + 2/3 ) du ϕn ( nu) du = 1/6 n n s 2+s/n2/3 from the above. Write u 1 cosh θ = 1 + 2/3 , β(θ) = sinh 2θ − θ. 2 2n Then it follows by using asymptotic formula Z Z ∞  √ 1 ∞ −(2n+1)β(θ) u 2 1 e dθ, du = (1 + o(1)) ) ϕ n(2 + n π θs n1/6 s n2/3 where cosh θs = 1 + s/2n2/3 . Since β 0 (θ) = 2(sinh θ)2 is increasing, we have for s → ∞ and n → ∞ Z 1 ∞ −(2n+1)β(θ) 1 e−(2n+1)β(θs ) . e dθ ≤ π θs 2πnβ 0 (θs ) Note an elementary inequality θ2 ≤ cosh θ − 1 ≤ (sinh θ)2 , 2 from which one can readily derive β 0 (θs ) = 2(sinh θs )2 ≥ 2(cosh θs − 1) s = 2/3 n and Z

θs

β(θs ) = 2 0 3/2

≥

(sinh x)2 dx ≥

2 θs

Z

θs

2 sinh xdx

0

s . 2n

Thus √ n

∞

√ (1 + o(1)) −(1+o(1))s3/2 ϕn ( nu)2 du ≤ e . 2/3 2πn1/3 s 2+s/n

Z

(3.14)

√ Similarly, the upper bound of (3.14) holds for the integral of ϕn−1 ( nu)2 , and so (3.12) is proven. Last, we turn to (iii). The proof is completely similar to (ii). Now we conclude the proposition. 

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Now we are ready to state and prove the celebrated Wigner semicircle law for the GUE. Define the empirical spectral distribution for normalized eigenvalues by n

Fn (x) =

1X 1(λk ≤√nx) , n

−∞ < x < ∞.

(3.15)

k=1

Proposition 3.2 gives the limit of the mean spectral density. In fact, we can further prove the following Theorem 3.2. We have as n → ∞ d

Fn −→ ρsc

in P,

(3.16)

where ρsc was defined by (1.15). Proof. The statement (3.16) means that for any bounded continuous function f , Z ∞ Z ∞ P f (x)ρsc (x)dx, n → ∞. (3.17) f (x)dFn (x) −→ −∞

−∞

Note that f in (3.17) can be replaced by any bounded Lipschitz function. Let f be a bounded 1-Lipschtiz function, we will prove the following claims Z ∞ n 1 X  λk  √ E f f (x)ρsc (x)dx (3.18) → n n −∞ k=1

and V ar

n 1 X  λ  k → 0. f √ n n

(3.19)

k=1

First, we prove (3.18). Note λ  Z ∞  x  k = f √ pn,1 (x)dx Ef √ n n −∞ Z ∞ = f (x)¯ pn (x)dx.

(3.20)

−∞

Fix a small δ > 0 and let δn = sn /n2/3 satisfy δn n2/3 → ∞ and δn n1/2 → 0. Write the integral on the righthand side of (3.20) as the sum of integrals Ik , k = 1, 2, 3, 4, over the sets A1 = {x : |x| < 2−δ}, A2 = {x : 2−δ ≤ |x| < 2 − δn }, A3 = {x : 2 − δn ≤ |x| < 2 + δn } and A4 = {x : 2 + δn ≤ |x| < ∞}.

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We will below estimate each integral, separately. First, it clearly follows from Proposition 3.2 (i) that Z Z I1 = f (x)¯ pn (x)dx → f (x)ρsc (x)dx. A1

A1

It remains to show that Ik , k = 2, 3, 4 are asymptotically as small as δ. Note f is bounded. Then Ik is bounded by the corresponding integral of p¯n (x) over Ak . Since sn → ∞, then according to Lemma 3.2 (i), we have for x ∈ A2 ,  √ 1 k+1 π n1/4 ϕn+k ( nx) = √ cos nα(θ) + θ− (1 + o(1)), 2 4 π sin θ where x = 2 cos θ, α(θ) = sin θ/2 − θ. Hence it follows Z Z
2 √ p¯n (x)dx = n1/4 ϕn−1 ( nx) dx A2 A2 r Z √ √ n−1 n1/2 ϕn ( nx)ϕn−2 ( nx)dx − n A2 = O(δ). To estimate the integral over A3 , we note (3.13) and use the bound in (3.11). Then Z p¯n (x)dx ≤ 2k¯ pn k∞ δn A3

≤ 4κ2 n1/2 δn → 0. To estimate the integral over A4 , we use Proposition 3.2 (ii) to get Z Z ∞   s  s  p¯n (x)dx ≤ p¯n 2 + 2/3 + p¯n −2 − 2/3 ds. n n I4 sn → 0. Combining the above four estimates together yields  λ  Z 2−δ 1 f (x)ρsc (x)dx + O(δ). = lim Ef √ n→∞ n −2+δ Letting δ → 0, we can conclude the proof of (3.18). Next we turn to the proof of (3.19). Observe n X  λ  k V ar f √ n k=1 h  λ 2   λ 2 i 1 1 − Ef √ = n Ef √ n n h  λ   λ    λ 2 i 1 2 1 + n(n − 1) Ef √ f √ − Ef √ . n n n

(3.21)

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Note  λ 2 Z ∞ 1 Ef √ f (x)2 p¯n (x)dx = n −∞ Z ∞ √ √ 1 = √ f (x)2 Kn ( nx, nx)dx n −∞ Z ∞Z ∞ √ √ f (x)2 Kn ( nx, ny)2 dxdy = −∞

(3.22)

−∞

and λ  λ  2 1 Ef √ f √ n n Z ∞Z ∞ √ √ f (x)f (y)npn,2 ( nx, ny)dxdy = −∞ −∞ Z ∞Z ∞
√ √ 1 f (x)f (y) det Kn ( nx, ny) dxdy = n − 1 −∞ −∞ Z 2 √ √ 1  ∞ f (x)Kn ( nx, nx)dx = n − 1 −∞ Z ∞Z ∞ √ √ 1 f (x)f (y)Kn ( nx, ny)2 dxdy. − n − 1 −∞ −∞

(3.23)

Substituting (3.22) and (3.23) into (3.21) yields n X  λ  k V ar f √ n k=1 Z ∞Z ∞ √ √ 1 (f (x) − f (y))2 Kn ( nx, ny)2 dxdy = 2 −∞ −∞ Z Z n ∞ ∞  f (x) − f (y) 2 = 2 −∞ −∞ x−y
2 √ √ √ √ ϕn ( nx)ϕn−1 ( ny) − ϕn ( ny)ϕn−1 ( nx) dxdy. Since f is 1-Lipschitz function, it follows by the orthogonality of ϕl n X  λ  k V ar f √ n k=1 Z ∞Z ∞
2 √ √ √ √ n ≤ ϕn ( nx)ϕn−1 ( ny) − ϕn ( ny)ϕn−1 ( nx) dxdy 2 −∞ −∞ = 1, (3.24) which implies the claim (3.19). Now we conclude the proof of Theorem 3.2.



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3.2

Fluctuations of Stieltjes transforms

Let An be the standard GUE matrix as in the Introduction. Consider the √ normalized matrix Hn = An / n and denote by λ1 , λ2 , · · · , λn its eigenvalues. Define the Green function 1 Gn (z) = (Gij (z))n×n = Hn − z and its normalized trace n

mn (z) =

1 1 1X 1 trGn (z) = tr = Gii (z). n n Hn − z n i=1

It obviously follows that Z

∞

mn (z) = −∞

1 dFn (x) = sFn (z) x−z

where Fn is defined by (3.15). In this section we shall first estimate Emn (z) and V ar(mn (z)) and then prove a central limit theorem for mn (z). Start with some basic facts and lemmas about the Green function and trace of a matrix. We occasionally suppress the dependence of functions on z when the context is clear, for example we may write Gij instead of Gij (z), and so on. Lemma 3.3. Let Hn = (Hij )n×n be a Hermitian matrix, Gn (z) = (Gij (z))n×n its Green function. Then it follows (i) matrix identity n

1 1X Gik Hki ; Gij = − + z z k=1

(ii) for z = a + iη where η 6= 0 and k ≥ 1 1 sup (Gkn )ij (z) ≤ k , |η| 1≤i,j≤n

1 1 trGkn ≤ k ; n |η|

(iii) differential relations ∂Gkl = −Gki Gil ∂Hii and for i 6= j ∂Gkl = −(Gki Gjl + Gkj Gil ), ∂ReHij

∂Gkl = −i(Gki Gjl − Gkj Gil ). ∂ImHij

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Proof. (i) trivially follows form the fact Gn (Hn − z) = In . To prove (ii), let U = (uij )n×n be a unitary matrix such that   λ1 0 · · · 0  0 λ2 · · · 0   ∗  (3.25) Hn = U  . . . . U .  .. .. . . ..  0 0 · · · λn Then (λ1 − z)−k 0  0 (λ − z)−k 2  Gkn = U  .. ..  . . 

0

0

··· ··· .. .

0 0 .. .

   ∗ U . 

· · · (λn − z)−k

Hence we have (Gkn )ij =

n X

uil (λl − z)−k u∗lj ,

l=1

from which it follows n X k 1 (Gn )ij ≤ 1 |uil ||u∗lj | ≤ k . |η|k |η| l=1

We conclude (ii). Finally, (iii) easily follows from the Sherman-Morrison equation 1 1 1 1 − = −δ A . Hn + δA − z Hn − z Hn + δA − z Hn − z



The next lemma collects some important properties that will be used below about Gaussian random variables. Lemma 3.4. Assume that g1 , g2 , · · · , gm are independent centered normal random variables with Egk2 = σk2 . Denote σ 2 = max1≤k≤m σk2 . (i) Stein equation: if F : Rm 7→ C is a differentiable function, then ∂F Egk F (g1 , · · · , gm ) = σk2 E (g1 , · · · , gm ). ∂gk (ii) Poincar´e-Nash upper bound: if F : Rm 7→ C is a differentiable function, then 2 E F (g1 , · · · , gm ) − EF (g1 , · · · , gm ) ≤ σ 2 E|∇F |2 where ∇F stands for the gradient of F . (iii) Concentration of measure inequality: if F : Rm 7→ C is a Lipschitz function, then for any t > 0
2 2 P F (g1 , · · · , gm ) − EF (g1 , · · · , gm ) > t ≤ e−t /2σ kF klip .

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Now we can use the above two lemmas to get a rough estimate for Emn (z) and V ar(mn (z)). Proposition 3.3. For each z with Imz 6= 0, it follows (i) Emn (z) = msc (z) + O(n−2 ), where msc (z) denotes the Stieltjes transform of ρsc , namely z 1p 2 msc (z) = − + z − 4; 2 2 (ii) E|mn (z) − Emn (z)|2 = O(n−2 ).

(3.26)

(3.27)

Proof. Start with the proof of (3.27). Note that mn is a function of independent centered normal random variables {Hii , 1 ≤ i ≤ n} and {ReHij , ImHij , 1 ≤ i < j ≤ n}. We use the Poincar´e-Nash upper bound to get n 1 hX ∂mn 2 X ∂mn 2 E E|mn − Emn |2 ≤ E + n i=1 ∂Hii ∂ReHij i<j X ∂mn 2 i E + (3.28) . ∂ImHij i<j It easily follows from the differential relations in Lemma 3.3 (iii) that n 1X ∂mn =− Gli Gil , ∂Hii n l=1

∂mn ∂ReHij

n 1X =− (Gli Gjl + Glj Gil ), n

∂mn 1 = −i ∂ImHij n

l=1 n X

(Gli Gjl − Glj Gil ).

l=1

In turn, according to Lemma 3.3 (ii), we have (3.27). Proceed to the proof of (3.26). First, use the matrix identity to get Emn n 1X EGii = n i=1 n

=

n

 1X  1 1X E − + Gik Hki n i=1 z z k=1

n n
1 X 1 X 1 EGii Hii + EGik ReHki + iImHki . (3.29) =− + z zn i=1 zn i6=k

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Also, we have by Lemma 3.3 (iii) 1 ∂Gii 1 E = − G2ii n ∂Hii n ∂Gik 1 1 E = − (G2ik + Gii Gkk ) = 2n ∂ReHki 2n 1 ∂Gik i = E = − (G2ik − Gii Gkk ). 2n ∂ImHki 2n

EGii Hii = EGik ReHki EGik ImHki

(3.30) (3.31) (3.32)

Substituting (3.30)-(3.32) into (3.29) immediately yields 1 1 Emn = − − Em2n . z z

(3.33)

According to (ii), Em2n − (Emn )2 = O(n−2 ). Hence it follows 1 1 Emn = − − (Emn )2 + O(n−2 ). z z Recall that msc (z) satisfies the equation

(3.34)

1 1 msc = − − m2sc . z z It is now easy to see Emn (z) = msc (z) + O(n−2 ), as desired.



Remark 3.1. (3.27) can be extended to any linear eigenvalue statistic (see (3.52) below) with differentiable test function. See Proposition 2.4 of Lytova and Pastur (2009). As a direct consequence of Proposition 3.3, we obtain Corollary 3.1. For each z with Imz 6= 0, P

mn (z) −→ msc (z),

n → ∞.

(3.35)

According to Theorem 1.14, the Stieltjes continuity theorem, (3.35) is in turn equivalent to saying that as n → ∞ d

Fn −→ ρsc

in P.

Thus we have derived the Wigner semicircle law using the Green function approach.

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In the following we shall be devoted to the refinement of estimates of Emn and V ar(mn ) given in Proposition 3.3. A basic tool is still Stein’s equation. As above, we will repeatedly use the Stein equation to get the precise coefficients in terms of n−2 . Main results read as follows Theorem 3.3. For each z with Imz 6= 0, it follows (i) 1 1 Emn (z) = msc (z) + + o(n−2 ); 2 5/2 n2 2(z − 4) (ii)
2 1 1 + o(n−2 ), E mn (z) − Emn (z) = 2 2 (z − 4) n2
Cov mn (z1 ), mn (z2 ) 1  1 z1 z2 − 4 p p = − 1 + o(n−2 ). 2(z1 − z2 )2 n2 z12 − 4 · z22 − 4

(3.36)

(3.37)

Proof. One can directly get (i) from (ii) by noting the equation (3.33). We shall mainly focus on the computation of (3.36), since (3.37) is similar. To do this, note n 1X Emn Gii Em2n = n i=1 n

=

n

 1 1X  1X Emn − + Gki Hik n i=1 z z k=1

n 1 X 1 Emn Gii Hii = − Emn + z zn i=1 1 X i X + Emn Gki ReHik + Emn Gki ImHik . zn zn i6=k

i6=k

Using Lemma 3.3 (iii) and some simple algebra we get 1 1 1 X Em2n = − Emn − Em3n − 3 EGij Gjk Gki . z z zn

(3.38)

i,j,k

Hence we have Em2n

n 1 1 1 X 3 2 − (Emn ) = − Emn − Emn − (Emn ) − 3 EGij Gjk Gki z z zn 2

i,j,k=1

2 = − Emn z

n
1 X Em2n − (Emn )2 − 3 EGij Gjk Gki zn


3 1 − E mn − (Emn ) . z

i,j,k=1

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Solving this equation further yields Em2n − (Emn )2 = −

n 1 1 X EGij Gjk Gki + o(n−2 ). (3.39) z + 2Emn n3 i,j,k=1

We remark that the summand in the righthand side of (3.39) is asymptotically as small as n−2 by Lemma 3.3 (ii). It remains to precisely estimate this summand below. For this, we first observe n n n  1  1X 1 X 1 X EGik Gki = EGik − δi,k + Gkl Hli n n z z i,k=1

i,k=1

l=1

1 1 = − Emn + z zn

n X

EGik Gkl Hli

i,j,k=1

n 1 2 1 X = − Emn − Emn Gik Gki z z n

1 2 1 = − Emn − Emn z z n

i,k=1 n X

EGik Gki + O(n−1 ),

i,k=1

and so we have n
Emn 1 X 1 + O(n−1 ) . EGik Gki = − n z + 2Emn

(3.40)

i,k=1

In the same spirit, we have n n 1 X 1 X 1 EGij Gjk Gki = − EGij Gji + n zn i,j=1 zn i,j,k=1

=−

n X

EGij Gjk Gkl Hli

i,j,k,l=1 n  X

n 2 1 X 1 1 EGij Gji − EGij Gji zn i,j=1 z n i,j=1

n 1 X 2 EGij Gjk Gki + O(n−1 ). − Emn z n i,j,k=1

Solving this equation and noting (3.40) yields n n n 2 i 1 X 1 + o(1) h 1 X 1 X EGij Gji EGij Gjk Gki = EGik Gki + n z + 2Emn n n i,j=1 i,j,k=1

i,k=1

zEmn + (Emn )2 (1 + o(1)) (z + 2Emn )3 1 + o(1) , = (z + 2Emn )3 =−

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where in the last step we used the fact zEmn + (Emn )2 = −1 + o(1). Next, we turn to prove (3.37). Since the proof is very similar to (3.36), we only give some main steps. It follows by the matrix identity Emn (z1 )mn (z2 ) n 1X = Emn (z1 )Gii (z2 ) n i=1 =

n n  1  1X 1 X Emn (z1 ) − + Gik (z2 )Hki n i=1 z2 z2 k=1

=−

1 1 Emn (z1 ) + z2 z2 n

n X

Emn (z1 )Gik (z2 )Hki .

(3.41)

i,k=1

Applying the Stein equation to mn (z2 ), n 1 X Emn (z1 )Gik (z2 )Hki = −Emn (z1 )mn (z2 )2 n i,k=1

n

−

1X EGkl (z1 )Gli (z1 )Gik (z2 ). n

(3.42)

i,k,l

Substituting (3.42) into (3.41) yields 1 1 Emn (z1 ) − Emn (z1 )mn (z2 )2 z2 z2 n X 1 EGkl (z1 )Gli (z1 )Gik (z2 ). − z2 n

Emn (z1 )mn (z2 ) = −

i,k,l

We now have Emn (z1 )mn (z2 ) − Emn (z1 )Emn (z2 ) 1 1 = − Emn (z1 ) − Emn (z1 )mn (z2 )2 − Emn (z1 )Emn (z2 ) z2 z2 n 1 X − EGkl (z1 )Gli (z1 )Gik (z2 ). z2 n i,k,l=1

By virtue of Proposition 3.3 and (3.33), it follows Emn (z1 )mn (z2 ) − Emn (z1 )Emn (z2 ) n 1 X 1 EGkl (z1 )Gli (z1 )Gik (z2 ) + o(n−2 ). (3.43) =− z2 + 2Emn (z2 ) n i,k,l=1

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It remains to estimate the summand of the righthand side of (3.43). To this end, note n 1 X EGkl (z1 )Glk (z2 ) n =

k,l=1 n X

1 n

=−

=−

k,l=1

n  1  1 X EGkl (z1 ) − δk,l + Glm (z2 )Hmk z2 z2 m=1

1 1 Emn (z1 ) + z2 z2

n X

EGkl (z1 )Glm (z2 )Hmk

k,l,m=1

n 1 1 1 X Emn (z1 ) − Emn (z1 ) Gkl (z1 )Glk (z2 ) z2 z2 n k,l=1

n 1 1 X − Emn (z2 ) Gkl (z1 )Glk (z2 ) z2 n k,l=1

=−

n 1 1 X 1 Emn (z1 ) − Emn (z1 ) EGkl (z1 )Glk (z2 ) z2 z2 n k,l=1

n 1 1 X − Emn (z2 ) EGkl (z1 )Glk (z2 ) + o(1) z2 n k,l=1

which immediately gives n 1 X Emn (z1 ) (1 + o(1)). EGkl (z1 )Glk (z2 ) = − n z2 + Emn (z1 ) + Emn (z2 ) k,l=1

Applying once again the matrix identity and Stein equation, we obtain n 1 X EGkl (z1 )Gli (z1 )Gik (z2 ) n i,k,l=1

=−

n 1 X EGkl (z1 )Glk (z1 ) z2 n k,l=1

−

n 1 X

1 E z2 n

k,l=1

n  1 X  Gkl (z1 )Glk (z1 ) E Gkl (z1 )Glk (z2 ) n k,l=1

n 1 X 1 Em (z ) EGkl (z1 )Gli (z1 )Gik (z2 ) − n 1 z2 n n

−

1 1 Emn (z2 ) z2 n n

i,k,l=1 n X i,k,l=1

EGkl (z1 )Gli (z1 )Gik (z2 ) + o(1).

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In combination, we have Emn (z1 )mn (z2 ) − Emn (z1 )Emn (z2 ) 1 + o(1) Emn (z1 ) =− z2 + 2Emn (z2 ) z2 + Emn (z2 ) + Emn (z1 )  1 Emn (z1 ) 1 . (3.44) 1− × z1 + 2Emn (z1 ) z2 + Emn (z2 ) + Emn (z1 ) n2 To simplify the righthand side of (3.44), we observe the following asymptotic formulae Emn (z) = msc (z)(1 + o(1)) and Emn (z2 ) Emn (z1 ) = (1 + o(1)). z2 + Emn (z2 ) + Emn (z1 ) z1 + Emn (z2 ) + Emn (z1 ) Thus a simple calculus now easily yields Emn (z1 )mn (z2 ) − Emn (z1 )Emn (z2 ) 1 + o(1) 2 p p p =p 2 2 2 z1 − 4 z2 − 4 z1 − 4 + z22 − 4 + z1 − z2 1 2 p ×p 2 2 z1 − 4 + z2 − 4 − (z1 − z2 ) n2 1 + o(1) 2 1 p p z12 − 4 z22 − 4 z1 z2 − 4 + (z12 − 4)(z22 − 4) n2 p z1 z2 − 4 − (z12 − 4)(z22 − 4) 1 1 + o(1) p = p 2 , (z1 − z2 )2 n2 2 z1 − 4 z22 − 4

=p

as desired.



We have so far proved a kind of law of large numbers for mn (z) and provided a precise estimate of Emn (z) and V ar(mn (z)). Having these, one may ask how mn (z) fluctuates around its average. In the rest of this section we will deal with such a issue. It turns out that mn (z) asymptotically follows a normal fluctuation. Moreover, we have Theorem 3.4. Define a random process by ζn (z) = n(mn (z) − Emn (z)), z ∈ C \ R. Then there is a Gaussian process Ξ = {Ξ(z), z ∈ C \ R} with the covariance structure  
1 z1 z2 − 4 p p − 1 Cov Ξ(z1 ), Ξ(z2 ) = 2 2(z1 − z2 ) z12 − 4 · z12 − 4 such that ζn ⇒ Ξ, n → ∞.

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Proof. We use a standard argument in the context of weak convergence of processes. That is, we will verify both finite dimensional distribution convergence and uniform tightness below. Start by proving the uniform tightness. As in (3.28), we have 1 E|mn (z1 ) − mn (z2 )|2 ≤ E|∇(mn (z1 ) − mn (z2 ))|2 n n 1 hX ∂(mn (z1 ) − mn (z2 )) 2 = E n i=1 ∂Hii X ∂(mn (z1 ) − mn (z2 )) 2 + ∂ReHij i<j X ∂(mn (z1 ) − mn (z2 )) 2 i (3.45) + . ∂ImHij i<j Observe the eigendecomposition (3.25). Then it follows ∂λk = uik u∗ik , ∂Hii ∂λk = uik u∗jk + u∗ik ujk ∂ReHij = 2Re(uik u∗jk ), ∂λk = i(u∗ik ujk − uik u∗jk ) ∂ImHij = 2Im(uik u∗jk ). Hence we have ∂(mn (z1 ) − mn (z2 )) ∂Hii n X ∂(mn (z1 ) − mn (z2 )) ∂λk = · ∂λk ∂Hii k=1

n

=

 1 1 1 X − uik u∗ik , n (λk − z1 )2 (λk − z2 )2 k=1

and so ∂(m (z ) − m (z )) 2 n 1 n 2 ∂Hii n  1 1 1 X  − ≤ 2 n (λk − z1 )2 (λk − z2 )2 k,l=1   1 1 × uik u∗ik uil u∗il . − 2 2 (λl − z¯1 ) (λl − z¯2 )

(3.46)

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Similarly, ∂(m (z ) − m (z )) 2 n 1 n 2 ∂ReHij n  1 1 4 X  − ≤ 2 2 2 n (λk − z1 ) (λk − z2 ) k,l=1   1 1 × Re(uik u∗jk )Re(uil u∗jl ) − (λl − z¯1 )2 (λl − z¯2 )2

(3.47)

and ∂(m (z ) − m (z )) 2 n 1 n 2 ∂ImHij n  1 1 4 X  − ≤ 2 n (λk − z1 )2 (λk − z2 )2 k,l=1   1 1 × − Im(uik u∗jk )Im(uil u∗jl ). (λl − z¯1 )2 (λl − z¯2 )2

(3.48)

Substituting (3.46)-(3.48) into (3.45) yields n  1 1 4 X  − E|mn (z1 ) − mn (z2 )|2 ≤ 3 2 2 n (λk − z1 ) (λk − z2 ) i,j,k,l=1   1 1 × − uik u∗jk u∗il ujl . (λl − z¯1 )2 (λl − z¯2 )2 Note by orthonormality n n X X uik u∗il = δk,l , ujk u∗jl = δk,l i=1

j=1 n X

uik u∗jk u∗il ujl = n.

i,j,k,l=1

We have n 2 1 1 4 X − E|mn (z1 ) − mn (z2 )| ≤ 3 n (λk − z1 )2 (λk − z2 )2 2

k=1

4 ≤ 2 6 |z1 − z2 |2 , n η from which we can establish the uniform tightness for mn (z). Proceed to proving finite dimensional distribution convergence. Fix z1 , z2 , · · · , zq ∈ C \ R. It is enough to prove d

(ζn (z1 ), · · · , ζn (zq )) −→ (Ξ(z1 ), · · · , Ξ(zq )),

n → ∞.

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Equivalently, for any c1 , · · · , cq q q X X d cl ζn (zl ) −→ cl Ξ(zl ), l=1

109

n → ∞.

l=1

Let Xn =

q X

cl ζn (zl ).

l=1

We shall prove that for any t ∈ R, EXn eitXn − itEXn2 EeitXn → 0. This will be again done using the Stein equation. For simplicity and clarity, we only deal with the 1-dimensional case below. In particular, we shall prove Eζn eitζn − itEζn2 Eeitζn → 0. Namely,
nEmn eitζn − nEmn Eeitζn − itn2 Em2n − (Emn )2 Eeitζn → 0. (3.49) Following the strategy in the proof of Proposition 3.3, it follows n X 1 1 it Emn eitζn = − Eeitζn − Em2n eitζn − E Gik Gkl Gli eitζn . z z zn i,k,l=1

We have by virtue of (3.33)

n n Emn eitζn − Emn Eeitζn = − Em2n eitζn − Em2n Eeitζn z n X it − E Gik Gkl Gli eitζn . (3.50) zn i,k,l=1

Likewise, it follows n X 1 1 1 itζn 3 itζn 2 itζn − Emn e − 3E Gik Gkl Gli eitζn Emn e = − Emn e z z zn i,k,l=1

−

it Emn zn3

n X

Gik Gkl Gli eitζn .

i,k,l=1

We have by virtue of (3.38) Em2n eitζn − Em2n Eeitζn
1 1 = − E(mn − Emn )eitζn − E m3n − Em3n eitζn z z n n  X it  X itζn E G G G e − E Gik Gkl Gli Eeitζn − 3 ik kl li zn −

i,k,l=1 n X

it Emn zn3

i,k,l=1

i,k,l=1

Gik Gkl Gli eitζn .

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Also, using some simple algebra and Proposition 3.3 yields
E m3n − Em3n eitζn = 3(Emn )2 E(mn − Emn )eitζn + o(n−2 ). In turn, this implies Em2n eitζn − Em2n Eeitζn 1 + 3(Emn )2 =− E(mn − Emn )eitζn z n X itEmn − E Gik Gkl Gli eitζn + o(n−2 ). zn3

(3.51)

i,k,l=1

Substituting (3.51) into (3.50) and solving the equation,  1 + 3(Emn )2  1− n(Emn eitζn − Emn Eeitζn ) z2 n  Em X 1 1 n = it − E Gik Gkl Gli Eeitζn + o(1). z2 z n i,k,l=1

Note by (3.34)  Em 1 1 + 3(Emn )2 n = −(z + 2Em ) − + o(1). n z2 z2 z In combination with (3.39), it is now easy to see that (3.49) holds true, as desired.  1−

To conclude this section, let us turn to linear eigenvalue statistics. This is a very interesting and well studied object in the random matrix theory. Let f : R → R be a real valued measurable function. A linear eigenvalue statistic with test function f is defined by n 1X Tn (f ) = f (λi ), (3.52) n i=1 where the λi ’s are eigenvalues of normalized GUE matrix Hn . As shown in Theorem 3.2, if f is bounded and continuous, then Z 2 P Tn (f ) −→ f (x)ρsc (x)dx, n → ∞. −2

This is a certain weak law of large numbers for eigenvalues. From a probabilistic view, the next natural issue is to take a closer look at the fluctuation. Under what conditions could one have asymptotic normality? As a matter of fact, this is usually a crucial problem in the statistical inference theory. As an immediate application of Theorem 3.4, we can easily derive a central limit theorem for a class of analytic functions.

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Theorem 3.5. Suppose that f : R → R is a bounded continuous function and analytic in a region including the real line. Then Z 2  
d f (x)ρsc (x)dx −→ N 0, σf2 , n → ∞ (3.53) n Tn (f ) − where

−2 2 the variance σf is given by Z 2Z 2 1 4 − xy 2 p σf = √ 4π 2 −2 −2 4 − x2 4

(f (x) − f (y))2 dxdy. (x − y)2 − y2

(3.54)

Proof. Without loss of generality, we may and do assume f is analytic in the region {z = x + iη : x ∈ R, |η| ≤ 1}. According to the Cauchy integral formula, Z f (z) 1 dz, f (x) = 2πi |z|=1 x − z which in turn implies Tn (f ) =

1 2πi

Z f (z)mn (z)dz. |z|=1

Hence it follows from Proposition 3.3 Z 1 ETn (f ) = f (z)Emn (z)dz 2πi |z|=1 Z
1 = f (z)msc (z) 1 + O(n−2 ) dz 2πi |z|=1 Z 2 f (x)ρsc (x)dx + O(n−2 ). = −2

In addition, it also follows from Theorem 3.4 Z

1 f (z)n mn (z) − Emn (z) dz n Tn (f ) − ETn (f ) = 2πi |z|=1 Z 1 d −→ f (z)Ξ(z)dz, 2πi |z|=1 where the convergence is a standard application of continuous mapping theorem. To get the variance, note the following integral identity   z1 z2 − 4 1 p p − 1 Cov(Ξ(z1 ), Ξ(z2 )) = 2(z1 − z2 )2 z12 − 4 · z12 − 4 Z 2Z 2 1 xy − 4 √ √ = 4π 2 −2 −2 (x − y)2 x2 − 4 x2 − 4  1 1  1 1  × − − dxdy. z1 − x z1 − y z2 − x z2 − y

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Therefore we have Z Z 1 2 σf = f (z1 )f (z2 )Cov(Ξ(z1 ), Ξ(z2 ))dz1 dz2 (2πi)2 |z1 |=1 |z2 |=1 Z 2Z 2 Z Z 1 1 xy − 4 p dxdy = √ 4π 2 −2 −2 (x − y)2 x2 − 4 y 2 − 4 (2πi)2 |z1 |=1 |z2 |=1   1 1 1  1 − − dz1 dz2 f (z1 )f (z2 ) z1 − x z1 − y z2 − x z2 − y Z 2Z 2 1 (f (x) − f (y))2 xy − 4 p = dxdy. √ 2 4π −2 −2 x2 − 4 y 2 − 4 (x − y)2 The proof is now complete.



It has been an interesting issue to study fluctuations of linear eigenvalue statistics for as wide as possible class of test functions. In Theorem 3.5, the analyticity hypothesis was only required to use the Cauchy integral formula. This condition can be replaced by other regularity properties. For instance, Lytova and Pastur (2009) proved that Theorem 3.5 is valid for a bounded continuous differentiable test function with bounded derivative. Johansson (1998) studied the global fluctuation of eigenvalues to manifest the regularity of eigenvalue distribution. In particular, assume that f : R 7→ R is not too large for large values of x: (i) f (x) ≤ L(x2 + 1) for some constant L and all x ∈ R; (ii) |f 0 (x)| ≤ q(x) for some polynomial q(x) and all x ∈ R; (iii) For each x0 , there exists an α > 0 such that f (x)ψx0 (x) ∈ H 2+α , where H 2+α is standard Sobolev space and ψx0 (x) is an infinitely differentiable function such that |ψx0 (x)| ≤ 1 and  1 |x| ≤ x0 , ψx0 (x) = 0 |x| > x0 + 1. Then (3.53) is also valid with σf2 given by. √ Z 2Z 2 0 f (x)f (y) 4 − x2 1 2 p dxdy. σf = 4π 2 −2 −2 (y − x) 4 − y 2

(3.55)

Here we note that the righthand sides of (3.54) and (3.55) are equal. 3.3

Number of eigenvalues in an interval

In this section we are further concerned with such an interesting issue like how many eigenvalues locate in an interval. Consider the standard GUE

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matrix An , and denote by λ1 , λ2 , · · · , λn its eigenvalues. For [a, b] ⊆ (−2, 2), define Nn (a, b) to be the number of normalized eigenvalues lying in [a, b]. Namely,  √ √ Nn (a, b) = ] 1 ≤ i ≤ n : n a ≤ λi ≤ n b . According to the Wigner semicircle law, Theorem 3.2, Z b Nn (a, b) P −→ ρsc (x)dx, n → ∞. n a In fact, using the asymptotic behaviors for Hermitian orthogonal polynomials as in Section 3.1, we can further have Proposition 3.4. b

Z

ρsc (x)dx + O(1)

ENn (a, b) = n

(3.56)

a

and

1 log n 1 + o(1) . V ar Nn (a, b) = 2 2π

(3.57)

Proof. (3.56) is trivial since the average spectral density function p¯n (x) converges uniformly to ρsc (x) in [a, b]. To prove (3.57), note the following variance formula Z b √ √
√ V ar(Nn (a, b)) = n Kn nx, nx dx a b

Z

Z

b

−n

Kn a a bZ ∞

Z

Kn

= n a

Z

√

√ √
2 nx, ny dxdy

b bZ a

+n

√
2 nx, ny dxdy

Kn

√
2 √ nx, ny dxdy

−∞

a

=: I1 + I2 . We shall estimate the integrals I1 and I2 below. The focus is upon I1 , since I2 is completely similar. A change of variables easily gives Z ∞ Z b−a
2 √ √ I1 = n dv Kn n(b − u), n(b − u + v) du b−a Z b−a

+n

0

Z dv

0

=: I1,1 + I1,2 .

v

Kn 0

√


2 √ n(b − u), n(b − u + v) du

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It is easy to control I1,1 from above. In fact, when v ≥ b − a √ √ Kn ( nx, ny)2
2 √ √ √ √ 1 ϕn ( nx)ϕn−1 ( ny) − ϕn ( ny)ϕn−1 ( nx) , ≤ 2 (b − a) where x = b − u, y = b − u + v. Hence we have by the orthogonality of ϕn and ϕn−1 Z ∞Z ∞ √ √ 1 n ϕn ( nx)ϕn−1 ( ny) I1,1 ≤ 2 (b − a) −∞ −∞
2 √ √ −ϕn ( ny)ϕn−1 ( nx) dxdy 2 ≤ . (b − a)2 Turn to estimating I1,2 . Note
√ lim Kn (x, y) = n ϕ0n (x)ϕn−1 (x) − ϕn (x)ϕ0n−1 (x) y→x

and kϕ0l k∞ ≤

kϕl k∞ ≤ κ,

√ lκ.

So, 1 n

Z

v

Z dv

n 0


2 √ √ n(b − u), n(b − u + v) du = O(1).

Kn 0

For the integral over (1/n, b − a), we use Lemma 3.2 to get
√ √ √ √ n1/2 ϕn ( nx)ϕn−1 ( ny) − ϕn ( ny)ϕn−1 ( nx)  h  π 1 π θ1 2 − cos nα(θ2 ) − cos nα(θ1 ) + = 2 1/4 2 1/4 π (4 − x ) (4 − y ) 2 4 4   θ2 π π i − cos nα(θ2 ) + − cos nα(θ1 ) − + O(n−1 ) 2 4 4 (4 − xy)1/2 1 = + O(n−1 ). 2π (4 − x2 )1/4 (4 − y 2 )1/4 Thus with x = b − u and y = b − u + v, Z b−a Z v
2 √ √ dv Kn n(b − u), n(b − u + v) du n 1 n

=

1 4π 2

0

Z

b−a

Z dv

1 n

+O(n−1 )

v

du 0

Z

b−a

4 − xy v 2 (4

Z dv

1 n

0

v

−

x2 )1/2 (4

1 du. v2

− y 2 )1/2

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Trivially, it follows b−a

Z

Z dv

1 n

0

v

1 du = O(log n). v2

We also note sup x,y∈(a,b)

(4 −

4 − xy ≤ Ca,b − y 2 )1/2

x2 )1/2 (4

for some positive constant Ca,b . Then for any ε > 0 Z b−a Z v 4 − xy du 2 dv = O(| log ε|). 2 )1/2 (4 − y 2 )1/2 v (4 − x 0 ε

(3.58)

On the other hand, it is easy to see 4 − xy = 1 + O(ε2 ), (4 − x2 )1/2 (4 − y 2 )1/2

0 < v < ε.

Hence we have Z ε Z v 4 − xy dv du 2 = (1 + O(ε2 ))(log n + | log ε|). (3.59) 2 )1/2 (4 − y 2 )1/2 1 v (4 − x 0 n Combining (3.58) and (3.59) together yields Z b−a Z v 4 − xy dv du 2 2 1 v (4 − x )1/2 (4 − y 2 )1/2 0 n Z ε Z v 4 − xy = dv du 2 2 )1/2 (4 − y 2 )1/2 1 v (4 − x 0 n Z b−a Z v 4 − xy + dv du 2 2 )1/2 (4 − y 2 )1/2 v (4 − x ε 0 = (1 + O(ε2 ))(log n + | log ε|).

 √ We remark that the linear eigenvalue statistic i=1 f (λi / n) has variance at most 1 whenever f is a 1-Lipschtiz test function, see (3.24). On the other hand, the counting function is not a 1-Lipschitz function. The Proposition 3.4 provides a log n-like estimate for the size of variance of Nn (a, b). Having the Proposition, one would expect the asymptotic normal fluctuations for Nn (a, b). Below is our main result of this section. Pn

Theorem 3.6. Under the above assumptions, as n → ∞ Z b   1 d q ρsc (x)dx −→ N (0, 1). Nn (a, b) − n 1 a 2π 2 log n

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The rest of this section is devoted to the proof of Theorem 3.6. In fact, we shall prove the theorem under a more general setting. To do this, we need to introduce some basic definitions and properties about determinantal point processes. Recall a point process X on R is a random configuration such that any bounded set contains only finitely many points. The law of X is usually characterized by the family of integer-valued random variables {NX (A), A ∈ B}, where NX (A) denotes the number of X in A. Besides, the correlation function is becoming a very useful concept in describing the properties of point processes. The so-called correlation function was first introduced to study the point process by Macchi (1975). Given a point process X , its k-point correlation function is defined by ρk (x1 , · · · , xk ) = lim

δ→0


1 P (xi − δ, xi + δ) ∩ X 6= ∅, 1 ≤ i ≤ k , (2δ)k

where x1 , · · · , xk ∈ R. Here we only considered the continuous case, the corresponding discrete case will be given in Chapter 4. It turns out that the correlation functions is a powerful and nice tool in computing moments of NX (A). In fact, it is easy to see Z
↓k E NX (A) = ρk (x1 , · · · , xk )dx1 · · · dxk , (3.60) A⊗k

where m↓k = m(m − 1) · · · (m − k + 1), and E

k Y i=1

Z ρk (x1 , · · · , xk )dx1 · · · dxk .

NX (Ai ) = A1 ×···×Ak

A point process X is said to be determinantal if there exists a kernel function KX : R × R 7→ R such that
ρk (x1 , · · · , xk ) = det KX (xi , xj ) k×k for any k ≥ 1 and x1 , · · · , xk ∈ R. The determinantal point processes have been attracting a lot of attention in the past two decades. More and more interesting examples have been found in the seemingly distinct problems. For instance, the GUE model An is a determinantal point process with X = {λ1 , λ2 , · · · , λn } and kernel function KX = Kn given by (3.7). Another well-known example is Poisson point process on R. Let P be a Poisson point process with intensity function %(x), then P can be viewed as a determinantal process having KP (x, y) = %(x)δx,y . Note that Poisson point process is an independent point process, that is, a two-point correlation function is equal

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to the product of two one-point correlation functions. However, a general determinantal point process is a negatively associated process, since ρ2 (x, y) ≤ ρ1 (x)ρ1 (y). Note that no claim is made about the existence or uniqueness of a determinantal point process for a given kernel K. To address these issues, we need to make some additional assumptions below. The kernel K is required to be symmetric and nonnegative definite, that is, K(x, y) = K(y, x) for every x, y ∈ R and det(K(xi , xj ))k×k ≥ 0 for any x1 , x2 , · · · , xk ∈ R. we also further assume that K is locally square integrable on R2 . This means that for any compact D ⊆ R, we have Z |K(x, y)|2 dxdy < ∞. D2

Then we may use K as an integral kernel to define an associated integral operator as Z Kf (x) = K(x, y)f (y)dy < ∞ R

for functions f ∈ L2 (R, dx) with compact support. For a compact set D, the restriction of K to D is the bounded linear operator KD on L2 (R) defined by Z KD f (x) = K(x, y)f (y)dy, x ∈ D. D

Thus KD is a self-adjoint compact operator. Let qnD , n ≥ 1 be nonnegative eigenvalues of KD , the corresponding eigenfunctions φD n forms a orthonor2 mal basis on L (D, dx). We say that KD is of trace class if ∞ X |qnD | < ∞. n=1

If KD is of trace class for every compact subset D, then we say that K is locally of trace class. The following two lemmas characterize the existence and uniqueness of a determinantal point process with a given kernel. Lemma 3.5. Let X be a determinantal point process with kernel KX . If ENX (A) < ∞, then EetNX (A) < ∞ for any t ∈ R. Consequently, for each compact set D, the distribution of NX (D) is uniquely determined by KX . Proof.

It easily follows
NX (A) EetNX (A) = E 1 + (et − 1) ∞ X
↓k (et − 1)k = E NX (A) . k! k=0

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Also, by (3.60) and the Hadamard inequality for nonnegative definite matrix Z
↓k
E NX (A) = det KX (xi , xj ) dx1 · · · dxk A⊗k Z k
k ≤ KX (x1 , x1 ))dx1 = ENX (A) . A

Therefore, we have EetNX (A) ≤

∞ X (et − 1)k k=0

k!

ENX (A)

For any compact set D, ENX (D) < ∞ since EetNX (D) < ∞ for all t ∈ R.


k

R D

< ∞. KX (x, x)dx < ∞, so 

Lemma 3.6. Assume that K is a symmetric and nonnegative definite kernel function such that the integral operator K is a locally trace class. Then K defines a determinantal point process on R if and only if the spectrum of K is contained in [0,1]. Proof.

See Theorem 4.5.5 of Soshnikov (2000).



Theorem 3.7. Let Xn , n ≥ 1 be a sequence of determinantal point processes with kernel KXn on R, let In , n ≥ 1 be a sequence of intervals on R. Assume that KXn · 1In define an integrable operator of locally trace class. Set Nn = NXn (In ). If V ar(Nn ) → ∞ as n → ∞, then N − ENn d pn −→ N (0, 1). V ar(Nn ) The theorem was first proved by Costin and Lebowitz (1995) in the very special case. They only considered the Sine point process with kernel KSine (x, y) = sin(x − y)/(x − y), and Widom suggested it would hold for the GUE model. Later on Soshnikov (2002) extended it to general determinantal random points fields, including Bessel, Airy point processes. The proof of the theorem is quite interesting. A basic strategy is to use the moment method, namely Theorem 1.8. Set Nn − ENn . Xn = p V ar(Nn ) Trivially, τ1 (Xn ) = 0, τ2 (Xn ) = 1 and τk (Xn ) =

γk (Nn ) , (V arNn )k/2

k ≥ 3.

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Then it suffices to show
γk (Nn ) = o γ2 (Nn )k/2 ,

k≥3

provided γ2 (Nn ) = V ar(Nn ) → ∞. Proof. For the sake of clarity, we write γk for γk (Nn ). A key ingredient is to express each γk in terms of correlation functions and so kernel functions of Xn . Start with γ3 . It follows from (1.17) γ3 = ENn3 − 3ENn2 ENn + 2(ENn )3 = E(Nn )3 + 3E(Nn )2 + ENn − 3E(Nn )2 ENn −3(ENn )2 + 2(ENn )3 . Also, we have by (3.60) and a simple algebra Z γ3 = 2 Kn (x1 , x2 )Kn (x2 , x3 )Kn (x3 , x1 )dx1 dx2 dx3 ⊗3 In Z Z Kn (x1 , x2 )Kn (x2 , x1 )dx1 dx2 + −3 Kn (x1 , x1 )dx1 . ⊗2 In

In

To obtain a general equation for γk , we need to introduce k-point cluster function, namely αk (x1 , · · · , xk ) =

X

l−1

(−1)

(l − 1)!

l Y

ρ|Gj | (¯ x(Gj )),

j=1

G

where 1 ≤ l ≤ k, G = (G1 , · · · , Gl ) is a partition of the set {1, 2, · · · , k}, |Gj | stands for the size of Gj , x ¯(Gj ) = {xi , i ∈ Gj }. Using M¨ obius inversion formula, we can express the correlation functions in terms of Ursell functions as follows l XY ρk (x1 , · · · , xk ) = α|Gj | (¯ x(Gj )). G j=1

Moreover, we have an elegant formula in the setting of determinantal point processes αk (x1 , · · · , xk ) X =(−1)k−1 KXn (x1 , xσ(1) )KXn (xσ(1) , xσ(2) ) · · · KXn (xσ(k) , x1 ), (3.61) σ

where the sum is over all cyclic permutations (σ(1), · · · , σ(k)) of (1, 2, · · · , k). Define Z αk (x1 , · · · , xk )dx1 · · · dxk , k ≥ 1. βk = ⊗k In

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Then it is not hard to see βk =

X

l−1

(−1)

(l − 1)!

X

(−1)l−1 (l − 1)!

l Y

E(Nn )↓|Gj |

j=1

G

=

ρ|Gj | (¯ x(Gj ))d¯ x(Gj )

⊗|Gj |

In

j=1

G

=

l Z Y

X Q τ 7→k

l Y k! E(Nn )↓τj , (−1)l−1 (l − 1)! τi !mτi ! j=1

where τ = (τ1 , · · · , τl ) → 7 k is an integer partition of k, mτi stands for the multiplicity of τi in τ . We can derive from (1.17) ∞ X βk k=1

k!

(et − 1)k = log

∞ X E(Nn )↓k

k!

k=0

(et − 1)k

∞ X γk

= log EetNn =

k=1

k!

tk .

(3.62)

Comparing the coefficients of the term tk at both sides of (3.62), we obtain γk =

k X βl l=1

l!

X τ1 +···+τl =k

k! . τ1 ! · · · τl !

Equivalently, γk = βk +

k−1 X

bk,j γj

(3.63)

j=1

where the coefficients bk,j are given by bk,1 = (−1)k (k − 1)!,

bk,k = −1,

k≥2

and bk,j = bk−1,j−1 − (k − 1)bk−1,j ,

2 ≤ j ≤ k − 1.

Since it follows from (3.63) γk = βk + (−1)k (k − 1)!γ1 +

k−1 X

bk,j γj ,

k ≥ 3,

j=2

then it suffices to show k/2


βk + (−1)k (k − 1)!γ1 = o γ2

,

k ≥ 3.

(3.64)

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To do this, use (3.61) to get XZ KXn (x1 , xσ(1) ) · · · KXn (xσ(k) , x1 )dx1 · · · dxk βk = (−1)k−1 σ k−1

= (−1)

⊗k In

Z (k − 1)! ⊗k In

KXn (x1 , x2 ) · · · KXn (xk , x1 )dx1 · · · dxk ,

and so Z βk + (−1) (k − 1)!γ1 = (−1) (k − 1)! KXn (x1 , x1 )dx1 In Z  KXn (x1 , x2 ) · · · KXn (xk , x1 )dx1 · · · dxk . − k

k

⊗k In

Define an integrable operator KIn : L2 (In , dx) → L2 (In , dx) by Z KIn f (x) = f (y)KIn (x, y)dy, x ∈ In . In

Then it follows βk + (−1)k (k − 1)!γ1 = (−1)k (k − 1)! trKIn − trKIkn = (−1)k (k − 1)!

k X





trKIl−2 KIn − KI2n . n

l=2

According to Lemma 3.6, we have
βk + (−1)k (k − 1)!γ1 ≤ k! trKIn − trKI2 n = k!γ2 , which gives (3.64). Now we conclude the proof of the theorem.



As the reader may see, Theorem 3.7 is of great universality for determiantal point processes in the sense that there is almost no requirement on the kernel function. However, the theorem itself does not tell what the expectation and variance of NXn (In ) look like. To have numerical evaluation of expectation and variance, one usually needs to know more information about the kernel function. In the case of GUE, the kernel function is given by Hermite orthogonal polynomials so that we can give precise estimates of expectation and variance. This was already done in Proposition 3.4. It is believed that Theorem 3.7 would have a wide range of applications. We only mention the work of Gustavsson (2005), in which he studied the kth greatest eigenvalue λ(k) of GUE model and used Theorem 3.7 to prove the λ(kn ) after properly scaled has a Gaussian fluctuation around its average as kn /n → a ∈ (0, 1). He also dealt with the case of kn → ∞ and

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kn /n → 0. These results are complement to the Tracy-Widom law for largest eigenvalues, see Section 4.1 and Figure 5.2. In the end of this section, we shall provide a conceptual proof of Theorem 3.7. This is based on the following expression for the number of points lying in a set as a sum of independent Bernoulli random variables. Let K be a kernel function such that the integral operator K is locally of trace class. Let X be a determinantal point process with K as its kernel. Let I be a bounded Borel set on R, then K · 1I is locally trace class. Denote by qk , k ≥ 1 the eigenvalues of K · 1I , the corresponding eigenfunctions φk form a orthonormal basis in L2 (I). Define a new kernel function K I by K I (x, y) =

∞ X

qk φk (x)φk (y),

k=1

which is a mixture of the qk and φk . It is evident that the point process X ∩I is determinantal. The following proposition implies that its kernel is given by K I . Proposition 3.5. It holds almost everywhere with respect to Lebesgue measure K(x, y) = K I (x, y). Furthermore, assume that ξk , k ≥ 1 is a sequence of independent Bernoulli random variables, P (ξk = 1) = qk ,

P (ξk = 0) = 1 − qk ,

then we have d

NX (I) =

∞ X

ξk .

(3.65)

k=1

Proof.

By assumption of trace class, Z X ∞ ∞ 2 X qk φk (x) dx = qk < ∞. I

k=1

k=1

P∞ This shows that the series k=1 qk2 φk (x) converges in L2 (I) and also that it converges pointwise for every x ∈ I \ I0 for some set I0 of zero measure. By the Cauchy-Schwarz inequality, ∞ ∞ ∞ X 2  X  X  qk φk (x)φk (y) ≤ qk φk (x)2 qk φk (y)2 . (3.66) k=n

Hence the series

k=n

P∞

k=1 qk φk (x)φk (y)

k=n

converges absolutely.

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Let f ∈ L2 (I). Write f in terms of the orthonormal basis {φk } to get for any x ∈ I \ I0 f (x) =

∞ Z X

 f (y)φk (y)dy φk (x)

I

k=1

and so Kf (x) =

∞ Z X k=1

I

Z =

f (y) I

Z =

 f (y)φk (y)dy Kφk (x) ∞ X

qk φk (y)φk (x)dy

k=1

f (y)K I (x, y)dy.

I

This implies that we must have K(x, y) = K I (x, y)

a.e.

Turn to prove (3.65). We shall below prove EetNX (I) = Eet

P∞

k=1

ξk

,

t ∈ R.

First, it is easy to see Eet

P∞

k=1

ξk

= =

∞ Y

Eetξk

k=1 ∞  Y

 1 + qk (et − 1)

k=1

= 1+

∞ X

X

qi1 · · · qik (et − 1)k .

k=1 1≤i1 <···
Second, to compute EetNX (I) , we use the following formula
∞ X E NX (I) k t tNX (I) (e − 1)k Ee = k! k=0 Z ∞ X
(et − 1)k = det K I (xi , xj ) k×k dx1 · · · dxk . (3.67) k! k R k=0

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For k ≥ 1,  q1 φ1 (x1 ) q2 φ2 (x1 ) · · · qn φn (x1 ) · · ·  q1 φ1 (x2 ) q2 φ2 (x2 ) · · · qn φn (x2 ) · · ·    =  .. .. .. ..   . . ··· . .  q1 φ1 (xk ) q2 φ2 (xk ) · · · qn φn (xk ) · · ·   φ1 (x1 ) φ1 (x2 ) · · · φ1 (xk )  φ2 (x1 ) φ2 (x2 ) · · · φ2 (xk )      .. .. ×  ...  . · · · .    φn (x1 ) φn (x2 ) · · · φn (xk )  


K I (xi , xj ) k×k

···

···

···

···

=: AB.

(3.68)

Then according to the Cauchy-Binet formula X
det K I (xi , xj ) k×k =

det(Ak Bk ),

(3.69)

1≤i1 <···
where Ak is a k × k matrix consisting of row 1, · · · , row k and column i1 , · · · , column ik from A, Bk is a k × k matrix consisting of column 1, · · · , column k and row i1 , · · · , row ik from B. Using the orthogonality of ϕi , we have Z det(Ak Bk )dx1 · · · dxk = k!qi1 · · · qik . (3.70) Rk

Combining (3.67), (3.69) and (3.70) together yields EetNX (I) = 1 +

∞ X

X

qi1 · · · qik (et − 1)k .

k=1 1≤i1 <···
Thus we prove (3.65), and so conclude the proof.



Proof of Theorem 3.7. Having the identity (3.65) in law, a classic Lyapunov theorem (see (1.9)) can be used to establish the central limit theorem for Nn . Indeed, applying the Proposition 3.5, we get an array {ξn,k , n ≥ 1, k ≥ 1} of independent Bernoulli random variables, so it sufP∞ fices to show the central limit theorem holds for the sums k=1 ξn,k . In turn, note P∞ 3 1 k=1 E|ξn,k − Eξn,k | P∞ ≤ →0 (V ar( k=1 ξn,k ))3/2 (V ar(Nn ))1/2 provided V ar(Nn ) → ∞. Thus the Lyapunov condition is satisfied.



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Logarithmic law

In this section we are concerned with the asymptotic behaviors of the logarithm of the determinant of the GUE matrix. Let An = (zij )n×n be the standard GUE matrix as given in the Introduction, denote its eigenvalues by λ1 , λ2 , · · · , λn . Then we have Theorem 3.8. As n → ∞,   1 1 1 d q log | det An | − log n! + log n −→ N (0, 1). 2 4 1 2 log n

(3.71)

The theorem is sometimes called the logarithmic law in literature. We rePn mark that log |det An | = i=1 log |λi | is a linear eigenvalue statistic with test function f (x) = log |x|. However, the function log |x| is not so nice that one could not directly apply the results discussed in Section 3.2. The theorem was first proved by Girko in the 1970s using the martingale argument, see Girko (1979, 1990, 1998) and references therein for more details. Recently, Tao and Vu (2012) provided a new proof, which is based on a tridiagonal matrix representation due to Trotter (1984). We shall present their proof below. Before that, we want to give a parallel result about Ginibre model . Let Mn = (yij )n×n be an n × n random matrix whose entries are all independent complex standard normal random variables. This is a rich and well-studied matrix model in the random matrix theory as well. Let ν1 , ν2 , · · · , νn be its eigenvalues, then the joint probability density function is given by n Y 2 1Y %n (z1 , · · · , zn ) = e−|zi | /2 , zi ∈ C. (3.72) |zi − zj |2 n i<j i=1 Define the bivariate empirical distribution function n 1X Fn (x, y) = 1(Reνi ≤x, Imνi ≤y) . n i=1 Then it follows d

Fn −→ %c

in P,

(3.73)

where %c stands for the uniform law in unit disk in the plane. We leave the proofs of (3.72) and (3.73) to readers. Other more information can be found in Ginibre (1965) and Mehta (2004). As far as the determinant, a classic and interesting result is

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Proposition 3.6. As n → ∞,   1 1 1 d q log | det Mn | − log n! + log n −→ N (0, 1). 2 4 1 4 log n Proof.

(3.74)

First, observe the following identity in law d

| det Mn | =

n 1 Y

2n/2

χ2i ,

(3.75)

i=1

where the χi is a chi random variable with index i, and all chi random variables are independent. Indeed, let y1 , y2 , · · · , yn denote row vectors of Mn . Then the absolute value of the determinant of Mn is equal to the volume of parallelnoid consisting of vectors y1 , y2 , · · · , yn . In turn, the volume is equal to |y1 | · |(I − P1 )y2 | · · · |(I − Pn−1 )yn |,

(3.76)

where Pi is an orthogonal projection onto the subspace spanned by vectors {y1 , y2 , · · · , yi }, 1 ≤ i ≤ n − 1. Note Pi is an idempotent projection with rank i, so Pi yi+1 is an i-variate complex standard normal random √ vector and√is independent of {y1 , y2 , · · · , yi }. Then letting χ2n = 2|y1 |, χ2(n−i) = 2|(I − Pi )yi+1 |, 1 ≤ i ≤ n − 1 conclude the desired identity. Second, note the χi has a density function 21−i/2 i−1 −x2 /2 x e , Γ(i/2)

x>0

then it is easy to get Eχki = 2k

Γ((i + k)/2) Γ(i/2)

and the following asymptotic estimates

1 1 1 + O i−2 , V ar(log χi ) = + O i−2 . E log χi = log i − 2 2i 2i In addition, for each positive integer k ≥ 1
E(log χi − E log χi )2k = O i−k . Lastly, note by (3.75) d

log | det Mn | = −

n X log 2 log χ2i . n+ 2 i=1

(3.74) now directly follows from the classic Lyapunov CLT for sums of independent random variables. 

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The proof of Proposition 3.6 is simple and elegant. The hypothesis that all entries are independent plays an essential role. It is no longer true for An since it is Hermitian. We need to adopt a completely different method to prove Theorem 3.8. Start with a tridiagonal matrix representation of An . Let an , n ≥ 1 be a sequence of independent real standard normal random variables, bn , n ≥ 1 a sequence of independent random variables with each bn distributed like χn . In addition, assume an ’s and bn ’s are all independent. For each n ≥ 1, construct a tridiagonal matrix   an bn−1 0 0 ··· 0  b  n−1 an−1 bn−2 0 · · · 0     0 bn−2 an−2 bn−3 · · · 0    (3.77) Dn =  .. .. . . . . ..  . . . . .   0 .    0 0 · · · b2 a2 b1  0 0 0 · · · b1 a1 Lemma 3.7. The eigenvalues of Dn are distributed according to (3.2). In particular, d

det An = det Dn .

(3.78)

Proof. We shall obtain the Dn in (3.77) from An through a series of Householder transforms. Write   z11 z1 An = z∗1 An,n−1 where z1 = (z12 , · · · , z1n ). Let w1 = 0,

w2 = −

wl = −

z21  1 |z21 | 1/2 (1 − ) |z21 | 2 α

zl1 , (2α(α − |z21 |))1/2

l≥3

where α > 0 and α2 = |z21 |2 + |z31 |2 + · · · + |zn1 |2 . Define the Householder transform by Vn = In − 2wn wn∗   1 0 ··· 0  0   = . ,  .. Vn,n−1  0

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where wn = (w1 , w2 , · · · , wn )τ . It is easy to check that Vn is a unitary matrix and   z11 (z , z , · · · , z1n )Vn,n−1  ∗  12 13   z12   ∗     z13 Vn An Vn =      Vn,n−1  .  Vn,n−1 An,n−1 Vn,n−1     ..  ∗ z1n



z11

α |zz12 21 |

0



∗   =  α |zz12 | . 21 Vn,n−1 An,n−1 Vn,n−1 0

To make the second entry in the first column nonnegative, we need to add one further configuration. Let Rn differ from the identity matrix by having (2,2)-entry e−iφ with φ chosen appropriately and form Rn Vn An Vn Rn∗ . Then we get the desired matrix   z11 α 0 ,  α Vn,n−1 An,n−1 Vn,n−1 0 where α2 = |z21 |2 + |z31 |2 + · · · + |zn1 |2 . Define an = z11 , bn−1 = α. Vn,n−1 is also a unitary matrix and is independent of An,n−1 , so Vn,n−1 An,n−1 Vn,n−1 is an n − 1 × n − 1 GUE matrix. Repeating the preceding procedure yields the desired matrix Dn . The proof is complete.  According to (3.78), it suffices to prove (3.71) for log | det Dn | below. Let dn = det Dn . It is easy to see the following recurrence relations dn = an dn−1 − b2n−1 dn−2

(3.79)

b2n−2 dn−3 .

dn−1 = an−1 dn−2 − (3.80) √ √ Let en = dn / n! and cn = (b2n − n)/ n. Note cn−k is asymptotically normal as n − k → ∞. So we deduce from (3.79) and (3.80)  cn−1 1  an en = √ en−1 − 1 + √ − en−2 + 1 (3.81) n n 2n  cn−2 1  an−1 en−3 + 2 , (3.82) en−1 = √ en−2 − 1 + √ − n n 2n where and in the sequel 1 , 2 denote a small negligible quantity, whose value may be different from line to line.

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In addition, substituting (3.82) into (3.81), we have  a cn−1 1  an an−1 an cn−2  n en = −1 − √ + + en−2 − √ + en−3 + 1 . n 2n n n n In terms of vectors, we have the following recurrence formula       e 1 1 en 1 n−2 = −I2 − √ Sn,1 + Sn,2 + , en−1 en−3 2 n n

(3.83)

where 

 cn−1 an −an−1 cn−2   1 an−1 an −an cn−2 . = I2 + 0 0 2

Sn,1 = Sn,2

Let rn2 = e22n + e22n−1 . It turns out that log rn satisfies a CLT after properly scaled. This is stated as Lemma 3.8. As n → ∞, log rn + 14 log n d q −→ N (0, 1). 1 2 log n Proof.

(3.84)

Use (3.83) to get rn2

 = (e2n , e2n−1 )

e2n



e2n−1  1 τ = (e2n−2 , e2n−3 ) I2 + √ (S2n,1 + S2n,1 ) 2n  e 1 τ 2n−2 τ + (S2n,1 S2n,1 − S2n,2 − S2n,2 ) + 0 , e2n−3 2n

where 0 denotes a small negligible quantity, whose value may be different from line to line. Define  
e2n−2 1 τ ξn = 2 √ (e2n−2 , e2n−3 ) S2n,1 + S2n,1 , e2n−3 rn−1 2n ηn =

1 2 rn−1 2n

(e2n−2 , e2n−3 )

τ S2n,1 S2n,1

− S2n,2 −

Then we have a recursive relation 2 rn2 = (1 + ξn + ηn + 0 )rn−1 .

τ S2n,2






e2n−2 e2n−3

 .

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Let Fn = σ{a1 , · · · , a2n ; b1 , · · · , b2n−1 }. Then it follows for n ≥ 1
E ξn |Fn−1 = 0,


2 E ξn2 |Fn−1 = , n


E ξn4 |Fn−1 = O(n−2 ) (3.85)

and
1 E ηn |Fn−1 = , 2n


E ηn2 |Fn−1 = O(n−2 ).

(3.86)

Using the Taylor expansion of log(1 + x) we obtain 2 log rn2 = log rn−1 + ξn + ηn −

ξn2 + 0 . 2

Let mn be a sequence of integers such that mn / log n → 1. Then log rn2 =

n X

ξn +

l=mn +1

n  X

ηl −

l=mn +1

 ξl2 2 + 0 + log rm . n 2

By the choice of mn , we have 2 log rm P n √ −→ 0. log n

Also, by the Markov inequality, (3.85) and (3.86) n   X 1 ξ2 1 P √ + 0 −→ 0. ηl − l + 2 2l log n l=m +1 n

Finally, by (3.85) and the martingale CLT we have n X 1 d √ ξl −→ N (0, 1). log n l=m +1 n

In combination, we have so far proven (3.84).



The above lemma describes asymptotically the magnitude of the vector (e2n , e2n−1 ). In order to obtain each component, we also need information about the phase of the vector. Lemma 3.9. Let θn ∈ (0, 2π) be such that (e2n , e2n−1 ) = rn (cos θn , sin θn ). Then as n → ∞, d

θn −→ Θ, where Θ ∼ U (0, 2π).

(3.87)

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Proof. Let us first look at the difference between θn and θn−1 . Rewrite (3.83) as     cos θn cos θn−1 rn = rn−1 (−I2 + Dn ) , sin θn sin θn−1 where 1 1 Dn = − √ Sn,1 + Sn,2 +  n n where  is a negligible matrix. It then follows       rn cos θn cos θn−1 cos θn−1 =− + Dn . sin θn−1 sin θn−1 rn−1 sin θn

(3.88)

Note (cos θn−1 , sin θn−1 ) and (− sin θn−1 , cos θn−1 ) form an orthonormal basis. It is easy to see       cos θn−1 cos θn−1 − sin θn−1 Dn = xn + yn , (3.89) sin θn−1 sin θn−1 cos θn−1 where the coefficients xn and yn are given by   cos θn−1 xn = (cos θn−1 , sin θn−1 )Dn sin θn−1 and  yn = (− sin θn−1 , cos θn−1 )Dn

cos θn−1 sin θn−1

 .

Substituting (3.89) back into (3.88) yields       rn cos θn cos θn−1 − sin θn−1 = (−1 + xn ) + yn . sin θn−1 cos θn−1 rn−1 sin θn Thus it is clear that tan(θn − θn−1 ) =

yn , −1 + xn

which in turn leads to θn − θn−1 = arctan

yn . −1 + xn

Next we estimate xn and yn . There is a constant ς > 0 such that xn = OP (n−ς ), In addition, for some ι > 1
E xn θn−1 = O(n−ι ),

yn = OP (n−ς ).



E yn θn−1 = O n−ι

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1 E yn2 θn−1 = + O n−ι . 2n Using the Taylor expansions of arctan x and (1 + x)−1 we obtain
θn − θn−1 = −(yn − xn yn ) + O yn3 , and so 
 k2 eikθn = eikθn−1 1 − ikyn + ikyn xn − yn2 + O yn2 (xn + yn ) . 2 Hence we have 
k2  Eeikθn = Eeikθn−1 1 − ikyn + ikyn xn − yn2 + O yn2 (xn + yn ) . 2 Moreover, using conditioning argument we get 
k2  Eeikθn = Eeikθn−1 1 − + O n−ι . (3.90) 4n Let mn → ∞ and mn /n → 0. Then repeatedly using (3.90) to get n  Y

Eeikθn =

1−

l=mn +1

→ 0,

n  X  k 2  ikθmn +O Ee l−ι 4l l=mn +1

n→∞

  Qn 2 where in the last limit we used ι > 1 and the fact that l=mn +1 1− k4l → 0   Pn 2 since l=mn +1 1 − k4l → ∞. Thus we complete the proof of (3.87).  Proof of Theorem 3.8. It easily follows from Lemma 3.9  1  → 0, n → ∞. P | cos θn | < log n This in turn implies log | cos θn | P √ −→ 0, log n

n → ∞.

(3.91)

On the other hand, we have log |e2n | = log rn + log | cos θn |. Then according to Lemma 3.8 and (3.91), log |e2n | + 41 log n d q −→ N (0, 1). 1 log n 2 The analog is valid for e2n−1 . Therefore we have proven Theorem 3.8.



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To conclude this section, we shall mention the following variants of logarithmic law. Given z ∈ R, we are interested in the asymptotic behaviors of the characteristic polynomials of An at z. We need to deal with two cases separately: either outside or inside the support of Wigner semicircle law. Theorem 3.9. If z ∈ R \ [−2, 2], then
√ n d log det(An − z n) − log n − nµz −→ N 0, σz2 , 2 where µz and σz2 are given by √ √z 2 − 4 + z 1 z − z2 − 4 √ + log µz = , 2 2z+ z −4 2 √ |z| + z 2 − 1 1 2 − log(z 2 − 1). σz = log 2 2 Proof. This is a corollary to Theorem 3.5. Given z ∈ R \ [−2, 2], define fz (x) = log |z − x|. This is analytic outside a certain neighbourhood of [−2, 2]. In addition, a direct and lengthy computation shows Z 2 fz (x)ρsc (x)dx µz = −2 √ √z 2 − 4 + z 1 z − z2 − 4 √ = + log 2 z + z2 − 4 2 and √ Z 2Z 2 0 fz (x)fz (y) 4 − x2 1 2 p dxdy σz = 4π 2 −2 −2 (y − x) 4 − y 2 √ |z| + z 2 − 1 1 = log − log(z 2 − 1).  2 2 Theorem 3.10. If z ∈ (−2, 2), then   √ n 1 n z2 q log det(An − z n) − log n − ( − 1) 2 2 2 1 2 log n d

−→ N (0, 1).

(3.92)

To prove Theorem 3.10, we need a well-estimated result on the power of the characteristic polynomial for the GUE by Krasovsky (2007). Proposition 3.7. Fix z ∈ (−2, 2). The following estimate holds √ 2α E det(An − z n) 2 2  z 2 α /2  n αn+α (z2 /2−1)αn e (1 + εα,n ) = C(α)2αn 1 − 4 2

(3.93)

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uniformly on any fixed compact set in the half plane Reα > −1/2. Here  Z α  1 1 C(α) = log Γ(s + )ds + α2 exp 2 1 2 Γ(α + 2 ) 0 2

= 22α

G(α + 1)2 , G(2α + 1)

(3.94)

where G(α) is Barnes’s G-function. The remainder term εα,n = O(log n/n) is analytic in α. The proof is omitted. For α positive integers (3.93) has been found by Br´ezin and Hikami (2000), Forrester and Frankel (2004). For such α, √ E| det(An − z n)|2α can be reduced to the Hermite polynomials and their derivatives at the points z. However, it is not the case for noninteger α. In order to obtain (3.93), Krasovsky (2007) used Riemann-Hilbert problem approach to compute asymptotics of the determinant of a Hankel matrix whose support is supported on the real line and possesses power-like singularity. Proof of Theorem 3.10. Start with computing expectation and variance √ √ 2α of log det(An −z n) . For simplicity, write M (α) for E det(An −z n) below, and set M (α) = A(α)B(α), where 2 2  z 2 α /2  n αn+α (z2 /2−1)αn e A(α) = 2αn 1 − 4 2

and B(α) = C(α)(1 + εα,n ). It obviously follows √ 2 E log det(An − z n) = M 0 (0) and 2  √ E log | det(An − z n)|2 = M 00 (0). Thus we need only to evaluate M 0 (0) and M 00 (0). A direct calculation shows   α2  n  z2 z 2  −1 n+ M 0 (α) = n log n + 2α log + 1− A(α)B(α) 2 2 2 4 +A(α)B 0 (α)

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and   n  z2 α2  z 2  M 00 (α) = 2 n log n + 2α log + −1 n+ 1− A(α)B 0 (α) 2 2 2 4   α2  z 2 2 n  z2 −1 n+ A(α)B(α) 1− + n log n + 2α log + 2 2 2 4  n 1 z 2  + 2 log + 1− A(α)B(α) + A(α)B 00 (α). 2 2 4 It is easy to see M (0) = A(0) = 1, and so B(0) = 1. Furthermore, by the analyticity of B(α) for Reα > −1/2 and using Cauchy’s theorem, we have B 0 (0) = C 0 (0) + O(n−1 log n),

B 00 (0) = C 00 (0) + O(n−1 log n).

Similarly, it follows from (3.94)  G0 (2α + 1)  G0 (α + 1) −2 . C 0 (α) = C(α) 4 log 2α + 2 G(α + 1) G(2α + 1) Note G0 (α + 1) 1 1 Γ0 (α) = log(2π) + − α + α G(α + 1) 2 2 Γ(α) 1 1 π2 = log(2π) − − (γ + 1)α + α + O(α3 ), 2 2 6 where γ is the Euler constant. So we have C 0 (0) = 0,

C 00 (0) = 4 log 2 + 2(γ + 1).

In combination, we obtain   z2 M 0 (0) = n log n + ( − 1)n + B 0 (0) 2 and  2   z2 z2 M 00 (0) = n log n + ( − 1)n + 2 n log n + ( − 1)n B 0 (0) 2 2  n 1 z 2  + 2 log + 1− + B 00 (0). 2 2 4 This in turn gives  √ 1 z2 E log det(An − z n) = (3.95) n log n + ( − 1)n + o(1) 2 2 and √
V ar log | det(An − z n)| =

z2 1 1 5 1 log n − − log 2 + γ + + o(1). 2 32 2 2 8

(3.96)

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Next we turn to the proof of (3.92). Define for t ∈ R  log | det(A − z √n)| − E log | det(A − z √n)|  n n p mn (t) = E exp t . √ V ar(log | det(An − z n)|) It is sufficient to prove mn (t) → et

2

/2

,

n → ∞.

(3.97)

Indeed, using (3.95) and (3.96) we have  n log n + ( z2 − 1)n    t 2 √ mn (t) = exp − t M √ (1 + o(1)) 2 log n 2 log n  t  2 (1 + o(1)). = et /2 C 2 log n As is known,  the Barne’s G-function is entire and G(1) = 1. It follows that C t/2 log n → 1 as n → ∞, then we get (3.97) as desired. The proof is complete.  3.5

Hermite β ensembles

In the last section of this chapter, we will turn to the study of the Hermite β Ensemble (HβE), which is a natural extension of the GUE. By the HβE we mean an n-point process in the real line R with the following joint probability density function pn,β (x1 , · · · , xn ) = Zn,β

Y

|xi − xj |β

n Y

2

e−xj /2 ,

(3.98)

j=1

1≤i<j≤n

where x1 , · · · , xn ∈ R, β > 0 is a model parameter and Zn,β =

Γ( β2 )n 1 Q (2π)n/2 n! nj=1 Γ( βj 2 )

by Selberg’s integral. This model was first introduced by Dyson (1962) in the study of Coulomb lattice gas in the early sixties. The formula (3.98) can be rewritten as pn,β (x1 , · · · , xn ) ∝ e−βHn (x1 ,··· ,xn ) , where Hn (x1 , · · · , xn ) =

n 1 X 2 1X x − log |xi − xj | 2β j=1 j 2 i6=j

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is a Hamiltonian quantity, β may be viewed as inverse temperature. The quadratic function part means the points fall independently in the real line with normal law, while the extra logarithmic part indicates the points repel each other. The special cases β = 1, 2, 4 correspond to GOE, GUE and GSE respectively. In the study of HβE, a remarkable contribution was made by Dumitriu and Edelman (2002), in which a tridiagonal matrix representation was discovered. Specifically speaking, let an , n ≥ 1 be a sequence of independent normal random variables with mean 0 and variance 2. Let bn , n ≥ 1 be a sequence of independent chi random variables, each bn having density function: 21−βn/2 βn−1 −x2 /2 x e , x > 0. Γ( βn 2 ) In addition, all an ’s and bn ’s are assumed tridiagonal matrix  an bn−1 0 b  n−1 an−1 bn−2  0 bn−2 an−2 1  Dn,β = √  . .. .. 2 .  .. .   0 0 ··· 0 0 0

to be independent. Define a ··· ··· bn−3 · · · .. .. . . b2 a2 · · · b1 0 0

0 0 0 .. .



    .    b1  a1

Then we have Theorem 3.11. The eigenvalues of Dn,β are distributed according to (3.98). As we see from Lemma 3.7, an explicit Householder transform can be used to produce Dn,2 from the GUE square matrix model. The general case will be below proved using eigendecomposition of a tridiagonal and the change of variables formula. Given two sequences of real numbers x1 , x2 , · · · , xn and y1 , y2 , · · · , yn−1 , construct a tridiagonal matrix Xn as follows   xn yn−1 0 0 ··· 0  y  n−1 xn−1 yn−2 0 · · · 0     0 yn−2 xn−2 yn−3 · · · 0    Xn =  . .. .. . . . . ..  . . . . .   .. .    0 0 · · · y2 x2 y1  0 0 0 · · · y1 x1

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(n)

(n)

(n)

(n)

Let λ1 , λ2 , · · · , λn be eigenvalues of Xn and assume that λ1 (n) · · · > λn . Write   (n) λ1 0 ··· 0   (n)  0 λ2 · · · 0  τ Xn = QΛQτ =: Q  .. . . .  Q  .. . ..   . .

(n)

> λ2

>

(3.99)

(n)

0 · · · λn

0

for eigendecomposition, where Q is eigenvector matrix such that QQτ = Qτ Q = In and the first row q = (q1 , q2 , · · · , qn ) is strictly positive. Note that once q1 , q2 , · · · , qn are specified, then other components of Q will be uniquely determined by eigenvalues and Xn . Conversely, starting from Λ and q, one can reconstruct the matrix Xn . Lemma 3.10. (n) λi

Y

−

(n)
λj

1≤i<j≤n

Qn−1 i yi = Qi=1 . n i=1 qi

Proof. We similarly define Xk using x1 , x2 , · · · , xk and y1 , y2 , · · · , yk−1 for 2 ≤ k ≤ n. Let Pk (λ) be the characteristic polynomial of Xk , and let (k) (k) (k) also λ1 , λ2 , · · · , λk be the eigenvalues in decreasing order. Then it is easy to see the following recursive formula 2 Pn (λ) = (xn − λ)Pn−1 (λ) − yn−1 Pn−2 (λ).

(3.100)

We can deduce from (3.100) that for any 1 ≤ j ≤ n − 1 n Y

(n) λi

−

(n−1)
λj

=

2 −yn−1

n−2 Y

(n−2)

λi

(n−1)


− λj

.

i=1

i=1

Hence it follows n n−1 YY

(n)

λi

(n−1)


− λj

2(n−1)

= (−1)n−1 yn−1

j=1 i=1

n−1 Y Y n−2

(n−2)

λi

(n−1)


− λj

j=1 i=1

= (−1)n(n−1)/2

n−1 Y

yl2l .

l=1

On the other hand, note the following identity n

qi2 Pn−1 (λ) X = , (n) Pn (λ) −λ i=1 λ i

(3.101)

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that is Pn−1 (λ) =

n X

Y

qi2

i=1

(n)

λl


−λ .

l6=i

This obviously implies for each 1 ≤ j ≤ n n−1 Y

(n−1)

λi

Y (n)
(n)
− λnj = qj2 λi − λj .

i=1

Hence it follows n n−1 Y Y j=1 i=1 n Y

qj2

=

j=1

i6=j

(n−1)

λi

Y

(n)


− λj

(n)

λi

(n)


− λj

i6=j

= (−1)n(n−1)/2

n Y

qj2

j=1

(n)

Y

λi

(n)
2

− λj

.

(3.102)

1≤i<j≤n

Combining (3.101) and (3.102) together yields Qn−1 2i Y yi (n) (n)
2 = Qi=1 λi − λj n 2 . i=1 qi 1≤i<j≤n The proof is complete.



Consider the eigendecomposition (3.105), the 2n − 1 variables x = (x1 , x2 , · · · , xn ),

y = (y1 , y2 , · · · , yn−1 )

can be put into a one-to-one correspondence with the 2n−1 variables (Λ, q). Let J denote the determinant of the Jacobian for the change of variables from (x, y) to (Λ, q). Then we have Lemma 3.11. Qn−1 i=1 yi Q J= . n qn i=1 qi Proof.

Observe the following identity n
−1 X In − λXn 11 = i=1

(3.103)

qi2 (n)

1 − λλi

.

Use the Taylor expansion of (1 − x)−1 to get ∞ n ∞ X X X
(n)k k k 1+ qi2 λi λk . λ Xn 11 = k=1

k=0 i=1

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Hence we have for each k ≥ 1 Xnk


11

=

n X

(n)k

qi2 λi

.

n X

(n)

i=1

In particular, xn =

qi2 λi

i=1 2 ∗ + yn−1 =

n X

(n)2

qi2 λi

i=1 2 ∗ + yn−1 xn−1 =

n X

(n)3

qi2 λi

(3.104)

i=1

··· ··· ··· n X (n)2n−1 2 qi2 λi ∗ + yn−1 · · · y12 x1 = i=1

where ∗ stands for what already appeared in the preceding equation. Taking differentials at both sides of equations in (3.104) and noting the Pn−1 fact qn2 = 1 − i=1 qi2 yields  (n)    dλ1 dxn  dλ(n)   dx   2   n−1   .   .   .   ..   .        A  dx1  = (B1 , B2 )  dλ(n) ,  n     dq1   dyn−1       .   ..   ..   .  dy1 dqn−1 where n−1 n−1  Y  Y 2 2 yi2 , 2yn−1 , 2yn−2 yn−1 · · · , 2y1 A = diag 1, yn−1 ,··· , yl2 , B1 =



i=1 (n)i 2qj (λj −



l=2

(n)i−1

B2 = iqj2 λj



λ(n)i n )

, 1≤i≤2n−1, 1≤j≤n−1



. 1≤i≤2n−1, 1≤j≤n

Hence a direct computation gives  ∂(x, y)  J = det ∂(λ(n) , q) Qn−1  Qn 4 1 i=1 yi i=1 qi Qn = Qn−1 i qn i=1 qi i=1 yi

Y

(n)

λi

(n)
4

− λj

.

1≤i<j≤n

Now according to Lemma 3.10, (3.103) holds as desired.



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Proof of Theorem 3.11. Denote by λ1,β , λ2,β , · · · , λn,β the eigenvalues of Dn,β . For clarity, we first assume that λ1,β > λ2,β > · · · > λn,β , and write   λ1,β 0 · · · 0  0 λ2,β · · · 0,   τ  Dn,β = QΛQτ =: Q  . (3.105) .. . . ..  Q  .. . . .  0

0 · · · λn,β

for eigendecomposition, where Q is eigenvector matrix such that QQτ = Qτ Q = In and the first row q = (q1 , q2 , · · · , qn ) is strictly positive. As remarked above, such an eigendecomposition is unique. Let T be a one-to-one correspondence between Dn,β and (Λ, Q), then the determinant of its Jacobian is given by (3.103). Hence the joint probability density p(λ, q) of (λ1,β , λ2,β , · · · , λn,β ) and q = (q1 , q2 , · · · , qn−1 ) is equal to n n−1 Y Y βi−1 2 1 2n−1 − 12 x2i e yi e−yi |J| Qn−1 βi n/2 (2π) i=1 Γ( 2 ) i=1 i=1 n n−1 Y βi−1 Qn−1 yi Y 1 2 n−1 1 2 − 2 λi i=1 Q yi e = Q n βi q (2π)n/2 n−1 ) Γ( n i=1 qi i=1 i=1 2 i=1

=

n n Y Y Y 1 2n−1 β 1 − 12 λ2i (λ − λ ) e qiβ−1 . (3.106) Q i j βi q (2π)n/2 n−1 n ) Γ( i=1 i=1 2 i=1 1≤i<j≤n

We see from (3.106) that (λ1,β , λ2,β , · · · , λn,β ) and q = (q1 , q2 , · · · , qn−1 ) are independent and so the joint probability density pn,β (λ) of (λ1,β , λ2,β , · · · , λn,β ) can be obtained by integrating out the variable q pn,β (λ1 , · · · , λn ) =

=

n Y Y 2 2n−1 1 e−λi /2 (λi − λj )β Q n−1 βi n/2 (2π) ) Γ( i=1 2 i=1 1≤i<j≤n Z n Y 1 · qiβ−1 dq1 · · · dqn Pn q:qi >0, i=1 qi2 =1 qn i=1 n Y Y Γ( β2 )n 2 1 β (λ − λ ) e−λi /2 , Q i j n βi n/2 (2π) i=1 Γ( 2 ) 1≤i<j≤n i=1

(3.107)

where we used Lemma 2.18 to compute the integral in the second equation. Finally, to obtain the joint probability density of unordered eigenvalue, we only need to multiply (3.107) by the factor 1/n!. The proof is now concluded. 

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The HβE is a rich and well studied model in random matrix theory. It possesses many nice properties similar to the GUE. In particular, there have been a lot of new advances in the study of asymptotic behaviours of point statistics since the tridiagonal matrix model was discovered. We will below quickly review some results related to limit laws without proofs. The interested reader is referred to original papers for more information. √ √ To enable eigenvalues q asymptotically fall in the interval (−2 n, 2 n),

we consider Hn,β =: β2 Dn,β . Denote by λ1,β , λ2,β , · · · , λn,β the eigenvalues of Hn,β , the corresponding empirical distribution function is n 1X √ Fn,β (x) = . 1 n i=1 (λi,β ≤ nx)

Dumitriu and Edelman (2002) used moment methods to prove the Wigner semicircle law as follows d

Fn,β −→ ρsc

in P.

In particular, for each bounded continuous function f , Z 2 n 1 X  λi,β  P −→ f (x)ρsc dx. f √ n i=1 n −2 Moreover, if f satisfy a certain regularity condition, then according to Johansson (1998), the central limit theorem holds. Namely, Z 2 n λ  X
d i,β 2 f √ −n f (x)ρsc (x)dx −→ N 0, σβ,f , n −2 i=1 2 where σβ,f is given by Z 2  1  2 2 −1 (f (2) + f (−2)) − f (x)ρ0sc (x)dx σβ,f = β 4 −2 Z 2Z 2 0 1 f (x)f (y)ρsc (x) − 2 dxdy. 2π β −2 −2 (x − y)ρsc (y) Following the line of the proof in Theorem 3.8, one could also prove the logarithmic law   1 1 1 d q log | det Hn,β | − log n! + log n −→ N (0, 1). 2 4 β log n 4

As for the counting functions of eigenvalue point process, it is worthy mentioning the following two results both at the edge and inside the bulk. Let un be a sequence of real numbers. Define for x ∈ R  Nn,β (x) = ] 1 ≤ i ≤ n : n1/6 (λi,β − un ) fall between 0 and x .

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Based on variational analysis, Ram´ırez, Rider and Vir´ag (2011) proved that √ under the assumption n1/6 (2 n − un ) → a ∈ R, d

Nn,β (x) −→ NAiryβ (x),

x ∈ R,

where Airyβ is defined as −1 times the point process of eigenvalues of the stochastic with parameter β, and NAiryβ (x) is the number of points between 0 and x. In the same spirit, Valk´ o and Vir´ag (2009) considered the eigenvalues around any location away from the spectral edge. Let un be a sequence of √ real numbers so that n1/6 (2 n − |un |) → ∞. Define for x ∈ R p  Nn,β (x) = ] 1 ≤ i ≤ n : 4n − u2n (λi,β − un ) fall between 0 and x , then d

Nn,β (x) −→ NSineβ (x),

x ∈ R,

where Sineβ is a translation invariant point process given by the Brownian carousel. As the reader may see, the point process from the HβE is no longer determinantal except in special cases. Thus Theorem 3.7, the CostinLebowitz-Soshnikov theorem, is not applicable. However, we can follow the strategy of Valk´ o and Vir´ag (2009) to prove the central limit theorem for the number of points of the HβE lying in the right side of the origin, see and (2010). Theorem 3.12. Let Nn (0, ∞) be the number of points of the HβE lying in the right side of the origin. Then it follows  1 n d q −→ N (0, 1). (3.108) Nn (0, ∞) − 1 2 log n 2 βπ We remark that the number Nn (0, ∞), sometimes called the index, is a key object of interest to physicists. Cavagna, Garrahan and Giardina (2000) calculated the distribution of the index for GOE by means of the replica method and obtained Gaussian distribution with asymptotic variance like log n/π 2 . Majumdar, Nadal, Scardicchio and Vivo (2009) further computed analytically the probability distribution of the number Nn [0, ∞) of positive points for HβE using the partition function and saddle point analysis. They computed the variance log n/βπ 2 + O(1), which agrees with the corresponding variance in (3.108), while they thought the distribution is not strictly Gaussian due to an usual logarithmic singularity in the rate function. The rest part of the section will prove Theorem 3.12. The proof relies largely on the new phase evolution of eigenvectors invented by Valk´o and

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p Vir´ ag (2009). Let Hn,β = 2/βDn,β . We only need to consider the number of positive eigenvalues of Hn,β . A key idea is to derive from the tridiagonal matrix model a recurrence relation for a real number Λ to be an eigenvalue, which yields an evolution relation for eigenvectors. Specifically, let sj = p n − j − 1/2. Define   d11 0 0 · · · 0  0 d22 0 · · · 0    On =  . . . . . ,  .. .. .. . . ..  0

0 0 · · · dnn

where d11 = 1, Let

dii =

bn−1−i di−1,i−1 , si−1

an−i Xi = √ , β

2 ≤ i ≤ n.

0≤i≤n−1

and Yi =

b2n−1−i − si , βsi+1

0 ≤ i ≤ n − 2.

Then  X0 s0 + Y0 0 ··· 0  s1 X1 s1 + Y1 · · · 0     s2 X2 · · · 0  On−1 Hn,β On =  0   . .. .. ..  ..  .. . . . .  0 0 0 · · · Xn−1 obviously have the same eigenvalues as Hn,β . However, there is a significant difference between these two matrices. The rows between On−1 Hn,β On are independent of each other, while Hn,β is symmetric so that the rows are not independent. Assume that Λ is an eigenvalue of On−1 Hn,β On , then by definition there exists a nonzero eigenvector v = (v1 , v2 , · · · , vn )τ such that On−1 Hn,β On v = Λv. Without loss of generality, we can assume v1 = 1. Thus, Λ is an eigenvalue if and only there exists an eigenvector vτ = (1, v2 , · · · , vn ) such that      1 1 X0 s0 + Y0 0 ··· 0  v2   s1 X1 s1 + Y1 · · · 0   v2            0 s2 X2 · · · 0    v3  = Λ  v3  .   .      . . .. .. . ..  ..   .. . ..   ..  . . 

0

0

0

· · · Xn−1

vn

vn

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It can be equivalently rewritten as  0     1 0   1  v   v2  0   2      v3  = Λ  0  v3  .       ..   ..   ...  .  .    vn 1  vn  0 

1  0   0   ··· 

X0 s0 + Y0 0 s1 X1 s1 + Y1 0 s2 X2 .. .. .. . . . 0 0 0 0

··· ··· ··· .. .

0 0 0 .. .

· · · Xn−1

Let v0 = 0, vn+1 = 0 and define rl = vl+1 /vl , 0 ≤ l ≤ n. Thus we have the following necessary and sufficient condition for Λ to be an eigenvalue in terms of evolution: ∞ = r0 ,

rn = 0

and rl+1 =

1 1 + Ysll

 −

1 Λ − Xl + rl sl

 ,

0 ≤ l ≤ n − 2.

(3.109)

Since the (Xl , Yl )’s are independent, then r0 , r1 , · · · , rn−1 , rn forms a Markov chain with ∞ as initial state and 0 as destination state, and the next state rl+1 given a present state rl will be attained through a random fractional linear transform. Next we turn to the description of the phase evolution. Let H denote the upper half plane, U the Poincar´e disk model, define the bijection i−z ¯ → U, ¯ U:H z→ , i+z which is also a bijection of the boundary. As r moves on the boundary ∂H = R ∪ {∞}, its image under U will move along ∂U. In order to follow the number of times this image circles U, we need to extend the action from ∂U to its universal cover, R0 = R, where the prime is used to distinguish this from ∂H. For an action T on R0 , the three actions are denoted by ¯ →H ¯ : z → z• T, U ¯ →U ¯ : z → z◦ T, R0 → R0 : z → z∗ T. H Let Q(α) denote the rotation by α in U about 0, i.e., ϕ∗ Q(α) = ϕ + α. For a, b ∈ R, let A(a, b) be the affine map z → a(z + b) in H. Furthermore, define ! Xl 1 , − Wl = A sl 1 + Ysll

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and 

Rl,Λ

Λ = Q(π)A 1, sl

 Wl ,

0 ≤ l ≤ n − 1.

With these notations, the evolution of r in (3.109) becomes rl+1 = rl• Rl,Λ ,

0≤l ≤n−1

and Λ is an eigenvalue if and only if ∞• R0,Λ · · · Rn−1,Λ = 0. For 0 ≤ l ≤ n define ϕˆl,Λ = π∗ R0,Λ · · · Rl−1,Λ ,

−1 −1 ϕˆ l,Λ = 0∗ Rn−1,Λ · · · Rl,Λ ,

then ϕˆl,Λ = ϕˆ l,Λ

mod 2π.

The following lemma summarizes nice properties about ϕˆ and ϕˆ , whose proof can be found in Valk´ o and Vir´ag (2012). Lemma 3.12. With the above notations, we have (i) rl,Λ• U = eiϕˆl,Λ ; (ii) ϕˆ0,Λ = π, ϕˆ n,Λ = 0; (iii) for each 0 < l ≤ n, ϕˆl,Λ is an analytic and strictly increasing in Λ. For 0 ≤ l < n, ϕˆ l,Λ is analytic and strictly decreasing in Λ; (iv) for any 0 ≤ l ≤ n, Λ is an eigenvalue of Hn,β if and only if ϕˆl,Λ − ϕˆ l,Λ ∈ 2πZ. √ √ Fix −2 < x < 2 and n0 = n(1 − x2 /4) − 1/2. Let Λ = x n + λ/2 n0 and recycle the notation rl,λ , ϕˆl,λ , ϕˆ ˆl,Λ , ϕˆ l,λ for the quantities rl,Λ , ϕ l,Λ . Note that there is a macroscopic term Q(π)A(1, Λ/sl ) in the evolution operator Rl,Λ . So the phase function ϕl,Λ exhibits fast oscillation in l. Let  √  x n Jl = Q(π)A 1, sl and s ρl =

nx2 /4 +i 2 nx /4 + n0 − l

s

n0 − l . nx2 /4 + n0 − l

Thus Jl is a rotation since ρl• Jl = ρl . We separate Jl from the evolution operator R to get   λ Rl,λ = Jl Ll,λ Wl , Ll,λ = A 1, √ . 2 n0 sl

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Note that for any finite λ, Ll,λ and Wl become infinitesimal in the n → ∞ limit while Jl does not. Let   1 Tl = A , −Re(ρl ) , Im(ρl ) then −1

Jl = Q(−2arg(ρl ))Tl where AB = B −1 AB. Define

Ql = Q(2arg(ρ0 )) . . . Q(2arg(ρl )) and ϕ ˆ l,λ = ϕ l,λ∗ Tl Ql−1 .

ϕl,λ = ϕˆl,λ∗ Tl Ql−1 ,

The following lemma is a variant of Lemma 3.12. Lemma 3.13. With the above notations, we have for 1 ≤ l ≤ n − 1 (i0 ) ϕ0,λ = π; (ii0 ) ϕl,λ and −ϕ l,λ are analytic and strictly increasing in λ and are also independent; 2 2 2 (iii0 ) with Sl,λ = T−1 l Lλ Wλ Tl+1 and ηl = ρ0 ρ1 · · · ρl , we have ∆ϕl,λ := ϕl+1,λ − ϕl,λ = ash(Sl,λ , −1, eiϕl,λ η¯l ); −1 (iv 0 ) ϕˆl,λ = ϕl,λ∗ Q−1 l−1 Tl ; 0 0 (v ) for any λ < λ we have a.s.  √ λ Nn x n + √ , 2 n0

= ] (ϕl,λ − ϕ l,λ ,

√ λ0 x n+ √ 2 n0




ϕl,λ0 − ϕ l,λ0 ] ∩ 2πZ .

The difference ∆ϕl,λ can be estimated as follows. Let Zl,λ = i• S−1 l,λ − i −1 = i• T−1 Tl − i l+1 (Lλ Wλ )

= vl,λ + Vl , where ρl+1 − ρl λ + vl,λ = − √ √ , Im(ρl ) 2 n0 n0 − l

Vl =

Xl + ρl+1 Yl √ . n0 − l

(3.110)

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Then ∆ϕl,λ = ash(Sl,λ , −1, z η¯)   i(1 + z¯η)2 2 = Re −(1 + z¯η)Z − Z + O(Z 3 ) 4 2 ImZ + η terms + O(Z 3 ), = −ReZ + 4 where we used Z = Zl,λ , η = ηl and z = eiϕl,λ . √ Lemma 3.14. Assume λ = λn = o( n). For l ≤ n0 , we have

1 1 E ∆ϕl,λ |ϕl,λ = x = bn + osc1 + O (n0 − l)−3/2 , n0 n0

(3.111)



1 1 an + osc2 + O (n0 − l)−3/2 , (3.112) E (∆ϕl,λ )2 |ϕl,λ = x = n0 n0

E |∆ϕl,λ |d |ϕl,λ = O (n0 − l)−3/2 , d > 2, (3.113) where

√ Re dρ n0 λ n0 Im(ρ2 ) dt √ − , + √ bn = Imρ 2 n0 − l 2β n0 − l an =

2n0 n0 (3 + Reρ2 ) + . β(n0 − l) β(n0 − l)

The oscillatory terms are  1 
q osc1 = Re (−vλ − i )e−ix ηl + Re ie−2ix ηl2 q , 2 4 i h
1 osc2 = pn Re e−ix ηl + Re qn (e−ix ηl + e−i2x ηl2 ) , 2 where 4n0 2n0 (1 + ρl ) pn = , qn = . β(n0 − l) β(n0 − l) Lemma 3.15. We have for 0 < l ≤ n0 (i) φˆl,∞ = π, φˆ l,∞ = −2(n − l)π; (ii) φl,∞ = (l + 1)π,

φ l,∞ = −2nπ + 3lπ;

(iii) ˆ φ l,0 = φl,0 + lπ,

−1 −1 φˆ l,0 = 0∗ Rn−1 · · · Rl .

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Proof.

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First, prove (i). Recall ϕˆl,∞ = π∗ R0,∞ · · · Rl−1,∞ ,

−1 −1 ϕˆ l,∞ = 0∗ Rn−1,∞ · · · Rl,∞ ,

where Rl,∞ = Q(π)A (1, ∞) Wl . Note the affine transformation A (1, ∞) Wl maps any z to ∞, and the image of ∞ under the M¨ obius transform U is −1, which in turn corresponds to π ∈ R0 . Thus it easily follows φˆl,∞ = π. As for φˆ l,∞ , note −1 ˆ φˆ l,∞ = φl+1,∞∗ Rl,∞ , −1 where R−1 l,∞ = Wl A (1, −∞) Q(−π). By the angular shift formula, we have −1 ˆ φˆ l,∞ = φl+1,∞∗ Wl A (1, −∞) − π   ˆ iφ −1 l+1,∞ = φˆ −π l+1,∞ + ash Wl A (1, −∞) , −1, e ! ˆ eiφl+1,∞ ◦ Wl−1 A (1, −∞) = φˆ l+1,∞ + arg[0,2π) −1◦ Wl−1 A (1, −∞) ! ˆ eiφl+1,∞ −arg[0,2π) −π −1 ˆ

iφl+1,∞ = φˆ ) − π, l+1,∞ + arg[0,2π) (1) − arg[0,2π) (−e

from which and the fact φˆ n,∞ = 0 we can easily derive φˆ n−1,∞ = −2π,

ˆ φˆ n−2,∞ = −4π, · · · , φl,∞ = −2(n − l)π.

Next we turn to the proof of (ii) and (iii). Since x = 0, then ρl = i, and so Tl is the identity transform and Ql−1 = Q(lπ) for each 0 < l ≤ n0 . Thus we have by the fact that Q is a rotation, ϕl,λ = ϕˆl,λ∗ Tl Ql−1

(3.114)

= ϕˆl,λ∗ Q(lπ) = ϕˆl,λ + lπ, where 0 ≤ λ ≤ ∞. Similarly, ϕ ˆ l,λ = ϕ l,λ + lπ.



Proof of Theorem 3.12. Let x = 0 and n0 = n − 1/2. Taking l = bn0 c = n − 1, we have by (3.110)
Nn (0, ∞) = ] (ϕn−1,0 − ϕ ϕn−1,∞ − ϕ n−1,0 , n−1,∞ ) ∩ 2πZ ,

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from which it readily follows
Nn (0, ∞) − 1 ϕn−1,∞ − ϕ − (ϕ − ϕ ) n−1,0 n−1,∞ n−1,0 ≤ 1. 2π Applying Lemma 3.15 to l = n − 1 immediately yields ϕn−1,∞ − ϕ n−1,∞ = 3π and −1 ϕ n−1,0 = 0∗ Rn−1,0 + (n − 1)π.

Also, it follows 0∗ R−1 P √ n−1,0 −→ 0. log n In combination, we need only to prove ϕ d √n−1,0 −→ N log n

 0,

4 β

 .

To this end, we shall use the following CLT for Markov chain. Recall that π = ϕ0,0 , ϕ1,0 , · · · , ϕn−1,0 forms a Markov chain. Let
zl+1 = ∆ϕl,0 − E ∆ϕl,0 |ϕl,0 . Then z1 , z2 , · · · , zn−1 forms a martingale difference sequence. The martingale CLT implies: if the following conditions are satisfied: (i) Bn :=

n−1 X

Ezl2 → ∞,

(3.115)

l=1

(ii) n−1
P 1 X E zl2 |ϕl−1,0 −→ 1, Bn

(3.116)

n−1
P 1 X E zl2 1(|zl |>εσn ) |ϕl−1,0 −→ 0, Bn

(3.117)

l=1

(iii) for any ε > 0

l=1

then we have n−1 1 X d √ zl −→ N (0, 1). Bn l=1

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151

We will next verify conditions (3.115) - (3.117)
by asymptotic estimates (3.111) - (3.113) for the increments E ∆ϕl,0 |ϕl,0 . Start with Bn . Note
E(∆ϕl,0 |ϕl,0 ) = O (n0 − l)−3/2

E(∆ϕl,0 )2 = E E (∆ϕl,0 )2 |ϕl,0
4 4 = + ERe (−1)l+1 e−iϕl,0 β(n0 − l) n0 − l
+O (n0 − l)−3/2 . Hence a direct computation yields Bn =

n−1 X

E(∆ϕl,0 )2 − E(E(∆ϕl,0 |ϕl,0 ))2

l=1

=

4 log n + O(1) → ∞ β

and n−1
1 X E zl2 |ϕl−1,0 − 1 Bn

=

=

l=1 n−1 X

1 Bn 1 Bn +

E(zl2 |ϕl−1,0 ) − Ezl2




l=1 n−1 X

E((∆ϕl−1,0 )2 |ϕl−1,0 ) − E(∆ϕl−1,0 )2




l=1

n−1
2
2 1 X E E(∆ϕl−1,0 |ϕl−1,0 ) − E(∆ϕl−1,0 |ϕl−1,0 ) Bn l=1

P

→ 0. It also follows from (3.113) that n−1 X

E|∆ϕl−1,0 |3 = O(1),

l=1

which in turn immediately implies the Lindeberg condition (3.117). Thus we have completed the proof of the theorem. 

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Random Uniform Partitions

4.1

Introduction

The theory of partitions is one of the very few branches of mathematics that can be appreciated by anyone who is endowed with little more than a lively interested in the subject. Its applications are found wherever discrete objects are to be counted or classified, whether in the molecular and the atomic studies of matter, in the theory of numbers, or in combinatorial problems from all sources. Let n be a natural number. A partition of n is a finite nonincreasing Pl sequence of positive integers λ1 ≥ λ2 ≥ · · · ≥ λl > 0 such that j=1 λj = n. Set rk = ]{1 ≤ j ≤ l : λj = k}. Trivially, ∞ X k=1

rk = l,

∞ X

krk = n.

k=1

As we remarked in Section 2.2, there is a close connection between partitions and permutations. The set of all partitions of n are denoted by Pn , and the set of all partitions by P, i.e., P = ∪∞ n=0 Pn . Here by convention the empty sequence forms the only partition of zero. Among the most important and fundamental is the question of enumerating various set of partitions. Let p(n) be the number of partitions of n. Trivially, p(0) = 0, and p(n) increases quite rapidly with n. In fact, p(10) = 42, p(20) = 627, p(50) = 204226, p(100) = 190569292, p(200) = 3972999029388. The study of p(n) dates back to Euler as early as in the 1750s, who proved many beautiful and significant partition theorems, and so laid the 153

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foundations of the theory of partitions. Many of the other great mathematicians have contributed to the development of the theory. The reader is referred to Andrews (1976), which is a first thorough survey of this field with many informative historic notes. It turns out that generating function is a powerful tool for studying p(n). Define the generating function of the p(n) by F(z) =

∞ X

p(n)z n .

(4.1)

n=0

Euler started the analytic theory of partitions by providing the explicit formula F(z) =

∞ Y k=1

1 . 1 − zk

(4.2)

We remark that on the one hand, for many problems it suffices to consider F(z) as a formal power series in z; on the other hand, much asymptotic work requires that F(z) be an analytic function of the complex variable z. The asymptotic theory starts 150 years after Euler, with the first celebrated letters of Ramanujan to Hardy in 1913. In a celebrated series of memoirs published in 1917 and 1918, Hardy and Ramanujan found (and was perfected by Radamacher) very precise estimates for p(n). In particular, we have Theorem 4.1. √
1 p(n) = √ e2c n 1 + o(1) , 4 3n √ where and in the sequel c = π/ 6.

(4.3)

The complete proof of Theorem 4.1 can be found in §2.7 of the book Postnikov (1988). Instead, we prefer to give a rough sketch of the proof, without justifying anything. First, using the Cauchy integral formula for the coefficients of a power series, we obtain from (4.1) and (4.2) Z π
1 r−n e−inθ F reiθ dθ. p(n) = 2π −π Choose θn > 0 and split the integral expression for p(n) into two parts: Z Z 
1  p(n) = + r−n e−inθ F reiθ dθ. 2π |θ|≤θn |θ|>θn

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If θn = n−3/4+ε , ε > 0, then it holds Z
1 p(n) ≈ r−n e−inθ F reiθ dθ. 2π |θ|≤θn By Taylor’s formula with two terms, for |θ| ≤ θn we have
0 θ2
00 log F(reiθ ) ≈ log F(r) + iθ log F(r) − log F(r) , 2 and so p(n) ≈

F(r) 2πrn

Z

0

e−iθ(n−(log F (r)) ) e−θ

2

(log F (r))00 /2

dθ.

|θ|≤θn

Up to this point,
0 r has been a free parameter.−vWe now choose r so that n + log F(r) = 0, i.e., we must choose r = e to satisfy the equation n=

∞ X k=1

k . ekv − 1

Note ∞ X

∞ k 1 X kv = v kv ekv − 1 v2 e −1 k=1 k=1 Z ∞ 1 x ≈ 2 dx v 0 ex − 1 c2 = 2. v √ 2 Thus we must take v so that n ≈ c /v 2 , i.e., v ≈ c/ n. For such a choice, Z 2 00 F(r) p(n) ≈ e−θ (log F (r)) /2 dθ n 2πr |θ|≤θn Z F(r) ∞ −θ2 (log F (r))00 /2 e dθ ≈ 2πrn −∞ F(r) = p , (4.4) n r 2π log(F(r))00

using the classical normal integral. To evaluate the value of (4.4), we need the following lemma, see Postnikov (1988). Lemma 4.1. Assume Rez > 0 and z → 0, staying within some angle lying in the right half-plane. Then log F(e−z ) =

1 z c2 + log + O(|z|). z 2 2π

(4.5)

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As a consequence, it easily follows with r = e−c/ n √ 1 1 log F(r) ≈ c n + log (4.6) 4 24n and 2 (4.7) (log F(r))00 ≈ n3/2 . c Substituting (4.6) and (4.7) into (4.4) yields the desired result (4.3). Another effective elementary device for studying partitions is graphical representation. To each partition λ is associated its Young diagram (shape), which can be formally defined as the set of point (i, j) ∈ Z2 such that 1 ≤ j ≤ λi . In drawing such diagrams, by convention, the first coordinate i (the row index) increases as one goes downwards, the second coordinate j (the column index) increases as one goes from the left to the right and these points are left justified. More often it is convenient to replace the nodes by unit squares, see Figure 2.1. Such a representation is extremely useful when we consider applications of partitions to plane partitions or Young tableaux. Sometimes we prefer the representation to be upside down in consistency with Descartes coordinate geometry. The conjugate of a partition λ is the partition λ0 whose diagram is the transpose of the diagram λ, i.e., the diagram obtained by reflection in the main diagonal. Hence the λ0i is the number of squares in the ith column of λ, or equivalently, ∞ X λ0i = rk . (4.8) k=i

In particular, λ01 = l(λ) and λ1 = l(λ0 ). Obviously, λ00 = λ. We have so far defined the set Pn of partitions of n and known how to count its size p(n). Now we want to equip a probability measure on this set. As we will see, this set bears many various natural measures. The first natural measure is certainly uniform, i.e., choose at random a partition with equal probability. Let Pu,n be the uniform measure defined by 1 Pu,n (λ) = , λ ∈ Pn , (4.9) p(n) where the subscript u stands for Uniform. The primary goal of this chapter is to study the asymptotic behaviours of a typical partition under (Pn , Pu,n ) as its size n → ∞. The first remarkable feature is that a typical Young diagram properly scaled has a limit shape. To be precise, define X ϕλ (t) = rk , t ≥ 0. (4.10) k≥t

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In particular, ϕλ (i) = λ0i , and ϕλ (t) is a nonincreasing step function such R∞ that 0 ϕλ (t)dt = n. Theorem 4.2. Under (Pn , Pu,n ) we have as n → ∞ 1 √ P sup √ ϕλ ( nt) − Ψ(t) −→ 0 n a≤t≤b where 0 < a < b < ∞ and Z ∞ 1 e−cu du = − log(1 − e−ct ). Ψ(t) = −cu 1 − e c t

(4.11)

(4.12)

We remark that the curve Ψ(t) was first conjectured by Temperley (1952), who studied the number of ways in which a given amount of energy can be shared out among the different possible states of an assembly. The rigorous argument was given by Vershik (1994, 1996). In fact, Vershik and his school has been recognized to be the first group who started a systematic study of limit shapes of various random geometric objects. We also note that the curve Ψ(t) has two asymptotic lines: s = 0 and t = 0, see Figure 4.1.

Fig. 4.1

Temperley-Vershik curve

From the probabilistic viewpoint, Theorem 4.2 is a kind of weak law of large numbers. Next, it is natural to ask what the fluctuation is of a typical Young diagram around the limit shape. That is the problem of the second order fluctuation. It turns out that we need to deal with two cases separately: at the edge and in the bulk. Let us first treat the edge case of ϕλ (k), k ≥ 0. Note that λ and λ0 have the same likelihood, then it follows by duality d

ϕλ (k) = λ0k = λk ,

k ≥ 1.

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Hence it is sufficient to study the asymptotic distribution of the λk ’s, the largest parts of λ. Let us begin with the following deep and interesting result, due to Erd¨ os and Lehner (1941). Theorem 4.3. As n → ∞, we have for each x ∈ R √  c  −x n Pu,n √ λ1 − log ≤ x −→ e−e . n c

(4.13)

Note that the limit distribution in the right hand side of (4.13) is the famous Gumbel distribution, which appears widely in the study of extremal statistics for independent random variables. Besides, one can further consider the jointly asymptotic distributions of the first m largest parts. Fristedt (1993) obtained the following Theorem 4.4. As n → ∞, we have for x1 > x2 > · · · > xm √   c n ≤ xi , 1 ≤ i ≤ m lim Pu,n √ λi − log n→∞ n c Z x1 Z xm m Y = ··· p(xi−1 , xi )dx1 · · · dxm , p0 (x1 ) −∞

−∞

(4.14)

i=2

where p0 and p are defined as follows p0 (x) = e−e

−x

−x

x∈R

,

and ( p(x, y) =

ee 0,

−x

−e−y −y

,

x > y, x ≤ y.

To understand the limit (4.14), we remark the following nice fact. Let η1 , η2 , · · · be a sequence of random variables with (4.14) as their joint distribution functions, then for x1 > x2 > · · · > xm
P ηm = xm ηm−1 = xm−1 , · · · , η1 = x1 = p(xm−1 , xm ). Hence the ηk ’s form a Markov chain with p(x, y) as the transition density. √ Next, let us turn to the bulk case, i.e., treat ϕλ ( nt) where 0 < t < ∞. Define
√ √ 1 Xn (t) = 1/4 ϕλ ( nt) − nΨ(t) , t > 0. n An interesting result is the following central limit theorem due to Pittel (1997).

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Theorem 4.5. Under (Pn , Pu,n ) we have as n → ∞ Xn ⇒ X

(4.15)

in terms of finite dimensional distributions. Here X(t), t > 0 is a centered Gaussian process with the covariance structure
Cov X(t1 ), X(t2 ) = σt22 − st1 st2 , t1 ≤ t2 (4.16) where σt2

Z

∞

= t

e−cu e−ct du = (1 − e−cu )2 c(1 − e−ct )

and ∞

ue−cu du (1 − e−cu )2 t
1 te−ct − 2 log 1 − e−ct . = −ct (1 − e ) c Z

st =

What happens if t = tn tends to infinity? It turns out that a similar central limit theorem holds when tn grows slowly. In particular, we have Theorem 4.6. Assume tn , n ≥ 1 is a sequence of positive numbers such that 1 tn → ∞, tn − log n → −∞. (4.17) 2c Let Xn (tn ) =


√ √ ectn /2 ϕλ ( ntn ) − nΨ(tn ) , n1/4

t>0

then under (Pn , Pu,n ), as n → ∞ d

Xn (tn ) −→ N (0, 1).

(4.18)

Note that ectn /2 /n1/4 goes to zero under the assumption (4.17). We have so far seen many interesting probability limit theorems for random uniform partitions. In the next two sections we shall provide rigorous proofs. A basic strategy is as follows. First, we will in Section 4.2 construct a larger probability space (P, Qq ) where 0 < q < 1 is a model parameter, under which the rk ’s are independent geometric random variables. Thus we can P directly apply the classical limit theory to the partial sums k rk . Second, we will in Section 4.3 transfer to the desired space (Pn , Pu,n ) using the fact Pu,n is essentially the restriction of Qq to Pn . It is there that we develop

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a conditioning argument, which is consistent with the so-called transition between grand and small ensembles in physics literatures. Having the convergence of finite dimensional distributions, one might expect a weak convergence of the processes (Xn (t), t > 0). To this end, it is required to check the uniform tightness condition, i.e., for any ε > 0   lim lim Pu,n sup |Xn (t) − Xn (s)| > ε = 0. δ→0 n→∞

|t−s|≤δ

However, we are not able to find a good way to verify such a condition. Instead, we shall in Section 4.4 state and prove a weaker stochastic equicontinuity condition: for any ε > 0
lim lim sup Pu,n |Xn (t) − Xn (s)| > ε = 0. δ→0 n→∞ |t−s|≤δ

This together with Theorem 1.16 immediately implies a functional central limit theorem holds for a certain class of integral statistics of Xn (t). We shall also give two examples at the end of Section 4.4. To conclude this chapter, we shall briefly discuss a generalized multiplicative random partitions induced by a family of analytic functions. Throughout the chapter, c1 , c2 , · · · denote positive numeric constants, whose exact values are not of importance. 4.2

Grand ensembles

In this section we shall study unrestricted random partitions with multiplicative measures. Let 0 < q < 1, define the multiplicative measure by Qq (λ) =

q |λ| , F(q)

λ ∈ P,

(4.19)

where |λ| denotes the size of the partition λ. Note by (4.19) and (4.1) X 1 X |λ| q Qq (λ) = F(q) λ∈P

λ∈P

∞ 1 X X n = q = 1. F(q) n=0 λ∈Pn

Thus we can induce a probability space (P, Qq ), which is called a grand ensemble with parameter q. Surprisingly, this Qq has an elegant property as follows.

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Lemma 4.2. Under (P, Qq ), r1 , r2 , · · · is a sequence of independent geometric random variables. In particular, we have Qq (λ ∈ P : rk = j) = (1 − q k )q jk ,

j = 0, 1, 2, · · · .

Proof. The proof is easy. Indeed, note λ = (1r1 , 2r2 , · · · ), so |λ| = P∞ k=1 krk . Thus we have Qq (λ) =

∞ Y

q krk (1 − q k ),

k=1

as desired.



This lemma will play a fundamental important role in the study of random uniform partitions. It will enable us to apply the classical limit theorems for sums of independent random variables. Denote by Eq expectation with respect to Qq . As a direct consequence, we have Eq rk =

qk , 1 − qk

V arq (rk ) =

qk (1 − q k )2

and Eq z rk =

1 − qk . 1 − zq k √ n

Under (P, Qq ), the size |λ| is itself a random variable. Let qn = e−c/ then it is easy to see µn := Eqn |λ| =

∞ X

,

kEqn rk

k=1

√

∞ ∞ X X ke−ck/ n kqnk √ = = k −ck/ n 1 − qn 1 − e k=1 k=1 Z ∞ √ ue−cu = n du + O( n) −cu 1 − e 0 √ = n + O( n),

where in the last step we used the fact Z ∞ ue−cu du = 1. 1 − e−cu 0

(4.20)

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Similarly, σn2

:= V arqn (|λ|) =

∞ X

k 2 V arqn (rk )

k=1

√

∞ X

∞ X k 2 qnk k 2 e−ck/ n √ = = (1 − qnk )2 (1 − e−ck/ n )2 k=1 k=1 Z ∞ u2 e−cu 3/2 du + O(n) = n (1 − e−cu )2 0 2 = n3/2 + O(n). c Fristedt (1993) obtained the following refinement.

(4.21)

Theorem 4.7. Under (P, Qqn ), |λ| normally concentrates around n. Namely,  2 |λ| − n d . −→ N 0, 3/4 c n Moreover, we have the local limit theorem

1 1 + o(1) . (4.22) Qqn |λ| = n = (96)1/4 n3/4 Proof.

By virtue of (4.20) and (4.21), it suffices to prove

|λ| − µn d −→ N (0, 1). (4.23) σn In turn, this will be proved using characteristic functions below. Let  ix|λ|  . fn (x) = Eqn exp σn Then it follows from Lemma 4.2 ∞  ix X  fn (x) = Eqn exp krk σn k=1

=

∞ Y k=1

1

1 − qnk − qnk eikx/σn

.

Observe the following elementary Taylor formulas
z2 + O |z|3 , |z| → 0 log(1 + z) = z + 2 and
x2 eix = 1 + ix − + O |x|3 , |x| → 0. 2

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163

We have  q k (eikx/σn − 1)  q k (eikx/σn − 1) 1 qn2k (eikx/σn − 1)2 log 1 − n =− n + k 1 − qn 1 − qnk 2 (1 − qnk )2  q 3k |eikx/σn − 1|3  +O n (1 − qnk )3 x2 k 2 qnk kqnk ix + · · =− σn 1 − qnk 2σn2 (1 − qnk )2  k 3 q 3k  n +O . (1 − qnk )3 Taking summation over k yields ∞ X k=1

∞ ∞  ix X kqnk x2 X k 2 qnk q k (eikx/σn − 1)  = − + log 1 − n 1 − qnk σn 1 − qnk 2σn2 (1 − qnk )2 k=1

k=1

∞  1 X k 3 qn3k  . +O 3 σn (1 − qnk )3

(4.24)

k=1

It follows by (4.21), ∞
1 X k 3 qn3k = O n−1/4 , 3 k 3 σn (1 − qn )

(4.25)

k=1

which implies that the above Taylor expansions are reasonable. Therefore we see from (4.24) and (4.25) that log fn (x) = − =−

∞ X k=1 ∞ X k=1

log

1 − qnk eikx/σn 1 − qnk

 q k (eikx/σn − 1)  log 1 − n 1 − qnk

x2 ixµn + o(1). − = σn 2 We now conclude the desired assertion (4.23). Next we turn to the proof of (4.22). To this end, we use the inverse formula for lattice random variables to get  |λ| n Qqn (|λ| = n) = Qqn = σn σ Z πσn n 1 e−ixn/σn fn (x)dx. (4.26) = 2πσn −πσn

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1/3

Let ρ(n) = πσn . Then for |x| < ρ(n), ∞  1X 2q k (1 − cos kx/σn )  log |fn (x)| = − log 1 + n 2 (1 − qnk )2 k=1  X c1 x2  1 log 1 + 2/3 ≤− 2 2/3 σn 2/3 σn /2
2/3

σn /2
≤ −c3 x2 . Thus we get 2

|fn (x)| ≤ e−c3 x , |x| ≤ ρ(n). o n 2/3 For |x| > ρ(n), let Sx = k : k ≤ σn , cos kx/σn ≤ 0 . Then log |fn (x)| ≤ −

 1 X 2q k (1 − cos kx/σn )  log 1 + n 2 (1 − qnk )2 k∈Sx

≤ −c4 σn2/3 , which implies sup

2/3

|fn (x)| ≤ e−c4 σn


= o σn−1 .

(4.27)

ρ(n)<|x|≤πσn

Next we shall estimate the integral in the right hand side of (4.26). Split the interval (−πσn , πσn ) into two disjoint subsets: {|x| ≤ ρ(n)} and {ρ(n) < |x| ≤ πσn }, and evaluate the integral value over each one. Since 2 e−ixn/σn fn (x) → e−x /2 , then by the control convergence theorem, Z Z ∞ √ 2 e−x /2 dx = 2π. e−ixn/σn fn (x)dx −→ −∞

|x|≤ρ(n)

Also, we have by (4.27) Z

e−ixn/σn fn (x)dx = o(1).

ρ(n)<|x|≤πσn

In combination, we get the desired assertion. Theorem 4.8. Under (P, Qqn ) we have as n → ∞ 1 √ P sup √ ϕλ ( nt) − Ψ(t) −→ 0 n a≤t≤b where 0 < a < b < ∞.



(4.28)

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Proof. We first prove the convergence in (4.28) for each fixed t > 0. Indeed, it follows by (4.20) √ 1 1 X 1 X qnk Eqn √ ϕλ ( nt) = √ Eqn rk = √ n n √ n √ 1 − qnk k≥ nt k≥ nt Z ∞ −cu e du + o(1) = Ψ(t) + o(1). = 1 − e−cu t Similarly, by (4.21)  1
√  V arqn √ ϕλ ( nt) = O n−1/2 . n Therefore according to the Markov inequality, we immediately have 1 √ P (4.29) √ ϕλ ( nt) − Ψ(t) −→ 0. n Turn to the uniform convergence. Fix 0 < a < b < ∞. For any ε > 0, there is an m ≥ 0 and a = t0 < t1 < · · · < tm < tm+1 = b such that max |Ψ(ti ) − Ψ(ti+1 )| ≤ ε.

0≤i≤m

Also, by virtue of the monotonicity of ϕλ , we have 1 1 √ √ sup √ ϕλ ( nt) − Ψ(t) ≤ 3 max √ ϕλ ( nti ) − Ψ(ti ) 0≤i≤m n n a≤t≤b + max |Ψ(ti ) − Ψ(ti+1 )|. 0≤i≤m

(4.30)

Hence it follows from (4.29) and (4.30) 1   √ Qqn sup √ ϕλ ( nt) − Ψ(t) > 4ε n a≤t≤b  1  √ ≤ m max Qqn √ ϕλ ( nti ) − Ψ(ti ) > ε 0≤i≤m n → 0, n → ∞. The proof is complete. Theorem 4.9. For any x ∈ R, we have as → ∞ √   c −x n ≤ x −→ e−e . Qqn √ λ1 − log n c



(4.31)

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√ √ Proof. Let An,x = n(log n/c + x)/c. Since λ1 is the largest part of λ, then it is easy to see √   c
n ≤ x = Qqn rk = 0, ∀k ≥ An,x . Qqn √ λ1 − log n c It follows by Lemma 4.2 Y

Qqn (rk = 0) =

k≥An,x

Y


1 − qnk .

(4.32)

k≥An,x

For each x ∈ R, qnk → 0 whenever k ≥ An,x . Hence we have as n → ∞, X X log(1 − qnk ) = −(1 + o(1)) qnk k≥An,x

k≥An,x dA

= −(1 + o(1))

e

qn n,x 1 − qn

→ −e−x , which together with (4.32) implies (4.31).



Theorem 4.10. For x1 > · · · > xm √  c  n lim Qqn √ λi − log ≤ xi , 1 ≤ i ≤ m n→∞ n c Z x1 Z xm m Y = ··· p(xi−1 , xi )dx1 · · · dxm , p0 (x1 ) −∞

−∞

i=2

where p0 and p are as in Theorem 4.4. Proof. For simplicity of notations, we only prove the statement in the √ √ case m = 2. Let An,x = n(log n/c + x)/c. Then it is easy to see √  c  n Qqn √ λi − log ≤ xi , i = 1, 2 c n  \  = Qqn {rk = 0} k>An,x2

+

X

 Qqn {rj = 1}

An,x2 <j≤An,x1

\

 {rk = 0} .

k>An,x2 ,k6=j

By Lemma 4.2, it follows for each An,x2 < j ≤ An,x1   \ Y Qqn {rj = 1} {rk = 0} = qnj (1 − qnk ). k>An,x2 ,k6=j

k>An,x2

(4.33)

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A simple calculus shows dAn,x2 e 1

X

qnj = qn

dAn,x1 e−dAn,x2 e

− qn

1 − qn

An,x2 <j≤An,x1

→ e−x2 − e−x1 ,

n → ∞.

Also, according to the proof of Theorem 4.9,  \  −x lim Qqn {rk = 0} = e−e , n→∞

Therefore it follows from (4.33) √  c n ≤ xi , lim Qqn √ λi − log n→∞ c n = e−e

x ∈ R.

k>An,x

−x2

+ (e−x2 − e−x1 )e−e

−x1

i = 1, 2



.

This is the integral of p0 (x1 )p(x1 , x2 ) over the region {(x1 , x2 ) : x1 > x2 }. The proof is complete.  Theorem 4.11. Under (P, Qqn ) Xn ⇒ G in terms of finite dimensional distributions. Here G is a centered Gaussian process with the covariance structure Z ∞ e−cu Cov(G(s), G(t)) = du, s < t. (1 − e−cu )2 t Proof. First, we can prove in a way completely similar to that of Theorem 4.7 that for each t > 0 d

Xn (t) −→ G(t). Next we turn to the 2-dimensional case. Assume 0 < s < t < ∞. Then for any x1 and x2 ,  X √ x1  rk − n(Ψ(s) − Ψ(t)) x1 Xn (s) + x2 Xn (t) = 1/4 n √ √ ns≤k< nt   √ x1 + x2 X + 1/4 (4.34) rk − nΨ(t) . n √ k≥ nt

Since two summands in the right hand side of (4.34) are independent and each converges weakly to a normal random variable, then x1 Xn (s)+x2 Xn (t)

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2 must converges weakly to a normal random variable with variance σs,t given by Z t Z ∞ e−cu e−cu 2 x21 du + (x + x ) du 1 2 −cu )2 (1 − e−cu )2 s (1 − e t Z ∞ Z ∞ e−cu e−cu 2 du + x du = x21 2 (1 − e−cu )2 (1 − e−cu )2 t s Z ∞ e−cu +2x1 x2 du. (1 − e−cu )2 t Therefore
d
Xn (s), Xn (t) −→ G(s), G(t) , where (G(s), G(t)) is jointly normally distributed with covariance Z ∞ e−cu du. Cov(G(s), G(t)) = (1 − e−cu )2 t The m-dimensional case can be analogously proved. 

To conclude this section, we investigate the asymptotic behaviours when tn tends to ∞. Theorem 4.12. Assume that tn is a sequence of positive numbers such that 1 tn → ∞, tn − log n → −∞. (4.35) 2c Then under (P, Qqn ),
d √ √ ectn /2 ϕλ ( ntn ) − nΨ(tn ) −→ N (0, 1). (4.36) 1/4 n √ Proof. First, compute mean and variance of ϕλ ( ntn ). According to the definition of ϕλ , X X √ qnk Eqn ϕλ ( ntn ) = Eqn rk = 1 − qnk √ √ k≥ ntn k≥ ntn Z ∞
√ e−cu du 1 + O(n−1/2 ) = n −cu 1 − e tn
√ = nΨ(tn ) 1 + O(n−1/2 ) and X X √ qnk V arqn ϕλ ( ntn ) = V arqn (rk ) = (1 − qnk )2 √ √ k≥ ntn k≥ ntn Z ∞
√ e−cu = n du 1 + O(n−1/2 ) −cu 2 ) tn (1 − e
√ −ctn −1/2 1 + O(n ) . = ne

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√ The condition (4.35) guarantees ne−ctn → ∞. Next, we verify the Lindeberg condition. For any ε > 0, X X X X j 2 qnjk (1 − qnk ) ≤ j 2 qnjk √ k≥ ntn j≥εσn

√ k≥ ntn j≥εσn

=

=

X

j2

X

qnjk √ j≥εσn k≥ ntn √ jd ntn e X q n . j2 1 − qnj j≥εσn

It is now easy to see √

jd ntn e 1 X 2 qn j →0 σn2 1 − qnj j≥εσn

under the condition (4.35). So the is satisfied, and we conclude (4.36).  4.3

Small ensembles

This section is devoted to the proofs of main results given in the Introduction. A basic strategy is to use conditioning argument on the event {|λ| = n}. The following lemma due to Vershik (1996) characterizes the relations between grand ensembles and small ensembles. Lemma 4.3. For any 0 < q < 1 and n ≥ 0, we have (i) Pu,n is the conditional probability measure induced on Pn by Qq , i.e., Qq |Pn = Pu,n ; (ii) Qq is a convex combination of measures Pu,n , i.e., ∞ 1 X Qq = p(n)q n Pu,n . F(q) n=0

Let Wn (λ) be a function of λ taking values in Rbn where bn < ∞ or bn = ∞. Wn can be regarded as a random variable in (P, Qqn ). When restricted to Pn , Wn is also a random variable in (Pn , Pu,n ). Denote by Qqn ◦ Wn−1 and Pu,n ◦ Wn−1 the induced measures by Wn respectively. The total variation distance is defined by
dT V Qqn ◦ Wn−1 , Pu,n ◦ Wn−1 = sup Qqn ◦ Wn−1 (B) − Pu,n ◦ Wn−1 (B) . B⊂Rbn

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Lemma 4.4. If there exists a sequence of subsets Bn ⊆ Rbn such that (i) Qqn ◦ Wn−1 (Bn ) → 1,

(4.37)

Q (|λ| = n|W = w ) q n n sup n − 1 → 0, Qqn (|λ| = n) wn ∈Bn

(4.38)

(ii)

then it follows
dT V Qqn ◦ Wn−1 , Pu,n ◦ Wn−1 → 0, Proof.

n → ∞.

(4.39)

Observe for any B ⊆ Rbn

Qqn ◦ Wn−1 (B) = Qqn ◦ Wn−1 (B ∩ Bn ) + Qqn ◦ Wn−1 (B ∩ Bnc ) and Pu,n ◦ Wn−1 (B) = Pu,n ◦ Wn−1 (B ∩ Bn ) + Pu,n ◦ Wn−1 (B ∩ Bnc ). Since Qqn ◦ Wn−1 (Bnc ) → 0 by (4.37), then we need only estimate
Qqn ◦ Wn−1 (B ∩ Bn ) − Qqn ◦ Wn−1 B ∩ Bn |λ| = n X
≤ Qqn (Wn = wn ) − Qqn Wn = wn |λ| = n wn ∈Bn

=

X wn ∈Bn

Q |λ| = n W = w
n n q Qqn (Wn = wn ) n − 1 . Qqn (|λ| = n)

(4.40)

It follows from (4.38) that the right hand side of (4.40) tends to 0. Thus we conclude the desired assertion (4.39).  Lemma 4.5. Assume that Kn is a sequence of positive integers such that X k 2 qnk = o(n3/2 ). (4.41) (1 − qnk )2 k∈Kn

Then for Wn : λ → (rk (λ), k ∈ Kn )
dT V Qqn ◦ Wn−1 , Pn ◦ Wn−1 −→ 0. Proof. We will construct Bn such that (i) and (ii) of Lemma 4.4 holds. First, observe that there is an an such that X k 2 qnk = o(a2n ), an = o(n3/4 ). (4.42) (1 − qnk )2 k∈Kn

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Define X n o X kxk − kEqn rk ≤ an . Bn = (xk , k ∈ Kn ) : k∈Kn

k∈Kn

Then by (4.42)  X  X
Qqn ◦ Wn−1 Bnc = Qqn krk − kEqn rk > an k∈Kn

≤ =

P

V arqn 1 X a2n

k∈Kn

k∈Kn




k∈Kn krk 2 an k 2 qnk → (1 − qnk )2

0.

It remains to show that
Qqn |λ| = n Wn = wn →1 Qqn (|λ| = n)

(4.43)

uniformly in wn ∈ Bn . Fix wn = (xk , k ∈ Kn ). Then by independence of the rk ’s   Qqn |λ| = n Wn = wn = Qqn

∞ X

 krk = n rk = xk , k ∈ Kn

k=1

= Qqn

 X

= Qqn

 X

krk = n −

k∈K / n

X

 kxk rk = xk , k ∈ Kn

k∈Kn

krk = n −

k∈K / n

X

 kxk .

(4.44)

k∈Kn

It follows by (4.21) and (4.42) ∞  X  X X V arqn krk = k 2 V arqn (rk ) − k 2 V arqn (rk ) k∈K / n

k=1

=

σn2 (1

k∈Kn

+ o(1)) → ∞.

Hence as in Theorem 4.7 one can prove that under (P, Qqn ) P / n k(rk − Eqn rk ) qk∈K −→ N (0, 1) P V arqn ( k∈K / n krk ) and so 1 X k(rk − Eqn rk ) −→ N (0, 1). σn k∈K / n

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Note that ∞  X X 1  n− Eqn rk − k(xk − Eqn rk ) → 0 σn k=1

k∈Kn

uniformly in (xk , k ∈ Kn ) ∈ Bn . Then using the inverse formula as in Theorem 4.7, we have  X  X Qqn krk = n − kxk k∈K / n

= Qqn

 X

k∈Kn

k(rk − Eqn rk ) = n −

k∈K / n

=

1 (96)1/4 n3/4

∞ X

Eqn rk −

k=1

X

 k(xk − Eqn rk )

k∈Kn


1 + o(1) .

(4.45)

Combining (4.44), (4.45) and (4.22) yields (4.43), as desired.



Now we are ready to prove Theorems 4.3, 4.4 and 4.6. √ √ Proof of Theorems 4.3. As in Theorem 4.9, let An,x = n(log n/c + x)/c. Define Kn = {k : k ≥ An,x }, then it is easy to see X k 2 qnk = o(n3/2 ) (1 − qnk )2 k∈Kn √ since An,x / n → ∞. Hence applying Lemma 4.5 to Wn : λ 7→ (rk (λ), k ∈ Kn ) yields Qqn (λ ∈ P : rk = 0, k ∈ Kn ) − Pu,n (λ ∈ Pn : rk = 0, k ∈ Kn ) → 0. According to Theorem 4.9, we in turn have for any x ∈ R √   c n Pu,n λ ∈ Pn : √ λ1 − log ≤ x = Pu,n (λ ∈ Pn : rk = 0, k ∈ Kn ) n c −→ e−e

−x

.

The proof is complete.  Proof of Theorem 4.4. Similar to the proof of Theorem 4.3.  √ Proof of Theorem 4.6 Define Kn = {k : k ≥ ntn }, then it is easy to see X k 2 qnk = o(n3/2 ) (1 − qnk )2 k∈Kn

since tn → ∞. Hence applying Lemma 4.5 to Wn : λ 7→ (rk (λ), k ∈ Kn ) yields Qqn (λ ∈ P : Wn ∈ B) − Pu,n (λ ∈ Pn : Wn ∈ B) → 0.

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where n  o √ ectn /2  X B = (xk , k ∈ Kn ) : 1/4 xk − nΨ(tn ) ≤ x . n k∈K n

We now obtain the desired (4.18) according to Theorem 4.12.  However, for any fixed t ≥ 0, the condition (4.41) is not satisfied by √ Kn = {k : k ≥ nt}. Thus one cannot directly derive Theorem 4.2 nor Theorem 4.5 from Lemma 4.5. The rest of this section is devoted to proving Theorems 4.2 and 4.5 following Pittel (1997), and the focus is upon the latter since the other can be proved in a similar and simpler way. For simplicity of notations, we only consider two dimensional case below. Assume 0 < t1 < t2 , we shall prove that for any x1 , x2 ∈ R
d (4.46) x1 Xn (t1 ) + x2 Xn (t2 ) −→ N 0, σx21 ,x2 , where σx21 ,x2 = (x1 , x2 )Σt1 ,t2



x1 x2

 .

Here Σt1 ,t2 is a covariance matrix of X given by (4.16). Indeed, it suffices to prove (4.46) holds for x2 X x1 X (rk − Eqn rk ) + 1/4 (rk − Eqn rk ). ξn (x, t) := 1/4 n n √ √ k≥ nt1

k≥ nt2

This will in turn be done by proving  σ2  x1 ,x2 Eu,n eξn (x,t) → exp , n → ∞. (4.47) 2 In doing this, a main ingredient is to show the following proposition. We need additional notations. Define for any 1 ≤ k1 < k2 < ∞ m(k1 , k2 ) =

kX 2 −1 k=k1

m(k2 ) =

∞ X k=k2

qnk , 1 − qnk

σ 2 (k1 , k2 ) =

qnk , 1 − qnk

σ 2 (k2 ) =

kX 2 −1 k=k1

∞ X k=k2

qnk , (1 − qnk )2

qnk (1 − qnk )2

and s(k1 , k2 ) =

kX 2 −1 k=k1

kqnk , (1 − qnk )2

s(k2 ) =

∞ X k=k2

kqnk . (1 − qnk )2

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Proposition 4.1. For u1 , u2 ∈ R  u  X u2 X 1 Eu,n exp 1/4 (rk − Eqn rk ) + 1/4 (rk − Eqn rk ) n n k ≤k
2

2

 u2
2  u22 2 c 1 σ 2 (k1 , k2 ) + 1/2 σ (k2 ) − 2 u1 s(k1 , k2 ) + u2 s(k2 ) = exp 1/2 4n 2n 2n ·(1 + o(1)). (4.48) The proof of Proposition 4.1 will consist of several lemmas. Note that for any 0 < q < 1 and z1 , z2 Y Y r Y Y Eq z1rk z2k = Eq z1rk Eq z2rk k1 ≤k
k1 ≤k
k≥k2

Y

=

k1 ≤k
k≥k2

1 − qk Y 1 − qk . 1 − z1 q k 1 − z2 q k k≥k2

On the other hand, it follows by Lemma 4.3 Eq

Y k1 ≤k
Y

z1rk

z2rk

k≥k2

∞ 1 X = p(n)q n Eu,n F(q) n=0

Y k1 ≤k
z1rk

Y

z2rk .

k≥k2

Thus we have for each 0 < q < 1 ∞ X

Y

p(n)q n Eu,n

z1rk

k1 ≤k
n=0

Y

= F(q)

k1 ≤k
1 − z1 q 1 − qk

Y

z2rk

k≥k2 k

Y k≥k2

1 − qk 1 − z2 q k

=: F (z; q),

(4.49)

where z = (z1 , z2 ). Note that the above equation (4.49) is still valid for all complex number q with |q| < 1. Hence using the Cauchy integral formula yields Z π Y Y r
1 rk k Eu,n z1 z2 = r−n e−inθ F z; reiθ dθ, (4.50) 2πp(n) −π k1 ≤k
k≥k2

where r is a free parameter. Lemma 4.6. 
F z, reiθ ≤ c5 F (z, r) exp

 2r(1 + r)(cos θ − 1) . 3 (1 − r) + 2r(1 − r)(1 − cos θ)

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Proof.

Observe an elementary inequality 1 1 exp(Req − |q|), ≤ 1−q 1 − |q|

175

|q| < 1.

We have for 0 < r < 1 ∞ X  F (z; reiθ ) ≤ F (z; r) exp rk (cos kθ − 1) k=1



= F (z; r) exp Re

1 1  − . 1 − reiθ 1−r

Also, it is easy to see Re

1 2r(1 + r)(cos θ − 1) 1 − = . iθ 1 − re 1−r (1 − r)3 + 2r(1 − r)(1 − cos θ)

The proof is complete.



We shall asymptotically estimate r−n F (z; r) below. Define for t and z (1)

s

(t, z) =

kX 2 −1 k=k1

ke−tk (1 − ze−tk )(1 − e−tk )

and s(2) (t, z) =

∞ X k=k2

ke−tk (1 −

ze−tk )(1

− e−tk )

.

Lemma 4.7. Let r = e−τ where 2   1 X (zb − 1)s(b) (τ ∗ , zb ) , τ = τ∗ 1 + 2n b=1

Then we have for z1 = eu1 / r

−n

√ n

√

and z2 = eu2 /

c τ∗ = √ . n

n

√  e2c n  −1/4 F (z; r) = 1 + O(n ) (24n)1/4  u  u21 2 1 · exp 1/4 m(k1 , k2 ) + 1/2 σ (k1 , k2 ) n 2n

  u u22 2 2 m(k2 ) + 1/2 σ (k2 ) 1/4 2n n c
2  · exp − 2 u1 s(k1 , k2 ) + u2 s(k2 ) . 4n · exp

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Proof.

Let

H(z; t) = nt +

kX ∞ 2 −1 X c2 1 − e−tk 1 − e−tk + . + log log t 1 − z1 e−tk 1 − z2 e−tk k=k2

k=k1

A simple calculus shows 2

Ht (z; t) = n −

c2 X − (zb − 1)s(b) (t, zb ) t2 b=1

and 2

Htt (z; t) =

2c2 X (b) − (zb − 1)st (t, zb ). t3 b=1

Making the Taylor expansion at τ , we have
1 H(z; τ ∗ ) = H(z; τ ) + Ht (z; τ )(τ ∗ − τ ) + Htt z; t˜ (τ ∗ − τ )2 2 ∗ ˜ for a t between τ and τ . Hence it follows
1 H(z; τ ) = H(z; τ ∗ ) − Ht (z; τ )(τ ∗ − τ ) − Htt z; t˜ (τ ∗ − τ )2 . (4.51) 2 We shall estimate each summand in the right hand side of (4.51) below. Begin with H(z; τ ∗ ). As in Theorem 4.7, we have by the Taylor expansions kX 2 −1 k=k1

k2 −1 kX 2 −1 1 − qnk (z1 − 1)2 X qn2k qnk log = (z − 1) + 1 1 − z1 qnk 1 − qnk 2 (1 − qnk )2 k=k1

k=k1



+O |z1 − 1|3

kX 2 −1

qn3k

k=k1

(1 − qnk )3



k2 −1 k2 −1
u1 X u1 X qnk qnk + + O n−1/4 k k 2 1/4 1/2 1 − q (1 − q ) n 2n n n k=k1 k=k1
u1 u1 2 = 1/4 m(k1 , k2 ) + 1/2 σ (k1 , k2 ) + O n−1/4 . n 2n

=

Similarly, ∞ X k=k2

log


u2 u2 1 − qnk = 1/4 m(k2 ) + 1/2 σ 2 (k2 ) + O n−1/4 . k 1 − z2 qn n 2n

This immediately gives √ u1 u1 H(z; τ ∗ ) = 2c n + 1/4 m(k1 , k2 ) + 1/2 σ 2 (k1 , k2 ) n 2n
u2 u2 + 1/4 m(k2 ) + 1/2 σ 2 (k2 ) + O n−1/4 . n 2n

(4.52)

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Turn to the second term Ht (z; τ ). Note for b = 1, 2
ub zb − 1 = 1/4 + O n−1/2 n and s(b) (τ ∗ , zb ) = O(n),

s(b) (τ, zb ) = O(n).

Then we have 2

−2 1 X (zb − 1)s(b) (τ ∗ , zb ) Ht (z, τ ) = n − n 1 + 2n 

b=1

−

2 X

(zb − 1)s(b) (τ, zb )

b=1

=

2 X



(zb − 1) s(b) (τ ∗ , zb ) − s(b) (τ, zb ) + O n1/2

b=1


= O n1/2 .

(4.53)

To evaluate the third term, note for any t˜ between τ and τ ∗
2 Htt (z; τ ) = n3/2 + O n5/4 c and 2 τ∗ X (zb − 1)s(b) (τ ∗ , zb ) τ∗ − τ = 2n b=1

=

2
c X ub s(b) + O n−1 . 7/4 2n b=1

We have 2 2

c X ub s(b) + O n−1/2 . Htt z; t˜ (τ ∗ − τ )2 = 2 2n b=1

Inserting (4.52)-(4.54) into (4.51) yields √ u1 u1 H(z; τ ) = 2c n + 1/4 m(k1 , k2 ) + 1/2 σ 2 (k1 , k2 ) n 2n u2 u2 + 1/4 m(k2 ) + 1/2 σ 2 (k2 ) n 2n
2
c (1) − 2 u1 s + u2 s(2) + O n−1/4 . 4n
Finally, with the help of τ − τ ∗ = O n−3/4 , we have
1 τ 1 log = log + O n−1/4 , 1/4 2 2π (24n)

(4.54)

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and so

√


ec n F(r) = F(e ) = 1 + O(n−1/4 ) . 1/4 (24n) To conclude the proof, we need only to note  c2  r−n F (z; r) = F(r) exp − + H(z; τ ) . τ −τ



To estimate the integral over (−π, π), we split the interval (−π, π) into two subsets: |θ| ≤ θn and |θ| ≥ θn , where θn = n−3/4 log n. The following lemma shows that the overall contribution to the value of integral in (4.50) made by large θ’s is negligible. Lemma 4.8. Assume r = e−τ is as in Lemma 4.7. Then Z
−n F (z; reiθ ) dθ ≤ r−n F (z; r) exp − c7 log2 n . r

(4.55)

|θ|≥θn

Proof.

If r = e−τ , then for all n ≥ 1

2r(1 + r)(cos θ − 1) c6 θ 2 . ≥ − (1 − r)3 + 2r(1 − r)(1 − cos θ) n−3/2 + θ2 n−1/2 By Lemma 4.6, Z Z   c6 θ 2 −n iθ −n F (z; re ) dθ ≤ r F (z; r) r exp − −3/2 dθ. n + θ2 n−1/2 |θ|≥θn |θ|≥θn To estimate the integral, we consider two cases separately: θn ≤ |θ| ≤ n−1/2 and |θ| > n−1/2 . Z Z   1/2 c6 θ2 dθ ≤ e−c6 n /2 dθ exp − −3/2 2 −1/2 n +θ n θn ≤|θ|≤n−1/2 θn ≤|θ|≤n−1/2 = exp(−c6 log2 n/3) and Z |θ|≥n−1/2

 exp −

 c6 θ2 dθ ≤ n−3/2 + θ2 n−1/2

Z

3/2 2

e−c6 n

θ /2

dθ

|θ|≥n−1/2

= exp(−c6 log2 n/3). In combination, (4.55) holds for a new constant c7 > 0.



Turn now to the major contribution to the integral value of small θ’s. Lemma 4.9. Assume r = e−τ is as in Lemma 4.7. Then Z
π r−n e−inθ F (z; reiθ )dθ = r−n F (z; r) 1/4 3/4 1 + o(1) . 6 n |θ|≤θn

(4.56)

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First, observe 
r−n e−inθ F z; reiθ = F(eτ −iθ ) exp −

179

Proof.

 c2 + H(z; τ − iθ) . (4.57) τ − iθ

Also, it follows from (4.5)  c2 
1 τ − iθ F eτ −iθ = exp + log + O(|τ − iθ|) . (4.58) τ − iθ 2 2π It is easy to see sup{|Htt (z; t)| : |t − τ | ≤ θn } = O(n2 ). Then the Taylor expansion at τ gives 1 H(z; τ − iθ) = H(z; τ ) − iHt (z; τ )θ − Htt (z; τ )θ2 + O(n2 θn3 ). (4.59) 2 Hence combining (4.57)-(4.59) together implies
r−n e−inθ F z; reiθ  
θ2 = r−n F (z; r) exp − iθHt (z, t) − Htt (z, t) 1 + o(1) , 2 whichZ in turn gives
r−n e−inθ F z; reiθ dθ |θ|≤θn

=r

−n

 
θ2 exp − iθHt (z, t) − Htt (z, t) dθ 1 + o(1) . 2 |θ|≤θn

Z F (z, r)

Note for a c8 > 0 Ht2 (z; τ ) = O(n−1/2 ), θn2 Htt (z, t) ≥ c8 log2 n. Htt (z; t) Hence itZ follows   θ2 exp − iθHt (z; τ ) − Htt (z; τ ) dθ 2 |θ|≤θn  Z ∞ Z   θ2 − = exp − iθHt (z; τ ) − Htt (z; τ ) dθ 2 −∞ |θ|>θn s  
2 2π H 2 (z; τ ) −1/2 = exp − t + O Htt e−θn Htt (z;τ )/2 Htt (z; τ ) 2Htt (z; τ )
π = 1/4 3/4 1 + o(1) , 6 n where in the second equation we used a standard normal integral formula. Likewise, it follows Z  
θ2 π exp − iθHt (z; τ ) − Htt (z; τ ) dθ = 1/4 3/4 1 + o(1) . 2 6 n |θ|≤θn HenceZ
π r−n e−inθ F (z, reiθ )dθ = r−n F (z, r) 1/4 3/4 1 + o(1) . 6 n  |θ|≤θn

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1/4

1/4

Proof of Proposition 4.1. Let z1 = eu1 /n and z2 = eu2 /n , and choose r = e−τ in (4.50). Combining (4.55) and (4.56) yields Z π Z Z 

r−n e−inθ F z; reiθ dθ = + r−n e−inθ F z; reiθ dθ −π

|θ|≤θn

=r

−n

F (z, r)

|θ|≥θn

π 61/4 n3/4


1 + o(1) .

Taking Theorem 4.1 and Lemma 4.7 into account, we conclude (4.48) as desired.  √ Proof of Theorem 4.5. Take u1 = x1 and u2 = x1 + x2 and k1 = n t1 √ and k2 = n t2 . Note Z t2 √ e−cu σ 2 (k1 , k2 ) = n du(1 + o(1)), −cu )2 t1 (1 − e 2

σ (k2 ) =

√

∞

Z n

t2

Z

t2

s(k1 , k2 ) = n t1

Z

∞

s(k2 ) = n t2

e−cu du(1 + o(1)), (1 − e−cu )2 ue−cu du(1 + o(1)), (1 − e−cu )2

ue−cu du(1 + o(1)). (1 − e−cu )2

Substituting these into (4.48) of Proposition 4.1, we easily get (4.47), as required. We conclude the proof.  4.4

A functional central limit theorem

In this section we shall first prove a theorem that may allow us to get the distributional results in the case when the functional of λ depends primarily on the moderate-sized parts. Then we shall use the functional central limit theorem to prove the asymptotic normality for character ratios and the log-normality for dλ . Introduce an integer-valued function l √n 1 m kn (t) = log , t ∈ (0, 1). c 1−t Let √ c t0 (n) = √ , t1 (n) = n−δ0 , t2 (n) = 1 − e−c/ n 2 n

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where δ0 ∈ (0, 1/8). Define ( P Yn (t) =

t n1/4

k≥kn (t) (rk

− Eqn rk ),

0,

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t ∈ [t0 (n), 1), 0 ≤ t < t0 (n) or t = 1.

Let Y (t), 0 ≤ t ≤ 1 be a separable centered Gaussian process with the covariance function given by  1 1 EY (s)Y (t) = (4.60) s(1 − t) − l(s)l(t) , 0 < s ≤ t < 1 c 2 where  1 t log t − (1 − t) log(1 − t) . l(t) = c The so-called functional central limit theorem reads as follows. Theorem 4.13. (i) With probability 1, Y (t) is uniformly continuous on [0, 1]. (ii) Yn converges to Y in terms of finite dimensional distributions. (iii) Let g(t, x) be continuous for (t, x) ∈ D := (0, 1) × R and such that |g(t, x)| ≤ c10

|x|γ − t)β

tα (1

(4.61)

for some γ > 0, α < 1 + γ/2, β < 1 + γ/6, uniformly for (t, x) ∈ D. Then under (Pn , Pu,n ) Z 1 Z 1 d g(t, Y (t))dt. g(t, Yn (t))dt −→ 0

0

We shall prove the theorem following the line of Pittel (2002) by applying the Gikhman-Skorohod theorem, namely Theorem 1.16. A main step is to verify the stochastic equicontinuity for Yn (t), 0 ≤ t ≤ 1. As the reader may notice, a significant difference between Yn and Xn is that there is an extra factor t in Yn besides parametrization. This factor is added to guarantee that Yn satisfies the stochastic equicontinuity property and so the limit process Y has continuous sample paths. Lemma 4.10. For u ∈ R  u Eqn exp 1/4 n  u2 = exp 2n1/2

X

 (rk − Eqn rk )

k1 ≤k
X k1 ≤k
 qnk 3/4−2δ + O(n ) , (1 − qnk )2

(4.62)

where the error term holds uniformly over nδ ≤ k1 < k2 ≤ ∞ with δ < 1/2.

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Proof. The proof is completely similar to that of Theorem 4.7. It follows by independence  u    u X Y Eqn exp 1/4 rk = Eqn exp 1/4 rk n n k ≤k
2

1

2

Y

=

k1 ≤k
1 − qnk 1 − eu/n1/4 qnk 1/4



X

= exp −

k1 ≤k
1 − eu/n qnk  log . 1 − qnk

Using the Taylor expansion, we obtain 1/4

X k1 ≤k

1/4 1 − eu/n qnk = eu/n − 1 log 1 − qnk

X k1 ≤k

2 1/4 1 + eu/n −1 2 +O |eu/n

1/4

qnk 1 − qnk

X k1 ≤k
− 1|3




qn2k (1 − qnk )2

X k1 ≤k
Note u/n1/4 3 e − 1

X k1 ≤k
qn3k . (1 − qnk )3

 |u|3 Z ∞  e−3cx qn3k = O dx k 3 −cx 2 √ 1/4 (1 − qn ) ) n k1 / n (1 − e
= O n3/4−2δ ,

and the contribution proportional to u3 that comes from the first two terms is of lesser order of magnitude. We conclude the proof.  Lemma 4.11. For u ∈ R  u Eu,n exp 1/4 n

X

 (rk − Eqn rk )

k1 ≤k
 u2 ≤ c11 exp 2n1/2

X k1 ≤k

 qnk 3/4−2δ + O n , (1 − qnk )2

(4.63)

where the error term holds uniformly over nδ ≤ k1 < k2 ≤ ∞ with δ > 3/8. Proof. This can be proved by a slight modification of Theorem 4.1. For any 0 < q < 1, ∞ P P 1 X n q p(n)Eu,n x k1 ≤k
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183

On the other hand, Eq x

P

k1 ≤k
rk

1 − qk . 1 − xq k

Y

=

k1 ≤k
Thus we have for any 0 < q < 1 ∞ X

q n p(n)Eu,n x

P

k1 ≤k
rk

Y

= F(q)

k1 ≤k
n=0

1 − qk . 1 − xq k

Indeed, the above equation holds for any complex number |q| < 1. Using the Cauchy integral formula, we obtain Z π P
1 k1 ≤k
Y k1 ≤k
Choose r = qn = e

√ −c/ n

Eu,n x

1 − qk . 1 − xq k

, with the help of (4.5) we further obtain

P

k1 ≤k
rk

Y

≤ c8

k1 ≤k
1 − qnk . 1 − xqnk

√ u/ n

Letting x = e  u Eu,n exp 1/4 n

k1

, we have by (4.62)   u X rk ≤ c8 exp 1/4 n ≤k
 u2 · exp 2n1/2

X

Eqn rk



k1 ≤k
X k1 ≤k

 qnk 3/4−2δ + O n , (1 − qnk )2

which immediately implies (4.63). The proof is complete.



Lemma 4.12. (i) For ε > 0, lim lim


sup Pu,n |Yn (t) − Yn (s)| > ε = 0.

δ→0 n→∞ |t−s|≤δ

(ii) For any m ≥ 1 and 0 < ρ < 1/12 − 2δ0 /3,
Eu,n |Yn (t)|m ≤ c9 tm/2 (1 − t)m/2 + n−mρ ,

t ≥ t0 (n).

(4.64)

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Proof. Observe that if t > t2 (n) then kn (t) > n so Yn (t) = 0; while if t < t0 (n) then Yn (t) = Op (n1/4−δ0 log n). So it suffices to prove (4.64) uniformly for t, s ∈ [t1 (n), t2 (n)]. Assume s < t. We use Lemma 4.11 to obtain  us  X Eu,n exp 1/4 (rk − Eqn rk ) n kn (s)≤k
 u2 s2 ≤ c9 exp 2n1/2 ≤ c10 exp

 u2 2c

X kn (s)≤k
 qnk k 2 (1 − qn )

 (t − s)

(4.65)

and Eu,n exp

 u(t − s) n1/4

≤ c11 exp ≤ c12 exp

k≥kn (t)

 u2 (t − s)2 2n1/2  u2 2c

 (rk − Eqn rk )

X

X k≥kn (t)

 qnk (1 − qnk )2

 (t − s) .

(4.66)

Note Yn (t) − Yn (s) =

t−s n1/4 −

X

(rk − Eqn rk )

k≥kn (t)

s

X

n1/4

(rk − Eqn rk ).

kn (s)≤k
It follows by the Cauchy-Schwarz inequality, (4.65) and (4.66)   Eu,n exp u(Yn (t) − Yn (s))   2us 1/2 X ≤ Eu,n exp − 1/4 (rk − Eqn rk ) n kn (s)≤k
  2u(t − s) · Eu,n exp n1/4 ≤ c13 exp

 2u2

1/2 (rk − Eqn rk )

X

k≥kn (t)

 (t − s) .

c A standard argument now yields

( Pu,n


|Yn (t) − Yn (s)| > ε ≤

2

e−cε /8(t−s) , ρ e−cn ε/8 ,

ε ≤ nρ (t − s), ε ≥ nρ (t − s).

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185

Therefore we have
2 ρ Pu,n |Yn (t) − Yn (s)| > ε ≤ e−cε /8δ + e−cn ε/8 uniformly for s, t ∈ [t1 (n), t2 (n)] with |t−s| ≤ δ. This verifies the stochastic equicontinuity property (4.64). We can analogously obtain ( 2
e−cx /8(t−s) , x ≤ nρ t(1 − t)/c, Pu,n |Yn (t)| > x ≤ ρ e−cn x/8 , x ≥ nρ t(1 − t)/c. Therefore it follows by integral formula by parts Z ∞
Eu,n |Yn (t)|m = m xm−1 Pu,n |Yn (t)| > x dx 0
≤ c15 tm/2 (1 − t)m/2 + n−mρ , as desired.



Proof of Theorem 4.13. Begin with the continuity of sample paths of Y (t). Note
2 E Y (t) − Y (s) = EY (t)2 − 2EY (s)Y (t) + EY (s)2  1 1 = t − s − (t − s)2 − (l(t) − l(s))2 c 2 1 ≤ (t − s). c Since Y (t) − Y (s) is Gaussian with zero mean, we have
4 3 E Y (t) − Y (s) ≤ 2 (t − s)2 . c This implies by Kolmogorov’s continuity criterion that there exists a separable continuous version of Y (·) on [0, 1]. Turn to the proof of (ii). The asymptotic normality directly follows from Theorem 4.5. Indeed, making
a change of time parameter, we see √ 1 1 ), 0 < t < 1 converges weakly to tX( 1c log 1−t ), that tXn
( cn log 1−t 0 < t < 1 in terms of finite dimensional distributions. If letting Y (t) = 1 tX( 1c log 1−t ), then a simple calculus shows that Y (t), 0 < t < 1 has the desired covariance structure (4.60). Finally, we show (iii). Fix g. Without loss of generality, we assume that α, β > 1. Introduce εm = 1/m, m ≥ 1, and break up the integration interval into three subsets: (0, εm ), (εm , 1 − εm ), and (1 − εm , 1). Let Z 1−εm Z 1 g(t, Yn (t))dt, Zn,m = Zn = g(t, Yn (t))dt 0

εm

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and 1

Z Z=

Z g(t, Y (t))dt,

1−εm

Zm =

0

g(t, Y (t))dt. εm

Then using Lemma 4.12 and Theorem 1.16, it is not difficult to check the following three statements: (a) for any ε > 0, lim lim Pn (|Zn − Zn,m | > ε) = 0;

m→∞ n→∞

(b) for each m ≥ 1, d

Zn,m −→ Zm ; (c) for any ε > 0, lim P (|Zm − Z| > ε) = 0,

m→∞

m → ∞.

Here we prefer to leave the detailed computation to the reader, see also Pittel (2002). Having (a), (b) and (c) above, Theorem 4.2, Chapter 1 of Billingsley d

(1999a) guarantee Zn −→ Z, which concludes the proof.  To illustrate, we shall give two examples. The first one treats the character ratios in the symmetric group Sn . Fix a transposition τ ∈ Sn . Define the character ratio by χλ (τ ) γτ (λ) = , λ ∈ Pn dλ where χλ be an irreducible representation associated with the partition λ 7→ n, dλ is the dimension of χλ , i.e., dλ = χλ (1n ). The ratio function played an important role in the well-known analysis of the card-shuffling problem performed by Diaconis and Shahshahani (1981). In fact, Diaconis and Shahshahani proved that the eigenvalues for this random walk are the character ratios each occurring with multiplicity d2λ . Character ratios also play a crucial role in the work on moduli spaces of curves, see Eskin and Okounkov (2001), Okounkov and Pandharipande (2004). The following theorem can be found in the end of the paper of Diaconis and Shahshahani (1981). Theorem 4.14. Under (Pn , Pu,n ),
d n3/4 γτ (λ) −→ N 0, στ2 , where στ2 is given by Z Z 4 1 1 EY (s)Y (t) 1−s 1−t στ2 = 4 log log dsdt c 0 0 s(1 − s)t(1 − t) s t where EY (s)Y (t) is given by (4.60).

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Proof.

187

Recall the following classic identity due to Frobenius (1903)  X 1 γτ (λ) = λ2k − (2k − 1)λk n(n − 1) k    0  1 X  λk λk = n
. − 2 2 2 k

d

It follows from the second equation that γτ (λ) = γτ (λ0 ) and Eu,n γτ (λ) = 0 since Pu,n (λ) = Pu,n (λ0 ). To prove the central limit theorem, we observe  X 0 λ02 − (2k − 1)λ k k = n + Un,1 + Un,2 + Un,3 , k

where Un,1 =

X


m(k)2 − 2km(k) ,

k

Un,2 =

X


2 ϕλ (k) − m(k) ,

k

Un,3 = 2

X



m(k) − k ϕλ (k) − m(k) .

k

It is easy to check that Un,1 = Op (n log2 n),

Un,2 = Op (n log2 n).

Turn to Un,3 . Switching to integration, we get Z ∞


Un,3 = 2 m(x) − x ϕλ (x) − m(x) dx + Op n log2 n . 0

√ Substituting x = n| log(1 − t)|/c and a simple calculus shows that Z ∞

m(x) − x ϕλ (x) − m(x) dx 0

=

n5/4 c2

Z 0

1


Yn (t) 1−t log dt + Op n3/4 log n . t(1 − t) t

Set g(t, x) =

x 1−t log dt. t(1 − t) t

Then g(t, x) obviously satisfies the condition (4.61) of Theorem 4.13 with parameters µ = 1 and α = β = 3/2. So we have Z 1 Z 1 Y (t) 1−t 1−t Yn (t) d log dt −→ log dt. t(1 − t) t t(1 − t) t 0 0

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Note that the limit variable is a centered Gaussian random variable with variance Z 1Z 1 EY (s)Y (t) 1−s 1−t log log dsdt, s t 0 0 s(1 − s)t(1 − t) where EY (s)Y (t) is given by (4.60). In combination, we obtain
d n3/4 γτ (λ) −→ N 0, στ2 , as desired.



The second example we shall treat is dλ . It turns out that the logarithm of dλ satisfies the central limit theorem. Introduce Z ∞ log | log x| κ(t) = dx. (4.67) (1 − t − tx)2 0 Theorem 4.15. Under (Pn , Pu,n ), 
1  1 d −→ N 0, σd2 . n log n + An log d − λ 3/4 2 n Here A and σd2 are given by Z 1 ∞ y log y A = 1 − log c + 2 dy c 0 ey − 1 and σd2 =

1 c2

Z

1

Z

1

EY (s)Y (t)κ(s)κ(t)dsdt, 0

(4.68)

0

where EY (s)Y (t) is given by (4.60). Numerically, σd2 = 0.3375 · · · . The theorem was first proved by Pittel (2002). The proof will use the following classic identities (see also Chapter 5): Q 1≤i<j≤l (λi − λj + j − i) Q dλ = n! (4.69) 1≤i≤l (λi − i + l)! and dλ =

n! n! , := Q Hλ ∈λ h

where h = λi − i + λ0j − j + 1, the hook length of the (i, j) square. It follows directly from (4.70) and (4.69) that Q 0 0 1≤i<j≤λ1 (λi − λj + j − i) dλ = dλ0 = n! Q . 0 1≤i≤λ1 (λi − i + λ1 )!

(4.70)

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Consequently, we obtain log dλ − log n! X X

= log λ0i − λ0j + j − i − log λ0i − i + λ1 ! 1≤i<j≤λ1

1≤i≤λ1

=: Mn − Nn .

(4.71)

The bulk of the argument consists of computing Mn and Nn . We need some basic estimates about λ0i − λ0j below. Let 0 < δ < 1/2 and define  K = (k1 , k2 ) : nδ ≤ k1 ≤ k2 − nδ . Lemma 4.13. Let

√ n 1 √ , log x>0 c 1 − ecx/ n and denote `(x, y) = `(x) − `(y) for any 0 < x ≤ y. Then we have (i)
m(k1 , k2 ) = 1 + O(n−δ ) `(k1 , k2 ) √ uniformly for (k1 , k2 ) ∈ K, and for xi = cki / n √  
n 1 1 σ 2 (k1 , k2 ) = 1 + O(n−δ ) − ; c 1 − e−x1 1 − e−x2 (ii)
σ(k1 , k2 ) = O n−(δ−a)/2 m(k1 , k2 ) √ uniformly for (k1 , k2 ) ∈ K and k1 ≤ a n log n/c where a < δ; (iii) σ 2 (k1 , k2 ) lim =1 n→∞ m(k1 , k2 ) √ uniformly for (k1 , k2 ) ∈ K and k1 ≥ a n log n/c where a < δ. `(x) =

Proof.

See Lemmas 1 and 2 of Pittel (2002).



Lemma 4.14. Given ε > 0 and 0 < a < δ < 1/2, there is an n0 (a, δ, ε) ≥ 1 such that (i) p
Qqn |λ0k1 − λ0k2 − m(k1 , k2 )| > σ(k1 , k2 ) ε log n ≤ n−ε/3 √ uniformly for (k1 , k2 ) ∈ K and k1 ≤ a n log n/c; (ii) p
Qqn |λ0k1 − λ0k2 − m(k1 , k2 )| > (σ(k1 , k2 ) + log n) ε log n ≤ n−ε/3 √ uniformly for (k1 , k2 ) ∈ K and k1 ≥ a n log n/c.

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Proof. Let η be small enough to ensure that eη qnk < 1 for all k ∈ [k1 , k2 ). √ For instance, select |η| = σ(k1 , k2 ) ε log n. We have     η2 Eqn exp η(λ0k1 − λ0k2 ) = exp ηm(k1 , k2 ) + σ 2 (k1 , k2 ) 2    X qn3k 3 . (4.72) · exp O |η| 3k (1 − qn ) k1 ≤k
Moreover, a delicate analysis shows the remainder term is indeed of order n−δ/2 log3/2 n = o(1). Using (4.72) and Markov’s inequality, we easily get p
Qqn |λ0k1 − λ0k2 − m(k1 , k2 )| ≥ σ(k1 , k2 ) ε log n  ε
 ≤ 2 exp − log n + O n−δ/2 log3/2 n 2 ≤ n−ε/3 . This concludes the proof of (i). Turn to (ii). Set √ ε log n |η| = . σ(k1 , k2 ) + log n Then |η| → 0. Using only the first order expansion, we obtain     Eqn exp η(λ0k1 − λ0k2 ) = exp (eη − 1)m(k1 , k2 )    X qn2k . · exp O η 2 2k (1 − qn )

(4.73)

k1 ≤k
Note X

η2

k1 ≤k
qn2k = O(n−a log n). (1 − qn )2k

Using (4.73) and the Markov inequality, and noting σ 2 (k1 , k2 )/m(k1 , k2 ) = 1 + o(1), we have p
Qqn |λ0k1 − λ0k2 − m(k1 , k2 )| ≥ (σ(k1 , k2 ) + log n) ε log n ≤ n−ε/3 . The proof is complete.



Lemma 4.14 can be used to obtain the following concentration-type bound for λ0i − λ0j under (Pn , Pu,n ).
Proposition 4.2. With Pu,n -probability 1 − O n−1/4 at least, 0
λk − λ0k − m(k1 , k2 ) ≤ 3 σ(k1 , k2 ) + log n (log n)1/2 (4.74) 1 2 uniformly for (k1 , k2 ) ∈ K. If, in addition, k1 ≤ an1/2 log n/c where a < δ, then the summand log n can be dropped.

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Proof. Qqn

191

Take ε = 9 in Lemma 4.14 to conclude 0
λk − λ0k − m(k1 , k2 ) ≥ 3(σ(k1 , k2 ) + log n)(log n)1/2 ≤ n−3 . 1 2

On the other hand, note by Lemma 4.2
Pu,n (B) = Qqn B |λ| = n Qqn (B) ≤ . Qqn (|λ| = n) Then according to (4.22)
Pu,n λ0k1 − λ0k2 − m(k1 , k2 ) ≥ 3(σ(k1 , k2 ) + log n)(log n)1/2 ≤ n−9/4 for each (k1 , k2 ) ∈ K and 1 ≤ k1 , k2 ≤ n. This immediately implies (4.74), as asserted.  Having these basic estimates, we are now ready to compute Mn and Nn . Let us start with Nn . Define µk = d`(k)e, X

Nn =


log µk − k + λ1 !,

1≤k≤λ1

Rn =

X



λ0k − m(k) log m(k) − k + λ1 ,

ln ≤k
where ln = [n ] and kn =

 √ n log n/c .

Lemma 4.15. Under (Pn , Pu,n ), we have for 1/8 < δ < 1/4 Nn = Nn + Rn + op (n3/4 ). Proof.

(4.75)

Let φ(x) = x log x − x. Then by the Stirling formula for factorial, Nn =

λ1 X


φ λ0k − k + λ1 + ∆(λ1 ),

(4.76)

k=1


where ∆(λ1 ) = Op n1/2 log2 n . Define ¯n = N

λ1 X


φ m(k) − k + λ1 .

k=1

¯n below. For this, we break up the sum in We shall compare Nn with N (1) (2) (3) (4.76) into Nn , Nn and Nn for k ∈ [1, ln ), k ∈ [ln , kn ] and k ∈ (kn , ∞) ¯n(1) , N ¯n(2) and N ¯n(3) . respectively. We similarly define N

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Observe that uniformly for 1 ≤ k ≤ λ1

log λ0k − k + λ1 = Op (log n), log m(k) − k + λ1 = Op (log n), and so

φ λ0k − k + λ1 = Op n1/2 log2 n ,



φ m(k) − k + λ1 = Op n1/2 log2 n .

Then we have
Nn(1) = Op n1/2+δ log2 n ,


¯n(1) = Op n1/2+δ log2 n , N

which implies

¯n(1) = Op n1/2+δ log2 n = op n3/4 . Nn(1) − N

(4.77)

(2) ¯n(2) . With the help of (4.74), we expand φ(x) at x = Turn to Nn − N m(k) − k + λ1 :



φ λ0k − k + λ1 = φ m(k) − k + λ1 + λ0k − m(k) log m(k) − k + λ1  σ 2 (k) log n  +Op . m(k) − k + λ1

It follows from Lemma 4.13 that the remainder term is controlled by Op (log n). Hence Nn(2)

¯n(2) = −N

kn X



λ0k − m(k) log m(k) − k + λ1

k=ln


+Op n1/2 log2 n .

(4.78)

As for the third term, we analogously use the Taylor expansion to obtain X
¯ (3) = Nn(3) − N λ0k − m(k) log x∗k , n k>kn

x∗k

where is between m(k) − k + λ1 and λ0k − k + λ1 . It follows from (4.74)
¯n(3) = op n3/4 . Nn(3) − N

(4.79)

Putting (4.77)-(4.79) together yields
¯n = Rn + op n3/4 . Nn − N To conclude the proof, we observe
¯n = Nn + Op n1/2 log n . N Now the assertion (4.75) is valid.



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We shall next turn to compute the Mn . Let g(y) = log(ey − 1), ∞

Z

e−y log |g(y) − g(x)|dy,

v(x) = −

x>0

0

Mn =

X


log µ(i, j) + j − i ,

1≤i<j≤λ1

Sn =

X



λ0k − m(k) v(yk ) + log(g(yλ1 ) − g(yk )) ,

2ln ≤k≤λ1 −ln

Pj−1 √ where yk = ck/ n and µ(i, j) = l=i µl . Lemma 4.16. Under (Pn , Pu,n ), we have
Mn = Mn + Sn + op n3/4 .

(4.80)

Proof. Denote K = {(k1 , k2 ) : ln ≤ k1 ≤ k2 − ln } ⊆ [1, λ1 ] × [1, λ1 ]. Obviously, it follows X

log λ0k1 − λ0k2 + k2 − k1 + Op n1/2+δ log2 n . (4.81) Mn = (k1 ,k2 )∈K

By (4.74), with high probability |λ0k1 − λ0k2 − m(k1 , k2 )| σ(k1 , k2 ) ≤3 m(k1 , k2 ) + k2 − k1 m(k1 , k2 ) + k2 − k1 = o(1) for all (k1 , k2 ) ∈ K. So uniformly for (k1 , k2 ) ∈ K

log λ0k1 − λ0k2 + k2 − k1 = log m(k1 , k2 ) + k2 − k1 λ0k1 − λ0k2 + m(k1 , k2 ) m(k1 , k2 ) + k2 − k1  σ 2 (k , k ) log2 n + log4 n  1 2 . (4.82) +Op (m(k1 , k2 ) + k2 − k1 )2 +

Take a closer look at each term in the right hand side of (4.82). First, observe X σ 2 (k1 , k2 ) log2 n + log4 n
= op n3/4 . (4.83) 2 (m(k1 , k2 ) + k2 − k1 ) (k1 ,k2 )∈K

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Second, a simple algebra shows X λ0k − λ0k + m(k1 , k2 ) 1 2 = Sn(1) + Sn(2) , m(k1 , k2 ) + k2 − k1

(4.84)

(k1 ,k2 )∈K

where Sn(1) =

2ln X

λ1
X λ0k − m(k)

k=ln

j=k+ln

1 m(j, k) + j − k

and Sn(2) =

λ1 X k=2ln


λ0k − m(k)

X ln ≤j≤λ1 ,|j−k|≥ln

1 . m(j, k) + j − k (1)

It follows from Proposition 4.2 that with high probability Sn is of smaller (2) order than n3/4 . To study Sn , we need a delicate approximation (see pp. 200-202 of Pittel (2002) for lengthy and laborious computation): X
1 = v(yk ) + log g(yλ1 ) − g(yk ) m(j, k) + j − k ln ≤j≤λ1 ,|j−k|≥ln
+Op n−1/2 log n . Then (4.84) becomes λ1 X




λ0k − m(k) v(yk ) + log(g(yλ1 ) − g(yk )) + op n3/4 .

(4.85)

k=2ln

Inserting (4.82) into (4.81) and noting (4.83) and (4.85), X
Mn = log m(k1 , k2 ) + k2 − k1 (k1 ,k2 )∈K λ1 X

+




λ0k − m(k) v(yk ) + log(g(yλ1 ) − g(yk )) + op n3/4 .

k=2ln

To conclude the proof, we observe X

log m(k1 , k2 ) + k2 − k1 = Op n1/2+δ log2 n (k1 ,k2 )∈K c

and X



log m(k1 , k2 ) + k2 − k1 − log µ(k1 , k2 ) + k2 − k1

1≤k1
≤2

X 1≤k1

1 = Op n1/2 log2 n . k2 − k1



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Lemma 4.17. Under (Pn , Pu,n ), we have
log dλ − log n! = log f (µ) + Tn + op n3/4 ,

(4.86)

where Q f (µ) =

1≤i<j≤λ1 (µi

Q

1≤i≤λ1 (µi

− µj + j − i) − i + λ1 )!

and Tn =

λ1 X


v(yk ) λ0k − m(k) .

(4.87)

k=2ln

Proof.

By (4.16), (4.75) and (4.80), we have log dλ − log n! = Mn − Nn = Mn − Nn + Sn − Rn + op n3/4
= log f (µ) + Sn − Rn + op n3/4 .




On the other hand, it trivially follows X
Sn − Rn − Tn = v(yk ) λ0k − m(k) k>λ1 −ln

X

+



log g(yλ1 ) − g(yk ) λ0k − m(k)

2ln ≤k≤λ1 −ln

X

−



log m(k) − k + λ1 λ0k − m(k) .

(4.88)

ln ≤k≤kn

Therefore we need only prove that the right hand side of (4.88) are negligible. First, according to Lemma 5 of Pittel (2002), there is a constant c6 > 0  1 c6 |v(x)| = c6 log x + (4.89) , |v 0 (x)| ≤ . x x We easily get X

v(yk )λ0k = O nδ λ1 log n = Op n1/2+δ log2 n . k>λ1 −ln

Also, it is even simpler to check X v(yk )m(k) = Op (n1/2 log2 n). k>λ1 −ln

Thus we have X k>λ1 −ln



v(yk ) λ0k − m(k) = op n3/4 .

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Second, observe the following simple facts: λ1 X λ0k = n k=1

and λ1 X

n c

Z

∞


1 dy + Op n1/2 log n −y 1−e 0
= n + Op n1/2 log n .

m(k) =

k=1

Then we easily have λ1 X

log



λ0k − m(k) = Op n1/2 log n .

(4.90)

k=1

On the other hand, it follows from (4.74) X
λ0k − m(k) = Op n3/4−ε , k∈[2l / n ,kn ]

which together with (4.90) in turn implies X

λ0k − m(k) = Op n3/4 .

(4.91)

2ln ≤k≤kn

Besides, we have by (4.74) X
λ0k − m(k) = Op n3/4 log n .

(4.92)

2ln ≤k≤kn

By the definition of g(x) and m(k), √ √ √

n n g(yλ1 ) − g(yk ) = m(k) − k + λ1 + 1 − e−cλ1 / n c c = m(k) − k + λ1 + Op (1). Therefore for k ≤ kn

log g(yλ1 ) − g(yk ) − log m(k) − k + λ1
c = log √ + Op n−1/2 log−1 n . n Thus by (4.91) and (4.92) X

log(g(yλ1 ) − g(yk )) − log(m(k) − k + λ1 ) λ0k − m(k) 2ln ≤k≤kn

c = log √ n +Op n

X


λ0k − m(k)

2ln ≤k≤kn

−1/2

log−1 n




X

0 λk − m(k)

2ln ≤k≤kn 3/4




= op n . The proof is complete.



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To proceed, we need to treat the constant term log fµ . Lemma 4.18. Under (Pn , Pu,n ), we have Z ∞
1 t log t log f (µ) = − n log n − n + op n3/4 . ct 2 e −1 0
Proof. Form a Young diagram µ =
µ(1), µ(2), · · · , µ(λ1 ) and denote its dual by ν = ν(1), ν(2), · · · , ν(µ(1)) , where ν(i) = max{1 ≤ k ≤ λ1 : µ(k) ≥ i},

1 ≤ k ≤ µ(1).

We remark that ν can be viewed as an approximation to the random diagram λ since µ is an approximation of λ0 . Now apply the hook formula (4.70) to the diagram µ to obtain X
log f (µ) = − log ν(i) − k + µ(k) − i + 1 . (4.93) i≤µ(1),k≤ν(i)

As a first step, we need to replace asymptotically µ(·) and ν(·) by `(·) in (4.93). Indeed, by the definition of `(·) and µ(·), it follows `(k) − 1 ≤ ν(k) ≤ `(k − 1),

1 < k ≤ µ(1).

Define  D = (i, k) : i ≤ min{µ(k), `(k)}, k ≤ min{ν(i), `(i)} , then X

log f (µ) = −



log ν(i) − k + µ(k) − i + 1 + Op n1/2 log2 n .

(i,k)∈D

Moreover, X
− log ν(i) − k + µ(k) − i + 1 (i,k)∈D

X

=−


log `(i) − k + `(k) − i + 1

(i,k)∈D

 |`(i) − ν(i)| |`(k) − µ(k)| + min{ν(i), `(i)} − k + 1 min{µ(k), `(k)} − i + 1 (i,k)∈D X

=− log `(i) − k + `(k) − i + 1 + Op n1/2 log2 n . +O

 X

(i,k)∈D


The same argument results in another Op n1/2 log2 n error term if we replace further D by D∗ = {(i, k) : i ≤ `(k), k ≤ `(i)}. Thus X

log f (µ) = − log `(i) − k + `(k) − i + 1 + Op n1/2 log2 n . (i,k)∈D ∗

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The next step is to switch the sum into an integral. Let  Hn = (x, y) : 0 < x, y ≤ `(1), x ≤ `(y), y ≤ `(x) , then Z Z


log `(x) − y + `(y) − x + 1 dxdy

log f (µ) = − (x,y)∈Hn

+O n

1/2


log2 n .

Furthermore, if letting  H∞ = (x, y) : x, y > 0, x ≤ `(y), y ≤ `(x) , then we have Z Z


log `(x) − y + `(y) − x + 1 dxdy

log f (µ) = − (x,y)∈H∞

+O n

1/2


log2 n .

(4.94)

To see this, make a change of variables `(y) − x `(x) − y , v= . u= n1/2 n1/2 Then in terms of u, v, the domain H∞ becomes {(u, v) : u ≥ 0, v ≥ 0}, and the inverse transform is √ √ n n ec(u+v) − 1 ec(u+v) − 1 x= log c(u+v) log , y = . c c e − ecv ec(u+v) − ecu So the Jacobian determinant is ! ∂x ∂x n ∂u ∂v . det ∂y ∂y = c(u+v) e −1 ∂u ∂v The difference between the integrals over H∞ and Hn is the double integral over H∞ \ Hn : Z Z
log `(x) − y + `(y) − x + 1 dxdy (x,y)∈H∞ \Hn

Z Z ≤n 0≤u≤n−1/2 ,v≥0
2

1/2

=O n

√ log( n(u + v) + 1) dudv ec(u+v) − 1

log n .

The last step is to explicitly calculate the double integral value over H∞ in (4.94). Via the substitutions, the integral is equal to Z ∞
t log t 1 n log n + n dt + O n1/2 , ct 2 e −1 0 as claimed. 

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To conclude, we shall show that the linearised weighted sum Tn in (4.87) admits an integral representation up to a negligible error term, and so converges in distribution to a normal random variable. Lemma 4.19. Under (Pn , Pu,n ), we have
Tn d −→ N 0, σT2 , 3/4 n where σT2 = σd2 given by (4.68). Proof. Start with an integral representation for Tn . Using the second inequality of (4.89), y − y 
k k−1 v(yk ) − v(x) = O = O n−δ yk uniformly for x ∈ [k − 1, k) and k ≥ 2ln . Also, it is easy to see m(k) − m(x) = O (eyk − 1)−1/2




uniformly for x ∈ [k − 1, k) and k ≥ 2ln . Therefore we have Z ∞ X  v(x) Tn = (rk − Eqn rk ) dx + ∆Tn , 2ln −1

(4.95)

k≥x

where ∆Tn is of order n−δ

∞ ∞ X X 0 log(yk + yk−1 ) + nδ λk − m(k) + . eyk − 1

k=2ln

k=2ln

By virtue of Proposition 4.2, the whole order is actually Op (n3/4−δ log1/2 n). Neglecting ∆Tn , we can equate Tn with the integral on the right side in (4.95). Furthermore, we also extend the integration to [xn , ∞), where √  n c  log 1 − √ . xn = − c 2 n Making the substitution √ x=

1 n log c 1−t

in the last integral and using the definition of the process Yn (t) we obtain Z
n3/4 1 v(− log(1 − t)) Tn = Yn (t)dt + O n3/4−ε . c t(1 − t) 0

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Now we are in a position to apply Theorem 4.13 to the function g(t, x) :=

v(− log(1 − t)) x, t(1 − t)

which clearly meets the condition (4.61) with µ = α = β = 1. Therefore Z Tn d 1 1 v(− log(1 − t)) Y (t)dt. −→ c 0 t(1 − t) n3/4 The limit variable in the right hand side is a centered Gaussian random variable with variance Z Z v(− log(1 − s)) v(− log(1 − t)) 1 1 1 EY (s)Y (t) dsdt. σT2 := 2 c 0 0 s(1 − s) t(1 − t) To conclude the proof, we note Z ∞ (1 − t) + log(ey − 1) dy e−y log log v(− log(1 − t)) = − t 0 = −t(1 − t)κ(t), where κ(t) is given by (4.67). The proof is complete.



Proof of Theorem 4.15. Putting Lemmas 4.17, 4.18 and 4.19 all together, we can conclude the proof.  4.5

Random multiplicative partitions

In this section we shall introduce a class of multiplicative measures as extension of uniform measure and describe briefly the corresponding limit shape and second order fluctuation around the shape. The reader is referred to Su (2014) for detailed proofs and more information. Consider a sequence of functions gk (z), k ≥ 1, analytic in the open disk D% = {z ∈ C : |z| < %}, % = 1 or % = ∞, such that gk (0) = 1. Assume further that (i) the Taylor series gk (z) =

∞ X

sk (j)z j

j=0

have all coefficients sk (j) ≥ 0 and (ii) the infinite product G(z) =

∞ Y k=1

gk (z k )

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201

converges in D% . Now define the measure Pm,n on Pn by
sk (j) Pm,n λ ∈ Pn : rk (λ) = j = Zm,n and Q∞ Pm,n (λ) =

k=1 sk (rk )

Zm,n

,

λ ∈ Pn ,

where Zm,n =

∞ X Y

sk (rk ).

λ∈Pn k=1

Here m in the subscript stands for multiplicative. We also define a family of probability measures Qm,q , q ∈ (0, %) on P in the following way: Q∞ sk (rk ) |λ| Qm,q (λ) = k=1 q , λ ∈ P. G(q) It is easy to see
sk (j)q kj , Qm,q λ ∈ P : rk (λ) = j = gk (q k )

j ≥ 0,

k≥1

and so different occupation numbers are independent. The measure Qm,q is called multiplicative. According to Vershik (1996), analog of Lemma 4.3 is valid for Qm,q and Pm,n . This will enable us make full use of conditioning argument. Note that the generating function G(z), along with its decomposition Q∞ G(z) = k=1 gk (z k ), completely determines such a family. It actually contain many important examples, see Vershik (1996), Vershik and Yakubovich (2006). A particularly interesting example is the G(z) is generated by β gk (z) = 1/(1 − z)k , β > −1. In such a special case, the convergence radius of gk and G is % = 1. We also write Qβ,q , Pβ,n for probabilities and Eβ,q , Eβ,n for expectations respectively. Set Gβ (z) =

∞ Y

1 . (1 − z k )kβ k=1

Vershik (1996), Vershik and Yakubovich (2006) treat Qβ,q and Pβ,n as generalized Bose-Einstein models of ideal gas; while in combinatorics and number theory they are well known for a long time as weighted partitions.

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Remark 4.1. P0,n corresponds to the uniform measure Pu,n on Pn , and Z0,n is the Euler function p(n): the number of partitions of n. In the case of β = 1, the Gβ (z) is the generating function for the numbers p3 (n) of 3-dimensional plane partitions of n (see Andrews (1976)): Y X 1 p3 (n)z n = . (1 − z k )k n≥0

k≥1

However, the P1,n is completely different from the uniform measure on 3-dimensional plane partitions of n. Vershik (1996), in an attempt to capture various limiting results concerning particular functionals in a unified framework, posed the question of evaluating the limit shape for ϕλ (t) under Pβ,n . In particular, we have Theorem 4.16. Assume β ≥ 0, let hn = ( Γ(β+2)ζ(β+2) )1/(β+2) . Consider n the scaled function  t  , t ≥ 0. ϕβ,n (t) = hβ+1 ϕλ n hn Then it follows ϕβ,n → Ψβ in the sense of uniform convergence on compact sets, where Ψβ is the function defined by Z ∞ β −u u e du. Ψβ (t) = 1 − e−u t More precisely, for any ε > 0 and 0 < a < b < ∞, there exists an n0 such that for n > n0 we have   Pβ,n λ ∈ Pn : sup |ϕβ,n (t) − Ψβ (t)| > ε < ε. a≤t≤b

Remark 4.2. The value of hn is in essence determined so that Eβ,q |λ| ∼ n, √ where q = e−hn . For β = 0, the scaling constants along both axes are c n. Moreover Ψ0 (t) is equal to Ψ(t) of (4.12). While for β > 0, two distinct scaling constants must be adapted. In fact, the value on the s axis is more compressed than the indices on the t axis. Also, it is worth noting Z ∞ β −u u e du < ∞ Ψβ (0) = 1 − e−u 0 by virtue of β > 0.

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Having the limit shape, we will continue to further study the second order fluctuation of Young diagrams around it. This will separately discussed according to two cases: at the edge and in the bulk. First, let us look at the asymptotic distribution of the largest part of a partition under Pβ,n . The following result, due to Vershik and Yakubovich (2006), is an extension of Erd¨ os and Lehner’s theorem Theorem 4.17.  −x An x  lim Pβ,n λ ∈ Pn : λ1 − = e−e , ≤ n→∞ hn hn where An =

β+1 β+1 β+1 log n + β log log n + β log − log Γ(β + 2)ζ(β + 2). β+2 β+2 β+2 d

Remark 4.3. As known to us, λ = λ0 , and so λ01 and λ1 have the same asymptotic distribution under (Pn , Pu,n ). But such an elegant property is no longer valid under (Pn , Pβ,n ). In fact, we have the following asymptotic normality for λ01 instead of Gumbel distribution. −(β+1)

Let σn2 = hn

µn,k =

and define for k ≥ 1

∞ X j=k

jβ

e−hn j , 1 − e−hn j

2 σn,k =

∞ X

jβ

j=k

e−hn j , (1 − e−hn j )2

where hn is as in Theorem 4.16. Theorem 4.18. (i) Under Pβ,n with β > 1, we have as n → ∞,
λ0k − µn,k d −→ N 0, κ2β (0) , σn where κ2β (0) = Γ(β + 1)ζ(β + 1, 0) −

Γ(β + 2)ζ 2 (β + 2, 0) ζ(β + 2)

and 1 ζ(r + 1, 0) := Γ(r + 1)

Z

∞

0

ur e−u du (1 − e−u )2

(ii) Under P1,n , we have as n → ∞, λ0k − µn,k d p −→ N (0, 1). σn | log hn |

for r > 1.

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Theorem 4.18 corresponds to the end of partitions. We consider the fluctuations in the deep bulk of partitions below. Let  1  t Xβ,n (t) = ϕλ ( ) − µn,d ht e , t ≥ 0. n σn hn Theorem 4.19. Under Pβ,n with β > −1, we have as n → ∞ (i) for each t > 0, d

Xβ,n (t) −→ Xβ (t), where Xβ (t) is a normal random variable with zero mean and variance κ2β (t) = σβ2 (t) −


2 1 2 σβ+1 (t) ; Γ(β + 3)ζ(β + 2)

(ii) for 0 < t1 < t2 < · · · < tm < ∞,
d
Xβ,n (t1 ), Xβ,n (t2 ), · · · , Xβ,n (tm ) −→ Xβ (t1 ), Xβ (t2 ), · · · , Xβ (tm ) ,
where Xβ (t1 ), Xβ (t2 ), · · · , Xβ (tm ) is a Gaussian vector with covariance structure 2 2
σβ+1 (s)σβ+1 (t) Cov Xβ (s), Xβ (t) = σβ2 (t) − , Γ(β + 3)ζ(β + 2)

s < t;

(iii) Each separable version of Xβ is continuous in (0, ∞). Next we give the limiting distribution of dλ after properly scaled. Theorem 4.20. Under (Pn , Pβ,n ) with β > 1, we have as n → ∞  
dλ d 2 − b −→ N 0, σ , h(β+3)/2 log n β,d n (n!)1/(β+2) 2 where the normalizing constant bn and limiting variance σβ,d are given by

bn =

∞

∞

k=1

k=1

∞

X X β+1X β+1 n, µn,k log n − µn,k log µn,k + µn,k − β+2 β+2 −

β+1 log Γ(β + 2)ζ(β + 2) β+2

k=1

∞
 X

µn,k − n



k=1

and 2 σβ,d

Z

∞

Z

= 0

0

∞


Cov Xβ (s), Xβ (t) log Ψβ (s) log Ψβ (t)dsdt.

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To conclude this chapter, we want to mention another interesting example of multiplicative measure which is given by the exponential generating function. Let a = (ak , k ≥ 1) be a parameter function determined by P g(x) = exp( k≥1 ak xk ). Define a probability Pa,n on Pn by Pa,n (λ) =

1 Za,n

Y ark k , rk !

λ ∈ Pn ,

k=1

where Za,n is the partition function. In terms of the form of parameter function, the measure Pa,n substantially differ from either Pu,n or Pβ,n . The reader is referred to Erlihson and Granovsky (2008) and the reference therein for the limit shape and functional central limit theorem for the fluctuation.

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Chapter 5

Random Plancherel Partitions

5.1

Introduction

In this chapter we shall consider another probability measure, namely Plancherel measure, in Pn and study its asymptotic properties as n → ∞. As in Chapter 4, our main concerns are again the fluctuations of a typical Plancherel partition around its limit shape. To start, let us recall the classic elegant Burnside identity X d2λ = n!, λ∈Pn

where dλ is the number of standard Young tableaux with a shape λ, see (2.13). This naturally induces a probability measure Pp,n (λ) =

d2λ , n!

λ ∈ Pn

where p in the subscript stands for Plancherel. Pp,n is often referred to as Plancherel measure because the Fourier transform
Fourier 2
L2 Sn , µs,n −→ L Sbn , Pp,n , is an isometry just like in the classical Plancherel theorem, where µs,n is the uniform measure on Sn and Sbn is the set of irreducible representations of Sn . Plancherel measure naturally arises in many representation-theoretic, combinatorial and probabilistic problems. To illustrate, we consider the length of longest increasing subsequences in Sn . For a given π = (π1 , π2 , · · · , πn ) ∈ Sn and i1 < i2 < · · · < ik , we say πi1 , πi2 , · · · , πik is an increasing subsequence if πi1 < πi2 < · · · < πik . Let `n (π) be the length of longest increasing subsequences of π. For example, let n = 10, 207

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π = (7, 2, 8, 1, 3, 4, 10, 6, 9, 5). Then `n (π) = 5, and the longest increasing subsequences are 1, 3, 4, 6, 9 and 2, 3, 4, 6, 9. The study of `n (π) dates back to Erd¨os and Szekeres in the 1930s. A celebrated theorem states every π of Sn contains an increasing and/or √ decreasing subsequence of length at least n (see Steele (1995, 1997)). This can be proved by an elementary pigeon-hole principle. But it also follows from an algorithm developed by Robinson, Schensted and Knuth (see Sagan (2000)) to obtain Young tableaux with the help of permutations. Let Tn be the set of standard Young tableaux with n squares. According to this algorithm, for any n ≥ 1 there is a bijection, the so-called RSK correspondence, between Sn and pairs of T, T 0 ∈ Tn with the same shape:
RSK Sn 3 π ←→ T (π), T 0 (π) ∈ Tn × Tn . The RSK correspondence is very intricate and has no obvious algebraic meaning at all, but it is very deep and allows us to understand many things. In particular, it gives an explicit proof of the Burnside identity (2.13). More interestingly, `n (π) is exactly the number of squares in the first row of T (π) or T 0 (π), namely `n (π) = λ1 (T (π)). Consequently,
|{π ∈ Sn : `n (π) = k}| µs,n π ∈ Sn : `n (π) = k = n! X d2λ = n! λ∈Pn :λ1 =k
= Pp,n λ ∈ Pn : λ1 = k . In words, the Plancherel measure Pp,n on Pn is the push-forward of the uniform measure µs,n on Sn . Thus the analysis of `n (π) is equivalent to a statistical problem in the geometry of the Young diagram. See an excellent survey Deift (2000) for more information. A remarkable feature is that there also exists a limit shape for random Plancherel partitions. Define for λ ∈ P ψλ (0) = λ1 ,

ψλ (x) = λdxe ,

x > 0.

that ψλ (x), x ≥ 0 is a nonincreasing step function such that RNote ∞ ψ (x)dx = |λ|. Also, ψλ0 (x) = ϕλ (x) where λ0 is a dual partition λ 0 of λ and ϕλ was defined by (4.10). The so-called limit shape is a function y = ω(x) defined as follows: 2 x = (sin θ − θ cos θ), y = x + 2 cos θ π where 0 ≤ θ ≤ 2π is a parameter, see Figure 5.1 below.

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209

ω curve

Logan and Shepp (1977) used the variational argument to prove √ √ Theorem 5.1. Under (Pn , Pp,n ), the rescaled function ψλ ( nx)/ n converges to the function ω(x) in a sense of weak convergence in a certain metric d. Here the metric d is defined by (1.15) of Logan and Shepp (1977). We remark that one cannot derive from Theorem 5.1 √
P 1 √ ψλ n x −→ ω(x) n for every x ≥ 0. Independently, Vershik and Kerov (1977, 1985) developed a slightly √ √ different strategy to establish a uniform convergence for ψλ ( nx)/ n. To state their results, it is more convenient to use the rotated coordinate system u = x − y,

v = x + y.

Then in the (u, v)-plane, the step function ψλ (x) transforms into a piecewise linear function Ψλ (u). Note Ψ0λ (u) = ±1, Ψλ (u) ≥ |u| and Ψλ (u) = |u| for sufficiently large u. Likewise, ω(x) transforms into Ω(u) (see (1.34) and Figure 1.4): √
2 u arcsin u2 + 4 − u2 , |u| ≤ 2 π (5.1) Ω(u) = |u|, |u| ≥ 2. Define √ 1 Ψn,λ (u) = √ Ψλ ( n u), n

∆n,λ (u) = Ψn,λ (u) − Ω(u).

(5.2)

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Theorem 5.2. Under (Pn , Pp,n ), 1 √ P sup √ Ψλ ( n u) − Ω(u) −→ 0, n −∞
n → ∞.

(5.3)

As an immediate corollary, we can improve Logan-Shepp’s result to a uniform convergence. Corollary 5.1. Under (Pn , Pp,n ), 1 √
P sup √ ψλ n x − ω(x) −→ 0, n 0≤x<∞

n → ∞.

(5.4)

We remark that ω(0) = 2 and ω(2) = 0. Compared to Ψ(x) of (4.12), ω(x) looks more balanced. This can be seen from the definition of Plancherel measure. Roughly speaking, the more balanced a Young diagram is, the more likely it appears. For instance, fix n = 10, and consider two partitions λ(1) = (110 ) and λ(2) = (1, 2, 3, 4). Then

1 256 1 Pp,10 λ(1) = , Pp,10 λ(2) = ≈ . 10! 1575 6 Corollary 5.2. Under (Pn , Pp,n ), λ P √1 −→ 2, n

λ0 P √1 −→ 2, n

n → ∞.

Consequently, `n (π) P √ −→ 2, n

n → ∞.

(5.5)

(5.5) provides a satisfactory solution to Ulam’s problem. The rest of this section shall be devoted to a rigorous proof of Theorem 5.2 due to and Kerov (1977). It will consist of a series of lemmas. A key technical ingredient is to prove a certain quadratic integral attains its minimum at Ω. Start with a rough upper bound. Lemma 5.1. √ √
Pp,n max{λ1 , λ01 } ≥ 2e n ≤ e−2e n .

(5.6)

Proof. We need an equivalent representation of `n (π). Let X1 , X2 , · · · , Xn be a sequence of i.i.d. uniform random variables on [0, 1]. Let `n (X) be the length of the longest increasing subsequences of X1 , X2 , · · · , Xn . Trivially, d

`n (π) = `n (X),

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from which we in turn derive d

d

λ1 = `n (π) = `n (X). Then it follows Pp,n (λ1 ≥ k) = P (`n (X) ≥ k)   n 1 ≤ . k k! In particular, √ √
Pp,n λ1 ≥ 2e n ≤ e−2e n .

We conclude the proof.



Next let us take a look at d2λ /n!. Lemma 5.2. As n → ∞  √  d2 Pp,n − log λ > 2c n → 0, n! √ where c = π/ 6 as in Chapter 4. Proof.

(5.7)

Denote by An the event in (5.7). Then by (4.3) X d2 λ Pp,n (An ) = n! λ∈An X √ ≤ e−2c n λ∈An

√ n

≤ p(n)e−2c

→ 0,

as desired.



Observe that it follows from the hook formula (4.70) n! d2λ = 2. n! Hλ So we have n! d2 − log λ = − log 2 n! Hλ = − log n! + 2 log Hλ X
= − log n! + 2 log λi − i + λ0j − j + 1 (i,j)∈λ

= − log n! + n log n + 2

X (i,j)∈λ


1 log √ λi − i + λ0j − j + 1 . n

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√ √ For simplicity of notations, write ψn,λ (x) for ψλ ( nx)/ n. Then
1 log √ λi − i + λ0j − j + 1 n Z Z
1 √ ψλ (x) − x + ψλ−1 (y) − y dxdy = log n Z Z 
1 ≥ log √ ψλ (x) − x + ψλ−1 (y) − y dxdy, n  −1 where  stands for the (i, j)th unit square, ψn,λ denotes the inverse of ψn,λ and the last inequality follows from the concavity property of logarithmic function. Hence we obtain d2 − log λ ≥ − log n! + n log n n! Z Z
−1 +2n log ψn,λ (x) − x + ψn,λ (y) − y dxdy 0≤y<ψn,λ (x)

=: nI(ψn,λ ) + n , where n = O(log n) and Z Z I(ψn,λ ) = 1 + 2 0≤y<ψn,λ (x)


−1 log ψn,λ (x) − x + ψn,λ (y) − y dxdy.

As a direct consequence of Lemma 5.2, it follows for any ε > 0
Pp,n I(ψn,λ ) > ε → 0.

(5.8)

Making a change of variables, we have Z Z

1 log(u − v) 1 − Ψ0n,λ (u) 1 + Ψ0n,λ (v) dudv I(ψn,λ ) = 1 + 2 v ε → 0.

(5.9)

Similarly, define J(Ω) = 1 +

1 2

Z Z



log(u − v) 1 − Ω0 (u) 1 + Ω0 (v) dudv. v
A remarkable contribution due to Vershik and Kerov (1977, 1985) is the following Lemma 5.3. With notations above, we have

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(i) J(Ω) = 0;

(5.10)

(ii) 1 J(Ψn,λ ) = − 4 Z +

∞

Z

Z

−∞

∞

log |u − v|∆0n,λ (u)∆0n,λ (v)dudv

−∞


|u| du. Ψn,λ (u) − |u| arccosh 2

|u|>2

(5.11)

Consequently, J(Ψn,λ ) ≥ −

1 4

Z

∞

−∞

Z

∞

log |u − v|∆0n,λ (u)∆0n,λ (v)dudv.

(5.12)

−∞

Proof.

Start with the proof of (5.10). Let Z x %0 (x) = − log |x|, %1 (x) = %0 (y)dy,

Z %2 (x) =

0

x

%1 (y)dy. 0

A simple calculus shows %1 (x) = x − x log |x|,

%2 (x) =

3x2 x2 − log |x| 4 2

and %1 (−x) = −%1 (x),

%2 (−x) = %2 (x).

Note Ω0 (u) = 1 for u ≥ 2 and Ω0 (u) = −1 for u ≤ −2. Then Z Z Z Z  1 2 u 1 2 2 %0 (u − v)dv Ω0 (u)du %0 (u − v)dudv + J(Ω) = 1 − 4 −2 −2 2 −2 −2 Z 2 Z 2  −2 %0 (u − v)du Ω0 (v)dv +

1 4

−2 2

Z

−2

Z

v 2

%0 (u − v)Ω0 (u)Ω0 (v)dudv.

−2

First, it is easy to see Z

2

Z

2

%0 (u − v)dudv = 2%2 (4). −2

−2

Also, it follows Z u %0 (u − v)dv = %1 (2 + u), −2

Z

2

%0 (u − v)du = %1 (2 − v). v

(5.13)

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To calculate the last three double integral in the right hand side of (5.13), set Z 2 H2 (u) = %1 (u − v)Ω00 (v)dv, −∞ < u < ∞. −2

Then 2 = π

H200 (u)

Z

2

−2

1 √ dv = 0, (u − v) 4 − v 2

−2 ≤ u ≤ 2

and so for each −2 ≤ u ≤ 2 H20 (u)

=

H20 (0)

2 = π

2

Z

log |v| √ dv = 0. 4 − v2

−2

This in turn implies Z

2

%1 (v)Ω00 (v)dv = 0

H2 (u) = H2 (0) = −

(5.14)

−2

since %1 (v) is odd. It follows by integration by parts Z 2 Z 2 Z  0 %0 (u − v)du Ω (v)dv = −2

2

%1 (2 − v)Ω0 (v)dv

−2

v

Z

2

%2 (2 − v)Ω00 (v)dv

= −%2 (4) + −2 Z 2

%2 (v)Ω00 (v)dv

= −%2 (4) + −2

= 2 − %2 (4), where we used (5.14) and the fact Z 2 %2 (v)Ω00 (v)du = 2. −2

Similarly, Z

2

−2

Z

u

−2

Z  %0 (u − v)dv Ω0 (u)du =

2

%1 (2 + u)Ω0 (u)du

−2

= %2 (4) − 2. Again, by (5.14) Z Z 2 %0 (u − v)Ω0 (v)dv = %1 (2 − u) − %1 (2 + u) +

2

−2

−2

= %1 (2 − u) − %1 (2 + u).

%1 (u − v)Ω00 (v)dv

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Hence we have Z Z 2Z 2 %0 (u − v)Ω0 (u)Ω0 (v)dudv = −2

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2


%1 (2 − u) − %1 (2 + u) Ω0 (u)du

−2

−2

= 4 − 2%2 (4). In combination, we have proven (5.10). Turn to the proof of (5.11). First, observe there is an a = an (may depend on λ) such that [−a, a] contains the support of ∆n,λ (u). Hence we have Z Z 1 a a J(Ψn,λ ) = 1 − %0 (u − v)dudv 4 −a −a Z Z  1 a u %0 (u − v)dv Ψ0n,λ (u)du + 2 −a −a Z Z  1 a a − %0 (u − v)du Ψ0n,λ (v)dv 2 −a v Z Z 1 a a + %0 (u − v)Ψ0n,λ (u)Ψ0n,λ (v)dudv. 4 −a −a A simple calculus shows Z aZ

a

%0 (u − v)dudv = 2%2 (2a), −a a

Z

−a

−a

u

Z

%0 (u − v)dv



Ψ0n,λ (u)du

Z

a

%1 (a + u)Ψ0n,λ (u)du

= −a

−a

= ρ2 (2a), Z

a

−a

Z v

a

%0 (u − v)du



Ψ0n,λ (v)dv

Z

a

=

%1 (a − v)Ψ0n,λ (v)dv

−a

= −ρ2 (2a). On the other hand, it follows Z Z 1 ∞ ∞ − log |u − v|∆0n,λ (u)∆0n,λ (v)dudv 4 −∞ −∞ Z Z 1 a a %0 (u − v)Ω0 (u)Ω0 (v)dudv = 4 −a −a Z Z 1 a a − %0 (u − v)Ω0 (u)Ψ0n,λ (v)dudv 2 −a −a Z Z 1 a a %0 (u − v)Ψ0n,λ (u)Ψ0n,λ (v)dudv. + 4 −a −a

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Note Ω00 (u) = 0 for |u| > 2 and so Z a %0 (u − v)Ω0 (u)du = %1 (a − v) − %1 (a + v) + H2 (v).

(5.15)

−a

Since H2 (v) = 0 for |v| ≤ 2, we have Z aZ a %0 (u − v)Ω0 (u)Ω0 (v)dudv −a −a Z a Z a 0 %1 (a + v)Ω0 (v)dv %1 (a − v)Ω (v)dv − = −a

−a −2

Z −

Z

H2 (u)du.

H2 (u)du + −a

a

2

Also, it is easy to see Z a Z %1 (a − v)Ω0 (v)dv = −%2 (2a) + −a

2

%2 (a − v)Ω00 (v)dv

−2 2

Z


%2 (a − v) − %2 (2 − v) Ω00 (v)dv

= 2 − %2 (2a) +

−2 aZ 2

Z = 2 − %2 (2a) + Z = 2 − %2 (2a) +

%1 (u − v)Ω00 (v)dvdu

−2

2 a

H2 (u)du, 2

and similarly Z a

%1 (a + v)Ω0 (v)dv = %2 (2a) − 2 +

a

−a

H2 (u)du. −a

−a

By (5.15), Z

−2

Z

Z

a

 %0 (u − v)Ω0 (u)du Ψ0n,λ (v)dv

−a

Z = −2ρ2 (2a) + 2

a

H2 (v)Ψ0n,λ (v)dv

Z

−2

H2 (v)Ψ0n,λ (v)dv.

+ −a

In combination, we get J(Ψn,λ ) Z Z 1 ∞ ∞ =− log |u − v|∆0n,λ (u)∆0n,λ (v)dudv 4 −∞ −∞ Z Z
0 1 −2 1 a H2 (u) Ψn,λ (u) − u du + H2 (u)(Ψn,λ (u) + u)0 dv. + 2 2 2 −a

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217

To proceed, note for u > 2 H20 (u) =

Z

2

ρ0 (u − v)Ω00 (v)dv

−2

Z

2

2 log(u − v) √ dv π 4 − v2 −2 u = −2 arccosh . 2 Thus by integration by parts and using the fact Ψn,λ (a) = a and H2 (2) = 0, Z Z
0
1 a 0 1 a H2 (u) Ψn,λ (u) − u du = − H2 (u) Ψn,λ (u) − u du 2 2 2 2 Z a
u = Ψn,λ (u) − u arccosh du. 2 2 =−

Similarly, it follows Z Z
0
1 −2 1 −2 0 H2 (u) Ψn,λ (u) + u du = − H2 (u) Ψn,λ (u) + u du 2 −a 2 −a Z −2 |u| du. = (Ψn,λ (u) + u) arccosh 2 −a In combination, we now conclude the proof of (5.11). Finally, (5.12) holds true since Ψn,λ (u) ≥ |u| for all u ∈ R .



The following lemma is interesting and useful since it introduces the Sobolev norm into the study of random partitions. Define Z ∞Z ∞ f (u) − f (v) 2 2 kf ks = dudv, u−v −∞ −∞ where s in the subscript stands for Sobolev. Lemma 5.4. Z

∞

Z

∞

− −∞

−∞

Proof.

log |u − v|f 0 (u)f 0 (v)dudv =

1 kf k2s . 2

Denote by H(f ) the Hilbert transform, namely Z ∞ f (u) du. H(f )(v) = v −u −∞

Then it is easy to see Z

∞

[)(ω) = H(f

ei2πωv H(f )(v)dv

−∞

= isgnω fb(ω),

(5.16)

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where fb is the Fourier transform of f . Then by integration formula by parts and the Pasval-Plancherel identity Z ∞ LHS of (5.16) = H(f )(v)f 0 (v)dv −∞ Z ∞ [)(ω)fb0 (ω)dω H(f = −∞ Z ∞ = isgnω fb(ω)iω fb(ω)dω −∞ Z ∞ 2 |ω| fb(ω) dω. = −∞

To finish the proof, we need a key observation due to Vershik and Kerov (1985) Z ∞ 2 1 |ω| fb(ω) dω = kf k2s . 2 −∞ The proof is complete.



Combining (5.12) and (5.16) yields J(Ψn,λ ) ≥

1 k∆n,λ k2s . 8

(5.17)

Now we are ready to give Proof of Theorem 5.2. In view of Lemma 5.5, we can and do consider only the case in which the support of ∆n,λ is contained in a finite interval, say, [−a, a]. We will divide the double integral into two parts: Z aZ a Z a 2 ∆n,λ (u) ∆n,λ (u) − ∆n,λ (v) 2 2 k∆n,λ ks = du, dudv + 4a 2 2 u − v −a −a −a a − u which together with (5.17) implies Z Z a ∆2n,λ (u)du ≤ a2 −a

a

∆2n,λ (u)

−a

a 2 − u2

du

a k∆n,λ k2s 4 ≤ 2aJ(Ψn,λ ). ≤

Also, since |∆0n,λ (u)| ≤ 2, then sup −a≤u≤a

 ∆n,λ ≤ 61/3

Z

a

∆2n,λ (u)du

−a 1/3

≤ (12a)

J(Ψn,λ )1/3 .

1/3

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By virtue of (5.9), it follows sup −a≤u≤a

P ∆n,λ (u) −→ 0.

The proof is complete.  We have so far proven that the limit shape exists and found its explicit form. To proceed, it is natural to look at the fluctuations of a typical Plancherel partition around the limit curve. This question was first raised by Logan and Shepp in 1977. The following words are excerpted from page 211 of Logan and Shepp (1977): It is of course natural to expect that for appropriate normalizing constants cn → ∞ (perhaps cn = n1/4 would do) the stochastic processes ¯ n (t) = cn (λn (t) − f0 (t)), λ

t≥0

would tend weakly to a nonzero limiting process W (t), t ≥ 0, as cn → ∞. It would be of interest to know what the process W is. It is clear only that W integrates pathwise to zero and that W (t) ≥ 0 for t ≥ 2. Perhaps W (t) = 0 for t ≥ 2 and is the Wiener process in 0 ≤ t ≤ 2 conditioned to integrate to zero over [0, 2] and to vanish at 0 and 2, but this is just a guess. This turns out to be an interesting and challenging problem. To see the fluctuation at a fixed point, we need to consider two cases separately: at the edge and in the bulk. At x = 0, ψλ (0) is equal to λ1 , the largest part of a partition. Around 2000, several important articles, say Baik, Deift and Johansson (1999), Johansson (2001), Okounkov (2000), were devoted to studying the asymptotic distribution of the λ1 after appropriately normalized. It was proved that √ λ1 − 2 n d −→ F2 , n → ∞ n1/6 where F2 is the Tracy-Widom law, which was first discovered by Tracy and Widom in the study of random matrices, see Tracy and Widom (1994, 2002). The analogs were proven to hold for each λk , k ≥ 2. By symmetry, one can also discuss the limiting distribution at x = 2. The graph of F2 is shown in Figure 5.2 below. The picture looks completely different in the bulk. It will be proved in Section 5.3 that for each 0 < x < 2,
d √ √ 1 √ ψλ ( nx) − nω(x) −→ ξ(x), n → ∞ 1 2π log n where ξ(x) is a centered normal random variable. Note that the normalizing √ constant log n is much smaller than n1/6 . In addition, we will also see
that ξ(x), 0 < x < 2, constitutes a white noise, namely Cov ξ(x1 ), ξ(x2 ) = 0

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Fig. 5.2

Tracy-Widom law

for two distinct points x1 and x2 . Thus, one cannot expect a kind of weak convergence of the stochastic process. However, we will in Section 5.2 establish a functional central limit theorem, namely Kerov’s integrated √ √ central limit theorem for ψλ ( nx) − nω(x). 5.2

Global fluctuations

In this section we shall establish an integrated central limit theorem, which √ is used to described the global fluctuation of Ψn,λ ( n u) around the limit shape Ω(u). Let uk (u), k ≥ 0 be a sequence of modified Chebyshev polynomials, i.e.,   [k/2] X j k−j uk (u) = (−1) uk−2j . (5.18) j j=0 Note uk (2 cos θ) =

sin(k + 1)θ sin θ

and Z

2

uk (u)ul (u)ρsc(u) du = δk,l . −2

Theorem 5.3. Define Z ∞
√ √ uk (u) Ψλ ( nu) − nΩ(u) du, Xn,k (λ) = −∞

λ ∈ Pn .

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Then under (Pn , Pp,n ) as n → ∞ Xn,k ,


d  2 k ≥ 1 −→ √ ξk , k+1

 k≥1 .

Here ξk , k ≥ 1 is a sequence of standard normal random variables, the convergence holds in terms of finite dimensional distribution. This theorem is referred to the Kerov CLT since it was Kerov who first presented it and outlined the main ideas of the proof in Kerov (1993). A complete and rigorous proof was not given by Ivanov and Olshanski (2002) until 2002. The proof uses essentially the moment method and involves a lot of combinatorial and algebraic techniques, though the theorem is stated in standard probability terminologies. We need to introduce some basic notations and lemmas. Begin with Frobenius coordinates. Let λ = (λ1 , λ2 , · · · , λl ) be a partition from P. Define a ¯i = λi − i,

¯bi = λ0 − i, i

i = 1, 2, · · · , `,

(5.19)

and 1 ai = λi − i + , 2

1 bi = λ0i − i + , 2

i = 1, 2, · · · , `,

(5.20)

where ` := `(λ) is the length of the main diagonal in the Young diagram of λ. The natural numbers {¯ ai , ¯bi , i = 1, · · · , `} are called the usual Frobenius coordinates, while the half integer numbers {ai , bi , i = 1, · · · , `} are called the modified Frobenius coordinates. We sometimes represent λ = (a1 , a2 , · · · , a` |b1 , b2 , · · · , b` ). Lemma 5.5. For any λ ∈ P Φ(z; λ) : =

=

∞ Y

z + i − 21 z − λi + i − i=1 ` Y z + bi . z − ai i=1

1 2

(5.21)

Proof. First, observe the infinite series in (5.21) is actually finite because λi = 0 when i is large enough. Second, the second product is an noncontractible function since the numbers a1 , a2 , · · · , a` , −b1 , −b2 , · · · , −b` are pairwise distinct. The identity (5.21) follows from a classical Frobenius lemma, see Proposition 1.4 of Ivanov and Olshanski (2002). 

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As a direct consequence, it follows log Φ(z; λ) =

∞ X p¯k (λ)

z −k ,

(5.22)


aki − (−bi )k .

(5.23)

k=1

k

where p¯k (λ) =

` X i=1

Motivated by (5.23), we introduce the algebra A over R generated by p¯1 , p¯2 , · · · . By convention, 1 ∈ A. Lemma 5.6. The generators p¯k ∈ A are algebraically independent, so that A is isomorphic to R[¯ p1 , p¯2 , · · · ]. Proof.

See Proposition 1.5 of Ivanov and Olshanski (2002).



Recall that the algebra of symmetric functions, denoted as F, is the graded algebra defined as the projective limit of Λn , where Λn is the algebra of symmetric polynomials in n variables defined in Section 2.2. Set F 3 pk 7→ p¯k ∈ A, we get an algebra isomorphism F 7→ A. We call it the canonical isomorphism, and call the grading in A inherited from that of F the canonical grading of A. For each λ ∈ P, we define the functions p˜2 , p˜3 , · · · by setting Z ∞
1 (5.24) p˜k (λ) = k(k − 1) uk−2 Ψλ (u) − |u| du. 2 −∞ Similarly, define Z ∞
1 p˜k (Ψn,λ ) = k(k − 1) uk−2 Ψn,λ (u) − |u| du 2 −∞ and Z ∞
1 p˜k (Ω) = k(k − 1) uk−2 Ω(u) − |u| du. (5.25) 2 −∞ Lemma 5.7. For each k ≥ 2 we have (i) ( (2m)! k = 2m, 2, p˜k (Ω) = (m!) 0, k = 2m + 1; (ii) q+1 q X X p˜k (λ) = xki − yik , λ ∈ P i=1

(5.26)

(5.27)

i=1

where the xi ’s are the local minima and the yj ’s are the local maxima of the function Ψλ and x1 < y1 < x2 < · · · < xq < yq < xq+1 , see Figure 5.3 below.

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Fig. 5.3

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223

Frobenius representation

Proof.

(i) Note by integration formula by parts Z ∞
00 1 uk Ω(u) − |u| du p˜k (Ω) = 2 −∞ Z ∞ 2 = uk √ du. π 4 − u2 −∞ (5.26) now easily follows. Turn to (ii). Note q+1 q X
00 X 1 Ψλ (u) − |u| = δxi − δyj − δ0 . 2 i=1 j=1 Then we have Z

∞


00 1 Ψλ (u) − |u| du 2 −∞ Z ∞ X q+1 q  X = uk δ xi − δyj − δ0 du uk

p˜k (Ψλ ) =

−∞

=

q+1 X i=1

as desired.

xki −

i=1 q X

j=1

yik ,

i=1



Lemma 5.8. The functions p˜2 , p˜3 , · · · belong to the algebra A. In particular, we have for any λ ∈ P,   [k/2] X k 1 p˜k+1 (λ) = p¯k−2j (λ), k ≥ 1. (5.28) 2j (2j + 1) 2j k+1 2 j=0

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Proof. Fix λ ∈ P and let x1 , · · · , xq+1 , y1 , · · · , yq be as in (5.27). Then according to Proposition 2.6 in Ivanov and Olshanski (2002), the following identity holds: Qq
yj
Φ z − 12 ; λ j=1 1 − z
= Qq+1
. xi Φ z + 21 ; λ i=1 1 − z This in turn implies   1  1  log Φ z − ; λ − log Φ z + ; λ 2 2 q+1 q    X X xi  yj − . log 1 − = log 1 − z z i=1 j=1 By (5.22), the left hand side of (5.29) equals   1  1  log Φ z − ; λ − log Φ z + ; λ 2 2 ∞  X p¯l 1  1 = (z − )−l − (z + )−l . l 2 2

(5.29)

(5.30)

l=1

Also, by Lemma 5.7, the right hand side of (5.29) equals ∞ X p˜k k=1

z −k .

k

(5.31)

By comparing coefficients of z −k in both (5.30) and (5.31), we easily get (5.28).  In a simpler way, (5.28) can be interpreted as p˜k+1 (λ) = p¯k + a linear combination of p¯1 , p¯2 , · · · , p¯k−1 , k+1

k ≥ 1.

Conversely, p¯k (λ) =

p˜k+1 (λ) + a linear combination of p˜2 , p˜2 , · · · , p˜k , k+1

k ≥ 1.

Hence the functions p˜2 (λ), p˜3 (λ), · · · are algebraically independent generators of the algebra A: A = R[˜ p2 , p˜3 , · · · ]. The weight grading of A is defined as wt(˜ pk ) = k,

k = 2, 3, · · · .

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Equivalently, the weight grading is the image of the standard grading of F under the algebra morphism: R[p1 , p2 , · · · ] = F 7→ A = R[˜ p2 , p˜3 , · · · ] p1 → 0,

pk → p˜k ,

k = 2, 3, · · · .

The weight grading induces a filtration in A, which we call the weight filtration and denote by the same symbol wt(·). In particular, wt(¯ pk ) = k + 1,

k≥1

since the weight of the top homogeneous component of p¯k is p˜k+1 /(k + 1). Define for each k ≥ 1 ( n−k ) n↓k χλ (k,1 , λ ∈ Pn , n ≥ k ] dλ pk (λ) = (5.32) 0, λ ∈ Pn , n < k where n↓k = n(n − 1) · · · (n − k + 1). Lemma 5.9. Fix k ≥ 1 and λ ∈ P. Then p]k (λ) equals the coefficient of z −1 in the expansion of the function 1 ↓k Φ(z; λ) 1 (5.33) − z− k 2 Φ(z − k; λ) in descending powers of z about the point z = ∞. Proof. We treat two cases separately. First, assume λ ∈ Pn where n < k. Then by definition, p]k (λ) = 0. Also, by Lemma 5.5, it is easy to see that the function of (5.33) is indeed a polynomial of z. So the claim is true. Next, consider the case n ≥ k. Recall the following formula due to Frobenius (see Example 1.7.7 of Macdonald (1995) and Ingram (1950)): p]k (λ) equals the coefficients of z −1 in the expansion of the function n Y 1 z − λi − n + i − k F (z) = − z ↓k k z − λi − n + i i=1
about z = ∞. Namely, p]k (λ) = −Res F (z), z = ∞ . A simple transformation yields
Φ z − n + 21 ; λ 1 ↓k
. F (z) = − (z − n) k Φ z − n + 12 − k; λ Note the residue at z = ∞ will not change under the shift z 7→ z + n − 1/2. Consequently,  1  1 ↓k Φ(z; λ) p]k (λ) = −Res − ,z = ∞ . z− k 2 Φ(z − k; λ) The proof is complete. 

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We shall employ the following notation. Given a formal series A(t), let [tk ]{A(t)} = the coefficient of tk in A(t). Lemma 5.10. The functions p]k (λ) belong to the algebra A. In particular, it can be described through the generators p¯1 , p¯2 , · · · as follows k n 1Y 1
p]k (λ) = [tk+1 ] − 1 − (j − )t k j=1 2 · exp Proof.

∞ X
o p¯j (λ)tj 1 − (1 − kt)−j . j j=1

This is a direct consequence of Lemmas 5.5 and 5.9.

(5.34) 

The expression (5.34) can be written in the form ∞ o n  X 1 k p¯j (λ)tj+1 (1 + ε1 (t)) p]k (λ) = − [tk+1 ] (1 + ε0 (t)) exp − k j=1 ∞ ∞ m o n X (−1)m  X 1 k p¯j (λ)tj+1 (1 + ε1 (t)) . = − [tk+1 ] (1 + ε0 (t)) k m! m=0 j=1

Here each εr (t) is a power series of the form c1 t + c2 t2 + · · · , where the coefficients c1 , c2 , · · · do not involve the generators p¯1 , p¯2 , · · · . We can now readily evaluate the top homogeneous component of p]k (λ) with respect to both the canonical grading and the weight grading in A. In the canonical grading, the highest term of p]k equals p¯k : p]k = p¯k + lower terms; while in the weight grading, the top homogeneous component of p]k has weight k + 1 and can be written as p˜k+1 p]k = + f (˜ p2 , · · · , p˜k ) + lower terms, (5.35) k+1 where f (˜ p2 , · · · , p˜k ) is a homogeneous polynomial in p˜2 , · · · , p˜k of total weight k + 1. Now we invert (5.35) to get Lemma 5.11. For k = 2, 3, · · · P X k ↓ ri Y ]
ri Q p˜k (λ) = pi−1 (λ) + lower terms , ri !

(5.36)

i≥2

where the sum is taken over all r2 , r3 , · · · with 2r2 +3r3 +· · · = k, and lower terms means a polynomial in p]1 , p]2 , · · · , p]k−2 of total weight ≤ k −1, where wt(p]i ) = i + 1.

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The proof is left to the reader. See also Proposition 3.7 of Ivanov and Olshanski (2002), which contains a general inversion formula. We next extend p]k in (5.32) to any partition ρ on P. Let |ρ| = r, define ( n−r ) n↓r χλ (ρ,1 , λ ∈ Pn , n ≥ r, ] dλ pρ (λ) = (5.37) 0, λ ∈ Pn , n < r. The following lemma lists some basic properties. The reader is referred to its proof and more details in Kerov and Olshanski (1993), Okounkov and Olshanski (1998), Vershik and Kerov (1985). Lemma 5.12. (i) For any partition ρ, the function p]ρ is an element of A. (ii) In the canonical grading, p]ρ (λ) = p¯ρ (λ) + lower terms, where λ = (1r1 , 2r2 , · · · ) and p¯ρ (λ) =

Y

p¯i (λ)ri .

i=1

p]ρ

form a basis in A. (iii) The functions (iv) For any partitions σ and τ , in the canonical grading p]σ p]τ = p]σ∪τ + lower terms.

(5.38)

We remark that the basis p]ρ is inhomogeneous both in the canonical grading and weight grading. For each f ∈ A, let (f )ρ be the structure constants of f in the basis of p]ρ . Namely, X (f )ρ p]ρ (λ). f (λ) = ρ

Define for any index set J ⊆ N X kρkJ = |ρ| + rj (ρ),

degJ (f ) = max kρkJ . ρ:(f )ρ 6=0

j∈J

We will be particularly interested in J = ∅, {1} and N below. For simplicity, denote kρk0 = kρk∅ ,

kρk1 = kρk1 ,

kρk∞ = kρkN

and deg0 (f ) = deg∅ (f ),

deg1 (f ) = deg1 (f ),

deg∞ (f ) = degN (f ).

Lemma 5.13. For any partition σ, p]σ p]1 = p]σ∪1 + |σ| · p]σ .

(5.39)

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Proof. This directly follows from the definition (5.37). Indeed, we need only to show for each λ ∈ Pn with n ≥ |σ| since otherwise both sides are equal to 0. When n = |σ| and λ ∈ Pn , p]σ (λ)p]1 (λ) = n · p]σ (λ) = |σ| · p]σ (λ). Next assume n ≥ |σ| + 1. Setting k = |σ|, then χλ (σ, 1n−k ) . dλ Hence the claim follows from a simple relation p]σ (λ)p]1 (λ) = n↓(k) · n

n↓k · n = n↓(k+1) + n↓k · k.



Lemma 5.14. For any partitions σ and τ , p]σ p]τ = p]σ∪τ + lower terms, where lower terms means a linear combination of kτ kN . Proof.

(5.40) p]ρ

with kρkN < kσkN +

Set p]σ p]τ =

X

(p]σ p]τ )ρ p]ρ .

ρ

We claim that only partitions ρ with kρkN ≤ kσkN + kτ kN can really contribute. Indeed, assume (p]σ p]τ )ρ 6= 0, and fix a set X of cardinality |ρ| and a permutation s : X → X whose cycle structure is given by ρ. Then according to Proposition 4.5 of Ivanov and Olshanski (2002) (see also Proposition 6.2 and Theorem 9.1 of Ivanov and Kerov (2001)), there must exist a quadruple {X1 , s1 , X2 , s2 } such that (i) X1 ⊆ X, X2 ⊆ X, X1 ∪ X2 = X; (ii) |X1 | = |σ| and x1 : X1 7→ X1 is a permutation of cycle structure σ; (iii) |X2 | = |τ | and x2 : X2 7→ X2 is a permutation of cycle structure τ ; (iv) denoting by s¯1 : X → X and s¯2 : X → X the natural extensions of s1,2 from X1,2 to the whole X. I.e., s¯1,2 is trivial on X \ X1,2 , then s¯1 s¯2 = s. Fix any such quadruple and decompose each of the permutations s, s1 , s2 into cycles. Let CN (s1 ) denote the set of all cycles of s1 , AN (s1 ) the subset of those cycles of s1 that entirely contained in X1 \ X2 , BN (s1 ) the subset of those cycles of s1 that have a nonempty intersection with X1 ∩ X2 . Then CN (s1 ) = AN (s1 ) + BN (s1 ). Define similarly CN (s2 ), AN (s2 ) and BN (s2 ), then we have CN (s2 ) = AN (s2 ) + BN (s2 ).

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Similarly again, let CN (s) denote the set of all cycles of s, BN (s) the subset of those cycles of s that intersect both X1 and X2 . Then CN (s) = AN (s1 ) + AN (s2 ) + BN (s). The claimed inequality kρkN ≤ kσkN + kτ kN is equivalent to |X| + |BN (s)| ≤ |X1 | + |BN (s1 )| + |X2 | + |BN (s2 )|.

(5.41)

To prove (5.41), it suffices to establish a stronger inequality: |BN (s)| ≤ |X1 ∩ X2 |.

(5.42)

To see (5.42), it suffices to show each cycle ∈ o ∈ BN (s) contains a point of X1 ∩ X2 . By the definition of BN (s), cycle o contains both points of X1 and X2 . Therefore there exist points x1 ∈ X1 ∩ c and x2 ∈ X2 ∩ o such that sx1 = x2 . By (iv), it follows that either x1 or x2 lies in X1 ∩ X2 . Thus the claim is true, as desired. Now assume (p]σ p]τ )ρ 6= 0 and kρkN = kσkN + kτ kN , then both BN (s1 ) and BN (s2 ) are empty, which implies X1 ∩ X2 = ∅. Therefore ρ = σ ∪ τ . Finally, by (5.38), (p]σ p]τ )σ∪τ = 1. It concludes the proof.



Lemma 5.15. (i) For any two partitions σ and τ with no common part, p]σ p]τ = p]σ∪τ + lower terms, where lower terms means terms with deg1 (·) < kσ ∪ τ k1 . (ii) For any partition σ ∈ P and k ≥ 2, if rk (σ) ≥ 1, then p]σ p]k = p]σ∪k + krk (σ)p](σ\k)∪1k + lower terms,

(5.43)

where lower terms means terms with deg1 (·) < kσk1 + k. Proof. (i) can be proved in a way similar to that of Lemma 5.14 with minor modification. Turn to (ii). Set X (p]σ p]τ )ρ p]ρ . p]σ p]τ = ρ

Again, only partitions ρ with kρk1 ≤ kσk1 + k can really contribute. We need below only search for partitions ρ such that (p]σ p]τ )ρ 6= 0 and kρk1 = kσk1 + k. As in Lemma 5.14, we get B1 (s1 ) = ∅,

B1 (s2 ) = ∅,

|B1 (s)| = |X1 ∩ X2 |.

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This means that X1 ∩ X2 = ∅ or X1 ∩ X2 entirely consists of common nontrivial cycles of the permutations s1 and s−1 2 . The first possibility, X1 ∩ X2 = ∅, means that ρ = σ ∪ (k). Furthermore, by (5.38), (p]σ p]τ )σ∪k = 1. The second possibility means X1 ⊇ X2 because s−1 reduces to a single 2 k-cycle which is also a k-cycle of s1 . This in turn implies that rk (σ) ≥ 1 and ρ = (σ \ k) ∪ 1k . It remains to evaluate (p]σ p]τ )ρ = krk (σ). Note that the number of ways to choose a k-cycle inside a k +r1 (σ)-point set equals (k +r1 (σ))!/k(r1 (σ))!. According to Proposition 6.2 and Theorem 9.1 of Ivanov and Kerov (2001), we know kzσ (k + r1 (σ))! zρ k(r1 (σ))! = krk (σ),

(p]σ p]τ )ρ =

where zλ is defined by (2.15) for a partition λ. The proof is complete.



In the preceding paragraphs we have briefly described the structure of algebra A and its three families of bases including {¯ pk }, {˜ pk } and {p]ρ }. Next we need to take average operation with respect to Pp,n for elements of A. A basic result is as follows. Lemma 5.16. Let |ρ| = r, n ≥ r. Then  ↓r n , ρ = (1r ), Ep,n p]ρ = 0, otherwise. Proof.

(5.44)

By (5.37) and (2.12) χλ (ρ, 1n−r ) dλ X χλ (ρ, 1n−r )dλ

Ep,n p]ρ = Ep,n n↓r =

n↓r n!

λ∈Pn

n↓r X = χλ (ρ, 1n−r )χλ (e) n! λ∈Pn  ↓r n , ρ = (1r ), = 0, otherwise. The proof is complete.



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To proceed, we shall prove a weak form of limit shape theorem. Theorem 5.4. Define Z

∞


uk Ψn,λ (u) − Ω(u) du,

Yn,k (λ) =

λ ∈ Pn .

−∞

Then for each k ≥ 0 P

Yn,k −→ 0,

n → ∞.

Proof.

Note by (5.24) and (5.25), Z ∞ Z
Yn,k (λ) = uk Ψn,λ (u) − |u| du − −∞

∞


uk Ω(u) − |u| du

−∞

 p˜ (λ)  2 k+2 = − p ˜ (Ω) . k+2 (k + 2)(k + 1) n(k+2)/2 Hence it suffices to prove for each k ≥ 2 p˜k (λ) P − p˜k (Ω) −→ 0, nk/2

n → ∞.

Equivalently, p˜k (λ) P −→ p˜k (Ω), n → ∞. nk/2 In turn, this will be done by checking as n → ∞ Ep,n p˜k (λ) −→ p˜k (Ω) nk/2

(5.45)

Ep,n p˜2k (λ) −→ p˜2k (Ω). nk

(5.46)

and

Expand p˜k in the basis of p]ρ p˜k (λ) =

X

(˜ pk )ρ p]ρ (λ),

(5.47)

ρ

where (˜ pk )ρ denotes the structure coefficient. Note by Lemmas 5.11 and 5.14 deg∞ (˜ pk ) = k so that the summation in (5.47) is over all ρ with kρk∞ ≤ k. Then it follows from (5.44) X Ep,n p˜k = (˜ pk )ρ Ep,n p]ρ kρk∞ ≤k

=

X 2r≤k

(˜ pk )1r n↓r .

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In addition, according to (5.36) and (5.39), if k = 2r (˜ pk )1r =

k ↓r . r!

(5.48)

Hence for k = 2m, (2m)! Ep,n p˜2m → , m n (m!)2 while for k = 2m + 1, Ep,n p˜k → 0. nk/2 This proves (5.45). Analogously, we can prove (5.46). Indeed, it follows from (5.47) X p˜2k (λ) = (˜ pk )ρ (˜ pk )σ p]ρ (λ)p]σ (λ) kρk∞ ≤k,kσk∞ ≤k

which in turn implies X

Ep,n p˜2k =

(˜ pk )ρ (˜ pk )σ Ep,n p]ρ p]σ .

kρk∞ ≤k,kσk∞ ≤k

Also, by (5.40), Ep,n p]ρ p]σ = Ep,n p]ρ∪σ + Ep,n lower terms where lower terms means a linear combination of p]τ ’s with kτ k∞ < kρk∞ + kσk∞ . Again by (5.48),  ↓r n , ρ ∪ σ = (1r ), ] Ep,n pρ∪σ = 0, otherwise. In summary, we have X Ep,n p˜2k =

(˜ pk )1r1 (˜ pk )1r2 n↓(r1 +r2 ) + Ep,n lower terms

2r1 ≤k,2r2 ≤k

In particular, we have Ep,n p˜22m =

 (2m)↓m 2 m!

n↓2m + O n2m−1




and
Ep,n p˜22m+1 = O n2m . Therefore it follows Ep,n p˜2k → nk The proof is now complete.

(

(2m)↓m m!

0,

2

,

k = 2m, k = 2m + 1. 

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Remark 5.1. One can derive the strong form of limit shape theorem, i.e., Theorem 5.2, using the equivalence between weak topology and uniform topology. The interested reader is referred to Theorem 5.5 of Ivanov and Olshanski (2002). Theorem 5.5. Define Zn,k (λ) =

p]k (λ) , nk/2

λ ∈ Pn .

Then under (Pn , Pp,n ) as n → ∞
d √ k ξk , Zn,k , k ≥ 2 −→

 k≥2 .

Here ξk , k ≥ 2 is a sequence of standard normal random variables, the convergence holds in terms of finite dimensional distribution. Proof. For simplicity of notations, we will mainly focus on the 1dimensional case. Namely, we shall below prove for each k ≥ 2 d √ Zn,k −→ k ξk , n → ∞. Adapt the moment method. Fix l ≥ 1, we need to check Ep,n (ηk )l → Eξkl , n → ∞ √ where ηk = Zn,k / k. This is equivalent to proving Ep,n hl (ηk ) −→ Ehl (ξk ),

n→∞

where hl is a classical Hermite √ orthogonal polynomial of order l with respect −x2 /2 to the weight function e / 2π, see (3.3). Trivially, by the orthogonality property, Ehl (ξk ) = 0. So we shall only prove Ep,n hl (ηk ) → 0.

(5.49)

p]kl p]k = p]kl+1 + klp]kl−1 ∪1k + lower terms,

(5.50)

Note by (5.43),

where lower terms means a term with deg1 (·) < k(l + 1). Also, by definition n↓kl χλ (k l−1 ∪ 1k , 1n−kl ) dλ n↓kl ] = ↓k(l−1) pkl−1 (λ), λ ∈ Pn . n

p]kl−1 ∪1k (λ) =

(5.51)

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√ Inserting (5.51) back into (5.50) and dividing by ( knk/2 )l+1 , we get n↓kl

1 lower terms, (5.52) ηkl−1 + √ k/2 ( kn )l+1 Recall that hl (x) is characterized by the recurrence relation ηkl ηk = ηkl+1 + l

n↓k(l−1) nk

xhl (x) = hl+1 (x) + lhl−1 (x) together with the initial data h0 = 1 and h1 (x) = x. Hence we make repeatedly use of (5.52) to yield 1 ηkl = hl (ηk ) + √ lower terms ( knk/2 )l where lower terms means a term with deg1 (·) < kl. In particular, it follows 1 lower terms. hl (ηk ) = ηkl + √ ( knk/2 )l Thus by (5.44) 1 Ep,n hl (ηk ) = Ep,n ηkl + √ Ep,n lower terms ( knk/2 )l = O(n−1/2 ), which proves (5.49) as desired. To treat m-dimensional case, we need to prove for any positive integers l2 , · · · , lm m m Y Y lk Ep,n ηk → Eξklk = 0, n → ∞. k=2

k=2

Equivalently, for hl2 , · · · , hlm m m Y Y Ep,n hlk (ηk ) → Ehlk (ξk ), k=2

n → ∞.

k=2

The details are left to the reader.



Now we are ready to prove Theorem 5.3. Define q1 = 0 and for any k ≥ 2
1 p˜k+1 (λ) − n(k+1)/2 p˜k+1 (Ω) , λ ∈ Pn . (5.53) qk (λ) = (k + 1)nk/2 Lemma 5.17. For any k ≥ 2, [(k−1)/2]   ] X k pk−2j (λ) qk (λ) = j n(k−2j)/2 j=0 +

1 nk/2

lower terms with deg1 (·) ≤ k − 1.

(5.54)

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Proof. Using Lemma 5.11, one can express p˜k+1 as a polynomial p]1 , p]2 , · · · up to terms of lower weight. In particular, we have [(k−1)/2]

X

p˜k+1 (λ) = (k + 1)

nj

j=0

  k ] p (λ) + n(k+1)/2 p˜k+1 (Ω) j k−2j

+lower terms with deg1 (·) ≤ k − 1.

(5.55)

A nontrvial point in (5.55) is the switch between two weight filtrations. Its proof is left to the reader. See also Proposition 7.3 of Ivanov and Olshanski (2002) for details.  Inverting (5.54) easily gives Lemma 5.18. For any k ≥ 2, p]k (λ) = nk/2

[(k−1)/2]

X

(−1)j

j=0

1

+

nk/2

  k−j k qk−2j (λ), j k−j

lower terms with deg1 (·) ≤ k − 1.

(5.56)

Proof. Recall the following combinatorial inversion formula due to Riordan (1968): assume that α0 , α1 , · · · ; β0 , β1 , · · · are two families of formal variables, then [k/2] 

αk =

X j=0

 k βk−2j , j

k = 0, 1, · · ·

m [k/2]

βk =

X

(−1)j

j=0

  k−j k αk−2j , j k−j

k = 0, 1, · · · .

Set α0 = α1 = 0, αk = qk (λ), k ≥ 2; β0 = β1 = 0, βk = p]k (λ)/nk/2 , k ≥ 2. If we neglect the lower terms in (5.54), then it obviously follows p]k (λ) = nk/2

[(k−1)/2]

X j=0

(−1)j

  k−j k qk−2j . j k−j

The appearance of remainder terms affect only similar remainder terms in the reverse relations. 

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Proof of Theorem 5.3. By (5.18) and integrating term by term, we obtain [k/2]

Xn,k (λ) =

X

(−1)j

j=0 [k/2]

=

X

(−1)j

j=0



k−j j



k−j j

Z

∞


√ √ uk−2j Ψλ ( nu) − nΩ(u) du

−∞

h Z

∞


√ √ uk−2j Ψλ ( nu) − n|u| du

−∞

Z

∞

−


i √ uk−2j n (Ω(u) − |u|) du

−∞ [k/2]

=

X j=0





2 k−j (k + 2 − 2j)(k + 1 − 2j) j  p˜  √ k+2−2j (λ) · (k+1−2j)/2 − n˜ pk+2−2j (Ω) . n

(−1)j

By the definition of (5.53) and noting     1 k−j k+1−j 1 = , j j k + 2 − 2j k+1−j we further get [k/2]

Xn,k (λ) =

X j=0

  k+1−j 2 qk+1−2j (λ). (−1) j k+1−j j

By (5.56), Xn,k (λ) =

2p]k+1 (λ) (k + 1)n(k+1)/2 1 + (k+1)/2 lower terms with deg1 (·) ≤ k. n

Since the remainder terms of negative degree do not affect the asymptotics, then we can use Theorem 5.5 to conclude the proof.  To conclude this section, we remark that Theorem 5.5 for character ratios is of independent interest. Another elegant approach was suggested by Hora (1998), in which a central limit theorem was established for adjacency operators on the infinite symmetric group. Still, Fulman (2005, 2006) developed the Stein method and martingale approach to prove asymptotic normality for character ratios.

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5.3

237

Fluctuations in the bulk

In this section we shall turn to the study of fluctuations of a typical Plancherel Young diagrams around their limit shape in the bulk of the partition spectrum. Here we use the term spectrum informally by analogy with the GUE, to refer to the variety of partition’s terms λi ∈ λ. Recall ψλ (x) and ω(x) introduced in Section 5.1 and define the random process √
√ Ξn (x) = ψλ n x − n ω(x), x ≥ 0, λ ∈ Pn . According to the Corollary 5.1, it follows under (Pn , Pp,n ) 1 P √ sup |Ξn (x)| −→ 0, n → ∞. n x≥0 This is a kind of weak law of large numbers. The following theorem describes the second order fluctuation of Ξn at each fixed 0 < x < 2. Theorem 5.6. Under (Pn , Pp,n )
Ξn (x) d √ −→ N 0, %2 (x) , log n

1 2π

for each 0 < x < 2, where %−2 (x) =

1 π

n→∞

(5.57)

arccos |ω(x)−x| . 2

The asymptotics of finite dimensional distributions of the random process Ξn (x) reads as follows. Theorem 5.7. Assume 0 < x1 < · · · < xm < 2, then under (Pn , Pp,n ),
d 1 √ Ξn (xi ), 1 ≤ i ≤ m −→ (ξi , 1 ≤ i ≤ m), n → ∞ 1 2π log n where ξi , 1 ≤ i ≤ m are independent normal random variables. Remark 5.2. (i) The work of this section, in particular Theorems 5.6 and 5.7, are motivated by Gustavsson (2005), in which he investigated the Gaussian fluctuation of eigenvalues in the GUE. There is a surprising similarity between Plancherel random partitions and GUE from the viewpoint of asymptotics, though no direct link exists between two finite models. (ii) Compared with the uniform random partitions, the normalizing con√ stant log n is much smaller than n1/4 , see Theorem 4.5. This means that Plancherel Young diagrams concentrated more stably around their limit shape. (iii) The random process Ξn (x) weakly converges to a Gaussian white noise in the finite dimensional sense. Thus one cannot expect a usual process convergence for Ξn in the space of continuous functions on [0, 2].

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(iv) As we will see, if xn → x where 0 < x < 2, then (5.57) still holds for Ξn (xn ), namely
Ξn (xn ) d √ −→ N 0, %2 (x) , log n

1 2π

n → ∞.

(5.58)

(v) Since %(x) → ∞ as x → 0 or 2, the normal fluctuation is no longer true at either 0 or 2. In fact, it was proved √ ψλ (0) − 2 n d −→ F2 , n → ∞ n1/6 where F2 is Tracy-Widom law. It is instructive to reformulate Theorems 5.6 and 5.7 in the rotated coordinates u and v. Recall Ψλ (u) and Ω(u) are rotated versions of ψλ (x) and ω(x). Define √ √ (5.59) Υn (u) = Ψλ ( n u) − n Ω(u), −∞ < u < ∞. We can restate Theorem 5.6 in the following elegant version, whereby— quite surprisingly—the normalization does not depend on the location in the spectrum. Theorem 5.8. Under (Pn , Pp,n ) Υn (u) d √ −→ N (0, 1), log n

1 π

n→∞

for −2 < u < 2. Proof. Fix −2 < u < 2 and assume λ ∈ Pn . A key step is to express the √ √ error Ψλ ( n u) − nΩ(u) in terms of ψλ and ω. Let local extrema consist of two interlacing sequences of points u ˇ1 < u ˆ1 < u ˇ2 < u ˆ2 < · · · < u ˇm < u ˆm < u ˇm+1 , where u ˇ ’s are the local minima and u ˆi ’s are the local maxima of the function √ i Ψλ ( n ·). Without loss of generality, we may and will assume that u is √ between u ˆk and u ˇk+1 for some 1 ≤ k ≤ m. Denote by n(xn , xn ) and √ √ √ √ n(x∗n , x∗n ) the projections of ( n u, Ψλ ( n u)) and n(u, Ω(u)) in the line u = v, respectively. Then we obviously have √ √ √ Ψλ ( n u) − n Ω(u) = 2 n(xn − x∗n ). According to Theorem 5.2, it follows P

xn − x∗n −→ 0,

n→∞

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and so 1 P xn −→ x := (Ω(u) + u). 2 √ √ On the other hand, if we let 2hn be the distance between n(xn , xn ) and √ √ ( n u, Ψλ ( n u)), then we have √ √ n(xn − x∗n ) = hn − n ω(x∗n ) √ √ (5.60) = hn − n ω(xn ) + n(ω(xn ) − ω(x∗n )). Now using the Taylor expansion for the function ω at xn and solving equation (5.60), we obtain √ √ hn − n ω(xn ) , n(xn − x∗n ) = 1 − ω 0 (˜ xn ) where x ˜n is between xn and x∗n . P Since xn , x∗n −→ x ∈ (0, 2), then it holds 1 ω(x) − x 1 1 P = arccos . −→ 1 − ω 0 (˜ xn ) 1 − ω 0 (x) π 2 √ Hence it suffices to prove hn − n ω(xn ) after properly scaled converges in distribution. Observe that √ √ ψλ ( n xn + 1) ≤ hn ≤ ψλ ( n xn ) since u is between u ˆk and u ˇk+1 . P Note xn −→ x ∈ (0, 2). Then for each subsequence {n0 } of integers there exists a further subsequence {n00 } ⊆ {n0 } such that xn00 → x a.e. Thus by (5.58) it holds √ √
ψλ ( n00 xn00 ) − n00 ω(xn00 ) d √ −→ N 0, %2 (x) . 1 00 2π log n By a standard subsequence argument, it holds √ √
ψλ ( n xn ) − n ω(xn ) d √ −→ N 0, %2 (x) . 1 2π log n
√ Similarly, since n ω(xn + √1n ) − ω(xn ) = Op (1), √ √
ψλ ( n xn + 1) − n ω(xn ) d √ −→ N 0, %2 (x) . 1 2π log n In combination, we have

√
n ω(xn ) d √ −→ N 0, %2 (x) . 1 2π log n

hn − We conclude the proof.



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Let us now turn to the proof of Theorems 5.6 and 5.7. A basic strategy is to adapt the conditioning argument—the Poissonization and dePoissonization techniques. Define the Poissonized Plancherel measure Qp,θ on P as follows:  d 2 λ Qp,θ (λ) = e−θ θ|λ| |λ|! ∞ X θn (5.61) = e−θ Pp,n (λ)1(λ∈Pn ) , λ ∈ P n! n=0 where θ > 0 is a model parameter. This is a mixture of a Poisson random variable with mean θ and the Plancherel measures. Let Ξθ (x) be given by (5.57) with n replaced by θ, namely √
√ (5.62) Ξθ (x) = ψλ θ x − θ ω(x), x ≥ 0, λ ∈ P. Theorem 5.9. Under (P, Qp,θ ),
Ξθ (x) d √ −→ N 0, %2 (x) , 1 log θ 2π for each 0 < x < 2.

θ→∞

Theorem 5.10. Assume 0 < x1 < · · · < xm < 2. Under (P, Qp,θ ),
d 1 √ Ξθ (xi ), 1 ≤ i ≤ m −→ (ξi , 1 ≤ i ≤ m), θ → ∞ 1 log θ 2π where ξi , 1 ≤ i ≤ m are independent normal random variables. Before giving the proof of Theorems 5.9 and 5.10, let us prove Theorems 5.6 and 5.7 with the help of the de-Poissonization technique. Lemma 5.19. For 0 < α < 1, define p θn± = n ± n(log n)α . Then uniformly in x ≥ 0 and z ∈ R, Qp,θn+ (λ ∈ P : ψλ (x) ≤ z) − εn ≤ Pp,n (λ ∈ Pn : ψλ (x) ≤ z) ≤ Qp,θn− (λ ∈ P : ψλ (x) ≤ z) + εn (5.63) where εn → 0 as n → ∞. Proof. We need only give the proof of the lower bound, since the upper bound is similar. Let X be a Poisson random variable with mean θn+ . Then we have EX = θn+ , V arX = θn+ and the following tail estimate p
εn := P |X − θn+ | > n(log n)α
= O log−α n .

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It follows by (5.61) Qp,X (λ) = EPp,X (λ),

λ∈P

and so for any event A ⊆ P Qp,X (A) = EPp,X (A),

A ⊆ P.

Note EPp,X (A) = EPp,X (A)1(X
(5.64)

In addition, set A = {λ ∈ P : ψλ (x) ≤ z}. Then using a similar argument to that of Lemma 2.4 of Johansson (1998b), A is a monotonoic event under (Pp,n , n ≥ 1), namely Pp,n+1 (A) ≤ Pp,n (A). Hence it follows EPp,X (A)1(X≥n) ≤ Pp,n (A). Combining (5.64) and (5.65) together implies the lower bound. Proof of Theorem 5.6. Set √ nx + xn = p , θn+

x− n

(5.65) 

√ nx =p θn−

− where θn± are as in Lemma 5.19. Trivially, x+ n , xn → x. Also, since 0 < α < 1, then it follows p √ θn± − n √ → 0, n → ∞. log n Note p p p √ √ θn± ω(x± log θn± ψλ ( n x) − n ω(x) ψλ ( θn± x± n)− n) √ p √ · = log n log n log θn± p √ ) − nω(x) θn± ω(x± √n (5.66) + log n

and as n → ∞ p √ θn± ω(x± n ω(x) n)− √ → 0, log n

p

log θn± √ → 1. log n
On the other hand, by Theorem 5.9, under P, Qp,θn± p p
ψλ ( θn± x± θn± ω(x± n)− n) d p −→ N 0, %2 (x) . ± 1 log θn 2π

(5.67)

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Hence it follows from (5.66) and (5.67) that under P, Qp,θn± √ √
ψλ ( n x) − n ω(x) d √ −→ N 0, %2 (x) . 1 log n 2π




Taking Lemma 5.19 into account, we now conclude the proof.  To prove Theorem 5.9, we will again apply the Costin-LebowitzSoshnikov theorem for determinantal point processes, see Theorem 3.7. To do this, we need the following lemma due to Borodin, Okounkov and Olshanski (2000), in which they proved the Tracy-Widom law for the largest parts. Set for λ = (λ1 , λ2 , · · · , λl ) ∈ P X (λ) = {λi − i, 1 ≤ i ≤ l}.

(5.68)

For k = 1, 2, · · · , the k-point correlation function ρk is defined by
ρk (x1 , x2 , · · · , xk ) = Qp,θ λ ∈ P : x1 , x2 , · · · , xk ∈ X (λ) , where x1 , x2 , · · · , xk are distinct integers. Lemma 5.20. ρk has a determinantal structure as follows:
ρk (x1 , x2 , · · · , xk ) = det Kθ (xi , xj ) 1≤i,j≤k , with the kernel Kθ of the form (√ J J −J J x 6= y θ x y+1x−yx+1 y ,
Kθ (x, y) = √ 0 Jx , x = y θ Jx0 Jx+1 − Jx+1 √
where Jm ≡ Jm 2 θ is the Bessel function of integral order m.

(5.69)

(5.70)

We will postpone the proof to Section 5.5. Now we are ready to give Proof of Theorem 5.9. Fix 0 < x < 2. It suffices to show that for any z∈R   %(x) p Qp,θ Ξθ (x) ≤ log θ z → Φ(z), θ → ∞, (5.71) 2π where Φ denotes the standard normal distribution function. Equivalently, it suffices to show that for any z ∈ R √ √
Qp,θ ψλ ( θ x) − d θ xe ≤ aθ → Φ(z), where √ aθ := aθ (x, z) =

θ(ω(x) − x) +

%(x) p log θ z. 2π

(5.72)

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Consider the semi-infinite interval Iθ := [aθ , ∞) and let Nθ be the number of points of λi − i ∈ X (λ) contained in Iθ . Using that the sequence λi − i is strictly decreasing, it is easy to see that relation (5.71) is reduced to √
Qp,θ Nθ ≤ d θ xe → Φ(z). (5.73) In this situation, one can apply the Costin-Lebowitz-Soshnikov theorem as in Section 3.4. Since the X (λ) is by Lemma 5.20 of determinantal, then N − Ep,θ Nθ d pθ −→ N (0, 1) V arp,θ (Nθ )

(5.74)

provided that V arp,θ (Nθ ) → ∞ as θ → ∞. In order to derive (5.73) from (5.74), we need some basic asymptotic estimation of the first two moments of the random variable Nθ . This will be explicitly given in Lemma 5.21 below. Module Lemmas 5.20 and 5.21, the proof is complete.  Lemma 5.21. Fix 0 < x < 2, z ∈ R and let Iθ = [aθ , ∞) be as in (5.72). Then as θ → ∞ √ z p log θ + O(1) Ep,θ Nθ = θ x − 2π and log θ (1 + o(1)). V arp,θ (Nθ ) = 4π 2 The proof of Lemma 5.21 essentially involves the computation of moments of the number of points lying in an interval for a discrete determinantal point process. Let k ∈ Z be a integer (possibly depending on the model parameter θ), and let Nk be the number of points of X (λ) lying in [k, ∞). Then we have by Lemma 5.20 Ep,θ Nk =

=

=

=

∞ X j=k ∞ X


Pp,θ λ ∈ P : j ∈ X (λ)

Kθ (j, j)

j=k ∞ ∞ X X j=k s=1 ∞ X

√
2 Jk+s 2 θ

√
2 (m − k)Jm 2 θ

m=k

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and V arp,θ (Nk ) =

∞ X

Kθ (i, i) −

i=k

=

∞ X

Kθ (i, j)2

i,j=k

∞ k−1 X X

Kθ (i, j)2

i=k j=−∞

=

∞ k−1 X X i=k

√

√ √ √
2 √ θ Ji (2 θ)Jj+1 (2 θ) − Ji+1 (2 θ)Jj (2 θ) . i−j j=−∞

To figure out these infinite sums, we need some precise asymptotic behaviours of Bessel functions in the whole real line. They behave rather differently in three regions so that we must take more care in treating the sums over two critical values. The lengthy computation will appear in the forthcoming paper, see Bogachev and Su (2015).

5.4

Berry-Esseen bounds for character ratios

This section is particularly devoted to the study of convergence rate of random character ratios. Define for n ≥ 2 (n − 1)χλ (1n−2 , 2) √ Wn (λ) = , λ ∈ Pn . 2dλ √ Note by (5.32) Wn (λ) = p]2 (λ)/ 2n. It was proved in Section 5.2 d

Wn −→ N (0, 1),

n→∞

using the moment method. Namely,
sup Pp,n Wn (λ) ≤ x − Φ(x) → 0,

n → ∞.

(5.75)

−∞<x<∞

Having (5.75), it is natural to ask how fast it converges. This was first studied by Fulman. In fact, in a series of papers, he developed a Stein method and martingale approach to the study of the Plancherel measure. In particular, Fulman (2005, 2006) obtained a speed of n−s for any 0 < s < 1/2 and conjectured the correct speed is n−1/2 . Following this, Shao and Su (2006) confirmed the conjecture to get the optimal rate. The main result reads as follows. Theorem 5.11. sup −∞<x<∞


Pp,n λ ∈ Pn : Wn (λ) ≤ x − Φ(x) = O(n−1/2 ).

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The basic strategy of the proof is to construct a Wn0 (λ) such that
0 Wn (λ), Wn (λ) is
an exchangeable pair and to apply the Stein method to Wn (λ), Wn0 (λ) . Let us begin with a Bratelli graph, namely an oriented graded graph G = (V, E). Here the vertex set V = P = ∪∞ n=0 Pn and there is an oriented edge from λ ∈ Pn to Λ ∈ Pn+1 if Λ can be obtained from λ by adding one square, denoted by λ % Λ, see Figure 5.4 below.

Fig. 5.4

Bratelli graph

Lemma 5.22. The Plancherel measure is coherent in G, namely Pp,n (λ) =

X dλ Pp,n+1 (Λ). dΛ

Λ:λ%Λ

Proof.

According to the hook formula (4.70), it suffices to prove X Hλ = 1. HΛ

Λ:λ%Λ

Let us compute the quotient Hλ /HΛ . Assume that the new square is located in the rth row and sth column of the diagram Λ. Since the squares outside the rth row or sth column have equal hook lengths in the diagrams λ and Λ, we have by the hook formula, s−1 Y hjs (λ) Y hri (λ) r−1 Hλ = , HΛ h (λ) + 1 j=1 hjs (λ) + 1 i=1 ri

where h (λ) denotes the hook length of  in λ.

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Next we want to express the quotient in terms of local extrema. Let λ have interlacing local extrema x1 < y1 < x2 < y2 < · · · < xq < yq < xq+1 and suppose the square that distinguishes Λ from λ is attached to the minimum xk of λ, see Figure 5.5.

Fig. 5.5

λ%Λ

Then it follows s−1 Y i=1

k−1 Y xk − ym hri (λ) = hri (λ) + 1 m=1 xk − xm

and r−1 Y j=1

hjs (λ) = hjs (λ) + 1

Thus we rewrite k−1 Y xk − ym Hλ = HΛ x − xm m=1 k

q+1 Y m=k+1 q+1 Y

m=k+1

xk − ym−1 . xk − xm

xk − ym−1 =: ak . xk − xm

It remains to check q+1 X

ak = 1.

(5.76)

k=1

To do this, note that these numbers coincides with the coefficients of the partial fraction expansion Qq q+1 X ak i=1 (u − yi ) = . Qq+1 u − xk i=1 (u − xi ) k=1 Multiplying both sides by u and letting u → ∞ yields (5.76). 

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(n)

(n)

To construct an exchange pair, we introduce a Markov chain X0 , X1 , (n) · · · , Xk , · · · with state space Pn and transition probability (n)
(n) pn (λ, µ) := P X1 = µ X0 = λ dµ = ]P (λ, µ), (5.77) (n + 1)dλ  where P (λ, µ) = τ ∈ Pn−1 , τ % λ, τ % µ . Lemma 5.23. (i) pn (λ, µ) is a well defined transition probability: X pn (λ, µ) = 1.

(5.78)

µ∈Pn

(ii) Pp,n is a stationary distribution of the Markov chain X (n) , namely X Pp,n (µ) = Pp,n (λ)pn (λ, µ). λ∈Pn

(iii) The Markov chain X

(n)

is , namely for any λ, µ ∈ Pn

Pp,n (λ)pn (λ, µ) = Pp,n (µ)pn (µ, λ). Proof. Note the following formula (see the note following the proof of Lemma 3.6 in Fulman (2005))
1 X ]P (λ, µ) = χµ (π)χλ (π) r1 (π) + 1 , (5.79) n! π∈Sn

where r1 (π) is the number of fixed points in π. Hence it follows from (5.77) X pn (λ, µ) µ∈Pn


1 X dµ χµ (π)χλ (π) r1 (π) + 1 (n + 1)dλ n! π∈Sn µ∈Pn  X 1 X
1 dµ χµ (π) χλ (π) r1 (π) + 1 . = (n + 1)dλ n!

=

X

π∈Sn

(5.80)

µ∈Pn

By (2.11),  1 X 1, dµ χµ (π) = 0, n! µ∈Pn

π = 1n , π 6= 1n .

(5.81)

Inserting into (5.80) easily yields (5.78), as desired. (ii) is a direct consequence of Lemma 5.22, while (iii) follows from (5.77) and the Frobenius formula. 

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Lemma 5.24. Given a λ ∈ Pn ,  EWn0 (λ) = 1 −

2  Wn (λ). n+1

(5.82)

Consequently, Ep,n Wn = 0. Proof.

(5.83)

By definition,
(n) (n) EWn0 (λ) = EWn X1 X0 = λ X = Wn (µ)pn (λ, µ) µ∈Pn

=

X
n−1 √ χµ 1n−2 , 2 ]P (λ, µ). (n + 1) 2dλ µ∈Pn

(5.84)

Substituting (5.79) and noting (2.12), then (5.84) becomes  X 1 X

n−1 √ χµ 1n−2 , 2 χµ (π) χλ (π) r1 (π) + 1 (n + 1) 2dλ π∈Sn n! µ∈Pn  2  = 1− Wn (λ). n+1 This completes the proof of (5.82). To see (5.83), note
Ep,n Wn = Ep,n Wn0 = Ep,n EWn0 (λ)  2  Ep,n Wn . = 1− n+1 The conclusion holds.



Lemma 5.25. E

Proof.


2 Wn0 (λ)


1 2(n − 1)(n − 2)2 χλ 1n−3 , 3 = 1− + · n n(n + 1) dλ
2 χ 1n−4 , 22 (n − 1)(n − 2)(n − 3) λ · . + 2n(n + 1) dλ

(5.85)

Similarly to (5.84), it follows X E(Wn0 (λ))2 = Wn2 (µ)pn (λ, µ) µ∈Pn


(n − 1)2 X χ2µ 1n−2 , 2 = ]P (λ, µ). 2(n + 1)dλ dµ µ∈Pn

(5.86)

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Substituting (5.79), (5.86) becomes

(n − 1)2 X  1 X χ2µ 1n−2 , 2 χµ (π) χλ (π) r1 (π) + 1 . 2(n + 1)dλ n! dµ π∈Sn

µ∈Pn

To proceed, we need the following equation (see Exercise 6.76 of Stanley (1999))
 −2 n 1 X χ2µ 1n−2 , 2 ](π), (5.87) χµ (π) = 2 n! dµ µ∈Pn

where ](π) is the number of pairs
of (σ, τ ) such that σ and τ come from the same conjugacy class 1n−2 , 2 and σ ◦ τ = π. See also Lemma 3.4 of Fulman (2005). Note σ ◦
τ = π can
assume
only values in three distinct conjugacy classes: 1n , 1n−3 , 3 , 1n−4 , 22 , and ](π), χλ (π) and r1 (π) are all class functions. It is easy to see  
n ] 1n = 2  
n ] 1n−3 , 3 = 2(n − 2) 2   
n − 2 n . ] 1n−4 , 22 = 2 2 In combination, we easily get the desired conclusion (5.85).



As a direct consequence, we obtain the following Corollary 5.3.
1 V arp,n (Wn ) = V arp,n Wn0 = 1 − . n The last lemma we need is to control the difference between Wn (λ) and Wn0 (λ). Lemma 5.26. Let ∆n (λ) = Wn (λ) − Wn0 (λ), then √  √ 4e 2  ≤ 2e−2e n . Pp,n |∆n (λ)| ≥ √ n Consequently, Ep,n |∆n (λ)|2 1(|∆n (λ)|≥4e√2/√n) = O(n−1/2 ).

(5.88)

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Proof.

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Recall the Frobenius formula √    0  2 X  λi λi Wn (λ) = − . n i 2 2 (n)

(n)

Given X0 = λ, X1 = µ only when µ is obtained from λ by moving a box from row i and column j to row s and column t. Then ∆n (λ) = Wn (µ) − Wn (λ) √
2 λi + λ0t − λs − λ0j . = n Hence we have

√ 2 2 max{λ1 , λ01 }. |∆n (λ)| ≤ n According to Lemma 5.1, we have (5.88), as desired. The proof is now complete.  Having the preceding preparation, we are now ready to prove Theorem 5.11. The proof is based on the following refinement of Stein’s result for exchangeable pairs. Theorem 5.12. Let (W, W 0 ) be an exchangeable pair of real-valued random variables such that E(W 0 |W ) = (1 − τ )W,

with 0 < τ < 1. Assume E(W 2 ) ≤ 1. Then for any a > 0, r  2 a3 1 sup |P (W ≤ x) − Φ(x)| ≤ E 1 − E(∆2 |W ) + 2τ τ −∞<x<∞ +2a + E∆2 1(|∆|>a) . Proof.

See Theorem 2.1 of Shao and Su (2006).



Proof of Theorem 5.11. This √ √ 5.12
is a direct application of Theorem to exchangeable pairs Wn , Wn0 . Set τn = 2/(n + 1), an = 4e 2/ n and ∆n = Wn − Wn0 . In view of (5.88), we need only prove 
2 1 E ∆2n Wn = O(n−1 ). Ep,n 1 − 2τn In fact, a simple algebra yields 
2 3n2 − 5n + 6 1 E ∆2n Wn . = Ep,n 1 − 2τn 4n3

(5.89)

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To see this, note 
2
1 1 Ep,n 1 − E ∆2n Wn = 1 − Ep,n E ∆2n Wn 2τn τn

2 1 + 2 Ep,n E ∆2n Wn . 4τn By Lemma 5.24, we have

E ∆2n Wn = E Wn2 + (Wn0 )2 − 2Wn Wn0 Wn  4 
= − 1 Wn2 + E (Wn0 )2 Wn , n+1 and so by Corollary 5.3,
1 4  1− . Ep,n E ∆2n Wn = n+1 n Again, by Lemma 5.25,
E ∆2n Wn = A + B + C + D where 1 , n 2(n − 1)(n − 2)2 χλ (1n−3 , 3) · , B= n(n + 1) dλ (n − 1)(n − 2)(n − 3)2 χλ (1n−4 , 22 ) C= · , 2n(n + 1) dλ  4  (n − 1)2 χ2 (1n−2 , 2) D= −1 · λ 2 . n+1 2 dλ A = 1−

What we next need is to compute explicitly Ep,n (A + B + C + D)2 . We record some data as follows. Ep,n AB = Ep,n AC = 0; Ep,n AD =

 (n − 1)2  4 −1 ; 2 n n+1



χλ 1n−3 , 3 χλ 1n−4 , 22 (n − 1)2 (n − 2)3 (n − 3)2 Ep,n BC = Ep,n · n2 (n + 1)2 dλ dλ 2 3 2 X
(n − 1) (n − 2) (n − 3) 1 = χλ 1n−3 , 3 χλ (1n−4 , 22 ) n2 (n + 1)2 n! λ∈Pn

= 0;

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χλ 1n−3 , 3 χ2λ 1n−2 , 2 (n − 1)3 (n − 2)2 (n − 3) Ep,n BD = − Ep,n · n(n + 1)2 dλ d2
2 λn−2
n−3 3 2 , 3 χλ 1 ,2 (n − 1) (n − 2) (n − 3) 1 X χλ 1 =− 2 n(n + 1) n! dλ λ∈Pn

2

=−

12(n − 1)(n − 2) (n − 3) ; n3 (n + 1)2



χλ 1n−4 , 22 χ2λ 1n−2 , 2 (n − 1)3 (n − 2)(n − 3)3 Ep,n · Ep,n CD = − 4n(n + 1)2 dλ d2
2 λn−2
n−4 2 , 2 χλ 1 ,2 (n − 1)3 (n − 2)(n − 3)3 1 X χλ 1 =− 2 4n(n + 1) n! dλ λ∈Pn

3

=−

2(n − 1)(n − 2)(n − 3) ; n3 (n + 1)2 χ2λ (1n−3 , 3) 4(n − 1)2 (n − 2)4 E p,n n2 (n + 1)2 d2λ 2 4 X 4(n − 1) (n − 2) 1 = χ2λ (1n−3 , 3) n2 (n + 1)2 n!

Ep,n B 2 =

λ∈Pn

12(n − 1)(n − 2)3 ; = n3 (n + 1)2 (n − 1)2 (n − 2)2 (n − 3)4 χ2 (1n−4 , 22 ) Ep,n λ 2 2 4n (n + 1) d2λ (n − 1)2 (n − 2)2 (n − 3)4 1 X 2 n−4 2 = χλ (1 ,2 ) 4n2 (n + 1)2 n!

Ep,n C 2 =

λ∈Pn

3

=

2(n − 1)(n − 2)(n − 3) ; n3 (n + 1)2


χ4λ 1n−2 , 2 (n − 1)4 (n − 3)2 Ep,n Ep,n D = 4(n + 1)2 d4λ
(n − 1)4 (n − 3)2 1 X χ4λ 1n−2 , 2 = 4(n + 1)2 n! d2λ λ∈Pn    2(n − 1)(n − 3)2  n + 4(n − 3) . 3 = n3 (n + 1)2 2 2

In combination, we obtain (5.89). The proof is now complete.



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Determinantal structure

The goal of this section is to provide a self-contained proof of Lemma 5.20 following the line of Borodin, Okounkov and Olshanski (2000). Let us first prove an equivalent form of Lemma 5.20. Let λ = (λ1 , λ2 , · · · , λl ) ∈ P. Recall the ordinary Fronebius coordinates are a ¯i = λi − i, ¯bi = λ0i − i, 1 ≤ i ≤ ` and the modified Frobenius coordinates are 1 1 ai = a ¯i + , bi = ¯bi + , 1 ≤ i ≤ `, 2 2 where ` is the length of the main diagonal in the Young diagram of λ. Define  F(λ) = ai , −bi , 1 ≤ i ≤ ` . (5.90) 1 This is a finite set of half integers, F(λ) ⊂ Z + 2 . Note F(λ) consists of equally many positive half integers and negative half integers. Interestingly, F(λ) have a nice determinantal structure under the Poissonized Plancherel measure. In particular, denote the k-point correlation function
%k (x1 , · · · , xk ) = Qp,θ λ ∈ P : x1 , · · · , xk ∈ F(λ) , where xi ∈ Z + 21 , 1 ≤ i ≤ k. Then we have Lemma 5.27.
%k (x1 , · · · , xk ) = det M (xi , xj ) k×k , where the kernel function  √  θ K+ (|x|,|y|) , xy > 0, |x|−|y| M (x, y) = √ K (|x|,|y|) −  θ xy < 0. |x|−|y| ,

(5.91)

Here K+ (x, y) = Jx− 21 Jy− 21 − Jx+ 21 Jy+ 21 , K− (x, y) = Jx− 21 Jy− 12 − Jx+ 21 Jy+ 21 . Its proof is based on the following three lemmas. The first one shows that the Poissonizaed Plancherel measure can be expressed by a determinant. Set ( 0, xy > 0, L(x, y) = 1 θ (|x|+|y|)/2 · xy < 0. x−y Γ(|x|+ 1 )Γ(|y|+ 1 ) , 2

2

Lemma 5.28. Qp,θ (λ) = where xi = ai ,

xi+` = −bi ,


det L(xi , xj ) 2`×2` det(1 + L) 1 ≤ i ≤ `.

,

(5.92)

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We remark that the numerator in (5.92) is a determinant of 2` × 2` matrix, while the denumerator det(1 + L) is interpreted as the X
det(1 + L) = det L(X) , X⊆Z+ 21

where det(L(X)) = 0 unless X consists of equally many positive and negative half integers. Proof.

Recall a classic determinant formula for dλ :   dλ 1 . = det |λ|! (¯ ai + ¯bj + 1)¯ ai !¯bj ! `×`

Thus letting  L(x1 , x`+1 ) L(x1 , x`+2 ) · · · L(x1 , x2` )  L(x2 , x`+1 ) L(x2 , x`+2 ) · · · L(x2 , x2` )    A= , .. .. .. ..   . . . . L(x` , x`+1 ) L(x` , x`+2 ) · · · L(x` , x2` ) 

we have    d 2 0 A λ = det . −A0 0 |λ|! It follows by definition of L
Qp,θ (λ) = Qp,θ {x1 , · · · , x2` }
= e−θ det L(xi , xj ) 2`×2` . As a direct consequence, X

eθ = eθ

Qp,θ (λ)

λ∈P

=

X


det L(X) = det(1 + L).

X

We now conclude the proof.



We shall below prove that the point process F(λ) is of determinantal. Let ML =

L . 1+L

Then we have Lemma 5.29. Given x1 , · · · , xk ∈ Z + 21 , %k (x1 , · · · , xk ) = det ML (xi , xj )


k×k

.

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Proof. Assume that g : Z + 12 → Z + 21 takes 0 except at finitely many points. According to Lemma 5.28, it follows Y X Y


1 + g(x) Qp,θ F(λ) = X Ep,θ 1 + g(x) = X⊆Z+ 21 x∈X

x∈F (λ)


det(L(X)) det(1 + L) X⊆Z+ 12 x∈X
X det (1 + g)L(X) = det(1 + L) X⊆Z+ 12

det 1 + L + gL = = det 1 + gML det(1 + L) X Y
g(x) det ML (X) , = =

X

Y

1 + g(x)

(5.93)

X⊆Z+ 21 x∈X

where in the last two equations we used the properties of Fredholm determinants. On the other hand, X Y Y


1 + g(x) = Ep,θ 1 + g(x) Qp,θ F(λ) = X X⊆Z+ 12 x∈X

x∈F (λ)

=

X

X

Y

g(x)Qp,θ F(λ) = X




X⊆Z+ 12 Y :Y ⊆X x∈Y

=

X

Y

=

Y

Qp,θ F(λ) = X




X:Y ⊆X

Y ⊆Z+ 21 x∈Y

X

X

g(x)

g(x)%(Y ).

(5.94)

Y ⊆Z+ 12 x∈Y

Thus comparing (5.93) with (5.94) yields %(X) = det ML (X) since g is arbitrary.




To conclude the proof of Lemma 5.27, we need only to determine ML above is exactly equal to M of (5.91). Lemma 5.30. ML (x, y) = M (x, y). √ Proof. Fix x, y ∈ Z + 12 and set z = θ. Note M and L are a function of z. We need to prove for all z ≥ 0, M + M L − L = 0.

(5.95)

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Obviously, (5.95) is true at z = 0. We shall below prove M˙ + M˙ L + M L˙ − L˙ = 0, ∂L ˙ where M˙ = ∂M ∂z and L = ∂z . It easily follows by definition ( 0, ˙ L= z |x|+|y|−1 sgn(x) Γ(|x|+ , 1 )Γ(|y|+ 1 ) 2

(5.96)

xy > 0, xy < 0.

2

To compute M˙ , we use the following formulas: x ∂ Jx (2z) = −2Jx+1 (2z) + Jx (2z) ∂z z x = 2Jx−1 (2z) − Jx (2z). z Then ( M˙ =

J|x|− 12 J|y|+ 21 + J|x|+ 12 J|y|− 21 , sgn(x)(J|x|− 12 J|y|− 21 − J|x|+ 12 J|y|+ 21 ),

xy > 0, xy < 0.

It remains to verify (5.96). To do this, recall the following identities: for any ν 6= 0, −1, −2, · · · and any z 6= 0 we have Γ(ν)Jν (2z) = z ν

∞ X

1 zm Jm (2z), m + ν m! m=0

Γ(ν)Jν−1 (2z) = z ν−1 − z ν

∞ X

1 zm Jm+1 (2z). m + ν m! m=0

Now the verification of (5.96) becomes a straightforward application, except for the occurrence of the singularity at negative integers ν. This singularity is resolved using the following identity due to Lommel sin πν . πz This concludes the proof of Lemma 5.30, and so Lemma 5.27. Jν (2z)J1−ν (2z) + J−ν (2z)Jν−1 (2z) =



Turn to the proof of Lemma 5.20. A key observation is the following link between X (λ) and F(λ) due to Frobenius (see (5.68) and (5.90)):  1 1  ∆ Z≤0 − . (5.97) F(λ) = X (λ) + 2 2 o n Given x1 , · · · , xk ∈ Z, denote X = x1 + 21 , · · · , xk + 12 . Divide X into   positive half integers and negative half integers: X+ = X ∩ Z≥0 + 21 and

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257

  X− = X ∩ Z≤0 − 12 . If X ⊆ X (λ) + 12 , then by (5.97 ), X+ ⊆ F(λ) and   there exists a finite subset S ⊆ Z≤0 − 21 \ X− such that S ⊆ F(λ). This implies  1 Qp,θ X ⊆ X (λ) + 2    1 \ X− : X+ ∪ S ⊆ F(λ) . (5.98) = Qp,θ ∃S ⊆ Z≤0 − 2 By exclusion-inclusion principle, the right hand side of (5.98) becomes   X (−1)|S| Qp,θ X+ ∪ S ⊆ F(λ)
S⊆ Z≤0 − 21 \X−

X

=

(−1)|S| %(X+ ∪ S)


S⊆ Z≤0 − 21 \X−

X

= S⊆

Z≤0 − 21


(−1)|S| det M (X+ ∪ S) ,

(5.99)




\X−

where in the last equation we used Lemma 5.27. Define a new as follows:  M (x, y), x ∈ Z≥0 + 12 ,    −M (x, y), x ∈ Z≥0 + 21 , y ∈ Z≤0 − 21 , M 4 (x, y) =  −M (x, y), x, y ∈ Z≤0 − 12 , x 6= y,   1 − M (x, x), x, y ∈ Z≤0 − 21 , x = y. Lemma 5.31. Given x, y ∈ Z,  1 1 M4 x + , y + = (x)(y)Kθ (x, y), 2 2
x+1 where (x) = sgn(x) . Proof.

It suffices to show  sgn(x)(x)(y)Kθ (x, y),  1  1 M x+ , y+ = Kθ (x, y),  2 2 1 − Kθ (x, y), Using the relation √
√
J−n 2 θ = (−1)n Jn 2 θ

x 6= y, x = y > 0, x = y < 0.

and the definition of M , one can easily verify the case x 6= y. Also, the claim remains valid for x = y > 0. It remains to consider the case x = y < 0. In this case, we have to show that  1 1 1 − M x + ,x + = J(x, x), x ∈ Z≤0 . 2 2

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Equivalently, 1 − J(k, k) = J(−k − 1, −k − 1),

k ∈ Z≥0 .

(5.100)

Note for any k ∈ Z, J(k, k) =

∞ X

(−1)

m=0

(2k + m + 2)↑m θk+m+1 , ((k + m + 1)!)2 m!

where we use the symbol (x)↑m = x(x + 1) · · · (x + m − 1). We need to show ∞ X (2k + m + 2)↑m θk+m+1 1− (−1)m ((k + m + 1)!)2 m! m=0 =

∞ X

(−1)l

nml=0

(−2k + l + 2)↑l θ−k+l . (Γ(−k + l + 1))2 l!

(5.101)

Examine the right hand side of (5.101). The terms with l = 0, 1, · · · , k − 1 vanish because then 1/Γ(−k + l + 1) = 0. The term with l = k is equal to 1. Next the terms with l = k + 1, · · · , 2k vanish because for these values of l, the expression (−2k+l)↑l vanishes. Finally, for l ≥ 2k+1, say l = 2k+1+m, (−1)l

(−2k + l)↑l θ−k+l θk+m+1 (m + 1)↑l m+1 = (−1) (Γ(−k + l + 1))2 l! ((k + m + 1)!)2 (2k + 1 + m)! (2k + m + 2)↑m θk+m+1 . = (−1)m+1 ((k + m + 1)!)2 m!

Thus we have proved (5.100).



Proof of Lemma 5.20 Fix X = (x1 , x2 , · · · , xk ) ⊆ Z. Then according to (5.27), (5.98) and (5.99), we have
ρk (x1 , · · · , xk ) = Qp,θ λ ∈ P : x1 , · · · , xk ∈ X (λ) X
= (−1)|S| det M (X+ ∪ S) . (5.102) S⊆(Z≤0 − 12 )\X−

To compute the alternating sum in (5.102), write   (X+ , X+ ) (X+ , S) , M (X+ ∪ S) =  (S, X+ )

(S, S)


where (X+ , X+ ) stands for the matrix M (xi + 21 , xj + 12 ) with xi + 1/2, xj + 1/2 ∈ X+ , the others are similar. Then by definition   (X+ , X+ ) (X+ , S) . M 4 (X+ ∪ S) =  −(S, X+ ) 1 − (S, S)

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259

A simple matrix determinant manipulation shows X

(−1)|S| det M (X+ ∪ S) = det M 4 (X+ ∪ S) . S⊆(Z≤0 − 21 )\X−

It follows in turn from Lemma 5.31 
1 1  det M 4 (xi + , xj + ) = det Kθ (xi , xj ) k×k . 2 2 k×k This concludes the proof.



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Bibliography

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Johansson, K. (1998b). The longest increasing subsequence in a random permutation and a unitary random matrix model, Math. Res. Lett. 5, 63-82. Johansson, K. (2001). Discrete orthogonal polynomials ensembles and the Plancherel measure, Ann. Math. 153, 259-296. Keating, J. P. and Snaith, N. C. (2000). Random matrix theory and ζ(1/2 + it), Commun. Math. Phys. 214, 57-89. Kerov, S. V. (1993). Gaussian limit for the Plancherel measure of the symmetric group, C. R. Acad. Sci. Paris, S´er. I Math. 316, 303-308. Kerov, S. V. and Olshanski, G. (1994). Polynomial functions on the set of Young diagrams, Comptes Rendus Acad. Sci. Paris S´er. I 319, 121-126. Killip, R. (2008). Gaussian fluctuations for β ensembles, Int. Math. Res. Not. 2008, 1-19. Killip, R. and Nenciu, I. (2004). Matrix models for circular ensembles, Int. Math. Res. Not. 2004, 2665-2701. Krasovsky, I. V. (2007). Correlations of the characteristic polynomials in the Gaussian unitary ensembles or a singular Hankel determinant, Duke Math. J. 139, 581-619. Ledoux, M. and Talagrand, M. (2011). Probability in Banach Spaces: Isoperimetry and Processes, Springer Verlag. Logan, B. F. and Shepp, L. A. (1977). A variational problem for random Young tableaux, Adv. Math. 26, 206-222. Lytova, A. and Pastur, L. (2009). Central limit theorem for linear eigenvalue statistics of random matrices with independent entries, Ann. Prob. 37, 17781840. Macchi, O. (1975). The coincidence approach to stochastic point processes, Adv. Appl. Prob. 7, 83-122. Macdonald, I. (1995). Symmetric Functions and Hall Polynomials, 2nd edition, Clarendon Press, Oxford. Majumdar, S. N., Nadal, C., Scardicchio, A. and Vivo, P. (2009). The index distribution of Gaussian random matrices, Phys. Rev. Lett. 103, 220603. McLeish, D. L. (1974). Dependent central limit theorems and Invariance principles, Ann. Probab. 2, 620-628. Mehta, L. A. (2004). Random Matrices, 3rd edition, Academic Press. Okounkov, A. (2000). Random matrices and random permutations, Int. Math. Res. Not. 2000, 1043-1095. Okounkov, A. (2001). Infinite wedge and random partitions, Selecta Math. 7, 57-81. Okounkov, A. (2003). The use of random partitions, XIVth ICMP, 379-403, World Sci. Publ., Hackensack, New Jersey. Okounkov, A. and Olshanski, G. (1998). Shifted Schur functions, St. Petersburg Math. J. 9, 239-300. Okounkov, A. and Pandharipande, R. (2005). Gromov-Witten theory, Hurwitz numbers, and matrix models, Algebraic geometry—Seattle 2005, Part I, 325414, Proc. Sympos. Pure Math., 80, Part I, Amer. Math. Soc., Providence, RI, 2009. Pastur, L. and Shcherbina, M. (2011). Eigenvalue Distribution of Large Random

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Index

Brown, 8 Brownian carousel, 143 Brownian motion, 31 Burnside identity, 44, 207, 208

Airy function, 92 Airy operator, 143 Airy point process, 118 algebra isomorphism, 222 Anderson, 28, 90 Andrews, 154, 202 angular shift formula, 149 aperiodic, 10 Ascoli-Arzel` a lemma, 30 average spectral measure, 36

canonical grading, 222, 226 canonical isomorphism, 222 Cantero, 60 Carleman condition, 12 Cauchy integral formula, 111, 154, 174 Cauchy random variable, 4 Cauchy theorem, 135 Cauchy transform, 25 Cauchy-Binet formula, 124 Cauchy-Schwarz inequality, 23, 94, 122, 184 Cavagna, 143 Chapman-Kolmogorov equation, 10 character, 40 character ratio, 186, 236, 244 character table, 41 characteristic function, 4, 162 characteristic polynomial, 34, 53, 59, 133 Chebyshev inequality, 4, 18 Chebyshev polynomial, 220 Chen, 19, 21 chi random variable, 126 Chow, 2 Chung, 2 Circular β ensemble, 36

Bai, 28 Baik, 219 Bao, 143 Barnes G-function, 134, 136 bell curve, 2 Bernoulli law, 1, 3 Bernoulli random variable, 3, 124 Berry-Esseen bound, 244 Bessel function, 242 Bessel point process, 118 Billingsley, 2, 29, 32, 186 binomial distribution, 4 binomial formula, 2 Blaschke product, 59 Bochner theorem, 5 Bogachev, 244 Borodin, 242, 253 Bose-Einstein model, 201 Bourgade, 69 Br´ezin, 134 Bratelli graph, 245 267

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Circular unitary ensemble, 33 CMV matrix, 64, 87 complete symmetric polynomial, 44 complex normal random variable, 33, 48, 50, 125 Concentration of measure inequality, 99 conjugacy class, 34, 39, 249 conjugate partition, 45 Conrey, 35 continual diagram, 28 converges in distribution, 3, 239 correlation function, 116, 253 Costin, 118 Costin-Lebowitz-Soshnikov theorem, 143, 242, 243 Cram´er rule, 62 Cram´er-Wald device, 56 cumulant, 14 cycle type, 38 De Moivre-Laplace CLT, 2 Deift, 48, 89, 90, 208, 219 determinantal point process, 91, 116, 122, 242, 243 Diaconis, 37, 48, 55, 186 dominance order, 42 Donsker invariance principle, 31 Dumitriu, 137, 142 Durrett, 2 Dyson, 36, 136 Edelman, 137, 142 eigendecomposition, 107, 137, 139 elementary symmetric polynomial, 44 empirical spectral distribution, 95 Erd¨ os, 158 ergodic, 10 Erlihson, 205 Eskin, 186 Euler constant, 135 Evans, 48, 55 exchangeable pair, 22, 245 exclusion-inclusion principle, 257 Favard theorem, 58

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Feller condition, 7 Feller-L´evy CLT, 6, 16, 31 Fischer, 2 five diagonal matrix model, 69 Forrester, 34, 81, 134 Four moment comparison theorem, 18 Fourier coefficient, 47 Fourier transform, 27, 37, 207, 218 fractional linear transform, 145 Frankel, 134 Fredholm determinant, 254, 255 Fristedt, 158, 162 Frobenius, 187 Frobenius coordinate, 221, 253 Frobenius formula, 247, 250 Fulman, 236, 244, 247, 249 Garrahan, 143 Gaussian unitary ensemble, 51, 89 geometric random variable, 161 Giardina, 143 Gibbs measure, 36 Gikhman, 32 Gikhman-Skorohod theorem, 32, 181 Ginibre, 125 Ginibre model, 125 Gioev, 89, 90 Girko, 125 Goldstein, 19, 21 Gram-Schmidt algorithm, 33, 57 grand ensemble, 160, 169 Granovsky, 205 Green function, 98 Guionnet, 28, 90 Gumbel distribution, 158, 203 Gustavsson, 121, 237 Haar measure, 33, 66 Hadamard inequality, 118 Hall, 9 Hankel matrix, 134 Hermite β ensemble, 136 Hermite orthogonal polynomial, 90, 92, 113, 233 Hermite wave function, 91 Hessenberg matrix, 59, 64

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Index

Heyde, 9 Hikami, 134 Hilbert space, 57 hook formula, 211, 245 Hora, 236 Householder transform, 66, 127, 137 Hughes, 69 idempotent projection, 126 Ingram, 225 integrating-out formula, 91 irreducible, 10, 39 Its, 48 Ivanov, 221, 222, 224, 227, 228, 230, 233, 235 Jacobian matrix, 65, 81 Johansson, 112, 142, 219, 241 Keating, 53 kernel function, 116, 257 Kerov, 209, 210, 212, 218, 221, 227, 228, 230 Kerov CLT, 221 Khinchine law, 3, 38 Killip, 69, 75, 79 Knuth, 208 Kolmogorov continuity criterion, 185 Krasovsky, 48, 133, 134 L´evy continuity theorem, 5 L´evy maximal inequality, 31 Lebesgue measure, 34, 89, 122 Lebowitz, 118 Ledoux, 6 left orthogonal L-polynomial, 60 Lehner, 158 lexicographic order, 41 Lindeberg condition, 7, 151, 169 Lindeberg replacement strategy, 16 Lindeberg-Feller CLT, 7 Logan, 209, 219 Lyapunov CLT, 74, 126 Lyapunov condition, 7, 124 Lytova, 101, 112

269

M¨ obius transform, 149 M¨ obius inversion formula, 119 Macchi, 116 Macdonald, 38, 225 Majumdar, 143 Marchenko-Pastur law, 14 Markov chain, 9, 145, 150, 158, 247 Markov inequality, 50, 55, 74, 165, 190 Markov property, 9 martingale, 8 martingale CLT, 8, 130, 150 martingale difference sequence, 8, 77, 150 matrix representation, 39 McLeish, 9 Mehta, 34, 125 moment generating function, 26 Moral, 60 multiplicative measure, 160, 200 Nadal, 143 Nenciu, 69, 75, 79 Nikeghbali, 69 normal random variable, 4, 240 Okounkov, 186, 219, 227, 242, 253 Olshanski, 221, 222, 224, 227, 228, 233, 235, 242, 253 Pandharipande, 186 partition, 41, 153 Pastur, 101, 112 Pasval-Plancherel identity, 218 Pittel, 158, 173, 181, 188, 189, 194 Plancherel measure, 207, 210, 240, 244 Plancherel theorem, 76, 207 Plancherel-Rotach formula, 92 Poincar´e disk model, 145 Poincar´e-Nash upper bound, 99 Poisson point process, 116 Poisson random variable, 4, 240 Poissonized Plancherel measure, 240, 253 polar coordinate, 48

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Random Matrices and Random Partitions

polytabloid, 43 positive recurrent, 10 Postnikov, 154, 155 power sum polynomial, 45, 49 Pr¨ afer phase, 59, 69 Prohorov theorem, 30 projective limit, 222 Rains, 81 Ram´ırez, 143 random Plancherel partition, 28, 207 random uniform partition, 161 recurrent, 10 reversible, 247 Rider, 143 Riemann zeta function, 53 Riemann-Hilbert problem, 134 Riesz transform, 28 right orthonormal L-polynomial, 60 Riordan, 235 Robinson, 208 Ross, 19 RSK correspondence, 208 Sagan, 38, 208 Scardicchio, 143 Schensted, 208 Schur orthonormality, 46 Schur polynomial, 45 Schur-Weyl duality, 46 selection principle, 18 Serfozo, 11 Shahshahani, 37, 186 Shao, 19, 21, 244, 250 Shepp, 209, 219 Sherman-Morrison equation, 99 Silverstein, 28 Simon, 48, 58 Skorohod, 32 slowly varying, 52 small ensemble, 169 Snaith, 53 Sobolev norm, 217 Sobolev space, 112 Soshnikov, 118 stationary distribution, 10, 247

Steele, 208 Stein, 19 Stein continuity theorem, 20 Stein equation, 19, 99, 102, 105 Stein method, 19, 236, 244, 245 Stieltjes continuity theorem, 27, 101 Stieltjes transform, 25, 100 Stirling formula, 3, 191 stochastic equicontinuity, 32, 160, 181 Su, 143, 200, 244, 250 symmetric group, 38 Szeg¨ o, 91 Szeg¨ o dual, 57 Szeg¨ o recurrence relation, 58 Szeg¨ o strong limit theorem, 48 tabloid, 42 Talagrand, 6 Tao, 18, 28, 125 Temperley, 157 Tracy, 219 Tracy-Widom law, 122, 219, 238, 242 transient, 10 transition density, 158 transition matrix, 9 transition probability, 9, 247 tridiagonal matrix model, 142, 144 Trotter, 125 Ulam problem, 210 uniform measure, 36, 66, 200, 207 uniform topology, 233 unitary, 33 Ursell function, 119 Valk´ o, 143, 146 Vandermonde determinant, 44, 90 Vel´ azquez, 60 Verblunsky coefficient, 58, 64, 69, 74, 79 Verblunsky theorem, 58 Vershik, 157, 169, 201–203, 209, 210, 212, 218, 227 Vir´ ag, 143, 144, 146 Vivo, 143 Vu, 18, 125

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Index

Wasserstein distance, 21 weak topology, 233 wedge product, 81 weight filtration, 225, 235 weight grading, 224, 226 Weyl formula, 34, 53 Widom, 219 Wieand, 55 Wigner semicircle law, 14, 95, 113, 133, 142 Yakubovich, 201, 203 Yor, 69 Young diagram, 28, 42, 156, 210, 221, 237, 253 Young tableau, 42 Zeitouni, 28, 90

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