Large Deviations for Random Matrices and a Conjecture of Lukic - PowerPoint PPT Presentation

Large Deviations for Random Matrices and a Conjecture of Lukic Jonathan Breuer Hebrew University of Jerusalem Joint work with B. Simon (Caltech) and O. Zeitouni (The Weizmann Institute) Western States Mathematical Physics Meeting, 13.2.2017

Szeg˝ o’s Theorem, Sum Rules, and Gems Given a probability measure with infinite support on the unit circle, d µ ( θ ) = w ( θ ) d θ + d µ sing , let { Φ n } ∞ n = 0 be the monic orthogonal polynomials w.r.t. µ . Then: α n Φ ∗ Φ ∗ n ( z ) = z n Φ n ( 1 / ¯ Φ n + 1 ( z ) = z Φ n ( z ) − ¯ n ( z ) z ) with the Verblunsky coefficients satisfying | α n | < 1 ∀ n .

Szeg˝ o’s Theorem, Sum Rules, and Gems Given a probability measure with infinite support on the unit circle, d µ ( θ ) = w ( θ ) d θ + d µ sing , let { Φ n } ∞ n = 0 be the monic orthogonal polynomials w.r.t. µ . Then: α n Φ ∗ Φ ∗ n ( z ) = z n Φ n ( 1 / ¯ Φ n + 1 ( z ) = z Φ n ( z ) − ¯ n ( z ) z ) with the Verblunsky coefficients satisfying | α n | < 1 ∀ n . There exists a (continuous) bijection between coefficient sequences and probability measures with infinite support (Verblunsky’s Theorem).

Szeg˝ o’s Theorem, Sum Rules, and Gems A spectral theory gem is an iff relation between properties of the sequence { α n } ∞ n = 0 and properties of the corresponding µ .

Szeg˝ o’s Theorem, Sum Rules, and Gems A spectral theory gem is an iff relation between properties of the sequence { α n } ∞ n = 0 and properties of the corresponding µ . One way of obtaining a gem is via a sum rule , i.e. an equation of the form � � a function of a finite number of α ’s = a function of components of µ j

Szeg˝ o’s Theorem, Sum Rules, and Gems Perhaps the most classical sum rule is Verblunsky’s formulation of Szeg˝ o’s Theorem (1935): � � ∞ log ( w ( θ )) d θ � 1 − | α j | 2 � log = 2 π, j = 0

Szeg˝ o’s Theorem, Sum Rules, and Gems Perhaps the most classical sum rule is Verblunsky’s formulation of Szeg˝ o’s Theorem (1935): � � ∞ log ( w ( θ )) d θ � 1 − | α j | 2 � log = 2 π, j = 0 implying the gem � � log ( w ( θ )) d θ | α j | 2 < ∞ ⇐ 2 π > − ∞ . ⇒

Szeg˝ o’s Theorem, Sum Rules, and Gems Perhaps the most classical sum rule is Verblunsky’s formulation of Szeg˝ o’s Theorem (1935): � � ∞ log ( w ( θ )) d θ � 1 − | α j | 2 � log = 2 π, j = 0 implying the gem � � log ( w ( θ )) d θ | α j | 2 < ∞ ⇐ 2 π > − ∞ . ⇒ • What if log w has a ‘weak’ singularity at a point? Can anything be said about the α n ’s?

Szeg˝ o’s Theorem, Sum Rules, and Gems In (’05) Simon proved the sum rule � � 2 π � ( 1 − cos ( θ )) log ( w ( θ )) d θ exp 2 π 0 � ∞ 2 ( 1 − | 1 + α 0 | 2 ) e | α j | 2 e − 1 2 | α j + 1 − α j | 2 1 � 1 − | α j | 2 � = e j = 0

Szeg˝ o’s Theorem, Sum Rules, and Gems In (’05) Simon proved the sum rule � � 2 π � ( 1 − cos ( θ )) log ( w ( θ )) d θ exp 2 π 0 � ∞ 2 ( 1 − | 1 + α 0 | 2 ) e | α j | 2 e − 1 2 | α j + 1 − α j | 2 1 � 1 − | α j | 2 � = e j = 0 � ∞ � ∞ � ∞ 2 ( 1 − | 1 + α 0 | 2 ) e ( j = 0 log ( 1 − | α j | 2 )) e j = 0 | α j | 2 e − 1 j = 0 | α j + 1 − α j | 2 1 = e 2 � ∞ � ∞ � ∞ � ∞ 2 ( 1 − | 1 + α 0 | 2 ) e ( j = 0 ( − | α j | 2 k )) e 1 j = 0 | α j | 2 e − 1 j = 0 | α j + 1 − α j | 2 = e k = 1 2

Szeg˝ o’s Theorem, Sum Rules, and Gems In (’05) Simon proved the sum rule � � 2 π � ( 1 − cos ( θ )) log ( w ( θ )) d θ exp 2 π 0 � ∞ 2 ( 1 − | 1 + α 0 | 2 ) e | α j | 2 e − 1 2 | α j + 1 − α j | 2 1 � 1 − | α j | 2 � = e j = 0 � ∞ � ∞ � ∞ 2 ( 1 − | 1 + α 0 | 2 ) e ( j = 0 log ( 1 − | α j | 2 )) e j = 0 | α j | 2 e − 1 j = 0 | α j + 1 − α j | 2 1 = e 2 � ∞ � ∞ � ∞ � ∞ 2 ( 1 − | 1 + α 0 | 2 ) e ( j = 0 ( − | α j | 2 k )) e 1 j = 0 | α j | 2 e − 1 j = 0 | α j + 1 − α j | 2 = e k = 1 2 Implying � ( 1 − cos ( θ )) log ( w ( θ )) d θ 2 π > − ∞ iff ∞ ∞ � � | α j | 4 < ∞ | α j + 1 − α j | 2 < ∞ and j = 0 j = 0

Szeg˝ o’s Theorem, Sum Rules, and Gems Based on this and on two more gems (Simon-Zlatoˇ s, ’05), Simon made the following Conjecture (Simon ’05) Fix θ 1 , θ 2 , . . . , θ k distinct in [ 0 , 2 π ) and m 1 , . . . , m k positive integers. Then � k � ( 1 − cos ( θ − θ j )) m j log ( w ( θ )) d θ 2 π > − ∞ j = 1 iff k � S − e − i θ j � m j α ∈ ℓ 2 � j = 1 and α ∈ ℓ 2 ( 1 + max j m j ) , where ( S α ) j = α j + 1 .

Szeg˝ o’s Theorem, Sum Rules, and Gems However, Lukic (’13) constructed a counterexample (with k = 2, θ 1 = 0 , , θ 2 = π , m 1 = 2 , , m 2 = 1) and made a modified Conjecture (Lukic) � k � ( 1 − cos ( θ − θ j )) m j log ( w ( θ )) d θ 2 π > − ∞ j = 1 iff k � S − e − i θ j � m j α ∈ ℓ 2 � j = 1 and, for each p = 1 , . . . , k, � ( S − e − i θ j ) m j α ∈ ℓ 2 m p + 2 j � = p

Szeg˝ o’s Theorem, Sum Rules, and Gems ◮ Lukic’s example satisfies ( S − 1 ) 2 ( S + 1 ) α ∈ ℓ 2 α ∈ ℓ 6 , and but � 2 π ( 1 − cos θ ) 2 ( 1 + cos θ ) log ( w ( θ )) d θ 2 π = − ∞ . 0

Szeg˝ o’s Theorem, Sum Rules, and Gems ◮ Lukic’s example satisfies ( S − 1 ) 2 ( S + 1 ) α ∈ ℓ 2 α ∈ ℓ 6 , and but � 2 π ( 1 − cos θ ) 2 ( 1 + cos θ ) log ( w ( θ )) d θ 2 π = − ∞ . 0 ◮ The above stated form of the conjecture is due to us. Lukic has a different, equivalent, formulation.

Szeg˝ o’s Theorem, Sum Rules, and Gems ◮ Lukic’s example satisfies ( S − 1 ) 2 ( S + 1 ) α ∈ ℓ 2 α ∈ ℓ 6 , and but � 2 π ( 1 − cos θ ) 2 ( 1 + cos θ ) log ( w ( θ )) d θ 2 π = − ∞ . 0 ◮ The above stated form of the conjecture is due to us. Lukic has a different, equivalent, formulation. ◮ Lukic (’16?) showed that in the case of k = 1, under the assumption that α has square-summable variation , the remaining two statements are equivalent.

Szeg˝ o’s Theorem, Sum Rules, and Gems ◮ Lukic’s example satisfies ( S − 1 ) 2 ( S + 1 ) α ∈ ℓ 2 α ∈ ℓ 6 , and but � 2 π ( 1 − cos θ ) 2 ( 1 + cos θ ) log ( w ( θ )) d θ 2 π = − ∞ . 0 ◮ The above stated form of the conjecture is due to us. Lukic has a different, equivalent, formulation. ◮ Lukic (’16?) showed that in the case of k = 1, under the assumption that α has square-summable variation , the remaining two statements are equivalent. ◮ How does one go about generating sum rules of arbitrary order?

Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault (’16) found a beautiful approach to proving Szeg˝ o’s Theorem (and many existing and new sum rules): The approach procceds through recognizing the two sides of the sum rule as different presentations of the rate function in a large deviation principle .

Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault (’16) found a beautiful approach to proving Szeg˝ o’s Theorem (and many existing and new sum rules): The approach procceds through recognizing the two sides of the sum rule as different presentations of the rate function in a large deviation principle . Large Deviations – Let { P n } ∞ n = 1 be a sequence of probability measures on some metric space, X . On an informal level, we say that { P n } ∞ n = 1 obey a large deviations principle (LDP) if the P n -probability to be near x 0 is ∼ e − v n I ( x 0 ) .

Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault (’16) found a beautiful approach to proving Szeg˝ o’s Theorem (and many existing and new sum rules): The approach procceds through recognizing the two sides of the sum rule as different presentations of the rate function in a large deviation principle . Large Deviations – Let { P n } ∞ n = 1 be a sequence of probability measures on some metric space, X . On an informal level, we say that { P n } ∞ n = 1 obey a large deviations principle (LDP) if the P n -probability to be near x 0 is ∼ e − v n I ( x 0 ) . The function I ( x ) is called the rate function , and the sequence v n is called the speed .

Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault (’16) found a beautiful approach to proving Szeg˝ o’s Theorem (and many existing and new sum rules): The approach procceds through recognizing the two sides of the sum rule as different presentations of the rate function in a large deviation principle . Large Deviations – Let { P n } ∞ n = 1 be a sequence of probability measures on some metric space, X . On an informal level, we say that { P n } ∞ n = 1 obey a large deviations principle (LDP) if the P n -probability to be near x 0 is ∼ e − v n I ( x 0 ) . The function I ( x ) is called the rate function , and the sequence v n is called the speed . It is not hard to see that the rate function is unique.

Sum Rules from Large Deviations More precisely, let X be a complete metric space and I : X → [ 0 , ∞ ] a lower semi-continuous function. Let v n be a positive sequence satisfying v n → ∞ .