Large Deviations for Random Matrices and a Conjecture of Lukic

Transcription

1 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,

2 Szegő s Theorem, Sum Rules, and Gems Given a probability measure with infinite support on the unit circle, dµ(θ) = w(θ)dθ + dµ sing, let {Φ n } n=0 be the monic orthogonal polynomials w.r.t. µ. Then: Φ n+1 (z) = zφ n (z) ᾱ n Φ n(z) with the Verblunsky coefficients satisfying Φ n(z) = z n Φ n (1/ z) α n < 1 n.

3 Szegő s Theorem, Sum Rules, and Gems Given a probability measure with infinite support on the unit circle, dµ(θ) = w(θ)dθ + dµ sing, let {Φ n } n=0 be the monic orthogonal polynomials w.r.t. µ. Then: Φ n+1 (z) = zφ n (z) ᾱ n Φ n(z) with the Verblunsky coefficients satisfying Φ n(z) = z n Φ n (1/ z) α n < 1 n. There exists a (continuous) bijection between coefficient sequences and probability measures with infinite support (Verblunsky s Theorem).

4 Szegő 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.

5 Szegő 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

6 Szegő s Theorem, Sum Rules, and Gems Perhaps the most classical sum rule is Verblunsky s formulation of Szegő s Theorem (1935): log ( 1 α j 2) = j=0 log(w(θ)) dθ 2π,

7 Szegő s Theorem, Sum Rules, and Gems Perhaps the most classical sum rule is Verblunsky s formulation of Szegő s Theorem (1935): log ( 1 α j 2) = j=0 log(w(θ)) dθ 2π, implying the gem αj 2 < log(w(θ)) dθ 2π >.

8 Szegő s Theorem, Sum Rules, and Gems Perhaps the most classical sum rule is Verblunsky s formulation of Szegő s Theorem (1935): log ( 1 α j 2) = j=0 log(w(θ)) dθ 2π, implying the gem αj 2 < log(w(θ)) dθ 2π >. What if log w has a weak singularity at a point? Can anything be said about the α n s?

9 Szegő s Theorem, Sum Rules, and Gems In ( 05) Simon proved the sum rule ( 2π exp (1 cos(θ)) log(w(θ)) dθ ) 0 2π ( = e 1 2 (1 1+α0 2 ) 1 αj 2) e α j 2 e 1 2 α j+1 α j 2 j=0

10 Szegő s Theorem, Sum Rules, and Gems In ( 05) Simon proved the sum rule ( 2π exp (1 cos(θ)) log(w(θ)) dθ ) 0 2π ( = e 1 2 (1 1+α0 2 ) 1 αj 2) e α j 2 e 1 2 α j+1 α j 2 j=0 = e 1 2 (1 1+α0 2) e ( j=0 log(1 α j 2 )) e j=0 α j 2 e 1 2 j=0 α j+1 α j 2 = e 1 2 (1 1+α0 2) e ( k=1 j=0( α j 2k )) e j=0 α j 2 e 1 2 j=0 α j+1 α j 2

11 Szegő s Theorem, Sum Rules, and Gems In ( 05) Simon proved the sum rule Implying iff ( 2π exp (1 cos(θ)) log(w(θ)) dθ ) 0 2π ( = e 1 2 (1 1+α0 2 ) 1 αj 2) e α j 2 e 1 2 α j+1 α j 2 j=0 = e 1 2 (1 1+α0 2) e ( j=0 log(1 α j 2 )) e j=0 α j 2 e 1 2 j=0 α j+1 α j 2 = e 1 2 (1 1+α0 2) e ( k=1 j=0( α j 2k )) e j=0 α j 2 e 1 2 (1 cos(θ)) log(w(θ)) dθ 2π > α j 4 < and j=0 α j+1 α j 2 < j=0 j=0 α j+1 α j 2

12 Szegő s Theorem, Sum Rules, and Gems Based on this and on two more gems (Simon-Zlatoš, 05), Simon made the following Conjecture (Simon 05) Fix θ 1, θ 2,..., θ k distinct in [0, 2π) and m 1,..., m k positive integers. Then k (1 cos (θ θ j )) m j log(w(θ)) dθ 2π > iff j=1 k ( S e iθ j ) mj α l 2 and where (Sα) j = α j+1. j=1 α l 2(1+max j m j ),

13 Szegő 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) iff k (1 cos (θ θ j )) m j j=1 k ( S e iθ j ) mj α l 2 j=1 log(w(θ)) dθ 2π > and, for each p = 1,..., k, (S e iθ j ) m j α l 2mp+2 j p

14 Szegő s Theorem, Sum Rules, and Gems Lukic s example satisfies but 2π 0 (S 1) 2 (S + 1)α l 2 and α l 6, (1 cos θ) 2 (1 + cos θ) log(w(θ)) dθ 2π =.

15 Szegő s Theorem, Sum Rules, and Gems Lukic s example satisfies but 2π 0 (S 1) 2 (S + 1)α l 2 and α l 6, (1 cos θ) 2 (1 + cos θ) log(w(θ)) dθ 2π =. The above stated form of the conjecture is due to us. Lukic has a different, equivalent, formulation.

16 Szegő s Theorem, Sum Rules, and Gems Lukic s example satisfies but 2π 0 (S 1) 2 (S + 1)α l 2 and α l 6, (1 cos θ) 2 (1 + cos θ) log(w(θ)) dθ 2π =. 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.

17 Szegő s Theorem, Sum Rules, and Gems Lukic s example satisfies but 2π 0 (S 1) 2 (S + 1)α l 2 and α l 6, (1 cos θ) 2 (1 + cos θ) log(w(θ)) dθ 2π =. 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?

18 Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault ( 16) found a beautiful approach to proving Szegő 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.

19 Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault ( 16) found a beautiful approach to proving Szegő 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 vni (x0).

20 Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault ( 16) found a beautiful approach to proving Szegő 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 vni (x0). The function I (x) is called the rate function, and the sequence v n is called the speed.

21 Sum Rules from Large Deviations Recently, Gamboa, Nagel and Rouault ( 16) found a beautiful approach to proving Szegő 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 vni (x0). 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.

22 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.

23 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. We say that the sequence of measures {P n } n=1 obeys a LDP with rate function I and speed {v n } n=1 if: For all closed sets F X lim sup n 1 v n log P n (F ) inf x F I (x). For all open sets U X lim inf n 1 v n log P n inf x U I (x).

24 Sum Rules from Large Deviations Here s a basic result: Theorem (Cramér s Theorem) Given a random variable ξ, let Λ(λ) = log E ( e λξ) be its cumulant generating functions and I (η) = sup (λη Λ(λ)) (1) λ R its Legendre transform. Let P N be the probability distribution for N 1 S N N 1 (ξ ξ N ), where {ξ j } j=1 are independent copies of ξ. Then P N has a LDP with speed N and rate function I.

25 Sum Rules from Large Deviations If f : X Y is a homeomorphism, then the push forward of {P n } n=1 on Y obeys a LDP with rate function I f 1.

26 Sum Rules from Large Deviations If f : X Y is a homeomorphism, then the push forward of {P n } n=1 on Y obeys a LDP with rate function I f 1. Thus, if the rate function on X is I and on Y it is J, then by uniqueness of the rate function I (x) = J(f (x)).

27 Sum Rules from Large Deviations If f : X Y is a homeomorphism, then the push forward of {P n } n=1 on Y obeys a LDP with rate function I f 1. Thus, if the rate function on X is I and on Y it is J, then by uniqueness of the rate function I (x) = J(f (x)). We want X = probability measures on the unit circle Y = Verblunsky coefficient sequences.

28 Sum Rules from Large Deviations If f : X Y is a homeomorphism, then the push forward of {P n } n=1 on Y obeys a LDP with rate function I f 1. Thus, if the rate function on X is I and on Y it is J, then by uniqueness of the rate function I (x) = J(f (x)). We want X = probability measures on the unit circle Y = Verblunsky coefficient sequences. A huge bonus is that the rate function is always nonnegative. Thus, one gets sum rules with two nonnegative sides, which makes it possible (but not necessarily easy) to get gems!

29 Sum Rules from Large Deviations If f : X Y is a homeomorphism, then the push forward of {P n } n=1 on Y obeys a LDP with rate function I f 1. Thus, if the rate function on X is I and on Y it is J, then by uniqueness of the rate function I (x) = J(f (x)). We want X = probability measures on the unit circle Y = Verblunsky coefficient sequences. A huge bonus is that the rate function is always nonnegative. Thus, one gets sum rules with two nonnegative sides, which makes it possible (but not necessarily easy) to get gems! What are the {P n } n=1??

30 Large Deviations for Random Matrices As it turns out, Szegő s Theorem follows from applying the above strategy to Haar measure on n n unitary matrices, or in other words, studying the CUE(n).

31 Large Deviations for Random Matrices As it turns out, Szegő s Theorem follows from applying the above strategy to Haar measure on n n unitary matrices, or in other words, studying the CUE(n). For a matrix chosen from CUE(n), any fixed vector is cyclic with probability one, and the corresponding spectral measures have the form dµ n = n w j δ θj. j=1

32 Large Deviations for Random Matrices As it turns out, Szegő s Theorem follows from applying the above strategy to Haar measure on n n unitary matrices, or in other words, studying the CUE(n). For a matrix chosen from CUE(n), any fixed vector is cyclic with probability one, and the corresponding spectral measures have the form dµ n = n w j δ θj. The θ s and w s are independent. The w s are uniformly distributed on { n j=1 w j = 1} and the θ s have the distribution 1 n! 1 j<k n j=1 e iθ j e iθ k 2 n j=1 dθ j 2π.

33 Large Deviations for Random Matrices Ben Arous and Guionnet ( 97) have shown that the random measure 1 n obeys a LDP with speed n 2. (They ve actually shown the analogous result on the line). n j=1 δ θj

34 Large Deviations for Random Matrices Ben Arous and Guionnet ( 97) have shown that the random measure 1 n obeys a LDP with speed n 2. (They ve actually shown the analogous result on the line). n j=1 To get a LDP for the spectral measure with speed n, this means that the θ j s are with very high probability very close to uniformly distributed. δ θj

35 Large Deviations for Random Matrices Ben Arous and Guionnet ( 97) have shown that the random measure 1 n obeys a LDP with speed n 2. (They ve actually shown the analogous result on the line). n j=1 To get a LDP for the spectral measure with speed n, this means that the θ j s are with very high probability very close to uniformly distributed. This and the independence allows one to prove a LDP for the spectral measure with speed n and rate function I (dµ) = 2π 0 δ θj log(w(θ)) dθ 2π.

36 Large Deviations for Random Matrices The distribution of the Verblunsky coefficients for the random spectral measure dµ was computed by Killip and Nenciu ( 04): The {α j } n 1 j=0 are independent with α n 1 distributed uniformly on the unit circle, and α j having density on the unit disk equal to n j 1 π ( 1 z 2 ) n j 2 d 2 z

37 Large Deviations for Random Matrices The distribution of the Verblunsky coefficients for the random spectral measure dµ was computed by Killip and Nenciu ( 04): The {α j } n 1 j=0 are independent with α n 1 distributed uniformly on the unit circle, and α j having density on the unit disk equal to n j 1 π ( 1 z 2 ) n j 2 d 2 z = n j 1 e (n j 2) log(1 z 2 ) d 2 z. π

38 Large Deviations for Random Matrices The distribution of the Verblunsky coefficients for the random spectral measure dµ was computed by Killip and Nenciu ( 04): The {α j } n 1 j=0 are independent with α n 1 distributed uniformly on the unit circle, and α j having density on the unit disk equal to n j 1 π ( 1 z 2 ) n j 2 d 2 z = n j 1 e (n j 2) log(1 z 2 ) d 2 z. π Taking n to infinity leads to a LDP with speed n and rate function J(α) = This proves Szegő s Theorem! log ( 1 α j 2). j=0

39 Large Deviations for Random Matrices Taking n on the coefficient side is subtle. The machinery of projective limits in large deviation theory can be employed to do this.

40 Large Deviations for Random Matrices Taking n on the coefficient side is subtle. The machinery of projective limits in large deviation theory can be employed to do this. The measure side of the sum rule is S(ν µ) the relative entropy of the limiting empirical measure, ν, with respect to µ. That a LDP exists with rate function given by minus the relative entropy is true for general random matrix ensembles of the form dp n (M) exp ( ntrv (M)) dm with nice enough V (dm being Haar measure).

41 Large Deviations for Random Matrices Taking n on the coefficient side is subtle. The machinery of projective limits in large deviation theory can be employed to do this. The measure side of the sum rule is S(ν µ) the relative entropy of the limiting empirical measure, ν, with respect to µ. That a LDP exists with rate function given by minus the relative entropy is true for general random matrix ensembles of the form dp n (M) exp ( ntrv (M)) dm with nice enough V (dm being Haar measure). Can we apply this strategy to get higher order Szegő Theorems?

42 Higher Order Szegő Theorems We know what the rate function on the measure side needs to be: I (µ) = k (1 cos (θ θ j )) m j j=1 log(w(θ)) dθ 2π

43 Higher Order Szegő Theorems We know what the rate function on the measure side needs to be: I (µ) = k (1 cos (θ θ j )) m j j=1 log(w(θ)) dθ 2π by the previous remark. = S(dν µ)

44 Higher Order Szegő Theorems We know what the rate function on the measure side needs to be: I (µ) = k (1 cos (θ θ j )) m j j=1 log(w(θ)) dθ 2π by the previous remark. = S(dν µ) Thus, we know what dν is!

45 Higher Order Szegő Theorems We know what the rate function on the measure side needs to be: I (µ) = k (1 cos (θ θ j )) m j j=1 log(w(θ)) dθ 2π by the previous remark. = S(dν µ) Thus, we know what dν is! Previous work shows that this is the limiting empirical measure for the ensemble with V (e iθ ) = 2 log e iθ e iϕ dν(ϕ).

46 Higher Order Szegő Theorems We know what the rate function on the measure side needs to be: I (µ) = k (1 cos (θ θ j )) m j j=1 log(w(θ)) dθ 2π by the previous remark. = S(dν µ) Thus, we know what dν is! Previous work shows that this is the limiting empirical measure for the ensemble with V (e iθ ) = 2 log e iθ e iϕ dν(ϕ). V turns out to be a trigonometric polynomial in θ.

47 Higher Order Szegő Theorems The coefficient side seems to be less straightforward, but isn t:

48 Higher Order Szegő Theorems The coefficient side seems to be less straightforward, but isn t: The ensemble is and V is a polynomial in M and M. dp n (M) exp ( ntrv (M)) dm We may write M in the CMV basis and so, since the LDP has speed n, we see that the rate function is simply the rate function for CUE + the limit of V (M) when n.

49 Higher Order Szegő Theorems The coefficient side seems to be less straightforward, but isn t: The ensemble is and V is a polynomial in M and M. dp n (M) exp ( ntrv (M)) dm We may write M in the CMV basis and so, since the LDP has speed n, we see that the rate function is simply the rate function for CUE + the limit of V (M) when n. Higher order sum rules are immediate!

50 Higher Order Szegő Theorems The coefficient side seems to be less straightforward, but isn t: The ensemble is and V is a polynomial in M and M. dp n (M) exp ( ntrv (M)) dm We may write M in the CMV basis and so, since the LDP has speed n, we see that the rate function is simply the rate function for CUE + the limit of V (M) when n. Higher order sum rules are immediate! Extracting a gem, however, is not!

51 A Partial Result Theorem (B-Simon-Zeitouni) If then (S 1) 2 (S + 1)α l 2, (S 1) 2 α l 4, α l 6, (1 cos (θ)) 2 (1 + cos (θ)) log(w(θ)) dθ 2π >.

52 A Partial Result The coefficient side of the sum rule is 1 (trm + trm 2 13 ) 2 trm3 log ( 1 α j 2) j=0 and showing this is finite translates to showing that I 2 = α j α j 1 2α j α j 2 + α j α j 3 + 2α 2 j are all finite. I 4 = α 2 j α 2 j 1 + 2α j α 2 j 1α j 2 α 2 j α j 1 α j 2 α j α j 1 α 2 j 2 α j α 2 j 1α j 3 α j α 2 j 2α j 3 + α 4 j I 6 = 1 3 α3 j α 3 j 1 + α j α 3 j 1α 2 j 2 + α 2 j α 3 j 1α j 2 α j α 2 j 1α 2 j 2α j α6 j

53 A Partial Result The coefficient side of the sum rule is 1 (trm + trm 2 13 ) 2 trm3 log ( 1 α j 2) j=0 and showing this is finite translates to showing that I 2 = α j α j 1 2α j α j 2 + α j α j 3 + 2α 2 j I 4 = α 2 j α 2 j 1 + 2α j α 2 j 1α j 2 α 2 j α j 1 α j 2 α j α j 1 α 2 j 2 α j α 2 j 1α j 3 α j α 2 j 2α j 3 + α 4 j I 6 = 1 3 α3 j α 3 j 1 + α j α 3 j 1α 2 j 2 + α 2 j α 3 j 1α j 2 α j α 2 j 1α 2 j 2α j α6 j are all finite. The other direction seems more diffiult. Computations become extremely messy.

54 A Partial Result The coefficient side of the sum rule is 1 (trm + trm 2 13 ) 2 trm3 log ( 1 α j 2) j=0 and showing this is finite translates to showing that I 2 = α j α j 1 2α j α j 2 + α j α j 3 + 2α 2 j I 4 = α 2 j α 2 j 1 + 2α j α 2 j 1α j 2 α 2 j α j 1 α j 2 α j α j 1 α 2 j 2 α j α 2 j 1α j 3 α j α 2 j 2α j 3 + α 4 j I 6 = 1 3 α3 j α 3 j 1 + α j α 3 j 1α 2 j 2 + α 2 j α 3 j 1α j 2 α j α 2 j 1α 2 j 2α j α6 j are all finite. The other direction seems more diffiult. Computations become extremely messy. New ideas???

55 Thank you for your attention