Brown University Analysis Seminar Eigenvalue Statistics for Ensembles of Random Matrices (especially Toeplitz and Palindromic Toeplitz) Steven J. Miller Brown University Providence, RI, September 15 th, 2004 http://www.math.brown.edu/ sjmiller/math/talks/talks.html
Collaborators Random Matrices Peter Sarnak Rebecca Lehman Yi-Kai Liu Chris Hammond John Ramey John Sinsheimer Jason Teich Inna Zakharevich Random Graphs Peter Sarnak Yakov Sinai Lior Silberman Dustin Steinhauer Randy Qian Felice Kuan Leo Goldmakher 1
Fundamental Problem General Formulation: Studying some system, observe events at t 1 t 2 t 3 Question: what rules govern the distribution of events? Often normalize by average spacing. 2
Examples Spacings Between Energy Levels of Heavy Nuclei. Spacings Between Eigenvalues of Large Matrices. Spacings Between High Zeros of L-Functions. 3
Origins of Random Matrix Theory Classical Mechanics: 3 Body Problem Intractable. Heavy nuclei like Uranium (200+ protons / neutrons) even worse! Fundamental Equation: Hψ n = E n ψ n E n are the energy levels Approximate with finite matrix. 4
Statistical Mechanics For each configuration, calculate quantity (say pressure). Average over all configurations. 5
Random Matrix Ensembles A = a 11 a 12 a 13 a 1N a 12. a 22. a 23 a 2N..... a 1N a 2N a 3N a NN = AT Let p(x) be a probability density, often assume moments finite. k th -moment = x k p(x)dx. Define R Prob(A)dA = 1 i<j N p(a ij )da ij. 6
Eigenvalue Questions Density of Eigenvalues: How many eigenvalues lie in an interval [a, b]? Spacings between Eigenvalues: How are the spacings between adjacent eigenvalues distributed? Note: study normalized eigenvalues. 7
Eigenvalue Distribution Key to Averaging: N Trace(A k ) = λ k i (A). i=1 By the Central Limit Theorem: N N Trace(A 2 ) = a ij a ji = i=1 N j=1 N i=1 j=1 N 2 1 N λ 2 i (A) N 2 i=1 a 2 ij Gives NAve(λ 2 i (A)) N 2 or λ i (A) N. 8
Eigenvalue Distribution (cont) δ(x x 0 ) is a unit point mass at x 0. To each A, attach a probability measure: Obtain: µ A,N (x) = 1 N N δ i=1 ( x λ ) i(a) 2 N k th -moment = x k µ A,N (x)dx = 1 N N i=1 = Trace(Ak ) 2 k N k 2 +1 λ k i (A) (2 N) k 9
Semi-Circle Law N N real symmetric matrices, entries i.i.d.r.v. from a fixed p(x). Semi-Circle Law: p mean 0, variance 1, other moments finite. With probability 1, µ A,N (x) 2 π 1 x 2 weakly Expected value of k th -moment of µ A,N (x) is R R Trace(A k ) 2 k N k 2 +1 i<j p(a ij )da ij 10
With probability 1, Method of Proof µ A,N (x) µ Semi-Circle (x) weakly. Will do this by showing E[M k (µ A,N )] M k (SC) and E[ M k (µ A,N ) E[M k (µ A,N )] 2 ] 0. 11
Proof: 2 nd -Moment Trace(A 2 ) = N i=1 N j=1 a ij a ji = N i=1 N j=1 a 2 ij. Substituting: 1 2 2 N 2 R N R i,j=1 Integration factors as a 2 ji p(a 11 )da 11 p(a NN )da NN a ij R a 2 ijp(a ij )da ij (k,l) (ij) k<l a kl R p(a kl )da kl = 1. Have N 2 summands, answer is 1 4. Key: Trace and Averaging Formulas. 12
Random Matrix Theory: Semi-Circle Law 0.025 Distribution of eigenvalues Gaussian, N=400, 500 matrices 0.02 0.015 0.01 0.005 0 1.5 1 0.5 0 0.5 1 1.5 500 Matrices: Gaussian 400 400 p(x) = 1 e x2 /2 2π 13
Random Matrix Theory: Semi-Circle Law 2500 The eigenvalues of the Cauchy distribution are NOT semicirular. 2000 1500 1000 500 0 300 200 100 0 100 200 300 Cauchy Distr: Not-Semicircular (Infinite Variance) p(x) = 1 π(1+x 2 ) 14
GOE Conjecture GOE Conjecture: N N Real Symmetric, entries iidrv. As N, the probability density of the distance between two consecutive, normalized eigenvalues universal. Only known if entries chosen from Gaussian. Consecutive spacings well approximated by Axe Bx2. 15
Uniform Distribution: p(x) = 1 2 3.5 x 104 The local spacings of the central 3/5 of the eigenvalues of 5000 300x300 uniform matrices, normalized in batches of 20. 3 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5000: 300 300 uniform on [ 1, 1] 16
Cauchy Distribution: p(x) = 1 π(1+x 2 ) 12000 The local spacings of the central 3/5 of the eigenvalues of 5000 100x100 Cauchy matrices, normalized in batches of 20. 10000 8000 6000 4000 2000 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5000: 100 100 Cauchy 3.5 x 104 The local spacings of the central 3/5 of the eigenvalues of 5000 300x300 Cauchy matrices, normalized in batches of 20. 3 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5000: 300 300 Cauchy 17
Fat Thin Families Need a family F A T enough to do averaging. Need a family THIN enough so that everything isn t averaged out. Real Symmetric Matrices have N(N+1) 2 independent entries. Examples of thin sub-families: Band Matrices Random Graphs Special Matrices (Toeplitz) 18
Band Matrices Example of a Band 1 Matrix: a 11 a 12 0 0 0 a 12 a 22 a 23 0 0 0 a 23 a 33 a 24 0............ a N 1,N 0 0 0 a N 1,N a NN For Band 0 (Diagonal Matrices): Density of Eigenvalues: p(x) Spacings b/w eigenvalues: Poissonian. 19
Random Graphs Degree of a vertex = number of edges leaving the vertex. Adjacency matrix: a ij vertex i to vertex j. = number edges from A = 0 0 1 1 0 0 1 0 1 1 0 2 1 0 2 0 These are Real Symmetric Matrices. 20
McKay s Law (Kesten Measure) Density of States for d-regular graphs f(x) = { d 2π(d 2 x 2 ) 4(d 1) x 2 x 2 d 1 0 otherwise d = 3. 21
McKay s Law (Kesten Measure) d = 6. Idea of proof: Trace lemma, combinatorics and counting, locally a tree. Fat Thin: fat enough to average, thin enough to get something different than Semi-circle. 22
d-regular and GOE 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.5 1. 1.5 2. 2.5 3. 3-Regular, 2000 Vertices: Graph courtesy of D. Jakobson, S. D. Miller, Z. Rudnick, R. Rivin 23
Toeplitz Ensembles A Toeplitz matrix is of the form b 0 b 1 b 2 b N 1 b 1 b 0 b 1 b N 2 b 2 b 1 b 0 b N 3....... b 1 N b 2 N b 3 N b 0 Will consider Symmetric Toeplitz. Main diagonal zero, N independent parameters. Normalize Eigenvalues by N. 24
Eigenvalue Density Measure µ A,N (x)dx = 1 N N i=1 δ ( x λ i(a) N ) dx. The k th moment of µ A,N (x) is M k (A, N) = 1 N k 2 +1 N i=1 λ k i (A). Let M k (N) = lim M k(a, N). N 25
N, M 0 (N) = 1. k = 0, 2 and k odd For k = 2: as a ij = b i j : M 2 (N) = 1 N 2 = 1 N 2 1 i 1,i 2 N 1 i 1,i 2 N E(b i1 i 2 b i 2 i 1 ) E(b 2 i 1 i 2 ). N 2 N times get 1, N times 0. Therefore M 2 (N) = 1 1 N. Trivial counting: odd moments 0. 26
M 2k (N) = 1 N k+1 Even Moments 1 i 1,,i 2k N Main Term: b j s matched in pairs, say E(b i1 i 2 b i2 i 3 b i2k i 1 ). b im i m+1 = b in i n+1, x m = i m i m+1 = i n i n+1. Two possibilities: i m i m+1 = i n i n+1 or i m i m+1 = (i n i n+1 ). (2k 1)!! ways to pair, 2 k choices of sign. 27
Main Term: All Signs Neg M 2k (N) = 1 N k+1 1 i 1,,i 2k N E(b i1 i 2 b i2 i 3 b i2k i 1 ). Let x 1,..., x k be the values of the i j i j+1 s. Let ɛ 1,..., ɛ k be the choices of sign. Define x 1 = i 1 i 2, x 2 = i 2 i 3,... Therefore i 2 = i 1 x 1 i 3 = i 1 x 1 x 2. i 1 = i 1 x 1 x 2k. x 1 + + x 2k = k (1 + ɛ j )η j x j = 0, η j = ±1. j=1 28
Even Moments: Summary Main Term: paired, all signs negative. M 2k (N) (2k 1)!! + O k ( 1 N ). Bounded by Gaussian. 29
The Fourth Moment M 4 (N) = 1 N 3 Let x j = i j i j+1. 1 i 1,i 2,i 3,i 4 N E(b i1 i 2 b i2 i 3 b i3 i 4 b i4 i 1 ) Case One: x 1 = x 2, x 3 = x 4 : i 1 i 2 = (i 2 i 3 ) and i 3 i 4 = (i 4 i 1 ). i 1 = i 3, i 2 and i 4 are arbitrary. Left with E[b 2 x 1 b 2 x 3 ]. N 3 N times get 1, N times get p 4 = E[b 4 x 1 ]. 30
Diophantine Obstruction M 4 (N) = 1 N 3 1 i 1,i 2,i 3,i 4 N E(b i1 i 2 b i2 i 3 b i3 i 4 b i4 i 1 ) Case Two: x 1 = x 3 and x 2 = x 4. i 1 i 2 = (i 3 i 4 ) and i 2 i 3 = (i 4 i 1 ). This yields i 1 = i 2 + i 4 i 3, i 1, i 2, i 3, i 4 {1,..., N}. If i 2, i 4 2N 3 and i 3 < N 3, i 1 > N: at most (1 1 27 )N 3 valid choices. 31
Fourth Moment: Answer Theorem: Fourth Moment: Let p 4 be the fourth moment of p. Then M 4 (N) = 2 2 ( ) 1 3 + O p 4. N 500 Toeplitz Matrices, 400 400. 1500 1250 1000 750 500 250-4 -2 0 2 4 32
Higher Moments M 6 (N) = 11 (Gaussian s is 15). M 8 (N) = 64 4 15 (Gaussian s is 105). For sixth moment, five configurations: Occur (respectively) 2, 6, 3, 3 and 1 time. Lemma: For 2k 4, lim N M 2k (N) < (2k 1)!!. 33
Lower Bound of High Moments unbounded support: a lower bound L 2k such that lim k 2k L 2k =. 1 N k+1e N i 1 =1 N i 2k =1 b i1 i 2 b i 2 i 3 b i 2k i 1. Matched in k pairs, matchings must occur with negative signs. Denote positive differences of i n i n+1 by x 1,..., x k. Let x j = i j i j+1 ; k are positive (negative) k + 1 degrees of freedom: k differences x j, and any index, say i 1. 34
Lower Bound of High Moments (II) i 2 = i 1 x 1 i 3 = i 1 x 1 x 2. i 2k = i 1 x 1 x 2k. Once specify i 1 and x 1 through x 2k, all indices are determined. If matched in pairs and each i j {1,..., N}, have a valid configuration, contributes +1. Problem: a running sum i 1 x 1 x m {1,..., N} 35
Lower Bound of High Moments (III) α ( 1 2, 1), I A = {1,..., A} where A = N k α. Choose each difference x j from I A : A k ways, to first order distinct. Put half of positive x j s and and non-matching negatives in first half ( x 1,..., x k ). Have not specified the order of the differences, just how many positive (negative) are in the first half / second half. Two different k-tuples of differences x j cannot give rise to the same configuration (if we assume the differences are distinct). 36
Lower Bound of High Moments (IV) Assume have ordered ( the ) positive differences in first half. There are k2! ways to relatively order corresponding negatives in second half. ( ) If ordered the negatives in first half, k2! ways to relatively order positives in second half. Still have freedom on how to intersperse positives and negatives in second half. "Most" configurations can be made valid by Central Limit Theorem. 37
Lower Bound of High Moments (V) Regard the x p s ( x n s) as iidrv from I A ( I A ) with mean 1 2 A ( 1 2A) and standard deviation 1 2 3 A. CLT: sum of the k 2 positive (negative) x ps ( x n s) in the first block converges to Gaussian, mean ka 4 ( ka 4 ) and standard deviation k2 A 2 3. N, k large, at least 4 1Ak have positive sum in [ ka ka 4 2 6, ka ] ka 4 + 2 6 (and similarly for negative sum). Freedom to choose how to intersperse the positives and negatives in the first and second halves. Keep running sum small: can ensure at most ( ) ka max 2A, 2 2. 6 Intersperse same way in second half. 38
If i 1 Lower Bound of High Moments (VI) [ ] 7 N 8, all indices in {1,..., N}. k α 1 2, N k α 1 2 Number of configurations giving 1 is at least ( ) ( ( ) 1 N 1 k 8 4 Ak )A k 1 (k/2)! 2. 2 k α 1 2 Divide by N k+1, recall A = N k α, main term: 1 32N k+1 N k+1 k kα 1 2 2k th root looks like ( ) e k 2 log k 2 πk 3 2 k 2 2 2π(k/2) = 16e (1+log 2)k e(1 α)k log k. e(1 α) log k e 1+log 2 > O(k 1 α ), proving the support is unbounded; 2k th root of Gaussian moment is k e. 39
Decay of Higher Moments 40
Poissonian Behavior? 1 0.8 0.6 0.4 0.2 1 2 3 4 Not rescaled. Looking at middle 11 spacings, 1000 Toeplitz matrices (1000 1000), entries iidrv from the standard normal. 41
Real Symmetric Palindromic Toeplitz b 0 b 1 b 2 b 3 b 3 b 2 b 1 b 0 b 1 b 0 b 1 b 2 b 4 b 3 b 2 b 1 b 2 b 1 b 0 b 1 b 5 b 4 b 3 b 2 b 3 b 2 b 1 b 0 b 6 b 5 b 4 b 3........... b 3 b 4 b 5 b 6 b 0 b 1 b 2 b 3 b 2 b 3 b 4 b 5 b 1 b 0 b 1 b 2 b 1 b 2 b 3 b 4 b 2 b 1 b 0 b 1 b 0 b 1 b 2 b 3 b 3 b 2 b 1 b 0 Extra symmetry seems to fix Diophantine Obstructions. Always have eigenvalue at 0. 42
Real Symmetric Palindromic Toeplitz (cont) 0.4 0.3 0.2 0.1-4 -2 2 4 500 Real Symmetric Palindromic Toeplitz, 1000 1000 Note the bump at the zeroth bin is due to the forced eigenvalues at 0. 43
Summary Ensemble Degrees of Freedom Density Spacings Real Symm O(N 2 ) Semi-Circle GOE Diagonal O(N) p(x) Poisson d-regular O(dN) Kesten GOE Toeplitz O(N) Toeplitz Poisson Palindromic Toeplitz O(N) Gaussian Red is conjectured Blue is new Maroon: Partial Results (first 9 moments) 44
Summary (Toeplitz) Converges (in some sense) to new universal distribution, independent of p; Moments bounded by those of Gaussian; Ratio tends to 0, but unbounded support; Can interpret as Diophantine Obstructions (Hammond-Miller) or volumes of Euler solids (Bryc-Dembo-Jiang). Real Symmetric Palindromic Toeplitz looks Gaussian (Miller-Ramey-Sinsheimer-Teich). 45