Determinantal Processes And The IID Gaussian Power Series

Transcription

1 Determinantal Processes And The IID Gaussian Power Series Yuval Peres U.C. Berkeley Talk based on work joint with: J. Ben Hough Manjunath Krishnapur Bálint Virág 1

2 Samples of translation invariant point processes in the plane: Poisson (left), determinantal (center) and permanental for K(z, w) = 1 π ez w 1 2 ( z 2 + w 2). Determinantal processes exhibit repulsion, while permanental processes exhibit clumping. 2

3 Determinantal Point Processes Let (Ω, µ) be a σ-finite measure space, Ω R d. One way to describe the distribution of a point process X on Ω is via its joint intensities. Definition: X has joint intensities ρ k, k = 1, 2,... if, for any mutually disjoint (measurable) sets A 1,...,A k, [ k ] E X A j = ρ k (x 1,...,x k )dµ j=1 j A j 3

4 In most cases of interest the following is valid (assume no double points): Ω is discrete and µ = counting measure: ρ k (x 1,...,x k ) is the probability that x 1,...,x k X. Ω is open in R d and µ = Lebesgue measure: ρ k (x 1,...,x k ) is lim ǫ 0 P(X has a point in each of B ǫ (x j )) (Vol(B ǫ )) k. 4

5 Now let K be the kernel of an integral operator K on L 2 (Ω) with the spectral decomposition K(x, y) = k λ kϕ k (x)ϕ k (y), where {ϕ k } k is an orthonormal set in L 2 (Ω). Definition: X is said to be a determinantal point process with kernel K if its joint intensities are ) ρ k (x 1,...,x k ) = det ((K(x i, x j )) 1 i,j k, (1) for every k 1 and x 1,...,x k Ω. 5

6 Key facts: (Maachi) A locally finite determinantal process with the Hermitian kernel K exists if and only if K is locally of trace class and 0 λ k 1 k. If K(x, y) = n k=1 ϕ k(x)ϕ k (y), then the total number of points in X is n, almost surely. Since the corresponding integral operator K on L 2 (Ω) is a projection, such processes are said to be determinantal projection process. 6

7 Karlin-McGregor (1958) Consider n independent simple symmetric random walks on Z started from i 1 < i 2 <... < i n where all the i j s are even. Let P i,j (t) be the t-step transition probabilities. Then the probability that at time t, the random walks are at j 1 < j 2 <... < j n and have mutually disjoint paths is P i1,j 1 (t)... P i1,j n (t) det P in,j 1 (t)... P in,j n (t) This is intimately related to determinantal processes. For instance, one can show that if t is even and we also condition the walks to return to i 1,...,i n, then the positions of the walks at any time s (1 s t) are determinantal. (See Johanson(2004) for this and more general results) 7

8 Uniform Spanning Tree Let G be a finite undirected graph. Let T be uniformly chosen from the set of spnning trees of G. Orient the edges of G arbitrarily. Let ě be the opposite orientation of e. For each directed edge e, let χe := 1 e 1ě denote the unit flow along e. l 2 (E) = {f : E R : f(e) = f(ě)} = span{ e=v χ e : where v is a vertex.} = span{ n χ e i : e 1,...,e n is an oriented cycle} i=1 It is easy to see that l 2 (E) =. Define I e := P χe, the orthogonal projection onto. Kirchoff (1847) proved that for any edge e, P[e T] = (I e, I e ). 8

9 Theorem: (Burton and Pemantle (1993)) The set of edges in T forms a determinantal process with kernel Y (e, f) := (I e, I f ). i.e., for any distinct edges e 1,...,e k P[e 1,...,e k T] = det[(y (e i, e j )) 1 i,j k ]. (2) 9

10 Ginibre Ensemble Let A be an n n matrix with i.i.d. standard complex normal entries. Then the eigenvalues of A form a determinantal process in C with the kernel K n (z, w) = 1 π e 1 2 ( z 2 + w 2 n 1 ) (zw) k. k=0 k! As n, we get a determinantal process with the kernel K(z, w) = 1 π e 1 2 ( z 2 + w 2 ) = 1 π e 1 2 ( z 2 + w 2 )+zw. k=0 (zw) k. k! 10

11 Construction of determinantal projection processes Define K H δ x ( ) = K(, x). The intensity measure of the process is given by µ H (x) = ρ 1 (x)dµ(x) = K H δ x 2 dµ(x). (3) Note that µ H (M) = dim(h), so µ H / dim(h) is a probability measure on M. We construct the determinantal process as follows. Start with n = dim(h), and H n = H. If n = 0, stop. Pick a random point X n from the probability measure µ Hn /n. Let H n 1 H n be the orthocomplement of the function K Hn δ x in H n. In the discrete case, H n 1 = {f H n : f(x n ) = 0}. Note that dim(h n 1 ) = n 1 a.s. Decrease n by 1 and iterate. 11

12 Proposition: The points (X 1,...,X n ) constructed by this algorithm are distributed as a uniform random ordering of the points in a determinantal process X with kernel K. Proof: Let ψ j = K H δ xj. Projecting to H j is equivalent to first projecting to H and then to H j, and it is easy to check that K Hj δ xj = K Hj ψ j. Thus, by (3), the density of the random vector (X 1,...,X n ) constructed by the algorithm equals n K Hj ψ j 2 p(x 1,...,x n ) =. j Note that H j = H ψ j+1,...,ψ n, and therefore V = n j=1 K Hj ψ j is exactly the repeated base times height formula for the volume of the parallelepiped determined by the vectors ψ 1,...,ψ n in the finite-dimensional vector space H L 2 (M). It is well-known that V 2 equals the determinant of the Gram matrix whose i, j entry is given by the scalar product of j=1 12

13 ψ i, ψ j, that is ψ i ψ j dµ = K(x i, x j ). We get p(x 1,...,x n ) = 1 n! det(k(x i, x j )), so the random variables X 1,...,X n are exchangeable. Viewed as a point process, the n-point joint intensity of {X j } n j=1 is n!p(x 1,...,x n ), which agrees with that of the determinantal process X. The claim now follows since X contains n points almost surely. 13

14 We have the following remarkable fact that connects the kernel K to the distribution of X: Theorem: (Shirai-Takahashi (2002)) Suppose X is a determinantal process on E with kernel K(x, y) = k λ kϕ k (x)ϕ k (y). Then L(X) = α(s)l (X(S)), (4) S N where X(S) is the determinantal process in E with kernel j S ϕ j(x)ϕ j (y) and α(s) = λ j (1 λ j). j S j S In particular the number of points in the process X has the distribution of a sum of independent Bernoulli(λ k ) random variables. 14

15 Proof: (HKPV) Assume K has finite rank i.e., take K(x, y) = n k=1 λ kϕ k (x)ϕ k (y). Otherwise we can approximate by finite rank kernels, and deduce the same for general K since the corresponding processes increase (stochastically) to the original process. Let I k, 1 k n be independent Bernoulli random variables with I k Bernoulli(λ k ). Then set K I (x, y) = n k=1 I kϕ k (x)ϕ k (y). K I is a random analogue of the kernel K. We want to prove m, x i s, [ )] ) E det ((K I (x i, x j )) 1 i,j m = det ((K(x i, x j )) 1 i,j m. (5) 15

16 Proof of (5): Take m = n first. Then we write K I (x 1, x 1 ) K I (x 1, x n ) = K I (x n, x 1 ) K I (x n, x n ) I 1 ϕ 1 (x 1 )... I n ϕ n (x 1 ) ϕ 1 (x 1 )... ϕ 1 (x n ) I 1 ϕ 1 (x 2 )... I n ϕ n (x 2 ) ϕ 2 (x 1 )... ϕ 2 (x n ) I 1 ϕ 1 (x n )... I n ϕ n (x n ) ϕ n (x 1 )... ϕ n (x n ). Hence det((k I (x i, x j ))) 1 i,j n = I 1...I n det(a A) where A is the second matrix on the right side above. On taking expectations we get [ )] E det ((K I (x i, x j )) 1 i,j n = λ 1...λ n det(a A). 16

17 Now we also have K(x 1, x 1 ) K(x 1, x n ) = K(x n, x 1 ) K(x n, x n ) λ 1 ϕ 1 (x 1 )... λ n ϕ n (x 1 ) ϕ 1 (x 1 )... ϕ 1 (x n ) λ 1 ϕ 1 (x 2 )... λ n ϕ n (x 2 ) ϕ 2 (x 1 )... ϕ 2 (x n ) λ 1 ϕ 1 (x n )... λ n ϕ n (x n ) ϕ n (x 1 )... ϕ n (x n ) From this we get ) det ((K(x i, x j )) 1 i,j n = λ 1...λ n det(a A).. 17

18 This proves that the two point processes X (determinantal with kernel K) and X I (determinantal with kernel K I ) have the same n-point joint intensity. But both these processes have at most n points. Therefore for every m, the m-point joint intensities are determined by the n-point joint intensities (zero for m > n, got by integrating for m < n). This proves the theorem. 18

19 Zeros of the i.i.d. Gaussian power series [Virág-P.]. Let f U (z) = a n z n n=0 Z U = zeros(f U ) (6) with {a n } complex Gaussian, density(re iθ ) = e r2. Theorem: (Hannay, Zelditch-Shiffman,...) iα z λ Law of Z U invariant under Möbius transformations z e 1 λz that preserve unit disk. 19

20 Euclidean analog: satisfies Thus, f C = n=0 [ a Cov[f C (z), f C (w)] = E n z n n n! k = n=0 zn w n n! = e z w. a n z n n!, (7) ] ā k w k k! (z+a)( w+ā) Cov[f C [ (z + a), f C (w + a)] = e ] = Cov e a 2 /2eāz f C (z), e a 2 /2eāw f C (w). Since Gaussian processes are determined by Cov(, ) this proves translation invariance of Law[zeros(f C )]. 20

21 Definition: Let p ǫ (z 1,...,z n ) denote the probability that a random function f has zeros in B ǫ (z 1 ),...B ǫ (z n ). Joint intensity of zeros (if it exists) is defined to be p(z 1,...,z n ) = lim ǫ 0 p ǫ (z 1,...,z n ) (πǫ 2 ) n (8) Theorem: (Hammersley) Let f be a Gaussian analytic function in a planar ( domain ) D, z 1,...,z n D, and consider the matrix A = Ef(z i )f(z j ). If A is non-singular then p(z 1,...z n ) exists and equals ( ) E f (z 1 ) f (z n ) 2 f(z1 )= =f(z n )=0 det(πa). 21

22 Theorem: (Virág - P.) The joint intensity of zeros for f U is [ ] p(z 1,...,z n ) = π n 1 det (1 z i z j ) 2 = det[k(z i, z j )] where K(z, w) = 1 π(1 z w) 2 is the Bergman kernel for U. i,j 22

23 Let Proof of Determinantal Formula T β (z) = z β 1 βz denote a Möbius transformation fixing the unit disk. Also, for fixed z 1,...,z n U denote (9) Υ(z) = n j=1 T zj (z). (10) Key facts: 1. Letf = f U and z 1,...,z n U. The distribution of the random function T z1 (z) T zn (z)f(z) is the same as the conditional distribution of f(z) given f(z 1 ) =... = f(z n ) = 0. 23

24 2. It follows that the conditional joint distribution of the random variables ( f (z k ) : k = 1,...,n ) given f(z 1 ) =... = f(z n ) = 0, is the same as the unconditional joint distribution of ( Υ (z k )f(z k ) : k = 1,...,n ). 3. Consider the n n matrices A jk = Ef(z j )f(z k ) = (1 z j z k ) 1, M jk = (1 z j z k ) 2. By the classical Cauchy determinant formula, det(a) = 1 (z k z j )(z k z j ) 1 z j z k k,j k<j n = Υ (z k ). (11) k=1 24

25 4. We also use Borchardt s identity: ( ) ( 1 perm x j +y det 1 k j,k x j +y k ) j,k = det ( 1 (x j +y k ) )j,k 2 setting x j = z 1 j and y k = z k and dividing both sides by j z2 j, gives that perm(a) det(a) = det(m). (12) 5. Finally, recall the Gaussian moment formula: If Z 1,...,Z n are jointly complex Gaussian random variables with covariance matrix C jk = EZ j Z k, then E ( Z 1 Z n 2) = perm(c). 25

26 From Hammersley s formula p(z 1,...,z n ) equals ( ) E f (z 1 ) f (z n ) 2 f(z 1 ),...,f(z n )=0 π n det(a). The numerator equals E ( f(z 1 ) f(z n ) 2) k Υ (z k ) 2 = perm(a) det(a) 2, where the last equality uses the Gaussian moment formula. Thus, p(z 1,...,z n ) = π n perm(a) det(a) = π n det(m). 26

27 Theorem 2: (Virág - P.) Let X k 1 r 2k 0 1 r 2k be independent. Then 1 X k and N r = Z U B(0, r) have same distribution. Corollary: Let h r = 4πr 2 /(1 r 2 ) (hyperbolic area). Then P(N r = 0) = e h r π 24 +o(h r) = e π2 /12+o(1) 1 r. (13) All of the above generalize to other simply connected domains with smooth boundary. ( ) E f D (z)f D (w) = 2πS D (z, w) (Szëgo Kernel) (14) 27

28 Denote q = r 2. Key to law of N r = Z U B(0, r) : ( ) Nr E = 1 p(z 1,...,z k )dz 1,...dz k k k! Euler s partition identity implies that has product form! B k r q (k+1 2 ) = (1 q)(1 q 2 )...(1 q k ) = γ k. k=0 E(1 + s) N r = γ k s k = (1 + q k s), (15) E k=0 ( Nr k ) s k = γ k s k (16) 28

29 Dynamics Let f U (t, z) = n a n (t)z n (17) with a n (t) performing Ornstein-Uhlenbeck diffusion, a n (t) = e t/2 W n (e t ). Suppose that the zero set of f U contains the origin. Movement of this zero satisfies stochastic differential equation dz = σdw (18) where 1 σ = f U(0) = c lim r r 2 z Z U 0< z <r z = c e 1/k z k. (19) k=1 29