A note on the extremal process of the supercritical Gaussian Free Field *

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1 Electron. Commun. Probab , no. 74, DOI: /ECP.v ISSN: X ELECTRONIC COMMUNICATIONS in PROBABILITY A note on the extremal process of the supercritical Gaussian Free Field * Alberto Chiarini Alessandra Cipriani Rajat Subhra Hazra Abstract We consider both the infinite-volume discrete Gaussian Free Field DGFF and the DGFF with zero boundary conditions outside a finite box in dimension larger or equal to 3. We show that the associated extremal process converges to a Poisson point process. The result follows from an application of the Stein-Chen method from [5]. Keywords: Extremal process; Gaussian free field; point processes; Poisson approximation; Stein Chen method. AMS MSC 2010: 60G15; 60G30; 60G55; 60G70; 82B41. Submitted to ECP on May 27, 2015, final version accepted on October 14, Supersedes arxiv: Introduction In this article we study the behavior of the extremal process of the DGFF in dimension larger or equal to 3. This extends the result presented in [9] in which the convergence of the rescaled maximum of the infinite-volume DGFF and the 0-boundary condition field was shown. It was proved there that the field belongs to the maximal domain of attraction of the Gumbel distribution; hence, a natural question that arises is that of describing more precisely its extremal points. In dimension 2, this was carried out by [6, 7] complementing a result of [8] on the convergence of the maximum; namely, the characterization of the limiting point process with a random mean measure yields as by-product an integral representation of the maximum. The extremes of the DGFF in dimension 2 have deep connections with those of Branching Brownian Motion [1, 2, 3, 4]. These works showed that the limiting point process is a randomly shifted decorated Poisson point process, and we refer to [15] for structural details. In d 3, one does not get a non-trivial decoration but instead a Poisson point process analogous to the extremal process of independent Gaussian random variables. To be more precise, we let E := [0, 1] d, + ] and V N := [0, n 1] d Z d the hypercube of volume N = n d. Let ϕ α α Z d be the infinite-volume DGFF, that is a centered Gaussian field on the square lattice with covariance g,, where g is the Green s function of the simple random walk. We define the following sequence of point processes on E: η n := ε α n, ϕα b N 1.1 a α V N N *The first author s research is supported by RTG The last author s research was supported by Cumulative Professional Development Allowance from Ministry of Human Resource Development, Government of India, and Department of Science and Technology, Inspire funds. TU Berlin, Germany. chiarini@math.tu-berlin.de Weierstrass Institute, Germany. Alessandra.Cipriani@wias-berlin.de Indian Statistical Institute, Kolkata, India. rajatmaths@gmail.com

2 where ε x, x E, is the point measure that gives mass one to a set containing x and zero otherwise, and b N := [ ] 2 log log N + log4π g0 log N 2, a N := g0b N log N Here g0 denotes the variance of the DGFF. Our main result is Theorem 1.1. For the sequence of point processes η n defined in 1.1 we have that η n d η, as n +, where η is a Poisson random measure on E with intensity measure given by d t e z d z where d t d z is the Lebesgue measure on E, and d is the convergence in distribution on M p E 1. The proof is based on the application of the two-moment method of [5] that allows us to compare the extremal process of the DGFF and a Poisson point process with the same mean measure. To prove that the two processes converge, we will exploit a classical theorem by Kallenberg. It is natural then to consider also convergence for the DGFF ψ α α Z d with zero boundary conditions outside V N. For the sequences of point measures ρ n := ε α n, ψα b N 1.3 a α V N N we establish the following Theorem: Theorem 1.2. For the sequence of point processes ρ n defined in 1.3 we have that ρ n d η, as n + in M p E, where η is as in Theorem 1.1. The convergence is shown by reducing ourselves to check the conditions of Kallenberg s Theorem on the bulk of V N, where we have a good control on the drift of the conditioned field, and then by showing that the process on the whole of V N and on the bulk are close as n becomes large. The outline of the paper is as follows. In Section 2 we will recall the definition of DGFF and the Stein-Chen method, while Section 3 and Section 4 are devoted to the proofs of Theorems 1.1 and 1.2 respectively. 2 Preliminaries 2.1 The DGFF Let d 3 and denote with the l -norm on Z d. Let ψ = ψ α α Z d be a discrete Gaussian Free Field with zero boundary conditions outside Λ Z d. On the space Ω := R Zd endowed with its product topology, its law P Λ can be explicitly written as P Λ d ψ = 1 exp 1 ψ α ψ β 2 d ψ α ε 0 ψ α. Z Λ 4d α Λ α, β Z d : α β =1 α Z d \Λ In other words ψ α = 0 P Λ -a. s. if α Z d \Λ, and ψ α α Λ is a multivariate Gaussian random variable with mean zero and covariance g Λ α, β α, β Z d, where g Λ is the Green s 1 M pe denotes the set of Radon point measures on E endowed with the topology of vague convergence. ECP , paper 74. Page 2/10

3 function of the discrete Laplacian problem with Dirichlet boundary conditions outside Λ. For a thorough review on the model the reader can refer for example to [16]. It is known [10, Chapter 13] that the finite-volume measure ψ admits an infinite-volume limit as Λ Z d in the weak topology of probability measures. This field will be denoted as ϕ = ϕ α α Z d. It is a centered Gaussian field with covariance matrix gα, β for α, β Z d. With a slight abuse of notation, we write gα β for g0, α β and also g Λ α = g Λ α, α. g admits a so-called random walk representation: if P α denotes the law of a simple random walk S started at α Z d, then gα, β = E α 1 {Sn=β. n 0 In particular this gives g0 < + for d 3. A comparison of the covariances in the infinite and finite-volume is possible in the bulk of V N : for δ > 0 this is defined as VN δ := {α V N : α β > δn, β Z d \V N. 2.1 In order to compare covariances in the finite and infinite-volume field, we recall the following Lemma, whose proof is presented in [9, Lemma 7]. Lemma 2.1. For any δ > 0 and α, β VN δ one has gα, β C d δn 1/d 2 d gvn α, β gα, β. 2.2 In particular we have, g VN α = g0 1 + O N 2 d/d uniformly for α V δ N. 2.2 The Stein-Chen method As main tool of this article we will use and restate here a theorem from [5]. Consider a sequence of Bernoulli random variables X α α I where X α Bep α and I is some index set. For each α we define a subset B α I which we consider a neighborhood of dependence for the variable X α, such that X α is nearly independent from X β if β I \B α. Set b 1 := p α p β, α I β B α b 2 := E [X α X β ], α I α β B α where b 3 := α I E [ E [X α p α H 1 ] ] H 1 := σ X β : β I \ B α. Theorem 2.2 [5, Theorem 2]. Let I be an index set. Partition the index set I into disjoint non-empty sets I 1,..., I k. For any α I, let X α α I be a dependent Bernoulli process with parameter p α. Let Y α α I be independent Poisson random variables with intensity p α. Also let W j := α I j X α and Z j := α I j Y α and λ j := E[W j ] = E[Z j ]. Then { 1/2 LW 1,..., W k LZ 1,..., Z k T V 2 min 1, 1.4 min λ j 2b 1 +2b 2 +b j k where T V denotes the total variation distance and LW 1,..., W k denotes the joint law of these random variables. ECP , paper 74. Page 3/10

4 3 Proof of Theorem 1.1: the infinite-volume case Proof. We recall that E = [0, 1] d, + ] and V N = [0, n 1] d Z d. To show the convergence of η n to η, we will exploit Kallenberg s theorem [11, Theorem 4.7]. According to it, we need to verify the following conditions: i for any A, a bounded rectangle 2 in [0, 1] d, and x, y], + ] E[η n A x, y]] E[ηA x, y]] = A e x e y. We adopt the convention e = 0 and the notation A for the Lebesgue measure of A. ii For all k 1, and A 1, A 2,..., A k disjoint rectangles in [0, 1] d and R 1, R 2,..., R k, each of which is a finite union of disjoint intervals of the type x, y], + ], P η n A 1 R 1 = 0,..., η n A k R k = 0 P ηa 1 R 1 = 0,..., ηa k R k = 0 = exp where ωd z := e z d z. k A j ω R j 3.1 Let us denote by u N z := a N z + b N. The first condition follows by Mills ratio 1 1t 2 e t 2 /2 2πt P N 0, 1 > t e t2 /2 2πt, t > j=1 More precisely E[η n A x, y]] = P ϕ α u N x, u N y] α na V N un x 2 e 2g0 2πuN x α na V N e x+o1 na V N N u N y 2 e 2g0 1 2πuN y e y+o1 N 1 u N y g0 log N1 + o 1 A e x e y. 3.4 Similarly, one can plug in 3.3 the reverse bounds of 3.2 to prove the lower bound, and thus condition i. To show ii, we need a few more details. Let k 1, A 1,..., A k and R 1,..., R k be as in the assumptions. Let us denote by I j = na j V N and I = I 1... I k. For α I j define X α := 1 { ϕα b N a N R j and p α := P ϕ α b N /a N R j. Choose now a small ɛ > 0 and fix the neighborhood of dependence B α := B α, log N 2+2ɛ I 3 for α I. Let W j := α I j X α and Z j be as in Theorem 2.2. By the simple observation that P η n A 1 R 1 = 0,..., η n A k R k = 0 = P W 1 = 0,..., W k = 0, 2 A bounded rectangle has the form J 1 J d with J i = [0, 1] a i, b i ], a i, b i R for all 1 i d. 3 Bx, r denotes a ball of radius r centered at x ECP , paper 74. Page 4/10

5 to prove the convergence 3.1, we can use Theorem 2.2 and show that the error bound on the RHS of 2.3 goes to 0. First we bound b 1 as follows. By definition of R 1, R 2,..., R k, there exists z R such that R j z, + ] for 1 j k. Hence for any 1 j k, for any α I j we have that ϕα b N p α = P a N R j Pϕ α > u N z 3.2 The bound is independent of α and j, therefore for some C > 0 u N z 2 e 2g0 g0. 2πuN z b 1 CNlog N d2+2ɛ e 2z N For b 2 note that it was shown in [9] that for z R and α β V N 2 κ3/2 Pϕ α > u N z, ϕ β > u N z N 2/2 κ max {e 2z 1 κ 1/2 {z 0, e 2z/2 κ 1 {z> Here we have introduced κ := P 0 H0 = + 0, 1 and H 0 = inf {n 1 : S n = 0. Observe that for any 1 j k, α I and β B α one has E[X α X β ] Pϕ α > u N z, ϕ β > u N z so that by 3.6 we can find some constant C > 0 such that b 2 C N κ/2 κ log N d2+2ɛ max {e 2z 1 {z 0, e 2z/2 κ 1 {z>0 0. The error is similar to the estimate obtained in [9, Equation 8]. Finally we need to handle b 3. From Section 2.2 we set for α I, H 1 := σ X β : β I \ B α and we define H 2 := σ ϕ β : β I \ B α. We observe that b 3 = α I E [ E [X α p α H 1 ] ] α I E [ E [X α H 2 ] p α ] since H 1 H 2 and using the tower property of the conditional expectation. Now denote by U α := Z d \ I \ B α. Let us abbreviate u N R j := {u N y : y R j. Then for α I j and 1 j k, by the Markov property of the DGFF [14, Lemma 1.2] we have that E [X α H 2 ] = P Uα ψ α + µ α u N R j P a. s. where ψ α α Z d is a Gaussian Free Field with zero boundary conditions outside U α and µ α = HI\Bα < +, S = β HI\Bα ϕ β. β I\B α P α Here H Λ := inf {n 0 : S n Λ, Λ Z d. Now as in [9, Equation 10] one can show, using the Markov property, that Var [µ α ] c sup gα, β β I\B α log N 21+ɛd 2 for some c > 0. Hence we get that there exists a constant c > 0 independent of α and j such that P µ α > u N z 1 ɛ c exp log N 2d 51+ɛ. 3.7 Recalling that R j z, + ] for all 1 j k, this immediately shows that for d 3 k E [ PUα ψ α + µ α u N R j p α 1{ µα >un z ] 0. 1 ɛ ECP , paper 74. Page 5/10

6 So to show that b 3 0 we are left with proving k E [ PUα ψ α + µ α u N R j p α 1{ µα un z ] 0. 1 ɛ 3.8 We now focus on the term inside the summation. For this, first we write R j = m l=1 w l, r l ] with < w 1 < r 1 < w 2 < < r m + for some m 1. Hence, we can expand the difference in the absolute value of 3.8 as follows: p α P Uα ψ α + µ α u N R j = = m l=1 m l=1 m l=1 Pϕ α u N w l, u N r l ] P Uα ψ α + µ α u N w l, u N r l ] Pϕ α > u N w l P Uα ψ α + µ α > u N w l Pϕ α > u N r l P Uα ψ α + µ α > u N r l 3.9 if r l = + for some l, we conventionally set Pϕ α > u N r l = 0 and similarly for the other summand. Using the triangular inequality in 3.8, it turns out that to finish it is enough to show that for an arbitrary w R, [ ] E PUα ψ α + µ α > u N w Pϕ α > u N w 1 { µα u N z 1 ɛ α I For this, first we show that on Q := { Pϕ α > u N w > P Uα ψ α + µ α > u N w T 1,2 = [ E Pϕ α > u N w P ] Uα ψ α + µ α > u N w 1 { µα u N z 1 ɛ 1 Q 0. α I 3.11 This follows from the same estimates of T 1,2 and Claim 6 of [9]. Indeed on Q { µ α u N z 1 ɛ α I α I Pϕ α > u N w P Uα ψ α + µ α > u N w g0 e u N w2 2g0 2πuN w o 1 guα αu N w e g0 g Uα α g0un w1 + o 1 un w 2 +o1 2g0 CN g0 e u N w2 2g0 2πuN w o 1 = o 1. Similarly one can show that on the complementary event Q c recall 3.11 for the definition of Q T 1,1 = ] E [ PUα ψ α + µ α > u N w Pϕ α > u N w 1 { µα un z 1 ɛ 1 Q c = o 1. α I This shows that b 3 0. Hence from Theorem 2.2 it follows that k PW 1 = 0,..., W k = 0 P Z j = 0 j=1 = o 1, ECP , paper 74. Page 6/10

7 having used the independence of the Z j s. Notice that by definition Z j is a Poisson random variable with intensity α I j P ϕ α b N /a N R j. Decomposing R j as a union of finite intervals and using Mills ratio, similarly to the argument leading to 3.4, one has P Z j = 0 exp A j ωr j recall ωr j = R j e z d z. Hence it follows that k PZ j = 0 exp j=1 which completes the proof of ii and therefore of Theorem 1.1. k A j ωr j, 3.12 j=1 4 Proof of Theorem 1.2: the finite-volume case We will now show the theorem for the field with zero boundary conditions. As remarked in the Introduction, since on the bulk defined in 2.1 we have a good control on the conditioned field, we will first prove convergence therein, and then we will use a converging-together theorem to achieve the final limit. We will first need some notation used throughout the Section: first, we consider ψ α α VN with law P N := P VN. We also use the shortcut g N, = g VN,. We will need the notation C + K E for the set of positive, continuous and compactly supported functions on E = [0, 1] d, + ]. We first begin with a lemma on the point process convergence on bulk. Define a point process on E by ρ δ n = ε α α VN δ n, ψα b N. 4.1 a N Lemma 4.1. Let δ > 0. On M p E, ρ δ d n ρ δ where ρ δ is a Poisson random measure with intensity d t [δ,1 δ] d e x d x 4. Proof. We will show i and ii of Page 4 and from which we will borrow the notation starting from now. [ i We begin with an upper bound on ẼN ρ δ n A x, y] ] : P N ψ α > u N x P N ψ α > u N y α na V δ N 3.2 α na V δ N e u N x 2 2g N α u N y2 gn α e 2g N α gn α 1 + o 1 2πuN x 2πuN y Lemma 2.1 = e u N x 2 2g01+cn u N y2 g01 + cn e 2g01+cn g0 1 + cn 2πuN x 2πuN y α na V δ N n + e x e y A [δ, 1 δ] d. 4.2 We stress that in the second step the error term c n := O n 2 d coming from Lemma 2.1 guarantees the convergence in the last line. The lower bound follows similarly. ii To show the second condition we again use Theorem 2.2. Let A 1,..., A k and R 1,..., R k be as in proof of Theorem 1.1. Let I j := na j V δ N and I = I 1 I k. For ɛ > 0 we are setting B α := B α, log N 21+ɛ I. Note that, albeit slightly different, we are using the same notations for the neighborhood of dependence and the index sets of Section 3, 4 d t [δ,1 δ] d is the restriction of the Lebesgue measure to [δ, 1 δ]d. ECP , paper 74. Page 7/10

8 but no confusion should arise. Observe that there exists z R such that for all 1 j k, R j z, ]; we have u N z 2 p α = P ψα b N N u N R j a P N ψ α > u N z 3.2 e 2g0 g0 N 2πuN z where we have also used the fact that g N α g0. The bound on b 1 cf. Theorem 2.2 follows exactly as in 3.5 and yields that, for some C > 0, b 1 CNlog N d2+2ɛ e 2z N 2 0. The calculation of b 2 can be performed similarly using the covariance matrix of the vector ψ α, ψ β, α β VN δ and Lemma 2.1. This gives that for some C, C > 0 independent of α, β VN δ b 2 C log N exp u N z O N 2 d/d g0 + gα β α I β B α C N κ/2 κ log N 2d1+ɛ max {e 2z 1 {z 0, e 2z/2 κ 1 {z>0 0 cf. [9, Equation 8]. Note the estimate for b 2 is exactly same as in the infinite volume case. We will now pass to b 3. We repeat our choice of H 1 = σ X β : β I \ B α and H 2 = σ ψ β : β I \ B α so that b 3 becomes k Ẽ N [ ] Ẽ N [X α p α H 1 ] k Ẽ N [ Ẽ N [X α H 2 ] p α ]. We define U α := V N \ I \ B α. By the Markov property of the DGFF Ẽ N [X α H 2 ] = P Uα ξ α + h α u N R j PN a. s. 4.3 for ξ α α Z d a DGFF with law P Uα and h α α Z d is independent of ξ. From [9, Equation 28] we can see that, for any α VN δ and N large enough such that B α, log N 21+ɛ V N, Var [h α ] = HI\Bα < +, S = β HI\Bα g N α, β This yields k Ẽ N β I\B α P α c sup g N α, β β I\B α log N. 21+ɛd 2 [ PUα ξ α + h α > u N R j p α 1{ hα >un z ] 0. 1 ɛ 4.4 It then suffices to show k [ Ẽ N PUα ξ α + h α > u N R j p α 1{ hα un z ] 0. 1 ɛ 4.5 One sees that the breaking up 3.9 can be performed also here replacing ϕ α and ψ α with their laws with ψ α and ξ α with their laws respectively, and µ α with h α. Accordingly, it is enough to show that [ Ẽ N PUα ξ α + h α > u N w P ] N ψ α > u N w 1 { hα u N z 1 ɛ α I ECP , paper 74. Page 8/10

9 for all w R. To this aim, we choose for any w R the event Q := { PN ψ α > u N w > P Uα ξ α + h α > u N w and we proceed as in 3.11 with the help of Lemma 2.1 to show 4.6. Given this, the convergence b 3 0 is finally ensured. Hence we can conclude that LW 1,..., W k LZ 1,..., Z k T V 0 where Z j are i. i. d. Poisson of mean p α. By Mills ratio, as in 4.2 we see that PZ j = 0 exp Aj [δ, 1 δ] d ωrj. From this it follows that the two conditions i and ii of Kallenberg s Theorem are satisfied, and thus we obtain the convergence to a Poisson point process with mean measure given in i. Proof of Theorem 1.2. M p E is a Polish space with metric d p : d p µ, µ = i 1 min { µf i µ f i, 1 2 i, µ, µ M p E for a sequence of functions f i C + K E cf. [12, Section 3.3]. Therefore we are in the condition to use a converging-together theorem [13, Theorem 3.5], namely to prove that d ρ n η it is enough to show the following: a ρ δ n d ρ δ, as n +. b ρ δ d η as δ 0. c For every ɛ > 0, lim lim P N dp ρn, ρ δ n > ɛ = n + δ 0 Note that by Lemma 4.1, a is satisfied. For f C + K E, the Laplace functional of ρδ is given by cf. [12, Prop. 3.6] Ψ δ f := E [ exp ρ δ f ] = exp 1 e ft,x d t [δ,1 δ] d e x d x. E Hence by the dominated convergence theorem we can exchange limit and expectation as δ 0 to obtain that Ψ δ f exp 1 e ft,x d t e x d x E and the right hand side is the Laplace functional of η at f. This shows b. Hence to complete the proof it is enough to show 4.7. Thanks to the definition of the metric d p it suffices to prove that for f C + K E and for ɛ > 0 lim sup δ 0 lim n + P N ρ n f ρ δ nf > ɛ = 0. Without loss of generality assume that the support of f is contained in [0, 1] d [z 0, + for some z 0 R. Choosing n large enough such that u N z 0 > 0 and g N α g0, we ECP , paper 74. Page 9/10

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