J Electr on i o u r nal o f P c r o ba bility Vol. 16 (2011), Paper no. 36, pages 1048–1064. Journal URL http://www.math.washington.edu/~ejpecp/ From the Pearcey to the Airy process M. Adler∗ † M. Cafasso ‡ P. van Moerbeke §¶ Abstract Putting dynamics into random matrix models leads to finitely many nonintersecting Brownian motions on the real line for the eigenvalues, as was discovered by Dyson. Applying scaling limits to the random matrix models, combined with Dyson’s dynamics, then leads to interesting, infinite-dimensional diffusions for the eigenvalues. This paper studies the relationship between two of the models, namely the Airy and Pearcey processes and more precisely shows how to approximate the multi-time statistics for the Pearcey process by the one of the Airy process with the help of a PDE governing the gap probabilities for the Pearcey process. Key words: Airy process, Pearcey process, Dyson’s Brownian motions. AMS 2010 Subject Classification: Primary 60K35; Secondary: 60B20. Submitted to EJP on December 19, 2010, final version accepted May 1, 2011. ∗ Department of Mathematics, Brandeis University, Waltham, Mass 02454, USA.adler@brandeis.edu The support of a National Science Foundation grant # DMS-07- 04271 is gratefully acknowledged ‡ Centre de Recherches Mathématiques, Université de Montréal C. P. 6128, succ. centre ville, Montréal, Québec, Canada H3C 3J7 and Department of Mathematics and Statistics, Concordia University 1455 de Maisonneuve W., Montréal, Québec, Canada H3G 1M8.cafasso@crm.umontreal.ca § Département de Mathématiques, Université Catholique de Louvain, 1348 Louvain-la-Neuve, Belgium and Brandeis University, Waltham, Mass 02454, USA. vanmoerbeke@math.ucl.ac.be ¶ The support of a National Science Foundation grant # DMS-07-04271, a European Science Foundation grant (MISGAM), a Marie Curie Grant (ENIGMA), Nato, FNRS and Francqui Foundation grants is gratefully acknowledged. † 1048 1 Introduction Putting dynamics into random matrix models leads to finitely many nonintersecting Brownian motions on R for the eigenvalues, as was discovered by Dyson [13]. Applying scaling limits to the random matrix models, combined with Dyson’s dynamics, then leads to interesting, infinite-dimensional diffusions for the eigenvalues. This paper studies the relationship between two of the models, namely the Airy and Pearcey processes and more precisely shows how to approximate the Pearcey process by the Airy process with the help of a PDE [2] governing the gap probabilities for the Pearcey process. The Airy process was introduced by Prähofer-Spohn [18] and further developed by K. Johansson [14, 15]. A simple non-linear 3rd order PDE for the transition probabilities for this process was found in [3]; see also [19, 20]. The Pearcey process was introduced in [21, 17] in the context of non-intersecting Brownian motions and plane partitions, also based on prior work on matrix models with external source [16, 10, 22, 23, 11, 12, 5, 7, 8, 9]. Consider n nonintersecting Brownian particles on the real line R, −∞ < x 1 (t) < ... < x n (t) < ∞ with (local) Brownian transition probability given by ( y−x)2 1 p(t; x, y) := p e− t , πt all starting from the origin x = 0 at time t = 0 and such that n 2 particles are forced to ± For very large n, the average mean density of particles has its support, for t ≤ 1 2 1 , 2 pn 2 at t = 1. on one interval centered about x = 0, and for < t ≤ 1 on two intervals symmetrically located about x = 0. The end points of the interval(s) of support describe a heart-shaped region in (x, t)-space, with a cusp at (x 0 , t 0 ) = (0, 1/2). The Pearcey process P (τ) is defined (see Figure 1) as the motion of these nonintersecting Brownian motions for large n, about (x 0 , t 0 ) = (0, 1/2) (i.e., near the cusp), with space microscopically rescaled by a factor of n−1/4 and time rescaled by a factor n−1/2 , in tune with the Brownian motion rescaling. A partial differential equation for the Pearcey process was found by Adler-van Moerbeke [4] and in [2] a much better version was obtained, namely a simple third order non-linear PDE for the transition probabilities; it was obtained by a scaling limit on a PDE for non-intersecting Brownian motions with target points. This PDE is related to the Boussinesq equation and its hierarchy; this is part of a general result on integrable kernels, as explained in the paper [1]. Near the boundary of the heart-shaped region of Figure 1, but away from the cusp, the local fluctuations behave as the so-called Airy process, which describe the non-intersecting Brownian motions with space stretched by the customary GUE edge rescaling n1/6 and time rescaled by the factor n1/3 , again in tune with the Brownian motion space-time rescaling. This paper shows how the Pearcey process statistics tends to the Airy process statistics when one is 2 moving out of the cusp x = 27 (3(t − t 0 ))3/2 very near the boundary, that is at a distance of (3τ)1/6 for τ very large, with τ being the Pearcey time. To be precise, in the two-time case, the times τi must be sufficiently near -in a very precise way- for the limit to hold. The main result of the paper can be summarized as follows: 1049 Theorem 1.1. Given finite parameters t 1 < t 2 , let both τ1 , τ2 → ∞, such that τ2 − τ1 → ∞ behaves in the following precise way: τ2 − τ1 2(t 2 − t 1 ) = (3τ1 )1/3 + t2 − t1 (3τ1 )1/3 2t 1 t 2 + 3τ1 −5/3 + O τ1 . (1) Let also E1 , E2 be two arbitrary finite intervals. The parameters t 1 and t 2 provide the Airy times in the following approximation of the Airy process by the Pearcey process: ( )! 2 2 \ P (τi ) − 27 (3τi )3/2 P ∩ −Ei = ; 1/6 (3τ ) i i=1 (2) ! ! 2 \ 1 . =P A (t i ) ∩ (−Ei ) = ; 1 + O 4/3 τ1 i=1 The same estimate holds as well for the one-time case. A similar (but different) result was then obtained in [6] for the one–time case in the situation when one is moving out of the cusp following both the two branches of the cusp simultaneously. It is also worth recalling that, in this paper as well as in [6], the statistics of the processes are studied on intervals rather than on general Borel subsets. The proof of this Theorem proceeds in two steps. At first, Proposition 2.1, stated later, shows that, using the scaling above, the Pearcey kernel tends to the Airy kernel. In a second step, we show, using the PDE for the Pearcey process [2] and Proposition 2.1, the result of Theorem 1.1. It is a fact that both the Airy and Pearcey processes are determinantal processes, for which the multitime gap probabilities is given by the matrix Fredholm determinant of the matrix kernel, which will be described here. The Airy process is a determinantal process, for which the multi-time gap probabilities1 are given by the matrix Fredholm determinant for t 1 < . . . < t ` , ` \ (3) P (A (t i ) ∩ Ei = ;) = det I − [χ Ei K tAt χ E j ] i j i=1 1≤i, j≤` of the matrix kernel in ~x = (x 1 , . . . , x ` ) and ~y = ( y1 , . . . , y` ), denoted as follows: p KA x , ~y ) d ~x d ~y t ,...,t (~ ` 1 := p K tA,t (x 1 , y1 ) d x 1 d y1 . . . 1 1 .. . p K tA,t (x ` , y1 ) d x ` d y1 . . . ` 1 p K tA,t (x 1 , y` ) d x 1 d y` 1 ` .. . p K tA,t (x ` , y` ) d x ` d y` ` (4) , ` where, for arbitrary t i and t j , the extended Airy kernel K tA,t (x, y) is given by (see Johansson [14, i j 15]) K tA,t (x, y) := K̃ tA,t (x, y) − 1I(t i < t j )pA (t j − t i , x, y), (5) i 1 j i j χ E is the indicator function for the set E 1050 n/2 # of particles n/2 t - p p n/2 n/2 t =1 t = 1/2 x Figure 1: The Pearcey process. where K̃ tA,t (x, i j Z ∞ e−λ(t i −t j ) A(x + λ)A( y + λ)dλ y) := 0 = 2 (2πi) 1 A Z Z 1 p (t, x, y) := p e 4πt dv Γ> du Γ< ( x− y )2 t3 − 4t − 2t 12 e−v 3 /3+ y v 3 e−u /3+xu 1 (v + t j ) − (u + t i ) ( y+x ) , for t > 0. The contour u ∈ Γ< consists of two rays emanating from the origin with angles θ1 and θ10 with the positive real axis, and the contour v ∈ Γ> also consists of two rays with angles θ2 and θ20 with the negative real axis, as indicated in Figure 2. As is well known, one may choose θ1 , θ2 , θ10 , θ20 = π/3. In particular, for t i = t j , this is the customary Airy kernel K tA,t i i A := K (x, y) := Z ∞ dλA(x + λ)A( y + λ) = A(x)A0 ( y) − A( y)A0 (x) x−y 0 . (6) The Pearcey process is determinantal as well, for which the multi-time gap probability is given by the Fredholm determinant for τ1 < . . . < τ` , ` \ P (P (τi ) ∩ Ei = ;) = det 1I − [χ Ei KτP τ χ E j ] (7) i i=1 1051 j 1≤i, j≤` ~ = (ξ1 , . . . , ξ` ) and η for the Pearcey matrix kernel in ξ ~ = (η1 , . . . , η` ), denoted as follows: Æ ~ η ~ dη ( ξ, ~ ) dξ ~ KP τ ,...,τ ` 1 := p KτP ,τ (ξ1 , η1 ) dξ1 dη1 . . . 1 1 .. . p KτP ,τ (ξ` , η1 ) dξ` dη1 . . . ` p KτP ,τ (ξ1 , η` ) dξ1 dη` 1 ` .. . p KτP ,τ (ξ` , η` ) dξ` dη` ` 1 (8) , ` where for arbitrary τi and τ j , (see Tracy-Widom [21]) KτP ,τ (ξ, η) = K̃τP ,τ (ξ, η) − 1I(τi < τ j )pP (τ j − τi ; ξ, η), i j i j with K̃τP ,τ (ξ, η) := − i j 1 4π2 Z Z dU X Y dV e− τjV2 V4 + 2 −V η 4 V −U e− τ U2 U4 + i2 −Uξ 4 (9) (ξ−η)2 1 pP (τ; ξ, η) := p e− 2τ , for τ > 0, 2πτ where X and Y are the following contours: X = X 1 ∪ X 2 consists of four rays emanating from the origin with angles σ1 , σ10 with the positive real axis and σ2 , σ20 with the negative real axis, as given in Figure 2. The contour Y consists of two rays emanating from the origin with angles τ, τ0 with the negative real axis; it is customary to pick σ1 = σ2 = σ10 = σ20 = π/4 and τ = τ0 = π/2. V ∈YE E E E U ∈ X 2 @I @ σ2 @ E OE U ∈ X1 E σ1 @τ E @Ep τ0 @ @ σ0 σ20 @ 1 @ U ∈ X2 R U ∈ X1 @ V ∈ Y v ∈ Γ> A AM A θ A u ∈ Γ< θ 1 A Ap A θ20 A θ10 A A AN u ∈ Γ< v ∈ Γ> A 2 π/8 < σ1 , σ2 , σ10 , σ20 < 3π/8 π/6 < θ1 , θ2 , θ10 , θ20 < π/2 3π/8 < τ, τ0 < 5π/8 Pearcey contour Airy contour Figure 2: Integration paths for the kernels. 1052 In [4, 2], it was shown that, given intervals (i) (i) Ei = (ξ1 , ξ2 ), the log of the probability Q = Q(τ1 , . . . , τ` ; E1 , . . . , E` ) := log P ` \ {P (τi ) ∩ Ei = ;} i=1 satisfies a non-linear PDE, which we describe here. Given the times and the intervals above, one defines the following operators: ∂τ := X ∂ i "τ := ∂ τi X ∂E := , X ∂ k ∂ ξk i τi i ∂ ∂ τi " E := , ∂E := , (i) ∂E i XX i X ∂ (i) ξk k (i) ∂ ξk i . (10) Then Q satisfies the Pearcey partial differential equation in its arguments and the boundary of the intervals2 : X ¦ © 1 2∂τ3Q + (2"τ + " E − 2)∂ E2Q − ( τi ∂E )∂τ ∂ E Q + ∂τ ∂ E Q, ∂ E2Q = 0. i ∂E 4 i 2 (11) From the Pearcey to the Airy kernel Define the rational functions Φ(x, t; u) := − h(x, t; u) := 1 4(3u)3 ux 4 − t (3u)2 + x − t2 3u + 4 3 tx + u 6 t2 x (12) 2 (x + 6t ), and the diagonal matrix S := eΦ(x,t 1 ;z 4 )−h(x,t 1 ;z 4 ) 0 0 eΦ(x,t 2 ;z 4 )−h(x,t 2 ;z 4 ) . We now state: Proposition 2.1. Given a parameter z → 0, define two times τ1 and τ2 blowing up like z −6 and depending on two parameters t 1 and t 2 , τi = 2 1 3z 6 (1 + 6t i z 4 ) + O(z 10 ). Here and below, given two arbitrary functions f , g and a vector field X , we denote { f , g}X := gX ( f ) − f X (g) 1053 (13) Given large ξi and η j , define new space variables x i and y j , using τi above, ξi = ηj = 2 27 2 27 (3τi )3/2 − (3τi )1/6 x i (14) 3/2 (3τ j ) 1/6 − (3τ j ) yj. With this space-time rescaling, the following asymptotic expansion holds for the Pearcey kernel in powers of z 4 , with polynomial coefficients in t 1 + t 2 , Æ Æ −1 4 A ~ η ~ η ~xd ~ y + O(z 8 ), S KP ( ξ, ~ ) d ξd ~ S = 1 + (t + t )O (z ) K (~ x , ~ y ) d (15) 1 2 1 τ ,τ t ,t 1 2 1 2 where O1 refers to a differential operator in ∂ /∂ x, ∂ /∂ y, with polynomial coefficients in x, y, t 1 ± t 2 and (t 1 − t 2 )−1 and varying with the four matrix entries of the matrix kernel. On general grounds, due to the Fredholm determinant formula, this would give Theorem 1.1, but −4/3 −2/3 with a much poorer estimate in (2), namely instead of O(τ1 ), the estimate would be O(τ1 ). The existence of the PDE for the Fredholm determinant of the Pearcey process enables us to boot−2/3 −4/3 strap the O(τ1 ) estimate to O(τ1 ), without tears! Also note that conjugating a kernel does not change its Fredholm determinant. Before proving Proposition 2.1, we shall need the following identities: Lemma 2.2. Introducing the polynomial Ψ(x, s; ω) := 3 ω4 + ω2 s2 − 4s(xω − ω3 ), 4 2 1 (16) the Airy kernel (6) and the (double) integral part of the extended Airy kernel (5), at t 1 = s, t 2 = −s, satisfy the following differential equations, Ψ(x, s, −∂ x ) − Ψ( y, s, ∂ y ) + 4s K A (x, y) (17) 1 = (x − y)(x + y + 6s2 )K A (x, y), 4 and 3 ∂ Ψ(x, s, −∂ x ) − Ψ( y, −s, ∂ y ) − s (∂ x − ∂ y ) K̃ A (x, y) s,−s 2 ∂s 1 = (x − y)(x + y + 6s2 )K̃ A (x, y). s,−s 4 (18) Proof. The operator on the left hand side of (17) reads Ψ((x, s, −∂ x ) − Ψ(( y, s, ∂ y ) + 4s 3 1 = 4s(1 + y∂ y − ∂ y3 + x∂ x − ∂ x3 ) + s2 (∂ x2 − ∂ y2 ) + (∂ x4 − ∂ y4 ). 2 4 (19) Using the first representation (6) of the Airy kernel, one checks using the differential equation for the Airy kernel, A00 (x) = xA(x) and thus A000 (x) = xA0 (x) + A(x) and A(iv) (x) = 2A0 (x) + x 2 A(x), 1054 and using differentiation by parts to establish the last equality, ( y∂ y − ∂ y3 ) + (x∂ x − ∂ x3 ) K A (x, y) Z∞ d = − dz z + 2 A(x + z)A( y + z) = −K A (x, y). dz 0 In order to take care of the other pieces in (19), one uses the second representation (6) of the kernel K A (x, y), yielding ∂ x2 − ∂ y2 K A (x, y) = (x − y)K A (x, y) ∂ x4 − ∂ y4 K A (x, y) = (x 2 − y 2 )K A (x, y). This establishes the first identity (17) of Lemma 2.2. The operator on the left hand side of the second identity (18) reads: 3 ∂ Ψ(x, s, −∂ x ) − Ψ( y, −s, ∂ y ) − s (∂ x − ∂ y ) 2 ∂s 1 4 3 3 ∂ = (∂ x −∂ y4 ) + s2 (∂ x2 −∂ y2 ) + 4s(x∂ x − ∂ x3 − y∂ y + ∂ y3 ) − s (∂ x −∂ y ), 4 2 2 ∂s of which we will evaluate all the different terms acting on the integral. Notice at first that, since the expression under the differentiation vanishes at 0 and ∞, one has Z∞ ∂ −2sz A 0= ze K (x + z, y + z) dz ∂z Z0∞ Z∞ dz e−2sz K A (x + z, y + z) − = zdz e−2sz A(x + z)A( y + z) 0 0 Z ∞ − 2s zdz e−2sz K A (x + z, y + z), 0 Again, using the second representation (6) of the kernel K A (x, y), one checks Z∞ A ∂ x − ∂ y K̃ (x, y) = −(x − y) dz e−2sz K A (x + z, y + z) s,−s ∂ x2 2 − ∂ y K̃ A s,−s 0 (x, y) = (x − y)K̃ A s,−s (x, y), and also s ∂ ∂s ∂ x −∂ y K̃ A s,−s (x, y) = 2s(x − y) Z ∞ zdz e−2sz K A (x + z, y + z) 0 and, using the differential equation x∂ x − ∂ x3 A(x) = −A(x) and (20), −2s x∂ x − ∂ x3 − y∂ y − ∂ y3 K̃ A (x, y) s,−s Z∞ zdz e−2sz K A (x + z, y + z) = −2s(x − y) 0 1055 (20) Using A(i v) (x) = 2A0 (x) + x 2 A(x) and the Darboux-Christoffel representation (6) of the Airy kernel, one checks ∂ x4 − ∂ y4 K̃ A (x, y) s,−s Z∞ e−2sz (A(i v) (x + z)A( y + z) − A(x + z)A(iv) ( y + z))dz = 0 = −2(x − y) Z +2(x − y) ∞ dz e−2sz K A (x + z, y + z) + (x 2 − y 2 )K̃ A (x, y) Z0 ∞ s,−s zdz e−2sz A(x + z)A( y + z). 0 Adding all these different pieces and using the expression for Z∞ zdz e−2sz K A (x + z, y + z), −2s(x − y) 0 given by (20), leads to the statement of Lemma 2.2. For the sake of notational convenience in the proof below, set p p P P Æ K11 (ξ1 , η1 ) dξ1 dη1 K12 (ξ1 , η2 ) dξ1 dη2 ~ ~) d ξ ~ dη KP ~= τ1 ,τ2 (ξ, η p p P P K21 (ξ2 , η1 ) dξ2 dη1 K22 (ξ2 , η2 ) dξ2 dη2 (21) P A and similarly for the Airy kernel KA t 1 t 2 ; the K̃i j and K̃i j refer, as before, to the (double) integral part. e A (x, y) and K e A (x, y), Proof of Proposition 2.1: Notice that acting with ∂ x and ∂ y on the kernels K 11 12 22 21 amounts to multiplication of the integrand of the kernels with −u and v respectively; i.e., u ↔ −∂ x and v ↔ ∂ y . Given the (small) parameter z ∈ R, consider the following (i, j)-dependent change of integration variables (U, V ) 7→ (u, v) in the four Pearcey kernels KiPj in (21), namely U= 1 3z 3 (1 + 3uz 4 )(1 + 3t i z 4 ), V= 1 3z 3 (1 + 3vz 4 )(1 + 3t j z 4 ), (22) together with the changes of variables (ξ, η, τi , τ j ) 7→ (x, y, t i , t j ), in accordance with (13) and (14), 1 1 τi = 6 (1 + 6t i z 4 ) + O(z 10 ), τ j = 6 (1 + 6t j z 4 ) + O(z 10 ) 3z 3z (23) 2 2 (3τi )3/2 − (3τi )1/6 x, η= (3τ j )3/2 − (3τ j )1/6 y. ξ= 27 27 P P P To be precise, for Kkk , one sets in the transformations above i = j = k, for K12 and K21 , one sets i = 1, j = 2 and i = 2, j = 1 respectively. Then, remembering the expressions Φ and Ψ, defined in 1056 (12) and (16), the following estimate holds for small z: 4 e−Φ( y,t j ;z ) e− e−Φ(x,t i ;z 4) τjV2 V4 + 2 −V η 4 4 e − U4 + τi U 2 −Uξ 2 = = e−z 4 e−z 4 Ψ(x,t e e Ψ( y,t j ;v) i ;u) v3 e− 3 + y v 3 e − u3 +xu 4 −z Ψ( y,t j ;∂ y ) −z 4 Ψ(x,t i ;−∂ x ) (1+ t i O(z 8 )+ t j O(z 8 )) (24) 3 e − v3 + y v e − u3 +xu 3 (1+ t i O(z 8 )+ t j O(z 8 )); replacing in the latter expression u and v by differentiations ∂ x and ∂ y has the advantage that upon doubly integrating in u and v, the fraction of exponentials in front can be taken out of the integration. Moreover, setting t 1 = t + s and t 2 = t − s, one also checks3 : e−Φ( y,t j ;z 4 ) p P p (τ − τ ; ξ, η) dξdη j i 4 e−Φ(x,t i ;z ) 4 p z t = pA (−2s, x, y) d x d y 1 + (x − y)(x + y + 6s2 ) + r(x, y, s) + O(z 8 ) . 4 s (25) The multiplication by the quotient of the exponentials on the left hand side of the expressions above will amount to a conjugation of the kernel, which will not change the Fredholm determinant. Setting t 1 = t + s and t 2 = t − s, with t = 0, one finds for the non-exponential part in the kernel dUdV dξdη V −U p d x d y dud v(1 + 27 (t i + t j )z 4 ) = + O(z 8 ) v + t j − u − t i + 3z 4 (v t j − ut i ) ¨ p i = 1, j = 1 d x d y dud v 4 8 1 ± 4sz + O(z ) for v−u i = 2, j = 2 = ¨ p i = 1, j = 2 d x d y dud v 1 ± 3z 4 s(u+v) + O(z 8 ) for v−u∓2s v−u∓2s i = 2, j = 1. p (26) Along the same vein as the remark in the beginning of the proof of this Theorem, one notices that multiplication of the integrand of the kernel K̃ A (x, y) with the fraction, appearing in the last 12 21 formula of (26), amounts to an appropriate differentiation, to wit: ± 3 3s(u + v) v − u ∓ 2s ←→ 3 ∂ s (∂ y − ∂ x ). 2 ∂s The precise expression for r(x, y, s) := (x − y)2 − 8s2 (x + y − 2s2 ) will be irrelevant in the sequel. 1057 (27) So we have, using (24), (25), (26), and the differential equation for K A of Lemma 2.2, e e −Φ( y,t 1 ;z 4 ) 2 −Φ(x,t 1 2 p K 11 (ξ, η) dξdη p − K A (x, y) d x d y P ;z 4 ) 22 t=0 p = z 4 ±4s + Ψ((x, ±s, −∂ x ) − Ψ(( y, ±s, ∂ y ) K A (x, y) d x d y + O(z 8 ) p z4 = (x − y)(x + y + 6s2 )K A (x, y) d x d y + O(z 8 ); 4 (28) the upper(lower)-indices correspond to the upper(lower)-signs. Using (27), and the differential equation for K̃ A of Lemma 2.2, 12 21 e e −Φ( y,t 2 ;z 4 ) 1 p p A K̃ 12 (ξ, η) dξdη − K̃ 12 (x, y) d x d y P −Φ(x,t 1 ;z 4 ) 2 21 21 t=0 3 ∂ p A 4 s (∂ y −∂ x ) + Ψ(x, ±s, −∂ x ) − Ψ( y, ∓s, ∂ y ) K̃ 12 (x, y) d x d y =z 2 ∂s 21 t=0 (29) + O(z 8 ) p z4 2 A = (x − y)(x + y + 6s )K̃ 12 (x, y) d x d y 4 21 + O(z 8 ). t=0 Also, using (25), e−Φ( y,t 2 ;z 4 ) p p A p (τ2 − τ1 ; ξ, η) dξdη − p (−2s, x, y) d x d y P e−Φ(x,t 1 ;z ) p z4 = (x − y)(x + y + 6s2 )pA (−2s, x, y) d x d y + O(z 8 ). 4 4 t=0 (30) 4 Since h( y, t i , z 4 ) − h(x, t j ; z 4 ) = − z4 (x − y)(x + y + 6s2 ) for arbitrary t i, j = t ± s at t = 0, and thus e−h( y,t i ;z e 4 ) −h(x,t j ;z 4 ) =1+ z4 4 (x − y)(x + y + 6s2 ) + O(z 8 ) (31) Then the following approximations follow upon combining the three estimates above (28), (29), (30) and using (31): e e e e −Φ( y,t 1 ;z 4 )+h( y,t 1 ;z 4 ) 2 −Φ(x,t 1 2 2 ;z 4 )+h(x,t 1 2 ;z 4 ) −Φ( y,t 2 ;z 4 )+h( y,t 2 ;z 4 ) 1 −Φ(x,t 1 2 1 ;z 4 )+h(x,t e−Φ( y,t 2 ;z 4 )+h( y,t 2 ;z 4 ) e−Φ(x,t 1 ;z 4 )+h(x,t 1 ;z 4) 1 2 p K 11 (ξ, η) dξdη p − K A (x, y) d x d y = O(z 8 ) P 22 p K̃ 12 (ξ, η) dξdη t=0 P ;z 4 ) 21 t=0 p pP (τ2 −τ1 ; ξ, η) dξdη 1058 p − K̃ A (x, y) d x d y = O(z 8 ) 12 t=0 21 p − pA (−2s, x, y) d x d y = O(z 8 ). The reader is reminded that this estimate sofar is done at t = 0. It then follows for t 6= 0 from the estimates (22), (23), (24), (25), (26), (27) that the right hand side of (15) is an asymptotic series in z 4 with polynomial coefficients in t 1 + t 2 = 2t, proving the claim about O1 . So far an important point was omitted, namely to analyze how the Pearcey contour turns in the limit into an Airy contour; a detailed description of the possible contours was given in Figure 2. The Pearcey-rays X 1 , with angles σ1 , σ10 ∈ (π/8, 3π/8) can be deformed into two acceptable rays θ1 , θ10 ∈ (π/6, π/2) for the Γ< -Airy contour, since the two intervals have a non-empty intersection. Also, since (3π/8, 5π/8)∩(π/6, π/2) 6= ;, the Pearcey Y -contour can be deformed into an acceptable v ∈ Γ> -Airy contour. Again, since the admissible interval (π/8, 3π/8) for σ2 and σ20 in the X 2 3 Pearcey contour contains π/6, one ends up in the limit integrating the function eu /3 along a contour of the form Γ> , which we may choose to have an angle of π/6 − δ with the negative real axis, for 3 small arbitrary δ > 0. In the sector Γ> , with 0 < θ2 = θ20 ≤ π/6 − δ, the function eu /3 decays 3 exponentially fast. Therefore, the contribution, due to the u-integration of eu /3 × (lower order terms) over Γ> , having an angle of π/6−δ with the negative real axis, vanishes by applying Cauchy’s Theorem in that sector. This ends the proof of Proposition 2.1. 3 From the Pearcey to the Airy statistics This section concerns itself with proving Theorem 1.1. Proof of Theorem 1.1: For any intervals E1 and E2 , the function 2 \ Q(τ1 , τ2 ; E1 , E2 ) = log P ! (P (τi ) ∩ Ei = ;) (32) i=1 satisfies the Pearcey PDE (11). Reparametrizing, without loss of generality, τ1 = τ + σ, τ2 = τ − σ, E1 = (ξ + η + µ, ξ + η − µ), E2 = (ξ − η + ν, ξ − η − ν), (33) leads to a manageable PDE for the function F (τ, σ; ξ, η, µ, ν) := Q(τ1 , τ2 ; E1 , E2 ), (34) namely the PDE: 2 ∂ 3F ∂ τ3 + 1 4 2(σ ∂ ∂σ −τ ∂ ∂ 2F +ν −2 ∂τ ∂ξ ∂η ∂µ ∂ν ∂ ξ2 ¨ « ∂ 3F ∂ 2F ∂ 2F −σ + , = 0. ∂ τ∂ ξ∂ η ∂ τ∂ ξ ∂ ξ2 ∂ )+ξ ∂ +η ∂ +µ ∂ ∂ (35) ξ To go from the Pearcey PDE (11) to the PDE (35), one notices that the operators (10), appearing in the Pearcey PDE (11), have simple expressions in terms of the variables τ, σ, ξ, η, µ, ν, ∂τQ = ∂F ∂τ , ∂E Q = ∂F ∂ξ , "E Q = ξ 1059 ∂ ∂ξ +η ∂ ∂η +µ ∂ ∂µ +ν ∂ ∂ν F, "τQ := σ ∂ ∂σ +τ ∂ ∂τ X F, τi ∂E Q = τ i i ∂ +σ ∂ξ ∂ ∂η F. The change of variables, considered in (13) and (14), combined with a linear change of variables, in parallel with (33), ¨ 1 t1 = t + s 4 τi = 6 (1 + 6t i z ) with t2 = t − s 3z ¨ (36) 2 Ẽ1 = (x + y + u, x + y − u) 3/2 1/6 (3τi ) − (3τi ) Ẽi with Ei = Ẽ2 = (x − y + v, x − y − v) 27 yields a z-dependent invertible map, T : (t, s, x, y, u, v) 7→ (τ, σ, ξ, η, µ, ν), (37) and thus the function F in (34) leads to a new z-dependent function G: F (τ, σ; ξ, η, µ, ν) = F (T (t, s, x, y, u, v)) =: G(t, s, x, y, u, v), 4 of which we compute, in principle, the series in z. From Proposition 2.1, p it follows that the z -term in the asymptotic expansion in (15) vanishes when t = 0; so, omitting d x d y, one has S for some kernel KPτ ,τ S−1 = KAt ,t 1 1 2 2 + z4 K1 + ∞ X z 4i Ki , with K1 = t H, 2 H, with the Ki polynomial in t. Then defining −1 L i := (I − KA t ,t ) Ki , 1 (38) 2 one finds: G(t, s, x, y, u, v) = F (τ, σ; ξ, η, µ, ν) 2 \ = log P {P (τi ) ∩ Ei = ;} (39) i=1 KPτ ,τ ) = log det(I − KA t ,t ) = log det(I − 1 2 E1 ×E2 where − z 4 Tr L1 − z 8 Tr(L2 + 1 L2) − . . . 2 1 =: G0 (s; Ẽ1 , Ẽ2 ) − G1 (t, s; Ẽ1 , Ẽ2 )z 4 − G2 (t, s; Ẽ1 , Ẽ2 )z 8 + O(z 12 ), 1 2 Ẽ1 × Ẽ2 KAt ,t ) G0 (s; Ẽ1 , Ẽ2 ) = log det(I − G1 (t, s; Ẽ1 , Ẽ2 ) = t Tr((I − 1 K 2 Ẽ1 × Ẽ2 A −1 t 1 ,t 2 ) H) , (40) Ẽ1 × Ẽ2 . Note G0 (s; Ẽ1 , Ẽ2 ) is t-independent, since the Airy process is stationary. Then setting F (τ, σ; ξ, η, µ, ν) = G(T −1 (τ, σ; ξ, η, µ, ν)) = G(t, s, x, y, u, v) 1060 into the PDE (35) yields, using differentiation by parts, a new PDE for G(t, s, x, y, u, v). Upon setting the expansion (39) of G for small z, G(t, s, x, y, u, v) = G0 (s; Ẽ1 , Ẽ2 ) − G1 (t, s; Ẽ1 , Ẽ2 )z 4 + O(z 8 ), into this new PDE leads to ∂ 3 G0 z 2 2s ∂ s∂ x 2 − ∂ 3 G1 + O(z 6 ) = 0, ∂ t∂ x 2 (41) 3 with G1 being polynomial in t. From the term ∂∂ τF3 in the PDE (35), it would seem like the leading term would be of order z −6 ; in fact, by explicit but non-enlightening calculations, there are two consecutive cancellations, so that the first non-trivial term has order z 2 , thus leading to a simple PDE connecting G1 with G0 . But since G0 is t-independent, from equation (41) we deduce ∂ ∂2 ∂ G − 2ts G (42) 1 0 =0 ∂ t ∂ x2 ∂s From (40), one finds 4 G1 − 2ts ∂ ∂s G0 = t Tr((I − = X̀ K −1 A t 1 ,t 2 ) H) Ẽ1 × Ẽ2 − 2s ∂ ∂s G0 t i ai (s; x, y, u, v), 1 which substituted back in (42), leads to the PDE’s for the ai , namely ∂ 2 ai (s; x, y, u, v) = 0, ∂x (43) implying ai (s; x, y, u, v) = x bi (s; y, u, v) + ci (s; y, u, v). (44) Letting the intervals Ẽ1 and Ẽ2 go to ∞, while keeping their relative position 2 y fixed and widths −2u and −2v fixed as well, is achieved by letting x → ∞, as follows from (33). Remember t 1 , t 2 , Ẽ1 , Ẽ2 from (36). But in the limit x → ∞, the expressions G0 , namely G0 (s; Ẽ1 , Ẽ2 ) = log det(I − KAt ,t ) 1 2 Ẽ1 × Ẽ2 tends to 0 exponentially fast using the exponential decay of the four components of the matrix Airy kernel (5); note that the term pA (t, x, y) tends to 0 as well, when x and y tend to ∞, due to the presence of e−t(x+ y)/2 . Next, we sketch the proof that G1 (t, s; Ẽ1 , Ẽ2 ) = t Tr((I − KAt ,t )−1H) 1 2 Ẽ1 × Ẽ2 tends to 0 exponentially fast, when x → ∞. Indeed, using the identity (R := the resolvent of the Airy kernel A t ,t ) K (I − 4 K The series is finite as A −1 t 1 ,t 2 ) K A t 1 ,t 2 1 KAt ,t (I − KAt ,t )−1 is 1 2 1 2 2 H = H + KAt ,t (I − KAt ,t )−1H = H + KAt ,t (I + R)H, 1 2 1 2 depends on t 2 − t 1 = −2s, G0 is t–independent and 1061 1 2 H, by Prop. 2.1 is a polynomial in t. one computes KAt ,t )−1K1) −1 = t Tr((I − KA t ,t ) H) + t Tr(KA = t Tr H t ,t H̃) G1 = Tr((I − 1 Ẽ1 × Ẽ2 2 1 Ẽ1 × Ẽ2 2 Ẽ1 × Ẽ2 1 2 Ẽ1 × Ẽ2 where H Ẽ1 × Ẽ2 KAt ,t H̃) Ẽ1 × Ẽ2 Tr Tr( 1 2 = = , with 2 X Z KAt ,t (I − KAt ,t )−1 1 2 1 2 K tAt (u, v)H̃ ii (v, u)dud v i i Ẽi × Ẽi ZZ K tAt (u, v)H̃21 (v, u)dud v 1 2 Ẽ1 × Ẽ2 + R := H ii (u, u)du i=1 Ẽi 2 ZZ X i=1 + H̃ := (I + R)H, ZZ K tAt (u, v)H̃12 (v, u)dud v. 2 1 Ẽ2 × Ẽ1 H Each of these integrals tend to 0 because each of them contains the Airy kernel and also because is obtained by acting with differential operators on the Airy kernel, as explained in section 2. This ends the proof that G1 (t, s; Ẽ1 , Ẽ2 ) → 0, when the Ẽi tend to ∞. This fact together with the form (44) of the ai imply bi (s; y, u, v) = ci (s; y, u, v) = 0 for i ≥ 1, and thus ai = 0 for i ≥ 1, implying G1 = 2ts ∂ ∂s G0 . 2 Summarizing, this implies that for Ei = 27 (3τi )3/2 − (3τi )1/6 Ẽi , substituting G1 = 2ts(∂ G0 /∂ s) into (39), ( )! 2 2 \ (3τi )3/2 P (τi ) − 27 ∩ − Ẽi = ; log P 1/6 (3τ ) i i=1 = G0 (s; Ẽ1 , Ẽ2 ) − 2z 4 ts ∂ G0 + O(z 8 ) ∂s = G0 (s − 2tsz 4 ; Ẽ1 , Ẽ2 ) + O(z 8 ) = log P ! 2 \ ¦ © A (t i (1 − t i z 4 )) ∩ (− Ẽi ) = ; + O(z 8 ), i=1 in view of the change of variables given in (36), one has that s−2tsz 4 = 21 (t 1 (1−t 1 z 4 )−t 2 (1−t 2 z 4 )). Then, substituting t i = ui (1 + ui z 4 ) into t i (1 − t i z 4 ) and τi = (1 + 6t i z 4 )/(3z 6 ) + O(z 10 ), as in (13), yields t i (1 − t i z 4 ) = ui − 2u3i z 8 − u4i z 12 and τi = 1 3z 6 + 2ui z2 + 2u2i z 2 + O(z 10 ) for i = 1, 2. 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