NONINTERSECTING BROWNIAN EXCURSIONS 3 Appear in One Form Or Another in the Literature [8, 13, 26], We Hope the Reader Will ﬁnd This Presentation Beneﬁcial

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[17] give explicit enumeration formulae for possible conﬁgurations of these walkers. If the BE paths are labeled X (τ) < < X (τ), 0 <τ< 1, then our main 1 · · · n focus is on the lowest path, X1(τ) and the highest path, Xn(τ). In particular, we show for each ﬁxed positive integer m, with 0 <τ < <τ < 1, that 1 · · · m P(X (τ ) x ,...,X (τ ) x ) 1 1 ≥ 1 1 m ≥ m and P(X (τ )

2 2 2 x /(2(1 τk )) y /(2τℓ) Hkℓ(x,y)= e− − − sτℓ(1 τk) − n 1 j+1/2 − τ (1 τ ) k − ℓ × τℓ(1 τk) jX=0 − x y p2j+1 p2j+1 , × 2τ (1 τ ) 2τ (1 τ ) k − k ℓ − ℓ p p where the pk are the normalized Hermite polynomials. Although rather complicated, it has the same ingredients as the kernel for the Gaussian unitary ensemble (GUE). Extended kernels such as these were first introduced by Eynard and Mehta [8] in the context of a certain Hermitian matrix model. In Section 2, we give a self-contained account of a class of extended kernels. Though the results NONINTERSECTING BROWNIAN EXCURSIONS 3 appear in one form or another in the literature [8, 13, 26], we hope the reader will find this presentation beneficial. In Section 3, we use this theory of extended kernels to derive our expressions for the entries of KBE(x,y). Using the (scalar) m = 1 kernel, we show in Section 4 that for τ (0, 1), ∈ s2 r(t) P(X1(τ) s 2τ(1 τ))= exp dt , ≥ − − 0 t q Z where r = r(s) is the solution to

2 2 2 2 1 2 (1.2) s (r′′) = 4(r′) (sr′ r) + (sr′ r) (4n + 1)(sr′ r)r′ + (r′) − − − − 4 satisfying the asymptotic condition r(s)= r s3/2 + O(s5/2), s 0, 0 → where 1 1 2n 4n(2n + 1) r = . 0 √ 2n n π 2 3 A similar result is obtained for P(X (τ) s 2τ(1 τ)). n ≤ − The differential equations are the so-calledp σ form of PainlevéV equations and the Fredholm determinants in these cases, where J = (0,x) or J = (x, ), are PainlevéV τ-functions [12]. To obtain∞ (1.2), we use results from [25] to obtain a system of differential equations whose solutions determine r(s) and then reduce these to the second order equation for r(s). In Section 5, we consider the limiting process, as n , of a suitably rescaled bottom path X (τ) near a fixed time τ (0, 1)→∞ and show that the 1 ∈ distributional limit is 1/2(τ), the Bessel process of [26] with parameter α = 1/2. (This is different fromB the three-dimensional Bessel process appearing in the literature, which is Brownian motion conditioned to stay positive forever. When one rescales BE near τ = 0, the three-dimensional Bessel process is the scaling limit.) This is a process whose finite-dimensional distribution functions are given in terms of an extended Bessel kernel KBes. Precisely, if <τ < <τ < , then −∞ 1 · · · m ∞ (1.3) P( (τ ) x ,..., (τ ) x ) = det(I KBesχ ), B1/2 1 ≥ 1 B1/2 m ≥ m − J where χJ = diag(χJk ) with Jk = (0,xk). For α = 1/2, the entries of the matrix kernel are 1 2 (τ τ )t2/2 e k− ℓ sin xt sinyt dt, if k ℓ, Bes π ≥ K (x,y)=  Z0 kℓ 2 2  ∞ (τk τℓ)t /2  e − sin xt sinyt dt, if k<ℓ.  −π Z1   4 C. A. TRACY AND H. WIDOM

Our result is that for ﬁxed τ (0, 1), if, in the probability ∈ P(X (τ ) x ,...,X (τ ) x ), 1 1 ≥ 1 1 m ≥ m we make the substitutions

τ(1 τ) τ(1 τ) xk − xk, τk τ + − τk, → s 2n → 2n then the resulting probability has the n limit (1.3). This is proved by →∞ establishing trace norm convergence of the extended kernels. As far as we know, this is the first appearance of the Bessel process in the literature as the distributional limit of a finite-n Brownian motion process. We expect the top path, Xn(τ), again suitably rescaled, to have the Airy process [13, 20] as its distributional limit. The reason is that the top path, as n , should not feel the presence of the “wall” at x = 0 and the Airy →∞ process is known to be the distributional limit of nonintersecting Brownian bridges. The bottom path does feel the presence of the wall and in the end this is why the Bessel process appears. We shall not derive the scaling limit at the top because it is the bottom scaling that gives something new, and is thus more interesting. For BE, a related random variable is the area A under the BE curve. (See [10, 18] and references therein.) Both the distribution of A and the moments of A are known (the moments are known recursively) and, perhaps surprisingly, numerous applications have been found [10, 19] outside their original setting. It is natural, therefore, to introduce An,L (An,H ) equal to the area under the lowest (highest) curve in n-nonintersecting BE. From our limit theorem specialized to the m = 1 case, we deduce in Section 5.4 the asymptotics of the first moments E(An,L) and E(An,H ). In particular, c E(A ) L , n , n,L ∼ √n →∞ where

π√2 c = ∞ det(I KBesχ ) ds 0.682808. L 16 − (0,s) ≃ Z0 Here, KBes is the m = 1 Bessel kernel,

2 1 1 sin(x y) sin(x + y) KBes(x,y)= sin xt sin ytdt = − . π 0 π x y − x + y Z − NONINTERSECTING BROWNIAN EXCURSIONS 5

2. Extended kernels.

2.1. Initial and final points all distinct. Although only a special case of the following derivation will be used in our discussion of nonintersecting BE, it might be useful for the future, and is no more difficult, to consider a rather general setting. n Suppose that Xi(τ) i=1 is a family of stationary Markov processes with continuous paths,{ with} common transition probability density P (x,y,τ), conditioned to be nonintersecting, to begin at a1, . . . , an at time τ =0 and to end at b1, . . . , bn at time τ =1. An extended kernel is a matrix kernel K(x,y) = (K (x,y)) depending on 0 <τ < <τ < 1 with the kℓ kl=1,...,m 1 · · · m following property. Given functions fk, the expected value of m n (1 + fk(Xi(τk))) kY=1 iY=1 is equal to det(I + Kf), where f denotes multiplication by diag(fk). In the special case where f = χ , this is the probability that for k = 1,...,m, no k − Jk path passes through the set Jk at time τk. For BE, this gives the statement made in the third paragraph of the Introduction, once we have computed the extended kernel for it. Here is how an extended kernel is obtained. Define the n n matrix A by × Aij = P (ai, bj, 1), define the m m matrix kernel E(x,y) by × P (x,y,τ τ ), if k<ℓ, (2.1) E (x,y)= ℓ k kℓ 0, − otherwise, and if A is invertible, define the m m matrix kernel H(x,y) by × n 1 (2.2) H (x,y)= P (x, b , 1 τ )(A− ) P (y, a ,τ ). kℓ j − k ji i ℓ i,jX=1 We shall show that H E is an extended kernel in the above sense. From Karlin and McGregor− [15] and the Markov property, we know the following. For paths starting at x0,i = ai at time τ = 0, the probability density that at times τk (k = 1,...,m + 1) the paths are at points xki (i = 1,...,n) is a constant times m

det(P (xk,i,xk+1,j,τk+1 τk))1 i,j n. − ≤ ≤ kY=0 In the above, we set xm+1,j = bj,τm+1 = 1. The expected value in question is obtained by multiplying m n (1 + fk(xki)) kY=1 iY=1 6 C. A. TRACY AND H. WIDOM by the probability density and integrating over all of the xki. We apply the general identity

det(ϕ (x ))n det(ψ (x ))n dµ(x ) dµ(x ) · · · j k j,k=1 · j k j,k=1 1 · · · n Z Z n = n! det ϕj(x)ψk(x) dµ(x) Z j,k=1 successively to the variables x1i,...,xmi and ﬁnd that the integral equals a constant times the determinant of the matrix with i, j entry m 1 − P (a ,x ,τ ) P (x ,x ,τ τ )P (x , b , 1 τ ) · · · i 1 1 k k+1 k+1 − k m j − m Z Z kY=1 m (1 + f (x )) dx dx . × k k 1 · · · m kY=1 If we set f = 0, the matrix is A by the semigroup property of P (x,y,τ) 1 and the expected value is 1, so the normalizing constant is (det A)− . Hence, if we replace P (ai,x,τ) by 1 Q(ai,x,τ)= (A− )ijP (aj,x,τ), Xj then the expected value equals the determinant of the matrix with i, j entry m 1 − Q(a ,x ,τ ) P (x ,x ,τ τ )P (x , b , 1 τ ) · · · i 1 1 k k+1 k+1 − k m j − m Z Z kY=1 m (1 + f (x )) dx dx , × k k 1 · · · m kY=1 which equals the determinant of I plus the matrix with i, j entry m 1 − Q(a ,x ,τ ) P (x ,x ,τ τ )P (x , b , 1 τ ) · · · i 1 1 k k+1 k+1 − k m j − m Z Z kY=1 m (1 + fk(xk)) 1 dx1 dxm. × " − # · · · kY=1 The bracketed expression may be written as a sum of products, f (x ) f (x ). k1 k1 · · · kr kr r 1 k1<

Q(a ,x ,τ )f (x ) · · · i k1 k1 k1 k1 Z Z P (x ,x ,τ τ )f (x ) P (x , b , 1 τ ) dx dx . × k1 k2 k2 − k1 k2 k2 · · · kr j − kr k1 · · · kr

If we let Uk,ℓ be the operator with kernel Uk,ℓ(x,y)= P (x,y,τl τk)fℓ(y), then the above may be written as the single integral −

Q(ai,x,τk1 )fk1 (x) (Uk1,k2 Ukr 1,kr P (x, bj, 1 τkr )) dx. · · · · − − Z (If r = 1, we interpret the operator product to be the identity.) Replacing the index k1 by k and changing notation, we see that the sum of all of these equals

Q(ai,x,τk)fk(x) Uk,k1 Uk1,k2 Ukr 1,kr P (x, bj, 1 τkr ) dx, · · · − − ! Z k r 0 k1,...,kr X X≥ X where the inner sum runs over all kr > > k1 > k. (Now, if r = 0, we interpret the operator product to be the identity.)· · · m Write Pj for the vector function P (x, bj, 1 τk) k=1 and U for the op- { − } m erator acting on vector functions with matrix kernel (Uk,ℓ(x,y))k,ℓ=1, where 1 r we set Uk,ℓ(x,y)=0 if k ℓ. Using the fact that (I U)− = r 0 U , we see that the expression in≥ large parentheses in the integral− equals≥ the kth 1 P component of the vector function (I U)− Pj. Denote it by T (k,x; j), and similarly denote the ﬁrst factor in the− integral by S(i; k,x). We may think of T (k,x; j) as the kernel of an operator T taking sequences into vector functions, n m n aj j=1 T (k,x; j)aj , { } → ( ) jX=1 k=1 while S(i; k,x) is the kernel of an operator taking vector functions to sequences, m n m ϕk(x) k=1 S(i; k,x)ϕk(x) dx . { } → ( ) Z kX=1 i=1 The integral is the kernel of the product ST , evaluated at i, j. We use the general fact that det(I +ST ) = det(I +TS). The operator TS, which takes vector functions to vector functions, has matrix kernel with k,ℓ entry n T (k,x; j)S(j; ℓ,y). jX=1 8 C. A. TRACY AND H. WIDOM

Since U is strictly upper-triangular, we have det(I U)=1, so − det(I + TS) = det(I U + (I U)TS). − − If we recall that T (k,x; j) is the kth component of (I U) 1 applied to the − − vector function P (x, bj, 1 τk) and recall the definition of S, we see that (I U)TS has k,ℓ{ entry − } − n P (x, b , 1 τ )Q(a ,y,τ )f (y) j − k j ℓ ℓ jX=1 n 1 = P (x, b , 1 τ )(A− ) P (y, a ,τ )f (y). j − k ji i ℓ ℓ i,jX=1 This equals Hkℓ(x,y)fℓ(y). Since U has k,ℓ entry Ekℓ(x,y)fℓ(y), we have shown that the expected value in question equals det(I + (H E)f). Thus H E is an extended kernel. − − 2.2. Initial and final points at zero. In the preceding derivation, we as- sumed that A was invertible. If any two ai are equal or any two bj are equal, this will certainly not hold. There are analogous, although more complicated, formulas if several of the ai or bj are equal. We consider here the simplest case where they are all equal to zero. We take (2.2) for the case when they are all different (when A is expected to be invertible) and compute the limit as all of the ai and bj tend to zero. There is a matrix function D0 = D0(c1,...,cn) such that for any smooth function f,

f(c1) f(0) f(c2) f ′(0) lim D0  .  =  .  . ci 0 . . →        (n 1)   f(cn)   f − (0)      Here, limc c is short for a certain sequence of limiting operations. The i→ matrix D0′ (the transpose of D0) acts in a similar way on the right on row vectors. Writing Da for D0(a1, . . ., an) and Db for D0(b1, . . ., bn), we use 1 1 A− = Db′ (DaADb′ )− Da and then take limits as all ai, bj 0. The matrix DaADb′ becomes the matrix B with i, j entry → i 1 j 1 Bij = ∂ − ∂ − P (ai, bj, 1) a =b =0, ai bj | i j the row vector P (x, b , 1 τ ) becomes the vector with jth component { j − k } b j 1 P (x, 1 τk)= ∂ − P (x, bj, 1 τk) b =0 j − bj − | j NONINTERSECTING BROWNIAN EXCURSIONS 9 and the column vector P (ai,y,τℓ) becomes the vector with ith component {a }i 1 P (y,τ )= ∂ − P (a ,y,τ ) . i l ai j ℓ |ai=0 Then if the matrix B is invertible, we obtain for the H-summand of the extended kernel, in this case, n b 1 a (2.3) H (x,y)= P (x, 1 τ )(B− ) P (y,τ ). kℓ j − k ji i ℓ i,jX=1 One would like B to be as simple as possible. Something one could have done before passing to the limit would be to take two sequences α and { i} βj of nonzero numbers and in (2.2), to replace A by the matrix (αiAijβj), replace{ } the function P (x, b , 1 τ ) by P (x, b , 1 τ )β and replace the j − k j − k j function P (ai,y,τℓ) by αiP (ai,y,τℓ). The kernel does not change. 2.2.1. An example. We apply (2.3) to the case of n nonintersecting Brow- nian bridges. The transition probability for Brownian motion is

1 (x y)2/2τ P (x,y,τ)= e− − √2πτ a2/2 b2/2 a b and we choose αi = e i and βj = e j . Then the modified A equals (e i j / 1 √2π), the modified B equals diag((i 1)!)/√2π, the modified B− equals √2π diag(1/(i 1)!) and − × − 1 2 b x /(2(1 τk )) P (x, 1 τk)= e− − j − 2π(1 τ ) − k (j 1)/2 p τk − x Hj 1 , × 2(1 τk) − 2τ (1 τ ) − k − k (i 1)/2 1 2 1 τ p y a y /(2τℓ) ℓ − Pi (y,τℓ)= e− − Hi 1 , √2πτℓ 2τℓ − 2τ (1 τ ) ℓ − ℓ where Hk(x) are the Hermite polynomials satisfying p k t2+2tx t e− = H (x). k! k k 0 X≥ In terms of the harmonic oscillator wave functions,

x2/2 1 x2/2 (2.4) ϕk(x)= e− pk(x)= e− Hk(x), √π1/22kk! we have 1 2 (1/2 τk )X Hkℓ(x,y)= e − k 2(1 τ )τ − k ℓ n 1 j/2 p − τ (1 τ ) 2 k ℓ (1/2 τℓ)Y − ϕj(Xk)ϕj(Yℓ)e− − ℓ , × τℓ(1 τk) jX=0 − 10 C. A. TRACY AND H. WIDOM where, for notational simplicity, we write Xk = x/ 2τk(1 τk) and Yℓ = y/ 2τ (1 τ ). The matrix kernel H(x,y) E(x,y) is called− the extended ℓ − ℓ − p Hermite kernel. For m = 1, H11(x,y) (with τ1 = τ) reduces to the Hermite p GUE kernel Kn (X,Y ). [We can neglect the two exponential factors since they do not change the value of the determinant. The factor 1/ 2τ(1 τ) also disappears when we multiply by the differential dY .] [In the usual− defini- p tion of the extended Hermite kernel, the times τk (0, 1) are replaced by 2ˆτ ∈ timesτ ˆk (0, ) defined by τk/(1 τk)= e k so that H E has k,ℓ entry KGUE(X,Y∈ ;τ ˆ∞ τˆ ), where − − n k − ℓ n 1 − jτˆ e ϕj(x)ϕj (y), ifτ ˆ 0,  ≥ (2.5) KGUE(x,y;τ ˆ)=  jX=0 n   ∞ jτˆ  e ϕj(x)ϕj (y), ifτ ˆ < 0. − j=n  X  The representation forτ ˆ < 0 follows from the Mehler formula ([7], Section 10.13 (22)).]

2.3. General case. In the recent literature, one finds investigations where not all starting and ending points are the same. In [2, 27], there is one starting point and two ending points and in [6], there are two starting points and two ending points. The previous discussion can be extended to the case where there are R different ai and S different bj. The different ai, we label a1, . . . , aR and assume that ar occurs mr times. Similarly, the different bj, we label b1, . . ., bS and assume that bs occurs ns times. Of course, mr = ns = n. It is convenient to use double indices ir and js where i = 0,...,m 1 and j = P Pr 0,...,n 1. We may think of A as consisting of m n blocks− A . s − r × s rs Once again, there is a matrix function Dc = Dc(c1,...,cn) such that

f(c1) f(c) f(c2) f ′(c) lim Dc  .  =  .  ci c . . →        (n 1)   f(cn)   f − (c)      and similarly for the action of Dc′ on row vectors. We then write 1 1 A− = Db′ (DaADb′ )− Da, where Da and Db are now block diagonal matrices,

Da = diag(Dar ), Db = diag(Dbs ), substitute into (2.2) and take the appropriate limits. The matrix entry Air,js is replaced by i 1 j 1 Bir,js = ∂ − ∂ − P (air, bjs, 1) a =a ,b =b , air bjs | ir r js s NONINTERSECTING BROWNIAN EXCURSIONS 11 while the row vector P (x, bjs, 1 τk) becomes the vector with js component { − }

bj j 1 P (x, 1 τk)= ∂ − P (x, bjs, 1 τk) b =b js − bjs − | js s and the column vector P (a ,y,τ ) becomes the vector with ir component { ir ℓ } ar j 1 P (y,τ )= ∂ − P (a ,y,τ ) . ir ℓ air ir ℓ |air =ar Thus (2.3) is now replaced by

bj 1 ar H (x,y)= P (x, 1 τ )(B− ) P (y,τ ). kℓ js − k js,ir ir ℓ ir,jsX 3. Nonintersecting Brownian excursions.

3.1. Preliminary result. Let Bτ denote standard Brownian motion (BM), so that

1 (x y)2/(2τ) Px(Bτ dy)= P (x,y,τ) dy = e− − dy, τ> 0, x,y R, ∈ √2πτ ∈ and the subscript x on P indicates that the BM was started at x. Let H = inf s : s> 0,B = a = hitting time of a. a { s } The basic result we use is [11, 21]

(3.1) Px(Bτ dy,H0 >τ) = (P (x,y,τ) P (x, y,τ)) dy =: P (x,y,τ) dy. ∈ − − − Equation (3.1) gives the probability that a Brownian path starting at x is in the neighborhood dy of y at time τ while staying positive for all times 0 < s<τ. This process has continuous sample paths [11]; therefore, by Karlin– McGregor [15], we can construct an n nonintersecting path version and apply the formalism of Section 2. It is convenient to write

2 (x2+y2)/(2τ) xy (3.2) P (x,y,τ)= e− sinh . − πτ τ r [The theory of Section 2 applies equally well to the n nonintersecting path version of reﬂected (elastic) BM, Y := B . In this case, τ | τ | P ( B dy) = (P (x,y,τ)+ P (x, y,τ)) dy x | τ |∈ − 2 (x2+y2)/(2τ) xy = e− cosh . πτ τ r We leave it as an exercise to construct the extended kernel.] 12 C. A. TRACY AND H. WIDOM

3.2. Extended kernel for Brownian excursion. Referring to (2.2) and the technique described after (2.3), we ﬁnd that for distinct ai and bi, n π b2/2 1 a2/2 Hkℓ(x,y)= P (x, bi, 1 τk)e i (A− )ijP (y, aj,τℓ)e j , 2 − − r i,j=1 − X where Aij = sinh(aibj). The even order derivatives of sinh x all vanish at x = 0, so to take ai, bj 0, we have to modify the procedure described in Section 2.2. (Otherwise,→ the matrix B so obtained would be noninvertible.) We now use the fact that there is a matrix function Dc, such that for any smooth odd function f, −

f(c1) f ′(0) f(c2) f ′′′(0) lim Dc,  .  =  .  . ci 0 − . . →        (2n 1)   f(cn)   f − (0)      Thus the B matrix is 2i 1 2j 1 Bij = ∂ − ∂ − sinh(aibj) a =b =0 = (2j 1)!δij . ai bj | i j − It follows that the limiting kernel has k,ℓ entry n 1 π − 1 (3.3) H (x,y)= P (x, 1 τ )P (y,τ ), kℓ 2 (2j + 1)! j k j ℓ r j=0 − X where 2j+1 y2/2 Pj(x,τ) = [∂y P (x,y,τ)e ] y=0 − | 1 x2/(2τ) = e− √2πτ 2j+1 ((1 τ)/(2τ))y2 +xy/τ 2j+1 ((1 τ)/(2τ))y2 xy/τ [∂ e− − ∂ e− − − ] × y − y |y=0 j+1/2 2 1 τ x2/(2τ) x = − e− H2j+1 . πτ 2τ 2τ(1 τ) r − Thus the H-summand of the extended kernel canp be written as

2 2 2 x /(2(1 τk )) y /(2τℓ) Hkℓ(x,y)= e− − − sτℓ(1 τk) − n 1 j+1/2 − τk(1 τℓ) x (3.4) − p2j+1 × τℓ(1 τk) 2τk(1 τk) jX=0 − − p y p2j+1 . × 2τ (1 τ ) ℓ − ℓ p NONINTERSECTING BROWNIAN EXCURSIONS 13

Recall that pk are the normalized Hermite polynomials. It follows from the Christoﬀel–Darboux formula (see, e.g., [7], Section 10.3 (10)) for Hermite polynomials that

n 1 − p2j+1(x)p2j+1(y) jX=0 √n p2n(x)p2n 1(y) p2n 1(x)p2n(y) = − − − 2 x y −

p2n(x)p2n 1(y)+ p2n 1(x)p2n(y) + − − . x + y

Setting τk = τ, X = x/ 2τ(1 τ) and Y = y/ 2τ(1 τ), we see that the diagonal elements are − − p p

n x2/(2(1 τ)) y2 /(2τ) H (x,y)= e− − − kk 2τ(1 τ) s − p2n(X)p2n 1(Y ) p2n 1(X)p2n(Y ) − − − × X Y (3.5) − p2n(X)p2n 1(Y )+ p2n 1(X)p2n(Y ) + − − X + Y 2 2 e(1/2 τ)X e (1/2 τ)Y = − − − KGUE(X,Y ) KGUE(X, Y ) . 2τ(1 τ) { 2n − 2n − } − More generally (recallp the notation of Section 2.2.1),

2 2 (1/2 τk )X (1/2 τℓ)Y e − k − − ℓ GUE Hkℓ(x,y) Ekℓ(x,y)= K2n (Xk,Yℓ;τ ˆk τˆℓ) − 2τℓ(1 τk) { − (3.6) − p KGUE(X , Y ;τ ˆ τˆ ) , − 2n k − ℓ k − ℓ }

GUE where Kn (x,y;τ ˆ) is deﬁned as in (2.5). (To establish this identity for k<ℓ, one uses the (slightly modiﬁed) Mehler formula,

∞ 2j+1 q ϕ2j+1(x)ϕ2j+1(y) jX=0 1 (1+q2)/(2(1 q2 ))(x2+y2) 2qxy = e− − sinh , π(1 q2) 1 q2 − − with x,y X ,Ypand q = [τ (1 τ )/τ (1 τ ]1/2.) → k ℓ k − ℓ ℓ − k 14 C. A. TRACY AND H. WIDOM

3.2.1. Density of paths. The density of paths at time τ is given by

ρn(x,τ)= H11(x,x) 1 (3.7) = KGUE(X, X) KGUE(X, X) 2τ(1 τ){ 2n − 2n − } − p n 1 2 − 2 = ϕ2j+1(X) 2τ(1 τ) − jX=0 and satisﬁes the normalizationp ∞ ρn(x,τ) dx = n. Z0 For n = 1, ρn reduces to the well-known density for Brownian excursion.

4. Distribution of lowest and highest paths at time τ .

4.1. Fredholm determinant representations. Recall that we assume the nonintersecting BE paths to be labeled so that for all 0 <τ< 1, X (τ) > > X (τ) > 0. n · · · 1 We are interested in the distribution of the lowest path, X1 and the highest path, Xn. From Section 2, we see that P P (X1(τ) x)= (X1(τ) x,...,Xn(τ) x) = det(I KχJ1 ), (4.1) ≥ ≥ ≥ − J1 = (0,s), and P P (Xn(τ)

Table 1 The expected areas of lowest and highest curves for n = 1,..., 9

n E(An,L) E(An,H ) π π 1 = 0.626657 . . . = 0.626657 . . . 8 8 2 5 (√2 1)q√π = 0.458859 . . . 5q√π = 1.107783 . . . 8 − 8 3 0.380211. . . 1.479479. . . 4 0.331955. . . 1.791454. . . 5 0.298441. . . 2.065097. . . 6 0.273407. . . 2.311582. . . 7 0.253784. . . 2.537567. . . 8 0.237865. . . 2.747386. . . 9 0.224612. . . 2.944040. . .

where Ψj(x)= √2ϕ2j+1(x) and where u v for functions u and v denotes the operator f u(v,f). It follows that ⊗ → n 1 (4.4) det(I Kχ ) = det(δ (Ψ , Ψ )) − , − J jk − j k j,k=0 where the inner product ( , ) is taken over the set J. Thus (4.4) gives ﬁnite · · determinant representations of P(X1(τ) x) and P(Xn(τ)

4.2. Integrable diﬀerential equations. In this section, we show that the Fredholm determinant of the integral operator with kernel

(4.7) K(x,y)χ (y) = [KGUE(x,y) KGUE(x, y)]χ (y) J 2n − 2n − J can be expressed in terms of a solution to a certain integrable diﬀerential equation when J = (0,s) and J = (s, ). To do this, we transform the kernel ∞ K(x,y) to the canonical form

ϕ(x)ψ(y) ψ(x)ϕ(y) (4.8) − x y − so that the theory of Fredholm determinants and integrable diﬀerential equations, as developed in [25], can be directly applied. If, instead of K(x,y), we use the kernel K(√x, √y) K (x,y)= , 0 2(xy)1/4 then

(4.9) det(I Kχ 2 ) = det(I K χ ). − (0,s ) − 0 (0,s)

By use of the Christoﬀel–Darboux formula, we see that K0 is of the form (4.8), with

1/4 1/4 1/4 1/4 (4.10) ϕ(x)= n x ϕ2n(√x), ψ(x)= n x− ϕ2n 1(√x), − where the ϕk are the harmonic oscillator wave functions (2.4). Using the well-known formulas for the action of multiplication by x and diﬀerentiation by x on the ϕk, we ﬁnd

1 1 (4.11) xϕ′(x) = ( x)ϕ(x)+ √nxψ(x), 4 − 2 1 1 (4.12) xψ′(x)= √nϕ(x) ( x)ψ(x). − − 4 − 2 These formulas will allow us to use the results of [25]. In the derivation of the next section, we write J for (0,s) and denote by R(x,y) the resolvent kernel of K χ , the kernel of (I K χ ) 1K χ . Of 0 J − 0 J − 0 J course, this also depends on the parameter s. A basic fact is that d (4.13) log det(I K χ )= R(s,s). ds − 0 J − Once R(s,s) is known, integration and exponentiation give the determinant. NONINTERSECTING BROWNIAN EXCURSIONS 17

4.2.1. Reduction to PainlevéV, J = (0,s). Following [25], we define the functions of x (which also depend on s since J does) 1 1 Q = (I K χ )− ϕ, P = (I K χ )− ψ − 0 J − 0 J and then the quantities (which depend on s only) q = Q(s), p = P (s), r = sR(s,s), u = (Q, ϕ), v = (Q, ψ), w(s) = (P, ψ), where the inner products are taken over J. The definitions are not important here. What is important are the relations among the various quantities, as established in [25]. Of course, we are interested in r. This function and its derivative are given in terms of the other functions by 2 1 r(s) = (w + √n)q + ( 2 s 2√nw)qp (4.14) − − + ( u + √ns + 2√nv)p2, − dr (4.15) = qp + √np2. ds − ((4.14) is the specialization of (2.27) of [25] with additional insertions described in (2.32); (4.15) is the specialization of (2.28) of [25] with additional insertions described in (2.35). In both cases, we use the formulas (4.11) and (4.11).) Specializing the differential equations of [25] to the case at hand, we find that dq (4.16) s = ( 1 1 s √nw)q + ( u + 2√nv + √ns)p, ds 4 − 2 − − dp (4.17) s = (w + √n)q ( 1 1 s √nw)p, ds − − 4 − 2 − du dv dw (4.18) = q2, = pq, = p2. ds ds ds ((4.16) and (4.17) are (2.25) and (2.26) of [25], resp., with additional insertions described in (2.31). We also used recursion relations (2.12) and (2.13) of [25], evaluated at x = s. Equations (4.18) are “universal equations” given by (2.16)–(2.18) of [25].) This is a system of five equations in five unknowns. One has the first integrals, discovered by a computer search, (4.19) sp(q √np)+ √n(u + 1 w) (1 + 2n)v + w(u 2√nv + nw) = 0, − 2 − − (√ns u + 2√nv)p2 + (√n + w)q2 (4.20) − + ( 1 s 2√nw)pq + v √nw = 0. 2 − − − 18 C. A. TRACY AND H. WIDOM

(A first integral for a system of differential equations is a relation stating that a particular expression in the variables is constant when the system is satisfied. Often, as here, the constant can be determined by using the initial conditions.) To verify either of these, we differentiate the left-hand side using (4.16)– (4.18) and find that the derivative equals zero. Thus the left-hand side is a constant. That the constant is zero follows from the easily verified fact that all quantities are o(1) as s 0. → In addition to (4.14) and (4.15), we compute r′′ using (4.16)–(4.18). These, together with the first integrals (4.19) and (4.20), give us five equations. Remarkably, from these, we can eliminate q, p, u, v and w and we obtain the second order equation (1.2) for r. (For the elimination, we used a Gröbner basis algorithm as implemented in Mathematica. A Gröbner basis of an ideal in a polynomial ring is a particular basis that is well suited for the elimination of variables. The reader is referred to the book by Cox, Little and O’Shea [5] for both the theory of Gröbner bases and many examples of their application to computational commutative algebra. Remarkably, although one would expect to need six equations to eliminate the five unknowns, we are in a nongeneric situation and these five equations suffice. This would not have been discovered without the use of the Gröbner basis. In our original derivation, we used only the first integral (4.19) and obtained a third order equation in r. Cosgrove [4] knew how to reduce this to the second order equation (1.2). This then led us to discover (4.20).) The solution we want satisfies the asymptotic condition (4.21) r(s)= r s3/2 + O(s5/2) as s 0, 0 → where 1 1 2n 4n(2n + 1) (4.22) r = . 0 √π 22n n 3 This condition follows from the fact that R(s,s) K0(s,s) as s 0. (Power counting suggests that cs1/2 is the leading term,∼ but a computation→ shows that c is zero. The s3/2 term is the leading one. Higher terms can be determined from the differential equation.)

4.2.2. PainlevéV representation of P(X1(τ) x). From (4.13) and (4.9), we conclude that ≥ s2 r(t) (4.23) P(X1(τ) s 2τ(1 τ)) = exp dt , ≥ − − 0 t q Z where r is the solution to (1.2) satisfying boundary condition (4.21). Using (1.2), it is routine to compute the small s expansion of r(s) and hence that NONINTERSECTING BROWNIAN EXCURSIONS 19 of (4.23), 2 2 64n2 + 32n + 9 P(X (τ) s 2τ(1 τ))=1 r s3 + (4n + 1)r s5 r s7 1 ≥ − − 3 0 25 0 − 735 0 q (4n + 1)(32n2 + 16n + 15) + r s9 8505 0 128(2n + 3)(n 1) + − r2s10 + O(s11), 275625 0 where r0 is defined as in (4.22). We remark that the number of BE curves, n, appears only as a parameter in the differential equation (1.2) and the boundary condition (4.21). Although we have explicit solutions in terms of finite n determinants, ex- pansions such as the one above are more easily derived from the differential equation representation.

4.2.3. Reduction to PainlevéV, J = (s, ). This differs from J = (0,s) in only minor ways. In this case, ∞ d log det(I K χ )= R(s,s), ds − 0 J so that with r defined as before,

∞ r(t) P(Xn(τ) s 2τ(1 τ))= exp dt . ≤ − − s2 t q Z Specializing the differential equations of [25], we find for J = (s, ), ∞ dq s = ( 1 1 s √nw)q + ( u + 2√nv + √ns)p, ds 4 − 2 − − dp s = (w + √n)q ( 1 1 s √nw)p, ds − − 4 − 2 − du dv dw = q2, = pq, = p2. ds − ds − ds − The connection between r and the above variables remains the same as that given in (4.14). One has the two first integrals sp(q √np)+ √n(u + 1 w) (1 + 2n)v + w(u 2√nv + nw) = 0, − − 2 − − (√ns u + 2√nv)p2 + (√n + w)q2 + ( 1 s 2√nw)pq v + √nw = 0. − 2 − − − Proceeding with the calculations as described above, we obtain

2 2 2 2 1 2 s (r′′) = 4(r′) (sr′ r) + (sr′ r) (4n + 1)(sr′ r)r′ + (r′) . − − − − − 4 20 C. A. TRACY AND H. WIDOM

5. Limit theorems as n → ∞. In this section, we obtain the scaling limit for the extended BE kernel at the bottom. We ﬁrst explain what this means and then state the result. Given a kernel L(x,y), the kernel resulting from substitutions x ϕ(x),y ϕ(y) is the scaled kernel → →

1/2 Lˆ(x,y) = (ϕ′(x)ϕ′(y)) L(ϕ(x), ϕ(y)), which deﬁnes a unitarily equivalent operator. If there is an underlying parameter and Lˆ has a limit Lˆ0 (pointwise or in some norm), then Lˆ0 is the scaling limit of L(x,y). If the convergence is in trace norm, then the limit of det(I L) equals det(I Lˆ ). Trace norm convergence is assured if the − − 0 underlying interval is bounded and if Lˆ(x,y) and its ﬁrst partial derivatives converge uniformly. The extended Bessel kernel KBES [24], Section 8, is the matrix kernel whose entries are given by

1 2 z (τk τℓ)/2 e − Φα(xz)Φα(yz) dz, if k ℓ, BES 0 ≥ Kkℓ (x,y)=  Z ∞ z2(τ τ )/2  e k− l Φ (xz)Φ (yz) dz, if k<ℓ,  − α α Z1  where Φα(z)= √zJα(z). In particular, Φ1/2(z)= 2/π sin z. We shall show the following. Take a ﬁxed τ (0, 1) and make the substi- p tutions ∈ τ(1 τ) τ(1 τ) (5.1) x − x, y − y → s 2n → s 2n and replacements τ(1 τ) (5.2) τ τ + − τ . k → 2n k Then the extended BE kernel has the extended Bessel kernel with parameter α = 1/2 as scaling limit as n , in trace norm over any bounded domain. In particular, we deduce the→∞ result stated in the introduction, that the ﬁnit-dimensional distribution functions for the BE process, when suitably normalized, converge to the corresponding distribution functions for the Bessel process.

5.1. Integral representation for the extended kernel. The extended BE kernel KBE has the form (H ) (E ), where E is given by (1.1) and kℓ − kℓ kℓ Hkℓ is given by (3.3). The limit theorem for the Hkℓ will be derived from an integral representation of (3.3) which, as in [3, 27], uses two diﬀerent integral representations for Pj . (We could use known integral representations for the NONINTERSECTING BROWNIAN EXCURSIONS 21

Hermite polynomials, but the method we use here is more general and also works in other cases.) The ﬁrst is easy. With integration over a curve around zero,

2 1 2 dt P (x,y,τ)ey /2 = P (x,t,τ)et /2 , − 2πi − t y Z − so

(2j + 1)! t2/2 dt Pj(x,τ)= P (x,t,τ)e 2πi − t2j+2 I 2 x2/2τ (2j + 1)! (1 τ)/(2τ)t2 xt dt = e− e− − sinh . πτ 2πi τ t2j+2 r I For the other, we take the Fourier transform on the y-variable. A computation gives

2 ∞ eisyP (x,y,τ)ey /2 dy − Z−∞ 2 x2/2(1 τ) τ/(2(1 τ))s2 ixs = e − e− − sinh . √1 τ 1 τ − − Taking inverse Fourier transforms and diﬀerentiating gives

2 x2/2(1 τ) Pj(x,τ)= e − π√1 τ − ∞ τ/(2(1 τ))s2 ixs 2j+1 e− − sinh ( is) ds × 1 τ − Z−∞ − i 2i x2/2(1 τ) ∞ τ/(2(1 τ))s2 xs 2j+1 = e − e − sinh s ds. π√1 τ i 1 τ − Z− ∞ − These give

1 2 2 x /(2(1 τk ))+y /(2(1 τℓ)) Hkℓ(x,y)= e− − − π2 (1 τ )(1 τ ) − k − ℓ i p ∞ τ /(2(1 τ ))t2+τ /(2(1 τ ))s2 xt e− k − k ℓ − ℓ sinh × i 1 τk I Z− ∞ − ys s 2n s ds dt sinh 1 . × 1 τ t − s2 t2 − ℓ − If we ensure that the s-contour passes to one side of the t-contour, then the summand 1 in the bracket contributes zero since the t-integral is zero. Thus we have

1 2 2 x /(2(1 τk ))+y /(2(1 τℓ)) Hkℓ(x,y)= e− − − π2 (1 τ )(1 τ ) − k − ℓ p 22 C. A. TRACY AND H. WIDOM

i ∞ τ /(2(1 τ ))t2+τ /(2(1 τ ))s2 (5.3) e− k − k ℓ − ℓ × i I Z− ∞ xt ys s 2n s ds dt sinh sinh . × 1 τ 1 τ t s2 t2 − k − ℓ − 5.2. Limit theorem for the bottom. If we ignore the denominator s2 t2 in (5.3), we have a product of two integrals for each of which we can ap−ply the method of steepest descent. If we think of x and y as small and k = ℓ with τk = τℓ = τ, then there are saddle points at i 2n(1 τ)/τ for both integrals. So we assume that all τ are near a single±τ and make− the changes k p of variable 2n(1 τ) 2n(1 τ) s − s, t − t. → s τ → s τ This is what suggested the substitutions (5.1) and replacements (5.2). Ob- serve that in the scaling limit, Ek,ℓ remains the same. For (5.3), we can ignore the outside factor involving x and y since remov- ing it does not change the determinant. An easy computation shows that after the various substitutions, the rest of Hkℓ(x,y) becomes i 1 2 2 1 2 1 2 ∞ n(s t +2 log s 2 log t) (τk +O(n− ))t /2+(τℓ+O(n− ))s /2 2 1 e − − − π + O(n− ) i I Z− ∞ xt ys s ds dt sinh sinh , × 1+ O(n 1) 1+ O(n 1) s2 t2 − − − 1 where the terms O(n− ) are independent of the variables. The saddle points are now at i for both integrals. The steepest descent curve for the s-integral would be± the imaginary axis and for the t-integral would be the lines Im t = 1, the lower one described left to right, the upper one right to left. But in± order to get to these steepest descent curves, we must pass the s-contour through part of the original t-contour. To be speciﬁc, assume that the t-contour was the unit circle described counterclockwise and that the s-contour passed to the right of the t-contour. Then if we replace the s-contour by the imaginary axis (after which we can replace the t-contour by the lines Im t = 1), we must add the contribution of the pole at s = t when t is in the right± half-plane. Thus the change of contour results in having to add i 1 1 2 xt yt (τℓ τk+O(n− ))t /2 − 1 e − sinh 1 sinh 1 dt. πi + O(n− ) i 1+ O(n− ) 1+ O(n− ) Z− This is 1 2 (τ τ )t2/2 1 e k− ℓ sin xt sin ytdt + O(n− ), π Z0 NONINTERSECTING BROWNIAN EXCURSIONS 23 uniformly for x and y bounded. It is easy to see that the remaining integral, the s-integral over the imag- 1/2 inary axis and the t integral over Im t = 1, is O(n− ) uniformly for x,y bounded. In fact, outside neighborhoods of± the points where s2 = t2, we may estimate the double integral by a product of single integrals, each of which 1/2 1 2 2 is O(n− ), so the product is O(n− ). There are four points where s = t , where both s and t are i or where one is i and the other i. If we set up polar coordinates in neighborhoods± based at any of these poi− nts, then the δnr2 integrand, aside from the last factor, is O(e− ) for some δ> 0, while the 1 last factor is O(r− ) because s and t (or s and t) are on mutually perpen- dicular lines. Therefore, because of the factor r−coming from the element of δnr2 1/2 area in polar coordinates, the integral is O( 0∞ e− dr)= O(n− ). Thus H (x,y) has the scaling limit kℓ R 1 2 (τ τ )t2/2 e k− ℓ sin xt sinytdt π Z0 uniformly for x and y bounded. When k<ℓ, we must subtract from this

Ekℓ(x,y)= P (x,y,τℓ τk), − − which equals the expression above, but with the integral taken over (0, ). So the eﬀect of the subtraction is to replace the expression by ∞

2 ∞ (τ τ )t2/2 e k− ℓ sin xt sin yt dt. −π Z1 This limiting matrix kernel is exactly the extended Bessel kernel KBes(x,y) with α = 1/2. We have shown that the k,ℓ entry of the scaled BE kernel converges to the corresponding entry of the extended Bessel kernel, uniformly for x,y bounded. A minor modiﬁcation of the computation shows that the corresponding result holds for derivatives with respect to x and y of all orders and this is more than suﬃcient to give trace norm convergence on any bounded set. This concludes the proof of the result stated at the beginning of the section.

5.3. Limit theorem for highest curve. As mentioned in the Introduction, we shall not derive the scaling limit at the top. But since below, we shall use the top scaling of the diagonal terms Hkk(x,y) to the Airy kernel, we point out that this is very easy. With the usual scaling X √2n + X/(21/2n1/6), Y √2n+Y/(21/2n1/6), the summand in (3.5) with→ the denominator X +Y is easily→ seen to converge in trace norm to zero over any interval (s, ), by the asymptotics of the Hermite polynomials. The rest of the kernel converges∞ to the Airy kernel in trace norm, by known results. 24 C. A. TRACY AND H. WIDOM

5.4. Asymptotics of E(An,L) and E(An,H ). From our limit theorems fol- low the asymptotics of E(An,L) and E(An,H ). For the lowest curve, we use formula (4.5), which tells us that π E(A )= ∞ det(I Kχ ) ds, n,L √ − (0,s) 4 2 Z0 where K is given by (4.3). It is given in terms of KBE with m =1 and arbitrary τ1 = τ by

K(x,y)= 2τ(1 τ)KBE( 2τ(1 τ)x, 2τ(1 τ)y) − − − times exponential factorsq which we mayq ignore. Hence,q the scaling we have derived shows that 1 x y 1 sin(x y) sin(x + y) K , KBes(x,y)= − 2√n 2√n 2√n → π x y − x + y − in trace norm on bounded sets. As in Section 4.2, instead of KBes, we use Bes K0 deﬁned by KBes(√x, √y) KBes(x,y)= . 0 2x1/4y1/4 Bes It is convenient to use K0 since we can make direct use of the diﬀerential equations for it derived in [24]. By a change of variable, we have π E(A )= ∞ det(I Kχ ) ds. n,L √ − (0,s/2√n) 8 2n Z0 Therefore, if we can take the limit under the integral sign (which we shall show below), then this implies that c (5.4) E(A ) L , n , n,L ∼ √n →∞ where

π ∞ Bes π ∞ 1 Bes c = det(I K χ 2 ) ds = det(I K χ ) dx. L √ − 0 (0,s ) √ √x − 0 (0,x) 8 2 Z0 16 2 Z0 After expressing this in terms of KBes, it becomes the formula quoted in the Introduction. Solving numerically the diﬀerential equation satisﬁed by the logarithmic Bes derivative of det(I K0 χ(0,x)) [24], followed by a numerical integration, gives − 1 ∞ det(I KBesχ ) dx 4.917948 √x − 0 (0,x) ≃ Z0 NONINTERSECTING BROWNIAN EXCURSIONS 25 and hence cL 0.682808. From Table 1, we compute √nE(An,L) for n = 5,..., 9: ≃ 0.667334, 0.669708, 0.671449, 0.672784, 0.673838. To justify taking the limit under the integral sign in

∞ det(I Kχ ) ds, − (0,s/2√n) Z0 it is enough to show two things:

(5.5) lim ∞ det(I Kχ ) ds = 0, n − (0,s/√n) →∞ Z√n log n

det(I Kχ(0,s/2√n)) (5.6) − δs = O(e− ) uniformly for some δ when s< √n log n. Combining (5.5) with an application of the dominated convergence theorem using (5.6) justiﬁes taking the limit under the integral sign. This completes the proof of (5.4). The eigenvalues of Kχ(0,s/2√n) lie in the interval [0, 1] because K is a projection operator. For (5.5), we use the fact that the largest eigenvalue of Kχ is at least that of 2ϕ ϕ χ , so (0,s/2√n) 1 ⊗ 1 (0,s/√n) s/2√n 2 ∞ 2 s2/4n det(I Kχ ) 1 2 ϕ (x) dx = ϕ (x) dx = O(e− ). − (0,s/2√n) ≤ − 1 1 Z0 Zs/2√n Integrating this over (√n log n, ) gives (5.5). ∞ For (5.6), we use the fact that since the spectrum of Kχ(0,s/2√n) lies in (0, 1), we have

trKχ det(I Kχ ) e− (0,s/2√n) . − (0,s/2√n) ≤ λ (This follows from the inequality 1 λ e− when λ [0, 1].) The trace equals − ≤ ∈

s/2√n K(x,x) dx. Z0 Now, the usual asymptotics for the Hermite polynomials for “small” x hold uniformly for x = O(nε) if ε< 1/6 ([1], (19.13.1) and (19.9.5)), so they certainly hold for x = O(√n log n). Therefore, when s = O(√n log n), the above integral, which equals s 1 x x K , dx, 2√n 2√n 2√n Z0 26 C. A. TRACY AND H. WIDOM also equals 1 s sin 2x 1 + o(1) dx δs π − 2x ≥ Z0 for some δ. This gives (5.6) and concludes the proof of (5.4). For the highest curve, we use the top scaling in trace norm of the m = 1 kernel, which is just H11(x,y), as described in Section 5.3. We find, for fixed τ, that Xˆn (a random variable independent of τ) defined by

X Xˆ n = 2√n + n 2τ(1 τ) 22/3n1/6 − has the limit distributionp F2, the limiting distribution of the largest eigenvalue in GUE [23]. Since 1 π π E(A )= E(X (τ)) dτ = √n + E(Xˆ ), n,H n 23/2 8(2n)1/6 n Z0 we see that as n , →∞ π cH 1/6 (5.7) E(An,H )= √n + + o(n− ), 23/2 n1/6 where π c = ∞ s dF (s). H 8 21/6 2 · Z−∞ Here,

∞ s dF (s)= 1.771086807411601 ..., 2 − Z−∞ as computed to high precision by M. Pr¨ahofer, so

c 0.619623767170 .... H ≃− Evaluating (5.7) for n = 5,..., 9 gives 2.00981, 2.26104, 2.49069, 2.70345, 2.90254, respectively. These numbers should be compared with the entries in the third column of Table 1.

Acknowledgments. The authors thank Chris Cosgrove for his invaluable assistance regarding the diﬀerential equation (1.2) and Neil O’Connell for elucidation of the other Bessel process. The ﬁrst author thanks Anne Boutet de Monvel for her invitation to visit, and her kind hospitality at, the Uni- versity of Paris 7, where part of this work was done. NONINTERSECTING BROWNIAN EXCURSIONS 27

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Department of Mathematics Department of Mathematics University of California University of California Davis, California 95616 Santa Cruz, California 95064 USA USA E-mail: [email protected] E-mail: [email protected]