Airy-kernel determinant on two large intervals

We find the probability of two gaps of the form $(sc,sb)\cup (sa,+\infty)$, $c0$, in the edge scaling limit of the Gaussian Unitary Ensemble of random matrices, including the multiplicative constant in the asymptotics.

We are interested in the large-s behaviour of the corresponding Fredholm determinant P Ai (sJ) = det(I − K Ai ) (sc,sb)∪(sa,+∞) .
The determinant P Ai (sJ) is the probability of 2 gaps (sc, sb) and (sa, +∞) in the edge scaling limit of the Gaussian Unitary Ensemble (GUE), see, e.g, [23,10]. Clearly, by rescaling s, we can consider the second interval to be (−s, +∞). In the case of one gap, (−s, +∞), the determinant P Ai (−s, +∞) = det(I − K Ai ) (−s,+∞) is the Tracy-Widom distribution [25] -the distribution of the largest eigenvalue of the GUE. The same determinant also describes the distribution of the longest increasing subsequence in a random permutation [3]. Its large s asymptotics were first considered by Tracy and Widom [25] in 1994, who observed that where u(x) is the Hastings-McLeod solution of the Painlevé II equation specified by the following asymptotic condition: The asymptotics of the logarithmic derivative (d/ds) log P Ai (−s, +∞) follow, up to a constant (which is in fact zero), from (4) and the known asymptotics of the Hastings-McLeod solution at −∞.
On the other hand, in the bulk of the spectrum of GUE, the probability of a gap (−s, s) is given by the Fredholm determinant det(I − K sine ) (−s,s) of the trace class operator K sine on L 2 (−s, s) with the sine kernel K sine (x, y) = sin(x − y) π(x − y) .
As we will see, the present work relates the results (5) and (7) in some sense. Since J consists of 2 intervals, we expect the appearance of Jacobi θ-functions in the asymptotics. This phenomenon was first observed in [11], where the authors considered the sine-kernel determinant on several large intervals, and found its asymptotics up to a multiplicative constant. The constant in the case of 2 intervals for the sine-kernel determinant was recently found in [19]. In the present work, following many ideas from [19], we establish the asymptotics of (1) including the relevant multiplicative constant. To describe our results, we introduce some notation. Let Figure 1: Cycles on the Riemann surface.
and consider the Riemann surface of the function p(z) 1/2 with branch cuts on R \ J (see Figure 1). Fix the first sheet of the surface by the condition p(z) 1/2 > 0 for z > a. Define the elliptic integrals around cycles where the cycles A 0 , A 1 , B 1 are depicted in Figure 1: parts represented by solid lines are on first sheet of the surface, while dotted line is on the second. Consider the function on the first sheet. Here q(z) = z 2 + q 1 z + q 0 is a polynomial such that z a q(ζ) p(ζ) 1/2 dζ = 2 3 and b c q(ζ) p(ζ) 1/2 dζ = 0.
We now state our result.
In the recent work [4], Blackstone, Charlier and Lenells have simultaneously and independently analyzed the large-s asymptotics of log P Ai (sJ). They found the expansion −α 2 s 3 − 1 2 log s + log θ 3 (s 3/2 Ω) + χ ′ + O(1/s) with an undetermined constant term χ ′ = χ ′ (a, b, c). (This analysis was then extended by the authors to the case of n gaps in the bulk of the Airy process in [5], and in the Bessel process in [6].) They followed the approach of [11], and used Riemann-Hilbert analysis to obtain the asymptotics of the derivative d ds log P Ai (sJ). To determine the multiplicative constant in the asymptotics of the determinant, one would need to integrate the logarithmic derivative over s. However, there is no appropriate initial point s 0 where the asymptotics of P Ai (s 0 J) would be independently known. For this reason the problem of determining the constant is different and requires additional ideas. As mentioned above, it was first solved for a single interval in the bulk of the spectrum of GUE. The solution in [21] and also the one in [14] involved representing the sine-kernel determinant as a double-scaling limit of a Toeplitz determinant (whose asymptotics at certain points are either known or can be independently determined) and then integrating a differential identity for Toeplitz determinants starting from this point. The differential identity can be found in the asymptotic form by a Riemann-Hilbert analysis. Similarly, the constant c Airy in (5) was determined in [12] by integrating a differential identity for a Hankel determinant and using the fact that the asymptotics of the Hankel determinant can be independently established at a certain point, thus providing an initial point for the integration.
It was observed in [19] that for the sine-kernel determinant on 2 intervals, one does not need to reduce the problem to Toeplitz determinants (although, of course, one still can), but rather it is easier to notice first that if the 2 intervals are far apart from each other in comparison with their width, then the determinant splits (to the main orders in s) into a product of 2 determinants, each on a single interval so that we can use the asymptotics (7). Then to complete the solution, one determines the asymptotic form of a differential identity with respect to the edges of the intervals and performs integration starting from a point where splitting into the product occurs.
Here we follow a similar approach to [19]. First, in Section 2, we establish a separation lemma, which states that if the length of the interval (c, b) is relatively small compared to b − a, and a is close to zero from the left then P Ai (sJ) is written (to main orders in s) as a product of a sine-kernel and Airy-kernel determinant for which we can use the asymptotics (7), (5), respectively. More precisely, we prove Lemma 3 (Separation of gaps). Set b = c + 2t 0 s 3/2 and a = − t 1 s , where t 0 = t 1 = (log s) 1/8 . Then as s → +∞, Remark 4. We can also choose different values for t 0 , t 1 , and slightly larger in s than (log s) 1/8 .
To solve the problem for arbitrary fixed c < b < a < 0, we proceed as follows. In Section 3, we formulate differential identities which express the derivatives with respect to the edges, d da log P Ai (sJ), d db log P Ai (sJ), in terms of a solution of a certain Riemann-Hilbert problem. In Section 4, this problem is solved asymptotically for large s and fixed c < b < a < 0, by the Deift-Zhou steepest descent method (as for the sine-kernel on 2 intervals, the solution involves θ-functions). In Section 4.6 we verify that this solution is extendable for variable edges up to the scaling regime of Lemma 3. We then substitute the solution into the differential identities and perform integration: first setting a = α − x, b = β + x, where α is close to zero and β is close to c in the sense of Lemma 3, we integrate d dx log P Ai (sJ) from x = 0 to x 0 such that a = α − x 0 . This fixes the desired value of a.
Then we integrate the identity d db log P Ai (sJ) to the general position of b and thus conlcude the proof of Theorem 1. Note that unlike the cases of just 1 gap, the integration of the differential identities for 2 gaps is technically involved: we have to use various identities for θ-functions (see Section 4.3) and averaging over fast oscillations. The details of this computation are presented in Section 5.
2 Separation of gaps: Proof of Lemma 3 In this section C j , j = 1, 2, . . . , will denote positive constants whose value may change from line to line.
Recall the kernel and denote We first observe the following.
Proposition 5. For z, z ′ ∈ (sc, sb), let Then, uniformly for x, y ∈ (0, 2), Proof. We will make use of the following expansions of the Airy function for large, negative argument (see, e.g., [1]): and for the derivative (Ai ′ (u) = d du Ai(u)) as z → +∞, where the principal branches of the roots are taken with cuts along (−∞, 0). Let us define, for Re (z) > 0, so that Since the Airy function is entire it follows, by definition, that Err Ai (z) is analytic in Re (z) > 0. Moreover it follows from differentiating Err Ai , making use of the expansions (32) and (33), that where for the second derivative we have used, in addition, the Airy equation Ai ′′ (z) = zAi(z). Let us denote , and C(z) = cos 2 3 Noting that we obtain, by means of (35) and its derivative, that For f, g denoting analytic functions, there are z * Applying this to the pairs f (z) = C(z), g(z) = Err Ai (−z) and f (z) = 1 (37), and using the estimates (36), we obtain (31), uniformly in x, y ∈ (0, 2). With the assistance of the above result we may now prove Lemma 1. Consider the kernel The corresponding determinant splits into the desired product of determinants up to a small error. We have Proposition 6. There exist constants C 0 , C 1 > 0 depending only on c such that Proof. First we note that Furthermore, using the change of variables, and Proposition 5, we write Let r j denote the j'th row of K sine and let e j denote the j'th row of the error matrix, so that e j = O t 2 0 s 3/2 , j = 1, . . . , k. Then we have Note that the determinants in the sum over j on the right hand side may be estimated by Hadamard's inequality. Let v ℓ denote the rows of a matrix. If all the matrix elements v ℓ m ≤ C, Using uniformity of the error term and the estimate we obtain This estimate and (43) imply (note that det( Since by (5), the estimate (47) implies the statement of the proposition.
On the other hand, the difference between the determinants corresponding to the kernels K Ai and K Ai is small: There exist constants C 2 , C 3 > 0 depending only on c such that Proof. We use another representation of the Airy-kernel, and the asymptotics of the Airy function at +∞, see, e.g., [1], and also the arguments similar to those in Proposition 5 to conclude that where , z, z ′ ∈ (sa, +∞).

Differential Identity
In this section we express, for p = a, b, c, the derivative d dp log det(I − K Ai ) in terms of a solution of a certain Riemann-Hilbert problem (RHP). First let us write the kernel of the operator K Ai in the form Note that 2 k=1 f k (z)g k (z) = 0. The operators of this form are known as integrable operators. They possess the following crucial properties (Lemmata 2.8 and 2.12 in [11]).
The resolvent R s of the operator K Ai is given by where R s can be expressed as where m(z) is the 2 × 2 matrix, the solution to the following RHP: The function m(z) possesses L 2 boundary values m + and m − as z approaches (c, b) ∪ (a, ∞) from above and below, respectively. These boundary values are related by the condition where the jump matrix Before proceeding with the derivation of the differential identity, we will now reduce this RHP to another one, with constant jumps. For that, we need the following model problem.

Airy model RH problem
Let y j (z) = ω j Ai(ω j z), ω = e The pre-factor normalises the determinant to unity at all points. For Γ = R ∪ R + e ± 2πi 3 (see Figure  2 it is known, see e.g. [8], [10], that Φ satisfies the following RHP with L 2 boundary values: where N 0 is given by The calculation to verify that Φ indeed satisfies (68) rests upon the well-known facts For further details we refer the reader to p.216 in [10]. The reason for introducing Φ is because the jump, V −1 (sz), of the m-RHP may be factorised in terms of Φ(sz) as follows:

RH problem for X
Making use of the Airy model problem, we transform the m-RHP into a form with constant jump matrices. We define the matrix X(z) in each region of C as per Figure 3. Using (70) and the m and Φ-RH problems, we find that X(z) satisfies the following problem, where Σ denotes the union of bold curves on Figure 3.

The Identity
We now proceed with the derivation of the differential identity. Consider the case when p = b, the identities at the points a, c are obtained similarly.
where, for the final equality, we used (59). The negative sign at R s (b, b) comes from the fact that b is the lower limit of integration. (For p = a, c, we have the opposite sign.) Now, by (60), we may write For definiteness, assume that the limit is taken from outside the lens and with Im z > 0. By (61), the definition of X and that of Φ in sector II, where we used the fact that det Φ(z) ≡ 1. Using it once again, we reduce (74) to Similarly, we obtain Substituting (75) and (76) into (73), we finally obtain where the limit is taken from outside the lens with Im z > 0; and therefore by (72) At the points a and c, we obtain the same result but with the opposite sign. Thus we have Lemma 8 (Differential identity). The Fredholm determinant (1) satisfies: where the + sign is taken if p = a, c and the − sign, if p = b. The limit is taken from outside the lens and with Im z > 0.

Solution to the Riemann-Hilbert problem for X
To solve the X-RHP for large s we, as usual, apply a series of transformations. The approach, known as the steepest descent method for Riemann-Hilbert problems, was first introduced by Deift and Zhou [15] and used and developed in many subsequent works. The first step is to normalise the exponential behaviour at ∞ by multiplying from the right by a suitable function. This process is set up as follows.

g-function and the RH problem for S
In the introduction, we defined Here , the branch of the root is chosen positive for positive arguments, and the branch cut is on (−∞, c) ∪ (b, a), c < b < a < 0; q(z) = z 2 + q 1 z + q 0 is the second degree polynomial. We require that g(z) satisfies conditions (11) and (12), which as we now show determine the coefficients q 0 , q 1 . We have Lemma 9. The function g(z) is analytic in C \ (−∞, a] and satisfies the jump conditions: The polynomial q(z) = z 2 + q 1 z + q 0 has coefficients given by (13). where Proof. Since p(z) + + p(z) − = 0, on (−∞, c) ∪ (b, a), it follows from (12) that On the other hand, for z ∈ (c, b), we have where the last equality follows from (12). Thus we have (80).
Write the expansion at infinity Setting and integrating, we obtain For some constant g 0 . It follows from the jump condition g + (z) + g − (z) = 0 on the half-line (−∞, c) that g 0 = 0. Thus, condition (11) holds with q 1 given in (13). The first expression in (13), follows immediately from (12). Next, taking the integral over the B 1 cycle (see Figure 1) on the Riemann surface of p(z) 1/2 , we have This allows one to write J 2 in terms of J 0 , J 1 , and thus we obtain the second equality in (13). A straightforward series expansion of g(z) at infinity verifies the values of α 1 , α 2 .

Now set
It follows from the lemma above and the X-RH problem, in (71), that S satisfies the following problem. For Σ as in Figure 4, we have The expansion of S(z) at infinity is found using (81) and the expansions of Φ and m at infinity. Up to the error term, it is the same in all sectors. We now construct approximate solutions (parametrices) for this S-RH problem: outside parametrix away from the points a, b, c, and local parametrices around these points, see Figure 5. This allows us to construct an asymptotic solution to the S-problem, and therefore to the X-problem.

Outside parametrix. Jacobi θ-functions
In this section we construct a parametrix for the S-RHP away from the points p = a, b, c, which models the behaviour of S(z) at infinity. We proceed as follows.

β model problem
Set , with branch cuts along (−∞, c) ∪(b, a) and such that β(z) > 0 when z → +∞. Consider the function where N 0 is given by (69). The function for some diagonal matrices N j , which may be written in terms of a, b, c. Moreover, since the boundary values of β satisfy β(z)

Abel map
Recall J 0 and τ as defined in (9) and (16) respectively, in the introduction. Consider the map u(z) defined by We note several relevant properties of u(z) which will be required in defining the outside parametrix.
and has the following properties. and where the coefficients u j may be written in terms of a, b, c.
Proof. It is immediate from the definition that on the + side of the cut Considering the integration around the A cycles (see Figure 1), we also note that From here the statements of the lemma easily follow.
Note that we consider u(z) on the first sheet of the Riemann surface of the function p(z) 1/2 (where we have p(x) 1/2 > 0, β(x) > 0, for x > a on the real line). The function u(z) maps the full Riemann surface to a torus (where θ-functions are defined).

Theta function parametrix
A θ-function parametrix for the sine-kernel determinant on several intervals was first constructed in [11]. We now modify the construction for our case.
Recall the function θ 3 (z) given by (17) and the constant Ω given by (16). Consider where d a constant to be determined. We will also write P ∞ (z; s) in the short hand notation where N jk are the entries of N(z) as defined in (87), and We prove: Proposition 11. P ∞ satisfies the following RHP. where Note that P ∞ (z) has the same jumps and behaviour at infinity (up to left-multiplication by a constant matrix), as S(z).
Proof. To check analyticity in C \ (−∞, a] one needs only to determine that no poles are introduced from the zeros of the theta functions in the denominators. In fact, as in [11], we determine the constant d so that those which do occur, are precisely cancelled by the zeros of N jk (z). We claim that the function β(z) − β(z) −1 has two zeros Indeed, we have that Since h(z) → +∞, if z → ±∞, we deduce that h has a zero, β * 1 in (c, b) and another β * 2 in (a, +∞). Since β(z) ± β(z) −1 = 0 ⇐⇒ β(z) 2 = ±1, the sign of the root (given our choice of branches) implies that β * 1 , β * 2 are the zeros of β(z) − β(z) −1 , whereas β(z) + β(z) −1 has no zeros (its zeros are on the second sheet of the Riemann surface).
It is well known (see, e.g., [26]) that θ 3 (z) has only one zero at z = 1 2 + τ 2 modulo its period lattice Λ = Z + τ Z. By Lemma 10, we easily see that u(z) maps C \ (−∞, a] onto (−1/2, 0) × (−τ /2, τ /2). Therefore θ(u(z)) = 0 ⇐⇒ z = c. Define then It follows that This proves that, for our specific choice of d, the zero of To see the analyticity of the diagonal terms P ∞ (z) 11 , P ∞ (z) 22 , we note that since 0 < d < 1/2 we have, using again the mapping u(z), that u(z) We now turn to the jump conditions. The jumps of P ∞ (z) are verified by means of (92)-(94), together with the quasi-periodicity properties of θ 3 (z) in (18) and the jumps for N(z). Indeed for the 11 entry, on the set (−∞, c) ∪ (b, a), we have where we have used θ 3 (z + 1) = θ 3 (z), and evenness of θ 3 (z). On the interval (c, b) we have Repeating this computation for each of the remaining 3 entries yields the jumps of P ∞ . To obtain the large z expansion of P ∞ (z) we note that P ∞ (z)N(z) −1 is analytic at infinity, and use the fact that by (95), u(z) → −1/2 as z → ∞.

Expansions of u(z), β(z), and some properties of θ jk
Below we will use the following easy to obtain expansions: and with The values of β 1,p are unimportant in what follows. By the quasiperiodicity relations (18) and the definition (16) of Ω, we obtain the following connections between the values of θ jk in (98) on the +-side of (−∞, a): Now considering the derivatives of the second relation in (18), we moreover obtain and where and all θ jk and their derivatives are evaluated at u + (b). Similarly, at u(a) = 0, we obtain

Identities for θ-functions and elliptic integrals
We now obtain a set of identities involving θ-functions which will be used below in the proof of Theorem 1. First, recall all four Jacobi θ-functions. These may be expressed in terms of θ 3 (z), defined in (17), as follows.
We note that θ 1 (z) is an odd function of z, whereas θ j (z), j = 2, 3, 4 are all even functions. θ-functions have the following periodicity properties: Denote by θ k , θ ′ k the values of θ-functions and their derivatives at zero (θ-constants), i.e., Lemma 12. Recall J 0 , τ , Ω, P ∞ (z; s) and d given by (9), (16), (96) and (102), respectively. The following identities hold. . . (123) Proof. The proof of this lemma is similar to the arguments in [19] wherein the authors derive analogous results in the case of the sine-kernel determinant. For the first three identities we proceed as follows. By the expansion of u(z) at a, we obtain Since θ 1 (0) = 0 is the only zero of θ 1 (z) modulo the lattice, it follows from Liouville's theorem that We evaluate the constant by taking z → ∞, making use of (114), and obtain Substituting here the value z = b, and making use of the identities (see, e.g., [26]) we obtain (i). Similarly, substituting z = c into (125), we obtain (ii). The difference of (i) and (ii) gives (iii).
One may expand the determinant at any point and deduce that all the terms apart from the constant one must be zero, yielding an infinite number of identities linking the derivatives of θ functions. In what follows, we will only need the constant terms in the expansion at z = b and z = a. A straightforward computation using expansions (104), (108) gives (in the notation of (97)) for z → b, where u 0,b , β 0,b are defined in (106) and (109), respectively. We note that Furthermore, using the identities (110), (111), we can rewrite (133) in the form Similarly, expanding (128) as z → a, one obtains the identity v) Consider the function η(z) defined by Note that this expression is similar to that of det P ∞ above. Indeed we may employ the same methods and deduce that it is identically equal to 0. On the other hand, by considering the constant coefficient (i.e. the coefficient of (z − b) 0 ) in its expansion for z → b, we obtain the desired identity (v). Vanishing of the term with (z − b) 1 in the expansion yields the identity (vi). Here we expanded the functions θ 1 θ 3 , β(z), used (91), (115), the oddness of θ 1 , the evenness of θ 3 , and the fact that Similarly, expanding the expression at the point z = a instead, one obtains (ix) and (x).

vii) Recall that
where we integrate along the cycles. Note that By means of direct differentiation, we obtain and Furthermore, expanding the integrand of along the cycles, we obtain Differentiating q 0 in (13) and using the above derivatives of J k , we find This identity will be used repeatedly in the proof of the theorem. The final ingredient is obtained by means of Riemann's period relations. Indeed these relations yield (by similar arguments as in the proof of Lemma 3.45 in [11] and Lemma 27 in [19]) Alternatively, to show (144), we can first reduce I 0 , I 1 , J 0 , J 1 to complete elliptic integrals (cf. next section) K, E, and then use the Legendre relation: The above facts, taken together, allow us to differentiate Ω directly and obtain (vii). Similarly, we have and Furthermore, With the help of these equations, (xi) easily follows.

Local parametrices
In this section we construct parametrices in small neighborhoods, U p , of the points p = a, b, c, see Figure 5 where their boundaries are depicted, which match the exterior parametrix to the main order as s → ∞ on the boundaries ∂U p . As in [11], [22], [8] these parametrices involve Bessel functions. Figure 5: Jump contour with neighbourhoods U p .
This will be used in subsequent analysis. First note that, since we have that Substituting these inequalities into the second form of q 0 in (13), we find that and To determine the sign of q(a) we require a sharper bound. Let us, first, write Then note the following standard reductions, with k = b−c a−c .
where K(k) and E(k) are the complete elliptic integrals of the first and second kind, respectively.
Note that 0 < k = b−c a−c < 1. With these expressions for J 0 and J 1 we find that q(a) takes the following form.
We want to show that q(a) < 0, i.e. that As the authors in [4] we will make use of the estimate This reduces the problem to seeing that But this is clear since, for 0 > a > b > c, we have The estimate (159) was proven in [24], in a different form, and also follows from log-convexity of the Gauss hypergeometric function [7]. We now provide a proof for the reader's convenience.
First we have that and so It, thus, remains to prove the estimate For 0 < α < β, let us denote by M(α, β) the arithmetic-geometric mean. This is defined as the common limit of the following sequences.
The required estimate, (163), now follows from (165) by noting that

Bessel model RH problem
First recall the Bessel model RH problem of [8], which is a slight modification of the corresponding problem in [22], which in turn, is related to the problem in [11].

Construction of the parametrices
Writing It is easy to verify that there exist an open disk U p centered at p (for each p = a, b, c) of a sufficiently small radius ǫ such that f p (z) is a conformal mapping of U p onto a neighborhood of zero. Expanding close to z = p, we have, in particular, Note that the signs of the quantities q(p) in (151) ensure the location of the cuts of f p (z) 1/2 in U p corresponds to the contour of the Ψ-RHP under the mapping ζ(z) = s 3 f p (z).
We now choose the exact form of the jump contour Σ S , for the S-RHP problem (86), in U p so that its image under the mapping ζ(z) is direct lines. This relates the construction to the Bessel problem. We look for local parametrices, with z ∈ U p , in the form where E p (z) are, analytic in U p , matrix-valued functions chosen so that P p (z; s) matches S(z) on ∂U p to the main order. We will now see that the following functions satisfy these conditions. Set and where in ± we choose +, if Im (z − p) > 0 and −, if Im (z − p) < 0.
We have Lemma 15. The functions E p (z) = E p (z; s) defined by (176), (177) are analytic in the respective neighbourhoods U p .
Proof. Note that (174) gives Consider U c . First, we verify that E c (z) has no jumps in U c . On (c − ǫ, c), we have using the jump conditions for P ∞ in (99), and the fact that by (80), g + (c) = −g − (c), On (c, c + ǫ), since g + (b) = g + (c), So that the singularity of E c (z) at c is not a branch point. Since f (z) and the matrix elements P ∞ jk (z) contain at worst a (z − c) −1/4 singularity each, it follows that the singularity of E c (z) at c is removable. Thus E c (z) is analytic in U c .
The functions E b (z) in U b , and E a (z) in U a are considered similarly. Now by the large ζ expansion of Ψ(ζ) and (173), we find that on the boundaries ∂U p for large s, uniformly in z, P p (z; s) = P ∞ (z; s)e −s 3/2 g ± (p)σ 3 I + 1 for p = a, c, and where as before, in ±, we choose +, if Im (z − p) > 0 and −, if Im (z − p) < 0. Since E p (z) are analytic, left-hand multiplication by them does not affect the jump conditions for P p (z), and it is easy to check that P p (z) satisfies the same jump conditions as S on Σ S ∪ U p .
Thus, we have shown Lemma 16. With Σ S and v S denoting the contour and jumps, respectively, of the S-RHP (86); the function P p (z; s) given by (175) in U p , p = a, b, c, has the following properties.

Small norm RH Problem for R
and let Σ R denote the contour in Figure 7. Then, R satisfies the following Riemann-Hilbert problem. where v S (z) = 1 0 e 2s 3/2 g(z) 1 .

R is analytic in
Indeed, the jump conditions easily follow from the RH problems for S(z), P ∞ (z) and P p (z). The condition at infinity follows from (86), (99), (88).
We will now show that all the jumps for the R-RHP problem are close to the identity for large s both in terms of L ∞ and L 2 norms, which guarantees the solvability of the problem for large s [10].
Let us first consider the jumps of R on Σ R \ (∂U a ∪ ∂U b ∪ ∂U c ). We claim that one may deform the contour Γ 1 ∪ Γ 3 ∪ Γ 4 ∪ Γ 6 , in the RH problem for S(z) (86), in order to have Re g(z) < 0 on this part of Σ R . In particular, we show that this implies we have that, for sufficiently large |z|, Re (g) < 0 in the sector, and hence (184) holds on Σ R for suffcieintly large |z|.
On the other hand it follows from (151) that and therefore Im q(z) Since Re g(x) = 0, for x ∈ (−∞, c) ∪ (b, a), using the Cauchy-Riemann equations we have, for 0 < y < δ, (δ sufficiently small) that Similarly, decreasing δ if need be, we show that Conditions (187) and (188) together with the boundedness (from infinity and zero) of θ-functions on the contour imply that the jump matrix satisfies for some ǫ > 0 uniformly on the lenses of the Σ R contour around (b, a), and also for sufficiently small |z − c| on the lenses around (−∞, c). It remains to obtain an L ∞ estimate on the rest of latter lenses: between a fixed small |z − c| and a fixed large |z − c| (above which we can use the estimate (185)). Consider the part of the lenses with Im z > 0 (for the Im z < 0 part we argue similarly). Clearly, it is sufficient to show that Let φ 1 > φ 2 > φ 3 > φ 4 > φ 5 denote the acute angles between the real line and the direct lines connecting x + iy, x ∈ (−∞, c), y > 0, and the points c < x 1 < b < a < x 2 , respectively. Then It is easily seen that and thus we obtain (189).
To summarize, we established the estimate on the lenses of the contour Σ R . It remains to consider the jumps of the R problem on the boundaries ∂U p . By means of Lemma uniformly on ∂U p , p = a, b, c. Thus, by standard theory (see, e.g., [10]), the R-problem is a small-norm problem for large s solvable by means of Neumann series. In what follows, we will need the first 2 terms. We then obtain where the estimate for R (1) (z) and the error term are uniform in z. As usual, we substitute these series into the R-RHP, and obtain that This problem is solved by the Plemelj-Sokhotski formula, so that Note that the integration along the circles ∂U p is in the negative direction. Below, we will find R (1) (z) by computing the residues. Retracing our transformations R → S → X yields the asymptotic solution to the RH problem for X(z).

Extension to the limiting regime of Lemma 3
We now show that the solution to the Riemann-Hilbert problem for fixed c < b < a < 0 extends to the regime where a → 0 − and b − c → 0 + , as in Lemma 3. We need to verify that the jumps of the R-RH problem remain small in L ∞ and L 2 norms. Then the R-RH problem is solvable by Neuman series by standard arguments for small norm RH problems on contracting contours. Let us first consider the jumps on ∂U a ∪ ∂U b ∪ ∂U c . We will now show that in the regime of Lemma 3, by choosing the diameters of U a , U b , U c appropriately, we have uniformly in z First, uniformly for a − b > ε > 0 as b → c we have and and, therefore, With this expression for q 0 we may expand q(p), p = a, b, c. Indeed, we find that uniformly for and Note that, for p = b, c, This is because we have diam(U p ) < min{|β * 1 − b|, |β * 1 − c|}, where β * j , j = 1, 2, are the zeros of β(z) − β(z) −1 (they satisfy (101)). We easily derive the expansions We may and do set diam(U p ) = ǫ ′ min{|β * 1 − b|, |β * 1 − c|}, with 0 < ǫ ′ < 1 2 . Note that the diameter of U p decreases in this regime. On the other hand, we have that The diameter of U a does not need to decrease, set diam(U a ) = ǫ ′ . By (173) and (174) and the above estimates for q(p), we have Recall also that Re g + (p) = 0, p = a, b, c. In view of Lemma 16,to prove (195) it remains to show that P ∞ (z) jk , z ∈ ∂U p , are uniformly bounded in the prescribed regime. By the first expansion in (202), the functions β(z) and β(z) −1 are uniformly bounded on ∂U b , ∂U c . Clearly, they are also bounded on ∂U a .
To investigate the behaviour of θ-functions, first analyze τ , and for that write Then it is easy to compute the integrals and obtain and, by using also (196), Thus iτ → −∞ logarithmically as b → c. By the series representation for θ 3 (ξ), this implies that the θ-functions would converge to unity if their argument was any and fixed or was real. However, since it is in general complex and depends on a, b, c, we need to show that θ(u(z) + (−1) j s 3/2 Ω + (−1) k d) in the numerators in P ∞ are bounded, and also that θ(u(z) ± d) in the denominators are bounded away from zero. From the definition of d in (102), we have using (196) and (202) that, as b → c, Therefore, Using (202), denote Thus, Arguing as in the case of d above, we obtain and |u(z)| ≤ ǫ for z ∈ ∂U a for some ǫ > 0. Choosing ǫ ′ > 0 sufficiently small, we see that these equations and (207) imply in particular that u(z) ± d is uniformly bounded away from the zero 1+τ 2 mod Λ of θ 3 (ξ) on ∂U a ∪ ∂U b ∪ ∂U c . We then obtain for some constants Thus P ∞ is bounded on ∂U a ∪ ∂U b ∪ ∂U c and therefore (195) holds. The analysis of the jumps of R on Σ R \ ∂U a ∪ ∂U b ∪ ∂U c is similar. (Note, in particular, that 0 < Im u(z) < |τ | on the contour.) We obtain To conlude, we see from (195) and (210) that in the regime of Lemma 3, where a − b > ε > 0, we have Therefore the R-RH problem is solvable in this regime, and we can write uniformly in z and uniformly in the range from fixed c < b < a < 0 to the regime (211).

Proof of Theorem 1
We start with the differential identity (79) for p = b, and write P Ai (sJ) = P Ai s (a, b), to indicate the values of a and b, where the limit is taken from outside the lens and with Im z > 0. We first concentrate on the case of c < b < a < 0 fixed, and then provide an extension to the asymptotic regime of Lemma 3. By the definitions of S(z) and R(z), we have that in the region U b ∩ {z : Im z > 0}, and outside the lens, By definition (175) of P b (z), in particular in the preimage ζ −1 (Ω 1 ) (which is outside the lens), and therefore Moreover, note that for z in a neighbourhood of b, where Note that the (uniform) estimate for the error term in (216), follows from the differentiability of the asymptotic expansion of R, which, in turn, follows from the uniform estimate for R, in (192), and Cauchy's integral formula. Starting with the (differentiable) expansions, see e.g. [1], for the Bessel functions where γ is Euler's constant, it is straightforward to verify that the function (169) Using this expression, the definition and the fact that P ∞ (z; s) is bounded, as s → ∞, we estimate the error term in lim z→b X −1 (z)X ′ (z), arising from that in (216), as follows Thus, in the region where With these we have the formula for fixed c < b < a < 0, where the limit is taken along a path in W b .
To extend (224) to the regime b → c, a → 0 of Lemma 3, we rely on the results of the previous section. Considering R −1 R ′ by using (213), we obtain and hence (cf. (220)) uniformly in the range from fixed c < b < a < 0 to the regime (211), where the limit is taken along a path in W b . In the next 3 sections, we prove the following lemmata (with the notation from the introduction).
Lemma 19. With ω = s 3/2 Ω, Using these results, we may now write the differential identity (226) in the following explicit form.
Proposition 20. (Asymptotic form of the differential identity for b) Let ε > 0 be fixed. Then where and Proof. By (226) and 3 lemmata above, we only need to prove (234). First we show that i.e. replacing lim z→b τ 3 (z) 21  First, let c < b < a < 0 be fixed. Then f is an analytic function of ω. Let f n denote its Fourier coefficients For n = 0, it follows from integration by parts that From (120) of Lemma 12, we see that dΩ db (a, b, c) is a strictly positive differentiable function of the parameters a, b, and c, bounded away from zero if c < b < a < 0 are bounded away from each other. Furthermore, the functions f n , ∂f n /∂b decrease with n exponentially. Therefore Now let c < b < a < 0 be in the range of the Proposition. We will show that which implies (235). By (120), (196), (200), as b → c uniformly for a − b > ε > 0. We note, in particular, that dΩ db remains bounded away from zero in the regime of the Proposition. To apply the above Fourier series arguments and deduce (240), it suffices to show that in the regime of the Proposition. Since and similarly for ∂ ∂b f n (a, b, c), we only need to show that One can use directly the definition of lim z→b τ 3 (z) 21 , but it is easier to consider its simplified form in Section 5.3 below. By (205), we have uniformly in ω, the estimates for the θ-functions and their derivatives θ ′ j (ω) = O(b − c). The analysis of the previous section shows that the matrix elements (98), θ jk (u(z)) are bounded on ∂U b , and so are their derivatives w.r.t. ω. Using (208), becomes O 1 s 3/2 min{|a|,b−c} after integration. But this follows from the above considerations similarly to (and simpler than) (240).
Using the analogous results for the differential identity at a, we follow the above methods and arrive in Section 5.4, similarly, at Proposition 21. (Asymptotic form of the differential identity for a) Let ε > 0 be fixed. Then uniformly for c < b < a 0 < a < a 1 < 0, where a 0 , where D(a, b) is given by (231), and Moreover, Because of the form of the error terms, we will need to move the edges a and b first simultaneously. To do this we will use the following Proposition which easily follows from the 2 ones above and some similar additional arguments. For fixed α, β, we denote a = α − x, b = β + x and remark that Proposition 22. (Asymptotic form of the differential identity for a = α − x, b = β + x) Let ε > 0 be fixed. Then uniformly for 0 ≤ where D(a, b) is given by (231), and Θ a , Θ a , Θ b , Θ b as in Propositions 21, 20 above. Moreover, .
With these propositions, we may prove Theorem 1. We will integrate to reach the desired values of a = a 0 , b = b 0 of the Theorem. Set α = − t 1 s , β = c+ 2t 0 s 3/2 , t 0 = t 1 = (log s) 1/8 . First, by Proposition 22, we integrate over x from x = 0 to x = α − a 0 , which fixes the desired value of a: Next, using Proposition 20, we integrate over b from the present value b = β + α − a 0 to b = b 0 : The sum of these expressions gives Note that log P Ai s (α, β) in (254) may be replaced with the expression in Lemma 3. Thus it remains only to expand the expression for D(α, β) when s → ∞. To this end we prove the following Proof. The formula is obtained by expanding all terms in D(a, b) with a = − t 1 s and b = c + 2t 0 s 3/2 , s → ∞. First, by definition of α 2 in (15) and the exansion of q 0 (198), Next, by (245), Finally, the expansions (199), (200), and (201) imply that Combining (256) In the following sections we prove the aforementioned lemmata 17, 18, and 19, and their analogues for the differential identity in a.
Remark 24. Identity (260) is suitable to prove that α 2 > 0. Note that this property is important since it determines exponential decay of the determinant itself. It is clear that the right hand side of (260) is positive for all values of a, b, and c. Since (cf Section 4.6) α 2 → − a 3 12 > 0 as b → c, we have that

The oscillatory term. Proof of Lemma 18.
We now study the term From the definition we have that It follows from the expansions of Ψ(ζ) in (218) and of f b (z) in (174), that the contribution to τ 2 from the first term in the above expression is Next, with the notation for P ∞ in (97), we obtain, using the fact that det P ∞ (z) = 1, det(P ∞ ) ′ (z) = 0 (see Lemma 12 (iv)), and where we omit the arguments for brevity. In this notation, we determine using the expansion of Ψ(ζ) in (218) that where we substituted (264). We first evaluate the term with B − C in this expression. Writing we have by (104) Replacing d dz θ jk by θ ′ jk du dz in the above expressions for B and C, we obtain by the expansion of β(z) in (108) and the identities for the values of θ jk at u + (b) in (110), where we used the identity (133) in the last equation. Then, by the expansion of f b (z) in (174), which cancels the last term in (269). We now evaluate 2A + i(B + C). We obtain as before, but now using also (112), that and similarly, Recalling also (134), we obtain Substituting this expression into (269) and using again (174) and the value of u 0,b in (106), we have Substituting this into (277) and writing out the total derivative, we obtain Lemma 18.
The above observations and the fact that det P ∞ = det F ≡ 1 imply where with, by recalling Lemma 16, where in ±, + is taken if Im ξ > 0, and − if Im ξ < 0. (In fact, g ± (b) = g ± (c).) Recall that ∆ 1 (ξ) (and hence ∆ 1 (ξ)) is a meromorphic function near each point a, b, c. In fact, it has a first order pole at these points. Note that since there is an additional pole at b introduced by the denominator in (281), we must expand ∆ 1 (ξ) up to the third term at b, but only up to the first term at a and c to compute the residues. Therefore, it is convenient to define where we denote ω = s 3/2 Ω. Then we have lim z→b τ 3 (z) 21 = T 2 (ω) + T 3 (ω). (290)
Substituting these expressions into (304), noting the signs of q(p) obtained earlier in (151), and using the simple identities we find

Evaluation of T 3 (ω)
In view of the definition (289), we will now simplify the expression for L(ξ) in (285), with ∆ 1 (ξ) given by (287). We use the following argument from [19] to decompose matrix products. In a neighborhood of b, define functions A j (z) and B j (z), j = 1, 2, by A 1 (z) = 1 2 θ 11 (u)(β(z) + β −1 (z))e ∓s 3/2 g + (b) + θ 12 (u)(β(z) − β −1 (z))e ±s 3/2 g + (b) , where the upper sign in ±, ∓ is taken for Im (z − b) > 0, and the lower, for Im (z − b) < 0. It follows from the jump relations (cf. proof of Proposition 11), that are analytic functions in U b . In this notation, we have This is a convenient way of expressing ∆ 1 (z) near b for the following reason. Since ∆ 1 (z) is meromorphic near b, its expansion at z = b contains only integer powers. On account of the 1/f b (z) 1/2 term, containing only half-powers, it follows from the analyticity of (310) that all 'cross-terms', containing products A j B k , must be zero. Thus we have the following decomposition: It is now a straightforward calculation to see that we may write (285) near b in the form By means of (104), (108), and the identities for θ jk in (110), (111), we expand β −1 (z)A j (z), z → b, for j = 1, 2, and determine that the analytic (by (310)) in U b function where all θ jk and their derivatives with omitted argument are evaluated at u + (b), and we used the identity (134) in the last equation. Notice that the z − b term is, up to a prefactor, is what we found earlier in (276). By the equations (278) and (279), Substituting the above expression into the first term in (313) and that in turn into (289), we find that 3 (z − b) 2 f b (z) 1/2 [B 1 (z)θ 22 (u + (b))e s 3/2 g + (b) + B 2 (z)θ 11 (u + (b))e −s 3/2 g + (b) ] 2 .
Since the meromorphic in U b function we immediately compute the first integral by taking the residue (recall the negative direction of the integration), and recalling the expressions (109), (106) for β 0,b , u 0,b , we obtain Recall that we are interested in the average of T 3 (ω) over ω. As was noticed in a similar situation in [19], it is easier to do the averaging of the second term in (317) before computing its residue. Furthermore, to average the first term, we can use the integral (A. 19) in Lemma 26 from [19]: By the heat equation, the identity θ ′ 1 = πθ 2 θ 3 θ 4 , and the identities of Lemma 12, we have that Thus, we obtain On the other hand, since ∂ ∂τ log θ 3 (ω) = 1 4πi and we obtain, using the expression for dτ /db from Lemma 12, Thus, integrating (317) and using (322) and (320), we obtain dω[B 1 (z)θ 22 (u + (b))e s 3/2 g + (b) + B 2 (z)θ 11 (u + (b))e −s 3/2 g + (b) ] 2 . (323) We now turn to computation of the average of the term with B's in this expression. First note that the analytic in U b function where we used (108) and wrote ) recalling (174). In the last expansion Expanding B j (z) similarly to A j (z) above, and using the identity (135), we obtain for the analytic (by (310)) in U b function β(z)[B 1 (z)θ 22 (u + (b))e s 3/2 g + (b) + B 2 (z)θ 11 (u + (b))e −s 3/2 g Taking the square of this expression, we may write The advantage of this representation is that we see (taking into account (324)) that only the first 2 terms on the r.h.s. may give a nonzero contribution to the residue in (323), and the second term is not squared. We now evaluate the average Substituting this into (323) and calculating the residue by (327) and (324) we obtain 22 (u + (b))e s 3/2 g + (b) + B 2 (z)θ 11 (u + (b))e −s 3/2 g + (b) ] 2 = 1 16 π 2 u 2 0,b + 3u 1,b − 3f 1,b /2 = − 1 16 For the last equation here we used (106), (325). On the other hand, differentiating q(b) w.r.t. b and using (143), we obtain With this we may write the r.h.s. of (338) as Substituting this expression into (323) we obtain

5.4
Asymptotic form of the differential identity at a. Proof of Proposition 21.
In this section we prove that the differential identity is symmetric in a and b as shown in Proposition 21. We start from the differential identity at a, namely where the limit is taken from outside the lens and with Im z > 0. By the definitions of S(z) and R(z; s), we have that X −1 X ′ = e s 3/2 g(z)σ 3 P −1 a R −1 R ′ P a e −s 3/2 g(z)σ 3 + e s 3/2 g(z)σ 3 P −1 a P ′ a e −s 3/2 g(z)σ 3 − s 3/2 g ′ (z)σ 3 .