Sine-kernel determinant on two large intervals

We consider the probability of two large gaps (intervals without eigenvalues) in the bulk scaling limit of the Gaussian Unitary Ensemble of random matrices. We determine the multiplicative constant in the asymptotics. We also provide the full explicit asymptotics (up to decreasing terms) for the transition between one and two large gaps.


Introduction
Let K s be the (trace class) operator on L 2 (A), A ⊂ R, with kernel K s (x, y) = sin s(x−y) π(x−y) . Consider the Fredholm determinant P s (A) = det(I − K s ) A . (1) In this paper, we take A to be the union of two intervals, and complete the description of the asymptotics of P s (A) as s → ∞, including the transition when the intervals merge into one. The determinant (1) is the probability that the set s π A = { s π x : x ∈ A} contains no eigenvalues of the Gaussian Unitary Ensemble of random matrices in the bulk scaling limit where the average distance between eigenvalues is 1. (Thus, if A is a union of intervals, they are called gaps.) Similar statements hold in other contexts: the sine-process with kernel K s (x, y) is the simplest, and one of the most common and well-studied determinantal processes appearing in random matrix theory, random partitions, etc.
In the case where A is an interval, which we can assume without loss 1 to be (−1, 1), the asymptotics of the logarithm of (1) have the form: where c 0 = 1 12 log 2 + 3ζ ′ (−1).
Here ζ ′ (z) is the derivative of Riemann's zeta function. The leading term − s 2 2 was found by Dyson in 1962 in one of his fundamental papers on random matrix theory [9]. Dyson used Coulomb gas arguments. The terms − s 2 2 − 1 4 log s were computed by des Cloizeaux and Mehta [4] in 1973 who used the fact that eigenfunctions of K s are spheroidal functions. The constant (3), known as the Widom-Dyson constant, was identified by Dyson [10] in 1976 who used the inverse scattering techniques and the earlier work of Widom [20] on Toeplitz determinants. The works [9], [4], and [10] are not fully rigorous. The first rigorous confirmation 1 P s (A) is invariant under translations of A, and rescaling results only in the appearance of a prefactor of s. Figure 1: Cycles on the Riemann surface Σ of the main term, i.e. the fact that log P s ((−1, 1)) = − s 2 2 (1 + o(1)), was given by Widom [21] in 1994. The full asymptotic expansion (2), apart from the expression (3) for c 0 , was proved by Deift, Its, and Zhou in a landmark work [6] in 1997, where the multi-interval case was also addressed. The authors of [6] used Riemann-Hilbert techniques to determine asymptotics of the logarithmic derivative d ds log P s (A), where A is one (or a union of several) interval(s). The asymptotics for P s (A) were then obtained in [6] by integrating the logarithmic derivative with respect to s. The reason the expression for c 0 was not established in [6] is that there is no initial integration point s = s 0 where P s (A) would be known explicitly. In [16], the author was able to justify the value of c 0 in (3) by using a different differential identity for associated Toeplitz determinants and again the result of Widom [20]. An alternative proof of (3) was given in [8], which was based on another differential identity for Toeplitz determinants. In [8], the result of [20] was also rederived this way. Both [16] and [8] relied on Riemann-Hilbert techniques. Yet another proof of (3) was given by Ehrhardt [11] who used a very different approach of operator theory.
If A is a union of several intervals, it was shown by Widom in [22] that d ds log P s (A) = −C 1 s + C 2 (x) + o(1), s → ∞, where C 1 > 0 and C 2 (x) is a bounded oscillatory function. The constant C 1 can be computed explicitly, but C 2 (x) is an implicit solution of a Jacobi inversion problem. This result was extended and made more explicit by Deift, Its, and Zhou in [6]. We will now present the solution of [6] in the case when A is the union of two intervals, which is relevant for the present work. As above, we assume without loss that Let p(z) = (z 2 − 1)(z − v 1 )(z − v 2 ), and consider the two-sheeted Riemann surface Σ of the function p(z) 1/2 . On the first sheet p(z) 1/2 /z 2 → 1 as z → ∞, while on the second, p(z) 1/2 /z 2 → −1 as z → ∞. The sheets are glued at the cuts (−1, v 1 ), (v 2 , 1). Each point z ∈ C \ ((−1, v 1 ) ∪ (v 2 , 1)) (including infinity) has two images on Σ. The Riemann surface Σ is topologically a torus. Let the elliptic integrals I j = I j (v 1 , v 2 ) > 0, J j = J j (v 1 , v 2 ) > 0 be given by x j dx p(x) 1/2 , j = 0, 1, 2, (5) where the loops (cycles) A 1 , B 1 are shown in Figure 1. The loops A 0 , A 1 lie on the first sheet, and the loop B 1 passes from one to the other: the part of it denoted by a solid line is on the first sheet, the other is on the second.
It follows that which gives an explicit expression for q(z) in terms of elliptic integrals. Note that (7) implies that ψ(z) has no residue at infinity. More precisely, we obtain as z → ∞ on the first sheet As shown in [6], G 0 > 0. Denote the holomorphic differential Clearly, it is normalized: Let τ = where the integration v 2 v 1 ψ(x)dx is taken on the first sheet, and where the last equation for Ω follows by Riemann's period relations (Lemma 3.45 in [6] for n = 1). Recall the definition (A.1) in Appendix A of the third Jacobian θ-function θ 3 (z; τ ). Deift, Its, and Zhou found in [6] that log P s ((−1, v 1 ) ∪ (v 2 , 1)) = −s 2 G 0 + G 1 log s + log θ 3 (sΩ; τ ) + c 1 with G 0 as in (10), and τ , Ω as given in (13). Constants G 1 , c 1 are independent of s. The constant G 1 is written in [6] in terms of a limit of an integral of a combination of θ-functions. The constant c 1 remained undetermined (for the same reason as given above in the case of one interval).
The main result of the present paper is the expression for c 1 , which completes the description of the asymptotics (14). We also find that the original expression for G 1 in [6] can be simplified, and we obtain that G 1 = −1/2 (see Appendix B). We also determine this coefficient −1/2 of log s in a different way, as a direct result of our computation of (14) which also produced c 1 . We describe this computation in more detail below in the introduction.
Remark 5. The rate −5/4 can be somewhat decreased with an appropriate change of the error term.
Remark 6. Using the translational invariance of det(I − K s ), we see by the shift of variable Thus Theorem 4 provides the asymptotics for P s ((−1, v 1 ) ∪ (v 2 , 1)) in the case when |v 1 − v 2 | > s −5/4 . In recent work [12], we obtained the asymptotics of P s ((−1, v 1 ) ∪ (v 2 , 1)) = P s ((α, −ν) ∪ (ν, β)) in the case of two gaps merging into one, i.e. where v 1 , v 2 are scaled with s in such a way that |v 1 − v 2 | ≤ 1/(s log 2 s) while being bounded away from ±1. We also showed implicitly that the asymptotics we obtained in that case uniformly connect to those of fixed v 1 < v 2 . Theorem 4 provides an explicit matching: More precisely, we showed in [12] that 3 Theorem 7. (Splitting of the gap (−1, 1) where G is the Barnes G-function, and where κ j is the leading coefficient of the Legendre polynomial of degree j orthonormal on the interval [−2, 2], given by The rest of notation in (21) is from Theorem 4. As s → ∞, uniformly for ν ∈ (ν 1 , ν 0 ), where sν 0 log ν −1 0 → 0, Thus we see that the asymptotic regime of Theorem 4 overlaps with that of Theorem 7 (for example, ν = s −6/5 belongs to both regimes), and comparing (19) with (23) we see an explicit matching. Taken together, these theorems describe the asymptotics for two large gaps and one large gap (note that (21) reduces to (2) when ν → 0 sufficiently rapidly) as well as the transition between them.
Our strategy to prove Theorem 1 relies on connecting the asymptotics for fixed v 1 < v 2 with another double-scaling regime, namely the one where v 1 approaches −1, and v 2 approaches 1. In this regime the scaled gaps, s(−1, v 1 ), s(v 2 , 1), although still growing with s, become small in comparison with the separation between them, and we show that in that case P s ((−1, v 1 ) ∪ (v 2 , 1)) splits to the main orders into the product of P s (−1, v 1 ) and P s (v 2 , 1). The advantage is that for each of the separate gaps we can use an appropriately rescaled asymptotics (2) which contains the constant c 0 . More precisely, we prove in Section 2 by elementary arguments the following Remark 9. The rate of increase of t, t = 1 2 (log s) 1/4 , can be replaced with a slower rate of growth with s, and the statement will still hold (cf. Theorem 10 below). Now we describe the steps of the proof of Theorem 1. First, we obtain in Section 3 an identity (equation (40) of Lemma 13) for the derivative ∂ ∂v 2 log P s ((−1, v 1 ) ∪ (v 2 , 1)) in terms of a certain Riemann-Hilbert (RH) problem, the Φ-RH problem. The fact that we use a differential identity with respect to one of the edges (v 2 ) of the gaps is crucial in allowing us to determine the constant c 1 .
We then give in Section 4.4 the asymptotic solution of the Φ-RH problem as s → ∞ with v 1 , v 2 fixed. This problem is very similar to that solved in [6], and its solution involves the Jacobian θ-functions (we give a collection of various useful properties of θ-functions in the Appendix A below). In Section 4.5, we show that the solution of the Φ-RH problem can be extended to the double-scaling range where v 2 is allowed to approach 1 at such a rate that (1 − v 2 )s → ∞ (by symmetry, also v 1 is allowed to approach −1 so that (1 + v 1 )s → ∞). It is this extension which eventually provides a connection with Lemma 8.
In Section 5, we then substitute the solution into our differential identity (see (157), (163)). In Proposition 16, we characterize the main asymptotic terms (equation (164)) in the differential identity using averaging with respect to fast oscillations.
A large part of our work, Sections 7,8,9, is to bring the expression (164) to an explicit form. This relies, apart from the use of standard formulae, on (specific to our setting) identities for θ-functions obtained in Lemma 15 of Section 4.2. As a result, we obtain an explicit form (187) for the non-small part (164) of the right-hand side of the differential identity (40).
We then, by Proposition 16, integrate the resulting identity with respect to v 2 from the point when v 2 = −v 1 is close to 1 to a fixed v 2 = −v 1 , and then, with v 1 fixed, over v 2 , so that at one of the integration limits we can use the result of Lemma 8. This proves Theorem 1. Thus the part 2c 0 of the constant c 1 in (15) comes from Lemma 8, while the rest of c 1 comes from the integration.
As a byproduct of our proof we also obtain the following extension of the asymptotics (14).
Then the asymptotics (14) hold with the error term O max Note that the use of Toeplitz determinants in [16], [8] was essential to determine the constant c 0 in the asymptotics for one gap. In this paper, however, we use Lemma 8 which, in turn, relies on the already known constant c 0 . 6 2 Separation of gaps: proof of Lemma 8 For w > 2 let With t as in Lemma 8 and v = s/(2t), we have By (2) and translational invariance, as t → ∞, Therefore, upon setting u = 2t, w = v, we obtain Lemma 8 as a direct consequence of the following lemma we now prove.
Lemma 11. Let u, w > 2. There exist absolute constants C 3 , C 4 > 0 such that We start with Proposition 12. Let m ∈ {0, 1, . . . } and B be an m + 1 × m + 1 matrix satisfying |B jk | ≤ u for all j, k = 1, . . . , m + 1. Let X be a set of indices j, k such that |B jk | < 1/w for all (j, k) ∈ X and set for a sufficiently large absolute constant C 1 > 0.
In particular, B = B (m+1) . Expanding B and B (1) in the first row we have where B (0)(jk) is the m × m matrix obtained by removing the j'th row and the k'th column from B = B (0) . Similarly, for any ℓ = 1, 2, . . . , m, expanding in the ℓ + 1 row, we have Inequalities (29), (30) imply Hadamard's inequality yields and so for some C 1 > 0.
Proof of Lemma 11. Let If we set with x j , y k ∈ A (w) , then B, B satisfy the conditions of Proposition 12 for some X. By (27) and the definition of the Fredholm determinant, we have for sufficiently large absolute constants C j > 0 The reason for introducing K is that the corresponding Fredholm determinant splits into the product of the determinants over A Combining this with the estimate (36) proves the lemma.

Differential Identity
Consider the following Riemann-Hilbert problem for a 2 × 2 matrix valued function Φ(w). Let Γ Φ be the contour shown in Figure 2, where as usual the + side of the contour is on the left w.r.t. the direction shown by the arrow, and the − side is on the right.
(b) Φ has L 2 boundary values Φ + (w), Φ − (w) as the point w ∈ Γ Φ is approached nontangentially from the + side, − side, respectively. These values are related by the jump condition Φ (c) As w → ∞, Remarks 1) As usual, we write for brevity 2) By general theory, see, e.g., [5], if this problem has a solution Φ(w), then the solution is unique.
Although it is unclear a-priori that it has a solution, we will show this in Section 4.4 for large s, which is the case we need. The rest of this section will be devoted to 2 different proofs of the following Lemma 13. (Differential identity) The Fredholm determinant (1) satisfies: where Φ ′ (z) = d dz Φ(z) and Φ −1

First proof of Lemma 13
The proof of identities of type (40) using the theory of integrable operators is standard [14,6,2,3]. We give an outline. First, we write the kernel of the (integrable) operator K s in the form Note that 2 j=1 λ j (z)µ j (z) = 0. The resolvent of the operator K s , has the property [6, Lemma 2.8] that the kernel of R s is of the form and moreover, 2 j=1 Λ j (z)M j (z) = 0. The functions Λ(z) and M(z) for z ∈ A can be written as [6, Lemma 2.12] where m(z) is the 2 × 2 matrix valued function which solves the following RHP (this is the m-RHP of [6] up to a slight modification: λ 2 , µ 2 are replaced by −λ 2 , −µ 2 , respectively): RH problem for m This problem is reduced to a constant jump problem by the transformation Indeed so defined ψ(z) satisfies (b) ψ(z) has L 2 boundary values related by the condition ψ (c) ψ(z) = (I + O(z −1 )) e iszσ 3 as z → ∞.

Differential identity for Toeplitz determinants
For the second proof of Lemma 13, we will first represent the Fredholm determinant det(I − K s ) A in terms of a special Toeplitz determinant and then obtain (40) as a limit of the corresponding differential identity for Toeplitz determinants. This way of proving Lemma 13 has a potential advantage of future applications to computing probabilities in the Circular Unitary Ensemble of random matrix theory, and to the theory of orthogonal polynomials.
Let J = J 1 ∪ J 2 be the union of two disjoint arcs J 1 and J 2 on the unit circle C. We parametrize the endpoints of J 1 by a 1 = e iφ 1 , a 2 = e iφ 2 and the endpoints of J 2 by b = e iφ 0 ,b = e −iφ 0 , see Figure  3. Let f be the indicator function of the set J: Consider the n-dimensional Toeplitz determinant with symbol f on the unit circle C: where the integration is in the counterclockwise direction.
If the end-points of the arcs vary with n as follows, φ 0 = 2s/n and φ j = 2v j s/n for j = 1, 2, then it is easily verified that We will now obtain a differential identity for D n (f ), and in the next subsection, by taking n → ∞ and using (49), will prove Lemma 13.
Since f is nonnegative, it follows from the multiple integral representation for Toeplitz determinants that D j (f ) > 0 for all j = 1, 2, . . . . Set D 0 (f ) = 1. Define the polynomials ψ 0 = 1/ √ f 0 , ψ j , j = 1, 2, . . . by where the leading coefficient χ j is given by These polynomials are orthonormal on J: For a given n ≥ 1, define the matrix-valued function Y = Y (z) in terms of the orthogonal polynomials: The function Y is a unique solution to the following RH Problem: (a) Y : C \ J → C 2×2 is analytic; (c) Y (z) = (I + O(1/z))z nσ 3 as z → ∞.
This fact was initially noticed in [13] for orthogonal polynomials on the real line and extended to the case of orthogonal polynomials on the unit circle in [1]. As in [16,7], we will use the orthogonal polynomials to obtain a differential identity for log D n (f ) in terms of the solution to the RH problem for Y . Namely, we have where F is given by (b) Let a 2 = a 1 = e iφ 2 . Then Proof. From the definition of the orthogonal polynomials it is clear that The orthogonality conditions imply that, with z = e iθ , and similarly, By (56)-(58) we obtain: The Christoffel-Darboux formula for orthogonal polynomials on the unit circle (see, e.g., equation (2.8) in [7]) states that On the other hand, using the following identity (equation (2.4) in [7]) and (52), we easily verify that Substitution of (60), (62) into (59) gives Since by orthogonality we obtain and proposition 14 (a) follows from (63). Part (b) is proved similarly.
As we are eventually interested in the limit n → ∞, we first reduce the Y RH problem to an approximate problem for Φ which does not contain the parameter n, and the dependence on n is in the error of approximation. Let We open the lenses around J 1 and J 2 , see Figure 4. Denote the edges of the lenses inside the unit disc by Γ In S , the edges of the lenses outside the unit disc by Γ Out S , and let inside the lenses, for |z| < 1, inside the lenses, for |z| > 1.
Then S satisfies the following RH problem: (c) As z → ∞, We assume that the lenses around J 1 and the contour part C \ J are contained within the set |z − 1| < 1/2. The following function M will approximate S for |z − 1| > 1/2: For |z − 1| < 1/2, we construct the following function Q. Let so that w(e 2it s n ) = t for any t, and define where Φ is the solution of the Φ RH problem at the beginning of the section. Let We have using (39), (70), , ǫ > 0, uniformly on the edges of the lenses. Thus, by standard small norm analysis (see, e.g., [5]), uniformly for z ∈ C. We now express F (a 2 ) from Proposition 14 in terms of elements of Φ. Tracing back the transformations, we see that as z approaches a 2 from the inside of the unit circle and being outside the lens, Using this, we obtain Taking the limit z → a 2 = exp(iφ 2 ) = exp(i2v 2 s/n) along this trajectory, we obtain as s, n → ∞ and s/n → 0. Substituting this into (53), recalling (49), and noting that dv 2 /dφ 2 = n/(2s), proves (40). The symmetric case identity (41) follows from (55). Thus we finished the proof of Lemma 13. We now solve the RH problem for Φ, compute the asymptotics of the r.h.s. of (40), integrate it, and use Lemma 8 at one of the integration limits to obtain Theorem 1.

Solution of the RH problem for Φ
Recall the definition of ψ(z) in (6), and for z ∈ C \ (−1, v 1 ) ∪ (v 2 , 1) on the first sheet of the Riemann surface Σ, let We see by (7) that φ(z) is a well defined function, analytic on Since by (7) ψ(z) has zero residue at infinity, ψ(z) = 1 + O(1/z 2 ) as z → ∞, and we have Let then S satisfies the following RH problem.
(c) As z → ∞, We need the conditions Im φ(z) < 0, Im φ(z) > 0, to hold uniformly on Γ Φ,L , Γ Φ,U , respectively, away from some fixed ǫ neighborhoods of the end-points for the corresponding jumps to be exponentially close to the identity. Since (77) is uniform as |z| → ∞, the conditions hold for |z| > W for some sufficiently large but fixed , the conditions hold on the contour as stated assuming (and we do this) that the angle between the parts of Γ Φ,L , Γ Φ,U emanating from ±1 and the real axis was chosen to be sufficiently small and the lens around (v 1 , v 2 ) was sufficiently narrow. Therefore as s → ∞, for some constant c > 0, uniformly on Γ Φ,L , Γ Φ,U away from fixed ǫ-neighborhoods of ±1, v 1 , v 2 .

Outside parametrix and θ-functions
Consider the following RH problem for the 2×2-matrix valued function N (z; ω) with a real parameter ω, which will give an approximate solution to the Φ RH problem away from the edge points ±1, v 1 , v 2 , when ω = sΩ. Later on we also construct approximate solutions (local parametrices) on a neighborhood of each edge point, and match them to the leading order with N (z; ω) on the boundaries of the neighborhoods. 1), N has L 2 boundary values related by the jump conditions: A more general problem with jumps on m intervals was solved in [6] in terms of multidimensional θ-functions. We now present the solution in our case of 2 intervals: also with branch cuts on (−1, v 1 ) ∪ (v 2 , 1), such that γ(z) → 1 as z → ∞ on the first sheet of the Riemann surface Σ.
Recall the definition of the holomorphic differential (11). Let u be the following analytic function with integration taken on the first sheet. Note that, mod Z, The function u(z) extends to the Riemann surface Σ and is then called the Abel map. It maps the Riemann surface onto the torus where θ-functions are defined. A simple calculation (see [6]) shows that the function γ(z) − γ(z) −1 has a single zero on (v 1 , v 2 ) on the first sheet, denote it byẑ, and no zeros on the second sheet. We havê Similarly, the function γ(z) + γ(z) −1 has no zeros on the first sheet and one zero on the second. Let with integration taken on the first sheet. Consider the third Jacobian θ-function has no other zeros on the Riemann surface. The function θ(u(z) + d) = 0 has only one zero on the Riemann surface located on the second sheet which coincides with the only zero of γ(z) + γ(z) −1 .
By an argument in [6] we have for some integer m. Consider the integral of ω along the closed contour composed of a large interval along the real axis and a semicircle in the upper half-plane. Then using (12) and the definition of τ in (13) we obtain in the case v 1 = −v 2 that u(∞) + d = 0 mod Z with u(z) considered on the first sheet. Therefore also in the general case of v 1 , v 2 , by continuity, The solution to the RH problem for N is given by with z on the first sheet. To see that N solves the RH problem for N , one makes several observations. First note that γ(z) is analytic on Secondly, as follows from (A.2) and the relations the matrix m has the jumps: The singularities of m cancel with the zeros of γ ± γ −1 . Furthermore, as z → ∞.

Identities for θ-functions
Our proof of Theorem 1 will use several identities satisfied by θ-functions. We present these identities now. Standard definitions and properties of theta-functions that we need are summarized in Appendix A.
The expansion of η 1 (z) as z → v 2 (using (150) below) shows that and since η 1 (v 2 ) = 1, we obtain Part (a) of the proposition. Now consider . (101) By the fact that is an elliptic function of ξ, and by (90) and (91), it follows that η 2 (z) is a meromorphic function, and again by cancellation of the poles from θ 3 (±u(z) + d) by the zeros of γ(z) ± γ(z) −1 , it follows that η 2 (z) in fact is entire. As z → ∞, η 2 (z) → 0 by (88) since θ 1 (0) = 0, and thus, η 2 (z) ≡ 0 by Liouville's theorem. We see from the expansion of Since this limit is zero, we obtain that which gives Part (b) of the proposition.
To prove part (c), we consider the coefficient of the first power z − v 2 in the expansion of By substituting the identity for g ′ (d) from Part (b) of the proposition into the right hand side of (104) and setting the resulting coefficient of z − v 2 equal to zero, we obtain Part (c). Finally, to prove Part (d), we consider By the same arguments as for η 1 and η 2 (and in addition by the fact that R + = −R − on A), it follows that η 3 is entire. By recalling the definition of u in (84), by (88), and by the definition of γ in (83), we obtain so that η 3 (z) is identically equal to this constant. Now consider the asymptotics of η 2 (z) as z → ∞.
We have from which we conclude that By substituting this into (106), we obtain for all z ∈ C. On the other hand, for z 0 ∈ {−1, v 1 , v 2 , 1}, from (105) by (85) and ellipticity, Equating this to (109) we obtain Part (d). To show Parts (e), (f), (g), we consider the function (as usual, theta functions written without argument stand for their values with argument zero) As before, we see that this function is identically constant. By evaluating at infinity, it is equal to Equating this constant to (111) we obtain the identity for all z: Evaluating it at z = 1 (recall from (85) that u(1) = 1/2 mod Z and recall the definition of θ j (z) from Appendix A), and using the identity θ ′ 1 = πθ 2 θ 3 θ 4 , we obtain Part (e). We similarly obtain Part (f) by evaluating (113) at z = v 1 . Finally, we obtain Part (g) by making use of the identity

Local parametrices
Our goal in this section is to construct a function P on a neighborhood of each point of the set , with the same jump conditions as S on these neighborhoods, and with an asymptotic behavior matching that of N to the main order on the boundaries of these neighborhoods. The first step is to recall the following model RH problem from [17] with an explicit solution in terms of Bessel functions.
(b) Ψ satisfies the jump conditions: For | arg ζ| < 2π/3, we have where I 0 and K 0 are Bessel functions, . For definitions and properties of Bessel functions see, e.g, [15]. Here the principal branch of ζ 1/2 with the cut along the negative real axis is chosen. For the explicit expression of the solution in the sector | arg ζ| > 2/3, see [17].
We have the following useful asymptotics as z → 0 for I 0 : We denote by U (p) fixed open nonintersecting balls containing (116) As z → p, we have the expansion Note that ζ (p) (z) is a conformal mapping of U (p) onto a neighborhood of zero. Observe also that ζ 0 > 0 for p = v 2 , −1, and ζ 0 < 0 for p = v 1 , 1, and so the contours in U (p) are mapped from the z-plane to the ζ-plane accordingly. We choose the exact form of the contours in the z-plane so that their images are direct lines. Keeping in mind our conventions for the root branches, we obtain By (7), (75) and the definition of Ω in (13), Let (120) where we have suppressed the superscript in ζ = ζ (p) , and the branch cut for ζ 1/4 is the same one as for ζ 1/2 . Using the jump conditions, it is straightforward to verify that E(z) has no jumps in U (p) , and since its singularity at p is removable, E(z) is analytic in the neighborhood U (p) , p = −1, v 2 .
Furthermore, it is easy to verify that P (z) satisfies the same jump conditions as S(z) in U (p) , Finally, using the condition (c) in the Ψ-RHP and (118), we obtain for w ∈ ∂U (p) uniformly on the boundary as s → ∞, where where ∓ is taken to be − on U (p) ∩ C + , and + on U (p) ∩ C − . Note that ∆ 1 (z) is meromorphic in U (p) , p = −1, v 2 , with the first order pole at z = p.
Similarly, for p = v 1 , 1, we define the local parametrix on U (p) by Here E(z) is analytic on U (p) , P (z) has the same jumps as S(z) in U (p) , p = v 1 , 1, and the same condition (122) holds with where ∓ is taken to be − on U (p) ∩ C + , and + on U (p) ∩ C − . As at v 2 , −1, ∆ 1 (z) in (125) is meromorphic in U (p) , p = v 1 , 1, with the first order pole at z = p.
Then R(z) is analytic for z ∈ C \ Γ R , where Γ R is as in Figure 5. We have By (81) and (122), it follows that as s → ∞, uniformly for z ∈ Γ R , and by the definition of S and N , we have as z → ∞. By standard small norm analysis, it follows that there is a solution to the RH problem for R for s sufficiently large, and that uniformly on compact sets as s → ∞. In particular, the existence of R also implies the existence of a solution to the RH problem for S for sufficiently large s, which also implies the existence of a solution to the RH problem for Φ. As usual, we expand R in the powers of the small parameter, 1/s in our case, to write where R 1 solves the following RH problem. R 1 (z) is analytic outside the clockwise oriented boundaries ∂U (p) of the neighborhoods U (p) , and R 1 (z) → 0 as z → ∞. The solution to this problem is given by where the integrals are taken with clockwise orientation. Taking (131) (one can obtain further terms in that expansion in a standard way) and tracing back the transformations R → S → Φ, we obtain an asymptotic solution of the Φ-RH problem.

Extension of the solution to the regimes
In our solution of the previous section, the end-points −1 < v 1 < v 2 < 1 were fixed. In this section, we show that the solution can be extended to the regime where v 2 not only can be fixed but can also approach 1 (and v 1 approach −1) sufficiently slowly as s → ∞. This will be needed for the proof of Theorem 1 below. More precisely, we fix ǫ > 0 and assume We let U (v 2 ) and U (1) have radius equal to c(1 − v 2 ), and similarly U (v 1 ) and U (−1) have radius equal to c(1 + v 1 ), for some fixed and sufficiently small c > 0. Note that the neighborhoods can now contract with growing s. As , for j = 0, 1, 2, and computing an additional term in the expansion we find by (9) that uniformly in the regime (133). By (8), By (134) and (116), uniformly in the regime (133) and also uniformly for z ∈ ∂U (p) , p ∈ T = {−1, v 1 , v 2 , 1}. Next we will show that N and N −1 are uniformly bounded on ∂U (p) for p ∈ T . As v 2 → 1 (under conditions (133)), we see from (83) that both γ(z) and γ −1 (z) are uniformly bounded also on ∂U (p) for p ∈ T .
We now consider θ-functions, and start with τ . For J 0 , we have as v 2 → 1, and since for any parameter t, it follows that Thus, since in the regime (133), so that we have −iτ → +∞. As −iτ → +∞, θ 3 θ 3 (ω) → 1 for any ω ∈ R. We also observe that as −iτ → +∞, the fraction is bounded uniformly under conditions (133) and over all ω ∈ [0, 1), for ξ bounded away from the zero of the θ-function 1+τ 2 modulo the lattice, and the same holds for derivatives of (140) with respect to ξ, ω, and τ . We now show that ξ = u(z) ± d remains bounded away from 1+τ 2 modulo the lattice for z ∈ ∂U (p) , p ∈ T .
Combining the statements about boundedness of m and γ and γ −1 , it follows that N (z) and N (z) −1 are uniformly bounded for z ∈ U (p) , p ∈ T , and thus by (135), the jump matrix J R (z) for R(z) on U (p) , p ∈ T , has the form as s → ∞, uniformly under conditions (133) and also uniformly for z ∈ U (p) , p ∈ T . The analysis of J R (z) on the rest of the jump contour is similar, and we obtain uniformly for (133) and uniformly on this part of the contour Thus we have a small norm problem for R, and as in the previous section we now obtain uniformly on compact sets under conditions (133). Therefore the solution of the Φ-Riemann-Hilbert problem for fixed v 1 , v 2 extends to the regime (133). Note, however, that the error terms are different from those in the previous section.

Preliminary asymptotic formula for the determinant
For ν = z − v 2 in a neighborhood of 0, we write the expansions of ζ(z), where −π < arg ν < π, and the branch cut is on (−∞, 0]. Similarly, we expand γ(z), m(z), and u(z), but with branches chosen such that 0 < arg ν < 2π, and the branch cut on [0, +∞). Here m jk are the matrix elements of m. Thus, arg ν in (149) and (150) are the same for Im ν > 0, but are different for Im ν < 0.
Using the definition of m and the jump conditions (92), we easily obtain the relations: We also find where ± means + for j = 1 and − for j = 2. Let By the definition of S in (78), R in (126) and X in (120), and the fact that φ(v 2 ) = 0, (154) With the notation of (149) and (150) where we are uninterested in the entries * , and ω = sΩ in m jj,k . We will now make use of the first identity (93) in Lemma 15, which, by the definitions of m jk,ℓ , we can write in the form Using this relation, we obtain by (154) and (155) for the r.h.s. of the differential identity of Lemma 13, where and we take ω = sΩ in m jj,k . Now the more explicit asymptotic expression of (157) is different (in the error term) for fixed v 1 , v 2 (Section 4.4) and for the double scaling regime of Section 4.5.

Remark 17.
In the proof, considering the effects of averaging w.r.t. ω = sΩ, we will show that (164) gives the main contribution, and the error terms are as presented.
Remark 18. Part (a) allows us to integrate over symmetric intervals from the position of 2 small ones at 1 and −1 (where Lemma 8 holds) to general symmetric intervals with a fixed 0 < V < 1. Part (b) allows then to move the V 2 edge to an arbitrary fixed position V 1 ≡ −V < V 2 < 1. Note that the condition −1 < V 1 < 0 here is not a loss of generality for det(I − K s ) A , since we can use the symmetry x → −x of the determinant. Part (c) allows us to to integrate to reach a scaling limit where V 2 = V 2 (s) can approach 1 provided s(1 − V 2 ) → ∞ and V 1 is fixed.
Finally, choose a V 1 (s) = −V 2 (s) such that 2t = (1 + V 1 )s → ∞ (in this case, Lemma 8 still holds by Remark 9), and then, if needed, move V 2 closer to 1 using Part (c). Then, if needed, one can use the symmetry x → −x, to reach an arbitrary situation with ( Proof. We first prove Part (b) of the proposition, then Part (c), and finally Part (a). By (159) and the differential identity (40), all we need to do for the proof of Part (b) is to show that, with V 2 = −V 1 , . This function is analytic in both ω and v 2 (v 2 is bounded away from v 1 and 1). Let f j denote its Fourier coefficients with respect to ω, so that For j = 0, it follows by integration by parts that In Proposition 23 (b) below we give an explicit formula for ∂ ∂v 2 Ω(v 2 , v 1 ), and in particular it is a strictly positive differentiable function bounded away from zero when v 2 is bounded away from v 1 and 1. Thus We now prove Part (c) of the proposition. Substituting (134) into the expression (215) for ∂Ω ∂v 2 in Proposition 23 below, and also using (209), we obtain and, in particular, ∂Ω ∂v 2 remains bounded away from 0. We now show that as v 2 → 1, for j = 0, uniformly under conditions (133), which proves Part (c) of the proposition by (163) and arguments similar to those we used in the proof of Part (b). Since and similarly for ∂ ∂v 2 f j (v 2 , v 1 ), it suffices to show that as v 2 → 1. It follows by the definition of W in (160), (146), (162), and the arguments of the previous section that as v 2 → 1 under conditions (133). We recall that To obtain the second one, we observe first that by (139), and by (144), as v 2 → 1, uniformly for z ∈ ∂U (p) , p ∈ T . It follows by (141) and (143) as v 2 → 1, uniformly for z ∈ U (p) , p ∈ T . Furthermore, by the definition (83), By (134) and (116), The above bounds taken together imply It follows by the definition of W in (160), (162), and boundedness of m jk that as v 2 → 1, uniformly under conditions (133). Since ∂ ∂ω ∂m ij (z) , which proves the second bound in (173), completing the proof of Part (c) of the proposition.
To show Part (a), we let v 2 = −v 1 = v, and take the limit s → ∞ such that ǫ < v < 1 − M s for some ǫ > 0 and a sufficiently large M. By (41), We observe that (163) is valid also for v 2 = −v 1 = v, and all that remains to finish the proof of Part (a) of the proposition is to consider the Fourier coefficients of f . In place of (168), we have By above arguments, it suffices to show that the right hand side of (183) is of order To do this we need the first bound in (173), which holds also for v 2 = −v 1 = v, and additionally we need to prove that as v → 1, and that d dv Ω(v, −v) remains bounded away from 0. Note that, using contour integration, and therefore The last derivative is thus bounded away from 0 by (170). In order to prove (184), we simply observe that the bounds obtained in (176)-(179) also hold for the derivatives with respect to v instead of v 2 , which yields , which proves (184) and thus Part (a) of the proposition.

Proof of Theorems 1 and 10
In the next 3 sections, we show that (164) in Proposition 16 can be written as where log |q(y)|. (188) We now use (187) to prove Theorems 1 and 10. First, we show that with V 2 fixed, and in all asymptotic regimes of Proposition 16, uniformly in integration regimes of Proposition 16, and so this part only contributes to the error term.
Using the differential equation (A.10) and (175), we write Also since by (139), we similarly obtain The estimates (190) Thus, applying Part (a) of Proposition 16 and using Lemma 8, we obtain To finish the proof of Theorem 1 in the symmetric case, we need to estimate G s; −1 + 2t s , 1 − 2t s . Using formulae (A.37), (A.36), we obtain in our case and so the term with s 2 in G s; −1 + 2t s , 1 − 2t The term log θ gives a contribution only to the error term, indeed, since by (A.36) we have that Finally, in this case and so Substituting (194), (195), (196), (198) into the expression (188) for G s; −1 + 2t s , 1 − 2t s , and that, in turn, into (193), we obtain asymptotics (14) with an error term o(1) and with G 1 and c 1 as in (15) in the case −v 1 = v 2 = V > 0. We then extend it to the general case of fixed −1 < v 1 < v 2 < 1 by now a straightforward application of Part (b) of Proposition 16. (In fact, for v 1 < 0, but the general case follows by a symmetry argument: see Remark 18.) Now since by [6], (14) (with the error term O(s −1 )) holds for some constants G 1 , c 1 , these must be equal to those in (15). This completes the proof of Theorem 1.
Given Theorem 1, we immediately obtain Theorem 10 by applying Part (c) of Proposition 16 and a symmetry argument as discussed in Remark 18. Thus, all that remains now is to verify (187). In Section 7 we consider the leading order term in (164), in Section 8 we consider the term involving (γ 2 0 Γ 2 + Γ 1 ), which yields the derivative of log θ(sΩ), and in Section 9 we consider the term with 1 0 W (ω)dω, which yields the constant. Thus we will prove the following 3 lemmata, which taken together imply (187).

Lemma 19.
Note that the r.h.s. here equals the partial derivative s ∂Ω where W (ω) is given in (160).

The leading order term. Proof of Lemma 19
Recall from (5) the notation for I j , J j , j = 0, 1, 2. We will calculate the derivatives ∂ ∂v 2 I j , j = 0, 1, 2 in terms of the integrals themselves. The crucial identity here is (207) below.
First, we have Therefore, and similarly, The last 2 equations imply From here and (203), We will now express the derivative ∂I 0 /∂v 2 in terms of I j 's. To this end, observe that Using this equation and (205) we have This and (206) imply By the formulas for x 1 and x 2 in (8) and (9), and the formula for ζ 0 in (149), we finish the proof of Lemma 19.

Remark 22.
We also observe for future reference that the arguments may be copied line for line and applied to the integrals (by instead considering an integral over a closed loop containing (v 1 , v 2 ) and different branches of the roots), and we obtain the analogues to (209) and (206):

The fluctuations. Proof of Lemma 20
We write the first subleading term in (164) in the form, using (149), (150) for ζ 0 , u 0 , , Our goal in this section is to prove the following proposition, of which Lemma 20 is an immediate corollary.
Proposition 23. There hold the identities: (c) Proof. To show Part (a) note that in the notation of (5) and therefore, using (209), (211), we have which gives Part (a) of the proposition by Riemann's bilinear relations (A.30). Part (b) follows from (13) and (209) by using (8), (9): We will now show Part (c). Substituting the definitions of m jj,k and Γ j into T 1 in (213), and using the identity (94) of Lemma 15, we write T 1 in the form We now show that T 1 (ω) has the same behavior as 2θ ′ (ω)/θ(ω) under the shift ω → ω + τ , and therefore their difference is an elliptic function. We obtain using (A.5) It is easily seen that f (ω) = f (ω + τ ) = f (ω + 1), so that f is elliptic. Furthermore, at the zero (1 + τ )/2 of θ(ω), by (A.6), and thus the expression in the square brackets in (220) vanishes as ω → (1 + τ )/2. So the pole of f at (1 + τ )/2 cannot have the order larger then 1. Thus f is an elliptic function with at most single first-order pole modulo the lattice, which means f is a constant. At ω = 0, by (94) of Lemma 15. Thus This immediately implies that the function is elliptic. To analyze its behavior at the pole, it is convenient to write T 1 in terms of θ 1 by (A.6), (A.7) with ν = ω − 1+τ 2 : It is obvious from this representation that the expression in the square brackets vanishes at ν = 0, and therefore the order of the pole of T 1 at ν = 0 is no larger than 1. Since the same is true for The value of this constant is easy to obtain by setting ω = 0: since both T 1 (0) = 0 (see (219)) and θ ′ (0) = 0, this value is 0, which proves Part (c).

The constant. Proof of Lemma 21
Recalling (160), we write the term with W in (164) where with the integrals traversed clockwise.
In this section we show (in subsection 9.1) that and (in subsection 9.2) that Substituting the last 2 equations into (222), we prove Lemma 21.

Evaluation of T 2
Our goal in this section is to obtain (224). We first compute T 2 (ω). By the definition of N in (89) and by (85) (Note that θ j (d) = 0, j = 1, 2, 3, 4, by the argument following (87). Moreover, θ 3 (ω) = 0 for ω ∈ R.) Thus, by (119), (123), (125), and the definition of m jj,0 in (152), a straightforward calculation yields where the upper sign is taken if Im z > 0, the lower if Im z < 0, and F j is given by To compute the residue of (227) at −1, we need to analyze ± γ(z) 2 √ ζ(z) at −1. It is meromorphic, and we need to determine the sign of its residue (the absolute value follows straightforwardly from the expansions of γ(z) and ζ(z)). Let x ∈ U (−1) , with x = −1 + ǫ, ǫ > 0, and x lying on the positive side of the cut. For such x, γ(x) 2 = i |γ(x) 2 | by (83), and by the expansion (117), and in particular the fact that ζ 0 is positive, we have that ζ(x) is positive.

Evaluation of T 3
Now consider T 3 . Our aim in this section is to prove (225). We write N in (89) in the form where ± means + for j = 1 and − for j = 2. Using the jump conditions (90), (91), we observe that (z − v 2 ) 1/4 A(z) and (z − v 2 ) −1/4 B(z) are analytic on U (v 2 ) . Since ∆ 1 (z) in (123) for p = v 2 is meromorphic on U (v 2 ) , all odd powers of roots (z − v 2 ) 1/2 in the expansion of (123) cancel, and it follows that for z ∈ U (v 2 ) and Im z > 0, Therefore Expanding A 1 (z) and A 2 (z) as z → v 2 , we obtain using (149), (150), and (151), with T 1 (ω) as defined in (213). By (216) in Proposition 23, this equals So that by the definition of T 3 in (223), we obtain computing the residue for the first term, where the integration is in the negative direction around v 2 , and where √ ζ and B 1 , B 2 are understood to be the analytic continuation from Im z > 0.
We now write the average where To compare with Lemma 20, we will now single out a contribution from Using the differential equation (A.10) and the fact that we can write Since, by (214), πu 2 0 = ∂|τ | ∂v 2 , we can rewrite (253) in the form Now by (A.19), Using the identity θ ′ 1 = πθ 2 θ 3 θ 4 , and the identities (97)-(99) of Lemma 15, we write here so that we can rewrite (257) in the form It remains to evaluate Q defined in (254). To simplify the computations, we first do the averaging over ω and only then compute the residue in this case.
As above for A(z), we expand B(z) to obtain and therefore as z → v 2 . Thus, upon changing the order of integration, where Since q(−ω) = q(1 − ω), we have that Applying (A.20) to evaluate 1 0 q(ω)dω, we obtain: Note that (A.13) gives for the derivative of f (z) Using this, we have, in particular, as z → v 2 , i.e. u → 0, Expanding also the other terms in (266), and also expanding u by (150), we obtain that, as z → v 2 , By applying (103) and (95) of Proposition 15, we simplify the combinations of the H j as follows: , which allows us to write (270) in the form Substituting this into (262) and calculating the residue, we obtain For the coefficients ζ 1 , u 1 in expansions (149) and (150) we easily obtain: so that On the other hand, by (244) and (247), and by (209), (9), These equations imply Substituting this into (274) for Q, and that, in turn, into (259), we obtain (225).
10 Slow merging of gaps. Proof of Theorem 4 10.1 Solution of the Φ-RH problem as v 2 − v 1 → 0.
We consider the asymptotics of the Φ-RH problem in the double-scaling regime where ν = v 2 −v 1 2 can approach zero with s → ∞ at a rate such that 2ν > 1 s 2−ε , for any fixed ε > 0. Let We need to evaluate the integrals I j in the limit ν → 0. To do this (and to make a comparison with [12] easier), we first change integration variable x = t + v 1 +v 2 2 , which maps (v 2 , 1) to (ν, β); we then split this interval into (ν, √ ν) ∪ [ √ ν, β) and use a change of variable y = t/ √ ν for integration over the first one. We then obtain: 5 Hence, by (9), Let the neighborhoods U (v 1 ) , U (v 2 ) have radius ν/3; they will be therefore contracting as ν → 0. We now evaluate the jumps J S (z) of S on the edges of the lenses Γ Φ,L ∪ Γ Φ,U . Recall from (81) that these jumps were exponentially close to the identity, in the case where v 1 and v 2 were fixed. For z ∈ Γ Φ,L ∪ Γ Φ,U and z bounded away from the points v 1 , v 2 , it is clear that the jumps are still exponentially close to the identity so that (81) holds, and we consider the case where z → v 1 +v 2 2 along the edges of the lenses. We substitute (282) into the definition of φ in (75) and obtain (taking as ν → 0, and z → v 1 +v 2 2 . Here ′ + ′ sign is taken on Γ Φ,U , and ′ − ′ sign is taken on Γ Φ,L , and thus Im φ(z) < 0, Im φ(z) > 0 on Γ Φ,L and Γ Φ,U respectively. Worsening somewhat the error term, we have that as s → ∞, uniformly for 2ν > s −2+ε and z ∈ Γ Φ,L ∪ Γ Φ,U . Next we consider the jumps of R on the boundary ∂U (p) for p ∈ T = {−1, v 1 , v 2 , 1}. Estimating φ(z) as above but now in the definition of ζ in (116), we obtain that as s → ∞, uniformly for 2ν > s −2+ε , O 1 s , uniformly on ∂U (1) and ∂U (−1) .
To estimate ∆(z), we need to consider N . We first observe that by the definition (83), γ(z), γ(z) −1 = O(1) uniformly on ∂U (p) for p ∈ T as ν → 0 . Using (281) and a simpler expansion for J 0 , we obtain and define By the inversion formula (A.11) for θ-functions, where We now show, in (291) below, that ∆(z), which enters the jump matrix for R, may be too large for certain parameter sets, which makes it necessary to modify the solution of the RH problem. First, a simple analysis of (87) shows that d → −1/2 as ν → 0. On the boundary of U (v 1 ) , U (v 2 ) , we have |u(z)| → 0, uniformly in z. Therefore, using the boundedness of γ, γ −1 on ∂U (p) for p ∈ T , and applying (288), we have for the 11 element of N on ∂U (v 1 ) if ω > 0 (and thus u(z) + d + ω = u(z) + d + ω for ν sufficiently small): for some constants C, C 1 , C 2 > 0. Similarly, we analyze the behavior of N 11 for ω < 0, the behavior of other matrix elements of N on ∂U (v 1 ) , as well as the behavior of N on ∂U (v 2 ) and ∂U (±1) . We find that the estimate (290) is the worst (note that, in fact, the estimates for N on ∂U (±1) are much better), and thus recalling (285), we have as s → ∞ and ν → 0, for z ∈ ∂U (vp ) . Thus if, for example, ν = 1 s and | sΩ | = 1/2 (which is a case we need to deal with since the splitting of the gap regime described in [12] breaks down in this limit), we cannot say that ∆ is small, and so the corresponding jump of R is not guaranteed to be close to the identity, and so we cannot claim solvability of the R-RH problem. However, it was shown in [12] for the case of the RH problem of [6] that we can modify the solution to ensure solvability for the range 2ν > s −5/4 . We now provide more details of that construction in the present case, and apply it for all values of sΩ . Let where k = ±1 is chosen such that −1/2 < t ≤ 1/2. Consider the following function: where the constant d ′ will be fixed later on, and we now take with branch cuts on (−1, v 1 ) ∪ (v 2 , 1), and positive as z → ∞ on the first sheet of the Riemann surface Σ. We have It is easy to verify that N (z) satisfies the same jump conditions as N : Furthermore, one verifies that δ(z) − δ(z) −1 has two zeros at z + , z − located on the first sheet and such that δ(z + ) = δ(z − ) = 1 and Set d ′ = u(z + ) + 1/2 + τ /2, then it follows by the properties of the Abel map u(z) (84) that θ(u(z) − d ′ ) has a single zero at z + , and θ(u(z) + d ′ ) has no zeros on the first sheet C \ A. Thus N (z − ) = I, and since det N extends to an entire function, det N (z) = 1 for z ∈ C. Considering the zeros and poles of the meromorphic function δ −2 − 1 on Σ, and using the Abel theorem, we have Using the change of integration variable x = t + v 1 +v 2 2 as above, we obtain (from now on always on the first sheet, so modulo Z) as ν → 0, whereǫ is real, satisfyingǫ → 0 as ν → 0. Similarly, Therefore by the definition of d ′ , and u(z − ) + d ′ = 1/2.
which is the asymptotics of m(z) 11 , m(z) 12 , and finally which is the asymptotics of m(z) 21 , m(z) 22 . Similarly, we analyze the case of −1/2 < t ≤ 0. Collecting the results together, we obtain uniformly on the closure of U (v 1 ) ∪ U (v 2 ) . By similar arguments, we obtain the same estimate also on the closure of U (1) ∪ U (−1) (in this case, |u(z) + 1/2| ≤ ǫ ′ , ǫ ′ > 0.) On the other hand, the definition of δ gives uniformly for z ∈ ∂U (p) as ν → 0, for p ∈ T = {−1, v 1 , v 2 , 1}. Thus, as ν → 0, uniformly on ∂U (p) for p ∈ T . Since the solution to the RH problem for N is unique, we have Define the new local parametrices by and let Then R(z) → 1 as z → ∞; and R(z) has jumps on Σ R , see Figure 5. By (121) and the expansion of ζ in (285), the jumps of R(z) on ∂U (p) have the form uniformly for z ∈ U (p) as s → ∞ for 2ν > s −2+ε . For the proof of Lemma 24 below, we will also require the finer estimate where where ∓ means + for Im z < 0 and − for Im z > 0. By (284), the jumps of R(z) on the rest of the contour are estimated as follows (we descrease c > 0 somewhat) as s → ∞, uniformly for 2ν > s −2+ε and for z ∈ Γ R,L ∪ Γ R,U . Thus R satisfies a small-norm problem and therefore has a solution for s sufficiently large and 2ν > s −2+ε , and as s → ∞, uniformly for 2ν > s −2+ε , and uniformly for z ∈ C \ Σ R . Since the RH problem for R has a unique solution, the RH problem for S (and hence for Φ) has a unique solution obtained by tracing back the transformations.
Proof. In this proof, ǫ stands for a sufficiently small positive constant whose value may vary from line to line.
In the previous section, we obtained the asymptotic solution of the S-RH problem in the regime s → ∞, 2ν > s −2+ǫ . By (126), R is also well defined in this regime, and thus (157) holds. We now aim to prove the analogue of (159), namely as s → ∞, uniformly for 2ν > s −5/4 , with the same notation as in (157), (159).
To finish the proof of the lemma we need to estimate the error of replacing W with its average value. From the definition (326) and the estimates above, we deduce By (281), as ν → 0.

Proof of Theorem 4
By (187) and Lemma 24, we see that to show that the expansion (14) holds in the asymptotic regime of Theorem 4 (with the error term O(s −1/9 )) it remains to prove that Since by (214), (281), and by (288) Also since by (209), we similarly obtain and therefore the term with y = v 2 in (B.6) is Now note (recall (8)) that so that all the other terms in the sum in (B.6) are also equal 2. Therefore