Critical edge behavior in unitary random matrix ensembles and the thirty
  fourth Painleve transcendent
Critical edge behavior in unitary random matrix
ensembles and the thirty fourth Painlevé transcendent
A.R. Its
Department of Mathematical Sciences
Indiana University – Purdue University Indianapolis
Indianapolis IN 46202-3216, U.S.A.
itsa@math.iupui.edu
A.B.J. Kuijlaars and J. Östensson
Department of Mathematics
Katholieke Universiteit Leuven
Celestijnenlaan 200B
3001 Leuven, Belgium
arno.kuijlaars@wis.kuleuven.be
ostensson@wis.kuleuven.be
June 3, 2018
Abstract
We describe a new universality class for unitary invariant random matrix ensem-
bles. It arises in the double scaling limit of ensembles of random n × n Hermitian
matrices Z−1n,N |detM |2αe−N Tr V (M)dM with α > −1/2, where the factor |detM |2α
induces critical eigenvalue behavior near the origin. Under the assumption that the
limiting mean eigenvalue density associated with V is regular, and that the origin
is a right endpoint of its support, we compute the limiting eigenvalue correlation
kernel in the double scaling limit as n,N → ∞ such that n2/3(n/N − 1) = O(1).
We use the Deift-Zhou steepest descent method for the Riemann-Hilbert problem
for polynomials on the line orthogonal with respect to the weight |x|2αe−NV (x).
Our main attention is on the construction of a local parametrix near the origin by
means of the ψ-functions associated with a distinguished solution of the Painlevé
XXXIV equation. This solution is related to a particular solution of the Painlevé
II equation, which however is different from the usual Hastings-McLeod solution.
2000 Mathematics Subject Classification: 15A52, 33E17, 34M55
http://arxiv.org/abs/0704.1972v1
2 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
1 Introduction and statement of results
1.1 Unitary random matrix models
For n ∈ N, N > 0, and α > −1/2, we consider the unitary random matrix ensemble
Z−1n,N | detM |2αe−N TrV (M) dM, (1.1)
on the space M(n) of n× n Hermitian matrices M , where V is real analytic and satisfies
V (x)
log(x2 + 1)
= +∞. (1.2)
This is a unitary random matrix ensemble in the sense that it is invariant under conju-
gation, M 7→ UMU−1, by unitary matrices U . As is well-known [11, 38], it induces the
following probability density on the n eigenvalues x1, . . . , xn of M
P (n,N)(x1, . . . , xn) = Ẑ
|xj|2αe−NV (xj)
|xi − xj |2. (1.3)
The eigenvalue distribution is determinantal with kernel Kn,N built out of the polynomials
pj,N(x) = κj,N x
j + · · · , κj,N > 0, orthonormal with respect to the weight |x|2αe−NV (x)
on R. Indeed, as shown by Dyson, Gaudin, and Mehta, see e.g. [11, 21, 38], for any
m = 1, . . . , n− 1, the m-point correlation function
R(n,N)m (x1, . . . , xm) ≡
(n−m)!
· · ·
P (n,N)(x1, . . . , xn) dxm+1 · · · dxn (1.4)
is given by
R(n,N)m (x1, . . . , xm) = det (Kn,N(xi, xj))1≤i,j≤m , (1.5)
where
Kn,N(x, y) = |x|α|y|αe−
N(V (x)+V (y))
pj,N(x) pj,N(y). (1.6)
In the limit n,N → ∞, n/N → 1, the global eigenvalue regime is determined by V as
follows. The equilibrium measure µV for V is the unique minimizer of
IV (µ) =
|x− y|dµ(x)dµ(y) +
V (x)dµ(x) (1.7)
taken over all Borel probability measures µ on R. Since V is real analytic we have that
µV is supported on a finite union of disjoint intervals [13], and it has a density ρV such
n,N→∞,n/N→1
Kn,N(x, x) = ρV (x), x 6= 0.
The limiting mean eigenvalue density is independent of α.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 3
The factor | detM |2α changes the local eigenvalue behavior near 0. This is reflected
in the local scaling limits of Kn,N around 0 that do depend on α. If 0 is in the bulk of
the spectrum and ρV (0) > 0, then instead of the usual sine kernel we get a Bessel kernel
depending on α [37]. If 0 is in the bulk and ρV (0) = ρ
V (0) = 0, ρ
V (0) > 0, then the local
scaling limits of the kernel near 0 are associated with the Hastings-McLeod solution of
the Painlevé II equation q′′ = sq + 2q3 − α [9].
In this paper we study the effect of α in case 0 is an endpoint of the spectrum which
is such that the density ρV vanishes like a square root at 0. For α = 0 the scaling limit is
the well-known Airy kernel, see the papers [6, 26, 39, 42] and also [3, 12], and so we are
asking the question: What is the α-generalization of the Airy kernel?
For α > −1/2, we have found a new one-parameter family of limiting kernels as stated
in Theorem 1.1 below. In Theorem 1.1 we also assume that the eigenvalue density ρV is
regular, which means the following.
• The function x 7→ 2
log |x − s|ρV (s)ds − V (x) defined for x ∈ R, assumes its
maximum value only on the support of ρV .
• The density ρV is positive on the interior of its support.
• The density ρV vanishes like a square root at each of the endpoints of its support.
Theorem 1.1 For every α > −1/2, there exists a one-parameter family of kernelsKedgeα (x, y; s)
such that the following holds. Let V be a real analytic external field on R such that its
mean limiting eigenvalue density ρV is regular. Suppose that 0 is a right endpoint of the
support of ρV so that for some constant c1 = c1,V > 0
ρV (x) ∼
|x|1/2 as x→ 0− . (1.8)
Then there exists a second constant c2 = c2,V > 0 such that
n,N→∞
(c1n)2/3
(c1n)2/3
(c1n)2/3
= Kedgeα (x, y; s) (1.9)
whenever n,N → ∞ such that
= L ∈ R (1.10)
and s = −c2,V L.
For α = 0, the limiting kernels reduce to the kernel
0 (x, y; s) =
Ai(x+ s)Ai′(y + s)− Ai′(x+ s)Ai(y + s)
x− y , (1.11)
which is the (shifted) Airy kernel from random matrix theory mentioned above, see also
Subsection 4.1 below. For α 6= 0, a new type of special functions is needed to describe
the limiting kernel Kedgeα (x, y; s). This description is given in the next subsections.
4 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Figure 1: Contour for the model RH problem.
1.2 The model RH problem
We describe Kedgeα (x, y; s) through the solution of a special Riemann-Hilbert (RH) prob-
lem, that we will refer to as the model RH problem.
The model RH problem is posed on a contour Σ in an auxiliary ζ-plane, consisting of
four rays Σ1 = {arg ζ = 0}, Σ2 = {arg ζ = 2π/3}, Σ3 = {arg ζ = π}, and Σ4 = {arg ζ =
−2π/3} with orientation as shown in Figure 1. As usual in RH problems, the orientation
defines a + and a − side on each part of the contour, where the +-side is on the left when
traversing the contour according to its orientation. For a function f on C \ Σ, we use f±
to denote its limiting values on Σ taken from the ±-side, provided such limiting values
exist. The contour Σ divides the complex plane into four sectors Ωj also shown in the
figure.
The model RH problem reads as follows.
Riemann-Hilbert problem for Ψα
(a) Ψα : C \ Σ → C2×2 is analytic.
(b) Ψα,+(ζ) = Ψα,−(ζ)
, for ζ ∈ Σ1,
Ψα,+(ζ) = Ψα,−(ζ)
e2απi 1
, for ζ ∈ Σ2,
Ψα,+(ζ) = Ψα,−(ζ)
, for ζ ∈ Σ3,
Ψα,+(ζ) = Ψα,−(ζ)
e−2απi 1
, for ζ ∈ Σ4.
(c) Ψα(ζ) = ζ
−σ3/4 1√
(I + O(1/ζ1/2))e−(
ζ3/2+sζ1/2)σ3 as ζ → ∞. Here σ3 =
( 1 00 −1 ) is the third Pauli matrix.
(d) Ψα(ζ) = O
|ζ |α |ζ |α
|ζ |α |ζ |α
as ζ → 0, if α < 0; and
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 5
Ψα(ζ) =
|ζ |α |ζ |−α
|ζ |α |ζ |−α
as ζ → 0 with ζ ∈ Ω1 ∪ Ω4,
|ζ |−α |ζ |−α
|ζ |−α |ζ |−α
as ζ → 0 with ζ ∈ Ω2 ∪ Ω3,
if α ≥ 0.
Here, and in what follows, the O-terms are taken entrywise. Note that the RH problem
depends on a parameter s through the asymptotic condition at infinity. If we want to
emphasize the dependence on s we will write Ψα(ζ ; s) instead of Ψα(ζ).
The model RH problem is not uniquely solvable. Indeed, if Ψα is a solution, then(
Ψα is also a solution for any η = η(s), and it turns out that this is the only freedom
we have (see Proposition 2.1).
Theorem 1.2 The model RH problem is solvable for every s ∈ R. Let Ψα be a solution
of the model RH problem and put
ψ1(x; s)
ψ2(x; s)
Ψα,+(x; s)
, for x > 0,
Ψα,+(x; s)e
−απiσ3
, for x < 0.
(1.12)
Then the limiting kernel Kedgeα (x, y; s) can be written in the “integrable form”
Kedgeα (x, y; s) =
ψ2(x; s)ψ1(y; s)− ψ1(x; s)ψ2(y; s)
2πi(x− y) . (1.13)
The function ψ2 depends on the particular choice of solution Ψα to the model RH
problem. Indeed, for any η we have that the mapping Ψα 7→
Ψα leaves ψ1 invariant
and changes ψ2 to ψ2 + ηψ1. However, this does not change the expression (1.13) for the
kernel Kedgeα (x, y; s).
It follows from (1.12) and part (c) of the model RH problem that ψ1 and ψ2 have the
asymptotic behavior
ψ1(x; s) =
2x1/4
x3/2−sx1/2(1 +O(x−1/2)), (1.14)
ψ2(x; s) =
ix1/4√
x3/2−sx1/2(1 +O(x−1/2)), (1.15)
as x→ +∞, and
ψ1(x; s) =
2|x|−1/4 cos
|x|3/2 − s|x|1/2 − απ − π/4
+O(x−3/4), (1.16)
ψ2(x; s) = −i
2|x|1/4 sin
|x|3/2 − s|x|1/2 − απ − π/4
+O(x−1/4), (1.17)
as x→ −∞.
Remark 1.3 The kernel Kedgeα (x, y; s) describes an edge effect for the random matrix
ensemble (1.1). If we assume that 0 is the rightmost point in the support of ρV , and if given
M we let λmax(M) denote its largest eigenvalue, then it follows under the assumptions of
Theorem 1.1, in particular the limit assumption (1.10), that
n,N→∞
(c1n)
2/3λmax ≤ t
= det
1− Kα,s|(t,∞)
, (1.18)
6 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
where Kα,s|(t,∞) is the trace class operator in L2(t,∞) with kernel Kedgeα (x, y; s). To
prove (1.18) one must show that the operator with kernel 1
(c1n)2/3
(c1n)2/3
(c1n)2/3
converges in the trace class norm on L2(t,∞) to the operator with kernel Kedgeα (x, y; s).
This requires good estimates on the rate of convergence in (1.9), which can be established
as in [12].
For α = 0, the kernel is the (shifted) Airy kernel, and the Fredholm determinant (1.18)
has an equivalent expression in terms of a special solution of the Painlevé II equation.
The resulting distribution is the famous Tracy-Widom distribution [42, 43]. It would be
very interesting to find an analogous expression for general α. The connection to the
model RH problem given in Theorem 1.2 can be used in obtaining such an expression,
following the approach of [5] and [27]. We are planning to address this question in a future
publication.
1.3 Connection with the Painlevé XXXIV equation
The model RH problem is related to a special solution of the equation number XXXIV
from the list of Painlevé and Gambier [29],
u′′ = 4u2 + 2su+
(u′)2 − (2α)2
. (1.19)
All solutions of (1.19) are meromorphic in the complex plane.
Theorem 1.4 Let Ψα(ζ ; s) be a solution of the model RH problem. Then
u(s) = −s
− i d
Ψα(ζ ; s)e
( 23 ζ
3/2+sζ1/2)σ3 1√
ζσ3/4
(1.20)
exists and satisfies (1.19). The function (1.20) is a global solution of (1.19) (i.e., it does
not have poles on the real line), and it does not depend on the particular solution of the
model RH problem.
The connection with the Painlevé XXXIV equation leads to the following characteri-
zation of ψ1 and ψ2.
Theorem 1.5 Let u be the solution of Painlevé XXXIV given by (1.20). Then there
exists a solution Ψα of the model RH problem so that the functions ψ1 and ψ2 defined by
(1.12) satisfy the following system of linear differential equations
ψ1(x; s)
ψ2(x; s)
u′/(2x) i− iu/x
−i(x+ s+ u+ ((u′)2 − (2α)2)/(4ux)) −u′/(2x)
ψ1(x; s)
ψ2(x; s)
(1.21)
and have asymptotics (1.14)–(1.17).
In fact we will prove that for ζ ∈ C \ Σ,
Ψα(ζ ; s) =
u′/(2ζ) i− iu/ζ
−i(ζ + s+ u+ ((u′)2 − (2α)2)/(4uζ)) −u′/(2ζ)
Ψα(ζ ; s) (1.22)
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 7
from which (1.21) readily follows in view of (1.12). We emphasize that (1.21) and
(1.22) hold for one particular solution of the model RH problem. Any other solution(
Ψα(ζ ; s) also satisfies a system of linear differential equations, but with matrix
u′/(2ζ) i− iu/ζ
−i(ζ + s+ u+ ((u′)2 − (2α)2)/(4uζ)) −u′/(2ζ)
. (1.23)
In order to make Theorem 1.5 a genuine, i.e., independent of the Ψα RH problem,
characterization of ψ1 and ψ2, we need an independent of formula (1.20) characterization
of the solution u(s) of equation (1.19). This can be achieved by indicating the asymptotic
behavior of u(s) as s → ∞, cf. the characterization of the Hastings-McLeod solution of
Painlevé II equation [28]. We discuss this issue in detail in the last section of the paper,
see in particular Proposition 4.1 and the end of Remark 4.2 where the possible asymptotic
characterizations of the solution u(s) are given.
1.4 Overview of the rest of the paper
In Section 2 we give the proofs of Theorem 1.1 and Theorem 1.2. We start by presenting
the RH problem for orthogonal polynomials on the line [23]. The eigenvalue correlation
kernel Kn,N can be explicitly expressed in terms of the solution of this RH problem [11, 15].
As in earlier papers, see e.g. [3, 4, 8, 9, 14, 15], we apply the Deift-Zhou steepest descent
method for RH problems, see [16]. For the local analysis near 0, we need the model RH
problem for Ψα(ζ ; s) as introduced in Subsection 1.2. We show, following the methodology
of [25], that the model RH problem has a solution for every s ∈ R. Then we follow the
usual steps in the steepest descent analysis for RH problems, which lead us to the proofs
of Theorem 1.1 and 1.2.
Section 3 is devoted to the proofs of Theorem 1.4 and Theorem 1.5. We start by
discussing the RH problem, in the form due to Flaschka and Newell, associated with the
Painlevé II equation q′′ = sq+2q3− ν. Following [2], we show that for a special choice of
monodromy data the Flaschka-Newell RH problem is related to the model RH problem.
The parameters in the Painlevé equations are related by ν = 2α + 1/2. The monodromy
data corresponds to a solution of Painlevé II which is different from the Hastings-McLeod
solution that has appeared more often in random matrix theory [4, 8, 9, 28, 42]. The
known results (asymptotics, Lax pair etc.) for the RH problem for Painlevé II are then
transferred to the model RH problem, and then used to complete the proofs of Theorems
1.4 and 1.5. In particular it gives rise to the special solution u of the Painlevé XXXIV
equation defined by (1.20).
In Section 4 we make some concluding remarks. For the important special cases α = 0
and α = 1, we show how the model RH problem can be explicitly solved in terms of
Airy functions, and how the limiting kernel Kedgeα (x, y; s) as well as the special Painlevé
XXXIV solution u can be explicitly computed in both cases. Our final remarks concern
the characterization of u, in the case of general α, through its asymptotic behavior at
infinity.
8 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
2 Proof of Theorem 1.1 and 1.2
2.1 The Riemann-Hilbert problem for orthogonal polynomials
The RH problem for orthogonal polynomials on the line, for our particular weight, is the
following (cf. [23]).
Riemann-Hilbert problem for Y
• Y : C \ R → C2×2 is analytic.
• Y+(x) = Y−(x)
1 |x|2αe−NV (x)
for x ∈ R\{0}, with R oriented from left to right.
• Y (z) = (I +O(1/z))
0 z−n
as z → ∞.
• If α < 0, then Y (z) = O
1 |z|2α
1 |z|2α
as z → 0. If α ≥ 0, then Y (z) = O ( 1 11 1 ) as z → 0.
The RH problem has the unique solution
Y (z) =
pn,N(z)
2πiκn,N
pn,N(s)|s|2αe−NV (s)
s− z ds
−2πi κn−1,N pn−1,N(z) −κn−1,N
pn−1,N(s)|s|2αe−NV (s)
s− z ds
 , (2.1)
where pj,N(x) = κj,N x
j + · · · is the orthonormal polynomial with respect to the weight
|x|2αe−NV (x). By (1.6) and the Christoffel-Darboux formula for orthogonal polynomials,
we have
Kn,N(x, y) = |x|α|y|αe−
N(V (x)+V (y))κn−1,N
pn,N(x) pn−1,N(y)− pn−1,N(x) pn,N(y)
x− y . (2.2)
Thus, using (2.1) and the fact that det Y ≡ 1, we may express the eigenvalue correlation
kernel directly in terms of Y :
Kn,N(x, y) =
2πi(x− y) |x|
α|y|αe− 12N(V (x)+V (y))
Y −1+ (y)Y+(x)
. (2.3)
The main idea for the proof of Theorems 1.1 and 1.2 is to apply the powerful steepest
descent analysis for RH problems of Deift and Zhou [16] to the RH problem satisfied by
Y . In the case at hand it consists of constructing a sequence of invertible transformations
Y 7→ T 7→ S 7→ R, where the matrix-valued function R is close to the identity. By
unfolding the above transformations asymptotics for Y and thus, in view of (2.3), for
Kn,N in various regimes may be derived. Our main attention will be devoted to the local
behavior of Y near 0. Around 0 we construct a local parametrix with the help of the
model RH problem, which we next discuss in more detail.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 9
2.2 The model RH problem
The model RH problem is not uniquely solvable.
Proposition 2.1 Let Ψα be a solution of the model RH problem. Then the following
hold.
(a) detΨα ≡ 1.
(b) For any η ∈ R (which may depend on s), we have that
Ψα also solves the
model RH problem.
(c) Any two solutions are related as in part (b), i.e., if Ψ
α and Ψ
α are two solutions
of the model RH problem, then Ψ
α for some η = η(s).
Proof. (a) We have that detΨα is analytic in C \ {0}, since all jump matrices have
determinant one. In case α < 0 we get from condition (d) of the RH problem that
detΨα(ζ) = O(|ζ |2α) as ζ → 0. Since 2α > −1 it follows that the singularity at the
origin is removable. In case α ≥ 0 we find from condition (d) of the RH problem that
detΨα(ζ) = O(1) as ζ → 0 in Ω1 ∪ Ω4. Thus the singularity at the origin cannot be
a pole. Since detΨα = O(|ζ |−2α) as ζ → 0, it cannot be an essential singularity either
and therefore the singularity at the origin is removable also in this case. Hence detΨα is
entire. From condition (c) of the RH problem it follows that detΨα(ζ) → 1 as ζ → ∞,
and so part (a) of the proposition follows from Liouville’s theorem.
(b) It is clear that
Ψα satisfies the conditions (a), (b), and (d) of the model RH
problem. To establish (c) it is enough to observe that
ζ−σ3/4
= ζ−σ3/4
2ζ1/2
= ζ−σ3/4
(I +O(1/ζ1/2))
as ζ → ∞.
(c) In view of part (a) we know that Ψ
α is invertible. Then Ψ
−1 is analytic
in C \ {0} and, by arguments similar to those in the proof of part (a), the singularity at
the origin is removable. As ζ → ∞ we get from condition (c) of the model RH problem
Ψ(2)α (ζ)(Ψ
α (ζ))
−1 = ζ−σ3/4
(I +O(ζ−1/2))
ζσ3/4
= I +O
ζ−1/2 ζ−1
1 ζ−1/2
The statement now follows from Liouville’s theorem. ✷
In the following we will need more information about the behavior at the origin of
functions satisfying properties (a), (b), and (d) of the model RH problem. The following
result is similar to Proposition 2.3 in [9].
Proposition 2.2 Let Ψ satisfy conditions (a), (b), and (d) of the RH problem for Ψα.
Then, with all branches being principal, the following hold.
10 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
• If α− 1
/∈ N0, there exist an analytic matrix-valued function E and constant matrices
Aj such that
Ψ(ζ) = E(ζ) ζασ3 Aj, for ζ ∈ Ωj . (2.4)
Letting vj denote the jump matrix for Ψ on Σj, we have
A1 = A4 v1, A1 = A2 v2, A3 = A4 v4, (2.5)
2 cosαπ
2 cosαπ
−eαπi e−απi
 . (2.6)
• If α− 1
∈ N0, then Ψ has logarithmic behavior at the origin: There exist an analytic
matrix-valued function E and constant matrices Aj such that
Ψ(ζ) = E(ζ)
ζα log ζ
0 ζ−α
Aj, for ζ ∈ Ωj . (2.7)
Letting vj denote the jump matrix for Ψ on Σj, we now have
A1 = A4 v1, A1 = A2 v2, A3 = A4 v4, (2.8)
0 e3πi/4
eπi/4 eπi/4
 . (2.9)
• In all cases it holds that detAj = 1 and
(A1)21 = (A4)21 = 0. (2.10)
Proof. The statement (2.10) is an immediate consequence of the explicit formulas for
the Aj ’s.
Consider the case α− 1
/∈ N0. Define E by (2.4), i.e., let
E(ζ) = Ψ(ζ)A−1j ζ
−ασ3, for ζ ∈ Ωj , (2.11)
with Aj as in (2.5), (2.6). Then E is analytic in C \ Σ. We now show that E is indeed
entire. The relations (2.5) and the condition (b) of the model RH problem imply that E
is analytic also on Σ1 ∪ Σ2 ∪ Σ4. Moreover, on Σ3
E−1− (ζ)E+(ζ) = ζ
− A3 v3A
Now, by (2.5), (2.6), and straightforward computation
A3 v3A
2 = A2 v2 v
1 v4 v3A
2 = e
2απiσ3 = ζ−ασ3− ζ
+ . (2.12)
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 11
Thus, E is analytic also on Σ3, and therefore in C \ {0}.
We next show that the singularity at 0 is removable. If α < 0, we see from (2.11) and
the condition (d) of the model RH problem, that as ζ → 0
E(ζ) = O
|ζ |α |ζ |α
|ζ |α |ζ |α
|ζ |−α 0
0 |ζ |α
1 |ζ |2α
1 |ζ |2α
so (since 2α > −1) the isolated singularity at 0 is indeed removable. If α ≥ 0 and ζ → 0
in Ω1 we find in the same way (also using (A1)21 = 0) that
E(ζ) = O
|ζ |α |ζ |−α
|ζ |α |ζ |−α
|ζ |−α 0
0 |ζ |α
so that E is bounded near 0 in Ω1 and thus 0 cannot be a pole. Since 0 cannot be an
essential singularity either, we conclude that the singularity is indeed removable.
In case α − 1
∈ N0 the proof is almost identical, only now the equation (2.12) is
replaced by
A3 v3A
2 = A2 v2 v
1 v4 v3A
−1 −2i
ζ−α − 1
ζα log ζ
ζα log ζ
0 ζ−α
. (2.13)
2.3 Existence of solution to the model RH problem
We will need that for s ∈ R the model RH problem indeed has a solution. To prove
existence of a solution to the model RH problem it suffices to prove existence of a (unique)
solution Ψ
(spec)
α to the special RH problem obtained when the asymptotics (c) at infinity
is replaced by the following stronger condition
Ψ(spec)α (ζ) = (I +O(1/ζ))ζ
−σ3/4 1√
ζ3/2+sζ1/2)σ3 , (2.14)
as ζ → ∞.
A key element in the proof of unique solvability of the RH problem for Ψ
(spec)
α is the
following vanishing lemma (cf. [25]).
Proposition 2.3 (vanishing lemma) Let α > −1/2, s ∈ R, and put θ(ζ) = θ(ζ ; s) =
ζ3/2+sζ1/2. Suppose that Fα satisfies the conditions (a), (b), and (d) in the RH problem
for Ψα but, instead of condition (c), has the following behavior at infinity:
Fα(ζ) = O(1/ζ)ζ
−σ3/4 1√
e−θ(ζ)σ3 , (2.15)
as ζ → ∞. Then Fα ≡ 0.
12 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Proof. The ideas of the proof are similar in spirit to those in Deift et al. [14]. Let Gα
be defined as follows:
Gα(ζ) =


Fα(ζ)e
θ(ζ)σ3
, for ζ ∈ Ω1,
Fα(ζ)e
θ(ζ)σ3
e2απie2θ(ζ) 1
, for ζ ∈ Ω2,
Fα(ζ)e
θ(ζ)σ3
−e−2απie2θ(ζ) 1
, for ζ ∈ Ω3,
Fα(ζ)e
θ(ζ)σ3 , for ζ ∈ Ω4.
(2.16)
Then Gα satisfies the following RH problem.
Riemann-Hilbert problem for Gα
(a) Gα : C \ R → C2×2 is analytic.
(b) Gα,+(ζ) = Gα,−(ζ)vGα(ζ) for ζ ∈ R \ {0}, where
vGα(ζ) =
e−2θ(ζ) −1
, for ζ > 0,
1 −e2απie2θ+(ζ)
e−2απie2θ−(ζ) 0
, for ζ < 0.
(2.17)
(c) Gα(ζ) = O(ζ
−3/4) as ζ → ∞.
(d) Gα has the following behavior near the origin: If α < 0,
Gα(ζ) = O
|ζ |α |ζ |α
|ζ |α |ζ |α
, as ζ → 0, (2.18)
and if α ≥ 0,
Gα(ζ) =
|ζ |−α |ζ |α
|ζ |−α |ζ |α
as ζ → 0, Im ζ > 0,
|ζ |α |ζ |−α
|ζ |α |ζ |−α
as ζ → 0, Im ζ < 0.
(2.19)
The jumps in (b) follow from straightforward computations which uses that θ+(ζ) +
θ−(ζ) = 0 for ζ < 0. The behavior (c) of Gα at infinity (uniformly in each sector) follows
directly from (2.15), (2.16), and the fact that Re θ(ζ) < 0 for ζ ∈ Ω2 ∪ Ω3. The behavior
(2.18) at the origin is immediate from the condition (d) of the RH problem for Fα, and
so is the behavior (2.19) if ζ → 0 with ζ ∈ Ω1 ∪ Ω4. To prove (2.19) if ζ → 0 with
ζ ∈ Ω2 ∪ Ω3, we need Proposition 2.2. Consider first the case α − 12 /∈ N0 and ζ ∈ Ω2.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 13
Then we have, using (2.16), (2.4), (2.5), and (2.10)
Gα(ζ) = Fα(ζ)e
θ(ζ)σ3
e2απie2θ(ζ) 1
= E(ζ)ζασ3A2
e2απi 1
eθ(ζ)σ3
= E(ζ)ζασ3A2
e2απi 1
e−θ(ζ)σ3
= E(ζ)ζασ3A1
e−θ(ζ)σ3 = E(ζ)ζασ3
e−θ(ζ)σ3 ,
where ∗ denotes an unspecified constant. Using the boundedness of E and θ at the origin,
we find (2.19) as ζ → 0 in the sector Ω2. The case ζ ∈ Ω3 is treated similarly. Using
(2.7), (2.8) instead of (2.4), (2.5), the same argument works in case α − 1
∈ N0. Note
that in spite of the logarithmic entry in (2.8), there are no logarithmic entries in (2.19).
Introduce the auxiliary matrix-valued function
Hα(ζ) = Gα(ζ) (Gα(ζ̄))
∗, ζ ∈ C \ R. (2.20)
Then Hα is analytic and
Hα(ζ) = O(ζ
−3/2), as ζ → ∞. (2.21)
From the condition (d) in the RH problem for Gα it follows that Hα has the following
behavior near the origin:
Hα(ζ) =
|ζ |2α |ζ |2α
|ζ |2α |ζ |2α
as ζ → 0, in case α < 0,
as ζ → 0, in case α ≥ 0.
(2.22)
Since α > −1/2, we see from (2.21) and (2.22) that each entry of Hα,+ is integrable over
the real line, and by Cauchy’s theorem and (2.21)
Hα,+(ζ) dζ = 0. (2.23)
That is, by (2.20),
Gα,+(ζ) (Gα,−(ζ))
∗ dζ = 0. (2.24)
Adding (2.24) to its Hermitian conjugate and using (2.17) we obtain
Gα,−(ζ) [vGα(ζ) + (vGα(ζ))
∗] (Gα,−(ζ))
Gα,−(ζ)
(Gα,−(ζ))
∗ dζ +
Gα,−(ζ)
2e−2θ(ζ) 0
(Gα,−(ζ))
∗ dζ.
(2.25)
14 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Here we also used that θ+(ζ) = −θ−(ζ) ∈ iR for ζ < 0, which holds because s is real.
The identity (2.25) implies that the first column of Gα,− vanishes identically on R. Thus,
in view of the form of the jump matrix in (2.17), the second column of Gα,+ vanishes
identically on R as well. It follows that the first column of Gα vanishes identically in the
lower half-plane, and the second column vanishes identically in the upper half-plane.
To prove that the full matrix Gα vanishes identically in both half-planes, we shall use
a Phragmen-Lindelöf type theorem due to Carlson [7, 41]. Define for j = 1, 2,
gj(ζ) =
(Gα)j1(ζ), for Im ζ > 0,
(Gα)j2(ζ), for Im ζ < 0.
(2.26)
The conditions of the RH problem for Gα yield that both g1 and g2 have analytic contin-
uation across (0,∞) and that they are both solutions of the following scalar RH problem.
Riemann-Hilbert problem for g
• g : C \ (−∞, 0] → C is analytic with jump
g+(ζ) = g−(ζ) e
−2απie2θ−(ζ), for ζ ∈ (−∞, 0). (2.27)
• g(ζ) = O(ζ−3/4) as ζ → ∞.
• g(ζ) = O(|ζ |−|α|) as ζ → 0.
We are going to prove that this RH problem has only the trivial solution.
Let g be any solution and define ĝ by
ĝ(ζ) =
g(ζ2), if Re ζ > 0,
g(ζ2)e−2απie−2(
ζ3+sζ), if Re ζ < 0, Im ζ > 0,
g(ζ2)e2απie−2(
ζ3+sζ), if Re ζ < 0, Im ζ < 0.
(2.28)
The jump property (2.27) ensures that ĝ is analytic across the imaginary axis.
Now define
h(ζ) =
1 + ζ
ĝ(ζ4/3), for Re ζ ≥ 0, (2.29)
with (as usual) the principal branches of the fractional powers. Then it can be checked
that h is analytic in Re ζ > 0, bounded for Re ζ ≥ 0, and satisfies
|h(ζ)| ≤ Ce−c|ζ|4, if ζ ∈ iR,
for some positive constants c and C. Hence, by Carlson’s theorem, h ≡ 0 in Re ζ ≥ 0.
Therefore g ≡ 0, and so g1 and g2 are both identically zero. It follows that the full matrix
Gα vanishes identically in both half-planes.
Thus Fα ≡ 0 by (2.16), and this completes the proof of the proposition. ✷
We now show how (unique) solvability of the RH problem for Ψ
(spec)
α can be deduced
from the above vanishing lemma.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 15
Proposition 2.4 The RH problem for Ψ
(spec)
α has a unique solution for every s ∈ R.
Proof. The idea of the proof is this: Given a solution Ψ
(spec)
α to the above RH prob-
lem, we show how to construct a solution mα to a certain normalized RH problem (i.e.,
mα(ζ) → I as ζ → ∞) characterized by a jump matrix v on a contour Σ̃, and con-
versely. To prove the proposition it therefore suffices to prove (unique) solvability of the
normalized RH problem. This can be done by utilizing the basic relationship between
normalized RH problems and singular integral equations. We recall briefly, in our setting,
some standard facts regarding this relationship. For further details, and proofs, the reader
is referred to the papers [17, 18, 19, 44], and to the appendix of [32].
Let C denote the Cauchy operator
Ch(ζ) =
s− ζ ds, h ∈ L
2(Σ̃), ζ ∈ C \ Σ̃, (2.30)
and denote by C±h(ζ), ζ ∈ Σ̃, the limits of Ch(ζ ′) as ζ ′ → ζ from the (±)-side of Σ̃. The
operators C± are bounded on L
2(Σ̃). Let
v(ζ) = (v−(ζ))
−1v+(ζ), ζ ∈ Σ̃, (2.31)
be a pointwise factorization of v(ζ) with v±(ζ) ∈ GL(2,C), and define ω± through
v±(ζ) = I ± ω±(ζ), ζ ∈ Σ̃. (2.32)
Our choice of factorization will imply that ω± ∈ L2(Σ̃) ∩ L∞(Σ̃). The singular integral
operator Cω : L
2(Σ̃) → L2(Σ̃), defined by
Cωh = C+(hω−) + C−(hω+), h ∈ L2(Σ̃), (2.33)
is then bounded on L2(Σ̃). Moreover, it makes sense to study the singular integral equation
(1− Cω)µ = I (2.34)
for µ ∈ I + L2(Σ̃). For if we write µ = I + h, then (2.34) takes the form
(1− Cω)h = CωI ∈ L2(Σ̃). (2.35)
Suppose that µ ∈ I + L2(Σ̃) is a solution of (2.34). Then, indeed
mα(ζ) = I + C(µ(ω+ + ω−))(ζ), ζ ∈ C \ Σ̃, (2.36)
solves the normalized RH problem. Thus, if we can prove that the operator 1 − Cω is
a bijection in L2(Σ̃), then solvability of the RH problem for mα, and hence of that for
(spec)
α , has been established. Bijectivity of 1 − Cω in L2(Σ̃) is proved in two steps. We
first show that, for an appropriate choice of ω = (ω−, ω+) in the above factorization,
1− Cω is Fredholm in L2(Σ̃) with index 0. Second, we show that the kernel of 1− Cω is
trivial. Now, it is a standard fact that ker (1 − Cω) = {0} if and only if the associated
homogeneous RH problem (for say m0α) has only the trivial solution. But the explicit
relation between Ψ
(spec)
α and mα also establishes a relation between solutions Fα and m
16 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Figure 2: Contour for the RH problem for mα.
of the associated homogeneous RH problems. In view of Proposition 2.3, which states
that Fα ≡ 0, the second step has thus already been accomplished.
We now establish the above mentioned relation between Ψ
(spec)
α and mα, derive the
RH problem satisfied by mα, and finally show that a factorization of v may be chosen so
that 1− Cω is Fredholm with index 0, cf. [25].
Let D = {ζ ∈ C | |ζ | < 1}. Set θ(ζ) = 2
ζ3/2 + sζ1/2 and
mα(ζ) =
(spec)
α (ζ)A
ζ−α −κα
ζα log ζ
, for ζ ∈ Ωj ∩ D,
(spec)
α (ζ)e
θ(ζ)σ3 1√
ζσ3/4, for ζ ∈ Ωj ∩ D
(2.37)
with {Aj}4j=1 being the matrices in Proposition 2.2, and where κα = 1 if α−1/2 ∈ N0 and
0 otherwise. By Proposition 2.2 it follows that mα is analytic in D. Let Σ̃ = Σ ∪ ∂D and
orient the components of Σ̃ as in Figure 2. This makes Σ̃ a complete contour, meaning that
C\Σ̃ can be expressed as the union of two disjoint sets, C\Σ̃ = Ω+∪Ω−, Ω+∩Ω− = ∅, such
that Σ̃ is the positively oriented boundary of Ω+ and the negatively oriented boundary of
Ω−. Let Σ̃j = Ωj ∩ ∂D.
Computations show that mα satisfies the following normalized RH problem. As in
Proposition 2.2 we use vj to denote the jump matrix on Σj in the model RH problem.
Riemann-Hilbert problem for mα
• mα : C \ Σ̃ → C2×2 is analytic.
• mα,+(ζ) = mα,−(ζ)v(ζ) for ζ ∈ Σ̃, where
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 17
v(ζ) =


I, for ζ ∈ Σ̃ ∩ D,
ζ−σ3/4 1√
e−θσ3vje
θσ3 1√
ζσ3/4, for ζ ∈ Σj ∩ D
, j ∈ {1, 2, 4},
I, for ζ ∈ Σ3 ∩ D
ζα κα
ζα log ζ
0 ζ−α
θσ3 1√
ζσ3/4, for ζ ∈ Σ̃j , j ∈ {1, 3},
ζ−σ3/4 1√
e−θσ3A−1j
ζ−α −κα
ζα log ζ
, for ζ ∈ Σ̃j , j ∈ {2, 4}.
• mα(ζ) = I +O(1/ζ) as ζ → ∞.
The analyticity of mα on Σ3 ∩ D
follows since θ+(ζ) + θ−(ζ) = 0 for ζ < 0.
It is important to note that v(ζ)− I decays exponentially as ζ → ∞ along Σ̃. Next
observe that, at any of the points 0, A, B, C,D of self-intersection of Σ̃ (see Figure 2),
precisely four contours come together. At a fixed point of self-intersection, say P , order
the contours that meet at P counterclockwise, starting from any contour that is oriented
outwards from P . Denoting the limiting value of the jump matrices over the jth contour
at P by v(j)(P ), we then have the cyclic relation
v(1)(P )
v(2)(P )
v(3)(P )
v(4)(P )
= I. (2.38)
This is trivial in case P = 0, and follows by direct computation in the other cases. We
remark that the cyclic relation (2.38) at C is a consequence of the relation (2.12) in the
case α − 1/2 6∈ N0, and of (2.13) in the case α − 1/2 ∈ N0 (see the proof of Proposition
2.2).
Outside small neighborhoods of the points of self-intersection we choose the trivial
factorization v+ = v, v− = I in (2.31), so that ω+ = v − I, ω− = 0 by (2.32). Using the
cyclic relations (2.38), we are then able to choose a factorization of v in the remaining
neighborhoods in such a way that ω+ is continuous along the boundary of each connected
component of Ω+, and similarly, ω− is continuous along the boundary of each connected
component of Ω−.
The exponential decay of v(ζ)− I as ζ → ∞ ensures that ω± ∈ L2(Σ̃)∩L∞(Σ̃). From
this it follows that 1− Cω is Fredholm in L2(Σ̃). Indeed, set
ω̃− = I − v−1− , ω̃+ = v−1+ − I. (2.39)
The choice of ω̃ = (ω̃−, ω̃+) is motivated by the relations
ω̃− ω− = ω̃− + ω−, ω̃+ ω+ = −(ω̃+ + ω+). (2.40)
A direct calculation, using C+ − C− = 1 and (2.40), shows that
(1− Cω)(1− Ceω) = 1 + T, (2.41)
where
Tf = C+((C−[f(ω̃+ + ω̃−)])ω−) + C−((C+[f(ω̃+ + ω̃−)])ω+) (2.42)
18 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
for f ∈ L2(Σ̃). Standard computations, using continuity of the functions ω+ resp. ω−
along the boundary of each connected component of Ω+ resp. Ω−, show that T is compact
in L2(Σ̃). Similar computations show that (1− Ceω)(1− Cω) = 1 + S, with S compact in
L2(Σ̃). So 1− Ceω is a pseudoinverse for 1− Cω, which is therefore Fredholm in L2(Σ̃).
It follows from general theory that the index of the operator 1−Cω equals the winding
number of det v along Σ̃, the latter being defined in the natural way. Now, since det v ≡ 1,
this is trivially zero. This completes the proof of Proposition 2.4. ✷
Remark 2.5 The RH problem for Ψ
(spec)
α is indeed solvable for all s ∈ C \D, where D is
a discrete set in C (disjoint from R according to Proposition 2.4), and the solution Ψ
(spec)
is meromorphic in s with poles in D. To see this, we first observe that the factorization
(2.31), (2.32) can be done so that ω± are both analytic in s. It follows that s 7→ 1−Cω is
an analytic map taking values in the Fredholm operators on L2(Σ̃). Since we know that
1−Cω is invertible for s ∈ R, we then get, by a version of the analytic Fredholm theorem
[44], that µ defined by (2.34) is meromorphic. Thus mα and hence Ψ
(spec)
α is meromorphic
in s.
2.4 Some preliminaries on equilibrium measures
Before we embark on the steepest descent analysis for the RH problem of Subsection 2.1,
we recall certain properties of equilibrium measures, see [11, 40]. We use the following
notation:
, Vt(x) =
V (x). (2.43)
As explained in the Introduction, we are interested in the case where n/N → 1 as n,N →
∞, which means that we are interested in t close to 1. For every t we consider the energy
functional IVt(µ) as in (1.7), and its minimizer µt.
The equilibrium measure dµt = ρt dx is characterized by the following Euler-Lagrange
variational conditions: There is a constant lt ∈ R such that
log |x− s|ρt(s) ds− Vt(x) + lt = 0, x ∈ suppµt, (2.44)
log |x− s|ρt(s) ds− Vt(x) + lt ≤ 0, x ∈ R \ supp µt. (2.45)
For t = 1, we have that the support of µV consists of a finite union of disjoint intervals,
see [13], say
supp µV =
[aj , bj ]
with a1 < b1 < a2 < · · · < ak < bk. Due to the assumption that the density ρV of µV is
regular, we have the following proposition.
Proposition 2.6 For every t in an interval around 1, we have that the density ρt of µt
is regular, and that supp µt consists of k intervals, say
supp µt =
[aj(t), bj(t)]
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 19
with a1(t) < b1(t) < a2(t) < · · · < ak(t) < bk(t). In this interval around 1, the functions
t 7→ aj(t) and t 7→ bj(t) are real analytic with a′j(t) < 0 and b′j(t) > 0.
Proof. See Theorem 1.3 (iii) and Lemma 8.1 of [36]. ✷
For the rest of the proof of Theorem 1.1 we shall assume that suppµV consists of
one interval. In the general case (when suppµV consists of k ≥ 2 intervals) one proceeds
analogously, but the parametrix away from the end points given in Subsection 2.5.4 must
then instead be constructed with the help of the θ-function of B-periods for the two-
sheeted Riemann surface y2 = Πkj=1[(z − aj)(z − bj)] obtained by gluing together two
copies of the slit plane C \
j=1[aj , bj] in the standard way [14, 37]. Since the formulas
will be more complicated in the multi-interval case, but do not contribute to the main
issue of the present paper, we chose to give the proof in full for the one-interval case only.
2.5 Steepest descent analysis
2.5.1 Preliminaries
We assume from now on that k = 1, so that supp µV consists of one interval which we
take as
supp(µV ) = [a, 0], a < 0.
Then there is δ1 > 0 such that µt is supported on one interval [at, bt] for every t ∈
(1− δ1, 1 + δ1), and its density ρt is regular. Hence ρt is positive on (at, bt) and vanishes
like a square root at the end points, and it takes the form [14]
ρt(x) =
(bt − x)(x− at)ht(x), for x ∈ [at, bt], (2.46)
where ht is positive on [at, bt], and analytic in the domain of analyticity of V . In addition,
ht depends analytically on t ∈ (1− δ1, 1 + δ1).
We are going to use the equilibrium measure µt in the first transformation of the RH
problem. We remark that in [8, 9, 10, 20] a modified equilibrium measure was used in the
steepest descent analysis of a RH problem at a critical point. It is likely that we could
have modified the equilibrium measure in the present situation as well, but the approach
with the unmodified µt also works, as we will see, and we chose to use it in this paper.
In the one-interval case one can show by explicit computation that
at = −
t(bt − at)ht(at)
t(bt − at)ht(bt)
, (2.47)
which indeed shows that d
at < 0 and
bt > 0. It follows that bt > 0 for t ∈ (1, 1 + δ1)
and bt < 0 for t ∈ (1− δ1, 1). In both cases we have at < 0.
We introduce two functions ϕt and ϕ̃t as follows. For z ∈ C \ (−∞, bt] lying in the
domain of analyticity of V (which we may restrict to be simply connected, without loss
of generality), we put
ϕt(z) =
((s− bt)(s− at))1/2ht(s) ds, (2.48)
20 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
and for z ∈ C \ [at,∞) also in the domain of analyticity of V ,
ϕ̃t(z) =
((s− bt)(s− at))1/2ht(s) ds. (2.49)
It follows from (2.48) that
ϕt(z) =
−at ht(bt)(z − bt)3/2χt(z), (2.50)
where χt is analytic in a neighborhood of bt and χt(bt) = 1. Taking
ft(z) =
ϕt(z)
−at ht(bt)
(z − bt)χ2/3t (z), (2.51)
we see that ft is analytic in a neighborhood of bt with ft(bt) = 0,
f ′t(bt) =
−at ht(bt)
6= 0, (2.52)
and ft(z) real for real values of z. Hence, in particular,
ft(0) > 0, if t < 1, f1(0) = 0, and ft(0) < 0, if t > 1. (2.53)
Moreover, ft → f1 as t → 1, uniformly in a neighborhood of 0. We choose a small disc
U (0) around 0 and δ2 > 0 sufficiently small, so that ft is a conformal map from U
(0) onto
a convex neighborhood of 0 for every t ∈ (1− δ2, 1 + δ2).
Similarly, there exists a disc U (a) centered at a < 0, and a δ3 > 0, so that
f̃t(z) =
ϕ̃t(z)
(2.54)
is a conformal map from U (a) onto a convex neighborhood of 0 for every t ∈ (1−δ3, 1+δ3).
We let δ0 = min(δ1, δ2, δ3) and we fix t ∈ (1− δ0, 1 + δ0). In what follows we also take
the neighborhoods U (0) and U (a) as above.
2.5.2 First transformation Y 7→ T
We introduce the so-called g-function:
gt(z) =
log(z − s) dµt(s) =
log(z − s) ρt(s) ds, z ∈ C \ (−∞, bt], (2.55)
where log denotes the principal branch. Then gt is analytic in C \ (−∞, bt]. Define T by
T (z) = e
nltσ3 Y (z) e−
nltσ3 e−ngt(z)σ3 , z ∈ C \ R, (2.56)
where lt is the constant from (2.44)–(2.45). By a straightforward calculation it then
follows that T has the following jump matrix vT on R (oriented from left to right):
vT (x) =
e−n(gt,+(x)−gt,−(x)) |x|2α en(gt,+(x)+gt,−(x)−Vt(x)+lt)
0 en(gt,+(x)−gt,−(x))
. (2.57)
Because of the identities, see [11, 14],
gt,+(x) + gt,−(x)− Vt(x) + lt = −2ϕt(x), for x > bt, (2.58)
gt,+(x) + gt,−(x)− Vt(x) + lt = −2ϕ̃t(x), for x < at, (2.59)
we see that the RH problem for T is the following.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 21
Figure 3: Opening of a lens around [at, 0].
Riemann-Hilbert problem for T
• T : C \ R → C2×2 is analytic.
• T+(x) = T−(x) vT (x) for x ∈ R, with
vT (x) =


1 |x|2α e−2nϕ̃t(x)
, for x < at,
e2nϕt,+(x) |x|2α
0 e2nϕt,−(x)
, for x ∈ (at, bt),
1 |x|2α e−2nϕt(x)
, for x > bt.
• T (z) = I +O(1/z) as z → ∞.
• If α < 0, then T (z) = O
1 |z|2α
1 |z|2α
as z → 0. If α ≥ 0, then T (z) = O ( 1 11 1 ) as z → 0.
2.5.3 Second transformation T 7→ S
The opening of lenses is based on the following factorization of vT on (at, bt):
vT (x) =
e2nϕt,+(x) |x|2α
0 e2nϕt,−(x)
|x|−2α e2nϕt,−(x) 1
0 |x|2α
−|x|−2α 0
|x|−2α e2nϕt,+(x) 1
Introduce a lens around the segment [at, 0] as in Figure 3 (recall that at < 0). In the
disc U (0) around 0 we take the lens such that z 7→ ζ = ft(z)− ft(0), see (2.51), maps the
parts of the upper and lower lips of the lens that are in U (0) into the rays arg ζ = 2π/3
and arg ζ = −2π/3, respectively. Similarly, in the disc U (a) we choose the lens so that
z 7→ ζ = f̃t(z), see (2.54), maps the parts of the upper and lower lips of the lens that are
in U (a) into the rays arg ζ = π/3, and arg ζ = −π/3, respectively. The remaining parts
of the lips of the lens are arbitrary. However, they should be contained in the domain of
analyticity of V , and we take them so that
Reϕt(z) < −c < 0
for z on the lips of the lens outside U (0) and U (a), with c > 0 independent of t.
It is important to note that the lens is around [at, 0], and not around [at, bt].
22 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Define S by
S(z) =


T (z), for z outside the lens,
T (z)
(−z)−2α e2nϕt(z) 1
, for z in the upper part of the lens,
T (z)
(−z)−2α e2nϕt(z) 1
, for z in the lower part of the lens.
(2.60)
Here the map z 7→ (−z)−2α is defined with a cut along the positive real axis. Then, from
(2.60) and the RH problem for T , we find that S is the unique solution of the following
RH problem.
Riemann-Hilbert problem for S
• S : C \ΣS → C2×2 is analytic, where ΣS consists of the real line and the upper and
lower lips of the lens, with orientation as in Figure 3.
• S+(z) = S−(z) vS(z) for z ∈ ΣS, where vS is given as follows. For t < 1, so that
bt < 0, we have
vS(z) =


1 |z|2α e−2nϕ̃t(z)
, for z ∈ (−∞, at),
0 |z|2α
−|z|−2α 0
, for z ∈ (at, bt),
0 |z|2α e−2nϕt(z)
−|z|−2α e2nϕt(z) 0
, for z ∈ (bt, 0),
1 |z|2α e−2nϕt(z)
, for z ∈ (0,∞),
(−z)−2α e2nϕt(z) 1
, for z on both lips of the lens,
while, for t ≥ 1, so that bt ≥ 0, we have
vS(z) =


1 |z|2α e−2nϕ̃t(z)
, for z ∈ (−∞, at),
0 |z|2α
−|z|−2α 0
, for z ∈ (at, 0),
e2nϕt,+(z) |z|2α
0 e2nϕt,−(z)
, for z ∈ (0, bt),
1 |z|2α e−2nϕt(z)
, for z ∈ (bt,∞),
(−z)−2α e2nϕt(z) 1
, for z on both lips of the lens.
• S(z) = I +O(1/z) as z → ∞.
• If α < 0, then S(z) = O
1 |z|2α
1 |z|2α
as z → 0. If α ≥ 0, then S(z) = O ( 1 11 1 ) as z → 0
from outside the lens and S(z) = O
|z|−2α 1
|z|−2α 1
as z → 0 from inside the lens.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 23
The next step is to approximate S by a parametrix P , consisting of three parts P (∞),
P (a), and P (0):
P (z) =
P (0)(z), for z ∈ U (0) \ ΣS,
P (a)(z), for z ∈ U (a) \ ΣS ,
P (∞)(z), for z ∈ C \ (U (0) ∪ U (a) ∪ (at, 0)),
(2.61)
where U (a) and U (0) are small discs centered at a and 0, respectively, that have been
introduced before. The parametrices P (∞), P (a) and P (0) are constructed below.
2.5.4 The parametrix P (∞)
The parametrix P (∞) is a solution of the following RH problem.
Riemann-Hilbert problem for P (∞)
• P (∞) : C \ [at, 0] → C2×2 is analytic.
• P (∞)+ (x) = P
− (x)
0 |x|2α
−|x|−2α 0
for x ∈ (at, 0), oriented from left to right.
• P (∞)(z) = I +O(1/z) as z → ∞.
The RH problem for P (∞) can be explicitly solved as in [9]. Take
D(z) = zα φ
2z − at
, for z ∈ C \ [at, 0], (2.62)
where φ(z) = z + (z − 1)1/2 (z + 1)1/2 is the conformal map from C \ [−1, 1] onto the
exterior of the unit circle. Then D+(x)D−(x) = |x|2α for x ∈ (at, 0). It follows that
D(∞)−σ3 P (∞)(z)D(z)σ3 satisfies the normalized RH problem with jump matrix ( 0 1−1 0 )
on (at, 0) (oriented from left to right), whose solution is well-known, see e.g. [11, 14], and
it leads to
P (∞)(z) = D(∞)σ3
(βt(z) + βt(z)
−1) 1
(βt(z)− βt(z)−1)
(βt(z)− βt(z)−1) 12 (βt(z) + βt(z)
D(z)−σ3 , (2.63)
for z ∈ C \ [at, 0], where
βt(z) =
z − at
, for z ∈ C \ [at, 0]. (2.64)
2.5.5 The parametrix P (a)
The parametrix P (a) is defined in the disc U (a) around a, where P (a) satisfies the following
RH problem.
24 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Riemann-Hilbert problem for P (a)
• P (a) : U (a) \ ΣS → C2×2 is analytic.
• P (a)+ (z) = P
− (z) vS(z) for z ∈ U (a) ∩ ΣS .
• P (a)(z)
P (∞)(z)
= I +O(n−1), as n→ ∞, uniformly for z ∈ ∂U (a) \ ΣS.
We seek P (a) in the form
P (a)(z) = P̂ (a)(z) enϕ̃t(z)σ3 (−z)−ασ3 , for z ∈ U (a) \ ΣS,
where (−z)−α is defined with a branch cut along [0,∞). Then P̂ (a) satisfies a RH problem
with constant jumps and can be constructed in terms of the Airy function in a standard
way; for more details see the presentation in [11].
2.5.6 The parametrix P (0)
The parametrix P (0), defined in the disk U (0) around 0, should satisfy the following RH
problem.
Riemann-Hilbert problem for P (0)
• P (0) : U (0) \ ΣS → C2×2 is continuous and analytic on U (0) \ ΣS .
• P (0)+ (z) = P
− (z) vS(z) for z ∈ ΣS ∩ U (0) (with the same orientation as ΣS).
• P (0)(z)
P (∞)(z)
= I+O(n−1/3), as n→ ∞, t→ 1 such that n2/3(t−1) = O(1),
uniformly for z ∈ ∂U (0) \ ΣS.
• P (0) has the same behavior near 0 as S has (see the RH problem for S).
A parametrix P (0) with these properties can be constructed using a solution Ψα of the
model RH problem of Subsection 1.2. The construction is done in three steps.
Step 1: Transformation to constant jumps. We seek P (0) in the form
P (0)(z) = P̂ (0)(z) enϕt(z)σ3 z−ασ3 , for z ∈ U (0) \ ΣS, (2.65)
where as usual z−α denotes the principal branch. It then follows from the RH problem
for P (0) that P̂ (0) should satisfy the following RH problem.
Riemann-Hilbert problem for P̂ (0)
• P̂ (0) : U (0) \ ΣS → C2×2 is continuous and analytic on U (0) \ ΣS .
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 25
• For z ∈ ΣS ∩ U (0), we have
+ (z) = P̂
− (z)×


, for z ∈ (0,∞) ∩ U (0),
e2απi 1
, for z in U (0) on the upper lip of the lens,
, for z ∈ (−∞, 0) ∩ U (0),
e−2απi 1
, for z in U (0) on the lower lip of the lens.
uniformly for z ∈ ∂U (0) \ ΣS.
• If α < 0, then
P̂ (0)(z) = O
|z|α |z|α
|z|α |z|α
as z → 0,
while if α ≥ 0 we have that
P̂ (0)(z) = O
|z|α |z|−α
|z|α |z|−α
as z → 0 from outside the lens, and
P̂ (0)(z) = O
|z|−α |z|−α
|z|−α |z|−α
as z → 0 from inside the lens.
Note that the jump matrices of P̂ (0) do not depend on t.
The reader may note the similarities between the above RH problem for P̂ (0) and the
RH problem for Ψα from Subsection 1.2. In the next step we show how we can use Ψα to
construct a solution of the RH problem for P̂ (0).
Step 2: The construction of P̂ (0) in terms of Ψα. Recall that ΣS in U
(0) was taken
such that z 7→ ft(z)− ft(0) maps ΣS ∩U (0) onto a subset of Σ, where Σ is the contour in
the RH problem for Ψα, see Subsection 1.2.
We choose any solution Ψα of the model RH problem and we define P̂
(0) by
P̂ (0)(z) = E(z) Ψα
3 (ft(z)− ft(0));n
3 ft(0)
, for z ∈ U (0) \ ΣS, (2.66)
where E = En,N is analytic in U
(0). Taking P (0) as in (2.65) with P̂ (0) as in (2.66) we find
that all the conditions of the RH problem for P (0) are satisfied, except for the matching
condition
P (0)(z)
P (∞)(z)
= I +O(n−1/3), (2.67)
as n→ ∞, t→ 1 such that n2/3(t− 1) = O(1).
26 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Step 3: Matching condition. To be able to satisfy (2.67) we have to take E in the
following way
E(z) = P (∞)(z) zασ3
3 (ft(z)− ft(0))
)σ3/4
, for z ∈ U (0) \ [at, 0],
(2.68)
where both branches are taken as principal. Clearly then E is analytic in U (0) \ [at, 0].
It turns out that E has analytic continuation to U (0). This follows by direct calculation,
but it relies on the fact that we chose [at, 0] as the jump contour for P
With the choice (2.68) for E, we now show that (2.67) is satisfied as well. By (2.65),
(2.66), we have for z ∈ ∂U (0) \ ΣS,
P (0)(z) = E(z) Ψα
3 (ft(z)− ft(0));n
3 ft(0)
enϕt σ3 z−ασ3
and we are interested in the behavior as n→ ∞, t→ 1 such that n2/3(t− 1) = O(1).
We show first that n2/3ft(0) remains bounded.
Lemma 2.7 Suppose n → ∞, t → 1 such that n2/3(t − 1) = O(1). Then n2/3ft(0)
remains bounded. More precisely, if n2/3(t− 1) → L ∈ R, then
n2/3ft(0) → −c2,V L = s, (2.69)
where
c2,V = (c1,V )
2/3dbt
(2.70)
and c1,V is the constant in (1.8).
Proof. It follows from (2.51), that
ft(0) =
−atht(bt)
(−bt)χ2/3t (0)
−ah1(0)
(t− 1)dbt
+O((t− 1)2) as t→ 1.
By (1.8) and (2.46), we have
c1,V =
−ah1(0), (2.71)
so that (2.69)–(2.70) indeed follows if n2/3(t− 1) → L. ✷
If we use the formula (2.47) for the t-derivative of bt at t = 1, then we find from (2.70)
c2,V = 2(−a)−1/2 c−1/31,V . (2.72)
Now we continue with the proof of (2.67).
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 27
Lemma 2.8 Suppose that n → ∞, t → 1 such that n2/3(t − 1) = O(1). Then (2.67)
holds.
Proof. In the proof all O-terms are for n→ ∞, t→ 1 such that n2/3(t− 1) is bounded.
By Lemma 2.7 the values n2/3ft(0) remain bounded. Since the asymptotic condition
(c) in the RH problem for Ψα is valid uniformly for s in bounded subsets of R, we find by
(2.65), (2.66), and (2.68)
P (0)(z) = E(z)
3 (ft(z)− ft(0))
)−σ3/4 1√
I +O(n−1/3)
× exp
−θ(n2/3(ft(z)− ft(0));n2/3ft(0))σ3
enϕtσ3z−ασ3
= P (∞)(z)(I +O(n−1/3)) exp
θ(n2/3(ft(z)− ft(0));n2/3ft(0))− nϕt
(2.73)
uniformly for z ∈ ∂U (0). As before we denote θ(ζ ; s) = 2
ζ3/2 + sζ1/2.
The next step is to evaluate the expression in the exponential factor. We have
θ(n2/3(ft(z)− ft(0));n2/3ft(0))− nϕt
(ft(z)− ft(0))3/2 − (ft(z))3/2
+ nft(0)(ft(z)− ft(0))1/2.
We will show that this is O(n−1/3) uniformly for z ∈ ∂U (0). To that end, it is enough to
show that
F (t, z) := (ft(z)− ft(0))3/2 − (ft(z))3/2 +
ft(0) (ft(z)− ft(0))1/2
= O((t− 1)2) as t→ 1, (2.74)
uniformly for z ∈ ∂U (0).
By (2.53), we have
F (1, z) = 0. (2.75)
Moreover,
F (t, z) =
(ft(z)− ft(0))
(ft(z)− ft(0))−
(ft(z))
ft(z)
ft(0)
(ft(z)− ft(0))
ft(0)
(ft(z)− ft(0))−
(ft(z)− ft(0)).
Let t = 1 and again use (2.53) and (2.51). Due to cancellations one finds
F (1, z) = 0. (2.76)
Since, in addition, F (t, z) is analytic in both variables and bounded with respect to z in
∂U (0), it follows from a Taylor expansion that F (t, z) = O((t− 1)2), as claimed in (2.74).
θ(n2/3(ft(z)− ft(0));n2/3ft(0))− nϕt = O(n−1/3),
28 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Figure 4: Contour for the RH problem for R.
so that (2.73) leads to
P (0)(z) = P (∞)(z)
I +O(n−1/3)
uniformly for z ∈ ∂U (0). Then (2.67) follows since P (∞)(z) and its inverse are bounded in
n and t, uniformly for z ∈ ∂U (0). ✷
This completes the construction of the parametrix P (0).
Remark 2.9 The local parametrix P (0) is constructed with the help of a solution Ψα of
the model RH problem. Since the solution Ψα is not unique (see Proposition 2.1), the
local parametrix is not unique. In what follows we can take any P (0) and it will not affect
the final results (Theorems 1.1 and 1.2).
2.5.7 Third transformation S 7→ R
Having P (∞), P (a), and P (0), we take P as in (2.61), and then we define
R(z) = S(z)P−1(z), for z ∈ C \ (∂U (0) ∪ ∂U (a) ∪ ΣS). (2.77)
Since S and P have the same jump matrices on U (0)∩ΣS, U (a)∩ΣS and (a, 0)\(U (0)∪U (a)),
we have that R is analytic across these contours. What remains are jumps for R on the
contour ΣR shown in Figure 4 with orientation that is also shown in the figure. Then, R
satisfies the following RH problem.
Riemann-Hilbert problem for R
• R : C \ ΣR → C2×2 is analytic.
• R+(z) = R−(z) vR(z) for z ∈ ΣR, where
P (∞) (P (0))−1, on ∂U (0),
P (∞) (P (a))−1, on ∂U (a),
P (∞) vS (P
(∞))−1, on ΣR \ (∂U (0) ∪ ∂U (a)).
(2.78)
• R(z) = I +O(1/z) as z → ∞.
Now let n → ∞, t → 1 such that n2/3(t − 1) = O(1). It then follows from the
construction of the parametrices (see in particular the RH problems for P (0) and P (a))
I +O(n−1/3), on ∂U (0),
I +O(n−1), on ∂U (a).
(2.79)
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 29
Furthermore, by regularity of the eigenvalue density, there is a constant c > 0 such that
Reϕt(z) > c > 0, for z ∈ ΣR ∩ (0,∞),
Re ϕ̃t(z) > c > 0, for z ∈ ΣR ∩ (−∞, a),
Reϕt(z) < −c < 0, for z ∈ ΣR \ (∂U (0) ∪ ∂U (a) ∪ R).
This implies (see the RH problem for S) that vS = I+O(e
−cn) uniformly on ΣR \ (∂U (0)∪
∂U (a)), so that by (2.78)
vR = I +O(e
−2cn) on ΣR \ (∂U (0) ∪ ∂U (a)). (2.80)
The O-terms in (2.79) and (2.80) are uniform on the indicated contours. In addition, it
follows from (2.58), (2.59), (2.55), and the growth condition (1.2) on V that for any C > 0
there exists r = r(C) > 1 such that ϕt(x) ≥ C log x for x ≥ r, and ϕ̃t(x) ≥ C log |x| for
x ≤ −r. Combined with (2.80) this implies that
||vR − I||L2(ΣR\(∂U (0)∪∂U (a))) = O(e
−2cn), as n→ ∞. (2.81)
Thus, by (2.79)–(2.81), as n → ∞ and t → 1 such that n2/3(t − 1) = O(1), the jump
matrix for R is close to I in both L2 and L∞ norm on ΣR, indeed
||vR − I||L2(ΣR)∩L∞(ΣR) = O(n−1/3). (2.82)
Standard estimates using L2-boundedness of the operators C± on L
2(ΣR) together with
the correspondence between RH problems and singular integral equations now imply that
R(z) = I +O(n−1/3), uniformly for z ∈ C \ ΣR, (2.83)
as n → ∞, t → 1 such that n2/3(t − 1) = O(1). To get the uniform bound (2.83) up to
the contour one needs a contour deformation argument. Again, see the presentation in
[11] for more details.
This completes the steepest descent analysis of the RH problem for Y .
2.6 Completion of the proofs of Theorem 1.1 and 1.2
Having completed the steepest descent analysis we are now ready for the proofs of Theorem
1.1 and 1.2. We start by rewriting the kernel (2.3) for x, y ∈ U (0) ∩ R according to the
transformations Y 7→ T 7→ S 7→ R that we did in the steepest descent analysis. To state
the result it is convenient to introduce B = Bn,N as
B(z) = R(z)E(z), for z ∈ U (0), (2.84)
where E and R are defined in (2.68) and (2.77). We also define for x, s ∈ R the column
vector
~ψα(x; s) =
ψ1(x; s)
ψ2(x; s)
Ψα,+(x; s)
, for x > 0,
Ψα,+(x; s)e
−απiσ3
, for x < 0,
(2.85)
cf. (1.12). We then have the following result.
30 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Lemma 2.10 Let x, y ∈ U (0) ∩ R. Then,
Kn,N(x, y) =
2πi(x− y)
3 (ft(y)− ft(0));n
3 ft(0)
))T ( 0 1
× B−1(y)B(x) ~ψα
3 (ft(x)− ft(0));n
3 ft(0)
. (2.86)
Proof. We start from the formula (2.3) for the eigenvalue correlation kernel. Using
(2.56) we obtain, for any x, y ∈ R,
Kn,N(x, y) = |x|α e
n(2gt,+(x)−Vt(x)+lt) |y|α e 12n(2gt,+(y)−Vt(y)+lt)
2πi(x− y)
T−1+ (y) T+(x)
. (2.87)
Using (2.58) and the fact that gt,+ = gt,− on (bt,∞), it follows that 2gt − Vt + lt = −2ϕt
on (bt,∞). Then, by analytic continuation, 2gt,+−Vt+ lt = −2ϕt,+ on all of R. Therefore
we can rewrite (2.87) as
Kn,N(x, y) = |x|α e−nϕt,+(x) |y|α e−nϕt,+(y)
2πi(x− y)
T−1+ (y) T+(x)
(2.88)
Now we analyze the effect of the transformations T 7→ S 7→ R on the expression
|x|αe−nϕt,+(x)T+(x) ( 10 ) in case x ∈ U (0) ∩ R. The result is that for x ∈ U (0) ∩ R,
|x|αe−nϕt,+(x)T+(x)
= B(x)Ψα,+
3 (ft(x)− ft(0));n
3 ft(0)
(2.89)
in case x > 0, and
|x|αe−nϕt,+(x)T+(x)
= B(x)Ψα,+
3 (ft(x)− ft(0));n
3 ft(0)
e−απiσ3
(2.90)
in case x < 0. Since the calculations for (2.89) are easier, we will only show how to obtain
(2.90). If x ∈ U (0) ∩ R and x < 0, then it follows from (2.60) that
|x|αe−nϕt,+(x)T+(x)
= |x|αe−nϕt,+(x)S+(x)
|x|−2αe2nϕt,+(x)
= S+(x)
|x|αe−nϕt,+(x)
. (2.91)
From (2.77), (2.61), (2.65), (2.68), and (2.84), we find that
S+(x) = B(x)Ψα,+
3 (ft(x)− ft(0));n
3ft(0)
enϕt(x)x−α
Inserting this into (2.91) and noting that x−α+ |x|α = e−απi we indeed obtain (2.90).
In a similar way, we find for y ∈ U (0) ∩ R,
|y|αe−nϕt,+(y)
T−1+ (y) =
Ψ−1α,+
3 (ft(y)− ft(0));n
3ft(0)
B−1(y),
(2.92)
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 31
in case y > 0, and
|y|αe−nϕt,+(y)
T−1+ (y) =
eαπiσ3Ψ−1α,+
3 (ft(y)− ft(0));n
3ft(0)
B−1(y),
(2.93)
in case y < 0. To rewrite (2.92) and (2.93) we use the following fact, which is easy to
check. If A is an invertible 2× 2 matrix having determinant 1, then
A−1 =
. (2.94)
If we apply (2.94) to Ψα,+ in (2.92) and (2.93), then we get
|y|αe−nϕt,+(y)
T−1+ (y) =
ΨTα,+
3 (ft(y)− ft(0));n
3ft(0)
B−1(y), (2.95)
in case y > 0, and
|y|αe−nϕt,+(y)
T−1+ (y) =
e−απiσ3ΨTα,+
3 (ft(y)− ft(0));n
3 ft(0)
B−1(y), (2.96)
in case y < 0. Then (2.86) follows if we insert (2.89), (2.90), (2.95), and (2.96) into (2.88)
and use the definition (2.85). ✷
As in Theorem 1.1 we now fix x, y ∈ R. We define
(c1n)2/3
, and yn =
(c1n)2/3
(2.97)
where c1 is the constant from (1.8).
In order to take the limit of (c1n)
−2/3Kn,N(xn, yn) we need one more lemma. Recall
that B = RE is defined in (2.84).
Lemma 2.11 Let n→ ∞, t→ 1 such that n2/3(t− 1) → L. Let x, y ∈ R and let xn and
yn defined as in (2.97), Then the following hold.
(a) n2/3ft(0) → s,
(b) n2/3(ft(xn)− ft(0)) → x and n2/3(ft(yn)− ft(0)) → y,
(c) B−1(yn)B(xn) = I +O
where the implied constant in the O-term is uniform
with respect to x and y.
Proof. (a) This follows from Lemma 2.7.
(b) By (1.8) and (2.46) we have c1 =
−ah1(0), so that f ′1(0) = c
1 by (2.52).
Taking note of the definitions (2.97), we then obtain part (b), since ft → f1 uniformly in
U (0).
32 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
(c) We have
R−1(yn)R(xn) = I +R
−1(yn)(R(xn)− R(yn))
= I +R−1(yn) (xn − yn)
R′(txn + (1− t)yn)dt. (2.98)
Recall that R is analytic in U (0), and that R(z) = I + O(n−1/3) by (2.83), uniformly in
U (0). Since detR ≡ 1, we find that R−1(yn) remains bounded as n → ∞. It also follows
from (2.83) and Cauchy’s theorem, that R′(z) = O(n−1/3) for z in a neighborhood of the
origin. By (2.98) we then obtain
R−1(yn)R(xn) = I +O
. (2.99)
Using analyticity of E in a neighborhood of the origin with E(z) = O(n1/6), see (2.68),
and the fact that detE ≡ 1, we obtain in the same way
E−1(yn)E(xn) = I +O
. (2.100)
The implied constants in (2.99) and (2.100) are independent of x and y.
Using (2.99), (2.100), and the fact that E(xn) = O(n
1/6) and E−1(yn) = O(n
1/6), we
obtain from (2.84)
B−1(yn)B(xn) = E
−1(yn)
E(xn)
= E−1(yn)E(xn) +O(n
1/6)O
O(n1/6)
= I +O
This completes the proof of part (c). ✷
Proof of Theorems 1.1 and 1.2. We let n,N → ∞, t = n/N → 1, in such a way that
n2/3(t− 1) → L. Then by parts (a) and (b) of Lemma 2.11, we have
~ψα(n
2/3(ft(xn)− ft(0));n2/3ft(0)) → ~ψα(x; s)
and similarly if we replace xn by yn. The existence of the limit (1.9) then follows easily
from Lemma 2.10 and part (c) of Lemma 2.11, which proves Theorem 1.1.
We also find that the limiting kernel Kedgeα (x, y; s) is given by
Kedgeα (x, y; s) =
2πi(x− y)
~ψα(y; s)
~ψα(x; s)
and so (1.13) follows because of (2.85). The model RH problem is solvable for every s ∈ R
by Proposition 2.4 and so we have also proved Theorem 1.2. ✷
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 33
S5 S6❄
Figure 5: Contour for the RH problem for ΨFNν .
3 Proof of Theorems 1.4 and 1.5
We prove Theorem 1.4 and Theorem 1.5 by first establishing, with the help of [2], a
connection between the model RH problem and the RH problem for Painlevé II in the
form due to Flaschka and Newell [22]. We can then use known properties of the RH
problem for Painlevé II to prove the theorems.
3.1 The Painlevé II RH problem
We review the RH problem for the Painlevé II equation q′′(s) = sq+2q3−ν, as first given
by Flaschka and Newell [22], see also [24] and [25]. We will assume that
ν > −1/2.
The RH problem involves three complex constants a1, a2, a3 satisfying
a1 + a2 + a3 + a1a2a3 = −2i sin νπ, (3.1)
and certain connection matrices Ej .
Let Sj = {w ∈ C | 2j−36 π < argw <
π} for j = 1, . . . , 6, and let ΣFN = C \
j Sj .
Then ΣFN consists of six rays ΣFNj for j = 1, . . . , 6, all chosen oriented towards infinity
as in Figure 5. The RH problem is the following.
Riemann-Hilbert problem for ΨFNν
• ΨFNν : C \ ΣFN → C2×2 is analytic,
• ΨFNν,+ = ΨFNν,−
on ΣFN1 ,
ΨFNν,+ = Ψ
on ΣFN2 ,
ΨFNν,+ = Ψ
on ΣFN3 ,
34 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
ΨFNν,+ = Ψ
on ΣFN4 ,
ΨFNν,+ = Ψ
on ΣFN5 ,
ΨFNν,+ = Ψ
on ΣFN6 .
• ΨFNν (w) = (I +O(1/w))e−i(
w3+sw)σ3 as w → ∞.
• If ν − 1
6∈ N0, then
ΨFNν (w) = B(w)
0 w−ν
Ej , for w ∈ Sj , (3.2)
where B is analytic. If ν ∈ 1
+ N0, then there exists a constant κ such that
ΨFNν (w) = B(w)
wν κwν logw
0 w−ν
Ej , for w ∈ Sj, (3.3)
where B is analytic.
The connection matrix E1 is given explicitly in [24, Chapter 5]. It is determined (up
to inessential left diagonal or upper triangular factors) by ν and the Stokes multipliers
a1, a2, and a3, except in the special case
+ n, a1 = a2 = a3 = i(−1)n+1, n ∈ Z, (3.4)
where an additional parameter c ∈ C ∪ {∞} is needed. For example, for ν 6∈ 1
+ N0 and
1 + a1a2 6= 0, we have
0 d−1
e−νπi − a2
1 + a1a2
−1 + a1a2
2 cos νπ
eνπi + a2
2 cos νπ
 , (3.5)
where d 6= 0 is arbitrary. In the special case (3.4), when E1 depends on the additional
parameter c ∈ C ∪ {∞}, by [24, Chapter 5, (5.0.21)] we may take E1 as
, if c ∈ C, while E1 =
if c = ∞. (3.6)
Assuming that the branch cuts for the functions in (3.2) and (3.3) are chosen along
argw = −π/6, we obtain the other connection matrices from E1 through the formula
Ej+1 = Ejv
j , j = 1, . . . , 5, (3.7)
where vFNj is the jump matrix on Σ
j . We shall refer to the Stokes multipliers a1, a2,
and a3, and in the special case (3.4) also to the additional parameter c, as the monodromy
data for Painlevé II. We note that in the special case (3.4) we have κ = 0 in (3.3).
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 35
The special case (3.4) has geometric interpretation. Indeed, (3.4) describes the singular
point of the algebraic variety (3.1), that is, the point at which the (complex) gradient of
the left-hand side of (3.1) vanishes. The singularity may be removed by attaching a copy
of the Riemann sphere (see also [30]).
The monodromy data does not depend on s. The RH problem is uniquely solvable,
except for a discrete set of s-values, and its solution ΨFNν depends on s through the asymp-
totic condition at infinity. We write ΨFNν (w; s) if we want to emphasize its dependence
on s. If we take
q(s) = 2i lim
ΨFNν (w; s)
w3+sw)σ3, (3.8)
then q satisfies the Painlevé II equation q′′ = sq + 2q3 − ν. In addition ΨFNν satisfies the
Lax pair for Painlevé II
Ψ = LΨ, L =
(−4iw2 − i(s + 2q2) 4wq + 2ir + ν
4wq − 2ir + ν
4iw2 + i(s+ 2q2)
, (3.9)
Ψ = PΨ, P =
−iw q
, (3.10)
where q = q(s) and r = r(s) = q′(s). In this way there is a one-to-one correspondence
between monodromy data and solutions of Painlevé II.
We also need the more precise asymptotic behavior
ΨFNν (w; s) =
H(s) q(s)
−q(s) −H(s)
+O(1/w2)
w3+sw)σ3 (3.11)
as w → ∞, where
H(s) = (q′(s))2 − sq2(s)− q4(s) + 2νq(s) (3.12)
is the Hamiltonian for Painlevé II. Note that H ′ = −q2.
We finally note that ΨFNν satisfies the symmetry property
ΨFNν (w; s) = σ1Ψ
ν (−w; s)σ1, (3.13)
where σ1 = (
1 0 ). Indeed, by a straightforward calculation (see also [24, Chapter 5])
we check that the function σ1Ψ
ν (−w; s)σ1 solves exactly the same RH problem as the
function ΨFNν (w; s). Unique solvability of the RH problem yields equation (3.13).
3.2 Connection with Ψα
The Hastings-McLeod solution of Painlevé II corresponds to the Stokes multipliers a1 =
−eνπi, a2 = 0, and a3 = e−νπi. This is not the solution that interests us here. We use
instead the solution corresponding to
a1 = e
−νπi, a2 = −i, a3 = −eνπi. (3.14)
For these Stokes multiplies (3.14) we obtain from (3.5) the following connection matrix
E1 in case ν 6∈ 12 + N0 (where we take d = (e
νπi − i)/(2 cos νπ))
eνπi − i
2 cos νπ
ieνπi + 1
2 cos νπ
−e−νπi 1
 . (3.15)
36 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
For ν ∈ 3
+ 2N0, it follows from (3.14) and the formulas in [24, Chapter 5, (5.0.18)] that
we can take
. (3.16)
If ν ∈ 1
+ 2N0, then we are in the special case (3.4). We then choose c = i, so that for
ν ∈ 1
+ 2N0 we have monodromy data
a1 = e
−νπi = −i, a2 = −i, a3 = −eνπi = −i, c = i. (3.17)
Lemma 3.1 For any ν > −1/2, we have that
(E2)21 = (E3)21 = 0. (3.18)
Proof. In all cases we may check from (3.14), (3.15), (3.16), (3.17), and (3.6) that the
second row of E1 is given by
−a1 1
. So by (3.7) we have that E2 = E1 (
a1 1 ) is upper
triangular. Then also E3 = E2 (
0 1 ) is upper triangular and therefore (3.18) holds. ✷
The following proposition holds for more general monodromy data, and it was estab-
lished in [2], see also [35]. For the reader’s convenience we present a detailed proof for
our particular case.
Proposition 3.2 ([2]) For α > −1/2, let ΨFN2α+1/2 be the unique solution of the RH prob-
lem for Painlevé II with parameter ν = 2α + 1/2 and monodromy data (3.14) in case
α 6∈ N0 (so that ν 6∈ 12 + 2N0), and monodromy data (3.17) in the special case α ∈ N0.
Then, for any η = η(s), we have that
Ψα(ζ ; s) =
η(s) 1
ζ−σ3/4
eπiσ3/4ΨFN2α+1/2(w;−21/3s)e−πiσ3/4 (3.19)
where w = eπi/22−1/3ζ1/2 with Imw > 0, is a solution of the model RH problem for Ψα
given in Subsection 1.2.
Proof. Because of Proposition (2.1) we may take η(s) = 0 without loss of generality.
Clearly Ψα is analytic on C\Σ. The correct asymptotics as ζ → ∞ follows immediately,
as well as the correct jumps across Σ1, Σ2, and Σ4. A little bit more work is needed to
check the jump across Σ3 = (−∞, 0) and the behavior at z = 0.
In order to analyze the jump across Σ3, we suppose that ζ ∈ Σ3. Then we have that
w+ ≡ eπi/22−1/3ζ1/2+ = −w− ≡ −eπi/22−1/3ζ
− (< 0),
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 37
and hence by (3.19) and the symmetry property (3.13),
Ψα,+(ζ ; s) = ζ
−σ3/4
eπiσ3/4ΨFN2α+1/2(w+;−21/3s)e−πiσ3/4
−σ3/4
eπiσ3/4ΨFN2α+1/2(−w−;−21/3s)e−πiσ3/4
−σ3/4
eπiσ3/4σ1Ψ
2α+1/2(w−;−21/3s)σ1e−πiσ3/4
−σ3/4
eπiσ3/4ΨFN2α+1/2(w−;−21/3s)e−πiσ3/4
−σ3/4
eπiσ3/4ΨFN2α+1/2(w−;−21/3s)e−πiσ3/4
= Ψα,−(ζ ; s)
This shows that Ψα has the correct jump across Σ3, and it follows that Ψα satisfies the
parts (a), (b), and (c) of the model RH problem.
Consider now a neighborhood of the point ζ = 0. We recall that (3.2) or (3.3) holds
with B(w) analytic at 0. A corollary of the symmetry property (3.13) is the equation
B(w) =
σ1B(−w)σ3, if ν 6∈ 12 + N0,
σ1B(−w)
1 O(w2ν)
as w → 0, if ν ∈ 1
+ N0,
which yields the formula (cf. [24, Chapter 5])
B(0)σ3 = σ1B(0).
The last relation, together with detB(0) = 1, in turn implies that B(0) can be represented
in the form
B(0) =
0 b−1
, b 6= 0.
If ζ ∈ Ωj then w ∈ Sπ(j), j = 1, 2, 3, 4, where π denotes the permutation
1 2 3 4
3 4 1 2
Therefore, for the function Ψα(ζ ; s) defined by equation (3.19) with η(s) = 0, we find that
(assuming that α 6∈ 1
Ψα(ζ ; s) = ζ
−σ3/4 1√
eπiσ3/4B(0)e−πiσ3/4
I +O(ζ1/2)
ζσ3/4ζασ3Ẽπ(j)
= ζ−σ3/4
0 b−1
I +O(ζ1/2)
ζσ3/4ζασ3Ẽπ(j)
= ζ−σ3/4
I +O(ζ1/2)
ζσ3/4ζασ3
0 b−1
Ẽπ(j)
= O(1)ζασ3
0 b−1
Ẽπ(j), as ζ → 0 in Ωj , (3.20)
38 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
where we have introduced the notation
Ẽj ≡ eπiσ3/4
eπi/22−1/3
)(2α+1/2)σ3
−πiσ3/4. (3.21)
From (3.20) it immediately follows that Ψα(ζ ; s) = O(ζ
−|α|) as ζ → 0, which is the
required behavior in the model RH problem if α < 0, or if α ≥ 0 and j ∈ {2, 3}. If α ≥ 0
and j ∈ {1, 4}, then π(j) ∈ {2, 3}, and it follows from Lemma 3.1 and (3.21) that
Ẽπ(j)
Eπ(j)
Then (3.20) also yields the required behavior of Ψα(ζ ; s) as ζ → 0 in Ω1 ∪ Ω4.
The calculation leading to (3.20) is valid for ν 6∈ 1
+N0, or α 6∈ 12N0. In fact it is also
valid if α ∈ N0, since then we are in the special case (3.4) where κ = 0 in (3.3) and so
no logarithmic terms appear. Logarithmic terms only appear if α ∈ 1
+ N0, and then a
similar calculation leads to
Ψα(ζ ; s) = O(1)ζ
0 b−1
1 O(log ζ)
Ẽπ(j),
with Ẽj again given by (3.21). Since α > 0, the required behavior as ζ → 0 then follows
in a similar way.
This completes the proof of the proposition. ✷
3.3 Differential equation
Recall that ΨFNν has the Lax pair (3.9) and (3.10). Then Ψα defined by (3.19) also satisfies
a system of differential equations. It will involve the solution q of the Painlevé II equation
with parameter ν = 2α + 1/2 and monodromy data (3.14) or (3.17). We put r = q′ and
U(s) = q2(s) + r(s) +
, (3.22)
V (s) = q2(s)− r(s) + s
. (3.23)
The functions U and V both satisfy the Painlevé XXXIV equation in a form similar to
(1.19), namely (cf. [24, Chapter 5]):
U ′′(s) =
(U ′(s))2
2U(s)
+ 2U2(s)− sU(s)− (2α)
2U(s)
, (3.24)
V ′′(s) =
(V ′(s))2
2V (s)
+ 2V 2(s)− sV (s)− (2α + 1)
2V (s)
. (3.25)
Then we obtain the following differential equations for Ψα.
Lemma 3.3 Let Ψα be given by (3.19).
(a) If η ≡ 0, then Ψα satisfies
Ψα(ζ ; s) = AΨα(ζ ; s), (3.26)
Ψα(ζ ; s) = BΨα(ζ ; s), (3.27)
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 39
where
−21/3q(−21/3s) + α
i− i2−1/3U(−21/3s)1
−iζ + i2−1/3V (−21/3s) 21/3q(−21/3s)− α
, (3.28)
−21/3q(−21/3s) i
−iζ 21/3q(−21/3s)
. (3.29)
(b) For general η we have that Ψα satisfies
Ψα(ζ ; s) =
η(s) 1
−η(s) 1
Ψα(ζ ; s), (3.30)
with A given by (3.28).
Proof. This follows by straightforward calculations from (3.9), (3.10), and (3.19). ✷
The Lax pair (3.26)–(3.27), after the replacement ζ 7→ ζ − s, becomes the Lax pair
from [2, 35]. Equations (3.24)–(3.25) can also be derived directly from the compatibility
conditions of the Lax pair (3.26)–(3.27) in a usual way.
It is a fact [33], that the solution q of the Painlevé II equation (with parameter
ν = 2α+1/2 and monodromy data (3.14) or (3.17)) has an infinite number of poles on the
positive real line, see also (4.29) below. If −21/3s is such a pole then ΨFN2α+1/2(·,−21/3s)
does not exist. So to be precise, if we assume that η is analytic on R, then (3.19) does not
define Ψα for values of s ∈ R which belong to the discrete set of values s where q(−21/3s)
has poles.
The relation (3.19) defines Ψα for all s ∈ R only if we are able to choose η so that all
the poles on the real line of the right-hand side of (3.19) cancel out. Such a choice of η
would require η itself to have poles at the poles of q(−21/3s).
We will describe two special choices for η. The first choice is such that (3.19) is equal
to the special solution Ψ
(spec)
α , which is characterized by the asymptotic condition (2.14).
From Proposition 2.4 we know that Ψ
(spec)
α exists for all s ∈ R, so that we can already
conclude that the special choice η = η(spec) will have poles at the poles of q(−21/3s), and
that the real poles of the right-hand side of (3.19) will indeed cancel out.
The second choice of η is made so that the differential equation (3.30) takes a nice
form. It will lead to the differential equation (1.21) for ψ1 and ψ2. This η is denoted η0,
and it is defined by the simple formula
η0(s) = i2
1/3q(−21/3s), (3.31)
from which it is already clear that it has poles at the poles of q(−21/3s). For the choice
(3.31) we can already check that the differential equation (3.30) leads to
Ψα(ζ ; s) = A0Ψα(ζ ; s) (3.32)
where
−η0 1
(α + iuη0)/ζ i− iu/ζ
−iζ + i(v + η20) + η0(2α + iuη0)/ζ −(α + iuη0)/ζ
, (3.33)
40 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
u(s) = 2−1/3U(−21/3s), (3.34)
v(s) = 2−1/3V (−21/3s). (3.35)
3.4 Special choice η(spec)
Lemma 3.4 Let H be the Hamiltonian for Painlevé II as in (3.12), with parameter ν =
2α + 1/2, and let
η(spec)(s) = i2−2/3
q(−21/3s) +H(−21/3s)
. (3.36)
Then the choice η = η(spec) in (3.19) leads to the special solution Ψ
(spec)
α of the model RH
problem characterized by (2.14).
Proof. It follows from (3.11) and (3.19) by straightforward computation, that
Ψα(ζ ; s) =
ζ−σ3/4
eπiσ3/4ΨFN2α+1/2(w;−21/3s)e−πiσ3/4
−η(spec) 1
0 i2−2/3(H − q)(−21/3s)
+ ζ−σ3/4
eπiσ3/4O(1/ζ)e−πiσ3/4
ζσ3/4
× ζ−σ3/4 1√
ζ3/2+sζ1/2)σ3 (3.37)
as ζ → ∞. From (3.37) it is clear that we need to take η = η(spec) in order to be able to
obtain (2.14). Thus the lemma follows. ✷
From the calculation (3.37) we also note that for any solution Ψα of the model RH
problem we have
Ψα(ζ ; s)e
(ζ3/2+sζ1/2)σ3
ζσ3/4
i2−2/3(H − q)(−21/3s)
+O(ζ−3/2)
(3.38)
as ζ → ∞. This property will be used later in the proof of Theorem 1.4.
Since the left-hand side of (3.38) is analytic in s for s ∈ R, it also follows from (3.38)
that H − q does not have poles on the real line. This and similar properties are collected
in the following lemma. Recall that U is given by (3.22).
Lemma 3.5 The following hold.
(a) H − q has no poles on the real line.
(b) U has no poles on the real line.
(c) U has a zero at each of the real poles of q and Uq has no poles on the real line.
(d) Uq takes the value ν − 1/2 at each of the real poles of q.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 41
Proof. (a) We noted already that part (a) follows from (3.38).
(b) Since H ′ = −q2, we have that
U(s) = q2(s) + q′(s) + s/2 = −(H − q)′(s) + s/2, (3.39)
and so it follows from part (a) that U has no poles on the real line either.
(c) Differentiating (3.22), we obtain
U ′ = 2qq′ + q′′ +
= 2qq′ + sq + 2q3 − ν + 1
= 2Uq − ν + 1
. (3.40)
Thus also Uq has no poles on the real line, which means that U has a zero at each of the
real poles of q.
(d) Using (3.40), we get
(Uq − ν + 1
)q = (U ′ − Uq)q = (Uq)′ − U(q2 + q′) = (Uq)′ − U(U − s/2). (3.41)
Since the right-hand side of (3.41) is analytic on the real line by parts (b) and (c), we
conclude that Uq−ν+ 1
has a zero at each of the real poles of q. This proves part (d). ✷
It is well-known and easy to check that each pole of q is simple and has residue +1 or
−1. Indeed, the Laurent series for q at a pole s0 has the form
q(s) =
s− s0
+ q1(s− s0) + · · · ,
where q−1 ∈ {−1, 1}. Using this, one easily verifies that either q2+ q′ or q2− q is analytic
at s0 (depending on the sign of the residue q−1). Our result that U = q
2 + q′ + s/2 is
analytic on R can then also be stated as follows.
Corollary 3.6 The solution q of the Painlevé II equation with parameter ν = 2α + 1/2
and monodromy data (3.14) or (3.17) has only simple poles on the real line, with residue
3.5 Special choice η0
As already announced we will also use the special choice η = η0 given by (3.31).
By (3.31) and (3.36) we have that
η0(s)− η(spec)(s) = i2−2/3
q(−21/3s)−H(−21/3s)
and so it follows from part (a) of Lemma 3.5 that η0 − η(spec) is analytic on the real line.
Since Ψ
(spec)
α exists for all s ∈ R, it follows that the solution of the model RH problem
associated with η0 exists for all s ∈ R as well, and it is analytic in s.
The differential equation for Ψα with η = η0 is given by (3.32) with A0 as in (3.33).
It then follows that A0 is analytic on the real line, and we will explicitly verify this by
rewriting its entries in terms of the function u from (3.34)
u(s) = 2−1/3U(−21/3s).
42 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
The analyticity of u is immediate from (3.34) and part (b) of Lemma 3.5. The analyticity
of uη0 follows from (3.34), (3.31) and part (c) of Lemma 3.5. Using also (3.40) we get
u′ = 2iuη0 + ν − 1/2 = 2iuη0 + 2α. (3.42)
Next, it follows from (3.22), (3.23), (3.34), (3.35), and (3.31) that
v(s) + η0(s)
2 = −u(s)− s. (3.43)
We can use (3.42) and (3.43) to eliminate η0 and v from the entries in A0, and we get
from (3.33) that
u′/(2ζ) i− iu/ζ
−iζ − i(u+ s)− i((u′)2 − (2α)2)/(4uζ) −u′/(2ζ)
. (3.44)
3.6 Proof of Theorem 1.4 and 1.5
After these preparations the proofs of Theorems 1.4 and 1.5 are short.
Proof of Theorem 1.4. From (3.24) and (3.34) it follows that u satisfies the Painlevé
XXXIV equation in the form (1.19).
From (3.38) it follows that
Ψα(ζ ; s)e
(ζ3/2+sζ1/2)σ3
ζσ3/4
= i2−2/3(H − q)(−21/3s)
which in view of (3.39) and (3.34) leads to (1.20). This proves Theorem 1.4. ✷
Proof of Theorem 1.5. Let Ψα be the solution of the model RH problem given by
(3.19) with η = η0 as in (3.31). Then
Ψα(ζ ; s) = A0Ψα(ζ ; s), (3.45)
with A0 given by (3.44). The differential equation (3.45) is valid for ζ ∈ C \ Σ. We can
take the limit ζ → x with x ∈ R\{0} to obtain a differential equation for Ψα,+(x; s), with
the same matrix A0 (but with ζ replaced by x). Using (1.12), we obtain the differential
equation (1.21) for ψ1 and ψ2. This completes the proof of Theorem 1.5. ✷
4 Concluding remarks
4.1 The case α = 0
The case α = 0 is classical and well understood. We know that K
0 (x, y; s) is the
(shifted) Airy kernel, see (1.11). We will show here how this follows from the calculations
from the previous section.
In the special case α = 0, we have ν = 1/2, and then the Painlevé II equation has
special solutions built out of Airy functions. To be precise if Ai and Bi are the standard
Airy functions, then for any C1 and C2, not both zero, we have that
q(s) =
C1Ai(−2−1/3s) + C2Bi(−2−1/3s)
(4.1)
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 43
is a solution of q′′ = sq + 2q3 − 1
. These are exactly the solutions that correspond to the
special Stokes multipliers a1 = a2 = a3 = −i. The corresponding solutions to the RH
problem were given by Flaschka and Newell [22, Section 3F(iv)]. For example, for w in
sector S1 we have (see also [24, Chapter 11])
ΨFN1/2 (w; s) =
1− iq(s)/w −2−1/3i/w
1 + iq(s)/w 2−1/3i/w
Ai(z) Bi(z)
Ai′(z) Bi′(z)
(4.2)
with z = −22/3w2 − 2−1/3s and α0 = 21/6
πeiπ/4. The expressions for ΨFNν (w; s) in the
other sectors follow by multiplying (4.2) by the appropriate jump matrices.
It follows from (4.1) and (4.2) that the extra parameter c in the monodromy data for
(4.1) is
iC1 − C2
C1 − iC2
. (4.3)
So if we take c = i as in (3.17) then C2 = 0 and the corresponding solution (4.1) is
q(s) =
log Ai(−2−1/3s) = −2−1/3Ai
′(−2−1/3s)
Ai(−2−1/3s) . (4.4)
Note that the solution (4.4) is special among all solutions (4.1) in its behavior for s→ −∞.
Indeed, from the asymptotic behavior for the Airy functions it follows that for (4.4) we
q(s) ∼ 1
2(−s)1/2 as s→ −∞,
while for the other solutions (4.1) we have
q(s) ∼ −1
2(−s)1/2 as s→ −∞.
So according to Proposition 3.2 we should be using q given by (4.4) and then define
Ψ0 as in (3.19). If ζ is in sector Ω3, then w = e
iπ/22−1/3ζ1/2 is in sector S1, so that by
(4.2)
ΨFN1/2 (w;−21/3s) =
1 + iη0(s)ζ
−1/2 −ζ−1/2
1− iη0(s)ζ−1/2 ζ−1/2
Ai(ζ + s) Bi(ζ + s)
Ai′(ζ + s) Bi′(ζ + s)
where η0(s) = i2
1/3q(−21/3) as in (3.31). Then (3.19) with η = η0 yields for ζ ∈ Ω3,
Ψ0(ζ ; s) =
η0(s) 1
0 ζ1/2
eπiσ3/4
1 + iη0(s)ζ
−1/2 −ζ−1/2
1− iη0(s)ζ−1/2 ζ−1/2
Ai(ζ + s) Bi(ζ + s)
Ai′(ζ + s) Bi′(ζ + s)
e−πiσ3/4
eπiσ3/4
Ai(ζ + s) Bi(ζ + s)
Ai′(ζ + s) Bi′(ζ + s)
e−πiσ3/4
Ai(ζ + s) + iBi(ζ + s) −(Ai(ζ + s)− iBi(ζ + s))
−i(Ai′(ζ + s) + iBi′(ζ + s)) i(Ai′(ζ + s)− iBi′(ζ + s))
44 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
Since (see e.g. formula (10.4.9) in [1])
Ai(z)± iBi(z) = 2e±πi/3Ai(e∓2πi/3z)
we can write Ψ0 in the more familiar form
Ψ0(ζ ; s) =
eπi/3Ai(e−2πi/3(ζ + s)) −e−πi/3Ai(e2πi/3(ζ + s))
−ie−πi/3Ai′(e−2πi/3(ζ + s)) ieπi/3Ai′(e2πi/3(ζ + s))
, for ζ ∈ Ω3.
(4.5)
For ζ ∈ Ω1 we find in a similar way (or by multiplying (4.5) on the right by ( 1 1−1 0 )), that
Ψ0(ζ ; s) =
Ai(ζ + s) eπi/3Ai(e−2πi/3(ζ + s))
−iAi′(ζ + s) −ie−πi/3Ai′(e−2πi/3(ζ + s))
, for ζ ∈ Ω1. (4.6)
Then it follows from (1.12) and (4.6) that for x > 0,
ψ1(x; s) =
2πAi(x+ s), ψ2(x; s) = −
2πiAi′(x+ s), (4.7)
and a similar calculation shows that (4.7) also holds for x < 0. Therefore, by (1.13)
0 (x, y; s) =
ψ2(x; s)ψ1(y; s)− ψ1(x; s)ψ2(y; s)
2πi(x− y)
Ai(x+ s)Ai′(y + s)− Ai′(x+ s)Ai(y + s)
x− y , (4.8)
which is indeed the (shifted) Airy kernel.
4.2 The case α = 1
The case α = 1 can be solved explicitly in terms of Airy functions as well. Let Ψ0 be a
solution of the model RH problem with parameter α = 0. Then for any matrix X = X(s),
it is easy to check that
Ψ1(ζ ; s) = (I −
X(s))Ψ0(ζ ; s) (4.9)
satisfies the conditions (a), (b), and (c) of the model RH problem for α = 1. For a special
choice of X we will have that the condition (d) is also satisfied.
Let’s take Ψ0 given by (4.6) for ζ ∈ Ω1. Then the condition (d) of the model RH
problem yields the following condition on X
(I − 1
X(s))
Ai(ζ + s)
−iAi′(ζ + s)
= O(ζ), as ζ → 0. (4.10)
The condition (4.10) is satisfied if and only if we take
X(s) =
Ai′(s)2 − sAi(s)2
Ai(s)
−iAi′(s)
Ai′(s) −iAi(s)
. (4.11)
Note that the denominator in (4.11) cannot be zero for s ∈ R. Indeed, its derivative is
−Ai(s)2, so that it is decreasing for s ∈ R, and since the limit for s → +∞ is equal to
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 45
0, it follows that Ai′(s) − sAi(s)2 > 0 for all s ∈ R. Note also that if we take the limit
x, y → 0 in (4.8), then
0 (0, 0; s) = Ai
′(s)2 − sAi(s)2. (4.12)
Using (1.12), (4.6), (4.9), (4.11), and (4.12), we obtain that
ψ1(x; s) =
2πAi(x+ s)−
Ai(x+ s)Ai′(s)− Ai(s)Ai′(x+ s)
x(Ai′(s)2 − sAi(s)2) Ai(s)
Ai(x+ s)− K
0 (x, 0; s)
0 (0, 0; s)
Ai(s)
ψ2(x; s) = −
2πiAi′(x+ s) +
Ai(x+ s)Ai′(s)−Ai(s)Ai′(x+ s)
x(Ai′(s)2 − sAi(s)2) Ai
Ai′(x+ s)− K
0 (x, 0; s)
0 (0, 0; s)
Ai′(s)
1 (x, y; s) =
ψ2(x; s)ψ1(y; s)− ψ1(x; s)ψ2(y; s)
2πi(x− y)
0 (x, y; s)−
0 (x, 0; s)K
0 (y, 0; s)
0 (0, 0; s)
. (4.13)
To compute the relevant solution u of the Painlevé XXXIV equation for α = 1, we
may assume that we have taken Ψ
0 in (4.9), and then use (1.20), (4.9), (2.14), and the
fact that u ≡ 0 for α = 0, to obtain that u(s) = iX ′12(s), which by (4.11) leads to
u(s) =
Ai(s)2
Ai′(s)2 − sAi(s)2
= − d
0 (0, 0; s). (4.14)
Its graph is shown in Figure 6.
One can verify from the explicitly known asymptotic formulas for Ai that
u(s) =
+O(s−7/2) as s→ +∞. (4.15)
On the negative real axis, u has an infinite number of zeros. These are the zeros of the
Airy function Ai, and an infinite number of additional zeros that interlace with the zeros
of Ai.
Equations (4.9) and (4.14) constitute the Schlesinger and (induced by it) Bäcklund
transformations, respectively, for the case of Painlevé XXXIV and applied to its zero
vacuum solution (for the general theory of Schlesinger transformations see [31]; see also
[24, Chapter 6]).
4.3 Asymptotic characterization of the Painlevé function u(s)
We finally want to characterize the solution u of the Painlevé XXXIV equation by its
asymptotic properties. Recall that u is connected to the solution of the Painlevé II
equation q′′ = sq + 2q3 − ν with ν = 2α+ 1/2 by the formulas
u(s) = 2−1/3U(−21/3s), U(s) = q2(s) + q′(s) + s
. (4.16)
46 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
–10 –8 –6 –4 –2 2 4 6 8 10
Figure 6: The solution of the Painlevé XXXIV equation for α = 1.
Assume that ν > −1/2. It is shown in [34] (see also [24, Chapters 5, 11]) that the
solution q(s) of the Painlevé II equation corresponding to the Stokes multipliers (3.14)
exhibits the following asymptotic behavior in the sector arg s ∈
q(s) =
[ν+1/2]∑
bn(−s)−3n/2 +O
s−3[ν+1/2]/2−1
+ c+(−s)−
(−s)3/2(1 +O(s−1/4)
as s→ ∞, arg s ∈
, arg(−s) ∈
, (4.17)
q(s) =
[ν+1/2]∑
bn(−s)−3n/2 +O
s−3[ν+1/2]/2
+ c−(−s)−
(−s)3/2(1 +O(s−1/4)
as s→ ∞, arg s ∈
, arg(−s) ∈
, (4.18)
where we have used the notation [r] for the integer part of the positive number r, i.e.
[r] ∈ N0, [r] ≤ r < [r] + 1.
The coefficients c+ and c− of the exponential terms, which oscillate on the respective
boundaries of the sector
, are given by the formulae
c+ = −
eπ(ν+
)i + 1
+ ν), (4.19)
c− = −
e−π(ν+
)i + 1
+ ν), (4.20)
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 47
where Γ denotes the Gamma function. Moreover, either the relations (4.17), (4.19) or the
relations (4.18), (4.20) can be taken as a characterization of the solution q(s).
Alternatively, the solution q(s) can be characterized by its comparison to the Boutroux
tri-tronquée solution q(tri−tronq)(s) of the Painlevée II equation, which is defined as the
unique solution satisfying the asymptotic condition
q(tri−tronq)(s) ∼
bn(−s)−3n/2,
as s→ ∞, arg s ≡ π + arg(−s) ∈
. (4.21)
The solution q(s) we are working with is the one whose asymptotic behavior as s→ −∞
is given by the equation
q(s)− q(tri−tronq)(s) = −e
−π(ν+ 1
)i + 1
× |s|− 32 ν− 14 e− 2
|s|3/2(1 +O(s−1/4)
, as s→ −∞. (4.22)
The coefficients bn of the asymptotic series in (4.17), (4.18), and (4.21) are determined
by substitution into the Painlevé II equation. Indeed, the following recurrence relation
takes place
b0 = 1, b1 =
bn+2 =
9n2 − 1
bmbn+2−m −
n+2−l∑
blbmbn+2−l−m.
(4.23)
Using relation (4.16) between the Painlevé II and Painlevé XXXIV functions we arrive
at the asymptotic characterization of the function u(s) of Theorem 1.5.
Proposition 4.1 The solution u(s) of the Painlevé XXXIV equation which appears in
Theorem 1.5 is uniquely characterized by one of the following asymptotic conditions
u(s) =
[2α+1]∑
− 3n+1
s−3[2α+1]/2−1
+ d+s
−3α+ 1
1 +O(s−1/4)
as s→ ∞, arg s ∈
, (4.24)
u(s) =
[2α+1]∑
− 3n+1
s−3[2α+1]/2−1
+ d−s
−3α+ 1
1 +O(s−1/4)
as s→ ∞, arg s ∈
, (4.25)
48 A.R. ITS, A.B.J. KUIJLAARS, and J. ÖSTENSSON
where
e±2απi − 1
2−6α−
3Γ(1 + 2α). (4.26)
Alternatively, the solution u(s) can be characterized by the asymptotic relation
u(s)− u(tri−tronq)(s) = −e
−2απi − 1
2−6α−
3Γ(1 + 2α)
× s−3α+ 12 e− 43s3/2
1 +O(s−1/4)
, as s→ +∞. (4.27)
The Painlevé XXXIV tri-tronquée solution u(tri−tronq)(s) is determined by the asymptotic
condition
u(tri−tronq)(s) ∼ α√
− 3n+1
2 , as s→ ∞, arg s ∈
−π, π
. (4.28)
Finally, the coefficients an of the asymptotic series above can be expressed in terms of the
coefficients bn defined in (4.23), with ν replaced by 2α + 1/2:
2 an = bn+1 −
3n− 2
k,m≥1;k+m=n+1
bkbm.
Remark 4.2 The leading asymptotics of the Painlevé II function q(s) as s → +∞ is
known (see [33]; see also [24, Chapter 10]). Unfortunately, the leading term is not enough
to derive the corresponding asymptotics as s → −∞ of the Painlevé XXXIV function
u(s). Indeed, the leading asymptotics of q(s) as s→ +∞ is of the form
q(s) ∼
s3/2 + χ
, (4.29)
(the phase χ is known) and it cancels out in the right-hand side of equation (3.22). The
better way to study the large negative s asymptotics of the function u(s) is via the direct
analysis of the model RH problem for Ψα. The case α = 1 shows that we might expect
oscillating behavior as s → −∞ (see Figure 6) and indeed, assuming that α− 1/2 6∈ N0,
we are able to show that
u(s) =
(−s)3/2 − απ
+O(1/s2), as s→ −∞. (4.30)
The proof of (4.30) will be given in a future publication. Moreover, we conjecture that
asymptotics (4.30) determines the solution u(s) uniquely.
Acknowledgements
Alexander Its was supported in part by NSF grant #DMS-0401009. Arno Kuijlaars is sup-
ported by FWO-Flanders project G.0455.04, by K.U. Leuven research grant OT/04/21,
by the Belgian Interuniversity Attraction Pole P06/02, by the European Science Foun-
dation Program MISGAM, and by a grant from the Ministry of Education and Science
of Spain, project code MTM2005-08648-C02-01. Jörgen Östensson is supported by K.U.
Leuven research grant OT/04/24.
CRITICAL EDGE BEHAVIOR IN RANDOM MATRIX ENSEMBLES 49
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http://arxiv.org/abs/nlin/0410009
	Introduction and statement of results
	Unitary random matrix models
	The model RH problem
	Connection with the Painlevé XXXIV equation
	Overview of the rest of the paper
	Proof of Theorem 1.1 and 1.2
	The Riemann-Hilbert problem for orthogonal polynomials
	The model RH problem
	Existence of solution to the model RH problem
	Some preliminaries on equilibrium measures
	Steepest descent analysis
	Preliminaries
	First transformation Y T
	Second transformation T S
	The parametrix P()
	The parametrix P(a)
	The parametrix P(0)
	Third transformation S R
	Completion of the proofs of Theorem 1.1 and 1.2
	Proof of Theorems 1.4 and 1.5
	The Painlevé II RH problem
	Connection with 
	Differential equation
	Special choice (spec)
	Special choice 0
	Proof of Theorem 1.4 and 1.5
	Concluding remarks
	The case = 0
	The case = 1
	Asymptotic characterization of the Painlevé function u(s)
