arXiv:1306.6892v1 [math-ph] 28 Jun 2013
Universality at the Edge for Unitary Matrix Models M. Poplavskyi Mathematics Division, B. Verkin Institute for Low Temperature Physics and Engineering National Academy of Sciences of Ukraine 47 Lenin Ave., Kharkiv 61103, Ukraine E-mail:
[email protected] Received August 5, 2012 Abstract Using the results on the 1/n-expansion of the Verblunsky coefficients for a class of polynomials orthogonal on the unit circle with n varying weight, we prove that the local eigenvalue statistic for unitary matrix models is independent of the form of the potential, determining the matrix model. Our proof is applicable to the case of four times differentiable potentials and of supports, consisting of one interval. Key words: unitary matrix models, local eigenvalue statistics, universality, polynomials orthogonal on the unit circle. Mathematics Subject Classification 2010: 15B52, 42C05.
1.
Introduction
We study a class of random matrix ensembles known as unitary matrix models. These models are defined by the probability law U + U∗ −1 dµn (U) , (1.1) pn (U) dµn (U) = Zn,2 exp −nTrV 2 1
where U = {Ujk }nj,k=1 is an n × n unitary matrix, µn (U) is the Haar measure on the group U(n), Zn,2 is the normalization constant, and V : [−1, 1] → R is a continuous function called the potential of the model. Denote eiλj the eigenvalues of the unitary matrix U. The joint probability density of λj , corresponding to (1.1), is given by (see [1]) 1 pn (λ1 , . . . , λn ) = Zn
Y
1≤j 0 uniformly on the set ΩA = {ζ, ξ : 1 ≤ ℑζ, ℑξ ≤ A, |ℜζ, ℜξ| ≤ A}
(3.3)
we have |Fn (ζ, ξ) − F (ζ, ξ)| ≤ εn ,
εn → 0, as n → ∞.
(3.4)
Then for any intervals I1 , I2 ⊂ R Z Z Z Z 2 lim dx dy |Kn (x, y)| = dx dy |K (x, y)|2 . n→∞
I1
I2
I1
I2
The lemma helps to prove the convergence of |Kn |2 to |K|2 . Similarly, we can check the convergence of Kn (t1 , t2 ) Kn (t2 , t3 ) . . . Kn (tl , t1 ) for any l ∈ N. To prove the second part of Theorem 1.5, we use another proposition from [12]. Proposition 3.2 Let ∆ ⊂ R be a system of disjoint intervals as in Theorem 1.5 and let Kn : L2 (∆) → L2 (∆) be a sequence of positive definite integral operators with kernels Kn (x, y) and K : L2 (∆) → L2 (∆) a positive definite integral operator with kernel K (x, y), such that for any l ∈ N, det {Kn (tj , tk )}lj,k=1 → det {K (tj , tk )}lj,k=1 weakly as n → ∞. Assume also that for any ∆ there exists C∆ such that Z Kn (s, s) ds ≤ C∆ . (3.5) ∆
Then, for the Fredholm determinants of Kn and K we have lim det (1 − Kn ) = det (1 − K) .
n→∞
We are going to use Lemma 3.1 for the scaled reproducing kernel of the system of OPUC. Let (3.6) Kn (x, y) = n−2/3 Kn(n) θ + xn−2/3 , θ + yn−2/3 1|x,y|≤cθ n2/3
for some small enough θ-dependent constant cθ . This will be sufficient in view of the following lemma (the analogue of Theorem 11.1.4, [13]) 8
Lemma 3.3 Let the model (1.1) satisfy conditions C1-C3. Then, for any n-independent ε > 0, there exists a constant dε > 0 such that Z Kn(n) (λ, λ) dλ ≤ Ce−ndε . σεc
(n)
Since the polynomials χk are functions of eiλ , it is more convenient to define a little bit different from (3.1) transformation and estimate the difference between it and (3.1). Hence, we consider the following transformation: ZZ 2 −4/3 G (z − λ) G (w − µ) Kn(n) (λ, µ) dλdµ, (3.7) Fn (z, w) = n [−π,π]
with
1 + eiz (3.8) 1 − eiz being the analogues of the Poisson and the Herglotz transformations. G (z) = ℜg (z) , and g (z) =
Proposition 3.4 It follows from the definition of g (z) that z g (z) = i cot , 2
g (z − λ) =
i sin x + sinh y , hence g (z) = −g (z). cosh y − cos x
For z = x + iy we have g (x + iy) = And for G (z) we get G (x + iy) =
eiλ + eiz . eiλ − eiz
sinh y , cosh y − cos x
G (z − λ) = ℑ cot
λ−z . 2
Moreover, G (z) is a Nevanlinna function and |g (z)|2 = −1 + 2 coth ℑz · G (z) .
(3.9)
The difference between the new transformation and the old one can be estimated in the following way: Proposition 3.5 Let z = θ + ζn−2/3 and w = θ + ξn−2/3 with |ζ| , |ξ| ≤ cθ n−2/3 and ℑζ, ℑξ ≥ 1. Then, |Fn (z, w) − 4Fn (ζ, ξ)| ≤ Cn−1/6 (|Fn (z, w)| + 1) . 9
(3.10)
The next step is to prove the convergence of Fn (z, w) to the transformation F (3.2) of the Airy kernel QAi (1.9). F can be calculated in terms of the Airy functions, thus we are concentrated on the calculations of Fn . First, using the properties of CMV matrices, we present Fn (z, w) in terms of the ”resolvent” of C (n) . After that we use the asymptotic behaviour of the Verblunsky coefficients, obtained in [9], to get an approximation of the ”resolvent”. The approximation will be given in terms of the Airy functions. Then we will estimate the error of the ”resolvent” approximation and prove the uniform bound (3.4). We start with a simple corollary from the spectral theorem and Proposition 3.4. Proposition 3.6 Let g (n) (z) = C (n) + eiz
C (n) − eiz
−1
,
be the ”resolvent” of the CMV matrix C (n) . Then, † 1 (n) g (n) (z) = −g (n) (z) , G(n) (z) := g (z) − g (n) (z) , 2 † g (n) (z) g (n) (z) = −I + 2 cot ℑz · G(n) (z) and
−4/3
Fn (z, w) = n
n−1 X
(n)
(n)
Gj,k (z) Gk,j (w) .
(3.11)
j,k=0
of all, First we would like to restrict the summation above by j, k ≤ M = 1/2 Cn log n with some constant C.
Lemma 3.7 There exists V -depended constants C such that under the conditions of Theorem 1.5 uniformly in ΩA of (3.3) we have −2/3
n
n X
j=M +1
(n)
Gn−j,n−j (z) ≤ Cn−1/12 log n. (n)
Now we present the approximation for the matrix elements Gn−j,n−k . Using the three-diagonal matrices expansion (2.2) of the C (n) , we can write the matrix g (n) as −1 g (n) (z) = M (n) e−iz/2 + L(n) eiz/2 M (n) e−iz/2 − L(n) eiz/2 . 10
From the definitions of M (n) and L(n) one can find their matrix elements (n)
(n)
(n)
(n)
(n)
Mn+k,n+k−1 = dn+k ρn+k , Mn+k,n+k = dn+k αn+k − dn+k+1αn+k+1 , (n) (n) (n) (n) (n) Ln+k,n+k−1 = dn+k+1ρn+k , Ln+k,n+k = dn+k+1αn+k − dn+k αn+k+1 , where dk = (1 + sk ) /2 and sk = (−1)k . Denote (n)
C± (z) = M (n) e−iz/2 ± L(n) eiz/2 . At the first step we derive the representation for the matrix elements of the (n) (n) inverse matrix of C− (z). Note that Cr− is three-diagonal and symmetric, and its entries are (n)
(n)
C−n+k,n+k−1 (z) = sn+k ρn+k en+k (z) , (n) (n) (n) C−n+k,n+k (z) = sn+k αn+k en+k (z) + sn+k αn+k+1 en+k+1 (z) with
z z − isk sin . 2 2 For the Verblunsky coefficients we use the result of [9]. ek (z) = cos
Lemma 3.8 Consider the system of orthogonal polynomials and the Verblunsky coefficients defined above. Let the potential V satisfy conditions C1–C3 above. Then, for any k, θ (n) −2/3 (n) k (n) + O (εn,k ) , cos − pθ xk n αn+k = (−1) s 2 θ θ (n) (n) ρn+k = sin + cot pθ xk n−2/3 + O (εn,k ) , 2 2 where s(n) = 1 or s(n) = −1 and (n) xk
−1/3
= kn
,
−4/3
εn,k = n
11
log n 1 +
√ π 2 with pθ = and P defined in (1.7). P (θ)
11
(n) xk
2
1|k| 0, the following asymptotics are uniform in the corresponding domains: 2 3/2 1 + O z −3/2 , |argz| < π − δ, Ai (z) = π −1/2 z −1/4 e− 3 z 2 3/2 π 2 Ai (−z) = π −1/2 z −1/4 sin 1 + O z −3/2 , |argz| < π − δ, z + 3 4 3 1 2 3/2 Ci (z) = π −1/2 z −1/4 e 3 z 1 + O z −3/2 , |argz| < π − δ, 3 π 2 3/2 i 3 z +i 2 4 1 + O z −3/2 , Ci (−z) = π −1/2 z −1/4 e |argz| < π − δ. 3 26
The main term for the derivatives can be obtained by direct differentiation of the asymptotics. The last proposition and the definition of the functions ψ± yield the asymptotic behaviour of them Proposition 5.3 The functions ψ± are entire in x and ζ and have the following asymptotic behaviour in x for ℑζ = ε > 0:
|ψ+ (x, ζ)| = 2 3/2 exp − |ℜXx,ζ | , x→∞ −1/4 −3/2 −1/2 3 π |Xx,ζ | 1 + O |Xx,ζ | n o exp a−1 b−2 ε |ℜX |1/2 , x → −∞ x,ζ
|ψ− (x, ζ)| = 2 3/2 exp |ℜXx,ζ | , x→∞ −1/2 −1/4 −3/2 3 (4π) |Xx,ζ | 1 + O |Xx,ζ | n o exp −a−1 b−2 ε |ℜX |1/2 , x → −∞ x,ζ
Proposition 5.4 For any non-negative integers s, q and any A ∈ R+ there exists a constant CA,s,q such that for any x ≥ −A and ζ ∈ ΩA q 2 Z∞ s ∂ |y| q Rζ (x, y) dy ≤ CA,s,q (1 + |x|)s+q−3/2 . I (s; q) = (5.5) ∂y −∞
P r o o f of Proposition 5.4. In view of equation (5.1), two extra derivatives in (5.4) give the extra factor of order |y|2 + |ζ|2 to the integrand. Therefore, 1/2 we start with I (s; 0). Since |Rζ (x, y)| ≤ CA e−cA |x−y| for x ≥ −A and ζ ∈ ΩA , we split the integral from (5.5) into two parts Z Z Z s s + ≤ Cs (x + |ζ| ) |Rζ (x, y)|2 dy I (s; 0) = |y−x|2|x|
1/2
(t + x)s e−cA t
dt.
(5.6)
t>2|x|
For the first integral we note that by the spectral theorem and the resolvent identity, Z∞ ℑRζ (x, x) |Rζ (x, y)|2 dy = . (5.7) ℑζ −∞
27
The asymptotic behaviour of ψ± from Proposition 5.3 implies |Rζ (x, x)| ≤ CA (1 + |x|)−1/2 ,
|ℑRζ (x, x)| ≤ CA (1 + |x|)−3/2 . (5.8) Combining (5.6) with (5.7) and (5.8), we obtain (5.5) with q = 0. In view of equation (5.1), it is sufficient to prove (5.4) only for q = 0, 1. If q = 1, similarly to the above argument, we split the integral into two parts. In the first term, integrating by parts, we have and
2 Z∞ Z∞ ∂ Rζ (x, y) dy = (c1 y + c2 ζ) |Rζ (x, y)|2 dy. ∂y
−∞
−∞
The r.h.s satisfies the necessary bound for q = 1, hence the proposition is proved. Proposition 5.5 Let h = n−1/3 , M = C0 n1/2 log n . Also, denote by xj = (1,2) (1,2) jh the equidistant set and zj = xj + δj h two shifted sets, with complex (1,2) shifts δj ≤ C for some absolute constant C. Then, M X (1) (2) h ℑRζ zj , zj ≤ C,
(5.9)
j=0
M X (1) (2) 1/2 h Rζ zj , zj ≤ C (Mh) ,
(5.10)
j=0
and for any non-negative integer p, d = 0 or 1 and k ≤ M d ∞ 2 X (1) (2) p ∂ h |xj | d Rζ zj , zk ≤ C (1 + |xk |)p+d−3/2 . ∂z j=0
(5.11)
(1,2) P r o o f of Proposition 5.5. Since zj − xj = O (h), |ℑRζ (x, x)| ≤
C (1 + |x|)−3/2 and derivatives of Rζ are bounded near the real line, we obtain that (1) (2) −3/2 ℑRζ zj , zj ≤ 2C (1 + |xj |) 28
for n > n0 with some integer n0 . Hence, M M X X (1) (2) (1 + |xj |)−3/2 ≤ C. h ℑRζ zj , zj ≤ Ch j=0
j=0
The second statement can be checked in a similar way. The proof of the third statement consists of several steps. First, we change zj by xj in (5.11). The error of this change is a combination of sums of higher derivatives with extra factors h. These sums are small, because for zj far from zk these derivatives admit the exponential bound, and for zj ∼ zk , in view of equation (5.1) and restriction |zk | ≤ Cn1/6 log n, every two extra derivatives will give us the sum as in (5.11) with the factor of order n−1/2 log n. After the change of zj by xj , we obtain the sum which can be estimated by the integral C
Z∞ 0
d 2 ∂ (2) xp d Rζ x, zk dx, ∂z
because of the smoothness and exponential decreasing of Rζ . And finally, the identity (5.7) and Proposition 5.4 yield (5.11). We used the identity (5.7) which is valid for real x, but it remains valid for complex x because the l.h.s and r.h.s of the (5.7) are entire functions equal at the real line. Acknowledgement. The author is grateful to Prof. M.V. Shcherbina for the problem statement and fruitful discussions.
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