Universality for conditional measures of the
sine process
Arno Kuijlaars
and
Erwin Miña-Dı́az
arXiv:1703:02349
Optimal Point Configurations and Orthogonal Polynomials,
Castro Urdiales, Cantabria, Spain, 22 April 2017
Sine point process
Determinantal point process with correlation kernel
sin π(x − y )
π(x − y )
Figure: Points from the sine point process
Figure: Points from Poisson point process
Orthogonal Polynomial ensembles
Sine point process is the limit of many OP ensembles,
in particular GUE
OP ensemble
N
Y
1 Y
w (xj )
(xk − xj )2
Zn j<k
j=1
Orthonormal polynomials ϕj (x; w ), j = 0, 1, . . ., give
KbN (x, y ; w ) =
N−1
X
ϕj (x; w )ϕj (y ; w )
j=0
KN (x, y ; w ) =
p
p
w (x) w (y ) KbN (x, y ; w )
OP ensemble is determinantal with correlation
kernel KN (x, y ; w ).
Universality
After proper rescaling, the OP kernel converges to the
sine kernel (in many cases).
For varying weights w (x) = e −NV (x)
1
x
y
∗
∗
−NV
KN x +
,x +
;e
NψV (x ∗ )
NψV (x ∗ )
NψV (x ∗ )
sin π(x − y )
→
as N → ∞
π(x − y )
ψV (x ∗ ) is the density of the equilibrium measure in
the external field V at the point x ∗ .
Universality limit holds if ψV (x ∗ ) > 0.
Rigidity
The sine point process is rigid
Rigidity
The sine point process is rigid
Suppose we observe the points outside a compact
interval.
With probability one we know the number of points
inside that interval.
Ghosh (2014), Ghosh and Peres (to appear)
Conditional measure
The conditional measure is the joint distribution of the
points inside the interval.
Configuration X from sine point process.
Suppose N points inside the interval I .
Bufetov (2016) showed that the conditional
measure is OP ensemble with weight
2
Y x
ρI ,X (x) =
1−
,
x ∈ I.
p
p∈X \I
Bufetov’s problem
Alexander Bufetov posed the following problem:
Problem
Let (pn )n∈Z be a sequence of real numbers with
frequency one and balanced. For R > 0 let
N = #{pn | |pn | ≤ R}.
Consider on [−R, R] the Nth OP ensemble with weight
ρR (x) =
Y n:|pn |≥R
Then
x
1−
pn
lim KN (x, y ; ρR ) =
R→∞
bounded intervals.
2
,
x ∈ [−R, R].
sin π(x − y )
π(x − y )
uniformly on
Conditions
Frequency one:
pn
=1
n→±∞ n
lim
Balanced:
X 1
pn
n:p 6=0
n
converges in principal value.
Our result
Theorem (K + Mina-Diaz)
Let (pn )n∈Z be a sequence of real numbers with
pn
= 1,
n→±∞ n
lim
X 1
pn
n:p 6=0
converges in principal value.
n
2
Y x
.
ρR (x) =
1−
pn
For R > 0, let
n:|pn |≥R
Then as N, R → ∞ with N/R → 2,
lim KN (x, y ; ρR ) =
N,R→∞
N/R→2
uniformly on bounded intervals.
sin π(x − y )
π(x − y )
Heuristics
We work with
wR (x) = ρR (Rx) =
Y |pn |≥R
Rx
1−
pn
2
,
x ∈ [−1, 1]
and we have to prove
2
2x 2y
sin π(x − y )
lim
KN
, ; wR =
N,R→∞ N
N N
π(x − y )
N/R→2
Compare with weights w (x) = e −NV (x)
1
x
y
sin π(x − y )
−NV
lim
KN
,
;e
=
N→∞ NψV (0)
NψV (0) NψV (0)
π(x − y )
ψV is density of equilibrium measure in external
field V
External field
For large R and N with N/R ≈ 2,
2
Y Rx
wR (x) =
1−
≈ e −NV (x) ,
pn
x ∈ [−1, 1]
|pn |≥R
with
V (x) = (1 + x) log(1 + x) + (1 − x) log(1 − x)
External field
For large R and N with N/R ≈ 2,
2
Y Rx
wR (x) =
1−
≈ e −NV (x) ,
pn
x ∈ [−1, 1]
|pn |≥R
with
V (x) = (1 + x) log(1 + x) + (1 − x) log(1 − x)
Equilibrium measure µV has constant density
1
ψV ≡ ,
2
on [−1, 1].
Lubinsky comparison technique
Lubinsky developed a technique to prove universality by
comparing with known weights. It is based on
inequalities for
KbN (x, y ; w ) =
N−1
X
ϕj (x; w )ϕj (y ; w )
j=0
If w1 ≤ w2 then
KbN (x, x; w2 ) ≤ KbN (x, x; w1 )
and
b
KN (x, y ; w1 ) − KbN (x, y ; w2 )
s
q
KbN (x, x; w2 )
≤ KbN (x, x; w1 )KbN (y , y ; w1 ) 1 −
KbN (x, x; w1 )
Technical estimates on wR
Proposition
Let α > 1 and 0 < β < 1. Then for R large enough
(depending on α and β)
x
wR (x) ≤ e −N (V ( α )+εR x ) ,
x ∈ [−1, 1],
and
(
e −N(V (αx)+εR x) ,
wR (x) ≥
0,
P
where the number εR = 2R
N
|pn |≥R
derivatives agree at x = 0.
Note εR → 0 as R → ∞.
x ∈ [−β, β],
elsewhere,
1
pn
is such that the
Slight modifications
x
1
+
(x) = √
e −N (V ( α )+εR x ) ,
wR,α
2
1−x
p
−
wR,α (x) = 1 − β −2 x 2 e −N(V (αx)+εR x) χ[−β,β] (x),
Then, for R large enough,
−
+
wR,α
(x) ≤ wR (x) ≤ wR,α
(x)
and
−
lim wR,α
R→∞
x R
+
= lim wR,α
R→∞
x R
=1
β = α−2 .
±
Universality for wR,α
Proposition
Universality at the origin:
1
sin π(x − y )
x
y
+
=
lim
K
,
;
w
,
N
R,α
+
R→∞ Ncα
Ncα+ Ncα+
π(x − y )
with constant cα+ satisfying
−
Similarly for wR,α
.
1
lim cα+ = .
α→1−
2
±
Universality for wR,α
Proposition
Universality at the origin:
1
sin π(x − y )
x
y
+
=
lim
K
,
;
w
,
N
R,α
+
R→∞ Ncα
Ncα+ Ncα+
π(x − y )
with constant cα+ satisfying
−
Similarly for wR,α
.
1
lim cα+ = .
α→1−
2
This is enough to complete the proof of our theorem by
playing with the Lubinsky comparison inequalities.
Thank you for your attention !!
Thank you for your attention !!
Thanks for the wonderful conference
Optimal Point Configurations and
Orthogonal Polynomials 2017
When will be the next one?