Monodromy dependence of Painlevé tau functions
Oleg Lisovyy
Institut Denis-Poisson, Université de Tours, France CIRM, 08/04/2019 collaborations with
- A. Its, A. Prokhorov, M. Cafasso, P. Gavrylenko
Monodromy dependence of Painlev tau functions Oleg Lisovyy - - PowerPoint PPT Presentation
Monodromy dependence of Painlev tau functions Oleg Lisovyy Institut Denis-Poisson, Universit de Tours, France CIRM, 08/04/2019 collaborations with A. Its, A. Prokhorov, M. Cafasso, P. Gavrylenko Example 1: Sine kernel Introduce sin x
2
dt ln τ(t) satisfies
4 e− t2 32
7 12 e3ζ′(−1) =
2
2
8 ,
2 |s − 1|−1
x + r 2 y → ∞,
4 2 3 8 τ± (t) ,
dt ln τ± (t) satisfy
4 .
1 12 e3ζ′(−1) = G
2
2
ν ,
form (z) = Gν ˆ
∞
−1
z + At z−t + A1 z−1 (4 simple poles 0, t, 1, ∞)
[Alday, Gaiotto, Tachikawa, ’09] [Gamayun, Iorgov, OL, ’12-13] [Iorgov, OL, Teschner, ’14] [Bershtein, Shchechkin, ’14]
n∈Z
λ,µ∈Y
0−θ2 t +|λ|+|µ|
0−θ2 t det (1 + K)
0−θ2 t
1−θ2 t
8
−1
− Π+
+ Π+Ψ−Π+
k=1 k (ln J)k (ln J)−k
± − Π± : H∓ → H± are integral operators
−1Ψ+
− ) is the problem to be solved
− + Ψ−1 + ∂zΨ+
− − Π−
+ Π+Ψ+ − Π+
+ ∂zΨ+.
ν ΘνGν with diagonal Θν
−1
− + Ψ−1 + ∂zΨ+
i
e Φ,
i
i
e Φ
i
e
ν
ν,i −
ν,e
1 2 Tr(S2−Θ2 0−Θ2 t )τ [J] .
1 t
t
1
1
t
3
Whittaker Bessel Gauss
1 2 Tr( ¯
S2−Θ2
1−Θ2 t ) τ
ν,i +
ν,e
ν,i −
ν,e
0 − Θ2 t
1 − Θ2 t
4
G(1−z), the parameters
k=1 (e2πi(νΣ−νk ) − e2πi(νΣ−λk ))
k=1 νk = 4 k=1 λk.
t z′
∞ − (θ1 ± σ)2
0 − (θt ∓ σ)2
2F1