[PPT] - Maximal angle of a system of self-repelling particles on the circle PowerPoint Presentation

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Maximal angle of a system of self-repelling particles

n the circle

Antoine Dahlqvist

Technische Universität Berlin

Berlin-Padova workshop October 25, 2014

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Dyson Brownian motion on the circle

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A random matrix model, a diffusion on U(N)

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Moment method

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Dyson Brownian motion on the circle

N ∈ N∗, ωt,1, . . . , ωt,N ∈ U, processes satisfying dωt,p = 1 √ N iωp,tdBt,p + 1 N (∇ log V(ωt))pdt, 1 ≤ p ≤ N, (Unitary Dyson) V(z1, . . . , zN) =

p<q

|zp − zq|2, Bt,1, . . . , Bt,N N independant standard real Brownian motions. Proposition The SDE (Unitary Dyson) admits a unique strong solution (ωt)t≥0 such that ω0,1 = . . . = ω0,N = 1 and a.s. for all t > 0, V(ωt) = 0.

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Dyson Brownian motion on the circle

FIGURE: −i log(ωx/20)0≤x≤100

Questions : if µt,N = 1

N

p=1 δωp, µt,N → µt, with supp(µt) = exp i[−θt, θt]? For

all t >0, a.s., ωt,max → exp[iθt]?

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Dyson Brownian motion on the circle

FIGURE: −i log(ωx/20)0≤x≤100

Questions : if µt,N = 1

N

p=1 δωp, µt,N → µt, with supp(µt) = exp i[−θt, θt]? For

all t >0, a.s., ωt,max → exp[iθt]?

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A random matrix model, a diffusion on U(N)

A scalar product ·, · on MN(C) : X, Y = NTr(X ∗Y). [GUE Matrix] : (Ht,N)t≥0 gaussian process on HN = {H :∈ MN(C) : H∗ = H} such that ∀X, Y ∈ HN, E[X, Ht,NY, Hs,N] = t ∧ s X, Y. 1 √ N     ... Yl,m,t Bp,t Yl,m,t ...     , Yl,m,t =

1 √ 2(B1 l,m,t + iB2 l,m,t)

(B1

l,m,t)l<m, (B2 l,m,t)l<m, (Bp,t)p : N2 independant standard real brownian

motions.

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A random matrix model, a diffusion on U(N)

A scalar product ·, · on MN(C) : X, Y = NTr(X ∗Y). [GUE Matrix] : (Ht,N)t≥0 gaussian process on HN = {H :∈ MN(C) : H∗ = H} such that ∀X, Y ∈ HN, E[X, Ht,NY, Hs,N] = t ∧ s X, Y. 1 √ N     ... Yl,m,t Bp,t Yl,m,t ...     , Yl,m,t =

1 √ 2(B1 l,m,t + iB2 l,m,t)

(B1

l,m,t)l<m, (B2 l,m,t)l<m, (Bp,t)p : N2 independant standard real brownian

motions.

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A random matrix model, a diffusion on U(N)

Set Kt,N = iHt,N, (Ut,N)t≥0 the solution of the MN(C)- valued SDE,

dUt,N

= Ut,N.dKt,N − 1

2Ut,Ndt = Ut,N ◦ dKt,N,

U0 = Id. Proposition i) A.s. for all t ≥ 0, Ut,N ∈ U(N) = {U ∈ MN(C) : UU∗ = Id}. ii) The eigenvalues (ωt,1, . . . , ωt,N) of Ut,N satisfy the equation (Unitary Dyson). Remark Λ : unitary diagonal matrices, X = U(N)/Λ, Θ : X × Λ − → U(N) (U.Λ, D) − → UDU−1 Then, Jac(Θ)(U.Λ, D) = V(D1,1, . . . , DN,N).

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A random matrix model, a diffusion on U(N)

Set Kt,N = iHt,N, (Ut,N)t≥0 the solution of the MN(C)- valued SDE,

dUt,N

= Ut,N.dKt,N − 1

2Ut,Ndt = Ut,N ◦ dKt,N,

U0 = Id. Proposition i) A.s. for all t ≥ 0, Ut,N ∈ U(N) = {U ∈ MN(C) : UU∗ = Id}. ii) The eigenvalues (ωt,1, . . . , ωt,N) of Ut,N satisfy the equation (Unitary Dyson). Remark Λ : unitary diagonal matrices, X = U(N)/Λ, Θ : X × Λ − → U(N) (U.Λ, D) − → UDU−1 Then, Jac(Θ)(U.Λ, D) = V(D1,1, . . . , DN,N).

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Moment method

µt,N = 1 N

λ∈Spec(Ut,N)

δλ. Moments :

U wnµt,N(dw) = 1

N Tr(Un t,N), n ∈ Z.

Theorem (P . Biane, F . Xu, T. Lévy) For all t ≥ 0, f ∈ C(U, R), a.s., µt,N(f)− →µt(f). µt unique measure on U, such that ∀n ∈ Z∗, a.s., 1 N Tr(Un

t,N)−

→

U

ωnµt(dω) = 1 |n|

|n|−1

k=0

(−|n|t)k k! |n| k + 1

e

−|n|t 2 . 7 / 16

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Moment method

µt,N = 1 N

λ∈Spec(Ut,N)

δλ. Moments :

U wnµt,N(dw) = 1

N Tr(Un t,N), n ∈ Z.

Theorem (P . Biane, F . Xu, T. Lévy) For all t ≥ 0, f ∈ C(U, R), a.s., µt,N(f)− →µt(f). µt unique measure on U, such that ∀n ∈ Z∗, a.s., 1 N Tr(Un

t,N)−

→

U

ωnµt(dω) = 1 |n|

|n|−1

k=0

(−|n|t)k k! |n| k + 1

e

−|n|t 2 . 7 / 16

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Moment method

dµt(x) = ρt(x)dx,

FIGURE: Plot of ρt(e2iπθ), Axes : X : angle : θ ∈] − 1

2, 1 2[, Y : time t ∈ [0, 6].

supp(ρt) = [−θt, θt] θ4 = π.

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Moment method

∀t ≥ 0, SN

t = Spec(Ut,N),

St = supp(µt). Theorem ( D., B. Collins,T. Kemp) ∀t ≥ 0, dHaus(SN

t , St) → 0, as N → ∞,

in probability. Problems : i) Back to Hn : 1

i log(Ut,N), i Ut,N−Id Ut,N+Id ∈ Hn sign moments, no hope

to study easily asymptotics of moments. ii) Contrary to the GUE case no independance between coefficients of U(N), no compatibility between the measures of Ut,N+1 and Ut,N, a priori, no recursion formula. iii) Stieltjes transform approach : leads to study the stability of a complex Burger equation ∂tHt,N(z) = zHt,N(z)∂zHt,N(z) + 1 N2 Rt,N(z), H0,N(z) = 1+z

1−z , Ht,N : {z : |z| < 1} → −iH.

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Moment method

∀t ≥ 0, SN

t = Spec(Ut,N),

St = supp(µt). Theorem ( D., B. Collins,T. Kemp) ∀t ≥ 0, dHaus(SN

t , St) → 0, as N → ∞,

in probability. Problems : i) Back to Hn : 1

i log(Ut,N), i Ut,N−Id Ut,N+Id ∈ Hn sign moments, no hope

to study easily asymptotics of moments. ii) Contrary to the GUE case no independance between coefficients of U(N), no compatibility between the measures of Ut,N+1 and Ut,N, a priori, no recursion formula. iii) Stieltjes transform approach : leads to study the stability of a complex Burger equation ∂tHt,N(z) = zHt,N(z)∂zHt,N(z) + 1 N2 Rt,N(z), H0,N(z) = 1+z

1−z , Ht,N : {z : |z| < 1} → −iH.

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Moment method

∀t ≥ 0, SN

t = Spec(Ut,N),

St = supp(µt). Theorem ( D., B. Collins,T. Kemp) ∀t ≥ 0, dHaus(SN

t , St) → 0, as N → ∞,

in probability. Problems : i) Back to Hn : 1

i log(Ut,N), i Ut,N−Id Ut,N+Id ∈ Hn sign moments, no hope

to study easily asymptotics of moments. ii) Contrary to the GUE case no independance between coefficients of U(N), no compatibility between the measures of Ut,N+1 and Ut,N, a priori, no recursion formula. iii) Stieltjes transform approach : leads to study the stability of a complex Burger equation ∂tHt,N(z) = zHt,N(z)∂zHt,N(z) + 1 N2 Rt,N(z), H0,N(z) = 1+z

1−z , Ht,N : {z : |z| < 1} → −iH.

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Moment method

Strategy for the upperbound : ∀ε > 0, t ≥ 0 to prove that a.s., lim sup 1 i log ωt,max ≤ θt + ε, it is sufficient to prove that ∀f ∈ C∞(U), ∃δ, K > 0, with |E[µt,N(f)] − µt(f)| ≤ KN−1−δ, (Speed order 1) and if supp(f) ∩ supp(µt) = ∅, Var(µt,N(f)) ≤ KN−3−δ. (Speed order 2) Choose f ∈ C∞ bump function with PN,ε = P(i−1 log(ωt,max) > θt + ε) ≤ P(Nµt,N(f) ≥ 1). Then, PN,ε ≤ (N−1 − E[µt,N(f)])−2Var(µt,N(f)) ≤ K ′N−1−δ.

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Moment method

Strategy for the upperbound : ∀ε > 0, t ≥ 0 to prove that a.s., lim sup 1 i log ωt,max ≤ θt + ε, it is sufficient to prove that ∀f ∈ C∞(U), ∃δ, K > 0, with |E[µt,N(f)] − µt(f)| ≤ KN−1−δ, (Speed order 1) and if supp(f) ∩ supp(µt) = ∅, Var(µt,N(f)) ≤ KN−3−δ. (Speed order 2) Choose f ∈ C∞ bump function with PN,ε = P(i−1 log(ωt,max) > θt + ε) ≤ P(Nµt,N(f) ≥ 1). Then, PN,ε ≤ (N−1 − E[µt,N(f)])−2Var(µt,N(f)) ≤ K ′N−1−δ.

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To get (Speed order 1), it is sufficient to prove that ∃m ∈ N∗, δ, K > 0, such that

E[
U

ωnµt,N(dω)] −

U

ωnµt(dω)

≤ K nm

N1+δ . N, d ∈ N∗, on the same probability space, consider (Ui

t,N)1≤i≤d,t≥0 i.i.d.

sequence of law (Ut,N)t≥0, (Ut,dN)t≥0 independant of it and set Ut,d,N =    U1

t,N

... Ud

t,N

   ∈ U(dN) For any P ∈ C[X], (dN)−1E[Tr(P(Ut,d,N))] = E[µt,N(P)]

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To get (Speed order 1), it is sufficient to prove that ∃m ∈ N∗, δ, K > 0, such that

E[
U

ωnµt,N(dω)] −

U

ωnµt(dω)

≤ K nm

N1+δ . N, d ∈ N∗, on the same probability space, consider (Ui

t,N)1≤i≤d,t≥0 i.i.d.

sequence of law (Ut,N)t≥0, (Ut,dN)t≥0 independant of it and set Ut,d,N =    U1

t,N

... Ud

t,N

   ∈ U(dN) For any P ∈ C[X], (dN)−1E[Tr(P(Ut,d,N))] = E[µt,N(P)]

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and E[

U

ωnµt,N(dω)] − E[

U

ωnµt,dN(dω)] = t d dsE[tr((Us,N,dUt−s,dN)n)]ds. Theorem (D.) For any n, N, d ∈ N∗ and t ≥ 0, |E[µt,N(X n)] − E[µt,dN(X n)]| ≤ tn4 d N2 . Corollary For any n ∈ Z, t, ε > 0, for N ≥ 2

1 ε ,

|E[µt,N(X n)] − E[µt(X n)]| ≤ 4tn4 3N2 .

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How to compute moments : σ ∈ Sn a permutation, M1, . . . , Mn ∈ MN(C), F(σ, M1, . . . , Mn) =

(i1,...,ik) cycle of σ

Tr(Mi1Mi2 . . . Mik ). Lemma ∀t ≥ 0, d dt e

nt 2 E[F(σ, UtM1, . . . , UtMn)] = − 1

N

1≤i<j≤n

E[F(σ(i j), M1, . . . , Mn)]. Consequences : d dsE[tr((Ut−s,N,dUs,dN)n)] = 1 N2

w∈F

tr(w(Ut−s,N,d, Us,dN, Πd)), (1) Fd,N familly of words with 3 letters, #F ≤ dn4, and Πd orthogonal projection CdN → CN.

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Var(tr(Un

t,N)) = n2

N2 t E[tr((Vs,NUt−s,N)n(Ws,NUt−s,N)n)]ds, (2) with (Ut,N)t≥0, (Vt,N)t≥0, (Wt,N)t≥0 i.i.d. sequence. For any function f ∈ C∞(U, R), N2Var(tr(Un

t,N)) =

t E[tr(f ′(Vs,NUt−s,N)f ′(Ws,NUt−s,N))]ds ≤ t E[tr(f ′(Vs,NUt−s,N)2)1/2tr(f ′(Ws,NUt−s,N)2)1/2]ds ≤ t E[tr(f ′(Vs,NUt−s,N)2)]1/2E[tr(f ′(Ws,NUt−s,N)2]1/2ds = tE[tr(f ′(Vs,NUt−s,N)2)] = tE[tr(f ′(Ut,N)2)]. Conclusion : (Speed order 1) ⇒ (Speed order 2)

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Var(tr(Un

t,N)) = n2

N2 t E[tr((Vs,NUt−s,N)n(Ws,NUt−s,N)n)]ds, (2) with (Ut,N)t≥0, (Vt,N)t≥0, (Wt,N)t≥0 i.i.d. sequence. For any function f ∈ C∞(U, R), N2Var(tr(Un

t,N)) =

t E[tr(f ′(Vs,NUt−s,N)f ′(Ws,NUt−s,N))]ds ≤ t E[tr(f ′(Vs,NUt−s,N)2)1/2tr(f ′(Ws,NUt−s,N)2)1/2]ds ≤ t E[tr(f ′(Vs,NUt−s,N)2)]1/2E[tr(f ′(Ws,NUt−s,N)2]1/2ds = tE[tr(f ′(Vs,NUt−s,N)2)] = tE[tr(f ′(Ut,N)2)]. Conclusion : (Speed order 1) ⇒ (Speed order 2)

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Var(tr(Un

t,N)) = n2

N2 t E[tr((Vs,NUt−s,N)n(Ws,NUt−s,N)n)]ds, (2) with (Ut,N)t≥0, (Vt,N)t≥0, (Wt,N)t≥0 i.i.d. sequence. For any function f ∈ C∞(U, R), N2Var(tr(Un

t,N)) =

t E[tr(f ′(Vs,NUt−s,N)f ′(Ws,NUt−s,N))]ds ≤ t E[tr(f ′(Vs,NUt−s,N)2)1/2tr(f ′(Ws,NUt−s,N)2)1/2]ds ≤ t E[tr(f ′(Vs,NUt−s,N)2)]1/2E[tr(f ′(Ws,NUt−s,N)2]1/2ds = tE[tr(f ′(Vs,NUt−s,N)2)] = tE[tr(f ′(Ut,N)2)]. Conclusion : (Speed order 1) ⇒ (Speed order 2)

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Thierry Lévy. Schur-Weyl duality and the heat kernel measure on the unitary group.

Adv. Math., 218(2) :537–575, 2008.

Benoît Collins, Antoine Dahlqvist and Todd Kemp, Strong convergence for the matrix unitary Brownian motion In preparation

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Thank you

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