Mean field Heisenberg models and random permutations Jakob E. Bj - - PowerPoint PPT Presentation

Mean field Heisenberg models and random permutations

Jakob E. Bj¨

• rnberg

University of Gothenburg, Sweden

Plan

◮ Heisenberg model ◮ probabilistic description ◮ results for complete graph

Heisenberg model (ferromagnet)

Finite graph G = (V , E), for example [−n, n]d ⊆ Zd or Kn

Heisenberg model (ferromagnet)

Finite graph G = (V , E), for example [−n, n]d ⊆ Zd or Kn H = −2

• xy∈E

Sx · Sy

Heisenberg model (ferromagnet)

Finite graph G = (V , E), for example [−n, n]d ⊆ Zd or Kn H = −2

• xy∈E

Sx · Sy where Sx · Sy = 3

j=1 Sj xSj y and

S1 = 1

2

1 1

• ,

S2 = 1

2

−i i

• ,

S3 = 1

2

1 −1

• Sj

x = Sj ⊗ IdV \{x}

Heisenberg model (ferromagnet)

Finite graph G = (V , E), for example [−n, n]d ⊆ Zd or Kn H = −2

• xy∈E

Sx · Sy where Sx · Sy = 3

j=1 Sj xSj y and

S1 = 1

2

1 1

• ,

S2 = 1

2

−i i

• ,

S3 = 1

2

1 −1

• Sj

x = Sj ⊗ IdV \{x}

• Conjecture. On Zd with d ≥ 3 there is a phase transition.

Heisenberg model (ferromagnet)

Finite graph G = (V , E), for example [−n, n]d ⊆ Zd or Kn H = −2

• xy∈E

Sx · Sy where Sx · Sy = 3

j=1 Sj xSj y and

S1 = 1

2

1 1

• ,

S2 = 1

2

−i i

• ,

S3 = 1

2

1 −1

• Sj

x = Sj ⊗ IdV \{x}

• Conjecture. On Zd with d ≥ 3 there is a phase transition.

Results:

◮ No phase-transition if d ≤ 2 (Mermin–Wagner) ◮ Phase-transition if G = Kn ← this talk!

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

1 2 3 4 5 6 7 8 9 10

G = {1, 2, . . . , 10} ⊆ Z :

1 2 3 9 10

ω = process of ‘crosses’

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

1 2 3 4 5 6 7 8 9 10 1

G = {1, 2, . . . , 10} ⊆ Z :

1 2 3 9 10

ω = process of ‘crosses’

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

1 2 3 4 5 6 7 8 9 10 1 3

G = {1, 2, . . . , 10} ⊆ Z :

1 2 3 9 10

ω = process of ‘crosses’

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

1 2 3 4 5 6 7 8 9 10 1 3 2

G = {1, 2, . . . , 10} ⊆ Z :

1 2 3 9 10

ω = process of ‘crosses’

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

1 2 3 4 5 6 7 8 9 10 3 4 1 7 5 2 6 9 10 8

G = {1, 2, . . . , 10} ⊆ Z :

1 2 3 9 10

ω = process of ‘crosses’

Interchange process (Harris ’72)

◮ Labelled particles at the vertices x ∈ V ◮ Swap particles at x ∼ y at rate 1, independently

1 2 3 4 5 6 7 8 9 10 3 4 1 7 5 2 6 9 10 8

G = {1, 2, . . . , 10} ⊆ Z :

1 2 3 9 10

ω = process of ‘crosses’ σt(ω) = (1, 3)(2, 6, 7, 4)(5)(8, 10, 9)

Coloured/weighted interchange process

Bicolour particles uniformly at random:

Coloured/weighted interchange process

Bicolour particles uniformly at random:

1 2 3 4 5 6 7 8 9 10

Coloured/weighted interchange process

Bicolour particles uniformly at random:

1 2 3 4 5 6 7 8 9 10

M = {all cycles monochromatic}

Coloured/weighted interchange process

Bicolour particles uniformly at random:

1 2 3 4 5 6 7 8 9 10

M = {all cycles monochromatic} σt(ω) = (1, 3)(2, 6, 7, 4)(5)(8, 10, 9) ⇒ M fails.

Coloured/weighted interchange process

Bicolour particles uniformly at random:

1 2 3 4 5 6 7 8 9 10

M = {all cycles monochromatic} σt(ω) = (1, 3)(2, 6, 7, 4)(5)(8, 10, 9) ⇒ M holds.

Coloured/weighted interchange process

Condition on ω: P(M) = E

• cycles γ

2(1

2)|γ|

= 2−|V |E[2ℓ(ω)] where ℓ(ω) = number of disjoint cycles.

Coloured/weighted interchange process

Condition on ω: P(M) = E

• cycles γ

2(1

2)|γ|

= 2−|V |E[2ℓ(ω)] where ℓ(ω) = number of disjoint cycles. Event A depending only on σ: P(A | M) = E[1 IA2ℓ(ω)] E[2ℓ(ω)] =: P2(A).

Interpretation as quantum spin system

Colours = +, − Colouring ↔ basis vector ⊗x∈V |σx of

x∈V C2

Interpretation as quantum spin system

Colours = +, − Colouring ↔ basis vector ⊗x∈V |σx of

x∈V C2

where |+ = ( 1

0 ) and |− = ( 0 1 )

Interpretation as quantum spin system

Colours = +, − Colouring ↔ basis vector ⊗x∈V |σx of

x∈V C2

where |+ = ( 1

0 ) and |− = ( 0 1 )

Can check: Txy := 2(Sx · Sy) + 1

2

acts by Txy ⊗z∈V |σz = ⊗z∈V |στ(z) τ = (x, y) transposition.

Interpretation as quantum spin system

Colours = +, − Colouring ↔ basis vector ⊗x∈V |σx of

x∈V C2

where |+ = ( 1

0 ) and |− = ( 0 1 )

Can check: Txy := 2(Sx · Sy) + 1

2

acts by Txy ⊗z∈V |σz = ⊗z∈V |στ(z) τ = (x, y) transposition. Heisenberg Hamiltonian H = −

• xy∈E

(Txy − 1)

Matrix exponential

Lemma

exp

• β

xy∈E(Txy − 1)

• = E

∗

(xy,t)∈ω Txy

• .

Matrix exponential

Lemma

exp

• β

xy∈E(Txy − 1)

• = E

∗

(xy,t)∈ω Txy

• .

Proof.

exp

• β
• xy∈E

(Txy − 1)

• = e−β|E|

∞

• k=0

βk k!

xy∈E

Txy k

Matrix exponential

Lemma

exp

• β

xy∈E(Txy − 1)

• = E

∗

(xy,t)∈ω Txy

• .

Proof.

exp

• β
• xy∈E

(Txy − 1)

• = e−β|E|

∞

• k=0

βk k!

xy∈E

Txy k =

• b∈E N

e∈E

e−β βke ke!

• e ke!

k! TbkTbk−1 · · · Tb1 ke = #copies of e in b.

Matrix exponential

Lemma

exp

• β

xy∈E(Txy − 1)

• = E

∗

(xy,t)∈ω Txy

• .

Proof.

exp

• β
• xy∈E

(Txy − 1)

• = e−β|E|

∞

• k=0

βk k!

xy∈E

Txy k =

• b∈E N

e∈E

e−β βke ke!

• e ke!

k! TbkTbk−1 · · · Tb1 ke = #copies of e in b.

Matrix exponential

Lemma

exp

• β

xy∈E(Txy − 1)

• = E

∗

(xy,t)∈ω Txy

• .

Proof.

exp

• β
• xy∈E

(Txy − 1)

• = e−β|E|

∞

• k=0

βk k!

xy∈E

Txy k =

• b∈E N

e∈E

e−β βke ke!

• e ke!

k! TbkTbk−1 · · · Tb1 ke = #copies of e in b. Configuration ω ↔ random product ∗

(xy,t)∈ω Txy.

Matrix exponential

Lemma

exp

• β

xy∈E(Txy − 1)

• = E

∗

(xy,t)∈ω Txy

• .

Proof.

exp

• β
• xy∈E

(Txy − 1)

• = e−β|E|

∞

• k=0

βk k!

xy∈E

Txy k =

• b∈E N

e∈E

e−β βke ke!

• e ke!

k! TbkTbk−1 · · · Tb1 ke = #copies of e in b. Configuration ω ↔ random product ∗

(xy,t)∈ω Txy.

β ↔ time

T´

• th’s representation of Heisenberg ferromagnet

(σ, σ)-diagonal element: σ| ∗Txy

• |σ =

1 if M happens,

• therwise.

T´

• th’s representation of Heisenberg ferromagnet

(σ, σ)-diagonal element: σ| ∗Txy

• |σ =

1 if M happens,

• therwise.

So tr

• eβ (Txy −1)

= E[2ℓ(ω)] = 2|V |P(M).

T´

• th’s representation of Heisenberg ferromagnet

(σ, σ)-diagonal element: σ| ∗Txy

• |σ =

1 if M happens,

• therwise.

So tr

• eβ (Txy −1)

= E[2ℓ(ω)] = 2|V |P(M).

Theorem (T´

• th ’93)

Let H = −2

xy Sx · Sy and Z(β) = tr(e−βH). Then the

correlation function S3

x S3 y = tr

• S3

x S3 y e−βH

Z(β) = 1

4P2(x ↔ y)

The Heisenberg ferromagnet

γ0 = cycle containing the origin

• Conjecture. Let G = [−n, n]d ⊆ Zd, d ≥ 3, and

m(β) = lim

k→∞ lim n→∞ P2(|γ0| > k)

The Heisenberg ferromagnet

γ0 = cycle containing the origin

• Conjecture. Let G = [−n, n]d ⊆ Zd, d ≥ 3, and

m(β) = lim

k→∞ lim n→∞ P2(|γ0| > k)

Then there is βc s.t. m(β) = 0 for β < βc, > 0 for β > βc.

Higher spins

Cycle-weighted interchange processes: Pθ(A) = E[1 IAθℓ(ω)] E[θℓ(ω)] , θ > 0

Higher spins

Cycle-weighted interchange processes: Pθ(A) = E[1 IAθℓ(ω)] E[θℓ(ω)] , θ > 0 Integer θ ↔ spin S = (θ − 1)/2

Higher spins

Cycle-weighted interchange processes: Pθ(A) = E[1 IAθℓ(ω)] E[θℓ(ω)] , θ > 0 Integer θ ↔ spin S = (θ − 1)/2 H = −

xy∈E(Txy − 1)

Higher spins

Cycle-weighted interchange processes: Pθ(A) = E[1 IAθℓ(ω)] E[θℓ(ω)] , θ > 0 Integer θ ↔ spin S = (θ − 1)/2 H = −

xy∈E(Txy − 1) ◮ S = 1 2: Txy = 2(Sx · Sy) + 1 2

Higher spins

Cycle-weighted interchange processes: Pθ(A) = E[1 IAθℓ(ω)] E[θℓ(ω)] , θ > 0 Integer θ ↔ spin S = (θ − 1)/2 H = −

xy∈E(Txy − 1) ◮ S = 1 2: Txy = 2(Sx · Sy) + 1 2 ◮ S = 1: Txy = (Sx · Sy)2 + (Sx · Sy) − 1

Higher spins

Cycle-weighted interchange processes: Pθ(A) = E[1 IAθℓ(ω)] E[θℓ(ω)] , θ > 0 Integer θ ↔ spin S = (θ − 1)/2 H = −

xy∈E(Txy − 1) ◮ S = 1 2: Txy = 2(Sx · Sy) + 1 2 ◮ S = 1: Txy = (Sx · Sy)2 + (Sx · Sy) − 1 ◮ etc

Complete graph G = Kn

V = {1, 2, . . . , n} and E = V

2

• all pairs

Complete graph G = Kn

V = {1, 2, . . . , n} and E = V

2

• all pairs

Scaling: β → β/n

Complete graph G = Kn

V = {1, 2, . . . , n} and E = V

2

• all pairs

Scaling: β → β/n X ε

n = 1

n

• |γ|≥εn

|γ|

Complete graph G = Kn

V = {1, 2, . . . , n} and E = V

2

• all pairs

Scaling: β → β/n X ε

n = 1

n

• |γ|≥εn

|γ| Theorem (Schramm ’05) θ = 1, ζ = max solution to 1 − z = e−βz lim

ε→0 lim n→∞ X ε n = ζ in probability

Complete graph G = Kn

V = {1, 2, . . . , n} and E = V

2

• all pairs

Scaling: β → β/n X ε

n = 1

n

• |γ|≥εn

|γ| Theorem (Schramm ’05) θ = 1, ζ = max solution to 1 − z = e−βz lim

ε→0 lim n→∞ X ε n = ζ in probability

β < 1 ⇒ no order n cycles β > 1 ⇒ large cycles

Large cycles for θ > 1

Cycles |γ1| ≥ |γ2| ≥ · · ·

Large cycles for θ > 1

Cycles |γ1| ≥ |γ2| ≥ · · · Theorem (B. ’15) θ > 1, β > θ. lim

ε→0 lim n→∞ Pθ(|γ1| ≥ εn) = 1.

Large cycles for θ > 1

Cycles |γ1| ≥ |γ2| ≥ · · · Theorem (B. ’15) θ > 1, β > θ. lim

ε→0 lim n→∞ Pθ(|γ1| ≥ εn) = 1.

Sketch proof:

Large cycles for θ > 1

Cycles |γ1| ≥ |γ2| ≥ · · · Theorem (B. ’15) θ > 1, β > θ. lim

ε→0 lim n→∞ Pθ(|γ1| ≥ εn) = 1.

Sketch proof: Let r = 1/θ ∈ (0, 1)

Large cycles for θ > 1

Cycles |γ1| ≥ |γ2| ≥ · · · Theorem (B. ’15) θ > 1, β > θ. lim

ε→0 lim n→∞ Pθ(|γ1| ≥ εn) = 1.

Sketch proof: Let r = 1/θ ∈ (0, 1) Colour loops independently

◮ red prob r ◮ white prob 1 − r

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n]

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n]

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n] Lemma: Given R, law of ωr is Poisson

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n] Lemma: Given R, law of ωr is Poisson

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n] Lemma: Given R, law of ωr is Poisson “Reason”: θℓ(ω)r #red(1 − r)#white

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n] Lemma: Given R, law of ωr is Poisson “Reason”: θℓ(ω)r #red(1 − r)#white = (θ − 1)#white

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n] Lemma: Given R, law of ωr is Poisson “Reason”: θℓ(ω)r #red(1 − r)#white = (θ − 1)#white So Pθ(|γ1| ≥ εn) ≥ Pθ(|γred

1

| ≥ εn) = Eθ[P′

1(|γ1| ≥ (εn N )N)]

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n] Lemma: Given R, law of ωr is Poisson “Reason”: θℓ(ω)r #red(1 − r)#white = (θ − 1)#white So Pθ(|γ1| ≥ εn) ≥ Pθ(|γred

1

| ≥ εn) = Eθ[P′

1(|γ1| ≥ (εn N )N)]

Under P′

1: ◮ N (red) vertices ◮ time β/n = (βN/n)/N

Large cycles for θ > 1

1 2 3 4 5 6 7 8 9 10

ω = ωr ∪ ωw ∪ ωm red lines R ⊆ V × [0, β/n] Lemma: Given R, law of ωr is Poisson “Reason”: θℓ(ω)r #red(1 − r)#white = (θ − 1)#white So Pθ(|γ1| ≥ εn) ≥ Pθ(|γred

1

| ≥ εn) = Eθ[P′

1(|γ1| ≥ (εn N )N)]

Under P′

1: ◮ N (red) vertices ◮ time β/n = (βN/n)/N

Use E[βN/n] = βr = β/θ > 1 and Schramm’s theorem.

Free energy

βc = θ?

Free energy

βc = θ? — probably not

Free energy

βc = θ? — probably not Define Zn(β, h) = E ℓ(ω)

• j=1
• eh|γj| + θ − 1

Free energy

βc = θ? — probably not Define Zn(β, h) = E ℓ(ω)

• j=1
• eh|γj| + θ − 1
• and the free energy

z(β, h) = lim

n→∞ 1 n log Zn(β, h) − h θ

Free energy

βc = θ? — probably not Define Zn(β, h) = E ℓ(ω)

• j=1
• eh|γj| + θ − 1
• and the free energy

z(β, h) = lim

n→∞ 1 n log Zn(β, h) − h θ

and “magnetization” z+(β) =

∂ ∂h+ z(β, h)|h=0.

Critical point

Theorem (B. ’16) θ ∈ {2, 3, 4, . . . } z+(β) = 0 if β < βc(θ) > 0 if β > βc(θ) where βc(θ) = θ if θ = 2 2 θ−1

θ−2

• log(θ − 1)

if θ > 2

Critical point

Theorem (B. ’16) θ ∈ {2, 3, 4, . . . } z+(β) = 0 if β < βc(θ) > 0 if β > βc(θ) where βc(θ) = θ if θ = 2 2 θ−1

θ−2

• log(θ − 1)

if θ > 2 At criticality: z+(βc) = 0 if θ = 2 but > 0 if θ > 2

Critical point

Theorem (B. ’16) θ ∈ {2, 3, 4, . . . } z+(β) = 0 if β < βc(θ) > 0 if β > βc(θ) where βc(θ) = θ if θ = 2 2 θ−1

θ−2

• log(θ − 1)

if θ > 2 At criticality: z+(βc) = 0 if θ = 2 but > 0 if θ > 2 Remarks

◮ θ = 2 done by T´

• th ’90 and Penrose ’91

Critical point

Theorem (B. ’16) θ ∈ {2, 3, 4, . . . } z+(β) = 0 if β < βc(θ) > 0 if β > βc(θ) where βc(θ) = θ if θ = 2 2 θ−1

θ−2

• log(θ − 1)

if θ > 2 At criticality: z+(βc) = 0 if θ = 2 but > 0 if θ > 2 Remarks

◮ θ = 2 done by T´

• th ’90 and Penrose ’91

◮ βc(θ) = critical value of random-cluster model G(n, p, q) with

q = θ and p = 1 − e−β/n.

Consequence for cycles

Corollary For any δ > 0 there is c > 0 s.t. Pθ(X ε

n > ζ + δ) ≤ e−cn,

where ζ = max root of

1−z 1+(θ−1)z = e−βz

Consequence for cycles

Corollary For any δ > 0 there is c > 0 s.t. Pθ(X ε

n > ζ + δ) ≤ e−cn,

where ζ = max root of

1−z 1+(θ−1)z = e−βz

• cf. size of the giant component of G(n, p, q).

Proof idea

Zn(β, h) = p−nP(M)

◮ M = {cycles monochromatic}

Proof idea

Zn(β, h) = p−nP(M)

◮ M = {cycles monochromatic} ◮ colours 1, 2, . . . , θ

Proof idea

Zn(β, h) = p−nP(M)

◮ M = {cycles monochromatic} ◮ colours 1, 2, . . . , θ ◮ probabilities p1 = peh and p2 = . . . = pθ = p.

Proof idea

Zn(β, h) = p−nP(M)

◮ M = {cycles monochromatic} ◮ colours 1, 2, . . . , θ ◮ probabilities p1 = peh and p2 = . . . = pθ = p.

Rewrite P(M) ≍ pn

λ⊢n

ehλ1 n λ

• P(σ ∈ Tλ)

λ = (λ1 ≥ · · · ≥ λθ ≥ 0) Tλ = Young subgroup

Some representation theory

Let Vλ = coset representation of Tλ

Some representation theory

Let Vλ = coset representation of Tλ Vλ =

µ⊢n KµλUµ

Some representation theory

Let Vλ = coset representation of Tλ Vλ =

µ⊢n KµλUµ

Alon–Kozma and Berestycki–Kozma: n λ

• P(σ ∈ Tλ) =
• µ⊢n

Kµλdµ exp β n n 2

• [r(µ) − 1]

Some representation theory

Let Vλ = coset representation of Tλ Vλ =

µ⊢n KµλUµ

Alon–Kozma and Berestycki–Kozma: n λ

• P(σ ∈ Tλ) =
• µ⊢n

Kµλdµ exp β n n 2

• [r(µ) − 1]
• dµ ≈

n

µ

• ≈ exp(−

i(µi n ) log(µi n ))

r(µ) ≈

i(µi n )2

Conclusion

Criterion for z+(β) > 0 in terms of maxima of φβ(x) = β 2

• θ
• i=1

x2

i − 1

• −

θ

• i=1

xi log xi

Conclusion

Criterion for z+(β) > 0 in terms of maxima of φβ(x) = β 2

• θ
• i=1

x2

i − 1

• −

θ

• i=1

xi log xi Maximizer at x = (1

θ, . . . , 1 θ) for β < βc, elsewhere for β > βc.

Conclusion

Criterion for z+(β) > 0 in terms of maxima of φβ(x) = β 2

• θ
• i=1

x2

i − 1

• −

θ

• i=1

xi log xi Maximizer at x = (1

θ, . . . , 1 θ) for β < βc, elsewhere for β > βc.

Open questions:

◮ Coupling with random-cluster-model G(n, p, q)?

Conclusion

Criterion for z+(β) > 0 in terms of maxima of φβ(x) = β 2

• θ
• i=1

x2

i − 1

• −

θ

• i=1

xi log xi Maximizer at x = (1

θ, . . . , 1 θ) for β < βc, elsewhere for β > βc.

Open questions:

◮ Coupling with random-cluster-model G(n, p, q)? ◮ X ε n → ζ?

Thank you!