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Self-avoiding walk on the hexagonal lattice Hugo Duminil-Copin, - - PowerPoint PPT Presentation

Self-avoiding walk on the hexagonal lattice Hugo Duminil-Copin, Universit e de Gen` eve May 2010 joint work with S. Smirnov Hugo Duminil-Copin, Universit e de Gen` eve Self-avoiding walk on the hexagonal lattice Consider the hexagonal


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Self-avoiding walk on the hexagonal lattice

Hugo Duminil-Copin, Universit´ e de Gen` eve May 2010 joint work with S. Smirnov

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Consider the hexagonal lattice H, and define cn to be the number of self-avoiding walks of length n starting at the origin.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Consider the hexagonal lattice H, and define cn to be the number of self-avoiding walks of length n starting at the origin.

a

We will consider walks starting and ending at mid-edges!

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Consider the hexagonal lattice H, and define cn to be the number of self-avoiding walks of length n starting at the origin.

a

We will consider walks starting and ending at mid-edges! Combinatorial question: How many such trajectories are there? In

  • ther words, what is the value of cn?

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Theorem (H. D-C, S. Smirnov, 2010) The connective constant µ satisfies µ := limn→∞ c

1 n

n

=

  • 2 +

√ 2.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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SLIDE 6

Theorem (H. D-C, S. Smirnov, 2010) The connective constant µ satisfies µ := limn→∞ c

1 n

n

=

  • 2 +

√ 2. Strategy : It is sufficient to prove that the partition function Z(x) is finite if and only if x >

  • 2 +

√ 2 =: xc, where Z(x) :=

  • γ : a→H

x−ℓ(γ)

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Theorem (H. D-C, S. Smirnov, 2010) The connective constant µ satisfies µ := limn→∞ c

1 n

n

=

  • 2 +

√ 2. Strategy : It is sufficient to prove that the partition function Z(x) is finite if and only if x >

  • 2 +

√ 2 =: xc, where Z(x) :=

  • γ : a→H

x−ℓ(γ) We restrict ourself to finite domains Ω and we weight walks by their winding.

mid-edge a

z Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Definition The winding WΓ(a, b) of a curve Γ between a and b is the rotation (in radians) of the curve between a and b.

Wγ(a, b) = 2π Wγ(a, b) = 0 a a b b

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Definition The winding WΓ(a, b) of a curve Γ between a and b is the rotation (in radians) of the curve between a and b.

Wγ(a, b) = 2π Wγ(a, b) = 0 a a b b

With this definition, we can define the parafermionic operator for a ∈ ∂Ω and z ∈ Ω: F(z) = F(a, z, x, σ) :=

  • γ⊂Ω: a→z

e−iσWγ(a,z)x−ℓ(γ).

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Definition The winding WΓ(a, b) of a curve Γ between a and b is the rotation (in radians) of the curve between a and b.

Wγ(a, b) = 2π Wγ(a, b) = 0 a a b b

With this definition, we can define the parafermionic operator for a ∈ ∂Ω and z ∈ Ω: F(z) = F(a, z, x, σ) :=

  • γ⊂Ω: a→z

e−iσWγ(a,z)x−ℓ(γ).

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Definition The winding WΓ(a, b) of a curve Γ between a and b is the rotation (in radians) of the curve between a and b.

Wγ(a, b) = 2π Wγ(a, b) = 0 a a b b

With this definition, we can define the parafermionic operator for a ∈ ∂Ω and z ∈ Ω: F(z) = F(a, z, x, σ) :=

  • γ⊂Ω: a→z

e−iσWγ(a,z)x−ℓ(γ). Lemma (Discrete integrals on elementary contours vanish) If x = xc and σ = 5

8, then F satisfies the following relation for every

vertex v ∈ V (Ω), (p − v)F(p) + (q − v)F(q) + (r − v)F(r) = 0 where p, q, r are the mid-edges of the three edges adjacent to v.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We write c(γ) for the contribution of the walk γ to the sum.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We write c(γ) for the contribution of the walk γ to the sum. One can partition the set of walks γ finishing at p, q or r into pairs and triplets of walks:

γ1 γ2 γ1 γ2 γ3

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We write c(γ) for the contribution of the walk γ to the sum. One can partition the set of walks γ finishing at p, q or r into pairs and triplets of walks:

γ1 γ2 γ1 γ2 γ3

In the first case: c(γ1) + c(γ2) = (q − v)e−iσWγ1(a,q)x−ℓ(γ1)

c

+ (r − v)e−iσWγ2(a,r)x−ℓ(γ2)

c

= (p − v)e−iσWγ1(a,p)x−ℓ(γ1)

c

  • ei 2π

3 e−i 5 8 · −4π 3

+ e−i 2π

3 e−i 5 8 · 4π 3

  • = 0

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We write c(γ) for the contribution of the walk γ to the sum. One can partition the set of walks γ finishing at p, q or r into pairs and triplets of walks:

γ1 γ2 γ1 γ2 γ3

In the first case: c(γ1) + c(γ2) = (q − v)e−iσWγ1(a,q)x−ℓ(γ1)

c

+ (r − v)e−iσWγ2(a,r)x−ℓ(γ2)

c

= (p − v)e−iσWγ1(a,p)x−ℓ(γ1)

c

  • ei 2π

3 e−i 5 8 · −4π 3

+ e−i 2π

3 e−i 5 8 · 4π 3

  • = 0

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We write c(γ) for the contribution of the walk γ to the sum. One can partition the set of walks γ finishing at p, q or r into pairs and triplets of walks:

γ1 γ2 γ1 γ2 γ3

In the first case: c(γ1) + c(γ2) = (q − v)e−iσWγ1(a,q)x−ℓ(γ1)

c

+ (r − v)e−iσWγ2(a,r)x−ℓ(γ2)

c

= (p − v)e−iσWγ1(a,p)x−ℓ(γ1)

c

  • ei 2π

3 e−i 5 8 · −4π 3

+ e−i 2π

3 e−i 5 8 · 4π 3

  • = 0

In the second case: c(γ1)+c(γ2) + c(γ3) = (p − v)e−iσWγ1(a,p)x−ℓ(γ1)

c

  • 1 + xcei 2π

3 e−i 5 8 · −π 3 + xce−i 2π 3 e−i 5 8 · π 3

  • = 0.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We deduce that the integral on any discrete contour vanishes. The next step is to choose the right contour.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We deduce that the integral on any discrete contour vanishes. The next step is to choose the right contour. ST,L ε ¯ ε β α

a

T cells L cells

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We deduce that the integral on any discrete contour vanishes. The next step is to choose the right contour. ST,L ε ¯ ε β α

a

T cells L cells

Sum the relation over all vertices in V (ST,L) (contour=boundary): 0 = −

  • z∈α

F(z) +

  • z∈β

F(z) + ei 2π

3

z∈ε

F(z) + e−i 2π

3

z∈¯ ε

F(z)

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We deduce that the integral on any discrete contour vanishes. The next step is to choose the right contour. Definition: Ax

T,L =

  • γ⊂ST,L: a→α\{a}

x−ℓ(γ), Bx

T,L =

  • γ⊂ST,L: a→β

x−ℓ(γ), E x

T,L =

  • γ⊂ST,L: a→ε∪¯

ε

x−ℓ(γ). ST,L ε ¯ ε β α

a

T cells L cells

Sum the relation over all vertices in V (ST,L) (contour=boundary): 0 = −

  • z∈α

F(z) +

  • z∈β

F(z) + ei 2π

3

z∈ε

F(z) + e−i 2π

3

z∈¯ ε

F(z)

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We know the winding on the boundary!

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We know the winding on the boundary! For instance, the winding on the boundary β equals 0, so:

  • z∈β

F(z) = Bx

T,L.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We know the winding on the boundary! For instance, the winding on the boundary β equals 0, so:

  • z∈β

F(z) = Bx

T,L.

The same reasoning gives:

  • z∈ε

F(z) = ei π

8

2 E x

T,L

  • z∈¯

ε

F(z) = e−i π

8

2 E x

T,L

  • z∈α

F(z) = F(a) +

  • z∈α\{a}

F(z) = 1 − cos 3π 8

  • Ax

T,L

Plugging these equalities into the previous relation, we obtain the claim.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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We know the winding on the boundary! For instance, the winding on the boundary β equals 0, so:

  • z∈β

F(z) = Bx

T,L.

The same reasoning gives:

  • z∈ε

F(z) = ei π

8

2 E x

T,L

  • z∈¯

ε

F(z) = e−i π

8

2 E x

T,L

  • z∈α

F(z) = F(a) +

  • z∈α\{a}

F(z) = 1 − cos 3π 8

  • Ax

T,L

Plugging these equalities into the previous relation, we obtain the claim. When x = xc, we have 1 = cos 3π 8

  • Axc

T,L + Bxc T,L + cos

π 4

  • E xc

T,L

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≤

  • 2 +

√ 2 (or equivalently Z(xc) = +∞)

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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SLIDE 26

µ ≤

  • 2 +

√ 2 (or equivalently Z(xc) = +∞) (1) you can let L go to infinity (exercice), cos 3π 8

  • Axc

T + Bxc T = 1

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≤

  • 2 +

√ 2 (or equivalently Z(xc) = +∞) (1) you can let L go to infinity (exercice), cos 3π 8

  • Axc

T + Bxc T = 1

(2) When considering both relations for T and T + 1, 0 = 1 − 1 = cos 3π 8

  • (Axc

T+1 − Axc T ) + Bxc T+1 − Bxc T

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≤

  • 2 +

√ 2 (or equivalently Z(xc) = +∞) (1) you can let L go to infinity (exercice), cos 3π 8

  • Axc

T + Bxc T = 1

(2) When considering both relations for T and T + 1, 0 = 1 − 1 = cos 3π 8

  • (Axc

T+1 − Axc T ) + Bxc T+1 − Bxc T

(3) From a walk entering in Axc

T+1 − Axc T , one can construct two

excursions of width T + 1, yielding Axc

T+1 − Axc T ≤

  • Bxc

T+1

  • 2. Thus,

Bxc

T+1 + cos

3π 8 Bxc

T+1

2 ≥ Bxc

T

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≤

  • 2 +

√ 2 (or equivalently Z(xc) = +∞) (1) you can let L go to infinity (exercice), cos 3π 8

  • Axc

T + Bxc T = 1

(2) When considering both relations for T and T + 1, 0 = 1 − 1 = cos 3π 8

  • (Axc

T+1 − Axc T ) + Bxc T+1 − Bxc T

(3) From a walk entering in Axc

T+1 − Axc T , one can construct two

excursions of width T + 1, yielding Axc

T+1 − Axc T ≤

  • Bxc

T+1

  • 2. Thus,

Bxc

T+1 + cos

3π 8 Bxc

T+1

2 ≥ Bxc

T

(4) It is an easy step to deduce that Bxc

T ≥ c/T and therefore

Z(xc) ≥

  • T>0

Bxc

T = +∞.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc)

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

(2) Any simple walk can be canonically decomposed into a sequence of excursions of widths T−i < · · · < T−1 and T0 > · · · > Tj.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

(2) Any simple walk can be canonically decomposed into a sequence of excursions of widths T−i < · · · < T−1 and T0 > · · · > Tj.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

(2) Any simple walk can be canonically decomposed into a sequence of excursions of widths T−i < · · · < T−1 and T0 > · · · > Tj.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

(2) Any simple walk can be canonically decomposed into a sequence of excursions of widths T−i < · · · < T−1 and T0 > · · · > Tj.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

(2) Any simple walk can be canonically decomposed into a sequence of excursions of widths T−i < · · · < T−1 and T0 > · · · > Tj.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

(2) Any simple walk can be canonically decomposed into a sequence of excursions of widths T−i < · · · < T−1 and T0 > · · · > Tj.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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µ ≥

  • 2 +

√ 2 (or equivalently Z(x) < +∞ for x > xc) (1) For x = xc, Bxc

T ≤ 1, so we deduce Bx T ≤ (xc/x)T for x > xc. In

particular

T>0(1 + Bx T) converges.

(2) Any simple walk can be canonically decomposed into a sequence of excursions of widths T−i < · · · < T−1 and T0 > · · · > Tj. (3) We obtain Z(x) ≤ 2

  • T−i<···<T−1

Tj<···<T0

  • j
  • k=−i

Bx

Tk

  • =
  • T>0

(1 + Bx

T)2 < +∞

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Conjecture (Nienhuis) There exists a constant A ∈ (0, +∞) such that: cn ∼ Anγ−1

  • 2 +

√ 2 n as n − → ∞ where γ = 43/32 is universal.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Conjecture (Nienhuis) There exists a constant A ∈ (0, +∞) such that: cn ∼ Anγ−1

  • 2 +

√ 2 n as n − → ∞ where γ = 43/32 is universal. Conjecture (Lawler, Schramm, Werner) The scaling limit of the self-avoiding walks is SLE(8/3).

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Extensions of the result If we can find the other half of holomorphicity of the parafermionic

  • bservable, we can obtain conformal invariance.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Extensions of the result If we can find the other half of holomorphicity of the parafermionic

  • bservable, we can obtain conformal invariance.

Probably possible to obtain further information: polynomial bounds

  • n the correction, etc...

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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Extensions of the result If we can find the other half of holomorphicity of the parafermionic

  • bservable, we can obtain conformal invariance.

Probably possible to obtain further information: polynomial bounds

  • n the correction, etc...

similar observable for O(n) models: for n ∈ [−2, 2], infinite correlation length at

  • 2 + √2 − n.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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SLIDE 44

Extensions of the result If we can find the other half of holomorphicity of the parafermionic

  • bservable, we can obtain conformal invariance.

Probably possible to obtain further information: polynomial bounds

  • n the correction, etc...

similar observable for O(n) models: for n ∈ [−2, 2], infinite correlation length at

  • 2 + √2 − n.

similar observable for random-cluster models: second (resp. first)

  • rder phase transition for q ∈ [1, 4) (resp. q ∈ (4, +∞)), complete

conformal invariance when q = 2.

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice

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SLIDE 45

Thank you

Hugo Duminil-Copin, Universit´ e de Gen` eve Self-avoiding walk on the hexagonal lattice