1Percolation

III Percolation and Random Walks on Graphs

1.4 Conformal invariance and SLE in d = 2

Instead of working on

Z

2

, we will now work on the triangular lattice

T

. We

consider site percolation. So every vertex is open (black) with probability

p

and

closed (white) with probability 1 − p, independently for different vertices.

Like for Z

2

, we can show that p = p

c

(T) =

1

2

.

Let

D

be an open simply connected domain in

R

2

with

∂D

a Jordan curve.

Let

a, b, c ∈ ∂D

be 3 points labeled in anti-clockwise order. Consider the triangle

T

with vertices

A

= 0,

B

= 1 and

C

=

e

iπ/3

. By the Riemann mapping theorem,

there exists a conformal map ϕ : D → T that maps a 7→ A, b 7→ B, c 7→ C.

Moreover, this can be extended to ∂D such that ϕ :

¯

D →

¯

T is a homeomor-

phism. If

x

is in the arc

bc

, then it will be mapped to

X

=

ϕ

(

x

) on the edge

BC of T .

Focus on the case

p

=

1

2

. We can again put a triangular lattice inside

D

,

with mesh size

δ

. We let

ac ↔ bx

to mean the event that there is an open path

in D joining the arc ac to bx.

Then by RSW (for T), we get that

P

δ

(ac ↔ bx) ≥ c > 0,

independent of

δ

. We might ask what happens when

δ →

0. In particular, does

it converge, and what does it converge to?

Cardy, a physicist studying conformal field theories, conjectured that

lim

δ→0

P

δ

(ac ↔ bx) = |BX|.

He didn’t really write it this way, but expressed it in terms of hypergeometric

functions. It was Carlesson who expressed it in this form.

In 2001, Smirnov proved this conjecture.

Theorem

(Smirnov, 2001)

.

Suppose (Ω

, a, b, c, d

) and (Ω

0

, a

0

, b

0

, c

0

, d

0

) are con-

formally equivalent. Then

P(ac ↔ bd in Ω) = P(a

0

c

0

↔ b

0

d

0

in Ω

0

).

This says percolation at criticality on the triangular lattice is conformally

invariant.

We may also take the dual of the triangular lattice, which is the hexagonal

lattice. Colour a hexagon black if the center is black, and similarly for whites.

Suppose we impose the boundary condition on the upper half plane that the

hexagons are black when x > 0, y = 0, and white when x < 0, y = 0.

Starting at (

x, y

) = (0

,

0), we explore the interface between the black and

white by always keeping a black to our right and white on our right. We can

again take the limit

δ →

0. What is this exploration path going to look like in

the limit δ → 0? It turns out this is an SLE(6) curve.

To prove this, we use the fact that the exploration curve satisfies the locality

property, namely that if we want to know where we will be in the next step, we

only need to know where we currently are.