III Percolation and Random Walks on Graphs
1.4 Conformal invariance and SLE in d = 2
Instead of working on
, we will now work on the triangular lattice
consider site percolation. So every vertex is open (black) with probability
closed (white) with probability 1 − p, independently for different vertices.
Like for Z
, we can show that p = p
be an open simply connected domain in
a Jordan curve.
a, b, c ∈ ∂D
be 3 points labeled in anti-clockwise order. Consider the triangle
= 1 and
. 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 :
T is a homeomor-
is in the arc
, then it will be mapped to
) on the edge
BC of T .
Focus on the case
. We can again put a triangular lattice inside
with mesh size
. 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
(ac ↔ bx) ≥ c > 0,
. We might ask what happens when
0. In particular, does
it converge, and what does it converge to?
Cardy, a physicist studying conformal field theories, conjectured that
(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.
, a, b, c, d
) and (Ω
) are con-
formally equivalent. Then
P(ac ↔ bd in Ω) = P(a
This says percolation at criticality on the triangular lattice is conformally
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 (
) = (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
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.