Cronograma del Proyecto

P[E₁n(z1)\E1n(z)]=P[Eτn(z2)\Eτn(z)]=P[Eτn2(z3)\En_τ2(z)]. Before proving this, we introduce the exploration process illustrated in Figure 5.18. Suppose that all vertices ‘just outside’ the arc A1Aτ2 (respectively, AτAτ2) ofTnare black (respectively, white). The exploration

path is defined to be the unique pathηnon the edges of the dual (hexago-

nal) graph, beginning immediately above A_τ2and descending to A1Aτsuch that: as we traverseηnfrom top to bottom, the vertex immediately to our left

(respectively, right), looking along the path from A_τ2, is white (respectively, black). When traversingηnthus, there is a white path on the left and a black

path on the right.

Proof. The event En₁(z1)\E₁n(z)occurs if and only if there exist disjoint paths l1, l2, l3ofTnsuch that:

(i) l1is white and joins s1to AτAτ2, (ii) l2is black and joins s2to A1Aτ2, (iii) l3is black and joins s3to A1Aτ.

5.7 Cardy’s formula 117

A₌A1

C₌A_τ2

B₌Aτ

Figure 5.18. The exploration pathηnstarted at the top vertex A_τ2and

stopped when it hits the bottom side A1Aτof the triangle.

See Figure 5.17 for an explanation of the notation. On this event, the exploration pathηnof Figure 5.18 passes through z and arrives at z along

the edge_hz3,ziofHn. Furthermore, up to the time at which it hits z, it lies

in the region ofH_nbetween l2and l1. Indeed, we may take l2(respectively, l1) to be the black path (respectively, white path) ofTnlying on the right

side (respectively, left side) ofηnup to this point.

Conditional on the event above, and with l1and l2given in terms ofηn

accordingly, the states of vertices ofT_nlying below l1∪l2are independent Bernoulli variables. Thus, the conditional probability of a black path l3 satisfying (iii) is the same as that of a white path. We make this measure- preserving change, and then we interchange the colours white/black to con- clude that: E₁n(z1)\E1n(z)has the same probability as the event that there exist disjoint paths l1, l2, l3ofTnsuch that:

(i) l1is black and joins s1to AτAτ2, (ii) l2is white and joins s2to A1Aτ2, (iii) l3is black and joins s3to A1Aτ.

This is precisely the event En_τ(z2)\Enτ(z), and the lemma is proved. We use Morera’s theorem in order to show the required analyticity. This theorem states that: if f : R_→Cis continuous on the open region R, and

H

γ f d z =0 for all closed curvesγ in R, then f is analytic. It is standard (see [210, p. 208]) that it suffices to consider trianglesγ in R. We may in

fact restrict ourselves to equilateral triangles with one side parallel to the

x -axis. This may be seen either by an approximation argument, or by an

argument based on the threefold Cauchy–Riemann equations

(5.54) ∂f ∂1 = 1 τ ∂f ∂τ = 1 τ2 ∂f ∂τ2,

where∂/∂j means the derivative in the direction of the complex number j . 5.55 Lemma. LetŴbe an equilateral triangle contained in the interior of T with sides parallel to those of T . Then

I Ŵ H₁n(z)d z₌ I Ŵ [H_τn(z)/τ] d z₊O(n−ǫ) = I Ŵ [H_τn2(z)/τ 2_{] d z} +O(n−ǫ),

whereǫis given in Lemma 5.48.

Proof. Every triangular facet of Tn (that is, a triangular union of faces)

points either upwards (in that its horizontal side is at its bottom) or down- wards. LetŴbe an equilateral triangle contained in the interior of T with sides parallel to those of T , and assume thatŴ points upwards (the same argument works for downward-pointing triangles). LetŴnbe the subgraph ofT_nlying withinŴ, so thatŴnis a triangular facet ofTn. LetDnbe the

set of downward-pointing faces ofŴn. Letηbe a vector ofR2such that: if

z is the centre of a face ofD_{n, then z+ηis the centre of a neighbouring}

face, that isη_{∈ {}i,iτ,iτ2_}/(n√3). Write

hn_j(z, η)₌P[En_j(z₊η)_\En_j(z)]. By Lemma 5.53, H₁n(z₊η)₋H₁n(z)₌hn₁(z, η)₋hn₁(z₊η,₋η) =hn_τ(z, ητ )₋hn_τ(z₊η,₋ητ ). Now, H_τn(z₊ητ )₋H_τn(z)₌hn_τ(z, ητ )₋hn_τ(z₊ητ,₋ητ ), and so there is a cancellation in

(5.56) I_ηn₌ X z∈Dn

[H₁n(z₊η)₋H₁n(z)]₋ X

z∈Dn

[H_τn(z₊ητ )₋H_τn(z)] of all terms except those of the form hn_τ(z′,₋ητ )for certain z′lying in faces ofTnthat abut∂Ŵn. There are O(n)such z′, and therefore, by Lemma 5.48,

5.7 Cardy’s formula 119 Consider the sum

Jn₌ 1 n(I

n

i +τIinτ+τ2Iinτ2), where I_jnis an abbreviation for the In

j/n√3of (5.56). The terms of the form H_jn(z)in (5.56) contribute 0 to Jn, since each is multiplied by

(1₊τ₊τ2)n−1₌0.

The remaining terms of the form H_jn(z₊η), H_jn(z₊ητ )mostly disappear also, and we are left only with terms H_jn(z′)for certain z′at the centre of upwards-pointing faces ofTnabutting∂Ŵn. For example, the contribution from z′if its face is at the bottom (but not the corner) ofŴnis

1 n (τ₊τ2)H₁n(z′)₋(1₊τ )H_τn(z′)_{= −}1 n[H n 1(z′)−Hτn(z′)/τ]. When z′is at the right (respectively, left) edge ofŴn, we obtain the same term multiplied byτ(respectively,τ2). Therefore,

(5.58)

I Ŵn

[H₁n(z)₋H_τn(z)/τ] d z_{= −}Jn₊O(n−ǫ)₌O(n−ǫ), by (5.57), where the first O(n−ǫ)term covers the fact that the z in (5.58) is a continuous rather than discrete variable. SinceŴandŴndiffer only around their boundaries, and the H_jnare uniformly H ¨older,

(5.59)

I Ŵ

[H₁n(z)₋H_τn(z)/τ] d z₌O(n−ǫ) and, by a similar argument,

(5.60)

I Ŵ

[H₁n(z)₋Hn

τ2(z)/τ2] d z=O(n−ǫ).

The lemma is proved.

As remarked after the proof of Lemma 5.48, the sequence(H₁n,H_τn,Hn τ2) possesses subsequential limits, and it suffices for convergence to show that all such limits are equal. Let(H1,Hτ,Hτ2)be such a subsequential limit. By Lemma 5.55, the contour integrals of H1, Hτ/τ, Hτ2/τ2around anyŴ are equal. Therefore, the contour integrals of the Gi in (5.52) around anyŴ

equal zero. By Morera’s theorem [2, 210], G1and G2are analytic on the interior of T , and furthermore they may be extended by continuity to the boundary of T . In particular, G1is analytic and real-valued, whence G1is a constant. By (5.50), G1(z)→1 as z→0, whence

Therefore, the real part of G2satisfies

(5.61) Re(G2)=H1−1₂(Hτ+Hτ2)=1₂(3H1−1), and similarly

(5.62) 2Re(G2/τ )=3Hτ −1, 2Re(G2/τ2)=3Hτ2−1. Since the Hj are the real parts of analytic functions, they are harmonic. It

remains to verify the relevant boundary conditions, and we will concentrate on the function H_τ2. There are two ways of doing this, of which the first specifies certain derivatives of the Hj along the boundary of T .

By continuity, H_τ2(C) = 1 and H_τ2 ≡ 0 on A B. We claim that the horizontal derivative,∂H_τ2/∂1, is 0 on AC∪BC. Once this is proved, it follows that H_τ2(z)is the unique harmonic function on T satisfying these boundary conditions, namely the function 2_|Im(z)_|/√3. The remaining claim is proved as follows. Since G2is analytic, it satisfies the threefold Cauchy–Riemann equations (5.54). By (5.61)–(5.62), (5.63) ∂Hτ2 ∂1 = 2 3Re 1 τ2 ∂G2 ∂1 =2₃Re 1 τ3 ∂G2 ∂τ = ∂_∂τH1. Now, H1 ≡ 0 on BC, and BC has gradientτ, whence the right side of (5.63)9equals 0 on BC. The same argument holds on AC with H1replaced by Hτ.

The alternative is slightly simpler, see [27]. For z_∈T , G2(z)is a convex combination of 1,τ,τ2, and thus maps T to the complex triangle T′with these three vertices. Furthermore, G2 maps∂T to∂T′, and G2(Aj)= j for j ₌1, τ, τ2_{. Since G}

2is analytic on the interior of T , it is conformal, and there is a unique such conformal map with this boundary behaviour, namely that composed of a suitable dilation, rotation, and translation of T . This identifies G2uniquely, and the functions Hj also by (5.61)–(5.62).

This concludes the proof of Cardy’s formula when the domain D is an equilateral triangle. The proof for general D is essentially the same, on noting that a conformal image of a harmonic function is harmonic. First, we approximate to the boundary of D by a cycle of the triangular lattice with meshδ. That G1(≡1) and G2are analytic is proved as before, and hence the corresponding limit functions H1, Hτ, Hτ2 are each harmonic with appropriate boundary conditions. We now apply conformal invariance. By the Riemann mapping theorem, there exists a conformal mapφfrom the inside of D to the inside of T that may be extended uniquely to their

9_{We need also that G}

2may be continued analytically beyond the boundary of T , see

5.8 The critical probability via the sharp-threshold theorem 121 boundaries, and that maps a (respectively, b, c) to A (respectively, B, C). The triple(H1◦φ−1,Hτ◦φ−1,Hτ2◦φ−1)solves the corresponding problem on T . We have seen that there is a unique such triple on T , given as above, and equation (5.47) is proved.

5.8 The critical probability via the sharp-threshold theorem