The travel time in a finite box in supercritical Bernoulli percolation

We consider the standard site percolation model on the three dimensional cubic lattice. Starting solely with the hypothesis that $\theta(p)>0$, we prove that, for any $\alpha>0$, there exists $\kappa>0$ such that, with probability larger than $1-1/n^\alpha$, every pair of vertices inside the box $\Lambda(n)$ are joined by a path having at most $\kappa(\ln n)^2$ closed sites.


Introduction
We consider the site percolation model on Z 3 . Each site is declared open with probability p and closed with probability 1 − p, and the sites are independent. One of the most important problems in percolation is to prove that, in three dimensions, there is no infinite cluster at the critical point. The most promising strategy so far seems to perform a renormalization argument [1]. The missing ingredient is a suitable construction helping to define a good block, starting solely with the hypothesis that θ(p) > 0. Our main result here is an estimate on the travel time in a finite box under the hypothesis that θ(p) > 0. For n ∈ N, we denote by Λ(n) the cubic box Λ(n) = [−n, n] 3 . Theorem 1.1. Let p be such that θ(p) > 0 and let α > 0. There exists a constant κ, depending on α and p, such that ∀n ≥ 2 P   every pair of vertices of the box Λ(n) are joined by a path in Λ(n) having at most κ(ln n) 2 closed sites This result can be recast in the language of first passage percolation. If we declare that the travel time is null through an open site and one through a closed site, and if we denote by T Λ(n) (x, y) the travel time between two points x, y in Λ(n), that is, the infimum of the travel time over all the paths joining x and y in Λ(n), then the above estimate can be rewritten as The bound κ(ln n) 2 is probably not optimal. If we start with the hypothesis that p > p c , then we get a bound of order κ ln n with the help of the slab technology. The goal of the game here is to see what we can get starting only with the hypothesis that θ(p) > 0. The proof relies essentially on the BK and the FKG inequalities for the probabilistic part (see [1]), and on a tiling of the sphere into 48 spherical triangles for the geometric part. The vertices of these triangles are the vertices of a Catalan solid called the disdyakis dodecahedron or the hexakis octahedron (see [2], page 54 top left for a picture, or [3]). This solid is the dual of one of the Archimedean solids, the great rhombicuboctahedron. We could write a proof using only cubes, however it would require more geometric computations. With the help of the tiling of the sphere into spherical triangles, we can build in a straightforward way a path converging at geometric speed to a prescribed target. The main point is that the spherical triangles have a diameter strictly less than one.

Basic notation
Two sites x, y of the lattice Z 3 are said to be connected if they are nearest neighbours, i.e., if |x − y| = 1. For x ∈ Z 3 , we denote by C(x) the open cluster containing x, i.e., the connected component of the set of the open sites containing x. If x is closed, then C(x) = ∅. Let A be a subset of Z 3 . We define its internal boundary ∂ in A and its external boundary ∂ out A by For x a point in Z 3 , the distance d(x, A) between x and A is defined as Recall that a path z 0 , . . . , z r is a sequence of sites such that each site is a neighbour of its predecessor: Let A be a subset of Z 3 . For x, y in A, we define the travel time T A (x, y) between x and y in A by For x in A and E a subset of A, we define the travel time between x and E in A by

An application of the BK inequality
A routine application of the BK inequality gives a control on the travel time until the infinite cluster, and this yields a control on the travel time to exit a finite domain.
Lemma 3.1. Let A be a finite subset of Z 3 and let x ∈ A. We have or, passing to the complementary event, .
is not empty, then ∂ out C(x), the outer boundary of the open cluster of x, contains a set of closed sites which separates x from ∞. We iterate next this argument. We set C 0 (x) = C(x) and we define successively, for k ≥ 0, It follows directly from this construction that Therefore, The set C k (x) becomes infinite when it meets the infinite open cluster. Whenever C k (x) is finite, its outer boundary ∂ out C k (x) contains a set of closed sites which separates x from ∞. This set realizes the event x ←→ ∞ . Moreover, by construction, the sets ∂ out C k (x), k ≥ 0, are pairwise disjoint. Thus we have Applying the BK inequality (see for instance [1]), we conclude that

Cubic boxes
We consider here the case of a cubic box Λ centered at the origin. Let F i , 1 ≤ i ≤ 6, be the faces of Λ. Each face F i is a square, which is itself the union of four squares F j i , 1 ≤ j ≤ 4. Each of these squares shares a vertex with a vertex of Λ and admits the center of F i as another vertex. We have and, by the FKG inequality, The last inequality is a consequence of the symmetry of the model, indeed the random variables T Λ (0, F j i ), 1 ≤ i ≤ 6, 1 ≤ j ≤ 4, are identically distributed. It follows then from lemma 3.1 that . We deal next with translated boxes. If Γ is a cubic box, we denote its faces by F i (Γ), 1 ≤ i ≤ 6, and we denote by F j i (Γ), 1 ≤ i ≤ 6, 1 ≤ j ≤ 4, the tiling of the faces of Γ into four squares. We consider the event E(Λ, k) defined as follows: for every cubic box Γ included in Λ, whose center is a point of Z 3 , whose sidelength is an integer, the center of Γ can be joined to each of the 24 squares on the faces of Γ with a path having at most k closed sites. More precisely, Proof. Let us estimate the probability of the complement of the event E(Λ, k): By translation invariance and symmetry, the probability inside the sum depends neither on x nor on i, j. The number of subboxes Γ included in Λ is bounded by |Λ| × diameter(Λ), so we conclude with the help of the previous estimate that We take now Λ = Λ(n) and k = c ln n. For any α > 0, we can choose the constant c sufficiently large so that the righthand side is smaller than cn −α for any n ≥ 1. Proof. We build iteratively a sequence travelling from x to the box Λ(3n/4). We start from y 0 = x. If x belongs to ∂ in Λ(n), we choose for y 1 a site in Λ(n) \ ∂ in Λ(n) such that |x − y 1 | ≤ 2. If x belongs to Λ(n) \ ∂ in Λ(n), we set y 1 = x. Suppose that y 0 , . . . , y m have been built in such a way that the following four conditions are satisfied for any l ∈ { 1, . . . , m − 1 }: • y l ∈ Λ(n) \ Λ(3n/4).
If y m belongs to Λ(3n/4), the construction terminates. Suppose that y m does not belong to Λ(3n/4). We will next find a site y m+1 so that the sequence y 0 , . . . , y m+1 satisfies the four conditions above. Let Γ be the largest cubic box centered at y m included in Λ(n). Let h be the smallest index such that d(y m , ∂ in Λ(n)) = d(y m , F h (Λ(n))) .
Since the event E(Λ(n), c ln n) occurs, there exists y m+1 in F j i (Γ) such that With this choice of y m+1 , the sequence y 0 , . . . , y m+1 satisfies the four conditions above. We prove next that the construction stops, i.e., that the sequence enters the box Λ(3n/4) after a finite number of steps. In fact, every three steps, the distance to the boundary of Λ(n) is doubling:

It follows that
If the construction has not stopped after m steps, then y m−1 is still outside of Λ(3n/4), thus d(y m−1 , ∂ in Λ(n)) ≤ n 4 + 1 .
These two inequalities imply that thus the construction stops at some step m * satisfying m * ≤ c ′ ln n, where c ′ is a positive constant. Now the point y m * is inside the box Λ(3n/4) and we have The site y m * belongs to the box Λ(3n/4) and its distance to the boundary of Λ(n) is larger or equal than n/4. By using a few more cubic boxes of side n/4 (nine boxes are certainly enough), we can join y m * to the box Λ(n/4) with a path having at most 9c ln n closed sites. We concatenate the two paths in order to obtain the desired estimate.

The tiling of the sphere
We denote by S the two dimensional sphere of R 3 . We consider the hyperplanes of equations: Let S be the set of the orthogonal symmetries with respect to these hyperplanes. These hyperplanes induce a tiling of the sphere S into 48 spherical triangles. We denote by T the collection of these triangles. The vertices of the triangles of T are the vertices of a convex polyhedron which is a Catalan solid, it is called the disdyakis dodecahedron or the hexakis octahedron [2,3]. The group of the isometries generated by S acts transitively on the collection T of spherical triangles. Let us consider one of these triangles, for instance the triangle having for vertices (1, 0, 0) , The longest arc of this triangle is the arc joining the vertices (1, 0, 0) and 1/ √ 3, 1/ √ 3, 1/ √ 3 and its length is arccos 1/ √ 3 < 0.96. Let r > 0. We define B r = (x, y, z) ∈ Z 3 : Let T belong to T . We define We have ∀y, z ∈ T r |y − z| ≤ 6 + r diameter(T ) ≤ 6 + 0.96r .
For any symmetry s in S, we have s B r ) = B r and s T r = s(T ) r . The percolation model is invariant under the action of S, therefore P 0 ←→ T r in B r = P 0 ←→ s(T ) r in B r and the above probability is the same for any triangle T in T . Moreover Proceeding as in the case of the cube, we have, for any r > 0, and, by the FKG inequality, By symmetry of the model, all the probabilities appearing in the product are equal. It follows then from lemma 3.1 that We deal next with translates of S. We consider the event F (Λ, k) defined as follows: for any x ∈ Λ ∩ Z 3 and any r > 0 such that x + B r ⊂ Λ and such that the boundary of x + B r intersects the lattice Z 3 , the site x can be joined to each of the 48 sets x + T r , T ∈ T , with a path having at most k closed sites. More precisely, Proposition 5.1. Let p be such that θ(p) > 0 and let α > 0. There exists a constant c, depending on α and p, such that Proof. The important point is to notice that the number of choices for the site x and the radius r is bounded by |Λ(n)| 2 . The rest of the proof is the same as proposition 4.1.
Proof. We build a sequence starting at x and which converges at geometric speed towards y, and which stops when it is at distance less than 600 from y. We start from y 0 = x. Suppose that y 0 , . . . , y m have been built in such a way that for any l ∈ { 1, . . . , m }: • y l ∈ Λ(n/4).
We build now y m+1 . Let r > 0 be such that y is on the boundary of y m + B r .
Since y and y m are in Λ(n/4), then y m + B r is included in Λ(n). If r < 600, then |y m − y| < 600 and the construction is finished. Suppose that r ≥ 600.
There exists T ∈ T such that y is in y m + T r . Since the event F (Λ(n), c ln n) occurs, then there exists y m+1 ∈ y m + T r such that T y m +B r (y m , y m+1 ) ≤ c ln n .
Since r ≥ 600, then |y m+1 − y| ≤ 0.97|y m − y| , and the sequence y 0 , . . . , y m+1 satisfies the required constraints. Since the sequence converges at geometric speed towards y, after at most c ′ ln n steps, where c ′ is a constant, it is at distance less than 600 from y and the construction terminates at some index m * ≤ c ′ ln n. Now we have T Λ(n) (x, y) ≤ 0≤m<m * T Λ(n) (y m , y m+1 ) + T Λ(n) (y m * , y) ≤ cc ′ (ln n) 2 + 1800 .
By enlarging the constants, we obtain the statement of the proposition.
This holds for any x, y in Λ(n), so we are done.