Divergence of non-random fluctuation in First Passage Percolation

We study non-random fluctuation in the first passage percolation on $\mathbb{Z}^d$ and show that it diverges for any dimension. We also prove the divergence of the non-random shape fluctuation, which was conjectured in [Yu Zhang. The divergence of fluctuations for shape in first passage percolation. {\em Probab. Theory. Related. Fields.} 136(2) 298--320, 2006].


Introduction
First Passage Percolation is a dynamical model of infection, which was introduced by Hammersley and Welsh [14]. The model has received much interests both in mathematics and physics because it has rich structures from the viewpoint of the random metric and it is related to the KPZ-theory [18]. See [2] on the background and related topics.
We consider the first passage percolation (FPP) on the lattice Z d with d ≥ 2. The model is defined as follows. The vertices are the elements of Z d . Let us denote by E d the set of edges: where we set |v − w| 1 = d i=1 |v i − w i | for v = (v 1 , · · · , v d ), w = (w 1 , · · · , w d ). Note that we consider non-oriented edges in this paper, i.e., {v, w} = {w, v} and we sometimes regard {v, w} as a subset of Z d with a slight abuse of notation. We assign a non-negative random variable τ e on each edge e ∈ E d , called the passage time of the edge e. The collection τ = {τ e } e∈E d is assumed to be independent and identically distributed with common distribution F .
Given a path γ, we define the passage time of γ as where [a] is the greatest integer less than or equal to a. Given two vertices v, w ∈ R d , we define the first passage time between vertices v and w as This g(x) is called the time constant. Note that, by the subadditivity, if x ∈ Z d , then g(x) ≤ ET (0, x) and moreover for any x ∈ R d , g(x) ≤ ET (0, x) + 2dEτ e . It is easy to check the homogeneity and convexity: g(λx) = λg(x) and g(rx , then g(x) > 0 for any x = 0 [19]. Therefore, if F (0) < p c (d), then g : R d → R ≥0 is a norm.
1.1. non-random fluctuation. Hammersley and Welsh [14] have proved that 1 N T (0, N e 1 ) converges to g(e 1 ) in probability when d = 2. This statement was strengthened by Kingman [16] as stated in (1.1). Since then, the rate of this convergence becomes one of the basic problems in this model. The difference T (0, x) − g(x) can be natrually divided into the random fluctuation part and non-random fluctuation part as follows: Let us briefly review the earlier works. It is widely believed that there exist universal constants in a suitable sense. This "universal" means that these values are independent of distributions. To sate the previous works precisely, we introduce four relevant quantities: Due to the works of Kesten [20], it is (the best currently) known that 0 ≤ χ(d) ≤χ(d) ≤ 1/2 under the condition that the second moment of τ is finite. On the other hand, Newman and Piza showed thatχ(2) ≥ 1/8 for useful distributions under an exponential moment condition [21], where useful distributions are defined in (1.3) below.
Let us move on to the previous researches on the non-random fluctuation. Alexander found the relationship betweenχ(d) andχ (d) and he provedχ (d) ≤ 1/2 with an exponential moment condition [1], which was later relaxed to a low moment condition [11]. For the lower bounds, it is proved that χ (d) ≥ −1 [20] andχ (d) ≥ −1/2 [3] with an exponential moment condition.
Remarkably, it was shown in [3] that χ(d) and χ (d) in (1.2) are actually the same under the assumption of the existence of χ(d) in a suitable sense. In fact, it is expected that they have the exactly same growth [12,15]. As a consequence, the above four quantities should be all the same, which are called the fluctuation exponent collectively. From the KPZ-theory, it is conjectured that χ(2)(= χ (2)) = 1/3. However for other dimensions, the values are unknown. Some physicists predicted that for sufficiently large dimension, χ(d) = 0 [9,13,22]. If it is correct, the further problem can be conceivable whether the random fluctuation and non-random fluctuation diverge or not. In this paper, we prove that the latter diverges for any dimension d ≥ 2, which is the first result around related models. Accordingly, we believe that the former does so.
We restrict our attention to the following class of distributions. A distribution F is said to be useful if where p c (d) and p c (d) stand for the critical probabilities for d-dimensional percolation and oriented percolation model, respectively and F − is the infimum of the support of F . Note that if F is continuous, i.e., P(τ e = a) = 0 for any a ∈ R, then F is useful.
In particular, by Jensen inequality, We take an arbitrary point x d ∈ ∂B d ∩ ∂B(0, R) (see Figure 1). Then x d satisfies the assumption in Theorem 1.
We will prove Theorem 1 as a corollary of Theorem 2.
Definition 1. For l > 0 and a subset Γ of R d containing the origin , let Although they do not coincide in general, the same proofs still work with a suitable modification and the results below hold even when we replace To consider the directional fluctuation, we define the following cone.
Definition 2. Given θ ∈ R d and r > 0, let where B(x, r) is the closed ball whose center is x and radius is r.
Note that if r > 2, L(θ, r) is the entire R d . Let us consider the divergence of the non-random shape fluctuation F (G(t), tB d ), which was predicted in Remark 2 of [23]. r). Then for any r > 0, there exists c > 0 such that for any sufficiently large t,

1.2.
Notation and terminology. This subsection collects useful notations and terminologies for the proof.
• It is useful to extend the definition of Euclidean distance d(·, ·) as When A = {x}, we write d(x, B).
• Let F − and F + be the infimum and supremum of the support of F , respectively: • We simply write log (2) x = log log x.

Proof of the divergence of the non-random fluctuation
The heuristic behind the proof for sup x∈Z d |ET (0, x) − g(x)| = ∞ is the following. Given is sufficiently large with some probability. However, this strategy does not work directly because ∆(x) is still complicated object. Instead, we first suppose that sup x∈Z d |ET (0, x) − g(x)| < ∞ and we will find a vertex where ∆(x) > 0 with probability greater than 1 , which leads to a contradiction.
2.1. Proof of Theorem 2. We begin with a basic observation of x d . Lemma 1. Let B ⊂ R d be a convex subset and x d ∈ ∂B. Suppose that there exists x 1 ∈ R d and r > 0 such that B(x 1 , r) ⊂ B and x d ∈ ∂B(x 1 , r). Let L be an unique tangent plane of ∂B(x 1 , r) at x d . Then there exists K > 0 such that for any t > 0 and y ∈ tL with |y − tx d | ≤ √ t, d(y, ∂(tB d )) ≤ K.
Proof. By the rotation and translation, it suffices to prove it in the case where d = 2, x 1 = re 2 and x d = 0 (See Figure 1). Then L = {(x, 0)| x ∈ R}. Note that ∂tB(x 1 , r) can be expressed by a function y = tr − t r 2 − (x/t) 2 and if |x| ≤ √ t, tr − t r 2 − (x/t) 2 ≤ K with some constant K > 0 independent of t. Since ∂tB d is between tL and ∂(tB(tx 1 , r)), d(y, ∂(tB d )) ≤ K follows.
Note that L is also a tangent plane of ∂B d at x d . Let K > 0 to be chosen later. Suppose that and we shall derive a contradiction. Then for any > 0, we can take a positive sequence {t n } ∞ n=1 such that t n ↑ ∞ as n → ∞ and for any n ∈ N.
One can find a finite subset S n of t n L such that the following hold: Given a, b, y ∈ R d , we define T (a, y, b) = T (a, y) + T (y, b), which is the first passage time from a to b passing through y. Lemma 2. Under the assumption of (2.2), if we take K > 0 sufficiently large independent of h, for any sufficiently large n ∈ N and y ∈ S n , Proof. Because g is a norm, the triangular inequality leads to g(2t n x d ) ≤ g(y) + g(2t n x d − y) for any y ∈ S n . By the reflection symmetry, we have B(2x d − x 1 , r) ⊂ {x ∈ R d | g(2x d − x) ≤ 1}. By Lemma 1, there exist y 1 ∈ t n B d and y 2 ∈ {x ∈ R d | g(2t n x d −x) ≤ t n } such that |y−y 1 |, |y−y 2 | ≤ K. Since g(x) ≤ 2dE[τ e ]|x| for any x ∈ R d , we obtain for sufficiently large n, g(y) + g(2t n x d − y) − 4dKE[τ e ] ≤ g(y 1 ) + g(y 2 ) ≤ 2t n = g(2t n x d ).
Lemma 4. If F is useful, there exsits δ > 0 and D > 0 such that for any v, w ∈ Z d , For a proof of this lemma, see Lemma 5.5 in [4].
Definition 3. Let c > 0 be a fixed constant. A y ∈ S n is said to be black if for any a, b ∈ B(y, c(log (2) P( y is good ) ≥ 3/8.

Lemma 5.
Under the assumption of (2.2), independent of the choice of S n , we have the following: for any sufficiently large n ∈ N, P( {y ∈ S n | y is good } ≥ S n /4) ≥ 1/8.
We define three events A 1 , A 2 , and A 3 as  Proof. We use the sublinear variance [5,6,10]: Under the assumption E[τ 2 e (log τ e ) + ] < ∞, there exists C > 0 depending only on F and d such that for any x ∈ R d , Then by the Chebyshev's inequality and the union bound, we have P(∃y ∈ S n such that max where C is a constant depending only on d and F . We set A = A 1 ∩ A 2 ∩ A 3 . Note that for sufficiently large n ∈ N, independent of the choice of S n , we have (2.15) P(A) ≥ 1/16.
Lemma 7. If we take h > 0 sufficiently small depending on c, for any y ∈ S n , the following holds: Proof. We use the resampling argument in [4]. Let τ * = {τ * e } e∈E d be independent copy of {τ e } e∈E d . We enlarge the probability space so that we can measure the event both for τ and τ * and we still denote the joint probability measure by P. We defineτ = {τ e } e∈E d as Note that the distributions of τ andτ are the same under P since τ and τ * are independent. Thus P(A y ) = P(Ã y ), whereÃ y is the same condition as A y forτ . We writeT (a, b) for the first passage time from a to b with respect toτ . We defineT (a, y, b) similarly. Since the right hand side of (2.16) equals to it suffices to show that the event in (2.17) impliesÃ y . To do this, we suppose that τ andτ are in this event.
Since {A y } y∈Sn are disjoint from each other, if we take c sufficiently small, by Lemma 7, we get (2.20) Since {y ∈ S n | y is good } ≥ Sn 4 on the event A, by (2.15), this is further bounded from below by S n 4 P(A)(log t n ) −c ≥ (log t n ) 1/32 .
If we take n sufficiently large, we have a contradiction and the proof of Theorem 2 is completed.
By the definition of G(t), we obtain for any t > k, Therefore, writing diam(B d ) = sup{d(x, y)| x, y ∈ B d }, we have which contradicts Theorem 2.
Therefore, for any m ∈ N, we can find a sequence {x