The expected number of critical percolation clusters intersecting a line segment

We study critical percolation on a regular planar lattice. Let $E_G(n)$ be the expected number of open clusters intersecting or hitting the line segment $[0,n]$. (For the subscript $G$ we either take $\mathbb{H}$, when we restrict to the upper halfplane, or $\mathbb{C}$, when we consider the full lattice). Cardy (2001) (see also Yu, Saleur and Haas (2008)) derived heuristically that $E_{\mathbb{H}}(n) = An + \frac{\sqrt{3}}{4\pi}\log(n) + o(\log(n))$, where $A$ is some constant. Recently Kov\'{a}cs, Igl\'{o}i and Cardy (2012) derived heuristically (as a special case of a more general formula) that a similar result holds for $E_{\mathbb{C}}(n)$ with the constant $\frac{\sqrt{3}}{4\pi}$ replaced by $\frac{5\sqrt{3}}{32\pi}$. In this paper we give, for site percolation on the triangular lattice, a rigorous proof for the formula of $E_{\mathbb{H}}(n)$ above, and a rigorous upper bound for the prefactor of the logarithm in the formula of $E_{\mathbb{C}}(n)$.


Background and statement of the main result
Consider critical bond percolation on Z 2 . Kovács, Iglói and Cardy [KIC12] studied the expected number of clusters which intersect the boundary of a polygon. The leading order is the size n of the boundary. The prefactor of this term is lattice dependent. Their main interest is in the first correction term (of order log n). Their motivation came from relations with entanglement entropy in a diluted quantum Ising model. Using indirect and non-rigorous methods from conformal field theory and the q-state Potts model (letting q → 1), they derived a (universal) formula for the prefactor of the logarithmic term.
A special case of their result is that of a line segment (treated in Section F of their paper). In their setup the line segment was placed in the full plane and they claim that the prefactor is equal to 5 √ 3 32π . Furthermore they refer to an earlier obtained result by Cardy in [Car01] (see also Yu, Saleur and Haas [YSH08]) where the line segment was placed on the boundary of the half-plane. In the latter case the claim is that the prefactor equals √ 3 4π . Also this latter result was obtained by non-rigorous arguments using q-state Potts models.
This motivated us to try to find rigorous and more direct proofs of these results (starting with the case of line segments). Since the prefactors are believed to be universal it is natural to consider the most well studied percolation model, site percolation on the triangular lattice with p = p c = 1/2.
Because conformal invariance plays a role, it is convenient to identify the plane with the set C of complex numbers. We embed the triangular lattice T in the halfplane H = {z : z ≥ 0} or the full plane C with vertex set {m + nj : m ∈ Z, n ∈ N ∪ {0}} (resp. {m + nj : m, n ∈ Z}), where j = e π 3 i . We denote the probability measure by P H (resp. P C ) and the expectation by E H (resp. E C ). For subsets A, B ⊂ C we denote by A ↔ B the event that there are open vertices x, y on the triangular lattice, with x ∈ A, y ∈ B, which are connected by a path of open vertices. With some abuse of notation we denote, for any x ∈ C, the set {x} by x. A cluster is a maximal collection of connected vertices. Consider the line segment [1, n] on R, containing n vertices. We are interested in where C G is the collection of clusters in the triangular lattice on the lattice G = H, C.
It is easy to derive the leading (of order n) term: see the Remark in Section 1.2. In the case of the half-plane we could obtain a rigorous proof for the earlier mentioned logarithmic correction term. In the case of the full plane we only obtained a logarithmic upper bound for the correction term. (We do not see a method how to prove the precise prefactor 5 √ 3 32π given in [KIC12]; even finding a non-trivial lower bound is, in our opinion, a challenging problem).
More precisely, our main contribution is a rigorous proof of the following:

Some introductory computations
We now describe the first steps of the strategy to derive the result above. This will also give some insight, where the log comes from. First rewrite the number of clusters as follows Remark: It is known that there is no infinite cluster almost surely, hence P G (k ↔ (−∞, 0]) → 0 as k → ∞. Therefore the above computation implies that the leading term of E G (n) is n(P G (1 ↔ (−∞, 0]) − 1 2 ). Let us introduce the following notation: That is, Hence Theorem 1 is equivalent to (a) lim n→∞ L H (n) = 4π . Take ε > 0. We will introduce M = M (n, ε) ∈ N and a sequence a(i) = a(i, n, ε) for 1 ≤ i ≤ M + 1, such that a(M + 1) = n.
With these values we split up the sum in L G (n) in the following terms. For all 1 ≤ i ≤ M , and Then The idea is now, roughly speaking, to choose a(i, n, ε) so that the ratio of two consecutive ones equals 1 + ε and choose M such that a(1, n, ε) goes to infinity as n → ∞, but is of a smaller order than log(n). Then obviously the term f 0 / log(n) is negligible. We will see that M is more or less of the order log(n)/ε. The existence of the limit lim n→∞ L G (n) would follow if we can show that, for ε close to zero, f i is approximately a constant times ε as n → ∞.
In the case that G = H, we will see in Section 3.1 that this strategy indeed leads to the existence, and even the value, of the limit of L H (n) as n → ∞. Unfortunately in the full-plane it only leads to the upper bound stated in Theorem 1 (b), as we will see in Section 3.2. Now we make the above choices precise. We define and Note that then a(1, n, ε) is of order log(n). To examine f i it is useful to rewrite it in terms of an expectation as follows. Let (5) 2 Ingredients from the literature In this section we state some results, which we will use in Section 3 to prove Theorem 1. First some additional notation. We use the following notation for the probabilities of so-called arm-events. Let, for m < n ∈ N and let π 3 (m, n) be the probability of having two disjoint closed paths, and an open path, from [−m, m] 2 to H \ [−n, n] 2 . The following lemma is well known (see for example Theorem 11, Proposition 14 and Theorem 24 in [Nol08]).
Lemma 2 There exist constants C 1 , C 2 > 0 and α ≤ 1/2 such that, for all m < n In fact, much more precise results for these probabilities are known, but will not be used in this paper.
In the rest of this section, for a simply connected domain D C and n ∈ N the notation nD denotes the set {n · u : u ∈ D}. For points a 1 , a 2 on the boundary of D we denote by [a 1 , a 2 ] the part of the boundary of D between a 1 and a 2 in the counter clockwise direction. Furthermore we generalize the notation slightly, namely by P D (and E D ) we will denote the probability measure for percolation restricted to the triangular lattice on D. In this setting two intervals [a 1 , a 2 ] and [a 3 , a 4 ] on the boundary are said to be connected if there are vertices x, y on the lattice inside D, which are connected by an open path, and are such that x has an edge which crosses [a 1 , a 2 ] and y has an edge which crosses [a 3 , a 4 ].
The first theorem is the famous Cardy's formula (proposed in [Car92]), which was proved by Smirnov in [Smi01].
Theorem 3 (Cardy's formula, [Smi01]) Let D C be a simply connected domain and φ : D → H a conformal map. Let a 1 , a 2 , a 3 , a 4 be ordered points on the boundary of D. We have where λ is the cross-ratio This theorem concerns crossing probabilities of generalized rectangles in one 'direction'. The following theorem gives a formula for probabilities of crossings in two directions. It is called after Watts, who proposed the formula in [Wat96]. The first rigorous proof was by Dubédat [Dub06]. An alternative proof was obtained by Schramm (see [SW11]).
Theorem 4 (Watts' formula, [Dub06,SW11]) Let D C be a simply connected domain and φ : D → H a conformal map. Let a 1 , a 2 , a 3 , a 4 be ordered points on the boundary of D. We have where λ is the cross-ratio (8).
The last theorem we state here concerns the expected number of crossing clusters of a rectangle. It was predicted by Cardy [Car01] and by Simmons, Kleban and Ziff [SKZ07]. A proof was given by Hongler and Smirnov in [HS11]. Here  N (nD, a 1 , a 2 , a 3 , a 4 ) N (nD, a 1 , a 2 , a 3 , a 4 where λ is the cross-ratio (8).

Proof of Theorem 1
Recall from the introduction that Theorem 1 is equivalent to 4π . Recall the definition (5) of T (i). We begin this section with a lemma which says that, to prove the convergence of L G (n) as n → ∞, it is sufficient to prove the convergence of ε −1 E G [T (i)].

Lemma 6
The following inequalities hold. Thus it is enough to prove that lim sup Hereto, note that it is also easy to see from the definition of M that, for fixed ε > 0 For all ε > 0 we have Next note that lim sup This together with (12) implies (11) and completes the proof of (9). The inequality in (10) follows in a similar way and we omit it.

Proof of Theorem 1 (a)
First note that it is easy to see that {T Therefore It is well-known from standard RSW arguments that P H (T (i) ≥ 1) goes, uniformly in i and n, to 0 as → 0, since the ratio between two consecutive a(i)'s goes to 1 as ε → 0. Hence the 'error term' (i.e. the second term in the r.h.s. of the equation array above) is negligible w.r.t. the main term (i.e. the first term in the r.h.s.). By this, Lemma 6, the fact that a(1) → ∞ as n → ∞, and the ratio between consecutive a(i)'s, it is sufficient to prove that where W k denotes the event that there is an open and a closed path from (−∞, 1] to [k, k(1 + ε)] and the closed path is below the open path.
Let W k be the event that there is an open and a closed path from (−∞, 1] to [k, k(1 + ε)]. (So, informally speaking, W k is the same as W k without the condition on which path is above or below). Using that (by duality), there is either an open path from [1, k] to [k(1 + ε), ∞) or a closed path from (−∞, 1] to [k, k(1 + ε)], we have The limits as k → ∞ of the first probability in the r.h.s. and the probability in the l.h.s. are obtained by Theorem 3 and Theorem 4 respectively, and we get Finally, letW k denote the event obtained from W k by replacing 'open' by 'closed' and vice versa. Since W k andW k have the same probability and W k =W k ∪ W k , we have Since W k ∩W k is contained in the disjoint occurrence of W k and the event that there is an open or closed path from (−∞, 1] to [k, k(1 + ε)], its probability is negligible (as k → ∞ and ε → 0) w.r.t. that of W k , and we get from (17) and (18) that · ε + o(ε).
As we saw (see the argument above (15)) this proves Theorem 1 (a).

Proof of Theorem 1 (b)
We will bound the relevant probabilities (concerning the full plane) by the probabilities of certain connection events in the half-plane. We do this by cutting along the real line from −∞ up to a(i + 1). Let us make the cutting precise. Let We have To bound P C (B(i)) we use the first inequality of Lemma 2 for those k in the definition of B(i) that are 'close to' 1 or a(i), and the other inequality in that lemma for the other k's. More precisely, we fix a constant β ∈ (0, 1), and let r(a(i)) := a(i) β . Then, P C (B(i)) ≤ 4π 1 (r(a(i)), a(i)) + 4 1 2 a(i) k=r(a(i))+1 where the factor 4 comes from symmetry considerations. Hence, there exist constants C 3 , C 4 > 0 such that Note that, since a(1) (the smallest of the a(i)'s) tends to ∞ as n → ∞, and C 3 (x β−1 ) α + C 4 x β tends to 0 as x → ∞, the contribution of P C (B(i)) to the r.h.s. of (9) is 0. , and one preventing it to touch the 'lower piece'; see e.g. Figure 3). So in this case we have S(i) = 2. More generally, a similar observation gives Thus it follows immediately, that To complete the proof we will use Theorem 5. Therefore we consider the domain C\L(i) and scale it by a(i). (As noted before, a(1) goes to ∞ as n → ∞). This gives the conformal rectangle C \ (−∞, 1 + ε) with 'corners' a 1 = 1 + , a 2 = 1 − , a 3 = 0 − , a 4 = 0 + (where, for x < 1+ε, x + and x − denote the 'copy' of x in the upper and the lower half-plane respectively). To apply Theorem 5 we need the cross-ratio, which can be computed as follows: Consider the conformal map ϕ(z) := i √ z − 1 − ε which maps C \ (−∞, 1 + ε) onto the upper half-plane. The cross-ratio is λ(ε) = (ϕ(1 + ) − ϕ(1 − ))(ϕ(0 + ) − ϕ(0 − )) (ϕ(1 + ) − ϕ(0 − ))(ϕ(0 + ) − ϕ(1 − )) .