Critical Exponents on Fortuin–Kasteleyn Weighted Planar Maps

In this paper we consider random planar maps weighted by the self-dual Fortuin–Kasteleyn model with parameter q∈(0,4)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${q \in (0,4)}$$\end{document}. Using a bijection due to Sheffield and a connection to planar Brownian motion in a cone we obtain rigorously the value of the annealed critical exponent associated with the length of cluster interfaces, which is shown to be 4πarccos2-q2=κ′8,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\frac{4}{\pi} \arccos \left( \frac{\sqrt{2 - \sqrt{q}}}{2} \right)=\frac{\kappa'}{8},$$\end{document}where κ′\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\kappa' }$$\end{document} is the SLE parameter associated with this model. We also derive the exponent corresponding to the area enclosed by a loop, which is shown to be 1 for all values of q∈(0,4)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${q \in (0,4)}$$\end{document}. Applying the KPZ formula we find that this value is consistent with the dimension of SLE curves and SLE duality.


Setup
Random surfaces have recently emerged as a subject of central importance in probability theory. On the one hand, they are connected to theoretical physics (in particular string theory) as they are basic building blocks for certain natural quantizations of gravity [24,9,20,11]. On the other hand, at the mathematical level, they show a very rich and complex structure which is only beginning to be unravelled, thanks in particular to recent parallel developments in the study of conformally invariant random processes, Gaussian multiplicative chaos, and bijective techniques. We refer to [15] for a beautiful exposition of the general area with a focus on relatively recent mathematical developments.
This paper is concerned with the geometry of random planar maps, which can be thought of as canonical discretisations of the surface of interest. The particular distribution on planar maps which we consider was introduced in [27] and is as follows. Let q < 4 and let n ≥ 1. The random map M n that we consider is decorated with a (random) subset T n of edges. T n induces a dual collection of edges T * n on the dual map of M (see Figure 1). Let m be a planar map with n edges, and t a given subset of edges of m. Then the probability to pick a particular (m, t) is, by definition, proportional to P(M n = m, T n = t) ∝ √ q , (1.1) where is the (total) number of loops induced by t, that is, the combined number of clusters in T n and T * n . In other words, given the map M n = m, the collection of edges T n follows the distribution of the self-dual Fortuin-Kastelyn model, which is in turn closely related to the critical q-state Potts model, see [2]. Accordingly, the map M n is chosen with probability proportional to the partition function of the Fortuin-Kastelyn model on it.
One reason for this particular choice is the belief (see e.g. [13]) that after Riemann uniformisation, in the scaling limit, such maps are closely related to Liouville quantum gravity. This is the random metric obtained by considering the Riemannian metric tensor e γh(z) dz, (1.2) where h(z) is an instance of the Gaussian free field. (We emphasise that a rigorous construction of the metric associated to (1.2) is still a major open problem.) The parameter γ ∈ (0, 2) is then believed to be related to the parameter q of (1.1) by the relation q = 2 + 2 cos 8π κ ; γ = 16 κ . (1.3) Note that when q ∈ (0, 4) we have that κ ∈ (4, 8) so that it is necessary to generate the Liouville quantum gravity with the associated dual parameter κ = 16/κ ∈ (0, 4). This ensures that γ = √ κ ∈ (0, 2), which is the non degenerate phase for the associated mass measure and Brownian motions, see [14,4,17].
Observe that when q = 1, the FK model reduces to ordinary bond percolation. Hence this corresponds to the case where M is chosen according to the uniform probability distribution on planar maps with n edges. This is a situation in which remarkably detailed information is known about the structure of the planar map. In particular, a landmark result due to Miermont [23] and Le Gall [21] is that, viewed as a metric space, and rescaling edge lengths to be n −1/4 , the random map converges to a multiple of a certain universal random metric space, known as the Brownian map. (In fact, the results of Miermont and Le Gall apply respectively to uniform quadrangulations with n faces and to p-angulation for p = 3 or p even, whereas the convergence result concerning uniform planar maps with n edges was established a bit later by Bettinelli, Jacob and Miermont [6]). Critical percolation on a related half-plane version of the maps has been analysed in a recent work of Angel and Curien [1], while information on the full plane percolation model was more recently obtained by Curien and Kortchemski [8].
The goal of this paper is to obtain detailed geometric information about the clusters of the self-dual FK model in the general case q ∈ (0, 4). As we will see, our results are in agreement with nonrigorous predictions from the statistical physics community. In particular, after applying the KPZ transformation, they correspond to Beffara's result about the dimension of SLE curves [3] and SLE duality (see the discussion below).

Main results
Let L n denote a typical loop, that is, a loop chosen uniformly at random from the collection of clusters (both primal and dual) induced by T n in (1.1). By loop we refer specifically to a collection of triangles surrounding a connected (primal or dual) component of T n or T * n (see Section 2.3). Each loop disconnects the map M n into two components, and we will call the inside (or interior) of L n the one which does not contain the root of M , and the outside the one which contains the root. (On the unlikely event that the root lies on the loop itself, toss a coin). We view the inside of the loop L n as a random map, rooted for the sake of discussion at an edge chosen uniformly on its boundary ∂L n , though this is unimportant. Our main results concern the limiting distribution of L n as n → ∞, for the local topology (or rather, since the map is rooted at a random edge, for the Benajmini-Schramm topology). Set Theorem 1.1. We have that L n → L in distribution for the Benjamini-Schramm topology, where L is a.s. a finite map. Furthermore, letting |L| denote its area and |∂L| denote its boundary length, the following inequalities hold. For any ε > 0, there exist constants c(ε), C(ε) depending solely on ε such that for all k ≥ 1,

5)
and c(ε)k −p 0 −ε ≤ P(|L| ≥ k) ≤ C(ε)k −p 0 +ε . (1.6) Remark 1.2. As we were finishing this paper, we learnt of the related work, completed independently and simultaneously, by Gwynne, Mao and Sun [19]. They obtain several scaling limit results, showing that various quantities associated with the FK clusters converge in the scaling limit to the analogous quantities derived from Liouville Quantum Gravity in [13]. Some of their results also overlap with the results above. In particular they obtain a slightly stronger version of (1.6) by showing that in addition, P(|L| ≥ k) is in fact regularly varying. Though both papers rely on the Sheffield bijection [27] and a connection to planar Brownian motion in a cone, it is interesting to note that the proof techniques are substantially different. The techniques in this paper are comparatively simple, relying principally on harmonic functions and appropriate martingale techniques.
Returning to Theorem 1.1, it is in fact not so hard to see that when rooted at a randomly chosen edge, the decorated maps (M n , T n ) themselves converge for the Benjamini-Schramm (local) topology. This is already implicit in the work of Sheffield [27] and properties of the infinite local limit (M ∞ , T ∞ ) have recently been analysed in a paper of Chen and Curien [7]. Among the properties that they obtain is a uniform exponential bound on the degree of the root. Together with earlier results of Gurel Gurevich and Nachmias [18], this implies for instance that random walk on M ∞ is a.s. recurrent.
Our results can also be phrased for the loop L * going through the origin in this infinite map M ∞ . However issues of biasing then arise, and thus the exponents are slightly different. For instance, we get where ≈ has the meaning of the inequalities appearing for instance in (1.5). For the area, our results suggest the following though we did not verify some technical lemmas which would be necessary to derive (1.8). (The authors of [19] have kindly indicated to us that (1.8), together with a regular variation statement, could probably also be deduced from their Corollary 5.3 with a few pages of work, using arguments similar to those already found in their paper).
In the particular case of percolation on the uniform infinite random planar map (UIPM) M ∞ , i.e. for q = 1, we note that our results give p 0 = 3/4, so that the typical boundary loop exponent is 1/p 0 = 4/3. This is consistent with the more precise asymptotics derived by Curien and Kortchemski [8] for a related percolation interface. Essentially their problem is analogous to the biased loop case, for which the exponent is, as discussed above, 1/p 0 − 1 = 1/3. This matches Theorem 1 in [8], see also Theorem 2 (ii) in [1] for the half-plane case. Likewise, the exponent for the area of L * (in the biased case) is 1 − p 0 = 1/4, which matches (i) in the same theorem of [1].

Isoperimetry, KPZ formula, bubbles and dimension of SLE
Cluster isoperimetry. The exponents in (1.7) and (1.8) suggest that for a large critical FK cluster on a random map, we have the following approximate relation between the area and the length: as |L * | → ∞. Though (1.8) is not rigorously established in this paper at this stage, a slightly different version of the isoperimetric relation (1.9) is nevertheless proved in Proposition 4.11. KPZ formula. We now discuss the consequences of the above isoperimetric property in connection with the KPZ relation. First recall that for a fixed or random independent set A with Euclidean scaling exponent x, its "quantum analogue" has a scaling exponent ∆, where x and ∆ are related by the formula See [14,25,5] for rigorous formulations of this formula at the continuous level. Concretely, this formula should be understood as follows. Suppose that a certain subset A within a random map of size N has a size |A| ≈ N 1−∆ . Then its Euclidean analogue within a box of area N (and thus of side length n = √ N ) occupies a size |A | ≈ N 1−x = n 1/2−x/2 . In particular, observe that the approximate (Euclidean) Hausdorff dimension of A is then 2 − 2x.
Dimension of SLE. Such considerations may be applied after revealing N triangles in the exploration of the planar map, assuming that the corresponding cluster satisfies |L| ≥ N , with the set A being the boundary of the cluster at that stage. (Note that this falls within the realm of application of the KPZ formula, since conditionally given the map M , the FK configuration is chosen independently of it). By (1.9), the (quantum) scaling exponent is thus We recall that q and γ are related via (1.3). Thus q = 2 + 2 cos 8π κ = 4 cos 2 4π κ using the angle doubling formula. Thus, since κ ∈ (4, 8), Plugging this into (1.10) gives us a Euclidean scaling exponent of This suggests that the Hausdorff dimension of the boundary of L, in the Euclidean setting, is This matches Vincent Beffara's rigorous computation of the Hausdorff dimension of the SLE(κ) curve [3]. Bubble isoperimetry. The isoperimetric relation (1.9) dealt with large critical clusters. We now address a different sort of isoperimetric relation, which concerns large filled in bubbles within a given critical cluster. To explain what this is, we first need to say a few words about Sheffield's bijection [27]. This provides us with a canonical way to explore the random planar map via a space-filling interface. A bubble is a part of the cluster formed between the beginning of the exploration process and the first cutedge which would disconnect the root from the next triangle to be explored. In other words, this is a pivotal edge for the root being connected to the next triangle to be discovered along the interface. (This terminology is the same as in Curien and Kortchemski [8]). A large, filled-in cluster, will consist of a looptree of bubbles, see Figure 2.) Let B denote the first bubble (attached to the root). Then we will see in Proposition 4.10 that, in a certain sense, Remarkably, this isoperimetric behaviour is independent of q (or equivalently of γ) and therefore corresponds with the usual Euclidean isoperimetry in two dimensions. Again, this has interesting consequences when applying the KPZ formula (1.10), which we now explain.
Dimension of dual SLE. Suppose we consider a large bubble B such that |B| ≈ N . Using the KPZ formula to the boundary ∂B should allow us to recover the Hausdorff dimension of the dual SLE, since ∂B corresponds to a piece of the outer boundary of the cluster as the cluster has been filled in, which (in the scaling limit) is believed to be described by an SLE curve with the dual parameter κ = 16/κ. This SLE duality was first conjectured by Duplantier and proved rigorously recently by Julien Dubédat [12] and Dapeng Zhan [28].
Here is the detail of the KPZ computations. From (1.11) we find that ∆ = 1/2, and hence from (1.10) we get Consequently, the (Euclidean) dimension of the outer boundary of a large cluster is therefore Again, this is consistent with Beffara's result on the Hausdorff dimension of the SLE curve with the dual parameter κ = 16/κ.

Hamburger/Cheeseburger word
We consider possibly infinite (and even possibly doubly-infinite) words {X i }, on the alphabet consisting of the five letters { C , H , C , H , F }. We think of such a word as describing a lastin, first-out queuing process with two products, C or cheeseburgers and H or hamburgers which are kept on a single stack. More precisely we associate to a word {X i } i≥0 the following process. We start with an empty stack of burgers and then discover the symbols {X 0 , . . .} one by one. When we see a C (resp. H ), we interpret it as the production of a cheeseburger (resp. hamburger) and we add it to the top of the stack. When we see a C (resp. H ), we interpret it as a cheeseburger (resp. hamburger) order and we remove from the stack the topmost or freshest cheeseburger (resp. hamburger). If there is no cheeseburger in the stack, we say it is an unfulfilled order. When we see an F , we interpret it as a free (or flexible) order and we remove the top burger (if it exists) from the stack regardless of its type. This is extended straightforwardly to the case of a doubly infinite word by considering a non empty (possibly infinite) initial stack. Given a finite word X = {X i } 1≤i≤n , we define the reduced word of X, which we writeX, as the concatenation of first all the unfulfilled orders in X then the burgers in the stack at the end of X (unfulfilled burgers) in the order in which they appear in X. This corresponds to reducing the word by applying the following rule : We call the order that eats a burger or the production that fulfils an order its match and we write φ(i) ∈ Z for the time at which this match occurs.

Map with a percolation model
Given a bi-infinite word X such that every symbol has a match, we can construct an infinite planar triangulation with coloured edges and vertices by the following procedure.
To each matched pair of marked symbols, we associate a quadrangle of green edges, crossed by either a black or red edge, and with labels on the green edges according to the rule described in figure 3. Finally we glue together each pair of green edges with the same label (with an orientation to keep the color of the vertices).
It is not immediately obvious but nevertheless true that the map we construct with this procedure is planar, as can for example been seen using the exploration described in the next section. Once this is observed, it is immediate that the sub-map made by green edges is a quadrangulation. Therefore using the so-called 'trivial' bijection, we have a pair of dual black and red maps. The black and red edges in the triangulation are dual subsets of the black and red maps, meaning that for every edge e of the black map, either e or its dual edge is in the triangulation, and vice-versa.
Notice that the labels on the green edges describe a natural way to explore the map step by step along a bi-infinite, space-filling path. We will describe more precisely this procedure later. We can also remark that the triangles of the map are in bijection with the symbols in the word. The F orders are somewhat special, and a certain choose has to be made in this bijection. Namely, there are two triangles associated with the flexible order and its match, but it is a matter of convention to decide which triangle to associate to the flexible order and which to its match.
We can draw loops (and an infinite path) crossing the green edges to separate the black clusters and the red clusters and it is easy to see that each F corresponds to a single loop, in a way that will be described more precisely later on.
Let the word X be an iid sequence of symbols taken with the following probability : The bijection described in [27] (the Hamburger-Cheeseburger bijection) is a measure preserving transformation between the FK weighted maps (M n , T n ) (described in Section 1) and finite sequences (X 1 , . . . , X 2n ) conditioned on X 1 , . . . , X 2n = ∅ (or in other words, the conditioning says that every symbol has a match). It can be shown (see [7,27]) that the local limit of such a conditioned sequence is X and that almost surely every symbol in X has a match. Thus, the (random) planar map obtained from X is the local limit of planar maps with n edges picked together with a FK-model as described in the introduction. In the following we will focus only on the case p < 1 2 , corresponding to q < 4.
Remark 2.1. The case where p > 1/2 (corresponding to q > 4) is believed to behave very differently. We note that Sheffield [27] obtained a phase transition for certain features of the model when p > 1/2.

An exploration procedure
In this section, we summarise the exploration procedure introduced by Sheffield [27] of a map given a bi-infinite sequence of symbols from the set of symbols { C , H , C , H , F }. The idea is to reveal the map, triangle by triangle, each triangle corresponds to either a cheeseburger production or a hamburger production or their corresponding match.
Going to the past. We start with the green root edge with a black primal vertex and a dual red vertex. Recall that the triangle corresponding to a hamburger order has two red dual vertices and one black primal vertex and the triangle corresponding to a cheeseburger has two black primal vertices and one red dual vertex. In every step, we reveal a triangle incident to a green edge. Further, the map revealed will be a triangulation of a polygon, whose boundary consists of a connected black segment, a connected red segment, the green root edge and a green edge where we reveal the next triangle. The green edges join together the red and black boundary segments as shown in Figure 4. To continue with the exploration, we first convert every F symbol into the corresponding C or H symbol. The exploration proceeds as follows.
• If we encounter a C , we add a cheeseburger triangle. If it is an H , we add a hamburger triangle.
• If we encounter a C , which do not correspond to any C symbol explored so far, we add a cheeseburger triangle. We do an analogous thing for an unmatched H .
• If the C we encounter has a match among the symbols revealed up to this point, we draw the green edge to match it with the cheeseburger triangle incident to the explored edge. We do an analogous thing for a H which has a match.
• If the match was originally an F , then we convert the type of the burger which this F symbol and its match correspond to as shown in Figure 4. Figure 4: The exploration process shown step by step for a sequence of symbols where the symbol explored is shown by an up arrow mark in each step. We go into the past. The grey triangles correspond to burgers which will remain unmatched throughout the exploration of the past. In the 7th and the 10th figure, a part of the interface (purple dashed line) breaks off into a separate loop. The last figure is the exploration path (in blue) without the loops breaking off. The fictitious edges which are part of the spanning tree which we changed because of the F symbols are drawn with dotted edges.
Each symbol in the word naturally corresponds to an arrow in the blue exploration path. The loop associated to the last F is the thick one on the left. The directions of the arrows indicate the order in which symbols appear from right to left.
• In each step, we add the exploration interface connecting the green edges of the triangle we add. Notice that matching an F symbol results in a parts of the exploration path breaking up into two parts to complete a new loop as shown in Figure 4.
As already mentioned, to every symbol in the doubly infinite sequence {X i } i∈Z , there is a.s. a unique match: if that symbol is a burger production then the match corresponds to the time at which the burger is eaten, while if it is a burger order, then the match corresponds to the burger which is being eaten. The matching procedure is formally an involution φ : Z → Z.
It is clear that every F symbol in the infinite word {X i } i∈Z corresponds to a loop and we can discover the loop by exploring the symbols to the left of the F symbol until we find its match. The interface separates the map into a finite and infinite component. The finite component is called the interior of the loop. The length of the loop is the number of triangles through which it passes. The number of triangles which lie completely in the interior of the loop plus the length of the loop is called the area inside the loop.
For every F symbol, its envelope is the F symbol, its match, and all the symbols in between them in the infinite sequence. The map corresponding to an envelope is the subset of triangles in the map corresponding to its symbols. An envelope is of type C F (resp. H F ) if the match of the F is a C (resp. H ).
Proposition 2.2. Condition on X 0 = F . Now reveal the map by exploring symbols to the left of X 0 one by one until we find the match of X 0 at time |φ(0)| (that is, we reveal the map corresponding to the envelope of X 0 ), and stop at that point. Then all the F symbols we explore before that point must find their respective match before we find the match of X 0 . Let L denote the map corresponding to the envelope of X 0 .
• The triangles through which the loop through X 0 denoted ∂L passes can be described as follows.
We start exploring from X −1 as described above until the match of X 0 . If we see anything other than F , we add one triangle to ∂L. If we see an F while exploring, we continue the exploration until we find its match and only one triangle from this whole operation is added to ∂L.
• Suppose during the exploration we are on a symbol which is not in the envelope of an explored F and the next symbol x we explore is an F . Then the triangles corresponding to the envelope of x is in the interior (resp. exterior) of L if its type is opposite (resp. same) as that of X 0 .
• The boundary of L (the triangles which are incident to a triangle outside L) consists of the triangles described by the reduced word X φ(0) . . . X 0 .
Proof. The fact that every F symbol we encounter must find its match before we find the match of X 0 is by definition: there cannot be an unmatched F symbol in the reduced word of the symbols explored. Also to see the first part of the proposition, notice that every envelope corresponds to a loop, hence a symbol corresponds to a triangle through which ∂L passes if and only if it is not inside an envelope of some other F symbol explored except the triangle corresponding to that F symbol itself.
It is easy to see that for any H F loop, all vertices of the triangles crossed by the loop and in the interior of the loop are primal (black) vertices and belong to the same cluster. Similarly, all vertices of the triangles crossed by the loop and in the exterior of the loop are dual (red) vertices in the same cluster. The same statement holds with primal replaced by dual for C F loops (see Figure 4). Now assume by symmetry that ∂L is a H F loop. When we encounter x, its loop must pass through a triangle adjacent to the triangle x explored immediately before x. If x is of type C F , then it is surrounded outside by primal vertices (see Figure 4). Hence the primal vertices of x common to that of the loop of x is in the interior of L. Hence the map m x corresponding to the envelope of x lies in the interior. This is because the triangles corresponding to the envelope of x forms a connected map and a strict subset of it being in the exterior of L would mean that L passes through m x which is impossible. The argument for the other cases follow similarly which we leave to the reader.
Going to the future. There is an analogous method of constructing a map while moving to the future of X 0 : {X i } i>0 . In this situation, one needs to add triangle corresponding to burger production and close them corresponding to orders. Notice that we might need to look in to the past to find out what the F symbols in the future of 0 matched to the past correspond to. We will not use this exploration in this paper, so we do not provide the details. Notice that using the exploration procedure for going in to the past or the future, we can explore the whole infinite map.

Preliminary lemmas 3.1 Forward-backward equivalence
In this section, we isolate some events whose probabilities we need to estimate. Modulo these estimates, we complete the proof of Theorem 1.1 at the end of this subsection.
Throughout the rest of the paper, we fix the following notations: Note that the value of p 0 is identical to the one in (1.4) (after applying simple trigonometric formulae).
For a time i ∈ Z, let ϕ(i) denote its match. Define E to be the event that X 0 ∈ { C , H } and X 0 is at the top of the burger stack at some point of time strictly greater than 0 before being eaten. On the event E, define {X 0 is at the top of the burger stack at time n}.
In other words, Σ is the first burger to the right of zero eaten by a flexible order F , and T is the time it takes for that burger to be eaten by such an order. Now Also let J T = X(0) . . . X(T ) and J T = X(Σ) . . . X(φ(Σ)) be the corresponding reduced words. Let S denote an arbitrary infinite burger stack.
In particular this bound is independent of the conditioning on S. It turns out that it is slightly easier to connect T , J T with that of the observables in the map. We prove in the next proposition that it is enough to prove Theorem 3.2 to obtain the tail exponents of T , J T . The following lemma will be also used when we estimate the isoperimetric relations for a bubble and its boundary in Section 4.5. Roughly, T represents the volume of the bubble and J T represents its boundary.
Proposition 3.4. Let T , J T , T, J T be as above. Let S i denote the burger stack when the ith burger is produced. Then for all borel sets A, B and n ≥ 1, there exists a constant c > 0 such that Proof. Let F T be the stopped sigma field generated by T . Observe that {T < ∞} on E. Thus which completes the proof for lower bound. For the upper bound first we need some estimates to know how long we need to wait until we find the first burger which is eaten by an F the first time it returns to the top of the stack. Conditioned on the past, this event has positive chance so one needs to look at logarithmic many burgers to have a polynomial chance to find one burger satisfying this property. To make this precise let β i denote the time when the ith burger is produced. Let F β i denote the stopped sigma field. Fix n ≥ 1. Let T i denote the time taken for the ith burger returns to the top of the stack for the first time. Let F i be the event that T i is infinite or X β i +T i +1 = F , and consider the event B n = ∩ n i=1 F i . Also let B i be the event that the ith burger produced not followed immediately by the sequence Also it is clear that B n ⊂ ∩ 1≤i≤n/2 B 2i . Hence using (3.1) iteratively, we obtain Notice that the probability of the event {F c i , T i ∈ A, J T i ∈ B} conditioned on F β i is at most that of {T ∈ A, J T ∈ B, E} conditioned on S i . This completes the proof.
Now we present a proposition which helps us suitably represent the law of T , J T . Proposition 3.6. The following random finite words have the same law • The envelope of the first F to the left of 0.
• The envelope of the first C or H to the right of 0 matched with a F .
• The envelope of the X 0 the symbol at 0, conditioned on X 0 being a F .
• The envelope of X 0 , conditioned on X 0 being matched with a F Furthermore this is the limit law as n → ∞ for the envelope of a F taken uniformly at random in from an iid sequence Z 1 , . . . , Z n distributed as X 0 and conditioned on Z 1 . . . Z n = ∅.
We call this law, the envelope of a typical loop.
Proof. This is immediate from the fact that the sequence is iid.
Let W be the set of words W = W 0 , . . . W n that end with a F and start by its match. Let Π W = 1 2 n−1 1 p(W i ). For the first case, since the sequence is iid, the law to the left of of the first F left of 0 is still iid so the probability for the envelope of F to be some W is exactly Π W . The same holds for the third case.
For the second and fourth case, observe if a burger is matched with a F , then then its envelope has to be a word in W. The probability to see a specific W (which include the type of the first burger) is then Π W /( W P i W ). However from the previous cases, we see that W P i W = 1 so we are done.
Consider the finite map (M n , T n ). Recall |∂L n | denotes the length of a uniformly picked loop from (M n , T n ). Let |L n | denote the number of triangles in the map which is in the component of the complement of ∂L n not containing the root if the loop ∂L n do not pass through the root.
where the convergence above is in law. The distribution of ∂L and L is defined as follows. Take an iid sequence {X i } conditioned on X 0 = F and consider its corresponding map. Then ∂L has the same law as the symbols corresponding to the loop through X 0 (as described in Proposition 2.2) and L has the same law as the symbols corresponding to its interior area (as described in Proposition 2.2). We shall call |∂L| the typical loop length and |L| the typical interior area.
Proof. Notice that there is a one to one correspondence between the number of F symbols in the finite word corresponding to (M n , T n ) except there is one extra loop. But since the number of F symbols in the finite word converges to infinity, the probability that we pick this extra loop converges to 0. The rest follows from Propositions 2.2 and 3.6.
Proof of Theorem 1.1. The convergence in the Benjamini-Schramm topology follows from Lemma 3.7. Also, via Lemma 3.7, we conclude that if we condition on X 0 = F and explore the map by looking at symbols to its left, then we will reveal the whole loop and the triangles belonging to the interior area by exploring until when we find the match of X 0 . To get a hand on the loop length and area we define the following way to look at the exploration which we call reduced walk. Start with (c 0 , h 0 ) = (0, 0). Suppose we have defined (c n , h n ).
• If we see a F , then we explore until we find its match. Notice that the reduced word that comes out of this part of the exploration is either of the form C C . . . C or H H . . . H depending on whether the match of the F is a H or C respectively. Either happens with equal probability. Let Z denote the number of symbols in the reduced word revealed in this step (the expectation of this quantity is χ − 1 of [27]). If we reveal H , define (c n+1 , h n+1 ) = (c n , h n ) + (0, Z). Otherwise, define (c n+1 , h n+1 ) = (c n , h n ) + (Z, 0).
Sheffield [27] computed the expected value of Z to be 1. From this, it is easy to see that each of c n or h n is a sum of n i.i.d. variables with mean 0 and a heavy tail with support [1, ∞). Furthermore it is easy to see that the tail is given by the tail of J T computed in Corollary 3.5. Now notice |∂L| = inf{k : h k < 0 or c k < 0}. In other words, we need to find the first time when this 2-dimensional walk leaves the positive quadrant. But it is easy to observe that this reduces to a one-dimensional problem. Indeed, in every step only one of the co-ordinate changes with probability 1/2 and the transitional probabilities are independent of the value of the other co-ordinate. So if the walk is run in continuous time with jump rates given by independent Poisson clocks of rate 1 for each co-ordinate, then the walks become completely independent. Thus if τ C is the time taken for just the C-walk to hit 0 then P(∂L > n) = (P(τ C > n)) 2 . For the C-walk, the time of the first return to 0 is well known (see, e.g., [16]) and in our case, using the tail bounds on J T from Corollary 3.5, we get To get a handle on A recall the notation T from Corollary 3.5. Observe that T denotes the total number of triangles revealed in the exploration and hence is definitely more than A. Thus the upper bound is easy: For the lower bound, we break up T as T = τ i=1 (T C i + T H i ) defined as follows. In every reduced walk exploration step, if the walk moves in the C-direction, then T C i denotes the number of triangles explored in this step otherwise T H i with probability 1/2 and each of them has the same distribution (though not independent). Hence we have The rest follows from the tail of T from Corollary 3.5.

Connection with random walk in cone
Given the sequence {X i } i∈Z , we can construct the sequence {Y i } i∈Z where we convert every F symbol in {X i } i∈Z into the corresponding C symbol or H symbol. Now define and C n = n i=1 c(Y i ). Also define D n to be the discrepancy count. Define Now define D n = n i=1 d(Y i ). Define α = 1 − 2p. Figure 6: The co-ordinate transformation we use.
Theorem 3.8 (Sheffield [27]). We have the following convergence uniformly in every compact interval where (B t ) t≥0 is the standard 2 dimensional Brownian motion.
A major tool to prove the above theorem is the following Lemma proved by Sheffield [27].
Lemma 3.9 (Sheffield [27]). The number of F symbols in Remark 3.10. It follows from the above that if S and S are two infinite burger stacks and X 0 , X 1 , . . . is a fixed semi-infinite word, then the processes (C n , D n ) constructed as in (3.6) and (3.7) using respectively S and S rescale asymptotically as n → ∞ to the same realisation of a Brownian motion.
Recall θ 0 = 2 arctan( 1/α). We now perform the following change of co-ordinates to (D n , C n ) for convenience.
• First scale the second coordinate by √ α to obtain (D n / √ α, C n ) n≥1 .
• Now apply a rotation counterclockwise to (D n / √ α, C n ) n≥1 by an angle θ 0 /2 to obtain (D n ,C n ) n≥1 .
• Let v 0 be the vector (cot θ 0 , 1). Now define Proposition 3.11. We have the following convergence uniformly in any compact interval where (B t ) t≥0 is a standard Brownian motion.
Proof. This follows from rotational invariance of Brownian motion and Donsker's invariance principle.
Definition 3.12. The process (R t , t ≥ 0) is known as a Brownian motion reflecting on the real axis in the direction v 0 .
Proposition 3.13. The events {T > n} ∩ E and {T * > n} ∩ E * have the same probability.
Proof. We can assume we are on the event X 0 = H since if X 0 = C we can interchange the role of cheeseburgers and hamburgers to arrive at the same conclusion. The key observation is that the walk exiting the cone is equivalent to X 0 being eaten and the walk reaching (0, 0) before exiting the cone means that X 0 has reached the top of the stack before being eaten. To see this, observe that when the walk hits the X-axis, then the reduced word consists of only H and when it hits the other wall {r, θ 0 ; r > 0} = {z : arg(z) = θ 0 }, there can only be C in the reduced word. If it is in the other wall, exiting the cone would mean a H or F order which would match with X 0 .
On the other hand, if we are on the X axis, then a cheeseburger order would only add a C to the reduced word. Finally, if the walk is at the origin without exiting, then this means there is no C or H in the reduced word. Thus the reduced word only consists of C orders and X 0 is at the top of the burger stack at that point. If T > n then it is easy to see that the walk {S * k } k≥1 has not reached (0, 0) before time n but the event E implies that it has not exited the cone C(θ 0 ) either because that would mean X(0) is eaten before time n. Hence E implies the walk must reach (0, 0) before exiting the cone. Hence If {T * > n} ∩ E * holds then the walk has not exited the cone before time n and after time n the burger reaches the top of the stack at some point. Hence {T * > n} ∩ E * ⊆ {T > n} ∩ E and we are done.

Random walk estimates
Recall the definition of T * from Proposition 3.13. In this section, we prove the following lemma.

Sketch of argument in Brownian case
To ease the explanations we will first explain heuristically how the exponent can be computed, discussing only the analogous question for a Brownian motion. To start with, consider the following simpler question. Let B be a planar Brownian motion and let S be the first time that B leaves the infinite cone C(θ 0 ) = {z = re iθ , r > 0, θ ∈ [0, θ 0 ]} of angle θ 0 rooted at the origin. For this we have: as t → ∞. To see why this might be the case, consider the conformal map z → z π/θ 0 . This sends the cone C(θ 0 ) to the upper-half plane. In the upper-half plane, the function z → (z) is harmonic with zero boundary condition. We deduce that, in the cone, gives a harmonic function with zero boundary condition on either side of the cone. Consequently, M t = f (B t∧S ) is a (local) martingale. Now, it is intuitive that conditionally on S ≥ t, the Brownian path will have travelled to distance roughly √ t in the cone. Therefore, the event {S ≥ t} is essentially the same as the event that the martingale reaches the value r π/θ 0 with r = √ t, before it reaches the value 0. By the gambler's ruin estimate, we find which is precisely (4.1).
Suppose now, that we are interested in the event E that the Brownian motion leaves the cone C(θ 0 ) through the ball of radius 1, that is, E = {|B S | = 1}. For this we can use the function z →f (z) := r −π/θ 0 sin πθ θ 0 ; z ∈ C(θ 0 ), which is harmonic for the same reason as above (note that the sign of the exponent in the power of r is now opposite of what it was before). Starting from distance r, we thus find, again by the gambler's ruin, P(E) ≈ r −π/θ 0 . Consequently, starting at distance say 2 from the origin, letting There are really two terms in the above identity: a first term t −p 0 , corresponding to what we call going out, where the process reaches distance r without leaving the cone, and another term t −p 0 corresponding to what we call coming in, where the process returns to a neighbourhood of the origin without leaving the cone. When the Brownian motion is reflected on the real axis, the situation is more complicated. Let P θ denote the law of Brownian motion reflected on the real axis with angle θ. If the refection was normal (θ = π/2) we could simply consider the (unreflected) Brownian motion in a cone obtained by reflecting C(θ 0 ) through the real axis. Of course this results in a cone with a double angle. Therefore, staying in C(θ 0 ) up to time t with normal reflection is really the same as staying in C(2θ 0 ) without reflection, and hence P π/2 (E; S ≥ t) ≈ t −p 0 by the above. When the angle of reflection is different from π/2, no such simple argument is possible. In fact we will see that individually, the probabilities for going out and coming in are not identical, though when combined they still yield (perhaps remarkably) the same result. Indeed, a first naïve approach is to consider the function z → g(z) := r −π/(2θ 0 ) cos πθ 2θ 0 ; z ∈ C(θ 0 ) This is indeed harmonic in the interior of C(θ 0 ). However, crucially, the harmonicity of this function is no longer enough to guarantee that g(B t ) is a local martingale under P θ 0 . As we will see in the next section (Lemma 4.2), an extra condition is required, namely (∇g).v 0 = 0, where v is a vector of angle θ (parallel to the direction of reflection). We establish in Lemma 4.3 that this forces us to consider different functions: we obtain which are the harmonic functions telling us the exponents, respectively, for going out and coming in. Note that when we add the two exponents we obtain 1 + (π/θ 0 − 1) = π/θ 0 = 2p 0 , as before. Dividing by two (as r = √ t), we obtain, at least heuristically, the results of Lemma 4.1. At the discrete levels, these estimates are respectively established in Lemmas 4.5 and 4.6. This strategy can of course be adapted (with simplifications) for Brownian motion, where even in this case the results might be new. The heart of the argument in the discrete case can be understood using solely the above Brownian estimates and Lemma 4.8. We point out that estimates for random walks in cone were derived for instance recently in [10], where they obtain very precise asymptotics. However we did not see how to adapt their estimates to our situation and instead relied on softer martingale arguments which give less precise information.
To carry out this strategy, we study harmonic functions in a wedge with reflecting boundary conditions. We digress a little bit to study them in the next subsection.

Harmonic functions in a cone
In this subsection, we study functions in a cone which are harmonic for reflected Brownian motion. The results are standard type but we could not locate a reference. The technique used to construct such harmonic functions are important because we use perturbations of such functions later to construct sub and superharmonic functions for a certain reduction of the walk S * n . First we find out a criterion for a function to be a sub/ super martingale with respect to reflected Brownian motion. Lemma 4.2. Let f be a function with ∆f ≥ 0 (resp. ∆f ≤ 0 ) and ∇f.v 0 ≥ 0 (resp. ∇f.v 0 ≤ 0) on the reflecting boundary {(r, 0) : r ≥ 0}. Let T (θ) denote the exit time from the cone C(θ) for R t . Then f (R t∧T (θ) ) is a local submartingale (resp. supermartingale) in C(θ).
For h, an exact similar computation as above will lead us to which gives us which gives w(0) = m − 2θ 0 π . From this we can again conclude via exactly the same argument as that of g that the only solution producing a nonnegative h in C(θ 0 ) is given by m = 1. We leave the rest to the reader.

Cone estimates for random walk
Now we resume the proof of Lemma 4.1. We start with a few definitions. Let T R := min{t : |S * t | ≥ R} and let T * x 0 := T * ∧ min{t ≥ 0 : |S * t | ≤ x 0 }, that is, T * x 0 is the smallest time at which the walk S * t either leaves the cone C(θ 0 ) or |S * t | becomes smaller than x 0 . Now fix an arbitrary infinite stack of burgers S. Let P x (.|S) denote the probability measure of the walk S * t starting from x conditioned on having an arbitrary infinite stack S before time 0. Define the points of R 2 on which the walk S * n is supported to be the lattice points. The following two lemmas will be the key to our estimates.
Lemma 4.6 (Coming back). Let R 0 , x 0 be as in Lemma 4.5. Fix R ≥ R 0 , θ ∈ [0, θ 0 ). For all ε > 0 there exist positive constants c(ε), C(ε) such that Proofs of Lemmas 4.5 and 4.6 are similar in nature and we use perturbations of the harmonic functions constructed in Section 4.2 to that end. However, the complexity of the proofs are technical in nature and all the fundamental conclusions have already been drawn in Section 4.2. The main idea is to approximate Brownian motion by large blocks of the walk S * , of an appropriate random length. This is similar to a strategy originally devised by McConnell [22], with some small but crucial differences. Recall the definitionS n = (D n ,C n ) from Section 3.2.
Definition 4.7. For ε > 0 we will define the following time-changed walk {Y i (ε)} i≥0 and stopping Notice that the stopping times are defined in term of the non-reflected walk while the positions we consider are those of the reflected walk. The reason for this is technical and appears in the next proof. At the same time this shows that Brownian estimates are sufficient using little more than the Central Limit Theorem and is thus the conceptual heart of the argument.
Proof. Remarkably, this statement is a direct consequence of the invariance principle and of Lemma 3.9. However to make it completely rigorous it requires a number of technical steps which makes it a bit long. We only prove the submartingale case as the supermartingale case is exactly similar.
Fix an arbitrary stack S. Conditioned on S we look at the walk S * started from nx (inside C(θ 0 )) for a large n. We show that the conditional expectation of the value of f at the point where S * exits the ball of radius ε|nx| from nx is strictly larger than f (nx). It is easy to see that this is enough to prove the lemma.
Let us denote, for any continuous function starting at zero P = (P The reflected path R(P) is defined by R t = P t − min 0≤s≤t P s v 0 . We denote by P x the translation of P such that P 0 = x. The definition of one step of the walk starting from x is then Y (P, x) = R(P x ) τ ε|x| (P 0 ) , i.e. look at the position of the reflected path started in x at the time where the non-reflected path is at distance ε|x| of its starting point.
By our assumptions, if B is a Brownian motion, f (R(B) t ) is a local submartingale so for any x, Y (B, x))] > f (x) (since for a fixed x, f remains bounded in the ball and hence we can assume that f (R(B) t ) is in fact a martingale). By continuity in x of this expectation and compactness, there exists some fixed δ such that For a fixed x in C(θ − ε) with |x| = 1, the function P → f (Y (P, x)) is continuous at almost all Brownian path so by the invariance principle, for all n large enough, In principle, how large n has to be chosen might depend on both x and the infinite stack S. However, note that by Remark 3.10 it doesn't in fact depend on S. Observe also that for any path P and fixed time t, R(P x ) t is a (1 + |v 0 |)-Lipschitz function of x hence, since the radius does not change either, Y (P, x) is Lipschitz as a function of x on the circle. In particular E[f (Y (S/n, x))] is uniformly continuous in x with a modulus of continuity independent of n. It thus suffices to look at a finite number of points x on the unit circle (about 1/δ of them) and so we conclude that for n larger than some universal n 0 (since δ is fixed),

Finally by homogeneity
as desired.
Proof of Upper bounds of Lemmas 4.5 and 4.6. Choose x 0 as in Lemma 4.8 and fix a lattice point x with |x| ≥ x 0 .
Proof of lower bounds of Lemmas 4.5 and 4.6. For the lower bound, we employ a similar strategy. Notice here we need an opposite chain of inequalities as that obtained in (4.13). Thus we need perturbations of the harmonic functions constructed in Section 4.2 to construct supermartingales with negative boundary conditions. To do this, we lay out the construction of the superharmonic functions first. We first consider a slighly smaller cone of angle C(θ 0 − ε) but with the same reflection angle θ 0 on the boundary {(r, 0) : r > 0}. There exists two homogeneous harmonic functions h 1 and g 1 which are analogous to h and g in Section 4.2 such that on the reflecting boundary ∇h 1 · v 0 = ∇g 1 · v 0 = 0 whose construction follows the same method as in Section 4.2. To be more precise, θ 1 := θ 0 − ε and Now consider the functions h 2 (r, θ) := h 1 (r, θ − ε/2) on the cone C(θ 1 + ε/2) and g 2 (r, θ) := g 1 (r, θ +ε/2) on the cone C(θ 1 −ε/2). Notice that the functions are positive in their respective cones with 0 boundary condition on the absorbing boundary. It is now easy to check that ∇h 2 · v 0 > 0 and ∇g 2 · v 0 > 0 on the reflecting boundary. This follows from the fact that we take the argument of cosine on the left hand side of (4.6) to be −π/2 + ε/2 for g 2 and π/2 − ε/2 for h 2 .
Then we make the functions strictly sub-harmonic by replacing the exponents of r by slightly bigger numbers so that the required sub-harmonicity follows from (4.3). To be more precise, we find δ = δ(ε) > 0 small enough such that still satisfies the inequalities ∇h 3 .v 0 > 0 and ∇g 3 .v 0 > 0. That we can do this follows from the expression of the dot product being 0 given by eq. (4.6). Notice in both cases, the cones on which the functions are positive has smaller opening angle than C(θ 0 ): it is θ 0 −ε/2 for g 3 and θ 0 −3ε/2 for h 3 . We now complete the final adjustment to g 3 and the rest of the construction is exactly the same for h 3 which we shall skip. Let θ 3 = θ 0 − ε/2. We modify slightly the functions near the absorbing boundary to get a negative boundary condition, see figure 7. By concavity of cosine near θ 3 , we can extend cosine by a straight line starting strictly before θ 3 in such a way that it still becomes negative before θ 0 . Adding a small second order term, we get a C 1 extension by a parabola with positive laplacian of cosine that becomes negative before θ 0 . We call this function φ 4 , it is clear from the expression of the spherical laplacian that for θ 4 close enough to the first 0 of φ 4 , the function h 4 = r p 3 φ 4 (θ) has a strictly positive laplacian (except one point) in the cone C θ 4 and negative values at θ 4 , which is what we needed.
There is still a small problem that the second derivative along the line where the parabola meets the cosine function does not exist. One way to get around this is to smoothen out the double derivative around a ball of radius ε/10 at that point (see Figure 7). Notice from the expression of the laplacian in eq. (4.3) that the double derivative of cosine increasing very fast to something positive can only help the laplacian to remain positive since the functions themselves are positive in the neighbourhood of radius ε/10. This solves the problem of the discontinuity of the derivative. As a final step, we choose ε > 0 small enough so that the at the starting point, the subharmonic function we construct will has positive value.
Finally, we need to find the appropriate sub-martingales. For the going out case, we need to look at the walk {Y k ((θ 0 − θ 4 )/2)} k≥1 . Choose x 0 , R 0 as in Lemma 4.8 and take R ≥ R 0 . Let δ = (θ 0 − θ 4 )/2. One now needs to look at T x 0 (Y (δ )) which is the first time Y k (δ ) either leaves C(θ 0 − 2δ ) or is at a distance smaller than x 0 (1 + δ ) from the origin as defined in Lemma 4.8.
) is a sub-martingale by construction. Using the negative boundary conditions, we can now obtain an opposite chain of inequalities like (4.13) as required.
For coming back, the proof has an analogous but repetitive strategy, so we leave out the details to the careful reader.
The next step is to show that on the event T * (x 0 ) > n, we can choose R to be roughly √ n with high probability. For this we need the following Lemma. Proof. We first observe that P 0 (T √ n log n < n|S) < exp(−c log 2 n) follows from Lemma 3.12 and eq. (15) of [27]. For the other part, notice from Proposition 3.11 that in the first n/ log 2 n steps, there is a constant probability c that the walk S * goes beyond distance √ n/ log n from the origin. Iterating this bound log 2 n times, we have the required bound.
Proof of Lemma 4.1. First choose R 0 , x 0 as in Lemmas 4.5 and 4.6. We will prove the lemma for large enough n and adjust the constants for smaller values of n.
Let S 1 be the burger stack at the first time (S * t ) t≥1 reaches a distance at least x 0 from origin within the first log 2 n steps if such an event occurs. Notice that the probability that the walk (S * t ) t≥1 is never at a distance more than x 0 from the origin within the first log 2 n steps is at most exp(−c log 2 n) for some constant c > 0 depending only on x 0 . Hence to prove the lemma, we need to show the following. Let x be any lattice point at a distance more than x 0 from the origin but has neighbours which are at a distance smaller than x 0 from the origin. It is enough to show that for all ε > 0, there exist positive constants c(ε), C(ε) such that Upper bound. Notice that on the event E at time T , X 0 is at the top of the burger stack and hence the reduced word J T consists of only X 0 and orders of the burger type opposite to that of X 0 . Suppose by symmetry that X 0 = C . Then the number of these unfulfilled H orders at time T is given by − 1 min 0≤k≤TCk , where is as shown in Figure 6. To see this, notice that up to time T , the reflected walk does not leave the cone, hence the number of extra H is the number of levels below 0 the unreflected walk has gone below the X-axis and the gap between levels is exactly . Now we break up the event as follows The required bound for the second term follows from the upper bound on T from the first part of the theorem. The rest follows from an exponential bound on max 0≤k≤n |C k | and max 0≤k≤n |D k | proved in [27], Lemma 3.12: for n ≥ 1 and a > 0, there exist positive constants c, c such that Plugging in a = log 2 n into (4.21) and using the fact that (D k ,C k ) are obtained by just rotating (D k , C k ), we get that the first term in (4.20) is bounded from above by c exp(−c log 2 n) which completes the proof.
Lower bound. The proof of the lower bound is very similar to the previous estimates. We first relate J T to the random walk in some cone and then use martingales techniques to control the random walk in the cone. By symmetry we again assume that X 0 = C . Recall the definition of the walk (D t ,C t ). Recall from the upper bound that is exactly the number of H in the reduced word X(0) . . . X(n) where, again, is as in Figure 6. At time T , the reduced word contains only H so Furthermore, by construction the random walkS n = (D n ,C n ) stayed inside the cone C J T where Observe that C j is a cone of angle θ 0 such that 0 is a point on the boundary located at distance |v 0 |j from the tip. Therefore for any fixed j, the probability that {J T = j} ∩ E is exactly the probability that the random walkS exits the cone C j through its tip. We will bound this exit probability using the same martingale technique as for the other estimates. The idea is the following. We first estimate the probability that the walk exits the ball of radius v 0 j/2 fromS 0 , which is similar to a Brownian motion exiting the same ball in the half-plane (see Figure 8). This has probability roughly j −1 . Then we ask the probability that the walk exits through the tip of the cone drawn in Figure 8. This has probability roughly j −2p 0 since the harmonic function in the cone with zero boundary conditions is r 2p 0 sin πθ θ 0 . Overall, this gives P(J T = j) ∩ E is roughly at least cj −1− π θ 0 . Now we fill in the details. The details are very similar as before and not much enlightening, but we still do it for completeness. Consider the following function where ε is chosen so that 1 + ε > π π−2ε . Such a function is subharmonic in the coneC bounded between angles −(π − θ 0 ) + ε and θ 0 − ε (cone shown by the red arc in Figure 8). We use the trick of replacing the sine function by a parabola near the end points as shown in Figure 7 to find a subharmonic function h 2 onC as well with negative boundary conditions.

Isoperimetric estimates
In this section we describe how one can precisely make sense of the isoperimetric estimates used to check the KPZ equations in Section 1.3.
Consider an i.i.d. sequence of symbols {X i } i∈Z and consider the corresponding map. Condition on X 0 = C , X −1 = H . The reasoning behind such a conditioning is merely to make the upcoming description a bit nicer and is of no more significance. If we start the exploration by going into the past as described in Section 2.3 and Figure 4, we see that the burgers corresponding to X 0 and X −1 will never find its match since its match is in {X i } i≥1 . Thus if we explore into the past forever, we obtain an infinite interface starting from the root with growing primal and dual cluster in its two sides.
In this setting, there is an obvious way to define the nth triangle the interface is in. Also there is a nice way to explore the map until we are in the nth triangle the interface through the root is in. This is exactly the exploration procedure used in the proof of Theorem 1.1: at every step, if we see an F , we explore until we find its match, or otherwise we explore for one step and reveal the corresponding triangle prescribed by the exploration procedure. This is what we called the reduced walk. Now notice that if we encounter an F , we add a map with a simple boundary (see Figure 9) and the boundary is formed by primal edges (resp. dual edges) if the F is of type C F (resp. H F ). We call this map, a bubble. The map inside the boundary also has a nice description: it is the map formed by exploring the symbols in an envelope of an F symbol of the correct type conditioned on the reduced word corresponding to the envelope having the same number of orders as the boundary edges.
Proposition 4.10. The number of triangles in a bubble produced at an above described exploration step has the same law as T and the number of boundary edges of the bubble has the same law as J T .
Proof. By exploring the map via the exploration procedure described in Section 2.3 and using Proposition 3.6, we see that the number of triangles added has law T since in each step one triangle is added. Further, any order left unfulfilled after the procedure must be on the boundary because it has to be fulfilled sometimes in the further past or in the future of 0. Also any unfulfilled order cannot be inside the boundary since it has to be fulfilled at some point in further past which is impossible. For instance, this would mean the blue interface in Figure 4) has to cross itself. Now the bubble isoperimetric estimates are clear from the tails of J T , T and its relation with the tails of T, J T .
If we want to compute the total number of triangles A n revealed in n such steps, we need to sum up n i.i.d. variables with tail given by that of T and hence A n scales roughy like n 1/p 0 . By

A bubble
Map expored so far. Figure 9: An illustration of a bubble. In the first step we encounter an F which is actually a cheeseburger order. Then we explore the whole envelope of this F and what we obtain is the grey map in the second figure. Finally the F is matched in the last figure and we close a bubble delimiting a light blue map as in the third figure. symmetry, A n is a sum A C n + A H n where A C n is the number of triangles by steps of the form C , H or an H F pair and the rest of the triangles is A H n . By symmetry A H n and A C n have the same distribution. Overall, we obtain a precise statement as follows.
We end with a few words about the geometry of the clusters. It is conjectured in [8] that the cluster containing the origin when rescaled properly would converge to a continuum object called a Stable looptree. They proved such a statement for the percolation model on random triangulations and conjectured that a similar picture exists for the O(n) loop model. Although the model we deal with in this paper is not precisely the same, they are neverthless believed to be in the same universality class. In Figure 2, we show how to use the hamburger-cheeseburger bijection and the exploration procedure described in this section to get an idea of the emerging looptree structure of the cluster. In particular, the bubbles correspond to loops in the cluster. In Figure 2  The grey and pink areas inside the loops correspond to a C . . . F and a H . . . F sequence respectively. We plan to address this question in future work.