samples/texts/2952501/page_2.md

As will be clearer in the proof of Theorem 3.1, this factor of 2 is really the stumbling block that prevents us from reaching the convergence to the Brownian map using only the bijective approach.

2.3 Convergence of the labeled trees

From what precedes, starting from a uniform random rooted, pointed planar Eulerian triangulation with $n$ black faces, we get a uniform random well-labeled tree $\mathcal{T}_n$ with $n$ edges. Let us now explain how we can make sense of taking a continuum scaling limit of the latter. We first define the contour process of $\mathcal{T}n$: let $e_0, e_1, \cdots, e{2n-1}$ be the sequence of oriented edges bounding the unique face of $\mathcal{T}_n$, starting with the root edge, and ordered counterclockwise around this face. Then let $u_i = e_i^-$ be the $i$-th visited vertex in this contour exploration, and set the contour process of $\mathcal{T}_n$ at time $i$:

$$C_n(i):=d_{{\mathcal{T}}_n}(u_0,u_i),0\le i\le 2n-1,C\_n(i)\; :=\; d\_\{\backslash mathcal\{T\}\_n\}(u\_0,\; u\_i),\; \backslash quad\; 0\; \backslash le\; i\; \backslash le\; 2n\; -\; 1,$$

with the convention that $u_{2n} = u_0$ and $C_n(2n) = 0$. We also extend $C_n$ by linear interpolation between integer times: for $0 \le s \le 2n$

$$C_n(s)=(1-\{s\})C_n(\lfloor s\rfloor )+\{s\}{C}_n(\lfloor s\rfloor +1),C\_n(s)\; =\; (1\; -\; \backslash \{s\backslash \})C\_n(\backslash lfloor\; s\; \backslash rfloor)\; +\; \backslash \{s\backslash \}C\_n(\backslash lfloor\; s\; \backslash rfloor\; +\; 1),$$

where ${s} = s - \lfloor s \rfloor$ is the fractional part of $s$. Thus, the contour process $C_n$ is a non-negative path of length $2n$, starting and ending at 0, with increments of 1 between integer times. We will use the rescaled contour process of $\mathcal{T}_n$:

$$C_{(n)}(t)=\frac{C_n(2nt)}{\sqrt{2n}},0\le t\le 1.C\_\{(n)\}(t)\; =\; \backslash frac\{C\_n(2nt)\}\{\backslash sqrt\{2n\}\},\; \backslash quad\; 0\; \backslash le\; t\; \backslash le\; 1.$$

We define similarly the rescaled label function of $\mathcal{T}_n$:

$$L_{(n)}(t)=\frac{L_n(2nt)}{n^{1\mathrm{/}4}},0\le t\le 1,L\_\{(n)\}(t)\; =\; \backslash frac\{L\_n(2nt)\}\{n^\{1/4\}\},\; \backslash quad\; 0\; \backslash le\; t\; \backslash le\; 1,$$

where, similarly, we start by defining $L_n(i)$ as the label of $u_i$ for $i \in {0, 1, \dots, 2n}$, then interpolate between integer times.

Finally, for a continuous, non-negative function $f : [0, 1] \to \mathbb{R}_+$ such that $f(0) = f(1) = 0$, for any $s, t \in [0, 1]$, we set

$$\check{f}(s,t)=\inf \{f(u)\mathrm{\mid }s\wedge t\le u\le s\vee t\}\mathrm{.}\backslash check\{f\}(s,t)\; =\; \backslash inf\backslash \{f(u)|s\; \backslash wedge\; t\; \backslash le\; u\; \backslash le\; s\; \backslash vee\; t\backslash \}.$$

Then we have the following result:

Theorem 2.12. [16] It holds that

$$(C_{(n)},L_{(n)})\underset{n}{\overset{(d)}{\to }}(\mathrm{\wp },Z),(2.3)(C\_\{(n)\},\; L\_\{(n)\})\; \backslash xrightarrow[n]\{(d)\}\; (\backslash wp,\; Z),\; \backslash qquad\; (2.3)$$

in distribution in $C([0, 1], \mathbb{R})^2$, where $\wp$ is a standard Brownian excursion, and, conditionally on $\wp$, $Z$ is a continuous, centered Gaussian process with covariance

$$\mathrm{Cov}(Z_s,Z_t)={\check{e}}_{s,t},s,t\in [0,1]\mathrm{.}\backslash operatorname\{Cov\}(Z\_s,\; Z\_t)\; =\; \backslash check\{e\}\_\{s,t\},\; s,t\; \backslash in\; [0,1].$$

As this convergence will be crucial to ultimately prove the convergence of Eulerian triangulations to the Brownian map, to describe and analyse these triangulations, we will need to use their oriented distances, instead of the usual graph distance.