samples/texts/2952501/page_11.md

To prove this, let us get back to the discrete level for a moment. Let $u_n, v_n$ be two vertices of $\mathcal{T}_n$ chosen independently and uniformly at random. As $\mathcal{T}_n$ re-pointed at $u_n$ has the same law as $\bar{\mathcal{T}}_n$, we have

$${\vec{d}}_n(u_n,v_n){\vec{d}}_n(o_n,v_n)\mathrm{.}(3.9)\backslash vec\{d\}\_n(u\_n,\; v\_n)\; \backslash stackrel\{(d)\}\{=\}\; \backslash vec\{d\}\_n(o\_n,\; v\_n).\; \backslash quad\; (3.9)$$

Similarly to the case of usual triangulations in [20], this implies the desired equality in distribution $D(U, V) \stackrel{(d)}{=} D(s_*, U)$. Indeed, set $U_n = \lceil(2n-1)U\rceil$ and $V_n = \lceil(2n-1)V\rceil$, which are both uniformly distributed over ${1, 2, \dots, 2n-1}$, so that

$$\frac{U_n}{n}\stackrel{n\to \mathrm{\infty }}{\to }U,\frac{V_n}{n}\stackrel{n\to \mathrm{\infty }}{\to }V\mathrm{.}\backslash frac\{U\_n\}\{n\}\; \backslash xrightarrow\{n\; \backslash to\; \backslash infty\}\; U,\; \backslash quad\; \backslash frac\{V\_n\}\{n\}\; \backslash xrightarrow\{n\; \backslash to\; \backslash infty\}\; V.$$

Then, from (3.4), we have

$${\vec{D}}_{(n)}(\frac{U_n}{n},\frac{V_n}{n})\stackrel{n\to \mathrm{\infty }}{\to }\vec{D}(U,V)\mathrm{.}\backslash vec\{D\}\_\{(n)\}\; \backslash left(\; \backslash frac\{U\_n\}\{n\},\; \backslash frac\{V\_n\}\{n\}\; \backslash right)\; \backslash xrightarrow\{n\; \backslash to\; \backslash infty\}\; \backslash vec\{D\}(U,\; V).$$

Now, from (3.9), we have that the distribution of $\vec{D}(U,V)$ is also the limiting distribution of

$$L_{(n)}\left(\frac{U_n}{n}\right)-\min L_{(n)}+1,L\_\{(n)\}\; \backslash left(\; \backslash frac\{U\_n\}\{n\}\; \backslash right)\; -\; \backslash min\; L\_\{(n)\}\; +\; 1,$$

so that $\vec{D}(U,V)$ has the same distribution as $Z_U - \inf Z$, which is also the distribution of $D(s_*, U)$, from (3.5). This concludes the proof. $\square$

4 Technical preliminaries

4.1 Consequences of the convergence of the rescaled labels

We now prove a few technical properties of $\vec{d}$ that stem from the convergence given in Theorem 2.12.

For any integer $n \ge 1$, let $\rho_n$ be the root vertex of the random triangulation $\mathcal{T}_n$, uniform over the rooted planar Eulerian triangulations with $n$ black faces. We denote by $\bar{\mathcal{T}}_n$, the triangulation $\mathcal{T}_n$ together with a distinguished vertex $o_n$, picked uniformly at random in $\mathcal{T}_n$. We then have the following result:

Proposition 4.1. The following convergence holds:

$$n^{-1\mathrm{/}4}\vec{d}(o_n,{\rho }_n)\underset{n\to \mathrm{\infty }}{\overset{(d)}{\to }}\sup Z\mathrm{.}n^\{-1/4\}\; \backslash vec\{d\}(o\_n,\; \backslash rho\_n)\; \backslash xrightarrow[n\backslash to\backslash infty]\{(d)\}\; \backslash sup\; Z.$$

Consequently, the sequence $(n^{-1/4}\vec{d}(o_n, \rho_n)){n\ge 1}$ is bounded in probability and bounded away from zero in probability, as well as the sequence $(n^{-1/4}\vec{d}(\rho_n, o_n)){n\ge 1}$.

Proof. Recall that $\bar{\mathcal{T}}_n$ is in correspondence with a random tree $\mathcal{T}_n$, uniform over the well-labeled plane trees with $n$ edges, whose labelling we denote by $l_n$. We have, from (2.2), that

$$\vec{d}(o_n,{\rho }_n)=-\underset{v\in V({\mathcal{T}}_n)}{\min }l(v)+1.\backslash vec\{d\}(o\_n,\; \backslash rho\_n)\; =\; -\; \backslash min\_\{v\; \backslash in\; V(\backslash mathcal\{T\}\_n)\}\; l(v)\; +\; 1.$$

Then, using the convergence of Theorem 2.12, we get that the quantity

$$n^{-1\mathrm{/}4}(-\underset{v\in V({\mathcal{T}}_n)}{\min }l(v)+1)n^\{-1/4\}\; \backslash left(\; -\; \backslash min\_\{v\; \backslash in\; V(\backslash mathcal\{T\}\_n)\}\; l(v)\; +\; 1\; \backslash right)$$

converges in distribution to $(-\inf Z) \stackrel{(d)}{=} \sup Z$.

This directly implies the bounds in probability for the sequence $(n^{-1/4}\vec{d}(o_n, \rho_n)){n\ge 1}$. For those pertaining to $(n^{-1/4}\vec{d}(\rho_n, o_n)){n\ge 1}$, recall that for any two vertices $u, v$ of an Eulerian triangulation, $\frac{1}{2}\vec{d}(u, v) \le \vec{d}(v, u) \le 2\vec{d}(u, v).$ $\square$