samples/texts/2952501/page_9.md

Theorem 3.1. Let $(\mathbf{m}_{\infty}, D^*)$ be the Brownian map. There exists some constant $c_0 \in [2/3, 1]$, such that the following convergence in distribution holds:

$$(C_{(n)},L_{(n)},{\vec{D}}_{(n)},{\overleftrightarrow{D}}_{(n)},D_{(n)})\underset{n\to \mathrm{\infty }}{\overset{(d)}{\to }}(\mathbb{e},Z,D^{\ast },D^{\ast },{\mathbf{c}}_0{D}^{\ast })\mathrm{.}(C\_\{(n)\},\; L\_\{(n)\},\; \backslash vec\{D\}\_\{(n)\},\; \backslash overleftrightarrow\{D\}\_\{(n)\},\; D\_\{(n)\})\; \backslash xrightarrow[n\; \backslash to\; \backslash infty]\{(d)\}\; (\backslash mathbb\{e\},\; Z,\; D^*,\; D^*,\; \backslash mathbf\{c\}\_0\; D^*).$$

Consequently, we have the following joint convergences

$$n^{-1\mathrm{/}4}\cdot (V({\mathcal{T}}_n),{\overleftrightarrow{d}}_n)\underset{n\to \mathrm{\infty }}{\overset{(d)}{\to }}({\mathbf{m}}_{\mathrm{\infty }},D^{\ast })n^\{-1/4\}\; \backslash cdot\; (V(\backslash mathcal\{T\}\_n),\; \backslash overleftrightarrow\{d\}\_n)\; \backslash xrightarrow[n\; \backslash to\; \backslash infty]\{(d)\}\; (\backslash mathbf\{m\}\_\{\backslash infty\},\; D^*)$$

$$n^{-1\mathrm{/}4}\cdot (V({\mathcal{T}}_n),d_n)\underset{n\to \mathrm{\infty }}{\overset{(d)}{\to }}{\mathbf{c}}_0\cdot ({\mathbf{m}}_{\mathrm{\infty }},D^{\ast }),n^\{-1/4\}\; \backslash cdot\; (V(\backslash mathcal\{T\}\_n),\; d\_n)\; \backslash xrightarrow[n\; \backslash to\; \backslash infty]\{(d)\}\; \backslash mathbf\{c\}\_0\; \backslash cdot\; (\backslash mathbf\{m\}\_\{\backslash infty\},\; D^*),$$

for the Gromov-Hausdorff distance on the space of isometry classes of compact metric spaces.

Note that we would like to have a statement similar to the one on $\overleftrightarrow{d}_n$ for $\vec{d}_n$. However, as $\vec{d}_n$ is not a proper distance, it does not induce a metric space structure on $V(T_n)$. Thus, we would need to generalize the Gromov-Hausdorff topology to spaces equipped with a non-symmetric pseudo-distance, to be able to write such a statement.

Proof. We admit here Theorem 1.2, that will be proven later in the paper: for every $\varepsilon > 0$, we have

$$P(\underset{(x,y)\in V({\mathcal{T}}_n)}{\sup }\mathrm{\mid }{d}_n(x,y)-{\mathbf{c}}_0{\vec{d}}_n(x,y)\mathrm{\mid }>\epsilon n^{1\mathrm{/}4})\stackrel{n\to \mathrm{\infty }}{\to }0.(3.1)P\backslash left(\; \backslash sup\_\{(x,y)\; \backslash in\; V(\backslash mathcal\{T\}\_n)\}\; |d\_n(x,y)\; -\; \backslash mathbf\{c\}\_0\; \backslash vec\{d\}\_n(x,y)|\; >\; \backslash varepsilon\; n^\{1/4\}\; \backslash right)\; \backslash xrightarrow\{n\; \backslash to\; \backslash infty\}\; 0.\; \backslash quad\; (3.1)$$

We proceed similarly to the case of usual triangulations in [20].

For this whole proof, we work with the pointed triangulation $\bar{\mathcal{T}}_n$, but, as all Eulerian triangulations with $n$ black faces have the same number of vertices, this does not introduce any bias for the underlying, non-pointed triangulation, so that the final statement also holds for $\mathcal{T}_n$.

We have, from Proposition 2.11, for any $0 \le i < j \le 2n$:

As noted before, if we did not have the global multiplicative factor of 2 in (3.2), we could then proceed as for usual triangulations and other well-known families of planar maps. Thus, the rest of this proof will consist in proving that Theorem 1.2 makes it possible to “get rid” of this cumbersome factor.

We claim that the sequence of the rescaled distances $(\vec{D}{(n)}(s,t)){s,t \in [0,1]}$ is tight. First note that, for any $s, s', t, t' \in [0,1]$, we have

$$\mathrm{\mid }{\vec{D}}_{(n)}(s,t)-{\vec{D}}_{(n)}(s^{\mathrm{\prime }},t^{\mathrm{\prime }})\mathrm{\mid }\le 2({\vec{D}}_{(n)}(s,s^{\mathrm{\prime }})+{\vec{D}}_{(n)}(t^{\mathrm{\prime }},t))\mathrm{.}(3.3)|\backslash vec\{D\}\_\{(n)\}(s,t)\; -\; \backslash vec\{D\}\_\{(n)\}(s\text{'},t\text{'})|\; \backslash le\; 2\; (\backslash vec\{D\}\_\{(n)\}(s,s\text{'})\; +\; \backslash vec\{D\}\_\{(n)\}(t\text{'},t)).\; \backslash quad\; (3.3)$$

Indeed, $\vec{D}_n$, like $\vec{d}_n$, satisfies the triangle inequality, so that

which gives (3.3) when taking into account that, while $\vec{D}{(n)}$ is not symmetric, we do have, for any $s,t \in [0,1]$, $\vec{D}{(n)}(s,t) \le 2\vec{D}_{(n)}(t,s)$. Then, using (3.2) and Theorem 2.12, we