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# There are no geodesic hubs in the Brownian sphere

Mathieu Mourichoux

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Mathieu Mourichoux. There are no geodesic hubs in the Brownian sphere. 2025. hal-04893236

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# There are no geodesic hubs in the Brownian sphere

Mathieu Mourichoux

## Abstract

A point of a metric space is called a $k$-hub if it is the endpoint of exactly $k$ disjoint geodesics, and that the concatenation of any two of these paths is still a geodesic. We prove that in the Brownian sphere, there is no $k$-hub for $k \ge 3$.

# 1 Introduction

The Brownian sphere ($S, D$) is a model of random geometry, that arises as the scaling limit of several models of random planar maps. In particular, it is the scaling limit of quadrangulations of the sphere with $n$ faces chosen uniformly at random [11, 19]. The Brownian sphere also comes with a volume measure $\mu$. In this work, we are interested in the existence of a family of exceptional points in the Brownian sphere, which are called **geodesic hubs**.

Recall that a geodesic $(\gamma(t))_{t \in [0, \tau]}$ in a metric space $(E, d)$ is a path $\gamma: [0, \tau] \mapsto E$ such that, for every $s, t \in [0, \tau]$, $d(\gamma(s), \gamma(t)) = |t - s|$. We say that a point $x \in E$ is a **geodesic hub** with at least $k$ arms, or a $k$-hub, if:

* there exists at least $k$ geodesics $\gamma_i : [0, \tau_i] \mapsto E$ such that $\gamma_i(0) = x$

* for every $1 \le i < j \le k$, $\gamma_i((0, \tau_i]) \cap \gamma_j((0, \tau_j]) = \emptyset$ and the path obtained by following $\gamma_i$ from $\gamma_i(\tau_i)$ to $x$ and then $\gamma_j$ from $x$ to $\gamma_j(\tau_j)$ is a geodesic.

The geodesics $(\gamma_i)_{1 \le i \le k}$ are called the arms of the $k$-hub, and we say that the $k$-uple $(x_i)_{1 \le i \le k}$ **borders a $k$-hub**. Of course, a $k$-hub is bordered by infinitely many points. We say that $x \in E$ is a **$k$-hub** if it is a $k$-hub, but not a $(k+1)$-hub. This notion was introduced in [13], in the course of studying some random fractal metric on $\mathbb{R}^2$. Note that there always exists 2-hubs in a geodesic space, since 2-hubs are just points in the interior of geodesics. The main contribution of this paper is to prove that 3$^+$-hubs do not exist in the Brownian sphere, confirming a prediction of [13].

**Theorem 1.1.** *Almost surely, there is no 3$^+$-hub in the Brownian sphere.*

Of course, this result implies that there is no $k$-hub for $k \ge 3$ either. Let us mention that exceptional points in the Brownian sphere, in particular geodesic stars, have already been studied in several works. Recall that a point $x \in E$ is a $k$-star if there exists $k$ disjoint geodesics emanating from $x$, and we say that $x$ is a $k$-star if it is a $k$-star but not a $(k+1)$-star. It was proved in [20, 6]

<figure><img src="image_3.png" /><figcaption>Figure 1: Illustration where $x$ is a 4-hub, and $(x_1, x_2, x_3, x_4)$ border a 4-hub. Each of the black paths between $x_i$ and $x_j$ for $i \neq j$ in $\{1, 2, 3, 4\}$ is a geodesic.</figcaption></figure>
that for $1 \le k \le 4$, the set of $k$-star in the Brownian sphere has Hausdorff dimension $5-k$, almost surely. However, the existence of 5-stars remains an open question. Similarly, it is not known if two geodesics can intersect each other at a single point which is in the interior of both geodesics.

Finally, the paper [13] studies several exceptional points for a random metric on $\mathbb{R}^2$, constructed from a Poisson process of roads. In particular, in stark contrast with the Brownian sphere, they prove that their model contains $k$-hubs up to $k = 4$.

Let us sketch the proof of Theorem 1.1:

* First, we prove that we can restrict ourselves to the study of $3^+$-hubs bordered by points distributed according to the volume measure $\mu$, and construct a variant of the Brownian sphere with three marked points. This essentially relies on results about approximations of geodesics from [20], and Bismut decomposition of a random labelled tree under $\mathbb{N}_0$.

* Then, we prove that the existence of a $3^+$-hub implies that with a positive probability, a geodesic passes through the apex of a Brownian slice (see section 2.4 for a definition of this space). To do so, we rely on the characterization of geodesics towards $x_*$ from [7], and the Palm formula.

* Then, we show that with a positive probability, there is no geodesic that passes through the apex of a Brownian slice, and we conclude with a $0-1$ argument. This part of the proof mostly relies on explicit formulas for Poisson point measures.

This paper is organized as follows. In Section 2, we introduce the notions of Brownian snake and Brownian spheres that will be used in this article. Then, Section 3 is devoted to the proof of Theorem 1.1. Finally, in Section 4, we use Theorem 1.1 to prove Proposition 2.2, which is of independent interest.

## Acknowledgements

I am grateful to my supervisor Grégory Miermont for his support, and for his careful reading of this paper. I would also like to thank Lou Le Bihan and Simon Renouf for their help in typing this work.

# 2 Preliminaries

In this section, we introduce the notion of Brownian sphere and Brownian slice. To do so, we first recall some basic notions about snake trajectories.

## 2.1 Snake trajectories

Here, we recall the definition and some basic notions about snake trajectories. A finite path is a continuous function $w: [0, \zeta] \to \mathbb{R}$, where $\zeta = \zeta(w) \ge 0$ is called the lifetime of $w$, and we set $\hat{w} = w(\zeta)$. We write $\mathfrak{M}$ for the set of all finite paths in $\mathbb{R}$, and for every $x \in \mathbb{R}$, we let $\mathfrak{M}_x := \{w \in \mathfrak{M} : w(0) = x\}$. The set $\mathfrak{M}$ is a Polish space when equipped with the distance

$$d(w, w') = |\zeta(w) - \zeta(w')| + \sup_{t \ge 0} |w(t \wedge \zeta(w)) - w'(t \wedge \zeta(w'))|.$$

Finally, we identify the point $x \in \mathbb{R}$ with the element of $\mathfrak{M}_x$ with zero lifetime.

**Definition 2.1.** Fix $x \in \mathbb{R}$. A snake trajectory starting from $x \in \mathbb{R}$ is a continuous mapping $s \mapsto \omega_s$ from $\mathbb{R}_+$ to $\mathfrak{M}_x$ which satisfies the following conditions:

* $\omega_0 = x$ and the quantity $\sigma(\omega) = \sup\{s \ge 0 : \omega_s \ne x\}$ is finite,
• For every $0 \le s \le s'$, we have $\omega_s(t) = \omega_{s'}(t)$ for every $t \in [0, \min_{s \le r \le s'} \zeta(\omega_r)]$.

The quantity $\sigma(\omega)$ is called the duration of the snake trajectory $\omega$. We will denote by $\mathfrak{S}_x$ the set of snake trajectories starting from $x \in \mathbb{R}$, and $\mathfrak{S} = \cup_{x \in \mathbb{R}} \mathfrak{S}_x$ the set of all snake trajectories. We will use the notation $W_s(\omega) = \omega_s$ and $\zeta_s(\omega) = \zeta(\omega_s)$. Note that a snake trajectory $\omega$ is completely determined by its lifetime function $s \mapsto \zeta_s(\omega)$ and its tip function $s \mapsto \tilde{W}_s(\omega)$ (see [17] for a proof). We also write $W_*(\omega) = \inf_{t \ge 0} \tilde{W}_t(\omega)$.

Given a snake trajectory $\omega \in \mathfrak{S}$, its lifetime function $\zeta(\omega)$ encodes a compact $\mathbb{R}$-tree, which will be denoted by $\mathcal{T}_\omega$. More precisely, if we introduce a pseudo-distance on $[0, \sigma(\omega)]$ by letting

$$d_{(\omega)}(s, s') = \zeta_s(\omega) + \zeta_{s'}(\omega) - 2 \min_{s \wedge s' \le r \le s \vee s'} \zeta_r(\omega),$$

then $\mathcal{T}_\omega$ is the quotient space $[0, \sigma(\omega)]/\{d_{(\omega)} = 0\}$ equipped with the distance induced by $d_{(\omega)}$. We write $p_\tau: [0, \sigma(\omega)] \to \mathcal{T}_\omega$ for the canonical projection, and root the tree $\mathcal{T}_\omega$ at $\rho_\tau := p_\tau(0) = p_\tau(\sigma(\omega))$. The tree $\mathcal{T}_\omega$ also comes with a volume measure, which is the pushforward of the Lebesgue measure on $[0, \sigma(\omega)]$ by the projection $p_\tau$. Finally, note that, because of the snake property, $W_s(\omega) = W_{s'}(\omega)$ if $p_\tau(s) = p_\tau(s')$. In particular, the mapping $s \to \tilde{W}_s(\omega)$ can be viewed as a function on the tree $\mathcal{T}_\omega$. In this article, for $u \in \mathcal{T}_\omega$ and $s \in [0, \sigma]$ such that $p_\tau(s) = u$, we will often use the notation $\ell_u = \tilde{W}_s(\omega)$.

We also define intervals on the tree $\mathcal{T}_\omega$ as follows. For every $s, t \in [0, \sigma]$ with $t < s$, we use the convention that $[s, t] = [s, \sigma] \cup [0, t]$. For every $u, v \in \mathcal{T}_\omega$, there is a smallest interval $[s, t]$ such that $p_\tau(s) = u$ and $p_\tau(t) = v$, and we define

$$[u, v] := \{p_{\tau}(r) : r \in [s, t]\}.$$

## 2.2 The Brownian snake excursion measure

In this subsection, we give the construction and some properties of the Brownian snake (see [8] for more details). For every $x \in \mathbb{R}$, we define a $\sigma$-finite measure on $\mathfrak{S}_x$, called the Brownian snake excursion measure and denoted as $\mathbb{N}_x$, as follows. Under $\mathbb{N}_x$:

1. The lifetime function $(\zeta_s)_{s \ge 0}$ is distributed according to the Itô measure of positive excursions of linear Brownian motion, normalized so that the density of $\sigma$ under $\mathbb{N}_x$ is $t \mapsto (2\sqrt{2\pi t^3})^{-1}$.

2. Conditionally on $(\zeta_s)_{s \ge 0}$, the tip function $(\tilde{W}_s)_{s \ge 0}$ is a Gaussian process with mean $x$ and covariance function:

$$K(s, s') = \min_{s \wedge s' \le r \le s \vee s'} \zeta_r.$$

The measure $\mathbb{N}_x$ is also an excursion measure away from $x$ for the Brownian snake, which is a Markov process in $\mathfrak{M}_x$. For every $t > 0$, we can define the conditional probability measure $\mathbb{N}_x^{(t)} = \mathbb{N}_x(\cdot | \sigma = t)$, which can also be constructed by replacing the Itô measure used to define $\mathbb{N}_x$ by the law of a Brownian excursion with duration $t$.

For every $y < x$, we have

$$\mathbb{N}_x(W_* < y) = \frac{3}{2(x - y)^2} \qquad (1)$$

(see [8] for a proof). Therefore, we can define the conditional probability measure $\mathbb{N}_x(\cdot | W_* < y)$. Moreover, one can prove that under $\mathbb{N}_x$ or $\mathbb{N}_x^{(t)}$, a.e, there exists a unique $s_* \in [0, \sigma]$ such that $\tilde{W}_{s_*} = W_*$ (see e.g. Proposition 2.5 in [16]).

Finally, these measures satisfy a scaling property. For every $\lambda > 0$ and $\omega \in \mathfrak{S}_x$, we define $\Theta_\lambda(\omega) \in \mathfrak{S}_{x\sqrt{\lambda}}$ by $\Theta_\lambda(\omega) = \omega'$ with

$$\omega'_s(t) := \sqrt{\lambda} \omega_s / \lambda^2 (t / \lambda), \quad \text{for } s \ge 0 \text{ and } 0 \le t \le \zeta'_s := \lambda \zeta_s / \lambda^2. \qquad (2)$$

Then, the pushforward of $\mathbb{N}_x$ by $\Theta_\lambda$ is $\lambda \mathbb{N}_{x\sqrt{\lambda}}$, and for every $t > 0$, the pushforward of $\mathbb{N}_x^{(t)}$ by $\Theta_\lambda$ is $\mathbb{N}_{x\sqrt{\lambda}}^{(\lambda^2 t)}$.
## 2.3 The Brownian sphere

Fix a snake trajectory $\omega \in \mathfrak{S}_0$ with duration $\sigma$. We introduce, for every $u, v \in \mathcal{T}_\omega$,

$$D^{\circ}_{(\omega)}(u, v) = \ell_u + \ell_v - 2 \max \left( \min_{r \in [u, v]} \ell_r, \min_{r \in [v, u]} \ell_r \right)$$

and

$$D_{(\omega)}(u, v) = \inf \left\{ \sum_{i=1}^{p} D^{\circ}_{(\omega)}(u_i, u_{i-1}) \right\} \quad (3)$$

where the infimum is taken over all integers $p \ge 1$ and sequences $u_0, \dots, u_p \in \mathcal{T}_\omega$ such that $u_0 = u$ and $u_p = v$. Note that $D_{(\omega)} \le D^{\circ}_{(\omega)}$.

Observe that $D^{\circ}_{(\omega)}(u, v) \ge |\ell_u - \ell_v|$, which translates into a simple (but very useful) bound:

$$D_{(\omega)}(u, v) \ge |\ell_u - \ell_v|. \quad (4)$$

The mapping $(u, v) \mapsto D_{(\omega)}(u, v)$ defines a pseudo-distance on $\mathcal{T}_\omega$. This allows us to introduce a quotient space $\mathcal{T}_\omega/\{D_{(\omega)} = 0\}$, which is equipped with the distance naturally induced by $D_{(\omega)}$. We can now apply the previous construction with a random snake trajectory.

**Definition 2.2.** The standard Brownian sphere is defined under the probability measure $N_0^{(1)}$ as the random metric space $S = \mathcal{T}/\{D = 0\}$ equipped with the distance $D$, and a volume measure $\mu$ which is the pushforward of the volume measure on $\mathcal{T}$ under the canonical projection $p_S: \mathcal{T} \to S$.

Observe that the labelling function $\ell$ can be defined on $S$. Therefore, for every $x \in S$, we let by $\ell_x$ stands for the label of $x$.

We also introduce the free Brownian sphere, which is defined in the same way replacing $N_0^{(1)}$ by $N_0$; even though this is not a random variable anymore, it is often more convenient to work with this object. Note that we can also see the standard Brownian sphere (or the free Brownian sphere) as a quotient of $[0, 1]$ (or $[0, \sigma]$). We will sometimes use this point of view, and we write $\mathbf{p}: [0, 1] \to S$ for the canonical projection. We also set $x_0 = \mathbf{p}(0)$.

As mentioned earlier, almost surely, there exists a unique element $u_* \in \mathcal{T}$ such that $\ell_{u_*} = \inf_{u \in \mathcal{T}} \ell_u = W_*$. Therefore, we write $x_* = p_S(u_*)$ and $\ell_* = \ell_{x_*}$. Note that the bound (4) together with the inequality $D \le D^\circ$ implies that almost surely, for every $x \in S$,

$$D(x, x_*) = \ell_x - \ell_*.$$

In particular, we have

$$D(x_0, x_*) = -\ell_*.$$

The following proposition, proved in [9], completely characterizes the points of $\mathcal{T}$ that are identified in the Brownian sphere.

**Proposition 2.1.** *Almost surely, for every $u, v \in \mathcal{T}$, we have*

$$D(u, v) = 0 \iff D^{\circ}(u, v) = 0.$$

In Section 4, we will prove that Theorem 1.1 implies the following result, which is of independent interest.

**Proposition 2.2.** Let $u, v \in \mathcal{T}$ such that

$$D^{\circ}(u, v) = D(u, v).$$

Then, $(x_*, p_S(u), p_S(v))$ are aligned, meaning that they are on a common geodesic.
## 2.4 Brownian slices

Here, we introduce the notion of Brownian slice, which plays a major role in the proofs to come. Fix a snake trajectory $\omega \in \mathcal{S}_0$ with duration $\sigma$. We define a pseudo-distance $\tilde{d}$ on $[0, \sigma]$ by

$$ \tilde{d}_{(\omega)}(s, t) = \hat{W}_s + \hat{W}_t - 2 \inf_{r \in [s \wedge t, s \vee t]} \hat{W}_r. $$

Then, similarly to what we did to construct the Brownian sphere, we can define a pseudo-distance $\tilde{D}^\circ(\omega)$ on $\mathcal{T}_\omega$:

$$ \tilde{D}^\circ(\omega)(u, v) = \inf\{\tilde{d}_{(\omega)}(s, t) : s, t \in [0, \sigma], p_{\mathcal{T}_\omega}(s) = u, p_{\mathcal{T}_\omega}(t) = v\}. \quad (5) $$

The difference with the distance $D^\circ$ of Section 2.3 is that we forbid “to go around the root of $\mathcal{T}_\omega$” when computing the distance. Finally, we can define another pseudo-distance on $\mathcal{T}_\omega$:

$$ \tilde{D}_{(\omega)}(u, v) = \inf_{u_0, \dots, u_p} \sum_{i=1}^{p} \tilde{D}^\circ(\omega)(u_i, u_{i-1}) $$

where the infimum is taken over every $p \in \mathbb{N}^*$ and sequences in $\mathcal{T}$ such that $u_0 = u$ and $u_p = v$.

**Definition 2.3.** The free Brownian slice is defined under the measure $\mathbb{N}_0$ as the metric space $\tilde{S} = \mathcal{T}/\{\tilde{D} = 0\}$, equipped with the distance $\tilde{D}$. We write $p_{\tilde{S}} : \mathcal{T} \to \tilde{S}$ for the canonical projection, $\tilde{p}$ for the projection $[0, \sigma] \to \tilde{S}$ and $\tilde{\rho} = \tilde{p}(0)$.

This space has already been studied in [11] to prove the convergence of quadrangulations toward the Brownian sphere, and in [1] to prove the convergence of quadrangulation with a boundary toward the Brownian disk (see also [6]). It is also the scaling limit of some models of random planar maps with geodesic boundaries.

Let us explain how this space is related to the Brownian sphere $S$. It was proved in [7] that almost surely, there exists a unique geodesic $\Gamma$ in $S$ between $x_0$ and $x_*$. Then, if we cut $S$ along the geodesic $\Gamma$, the resulting space is a Brownian slice. In particular, this space has a boundary made of two geodesic segments, which correspond to the geodesic $\Gamma$ that has been cut (see [11, Section 3.2] for more details).

Note that because $d \le \tilde{d}$, we have $D \le \tilde{D}$. Furthermore, $\tilde{S}$ has the same scaling property as the Brownian sphere.

## 2.5 Coding labelled trees with triples

Here, we briefly explain how to encode a labelled tree by a triple $(X, \mathcal{N}_l, \mathcal{N}_r)$. We refer to [12, Section 2.4] for more details.

Consider a triple $(X, \mathcal{N}_l, \mathcal{N}_r)$, where

* $X = (X_t)_{t \in [0, h]}$ is a random path,

* $\mathcal{N}_l$ and $\mathcal{N}_r$ are two random point measures on $[0, h] \times \mathcal{S}$.

Then, under some natural assumptions, one can define a labelled tree $\mathcal{T}$ from this triple, made of a spine of length $h$, and where each atom $(t_i, \omega_i)$ of $\mathcal{N}_l$ (respectively $\mathcal{N}_r$) represents a labelled subtree isometric $\mathcal{T}_{\omega_i}$ branching off the left side (respectively the right side) of the spine at height $t_i$. Moreover, the labels on the spine are given by the process $X$. This tree is rooted at the bottom of the spine, and has a distinguished point, which is the top of the spine.

One can also define an exploration process for the tree $\mathcal{T}$, which allows us to define intervals on this tree. Furthermore, it is possible to represent the labelled tree $\mathcal{T}$ by a snake trajectory $\omega \in \mathcal{S}$ such that $\mathcal{T}_{\omega} = \mathcal{T}$. Therefore, one can construct a random metric space $(S, d)$ and a projection $p_S : \mathcal{T} \to S$ from any admissible triple $(X, \mathcal{N}_l, \mathcal{N}_r)$, as explained in Section 2.3 and 2.4.
# 3 Proof of Theorem 1.1

As mentioned in the introduction, the proof will consist of three steps.

## 3.1 Marking three points in the Brownian sphere

We start by giving a construction of a Brownian sphere with a distinguished triple $(x_0, x_1, x_*)$ of typical points, and prove that we can restrict our study to this model. First, we show that the set of compact metric spaces with three distinguished points which bordered a $3^+$-hub is Borel. We refer to [18, Section 6.4] for details about the marked Gromov-Hausdorff topology.

**Lemma 3.1.** Let $\mathbb{M}^{***}$ be the set of isometry classes of triply-pointed compact metric spaces, equipped with the Gromov-Hausdorff topology. Then, the set

$$ \mathcal{H} = \{(M, x_1, x_2, x_3) \in \mathbb{M}^{***}, M \text{ is a geodesic space and } (x_1, x_2, x_3) \text{ borders a } 3^+\text{-hub}\} $$

is a Borel set.

*Proof.* First, recall that the set of geodesic spaces is closed in $\mathbb{M}$ (see [2, Theorem 7.5.1]). Therefore, in what follows, every space considered is a geodesic space.

For every $n, m \ge 1$, let $\mathcal{H}_{n,m}$ be the set of $(M, x_1, x_2, x_3) \in \mathbb{M}^{***}$ such that

* for every $i \neq j$ in $\{1, 2, 3\}$,

$$ d(x_i, x_j) > \frac{1}{m}, $$

* There exists $w \in \mathbb{M}$ such that for every $i \neq j$ in $\{1, 2, 3\}$,

$$ d(x_i, w) + d(w, x_j) < d(x_i, x_j) + \frac{1}{n}. \qquad (6) $$

The first condition guarantees that $(x_1, x_2, x_3)$ are disjoint points, and the second one means that they almost border a $3^+$-hub. Observe that $\mathcal{H}_{n,m}$ is an open set. Therefore, the set

$$ \bigcup_{m \ge 1} \bigcap_{n \ge 1} \mathcal{H}_{n,m} $$

is a Borel set. Moreover, we clearly have

$$ \mathcal{H} \subset \bigcup_{m \ge 1} \bigcap_{n \ge 1} \mathcal{H}_{n,m} $$

(we can choose $w$ in (6) as the $3^+$-hub bordered by $(x_1, x_2, x_3)$). Let us show a converse inclusion, which will give the desired result. Fix $(M, x_1, x_2, x_3) \in \mathbb{M}^{***}$ and suppose that there exists $m \ge 1$ such that

$$ M \in \bigcap_{n \ge 1} \mathcal{H}_{n,m}. $$

For every $n \ge 1$, consider $w_n \in \mathbb{M}$ such that (6) holds with this choice. By compactness, we can suppose that the sequence $(w_n)_{n \ge 1}$ converges toward some element $w_\infty \in \mathbb{M}$. Moreover, for every $i \neq j$ in $\{1, 2, 3\}$, we have

$$ d(x_i, w_\infty) + d(w_\infty, x_j) = d(x_i, x_j). $$

Since $x_1, x_2$ and $x_3$ are disjoint elements, we can easily deduce from this equality that $w_\infty$ is a $3^+$-hub, which gives $M \in \mathcal{H}$, and conclude the proof. $\square$

Now, we show that we just need to consider $3^+$-hubs bordered by typical points.
**Proposition 3.2.** Let $(x_0, x_1, x_2)$ be three points of the standard Brownian sphere $\mathcal{S}$ distributed according to the volume measure $\mu$. Then,

$$ \mathbb{P}(\text{There exists a } 3^+\text{-hub}) > 0 \quad \text{if and only if} \quad \mathbb{P}((x_0, x_1, x_2) \text{ borders a } 3^+\text{-hub}) > 0. $$

*Proof.* Since $\{\text{There exists a } 3^+\text{-hub}\} \supset \{(x_0, x_1, x_2) \text{ borders a } 3^+\text{-hub}\}$, one implication is straightforward. Conversely, suppose that there exists a $3^+$-hub with a positive probability. If a triple $(u, v, w)$ borders such a hub, by a result of approximations of geodesics [20, Theorem 1.7], there exist neighborhoods $(U, V, W)$ of $(u, v, w)$ such that for every $x_0 \in U, x_1 \in V, x_2 \in W$, the triple $(x_0, x_1, x_2)$ borders a $3^+$-hub. This proves the result, since these neighborhoods have a strictly positive $\mu$-measure. $\square$

Then, we will define the random trees and surfaces that we will be dealing with. For every $a > 0$, let $B^{(a)} = (B_t^{(a)})_{t \in [0, a]}$ be a Brownian motion starting from 0 of duration $a$, and given $B^{(a)}$, let $\mathcal{N}_l^{(a)}$ and $\mathcal{N}_r^{(a)}$ be two independent Poisson point measures on $[0, a] \times \mathcal{S}$, with intensity

$$ 21_{[0,a]}(t) \mathbb{N}_{B_t^{(a)}}(d\omega) dt. $$

As explained in Section 2.5, we can associate a random labelled tree $\mathcal{T}_a$ to the triple $(B^{(a)}, \mathcal{N}_l^{(a)}, \mathcal{N}_r^{(a)})$. This tree has two distinguished points, called $\rho_0$ and $\rho_a$, which are respectively the bottom and the top of the spine. Let $\mathcal{S}_a$ be the random metric space associated to $\mathcal{T}_a$, and $p_{\mathcal{S}_a} : \mathcal{T}_a \to \mathcal{S}_a$. This space comes with three distinguished points, which are

$$ x_0 = p_{\mathcal{S}_a}(\rho_0), \quad x_a = p_{\mathcal{S}_a}(\rho_a), \quad x_* = p_{\mathcal{S}_a}(u_\infty). $$

Now, we will explain how these trees and spaces are related to the measure $\mathbb{N}_0$.

Arguing for $\mathbb{N}_0(d\omega)$, for every $s \in (0, \sigma)$, we can encode the labelled subtrees branching off the ancestral line of $p_{\mathcal{T}}(s)$ by two point measures $\mathcal{P}_l^{(s)}$ and $\mathcal{P}_r^{(s)}$. More precisely, we consider the connected components $(u_i, v_i), i \in I$ of the open set $\{r \in [0, s] : \zeta_r(\omega) > \min_{t \in [r, s]} \zeta_t(\omega)\}$. For every $i \in I$, we can define a snake trajectory $\omega^i$ of duration $\sigma(\omega^i) = v_i - u_i$, by setting for every $r \in [0, \sigma(\omega^i)]$,

$$ \omega_r^i(t) = \omega_{u_i+r}(\zeta_{u_i}(\omega) + t), \quad \text{for } 0 \le t \le \zeta_{\omega_r^i} = \zeta_{u_i+r}(\omega) - \zeta_{u_i}(\omega). $$

Then, we can define a point measure $\mathcal{P}_l^{(s)}$ by

$$ \mathcal{P}_l^{(s)} = \sum_{i \in I} \delta(\zeta_{u_i}, \omega^i). $$

Similarly, one can define the point measure $\mathcal{P}_r^{(s)}$, by replacing $[0, s]$ by $[s, \sigma]$. The following proposition (which a consequence of [15, Proposition 3.5] and [15, Lemma 3.7]) makes the link between the trees $\mathcal{T}_a$ and the labelled tree associated to the measure $\mathbb{N}_0$.

**Proposition 3.3.** Let $M_p(\mathbb{R}_+ \times \mathcal{S})$ be the set of point measures on $\mathbb{R}_+ \times \mathcal{S}$. Then, for any non-negative Borel measurable function $F$ on $\mathcal{W} \times M_p(\mathbb{R}_+ \times \mathcal{S})^2$,

$$ \mathbb{N}_0 \left( \int_0^\sigma F(W_s, \mathcal{P}_l^{(s)}, \mathcal{P}_r^{(s)}) ds \right) = \int_0^\infty \mathbb{E} \left[ F(B^{(a)}, \mathcal{N}_l^{(a)}, \mathcal{N}_r^{(a)}) \right] da. $$

**Remark.** This decomposition is very similar to the one of [12, Proposition 2].

We can finally prove that we can restrict our study to the non-existence of $3^+$-hubs in $\mathcal{S}_1$.

**Proposition 3.4.** There exists a $3^+$-hub in the standard Brownian sphere with a positive probability if and only if with a positive probability, $(x_0, x_1, x_*)$ borders a $3^+$-hub in $\mathcal{S}_1$.
<figure><img src="image_4.png" /><figcaption>Figure 2: Representation of $T_0$, $\hat{T}_0$, $T_1$ and $\hat{T}_1$ in the random tree $\mathcal{T}_1$. Note that the two blue (respectively pink, green) portions correspond to the same path in $S_1$.</figcaption></figure>

*Proof.* First, by scaling, we can replace the standard Brownian sphere by a free Brownian sphere under $\mathbb{N}_0$. Then, by [6, Proposition 3], for every non-negative measurable function $F$ on the space of three-pointed measure metric spaces, we have

$$ \mathbb{N}_0 \left( \frac{1}{\sigma} \frac{1}{\sigma} \iint \int F(S, x, y, z) \mu(dx) \mu(dy) \mu(dz) \right) = \mathbb{N}_0 \left( \int F(S, x_*, x_0, y) \mu(dy) \right). $$

By Proposition 3.3, this gives

$$ \mathbb{N}_0 \left( \frac{1}{\sigma} \frac{1}{\sigma} \iint \int F(S, x, y, z) \mu(dx) \mu(dy) \mu(dz) \right) = \int_0^\infty \mathbb{E}[F(S_a, x_*, x_0, x_a)] da. \quad (7) $$

By Lemma 3.1, we can take $F(E, x, y, z) = 1_{\{(x,y,z) \text{ borders a } 3^+\text{-hub in } E\}}$. For this choice of $F$ and by Proposition 3.2, we see that the left-hand side of (7) is positive (in fact, infinite) if and only if there exists a $3^+$-hub in a standard Brownian sphere with positive probability. Furthermore, by scaling arguments, the right-hand side of (7) is positive if and only if there is a positive probability that $(x_0, x_1, x_*)$ border a $3^+$-hub in $S_1$, which concludes the proof. $\square$

## 3.2 Identifying the coalescence point

By Proposition 3.4, we want to prove that almost surely, $(x_0, x_1, x_*)$ does not border a $3^+$-hub. In this section, we study the geodesic network bordered by $(x_0, x_1, x_*)$ in $S_1$. More precisely, we identify where the geodesics from $x_0$ and $x_1$ to $x_*$ merge, and show how to study these geodesics near their merging point.

To lighten notations, we set $(B^{(1)}, \mathcal{N}_l^{(1)}, \mathcal{N}_r^{(1)}) = (B, \mathcal{N}_l, \mathcal{N}_r)$, and write $I, J$ for sets indexing the atoms of $\mathcal{N}_l^{(1)}$ and $\mathcal{N}_r^{(1)}$. We still denote by $u_*$ the element of $\mathcal{T}_1$ with minimal label. As mentioned previously, the random surface $S_1$ comes with three distinguished points $(x_0, x_1, x_*)$, which are

$$ x_0 = p_{S_1}(\rho_0), \quad x_1 = p_{S_1}(\rho_1), \quad x_* = p_{S_1}(u_*). $$

In what follows, we will abuse notations by considering infimum and supremum in intervals of $\mathcal{T}_1$. However, as explained in Section 2.5, since $\mathcal{T}_1$ can be represented by a snake trajectory, these infimums and supremums are in fact infimums and supremums in some interval $[0, \sigma]$. For every $s \in [0, -\ell_*$], we define

$$ T_0(s) = \inf\{u \in [\rho_0, u_*], \ell_u = -s\} \quad \text{and} \quad \hat{T}_0(s) = \sup\{u \in [u_*, \rho_0], \ell_u = -s\}. $$
Similarly, for every $s \in [-\ell_{\rho_1}, -\ell_*)$, we set

$$T_1(s) = \inf\{u \in [\rho_1, u_*], \ell_u = -s\} \quad \text{and} \quad \widehat{T}_1(s) = \sup\{u \in [u_*, \rho_1], \ell_u = -s\}.$$

Then, we define

$$\gamma_0(t) = p_S(T_0(t)) = p_S(\widehat{T}_0(t)) \quad \text{for } 0 \le t \le -\ell_*,$$

$$\gamma_1(t) = p_S(T_1(t)) = p_S(\widehat{T}_1(t)) \quad \text{for } -\ell_{\rho_1} \le t \le -\ell_*.$$

Using (4) and the inequality $D \le D^\circ$, it is easy to see that $\gamma_0$ (respectively $\gamma_1$) is a geodesic from $x_0$ to $x_*$ (respectively $x_1$ to $x_*$). Moreover, by the main result of [7], the geodesics $\gamma_0$ and $\gamma_1$ are almost surely the unique such geodesics (the result is stated for the Brownian sphere $S$, but it also holds for $S_1$ by (7) and scaling).

Without loss of generality, suppose that $u_* \in [\rho_0, \rho_1]$. Define $u_{**}$ as the unique element of $[\rho_1, \rho_0]$ such that $\ell_{u_{**}} = \inf\{\ell_u, u \in [\rho_1, \rho_0]\}$. In particular, let $i_*$ and $j_*$ be the indices of the unique atoms of $N_l$ and $N_r$ that contains the elements of minimal label on each side. Let us denote these elements by $u_{i_*}$ and $u_{j_*}$. Then, $u_*$ is the element such that

$$\ell_{u_*} = \ell_{u_{i_*}} \wedge \ell_{u_{j_*}},$$

and $u_{**}$ is the one that satisfies

$$\ell_{u_{**}} = \ell_{u_{i_*}} \vee \ell_{u_{j_*}}.$$

Also, set $\ell_{**} = \ell_{u_{**}}$ and $x_{**} = p_{S_1}(u_{**})$. Observe that for every $t \in [0, -\ell_{**})$, $s \in [-\ell_{\rho_1}, -\ell_{**})$,

$$D^\circ(\widehat{T}_0(t), T_1(s)) > 0,$$

whereas, for $t \in [-\ell_{**}, -\ell_*)$,

$$D^\circ(\widehat{T}_0(t), T_1(t) = 0.$$

Therefore, by Lemma 2.1, for every $t \in [0, -\ell_{**})$, $s \in [0, -\ell_{**} + \ell_{\rho_1})$,

$$\gamma_0(t) \neq \gamma_1(s))$$

whereas for every $t \in [-\ell_{**}, -\ell_*)$,

$$\gamma_0(t) = \gamma_1(t + \ell_{\rho_1}).$$

In particular, the geodesics $\gamma_0$ and $\gamma_1$ coalesce at $x_{**}$.

Let $\gamma_{0,1}$ be the path obtained by following $\gamma_0$ from $x_0$ to $x_{**}$, and then $\gamma_1$ from $x_{**}$ to $x_1$ (see Figure 3). Since there is almost surely a unique geodesic between $x_0$ and $x_1$, the triple $(x_*, x_0, x_1)$ borders a 3-hub if and only if $\gamma_{0,1}$ is a geodesic.

As mentioned previously, we need to study the geodesics $\gamma_0$ and $\gamma_1$ “near $x_{**}$”. To do so, let $W_{\min}$ be the atom that contains $u_{**}$. The following proposition characterizes the law of $W_{\min}$.

**Proposition 3.5.** The law of $W_{\min}$ given $B$ is absolutely continuous with respect to $\int_0^1 N_{B_s}(dW)$.

*Proof.* Recall that $i_*$ and $j_*$ be the indices of the unique atoms of $N_l$ and $N_r$ that contain the minimum on each side. For every $w \in \mathfrak{W}_0$, let $M(dt, d\omega)$ and $M'(dt, d\omega)$ be two independent Poisson point measures on $\mathbb{R}_+ \times C(\mathbb{R}_+, \mathfrak{W})$, defined on some probability space with probability measure $\Pi_w$, with intensity :

$$21_{[0, \zeta_{(w)}]}(t)dtN_{w(t)}(d\omega).$$
<figure><img src="image_5.png" /><figcaption>Figure 3: Illustration of the path $\gamma_{0,1}$. We need to determine whether it can be a geodesic.</figcaption></figure>

Using Palm's formula and (1), we obtain :

$$
\begin{align*}
\mathbb{E}[F(W_{i_*}, W_{j_*}) | B] &= \mathbb{E} \left[ \sum_{i \in I, j \in J} F(W_i, W_j) \mathbb{1}_{i=i_*, j=j_*} \middle| B \right] \\
&= \int_0^1 \int_0^1 ds ds' \int_{\mathcal{G} \times \mathcal{G}} \mathbb{N}_{B_s}(dW_1) \mathbb{N}_{B_{s'}}(dW_2) F(W_1, W_2) \\
&\qquad \times \Pi_B \left[ \mathcal{M}((t_i, \omega_i) : (\omega_i)_* < (W_1)_*) = 0, \mathcal{M}'((t'_j, \omega'_j) : (\omega'_j)_* < (W_2)_*) = 0 \right] \\
&= \int_0^1 \int_0^1 ds ds' \int_{\mathcal{G} \times \mathcal{G}} \mathbb{N}_{B_s}(dW_1) \mathbb{N}_{B_{s'}}(dW_2) F(W_1, W_2) \exp \left( -3 \int_0^1 \left( \frac{1}{(B_u - (W_1)_*)^2} + \frac{1}{(B_u - (W_2)_*)^2} \right) du \right).
\end{align*}
$$

This gives us the joint law of $(W_{i_*}, W_{j_*})$. In particular, we have

$$
[\mathbb{E}[F(W_{\min}) | B] = 2 \int_{0}^{1} \int_{0}^{1} ds ds' \int_{\mathcal{G} \times \mathcal{G}} \mathbb{N}_{B_s}(dW_1) \mathbb{N}_{B_{s'}}(dW_2) F(W_1) \mathbb{1}_{(W_1)_* < (W_2)_*} \\
\times \exp \left( -3 \int_{0}^{1} \left( \frac{1}{(B_u - (W_1)_*)^2} + \frac{1}{(B_u - (W_2)_*)^2} \right) du \right),
$$

which gives the result.

<figure><img src="image_6.png" /></figure>

Let $\mathcal{T}_{\min}$ be the labelled subtree associated to $W_{\min}$. The following lemma guarantees that $p_{S_1}(\mathcal{T}_{\min})$ contains a non-trivial portion of $\gamma_0$ and $\gamma_1$.

**Lemma 3.6.** *Almost surely,*

$$
\gamma_0 \cap p_{S_1}(\mathcal{T}_{\min}) \neq \{x_{**}\} \quad \text{and} \quad \gamma_1 \cap p_{S_1}(\mathcal{T}_{\min}) \neq \{x_{**}\}
$$

*Proof.* Without loss of generality, we can suppose that $x_{**} \in p_{S_1}(\mathcal{N}_r)$. Set $\underline{B} = \inf_{s \in [0,1]} B_s$. Note that conditionally on $B$, for every $\varepsilon > 0$, the number of atoms $W_i$ of $\mathcal{N}_r$ such that $(W_i)_* < \underline{B} - \varepsilon$ follows a Poisson distribution with parameter

$$
2 \int_{0}^{1} \mathbb{N}_{B_{s}}(W_{*} < \underline{B} - \varepsilon) ds = 3 \int_{0}^{1} \frac{1}{(B_{s} - \underline{B} + \varepsilon)^{2}} ds < +\infty.
$$
<figure><img src="image_7.png" /><figcaption>Figure 4: The same illustration as Figure 2, but viewed in $S_1$. The black path corresponds to the projection of the spine of $\mathcal{T}_1$, and the yellow (respectively gray) area is the projection of the atom containing $u_*$ (respectively $u_{**}$).</figcaption></figure>

In particular, this implies that there is no accumulation point in the set $\{(W_i, t_i), (W_i)_* < \underline{B}\}$. Since $W_{\min}$ belongs to this set, this means that almost surely, there exists $\delta > 0$ such that for every atom $W_i \in \mathcal{N}_r$ (which is different from $W_{\min}$), we have $(W_i)_* > (W_{\min})_* + \delta$. Therefore, for every $0 < \varepsilon < \delta$, we have

$$ \hat{T}_0(-\ell_{**} - \varepsilon) \in \mathcal{T}_{\min} \quad \text{and} \quad T_1(-\ell_{**} - \varepsilon) \in \mathcal{T}_{\min}, $$

which concludes the proof. □

Let $\tilde{S}$ be the random slice associated to $W_{\min}$. In what follows, we will study the geodesics $\gamma_0$ and $\gamma_1$ “restricted to $\tilde{S}$”. More precisely, if $0 < a_0 < b_0$ and $0 < a_1 < b_1$ are such that $\gamma_0|_{[b_0-a_0,b_0]} \subset p_S(\mathcal{T}_{\min})$ and $\gamma_1|_{[b_1-a_1,b_1]} \subset p_S(\mathcal{T}_{\min})$, with $p_{S_1}(\gamma_0(b_0)) = p_{S_1}(\gamma_1(b_1)) = x_{**}$, we set

$$ \tilde{\gamma}(t) = \begin{cases} p_{\tilde{S}}(T_0(\ell_{**} + t - a_0)) & \text{if } t \in [0, a_0] \\ p_{\tilde{S}}(T_1(\ell_{**} - (t - a_0))) & \text{if } t \in [a_0, a_0 + a_1]. \end{cases} $$

Since $\tilde{\gamma}$ is a sub-path of $\gamma_{0,1}$, we already know that if $(x_*, x_0, x_1)$ borders a 3$^+$-hub in $S_1$, then $\tilde{\gamma}$ is a geodesic in $S_1$. However, it is not clear it is also a geodesic in $\tilde{S}$, which is the content of the following proposition.

**Proposition 3.7.** Suppose that $(x_*, x_0, x_1)$ borders a 3$^+$-hub in $S_1$. Then, $\tilde{\gamma}$ is a geodesic in $\tilde{S}$.

*Proof.* First, note that $\tilde{\gamma}|_{[0,a_0]}$ (respectively $\tilde{\gamma}|_{[a_0,a_0+a_1]}$) is just a portion of the geodesic $\tilde{\gamma}^{(r)}$ (respectively $\tilde{\gamma}^{(l)}$). Therefore, we only need to compute the distance between $\tilde{\gamma}(s)$ and $\tilde{\gamma}(t)$ for every $s \in [0, a_0]$ and $t \in [a_0, a_0 + a_1]$. Using the inequality $\tilde{D} \ge D$ and the fact that $\gamma_0$ and $\gamma_1$ border a 3$^+$-hub, we have

$$ \tilde{D}(\tilde{\gamma}(s), \tilde{\gamma}(t)) \ge D(\gamma_0(s + (b_0 - a_0)), \gamma_1(b_0 - (t - a_0))) = t - s. $$

On the other hand, the bound $\tilde{D} \le \tilde{D}^\circ$ gives

$$ \tilde{D}(\tilde{\gamma}(s), \tilde{\gamma}(t)) \le \tilde{D}^\circ(T_0(-\ell_{**} + s - a_0), T_1(-\ell_{**} - (t - a_0)) = t - s, $$

which concludes the proof. □
## 3.3 Geodesics in the Brownian slice

In this section, we study geodesics in the Brownian slice to prove Theorem 1.1. By Proposition 3.7, $x_{**}$ is a 3$^+$-hub in the Brownian sphere if and only if $x_{**}$ is in the interior of a geodesic in the random slice associated to $W_{\min}$. Therefore, the rest of this paper is devoted to prove that this does not happen, almost surely. By scaling, it is enough to prove this result for a Brownian slice under $N_1(\cdot | W_{*} = 0)$, for which we have a construction based on a spine decomposition and well suited to our problem. We will prove the following result.

**Theorem 3.8.** Consider a Brownian slice $\tilde{S}$ distributed under $N_1(\cdot | W_{*} = 0)$. Then, almost surely, there is no geodesic that passes through $x_{*}$.

Note that Theorem 3.8 implies Theorem 1.1. Indeed, by scaling and Proposition 3.5, Theorem 3.8 implies that no geodesic passes through $x_{**}$ in the slice associated to $W_{\min}$.

To prove this theorem, we will use a spine decomposition of the labelled tree $\mathcal{T}$ under $N_1(\cdot | W_{*} = 0)$, that we recall here. This construction is a consequence of results in [5], and was already used in [4, 3, 10].

Consider a triple $(R, N_l, N_r)$ defined on a probability space $(\Omega, \mathcal{F}, \mathcal{P})$, where:

* $R = (R_t)_{t \in [0, \tau_0]}$ is a Bessel process of dimension -5, starting at 1 and stopped when it reaches 0,

* Given $R$, $N_l$ and $N_r$ are two independent Poisson point measure, with intensity

$$21_{[0, \tau_0]} 1_{\omega_* > 0} N_{R_t} (d\omega) dt.$$

Then, the random labelled tree $\mathcal{T}$ associated to this triple, as explained in Section 2.5, is distributed as the random tree encoded by $N_1(\cdot | W_{*} = 0)$. Consequently, the random slice $\tilde{S}$ encoded by $(R, N_l, N_r)$ (as explained in Section 2.4) is distributed as a Brownian slice under $N_1(\cdot | W_{*} = 0)$.

For every $0 \le \beta \le 1$, set

$$\tau_{\beta} = \inf\{t \ge 0 : R_t = \beta\}.$$

Note that the spine of $\mathcal{T}$, which can be identified with the interval $[\tau_1, \tau_0]$, has a random length. The benefit of this construction is that $p_{\tilde{S}}(\tau_0) = x_{*}$. In particular, the point $u_{*}$ of the tree $\mathcal{T}$ is exactly the top of the spine, which is identified with $\tau_0$. We also recall a particular case of Nagazawa's time reversal theorem (see [21, Theorem VII 4.5] and [21, Exercise XI 1.23]). Let $X = (X_t)_{t \in [0, S_1]}$ be a Bessel process of dimension 9, starting from 0 and stopped at its last hitting time of 1, denoted by $S_1$. Then, the processes

$$(R_{\tau_0 - t})_{t \in [0, \tau_0]} \quad \text{and} \quad (X_t)_{t \in [0, S_1]} \qquad (8)$$

have the same law.

We start by proving a weaker version of Theorem 3.8.

**Proposition 3.9.** We have

$$\mathbb{P}(x_{*} \text{ is not in the interior of a geodesic}) > 0.$$

*Proof.* In what follows, we write $[\cdot, \cdot]_{\mathcal{T}}$ to emphasize that we consider intervals on the tree $\mathcal{T}$. For any $x, y \in \tilde{S}$ and $u, v \in \mathcal{T}$ such that $p_{\tilde{S}}(u) = x$ and $p_{\tilde{S}}(v) = y$, we have

$$\tilde{D}(x, x_{*}) = \tilde{D}^{\circ}(u, u_{*}) = \ell_{x}, \quad \tilde{D}(y, x_{*}) = \tilde{D}^{\circ}(v, u_{*}) = \ell_{y}.$$

Therefore, if a geodesic between $x$ and $y$ passed through $x_{*}$, we would have

$$\tilde{D}(x, y) = \tilde{D}^{\circ}(u, v) = \ell_{u} + \ell_{v}.$$
Recall that $\tau_1$ identified with the bottom of the spine. First, note that if $u, v \in ]\tau_1, \tau_0]\tau$, we have

$$ \tilde{D}^{\circ}(u, v) = \ell_u + \ell_v - 2 \max \left( \min_{w \in [u, v]_{\tau}} \ell_w, \min_{w \in [v, u]_{\tau}} \ell_w \right) > \ell_u + \ell_v. $$

This inequality still holds if $u, v \in ]\tau_0, \tau_1]\tau$. Therefore, excluding these two cases, we just need to prove the result for $\tau_0 \in [u, v]\tau$. Then, observe that for topological reasons, the geodesic $\gamma$ between $u$ and $v$ must cross the curve $p_{\tilde{S}}((\tau_t)_{0\le t \le 1})$ (which is the projection of the spine). We will prove that $x_*$ is never the best point to cross this curve.

Fix $0 < \beta < 1$, and suppose that $u \in [\tau_1, \tau_\beta]\tau$ and $v \in [\tau_\beta, \tau_1]\tau$. Then,

$$ \tilde{D}(u, v) \le \tilde{D}^{\circ}(u, \tau_{\beta}) + \tilde{D}^{\circ}(\tau_{\beta}, v) = \ell_u + \beta - 2 \min_{w \in [u, \tau_{\beta}]_{\tau}} \ell_w + \ell_v + \beta - 2 \min_{w \in [\tau_{\beta}, v]_{\tau}} \ell_w. \quad (9) $$

In particular, if there exists $0 < \beta < 1$ such that

$$ 2 \min_{w \in [\tau_1, \tau_\beta]_\tau} \ell_w \wedge 2 \min_{w \in [\tau_\beta, \tau_1]_\tau} \ell_w > \beta \quad (10) $$

then by (9),

$$ \tilde{D}^{\circ}(u, v) < \ell_u + \ell_v $$

which would imply that the geodesic does not pass through $x_*$. Therefore, we need to show that the inequality (10) holds for some $\beta$ arbitrary small, almost surely. To this end, for every $n \in \mathbb{N}$, we introduce the event

$$ E_n = \left\{ \min_{w \in [\tau_1, \tau_{2-n}]_{\tau} \cup [\tau_{2-n}, \tau_1]_{\tau}} \ell_w > 2^{-n-1} \right\}. $$

The previous discussion implies that

$$ \{\limsup E_n\} \subset \{\text{No geodesic passes through } x_*\}. \quad (11) $$

Using properties of Poisson point measures, formula (1) and the time reversal property (8), we have

$$ \begin{aligned} \mathbb{P}(E_n) &= \mathbb{E} \left[ \exp \left( -4 \int_{\tau_1}^{\tau_{2-n}} \mathbb{N}_{R_t} (0 < W_* < 2^{-n-1}) dt \right) \right] \\ &= \mathbb{E} \left[ \exp \left( -6 \int_{\tau_1}^{\tau_{2-n}} \frac{1}{(R_t - 2^{-n-1})^2} - \frac{1}{(R_t)^2} dt \right) \right] \\ &= \mathbb{E} \left[ \exp \left( -6 \int_{S_{2-n}}^{S_1} \frac{1}{(X_t - 2^{-n-1})^2} - \frac{1}{(X_t)^2} dt \right) \right] \\ &= \mathbb{E} \left[ \exp \left( -6 \int_{S_1}^{S_{2n}} \frac{1}{(X_t - 1/2)^2} - \frac{1}{(X_t)^2} dt \right) \right] \end{aligned} $$

where $S_t$ stands for the last hitting time of $t$ by $X$ (the last equality follows from the scaling properties of Bessel processes). Note that the integral is finite for every $n \ge 1$, due to the well-known fact that for every $\varepsilon \in (0, 1/6)$ and $t$ large enough, $X_t > t^{1/2-\varepsilon}$ a.s. Moreover, for every $n \le m$, $\mathbb{P}(E_n) \ge \mathbb{P}(E_m)$. In this situation, we cannot use the Borel-Cantelli lemma to conclude because the events $(E_n)_{n \ge 1}$ are not independent. However, for $n > m$, we have

$$ E_n \cap E_m = \left\{ \inf_{w \in [\tau_1, \tau_{2-m}]_{\tau} \cup [\tau_{2-m}, \tau_1]_{\tau}} \ell_w > 2^{-m-1} \right\} \cap \left\{ \inf_{w \in [\tau_{2-m}, \tau_{2-n}]_{\tau} \cup [\tau_{2-n}, \tau_{2-m}]_{\tau}} \ell_w > 2^{-n-1} \right\}. $$
Using the fact that the processes $(R_t)_{\tau_1 \le t \le \tau_{2-m}}$ and $(R_t)_{\tau_{2-m} \le t \le \tau_{2-n}}$ are independent, we obtain

$$
\begin{align*}
\mathbb{P}(E_n \cap E_m) &= \mathbb{E} \left[ \exp \left( -6 \int_{\tau_1}^{\tau_{2-m}} \frac{1}{(R_t - 2^{-m-1})^2} - \frac{1}{(R_t)^2} dt \right) \exp \left( -6 \int_{\tau_{2-m}}^{\tau_{2-n}} \frac{1}{(R_t - 2^{-n-1})^2} - \frac{1}{(R_t)^2} dt \right) \right] \\
&= \mathbb{E} \left[ \exp \left( -6 \int_{\tau_1}^{\tau_{2-m}} \frac{1}{(R_t - 2^{-m-1})^2} - \frac{1}{(R_t)^2} dt \right) \right] \mathbb{E} \left[ \exp \left( -6 \int_{\tau_{2-m}}^{\tau_{2-n}} \frac{1}{(R_t - 2^{-n-1})^2} - \frac{1}{(R_t)^2} dt \right) \right] \\
&= \mathbb{P}(E_m) \mathbb{E} \left[ \exp \left( -6 \int_{S_1}^{S_{2n-m}} \frac{1}{(X_t - 1/2)^2} - \frac{1}{(X_t)^2} dt \right) \right] \\
&= \mathbb{P}(E_m) \mathbb{P}(E_{n-m})
\end{align*}
$$

which we rewrite as $C_{n,m}\mathbb{P}(E_n)\mathbb{P}(E_m)$, where

$$C_{n,m} = \frac{\mathbb{P}(E_{n-m})}{\mathbb{P}(E_n)}.$$

Define

$$p_{\infty} = \lim_{k \to \infty} \mathbb{P}(E_k) = \mathbb{E} \left[ \exp \left( -6 \int_{S_1}^{\infty} \frac{1}{(X_t - 1/2)^2} - \frac{1}{(X_t)^2} dt \right) \right].$$

Since the integral is finite, we have $p_{\infty} > 0$, and

$$C_{n,m} = \frac{\mathbb{P}(E_{n-m})}{\mathbb{P}(E_n)} \le \frac{1}{p_{\infty}} < \infty.$$

Therefore, by the Kochen-Stone lemma (see [14] and [22]),

$$\mathbb{P}(\limsup E_n) > 0.$$

This gives the result, using (11). □

To conclude the proof of Theorem 3.8, we rely on a 0-1 law argument. By (8), the random slice $\tilde{S}$ can be constructed from the triple $(X, \tilde{N}_l, \tilde{N}_r)$, where

* $X = (X_t)_{t \in [0, S_1]}$ is a Bessel process of dimension 9, starting at 0 and stopped when it reaches 1 for the last time,

* Given $X, \tilde{N}_l$ and $\tilde{N}_r$ are two independent Poisson point measure, with intensity

$$21_{[0,S_1]} 1_{\omega_* > 0} \mathbb{N}_{X_t}(d\omega) dt.$$

For the rest of this paper, we will work with this construction. Let $I$ and $J$ be sets indexing the atoms of the Poisson point measures $\tilde{N}_l$ and $\tilde{N}_r$, so that

$$\tilde{N}_l = \sum_{i \in I} \delta_{(t_i, \tau_i)} \quad \text{and} \quad \tilde{N}_r = \sum_{j \in J} \delta_{(t_j, \tau_j)}.$$

Let $T_1 = \inf\{t \ge 0, X_t = 1\}$. Then, set $\mathcal{F}_n = \sigma((X_t)_{0 \le t \le 1/n \wedge T_1}, (\mathcal{T}_i)_{0 \le t_i \le 1/n \wedge T_1}, (\mathcal{T}_j)_{0 \le t_j \le 1/n \wedge T_1})$.

**Lemma 3.10.** The $\sigma$-algebra

$$\bigcap_{n \ge 1} \mathcal{F}_n$$

is trivial.
Proof. Let $\mathcal{N}_l^*$ and $\mathcal{N}_r^*$ be two independent Poisson point measures on $\mathbb{R}_+ \times \mathcal{S}$, with intensities
$2\mathbb{N}_0(d\omega)dt$,

and let $I^*$ and $J^*$ be sets indexing the atoms of these measures. As $(X, \hat{\mathcal{N}}_l, \hat{\mathcal{N}}_r)$ can be expressed as
a measurable function of $(X, \mathcal{N}_l^*, \mathcal{N}_r^*)$, it is enough to prove the result with $\mathcal{F}_n^*$ instead of $\mathcal{F}_n$, where

$$
\mathcal{F}_n^* = \sigma((X_t)_{0 \le t \le 1/n \wedge T_1}, (\mathcal{T}_i)_{i \in I^*, 0 \le t_i \le 1/n}, (\mathcal{T}_j)_{j \in J^*, 0 \le t_j \le 1/n}).
$$

We will use the following result (see [21, Exercise II.2.15]): consider some $\sigma$-algebra $\mathcal{H}$ and $\mathcal{G}_0 \subseteq \mathcal{G}_1 \subseteq \dots$ such that $\mathcal{H}$ and $\mathcal{G}_0$ are independent. Then,

$$
\sigma(\mathcal{H}, (\mathcal{G}_i)_{i \in \mathbb{N}}) = \bigcap_{i \in \mathbb{N}} \sigma(\mathcal{H}, \mathcal{G}_i). \qquad (12)
$$

Set

$$
\begin{align*}
\mathcal{H}_n &= \sigma((X_t)_{0 \le t \le 1/n \wedge T_1}), \\
\mathcal{G}_n &= \sigma((\mathcal{T}_i)_{i \in I^*, 0 \le t_i < 1/n}), \\
\mathcal{G}'_n &= \sigma((\mathcal{T}_j)_{j \in J^*, 0 \le t_j < 1/n}),
\end{align*}
$$

and define

$$
\mathcal{H} = \bigcap_{n \ge 1} \mathcal{H}_n, \quad \mathcal{G} = \bigcap_{n \ge 1} \mathcal{G}_n, \quad \mathcal{G}' = \bigcap_{n \ge 1} \mathcal{G}'_n.
$$

By construction, for every $n, m, k \ge 1$, $\mathcal{H}_n, \mathcal{G}_m$ and $\mathcal{G}'_k$ are independent. First, we can apply (12) twice with $\mathcal{G}$ and $(\mathcal{G}'_n)_{n \ge 1}$, which gives

$$
\sigma(\mathcal{G}, \mathcal{G}') = \bigcap_{n \ge 1} \sigma(\mathcal{G}, \mathcal{G}'_n) = \bigcap_{n \ge 1} \bigcap_{m \ge 1} \sigma(\mathcal{G}_m, \mathcal{G}'_n).
$$

Then, we can apply (12) a couple more times to $\sigma(\mathcal{G}, \mathcal{G}')$ and $(\mathcal{H}_n)_{n \ge 1}$, and we obtain

$$
\sigma(\mathcal{H}, \mathcal{G}, \mathcal{G}') = \bigcap_{n \ge 1} \sigma(\mathcal{H}_n, \sigma(\mathcal{G}, \mathcal{G}')) = \bigcap_{n,m,h \ge 1} \sigma(\mathcal{H}_n, \mathcal{G}_m, \mathcal{G}'_h).
$$

However, since our sequences of $\sigma$-algebras are decreasing, one can easily check that

$$
\bigcap_{n,m,h \ge 1} \sigma(\mathcal{H}_n, \mathcal{G}_m, \mathcal{G}'_h) = \bigcap_{n \ge 1} \sigma(\mathcal{H}_n, \mathcal{G}_n, \mathcal{G}'_n),
$$

which gives

$$
\sigma(\mathcal{H}, \mathcal{G}, \mathcal{G}') = \bigcap_{n \ge 1} \mathcal{F}_n^*.
$$

However, since a Bessel process of dimension 9 is the norm of a Brownian motion in dimension 9, $\mathcal{H}$ is trivial by Blumenthal 0-1 law. Similarly, independence properties of Poisson point measures imply that $\mathcal{G}$ and $\mathcal{G}'$ are also trivial. Therefore, $\sigma(\mathcal{H}, \mathcal{G}, \mathcal{G}')$ is also trivial, which completes the proof. $\square$

We claim that for every $n \ge 1$, the event

$$
\{\text{There exists a geodesic that passes through } x_*\} \qquad (13)
$$

belongs to $\mathcal{F}_n$. To see this, set

$$
\theta_n = [0, X_{1/n \wedge T_1}] \cup \bigcup_{i \in I \cup J, 0 \le t_i < 1/n \wedge T_1} \mathcal{T}_i.
$$

This set is a random subtree of $\mathcal{T}$, and is measurable with respect to $\mathcal{F}_n$. We will need the following lemma.
**Lemma 3.11.** Almost surely, for every $n \ge 1$, $p_{\tilde{S}}(\theta_n)$ contains a neighborhood of $x_*$ in $\tilde{S}$.

*Proof.* We argue by contradiction. If the statement did not hold, we could find $n_0 \in \mathbb{N}$ and a sequence $(u_n)_{n \in \mathbb{N}} \in \mathcal{T}$ such that for every $n \in \mathbb{N}$,

$$D(x_*, p_{\tilde{S}}(u_n)) \le 1/n \quad \text{and} \quad u_n \notin \theta_{n_0}.$$

By compactness, we can suppose that the sequence $(u_n)$ converges toward some element $u \in \text{Cl}(\mathcal{T} \setminus \theta_{n_0})$ such that $D(x_*, p_{\tilde{S}}(u)) = 0$. However, by Lemma 2.1, this is not possible, which concludes. $\square$

We can finally prove the main result of this section.

*Proof of Theorem 3.8.* By Lemma 3.11, the event

$$\{\text{There exists a geodesic that passes through } x_*\} \quad (14)$$

belongs to $\bigcap_{n \ge 1} \mathcal{F}_n$. However, by Lemma 3.10 this $\sigma$-algebra is trivial. Moreover, by Proposition 3.9, the probability of the event (14) is strictly less than 1. Therefore, it has probability 0, which concludes the proof. $\square$

# 4 Proof of Proposition 2.2

In this section, we use Theorem 1.1 to prove Proposition 2.2.

*Proof of Proposition 2.2.* We argue by contradiction, by proving that if the statement did not hold, there would be a 3+-hub in the Brownian sphere with positive probability. First, we know that almost surely, $x_*$ is not in the interior of a geodesic (see [7, Corollary 7.7]). Therefore, if $(x_*, p_S(u), p_S(v))$ are aligned, then either $p_S(u)$ is on a simple geodesic from $p_S(v)$, or $p_S(v)$ is on a simple geodesic from $p_S(u)$.

Consider $u, v \in \mathcal{T}$ such that $D^\circ(u, v) = D(u, v)$, and suppose that $(x_*, p_S(u), p_S(v))$ are not aligned. Without loss of generality, suppose that $u_* \notin [u, v]$, and as previously, define

$$u_{**} = \inf \left\{ w \in [u, v] : \ell_w = \inf_{z \in [u, v]} \ell_z \right\}.$$

Note that $u_{**}$ is different from $u$ and $v$, otherwise $(x_*, p_S(u), p_S(v))$ would be aligned. Set $\ell_{**} = \ell_{u_{**}}$. Define, for $t \in [0, D(u, v)]$,

$$U(t) = \begin{cases} \inf \{w \in [u, u_{**}] : \ell_w = \ell_u - t\} & \text{if } t \in [0, \ell_u - \ell_{**}], \\ \sup \{w \in [u_{**}, v] : \ell_w = 2\ell_{**} - \ell_u + t\} & \text{if } t \in [\ell_u - \ell_{**}, \ell_u + \ell_v - 2\ell_{**}]. \end{cases}$$

and

$$\gamma_{u,v}(t) = p_S(U(t)).$$

This path corresponds to following a simple geodesic from $p_S(u)$ up to $p_S(u_{**})$, and then another simple geodesic from $p_S(v)$, in reverse direction (this is very similar to the path $\gamma_{0,1}$, see Figure 3). Let us show that $\gamma_{u,v}$ is a geodesic between $p_S(u)$ and $p_S(v)$. First, note that the restriction of $\gamma_{u,v}$ to $[0, \ell_u - \ell_{**}]$ (respectively $[\ell_u - \ell_{**}, \ell_u + \ell_v - 2\ell_{**}]$) is a portion of a simple geodesic. Therefore, we just need to show that for every $t \in [0, \ell_u - \ell_{**}]$, $s \in [\ell_u - \ell_{**}, \ell_u + \ell_v - 2\ell_{**}]$,

$$D(\gamma_{u,v}(t), \gamma_{u,v}(s)) = s - t. \quad (15)$$

By the triangle inequality, we have

$$\begin{aligned} D^\circ(u, v) &= D(u, v) \le D(p_S(u), \gamma_{u,v}(t)) + D(\gamma_{u,v}(t), \gamma_{u,v}(s)) + D(p_S(v), \gamma_{u,v}(s)) \\ &= D^\circ(u, U(t)) + D(\gamma_{u,v}(t), \gamma_{u,v}(s)) + D^\circ(U(s), v). \end{aligned}$$
Since

$$D^{\circ}(u, v) - D^{\circ}(u, U(t)) - D^{\circ}(U(s), v) = \ell_{U(t)} + \ell_{U(s)} - 2\ell_{**} = D^{\circ}(U(t), U(s)),$$

we have

$$D^{\circ}(U(t), U(s)) \leq D(\gamma_{u,v}(t), \gamma_{u,v}(s)).$$

Since the converse inequality always holds, the previous line is an equality. Finally, note that

$$D^{\circ}(U(t), U(s)) = \ell_{U(t)} + \ell_{U(s)} - 2\ell_{**} = \ell_u - t + 2\ell_{**} - \ell_u + s - 2\ell_{**} = s - t,$$

which gives (15). However, this implies that $u_{**}$ is a 3$^+$-hub, which is in contradiction with Theorem 1.1. Therefore, $(x_{*}, p_{S}(u), p_{S}(v))$ are aligned, which concludes the proof. □

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