Note that any element $g \in G$ satisfies $g^q = 1$ hence, the minimal polynomial $\mu_g$ of $g$ divides $X^q - 1$. Since $q \neq \text{char}(k)$, $X^q - 1$ has only simple roots so $\mu_g$ has only simple roots and $g$ is diagonalizable with eigenvalues in $U_q(k)$. Moreover, since $G$ is finite abelian and all its elements are diagonalisable, there exists a basis $\mathcal{B} = (e_1, \dots, e_n)$ of $V$ which simultaneously diagonalises every element of $G$. Let us denote by $\lambda(g) \in U_q(k)^n$ the element $\lambda(g) = (\lambda_1(g), \dots, \lambda_n(g))$, where $g(e_k) = \lambda_k(g)e_k$. This defines an injective map $G \to U_q(k)^n$, $g \mapsto \lambda(g)$. It follows that $|G| = q^N \le |U_q(k)^n| = q^n$, hence $N \le n = \text{dim}(V)$. $\square$

**Proposition 8.20.** The group $S_f(X)$ is not linear.

*Proof.* Let $N \in \mathbb{N}^*$ and $q$ be a prime number. Since $X$ is infinite, one can choose $q$-cycles $\sigma_1, \dots, \sigma_N$ in $S_f(X)$ with pairwise disjoint supports. It is then easy to check that there is an injective morphism of groups defined by

$$(\mathbb{Z}/q\mathbb{Z})^N \to S_f(X) \quad ; \quad (x_1, \dots, x_n) \mapsto \sigma_1^{x_1} \cdots \sigma_N^{x_N}.$$

Hence, the proof follows from Lemma 8.19. $\square$

For any subset $F \subseteq X$, let us denote by $\Sigma(F)$ the pointwise stabilizer of $F$ in $S_f(X)$, and remark that whenever $F$ is finite, the subgroup $\Sigma(F)$ is infinite, in fact isomorphic to $S_f(X)$ itself. We will abbreviate $\Sigma(\{x\})$ as $\Sigma(x)$. For any $k \ge 1$, let $X^{(k)}$ denote the set of $k$-tuples of pairwise distinct points in $X$.

**Lemma 8.21.** Let $F$ be a non-empty subset of $X$. The following hold:

(1) the stabilizer $\Sigma(F)$ is a core-free subgroup of $S_f(X)$, with infinite index;

(2) if $F$ is finite, then $\Sigma(F)$ is not highly core-free in $S_f(X)$;

(3) if $F$ is finite, then $\Sigma(F)$ is $s$-normal in $S_f(X)$.

*Proof.* (1) Let $x \in F$. Since the action $X \frown S_f(X)$ is transitive (even highly transitive) and faithful we have $\bigcap_{g \in S_f(X)} g^{-1}\Sigma(x)g = \bigcap_{g \in S_f(X)} \Sigma(x \cdot g) = \bigcap_{y \in X} \Sigma(y) = \{1\}$, hence $\Sigma(x)$ is core-free in $S_f(X)$. A fortiori, $\Sigma(F)$ is core-free in $S_f(X)$. Let us denote by $\tau_y \in S_f(X)$ the transposition $\tau_y = (x \ y)$ for $y \neq x$. The subgroup $\Sigma(x)$ has infinite index since, for all $y, z \in X \setminus \{x\}$, one has $\tau_y^{-1}\tau_z \in \Sigma(x) \Leftrightarrow y = z$. A fortiori, $\Sigma(F)$ has infinite index in $S_f(X)$.

(2) Let us write $F = \{x_1, \dots, x_k\}$, with $\bar{x} = (x_1, \dots, x_k) \in X^{(k)}$. The action $X^{(k)} \frown S_f(X)$ is transitive, since the action $X \frown S_f(X)$ is highly transitive, and the stabilizer of $\bar{x}$ is $\Sigma(F)$. Consequently, the action $\Sigma(F)\backslash S_f(X) \frown S_f(X)$ is conjugate to $X^{(k)} \frown S_f(X)$. Now, $X^{(k)} \frown S_f(X)$ is not strongly faithful, since taking $k+1$ permutations with pairwise disjoint and finite supports in $X$, every point in $X^{(k)}$ will be fixed by at least one of them. Consequently, $\Sigma(F)\backslash S_f(X) \frown S_f(X)$ is not highly faithful.

(3) For any $g \in S_f(X)$, we have $\Sigma(F) \cap g^{-1}\Sigma(F)g = \Sigma(F \cup F \cdot g)$, and $F \cup F \cdot g$ is still finite, hence $\Sigma(F) \cap g^{-1}\Sigma(F)g$ is infinite. $\square$

**Proposition 8.22.** Let $Y$ and $Z$ be two distinct infinite proper subsets of $X$, let $\tau : Y \to Z$ be a bijection, and let $\vartheta : \tau_* : S_f(Y) \to S_f(Z)$ be the isomorphism defined by $\vartheta(\sigma) = \tau^{-1}\sigma\tau$. Then, the HNN extension $\Gamma = \text{HNN}(S_f(X), S_f(Y), \vartheta)$ has the following properties:

(1) it admits an action which is both highly transitive and highly faithful;

(2) it is not linear;

(3) if $Y$ and $Z$ are both cofinite, then it is not acylindrically hyperbolic;
(4) if Y and Z are both cofinite, then for every bounded subtree $\mathcal{B}$ of its Bass-Serre tree, the pointwise stabilizer $\Gamma_{\mathcal{B}}$ is not highly core-free in a vertex stabilizer $\Gamma_u$.

*Proof.* (1) Note that $S_f(Y) = \Sigma(X \setminus Y)$. Thus, $S_f(Y)$ is a core-free subgroup of $S_f(X)$ by Lemma 8.21. Hence, Corollary F, or in this case Corollary 7.6, applies.

(2) Follows from Proposition 8.20.

(3) As Y and Z are both cofinite and since the intersection of finitely many cofinite subsets is cofinite, the powers $\tau^n$, defined by composition of partial bijections in X (for $n \in \mathbb{Z}$) all have a cofinite domain and a cofinite range, that we will denote by $Y_n$ and $Z_n$ respectively. Let U be any cofinite subset of X. For any $g \in S_f(X)$, one has $S_f(U) \cap g^{-1}S_f(U)g = S_f(U \cap U \cdot g)$, and $U \cap U \cdot g$ is still cofinite. Moreover, for any $n \in \mathbb{Z}$, one has $S_f(U) \cap t^{-n}S_f(U)t^n = S_f(U \cap U \tau^n)$, where $t \in \Gamma$ is the stable letter. Note that $U \tau^n$ is cofinite since the bijection $\tau^n$ realizes a bijection between $U \cap Y_n$ and $U \tau^n$, so that the subset $U \tau^n$ is cofinite in $Z_n$, hence cofinite. Therefore, $U \cap U \tau^n$ is cofinite.

Then, given $\gamma \in \Gamma$, one can write $\gamma = \gamma_1 \cdots \gamma_k$, where each $\gamma_j$ is either a power of the stable letter $t$, or an element of $S_f(X)$, and an easy induction based on previous facts shows that $S_f(Y) \cap \gamma^{-1}S_f(Y)\gamma$ contains $S_f(V)$ for some cofinite set $V$. This proves that $S_f(Y)$ is an s-normal subgroup in $\Gamma$.

Furthermore, $S_f(Y)$ is an amenable group, hence it is not acylindrically hyperbolic by Proposition 8.2. Finally, Proposition 8.1 implies that $\Gamma$ is not acylindrically hyperbolic.

(4) Up to conjugating and to enlarging $\mathcal{B}$, we may and will assume without loss of generality that the stabilizer $\Gamma_u$ is $S_f(X)$, and that $\Gamma_e = S_f(Y)$ for some edge $e$ in $\mathcal{B}$. Since $\Gamma$ acts transitively on the positive edges, there exists $\gamma_1, \dots, \gamma_k \in \Gamma$ such that

$$ \Gamma_{\mathcal{B}} = S_f(Y) \cap \bigcap_{j=1}^{k} \gamma_j^{-1} S_f(Y) \gamma_j. $$

As in the proof of (2), we see that there exist cofinite sets $V_1, \dots, V_k$ such that the intersection $S_f(Y) \cap \gamma_j^{-1}S_f(Y)\gamma_j$ contains $S_f(V_j)$ for every $j$, hence $\Gamma_{\mathcal{B}}$ contains $S_f(\bigcap_{j=1}^k V_j)$, where $\bigcap_{j=1}^k V_j$ is cofinite. Now, $S_f(\bigcap_{j=1}^k V_j)$ is not highly core-free in $\Gamma_u = S_f(X)$ by Lemma 8.21, hence $\Gamma_{\mathcal{B}}$ is not either. $\square$

**Remark 8.23.** When both Y and Z are cofinite in X, this proposition provides more explicit new examples of highly transitive groups, since items (1), (3) and (4) show that the results from [MO15, HO16, FMS15, GGS20] do not apply.

**Remark 8.24.** In the context of Proposition 8.22, notice that $\Gamma$ obviously admits a highly transitive action when $\tau$ can be extended to a permutation $\tilde{\tau} \in S(X)$. Indeed, the $\Gamma$-action defined by $t \mapsto \tilde{\tau}$ and $\sigma \mapsto \tilde{\sigma}$ for $\sigma \in S_f(X)$ is highly transitive since its restriction to $S_f(X)$ already is (in the terminology of Section 3, this action corresponds to the global pre-action $(X, \tilde{\tau})$). Nevertheless, the $\Gamma$-action we obtain factors through the semi-direct product $S_f(X) \rtimes \langle \tilde{\tau} \rangle$, which is amenable while $\Gamma$ is not; hence the $\Gamma$-action is not faithful. Furthermore, such an extension to a permutation $\tilde{\tau}$ is not possible when $X-Y$ and $X-Z$ have different cardinalities.

### 8.2.2 Examples of HNN extensions over $S_f(\mathbb{Z}) \rtimes \mathbb{Z}$

Let us now move to a modification of former examples to get groups which are moreover finitely generated. For these examples, we consider the permutation $s \in S(\mathbb{Z})$ given by $k \cdot s = k + 1$. It is straightforward to check that the subgroup $\langle S_f(\mathbb{Z}), s \rangle < S(\mathbb{Z})$ is finitely generated, and isomorphic to a
semi-direct product of the form $S_f(\mathbb{Z}) \times \mathbb{Z}$. As before, for any subset $F \subseteq \mathbb{Z}$, let us denote by $\Sigma(F)$ the pointwise stabilizer of $F$ in $S_f(\mathbb{Z})$. For some purposes, we will need the action

$$ \mathbb{Z} \times \mathbb{Z} \frown \langle S_f(\mathbb{Z}), s \rangle, \quad (k,l) \cdot s^ng := (ks^ng, ls^n) = ((k+n)g, l+n) $$

for $g \in S_f(\mathbb{Z})$ and $n \in \mathbb{Z}$. Notice that this action is faithful, as is the action $\mathbb{Z} \frown \langle S_f(\mathbb{Z}), s \rangle$. Moreover, given a subset $F \subseteq \mathbb{Z}$, we observe that the pointwise stabilizer in $\langle S_f(\mathbb{Z}), s \rangle$ of the subset $F \times \{0\} \subset \mathbb{Z} \times \mathbb{Z}$ is the subgroup $\Sigma(F) < S_f(\mathbb{Z})$.

**Lemma 8.25.** Let $F$ be a non-empty subset of $\mathbb{Z}$. The following hold:

(1) the stabilizer $\Sigma(F)$ is a core-free subgroup of $\langle S_f(\mathbb{Z}), s \rangle$, with infinite index;

(2) if $F$ is finite, then $\Sigma(F)$ is not highly core-free in $\langle S_f(\mathbb{Z}), s \rangle$;

*Proof.* (1) The group $\Sigma(F)$ is already core-free and has infinite index in $S_f(\mathbb{Z})$ by Lemma 8.21.

(2) Let us write $F = \{x_1, \dots, x_k\}$, with $x_1, \dots, x_k$ pairwise distinct, and set

$$ \bar{x} = ((x_1, 0), \dots, (x_k, 0)) \in (\mathbb{Z} \times \mathbb{Z})^{(k)}. $$

Let us denote by $\Omega$ the orbit of $\bar{x}$ under $\langle S_f(\mathbb{Z}), s \rangle$. As the action $\mathbb{Z} \frown S_f(\mathbb{Z})$ is highly transitive, $\Omega$ is the union $\cup_{n \in \mathbb{Z}} (\mathbb{Z} \times \{n\})^{(k)}$. Furthermore, the stabilizer of $\bar{x}$ is the pointwise stabilizer of $F \times \{0\}$, that is, $\Sigma(F)$. Consequently, the action $\Sigma(F) \setminus \langle S_f(\mathbb{Z}), s \rangle \frown \langle S_f(\mathbb{Z}), s \rangle$ is conjugate to $\Omega \frown \langle S_f(\mathbb{Z}), s \rangle$. Now, $\Omega \frown \langle S_f(\mathbb{Z}), s \rangle$ is not strongly faithful, since taking $k+1$ elements of $S_f(\mathbb{Z})$ with pairwise disjoint supports, every point in $\Omega = \cup_{n \in \mathbb{Z}} (\mathbb{Z} \times \{n\})^{(k)}$ will be fixed by at least one of them. Consequently, $\Sigma(F) \setminus \langle S_f(\mathbb{Z}), s \rangle \frown \langle S_f(\mathbb{Z}), s \rangle$ is not highly faithful. $\square$

Using Lemma 8.25 we can prove the following Proposition exactly as we proved Proposition 8.22.

**Proposition 8.26.** Let $Y$ and $Z$ be two distinct infinite proper subsets of $\mathbb{Z}$, let $\tau: Y \to Z$ be a bijection, and let $\vartheta = \tau_*: S_f(Y) \to S_f(Z)$ be the isomorphism defined by $\vartheta(\sigma) = \tau^{-1}\sigma\tau$. Then, the HNN extension $\Gamma = \text{HNN}(\langle S_f(\mathbb{Z}), s \rangle, S_f(Y), \vartheta)$ has the following properties:

(1) it admits an action which is both highly transitive and highly faithful;

(2) it is finitely generated and not linear;

(3) if $Y$ and $Z$ are both cofinite in $\mathbb{Z}$, then it is not acylindrically hyperbolic;

(4) if $Y$ and $Z$ are both cofinite in $\mathbb{Z}$, then for every bounded subtree $\mathcal{B}$ of its Bass-Serre tree, the pointwise stabilizer $\Gamma_{\mathcal{B}}$ is not highly core-free in a vertex stabilizer $\Gamma_u$.

**Remark 8.27.** Again, when both $Y$ and $Z$ are cofinite in $\mathbb{Z}$, this proposition provides explicit new examples of groups which are highly transitive.

Note that when the complements of $Y$ and $Z$ have the same cardinality, these groups admit a natural highly transitive action, but it fails to be faithful. Indeed, the subgroup $\langle S_f(\mathbb{Z}), t \rangle < \Gamma$ is isomorphic to the HNN extension $\text{HNN}(S_f(\mathbb{Z}), S_f(Y), \vartheta)$, and the action of this subgroup is not faithful by Remark 8.24.
### 8.2.3 Examples of amalgams

We now switch to the context of amalgams. We will need a refinement of Lemma 8.15.

**Lemma 8.28.** Let us consider an amalgam $\Gamma = \Gamma_1 *_{\Sigma} \Gamma_2$. Assume there is a collection $\mathcal{C}$ of infinite subgroups of $\Sigma$ such that, for every $\Sigma' \in \mathcal{C}$ and every $\gamma \in \Gamma_1 \cup \Gamma_2$, the intersection $\Sigma' \cap \gamma^{-1}\Sigma'\gamma$ contains an element of $\mathcal{C}$. Then, for every $\Sigma' \in \mathcal{C}$ and every $\gamma \in \Gamma$, the intersection $\Sigma' \cap \gamma^{-1}\Sigma'\gamma$ contains an element of $\mathcal{C}$. In particular, all $\Sigma' \in \mathcal{C}$ are s-normal subgroups of $\Gamma$.

*Proof.* Let $\Sigma_0$ be any element of $\mathcal{C}$, and let $\gamma$ be any element of $\Gamma$, that we write as a product $\gamma = \gamma_1 \cdots \gamma_n$ of elements of $\Gamma_1$ or $\Gamma_2$. Let us prove by induction that, for $k = 0, \dots, n$, there exists $\Sigma_k \in \mathcal{C}$ which is contained in $\Sigma_0 \cap (\gamma_1 \cdots \gamma_k)^{-1}\Sigma_0(\gamma_1 \cdots \gamma_k)$.

For $k = 0$, the group $\Sigma_0$ has been chosen in $\mathcal{C}$, and it coincides with $\Sigma_0 \cap (\gamma_1 \cdots \gamma_k)^{-1}\Sigma_0(\gamma_1 \cdots \gamma_k)$ in this case. Then, for $k \ge 1$, the subgroup $\Sigma_{k-1} \cap \gamma_k^{-1}\Sigma_{k-1}\gamma_k$ contains some $\Sigma_k \in \mathcal{C}$. Moreover, one has $\Sigma_{k-1} \subseteq \Sigma_0 \cap (\gamma_1 \cdots \gamma_{k-1})^{-1}\Sigma_0(\gamma_1 \cdots \gamma_{k-1})$ by induction hypothesis, hence

$$ \Sigma_k \subseteq \Sigma_{k-1} \cap \gamma_k^{-1} \Sigma_{k-1} \gamma_k \\ \subseteq \Sigma_0 \cap (\gamma_1 \cdots \gamma_{k-1})^{-1} \Sigma_0 (\gamma_1 \cdots \gamma_{k-1}) \cap \gamma_k^{-1} \Sigma_0 \gamma_k \cap (\gamma_1 \cdots \gamma_k)^{-1} \Sigma_0 (\gamma_1 \cdots \gamma_k) $$

whence $\Sigma_k \subseteq \Sigma_0 \cap (\gamma_1 \cdots \gamma_k)^{-1}\Sigma_0(\gamma_1 \cdots \gamma_k)$.

Finally, for $k = n$, we get $\Sigma_n \subseteq \Sigma_0 \cap \gamma^{-1}\Sigma_0\gamma$ with $\Sigma_n \in \mathcal{C}$, as desired. $\square$

<figure><img src="image_9.png" /></figure>

**Proposition 8.29.** Let $X, Y, Z$ be infinite countable sets such that $Z$ is proper subset of $X \cap Y$. Then, the amalgam $\Gamma = S_f(X) *_{S_f(Z)} S_f(Y)$ has the following properties:

(1) it admits an action which is both highly transitive and highly faithful;

(2) it is not a linear group;

(3) if $Z$ is cofinite in both $X$ and $Y$, then it is not acylindrically hyperbolic;

(4) if $Z$ is cofinite in $X$ (resp. $Y$), then $S_f(Z)$ is not highly core-free in $S_f(X)$ (resp. $S_f(Y)$).

*Proof.* Lemma 8.21 and Corollary 7.7 imply (1). Proposition 8.20 implies (2) while (4) follows from Lemma 8.21. Let us prove (3). Let $\mathcal{C}$ be the collection of subgroups of the form $S_f(U)$ where $U$ is cofinite in $Z$ (hence cofinite in both $X$ and $Y$). For every $\Sigma' = S_f(U) \in \mathcal{C}$ and every $\gamma \in S_f(X) \cup S_f(Y)$, one can check as in former proofs that the intersection $\Sigma' \cap \gamma^{-1}\Sigma'\gamma$ contains an element of $\mathcal{C}$. Hence, Lemma 8.28 implies that $S_f(Z)$ is s-normal in $\Gamma$. Since $S_f(Z)$ is amenable it is not acylindrically hyperbolic hence $\Gamma$ is not acylindrically hyperbolic. $\square$

<figure><img src="image_10.png" /></figure>

Again, one can easily modify the previous examples to get groups which are moreover finitely generated. The proof of the following Proposition is exactly the same as the proof of Proposition 8.29 (by using Lemma 8.25). We can now provide one last new class of highly transitive examples.

**Proposition 8.30.** Let $Z$ be an infinite proper subset of $\mathbb{Z}$ and consider the amalgam $\Gamma := \langle S_f(\mathbb{Z}), s \rangle *_{S_f(Z)} \langle S_f(\mathbb{Z}), s \rangle$. The following holds.

(1) $\Gamma$ admits an action which is both highly transitive and highly faithful;

(2) $\Gamma$ is finitely generated and not linear;

(3) If $Z$ is cofinite in $\mathbb{Z}$, then $\Gamma$ is not acylindrically hyperbolic and $S_f(Z)$ is not highly core-free in $\langle S_f(\mathbb{Z}), s \rangle$.
## 8.3 Faithful actions which are non-topologically free on the boundary

Although our main result provides a complete characterization of high transitivity for groups admitting a faithful minimal action of general type on a tree, one may wonder if Corollaries 7.6 and 7.7 can hold in a wider context, namely, if the core-freeness assumption of the edge group in a vertex group can be weakened to core-freeness in the whole group. We will see that it is not the case.

Thanks to the quoted result from [LBMB22], this amounts to finding examples of amalgams and HNN extensions whose action on their Bass-Serre trees are minimal of general type and faithful, but the action on the boundary is not topologically free. By Bass-Serre theory, we essentially need to find faithful edge transitive actions on trees without inversions which are not topologically free on the boundary, but which in the amalgam case have two vertex orbits, while in the HNN case they have only one vertex orbit.

Our examples belong to a class which was explored in depth by Le Boudec [LB16, LB17], generalizing a construction of Bader-Caprace-Gelander-Mozes [BCGM12] which takes its roots in the work of Burger-Mozes [BM00]. Such examples already appeared in Le Boudec and Matte-Bon's work on high transitivity, so our only contribution here is to point out that some of those naturally decompose as amalgams or HNN extensions. We will focus on specific easy examples instead of seeking large generality. For more examples, we refer the reader to Ivanov's recent work [Iva20].

### 8.3.1 An example of amalgam

Let $\mathcal{T}_d$ be a $d$-regular tree of finite degree $d \ge 3$. As in [BM00], let us fix a coloring on the set of edges $c : E(\mathcal{T}_d) \to \{1, \dots, d\}$ such that:

* every edge has the same color as its antipode;

* for any vertex $v$, the restriction of $c$ to the star $st(v)$ is a bijection onto $\{1, \dots, d\}$.

For any vertex $v$, any automorphism $g \in \mathrm{Aut}(\mathcal{T}_d)$ induces a bijection $g_v : st(v) \to st(gv)$, which itself induces a permutation $\sigma(g, v) \in S_d$, where $S_d = \mathrm{Sym}(\{1, \dots, d\})$. Let $C_d$ be a cyclic subgroup of $S_d$ generated by a $d$-cycle. Consider the group, coming from [LB16],

$$G = G(C_d) = \{g \in \mathrm{Aut}(\mathcal{T}_d) : \sigma(g, v) \in C_d \text{ for all but finitely many vertices}\}.$$

**Remark 8.31.** The group $G(C_d)$ is countable. Indeed, given an automorphism $g \in G(C_d)$ and any edge $e$ such that $\sigma(g, r(e))$ has to be in $C_d$, the permutation $\sigma(g, r(e))$ is determined by the color of $g(e)$. It follows that $g$ is completely determined by its restriction to any finite subtree containing at least one edge and all stars at vertices $v$ such that $\sigma(g, v) \notin C_d$.

In order to forbid inversions, recall there is a natural equivalence relation $\mathcal{R}_{even}$ on $V(\mathcal{T}_d)$ which relates any two vertices at even distance from each other, and that this equivalence relation is preserved by any automorphism of $\mathcal{T}_d$. We then let

$$\Gamma = G^+ = \{g \in G : g \text{ does not exchange the two classes of } \mathcal{R}_{even}\}.$$

It is fairly easy to see that the action $\Gamma \curvearrowright \mathcal{T}_d$ is transitive on undirected edges, hence minimal, of general type, and without inversion. Let us now fix some edge $e_0$ from $v_1$ to $v_2$, and consider the stabilizers $\Gamma_1$, $\Gamma_2$ and $\Sigma$ of $v_1$, $v_2$ and $e_0$ respectively (in $\Gamma$). By Bass-Serre theory, we have the following.

**Remark 8.32.** The morphism $\Gamma_1 *_{\Sigma} \Gamma_2 \to \Gamma$ given by inclusions is an isomorphism, and $\mathcal{T}_d$ is the Bass-Serre tree of $\Gamma_1 *_{\Sigma} \Gamma_2$.
The following result summarizes well-known properties of $\Gamma$ showing that the hypothesis that $\Sigma$ is core-free in $\Gamma_1$ or $\Gamma_2$ cannot be relaxed in Corollary 7.7. We provide a proof for the reader's convenience.

**Proposition 8.33.** With the above notations:

(1) $\Sigma$ is core-free in $\Gamma$, and the amalgam $\Gamma_1 *_{\Sigma} \Gamma_2$ is non-degenerate;

(2) the $\Gamma$-action on $\partial\mathcal{T}_d$ is not topologically free;

(3) $\Gamma$ is not highly transitive;

(4) $\Gamma$ is icc.

*Proof.* (1) The $\Gamma$-action on $\mathcal{T}_d$ is faithful (by definition) and of general type, so that $\Sigma$ is core-free in $\Gamma$, and the amalgam $\Gamma_1 *_{\Sigma} \Gamma_2$ is non-degenerate.

(2) Consider the half-tree $\mathcal{H}$ associated to $e_0$. It suffices to prove that the pointwise stabilizer of $\mathcal{H}$ is a non-trivial group.

To do so, we follow the proof of [LB17, Theorem C]. First, we take a non-trivial permutation $\sigma \in S_d$ which fixes the color of $e_0$ (this exists since $d \ge 3$; notice it lives in $S_d \setminus C_d$). Then, we define a non-trivial automorphism $\gamma \in \Gamma$ fixing $\mathcal{H}$ pointwise as follows.

* The restriction of $\gamma$ to $\mathcal{H}$ is the identity.

* Then, we let $\gamma$ act on $\text{st}(v_1)$ so that $\sigma(g, v_1) = \sigma$ (this is possible since $\sigma$ fixes the color of $e_0$, and will guarantee that $\gamma$ is non-trivial since $\sigma$ is non-trivial).

* Then, we extend the action inductively: given any vertex $w$ outside $\mathcal{H}$, we set $w'$ to be the unique neigbour of $w$ which is closer to $\mathcal{H}$ than $w$, and define the $\gamma$-action on $\text{st}(w)$ in terms of the (previously defined) $\gamma$-action on $\text{st}(w')$. Namely, denoting by $e_w$ the edge from $w$ to $w'$, the $\gamma$-action on $\text{st}(w')$ provides the edge $\gamma e_w$. Then there is a unique element $\sigma_w \in C_d$ sending $c(e_w)$ onto $c(\gamma e_w)$, and we let $\gamma$ act on $\text{st}(w)$ so that $\sigma(g, w) = \sigma_w$ (note that $\sigma(g, w) = \sigma(g, w')$ as soon as $\sigma(g, w')$ was already in $C_d$).

(3) The $\Gamma$-action on $\mathcal{T}_d$ is minimal, by edge-transitivity, and of general type. Consequently, [LBMB22, Corollary 1.5] applies, and the transitivity degree of $\Gamma$ is at most 2.

(4) Let $\gamma_0$ be a non-trivial element of $\Gamma$ and $\xi$ be a point in $\partial\mathcal{T}_d$ which is not fixed by $\gamma_0$. Given edges $e, e'$ with sources in the same class of vertices, such that $e$ is on the geodesic $[\xi, s(e')]$, there exists $\gamma \in \Gamma$ such that $\gamma e = e'$. This $\gamma$ is a hyperbolic element whose axis contains $e$ and $e'$. Moreover, if we choose $e$ close enough to $\xi$, then the repelling point $\xi^-$ of $\gamma$ in $\partial\mathcal{T}_d$ is close enough to $\xi$ so that $\gamma_0 \xi^- \neq \xi^-$. Now, since $\xi^-$ is not fixed by $\gamma_0$, the set of fixed points of $\gamma^n \gamma_0 \gamma^{-n}$ moves into smaller and smaller neighborhoods (in $\mathcal{T}_d \cup \partial\mathcal{T}_d$) of the attracting point of $\gamma$ as $n \to +\infty$. Thus, the set $\{\gamma^n \gamma_0 \gamma^{-n} : n \ge 1\}$ is infinite. $\square$

### 8.3.2 An example of HNN extension

In the previous example, notice that even the subgroup

$$U(\text{id}) = \{g \in \text{Aut}(\mathcal{T}_d) : \sigma(g, v) = \text{id} \text{ for every vertex } v\}$$

includes inversions, so that one cannot easily find a subgroup of $G$ without inversion and acting transitively on $V(\mathcal{T}_d)$. Hence, we slightly modify the construction in order to get an example of HNN extension.
Let $\mathcal{T}_{d,d}$ be a $(d,d)$-biregular tree, where $d \ge 2$ (by this, we mean an oriented tree in which every star $\mathrm{st}(v)$ contains exactly $d$ positive edges and $d$ negative edges). Let us denote $\mathrm{st}(v)^+ = \mathrm{st}(v) \cap E(\mathcal{T}_{d,d})^+$ and $\mathrm{st}(v)^- = \mathrm{st}(v) \cap E(\mathcal{T}_{d,d})^-$, and fix a coloring on the set of edges $c : E(\mathcal{T}_{d,d}) \to \{1, \dots, 2d\}$ such that:

* every edge has the same color as its antipode;

* for any vertex $v$, the restriction of $c$ to the star $\mathrm{st}(v)$ is a bijection onto $\{1, \dots, 2d\}$;

* for any vertex $v$, the image $c(\mathrm{st}(v)^+)$ is either $\{1, \dots, d\}$ or $\{d+1, \dots, 2d\}$.

By Aut($\mathcal{T}_{d,d}$), we mean the group of automorphisms of $\mathcal{T}_{d,d}$ preserving the orientation. For any vertex $v$, any automorphism $g \in \mathrm{Aut}(\mathcal{T}_{d,d})$ induces bijections $g_v^\pm : \mathrm{st}(v)^\pm \to \mathrm{st}(gv)^\pm$, which themselves induces a permutation $\sigma(g, v) \in S_{2d}$, where $S_{2d} = \mathrm{Sym}(\{1, \dots, 2d\})$, preserving the partition $\{1, \dots, d\} \sqcup \{d+1, \dots, 2d\}$. Let $F_d$ be a the subgroup of $S_{2d}$ generated by the commuting elements

$$ \sigma_1 = (1 \ 2 \ \dots \ d)(d+1 \ d+2 \ \dots \ 2d), $$

$$ \sigma_2 = (1 \ d+1)(2 \ d+2)\dots(d \ 2d). $$

Then, the group,

$$ \Gamma = G(F_d) = \{g \in \mathrm{Aut}(\mathcal{T}_{d,d}) : \sigma(g, v) \in F_d \text{ for all but finitely many vertices}\} $$

is countable (this is not hard to prove, using that $F_d$ acts freely on $\{1, \dots, 2d\}$).

It is fairly easy to see that the action $\Gamma \curvearrowright \mathcal{T}_{d,d}$ is transitive on positive edges, hence minimal, of general type. It is moreover without inversion since it preserves the orientation. Let us fix some vertex $v$, some positive edges $e_1, e_2$ such that $r(e_1) = v = s(e_2)$, and some automorphism $\tau \in \Gamma$ such that $\tau(e_1) = e_2$. Now, consider the stabilizers $H = \Gamma_v$ and $\Sigma = \Gamma_{e_2}$, and the isomorphism $\vartheta : \Sigma \to \Gamma_{e_1}$ given by $\vartheta(\sigma)(x) = \tau^{-1}\sigma\tau(x)$. By Bass-Serre theory, we have the following.

**Remark 8.34.** The morphism $\mathrm{HNN}(H, \Sigma, \vartheta) \to \Gamma$ given by inclusions is an isomorphism, and $\mathcal{T}_{d,d}$ is the Bass-Serre tree of $\mathrm{HNN}(H, \Sigma, \vartheta)$.

The following result summarizes well-known properties of $\Gamma$ showing that the hypothesis that one of the subgroups $\Sigma, \vartheta(\Sigma)$ is core-free in $H$ cannot be relaxed in Corollary 7.6. We omit the proof, which is similar to the one of Proposition 8.33.

**Proposition 8.35.** With the above notations:

(1) $\Sigma$ is core-free in $\Gamma$, and the $\mathrm{HNN}(H, \Sigma, \vartheta)$ is non-ascending;

(2) the $\Gamma$-action on $\partial\mathcal{T}_{d,d}$ is not topologically free;

(3) $\Gamma$ is not highly transitive;

(4) $\Gamma$ is icc.
# 9 Other types of actions and necessity of the minimality assumption

We now discuss various natural extensions of Theorem A by considering other types of actions (recall that group actions on trees are classified in five different types, see Section 2.3). As we will see, non-general type actions which are topologically free on the boundary seem to play no role regarding high transitivity. We will also see that the minimality hypothesis in Theorem A cannot be avoided.

Let us recall that a group action by homeomorphisms on a topological space is called **minimal** when every orbit is dense. Note that given a group action on a tree, the minimality of the action on the boundary implies the minimality of the action on the tree but the converse does not hold: for instance the standard $\mathbb{Z}$-action on itself is minimal but the action on the boundary is not since it has two distinct fixed points.

**Proposition 9.1.** Every residually finite group admits an elliptic faithful action on a tree with non-empty boundary such that the action on the boundary is both free and minimal.

*Proof.* Let $\Gamma$ be a residually finite group, let $(\Gamma_n)_{n \ge 0}$ be a decreasing chain of finite index normal subgroups with trivial intersection, where $\Gamma_0 = \Gamma$. Then the disjoint union of the coset spaces $\Gamma/\Gamma_n$ has a natural tree structure where we connect each $\gamma\Gamma_{n+1}$ to $\gamma\Gamma_n$, and the boudary is non-empty. Since the action is transitive on each level of the tree, this action is minimal on the boundary. It is free because the subgroups $\Gamma_n$ are normal and intersect trivially. Finally it is elliptic because the vertex $\Gamma$ is fixed (moreover, the only non-trivial invariant subtrees are balls around $\Gamma$). $\square$

Since there are both highly transitive residually finite groups (such as $\mathbb{F}_2$) and non-highly transitive residually finite groups (such as $\mathbb{Z}$), we see that there is no hope for a classification of the transitivity degree of groups admitting an elliptic faithful action on a tree such that the induced action on the boundary is free and minimal. We can use also this construction in order to show that the minimality assumption for the action on the tree is needed in Theorem A.

**Proposition 9.2.** Let $\Gamma$ be a non-abelian free group. Then for every residually finite group $\Lambda$, the group $\Gamma \times \Lambda$ admits an action of general type on a tree which is topologically free on the boundary.

*Proof.* Since $\Gamma$ is free, we have a free $\Gamma$-action on a tree $\mathcal{T}_1$ which is of general type because $\Gamma$ is not abelian. Let $\Lambda \curvearrowright \mathcal{T}_2$ be an action provided by the previous proposition, let $o$ be its unique fixed point. Our new tree $\mathcal{T}$ is obtained by gluing over each vertex of $\mathcal{T}_1$ a copy of $\mathcal{T}_2$ at its origin $o$. To be more precise, the vertex set is $V(\mathcal{T}) = V(\mathcal{T}_1) \times V(\mathcal{T}_2)$, and on the vertex set $\mathcal{T}_1 \times \{o\}$ we put a copy of the edges of $\mathcal{T}_1$, while for each $v \in V(\mathcal{T}_1)$, we put a copy of the edges of $\mathcal{T}_2$ on the vertex set $\{v\} \times \mathcal{T}_2$.

Then the $\Gamma \times \Lambda$ action on $V(\mathcal{T})$ given by $(\gamma, \lambda) \cdot (x_1, x_2) = (\gamma \cdot x_1, \lambda \cdot x_2)$ is an action by automorphisms on our new tree $\mathcal{T}$. Noting that each half-tree contains a copy of a half-tree of $\mathcal{T}_2$, it is not hard to check that this action is moreover topologically free. Moreover, it is of general type since the $\Gamma$-action on $\mathcal{T}_1$ was of general type. $\square$

The previous proposition will allow us to show that in our main result, the hypothesis of minimality of the action on the tree is needed, using the well-known fact that non-trivial product groups cannot be highly transitive. For the convenience of the reader, we provide a proof of the latter fact via the following stronger result.

**Proposition 9.3.** Let $G$ be a non-abelian topologically simple group. If $\Gamma$ is a dense subgroup of $G$, then the centralizer of every non-trivial element of $\Gamma$ is core-free.
**Proof.** Let $\gamma \in \Gamma \setminus \{1_\Gamma\}$. If the centralizer of $\Gamma$ is not core-free, then there exists a normal subgroup $N \le \Gamma$ such that every element of $N$ commutes with $\gamma$. Since $G$ is topologically simple, $N$ is dense in $G$, so by continuity of group multiplication we conclude that every element of $G$ commutes with $\gamma$. In particular, $G$ has a non-trivial center, which contradicts the topological simplicity of $G$ since $G$ is not abelian. $\square$

**Corollary 9.4.** Let $\Gamma$ be a highly transitive group. Then the centralizer of every non-trivial element of $\Gamma$ must be core-free, and $\Gamma$ cannot be decomposed as a non-trivial direct product.

**Proof.** If $\Gamma$ is highly transitive, it can be embedded as a dense subgroup of $S(X)$ for some infinite set $X$, and since the latter is topologically simple, the conclusion follows from the previous result. Moreover, if $\Gamma$ could be decomposed as a non-trivial direct product $\Gamma_1 \times \Gamma_2$, then if $\gamma \in \Gamma_1 \setminus \{1_{\Gamma_1}\}$, the element $(\gamma, 1_{\Gamma_2})$ would commute with every element of the non-trivial normal subgroup $\{1_{\Gamma_1}\} \times \Gamma_2$, a contradiction. $\square$

Applying Proposition 9.2 for instance to $\Gamma = \mathbb{F}_2$ and $\Lambda = \mathbb{Z}$, we see that the finitely generated group $\mathbb{F}_2 \times \mathbb{Z}$ admits a (faithful) action of general type on a tree $\mathcal{T}$ which is topologically free on the boundary, although the group $\mathbb{F}_2 \times \mathbb{Z}$ is not highly transitive because it decomposes as a direct product. We see moreover that when restricting to the minimal component of this action, we will lose the faithfulness of the action (in particular the topological freeness), which is why Theorem A cannot be applied.

We now move on to showing that no general classification can be hoped for in the case of parabolic actions (note that minimal parabolic actions only arise for non-finitely generated groups).

**Lemma 9.5.** Every non-finitely generated group admits a faithful parabolic action on a tree.

**Proof.** Since $\Gamma$ is countable, it can be written as a countable increasing union of finitely generated subgroups $\Gamma = \cup_{n \in \mathbb{N}} \Gamma_n$, where $\Gamma_0 = \{1\}$. We now put a tree structure on the vertex set $\sqcup_n \Gamma/\Gamma_n$ by connecting each $\gamma \Gamma_n$ to $\gamma \Gamma_{n+1}$. $\Gamma$ acts on this tree by left translation. Since $\Gamma_0 = \{1\}$, this action is faithful. Note that every group element $g \in \Gamma$ is elliptic (with fixed point $\Gamma_n$, where $n \in \mathbb{N}$ is such that $g \in \Gamma_n$). Since $\Gamma$ is not finitely generated, this action has no global fixed point, so we have a parabolic action. $\square$

Note that the $\Gamma$-invariant subtrees of the action constructed above are exactly those of the form $\sqcup_{n \ge m} \Gamma/\Gamma_n$ for some $m \ge 0$. So this action has no minimal globally invariant subtree, as opposed to what happens for finitely generated groups.

**Proposition 9.6.** Let $\Gamma$ be a non-finitely generated group, let $\Lambda$ be a residually finite infinite group. Then $\Gamma \times \Lambda$ admits a parabolic action on a tree such that the action on the boundary is topologically free.

**Proof.** As before we use the $\Gamma$-action on a tree $\mathcal{T}_1$ provided by the previous lemma. Let $\Lambda \curvearrowright \mathcal{T}_2$ be an elliptic action provided by Proposition 9.1, whose unique fixed point is denoted by $o \in V(\mathcal{T}_2)$. This time, we glue a copy of $\mathcal{T}_2$ on each terminal vertex (that is, each vertex of the form $\gamma\Gamma_0$) in the tree $\mathcal{T}_1$, thus yielding a $\Gamma \times \Lambda$ action which is easily seen to be parabolic. Moreover, the action on the boundary is topologically free. Note furthermore that the action on the boundary has a unique fixed point, corresponding to the unique element of the boundary of $\mathcal{T}_1$. $\square$

As an example, we can take for $\Gamma$ the group of finitely supported permutations, and for $\Lambda$ the group $\mathbb{Z}$, and we get a non-highly transitive group with a parabolic action which is topologically free on the boundary, and which has no minimal component. We do not know if there is a highly transitive group
with a parabolic action which is topologically free on the boundary.

Let us now treat the quasi-parabolic case. First, note that the Baumslag-Solitar groups $BS(1,n)$ for $n \ge 2$ provide examples of groups admitting an action on a tree which is minimal and quasi-parabolic (since it is an ascending HNN extension, cf. Sec. 2.5) and topologically free on the boundary (by Lemma 8.6), but which are not highly transitive since they are solvable.

**Remark 9.7.** Another example of a non-highly transitive group with a quasi-parabolic minimal action on a tree which is topologically free on the boundary is provided by Thompson's group $F = \langle x_0, x_1, x_2, \dots | x_k^{-1}x_nx_k = x_{n+1} \text{ for all } k < n \rangle$. $F$ is not highly transitive since, by [LBMB22, Corollary 5.3], it has transitivity degree at most 2. Let $H$ be the subgroup generated by $\{x_i : i \ge 1\}$ (which is isomorphic to $F$), and $\vartheta$ be the endomorphism which takes $x_i$ to $x_{i+1}$ and observe that $F = \text{HNN}(H, H, \vartheta)$. By Section 2.5 the action of $F$ on the associated Bass-Serre tree is quasi-parabolic and minimal. Moreover, since $x_0^{-k}\vartheta(H)x_0^k = \langle x_n | n \ge k+2 \rangle$ for all $k \ge 1$, it is not difficult to check that $\vartheta(H)$ is core-free in $H$ hence, the action is also topologically free on the boundary.

Coupled with the previous examples, the following proposition shows that for groups admitting minimal quasi-parabolic actions on a tree, the topological freeness of the action does not play a role in their high transitivity.

**Proposition 9.8.** *The finitely generated group $\Gamma = S_f(\mathbb{Z}) \times \mathbb{Z}$ is highly transitive and finitely generated but admits a minimal quasi-parabolic action on a tree which is topologically free on the boundary.*

*Proof.* We have already observed that $\Gamma$ is highly transitive thanks to its natural action on $\mathbb{Z}$. We will obtain our desired action on a tree by showing that it can be written as an *ascending* HNN extension. Note that $\Gamma$ is a semi-direct product, so it does have a natural HNN extension decomposition, but this decomposition provides a linear action so we need another one.

Denote by $\tau$ the translation on $\mathbb{Z}$. Let $\vartheta$ be the corresponding inner automorphism of $S(\mathbb{Z})$, i.e. $\vartheta(\gamma) = \tau^{-1}\gamma\tau$. Consider the subgroups $H = \Sigma = S_f(\mathbb{N})$ which we view as subgroups of $S_f(\mathbb{Z}) < \Gamma$. Note that $\vartheta(\Sigma)$ is the stabilizer of 0 in $S_f(\mathbb{N})$ i.e. $\vartheta(\Sigma) = \Sigma(0) < S_f(\mathbb{N})$ with the notations of Section 8.2. We claim that $\Gamma = \text{HNN}(H, \Sigma, \vartheta)$.

First, since $\vartheta(h) = \tau^{-1}h\tau$ for all $h \in H$, we have a quotient map $\pi : \text{HNN}(H, \Sigma, \vartheta) \to \Gamma$ given by $t \mapsto \tau$ and $h \mapsto h$ for every $h \in H$. To show that $\pi$ is injective, we use the fact that the HNN extension is ascending: since $H = \Sigma$, for every $g \in \text{HNN}(H, \Sigma, \vartheta)$, there exists $k \in \mathbb{Z}$ such that $t^{-k}gt^k = ht^n$ with $h \in H$ and $n \in \mathbb{Z}$ (it suffices to take $k$ sufficiently large). If $g \ne 1$, one must have $h \ne \text{id}$ or $n \ne 0$, and it is clear that $\pi(g) \ne 1$ in both cases.

So we do have $\Gamma = \text{HNN}(H, \Sigma, \vartheta)$, in particular it is an ascending non-degenerate HNN extension. So as explained in Section 2.5, its action on its Bass-Serre tree is minimal and quasi-parabolic. Finally, $\vartheta(\Sigma) = \Sigma(0)$ is core-free in $H = S_f(\mathbb{N})$ by Lemma 8.21, so by Lemma 7.3 we conclude that the action on the boundary is topologically free as wanted. $\square$

**Remark 9.9.** One can also construct a tree $\mathcal{T}$ directly, with vertex set $V = \bigsqcup_{k \in \mathbb{Z}} S_f(\mathbb{Z})/H_k$, where $H_k := S_f(\mathbb{Z}_{\ge -k})$, and positive edges corresponding to inclusions $\sigma H_k \subseteq \sigma H_{k+1}$. The reader can verify that $\Gamma = S_f(\mathbb{Z}) \times \mathbb{Z}$ acts on $\mathcal{T}$ via

$$ (\sigma, n) \cdot \sigma' H_k := \sigma \tau^n (\sigma' H_k) \tau^{-n} = \sigma (\tau^n \sigma' \tau^{-n}) H_{k+n}, $$

where $\tau$ is still the translation on $\mathbb{Z}$, and then check “by hand” that this action has all the properties announced in the above proposition.
We finally mention the lineal case. Note that in this case, minimal actions are not interesting with
respect to high transitivity since no subgroup of the automorphism group of the biinfinite line is highly
transitive.

In the elliptic case, a natural weakening of the minimality assumption was provided by asking that the action on the boundary is minimal. Here however, this is still too strong a condition since any lineal action will have the two ends corresponding to the axis as an invariant set. We thus replace it by topological transitivity (which means the existence of a dense orbit) and observe that in this setup, there seems to be no connection between high transitivity and lineal actions.

**Proposition 9.10.** Let $\Gamma$ be a residually finite group, then the group $\Gamma \times \mathbb{Z}$ admits a lineal action on a tree which is both topologically free and topologically transitive on the boundary.

*Proof.* Let $\mathcal{T}$ be a tree equipped with an elliptic $\Gamma$-action which is minimal and free on the boundary as provided by Proposition 9.1, let $o$ be the fixed point. As in the previous constructions, we then glue a copy of $\mathcal{T}$ at the vertex $o$ on top of every element of $\mathbb{Z}$, thus obtaining a tree with a natural $\Gamma \times \mathbb{Z}$-action which is both topologically free and topologically transitive on the boundary: every element of the boundary which does not belong to the two element set $\partial\mathbb{Z}$ has a dense orbit and is fixed by no nontrivial group element. $\square$

The previous proposition provides us many non-highly transitive groups with a lineal action on a
tree which is both topologically free and topologically transitive on the boundary.

In the opposite direction, the group $S_f(\mathbb{Z}) \times \mathbb{Z}$ provides us an example of a highly transitive group satisfying the assumptions of the previous proposition, thus showing that lineal actions which are topologically free and topologically transitive on the boundary do not play a role in high transitivity.

**Proposition 9.11.** The highly transitive group $S_f(\mathbb{Z}) \times \mathbb{Z}$ admits a lineal action on a tree which is both topologically free and topologically transitive on the boundary.

*Proof.* Let us construct a tree $\mathcal{T}$ as follows. We start with $\mathbb{Z}$, seen as a bi-infinite line. Then, for every $(\sigma, k) \in S_f(\mathbb{Z}) \times \mathbb{Z}$, we consider an infinite ray $\mathcal{R}_{(\sigma,k)}$, and link its origin $o_{(\sigma,k)}$ to the vertex $k \in \mathbb{Z}$ by an edge. The boundary $\partial\mathcal{T}$ consists of points $\xi_\gamma$ for $\gamma \in S_f(\mathbb{Z}) \times \mathbb{Z}$ (the extremities of the rays, which are isolated in $\partial\mathcal{T}$) and two accumulation points $\eta_\pm$.

The group $S_f(\mathbb{Z}) \times \mathbb{Z}$ acts on $\mathcal{T}$ by $(\sigma, k) \cdot k' = k + k'$ and by $\gamma \cdot \mathcal{R}_{\gamma'} = \mathcal{R}_{\gamma\gamma'}$ (it permutes the rays).
The induced action on $\partial\mathcal{T}$ fixes $\eta_\pm$ and is transitive free on $\{\xi_\gamma : \gamma \in S_f(\mathbb{Z}) \times \mathbb{Z}\}$. Thus, it is both
topologically free and topologically transitive.
$\square$

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AUTHORS

Pierre Fima

Université Paris Cité and Sorbonne Université, CNRS, IMJ-PRG,
F-75013 Paris, France.

pierre.fima@imj-prg.fr

François Le Maître

Université Paris Cité and Sorbonne Université, CNRS, IMJ-PRG,
F-75013 Paris, France.

françois.le-maitre@imj-prg.fr

Soyoung Moon

Université de Bourgogne, Institut Mathématiques de Bourgogne, CNRS, BP 47870,
21078 Dijon cedex France.

soyoung.moon@u-bourgogne.fr

Yves Stalder

Université Clermont Auvergne, CNRS, LMBP,
F-63000 Clermont-Ferrand, France.

yves.stalder@uca.fr