Proof. Since $H = G^+_{(i)}(\{0\},\{0\})$ is a subgroup of $G^+_{(i)}(Y_0,Y_1)$ , it suffices to prove that $\mathrm {Fix}_G(\mathcal {T})$ does not fix any vertices outside of $\mathcal {T}$ . If $\mathcal {T}$ has a single vertex, the result is trivial. Now, suppose that $\mathcal {T}$ contains at least one edge. For each leaf w of $\mathcal {T}$ , consider the unique vertex v of T that is adjacent to w and let

$$ \begin{align*}T(v,w)=\{x\in V(T): d_T(x,v) < d_T(x,w)\}.\end{align*} $$

By choice of w, one has that $\mathcal {T}\subseteq T(v,w)\cup \{w\}$ and since H satisfies Tits’ independence property, $\mathrm {Fix}_H(T(v,w))$ has a local action $\operatorname {\mathrm {Alt}}(d-1)$ at w, where w has degree d. In particular, since $d\geq 4$ , $\mathrm {Fix}_H(\mathcal {T})\subseteq \mathrm {Fix}_H(T(v,w)\cup \{w\})=\mathrm {Fix}_H(T(v,w))$ does not fix any neighbour of w other than v and thus does not fix any vertices outside of $T(v,w)\cup \{w\}$ . Since any vertex of T that does not belong to $\mathcal {T}$ must be outside of one of these $T(v,w)\cup \{w\}$ , it follows that $\mathrm {Fix}_H(\mathcal {T})$ does not fix any vertices outside of $\mathcal {T}$ and the result follows.

Proof. To shorten the proof and to make the argument clearer, parts of the reasoning are proved in Lemmas 4.8 and 4.9 below. Since the proof is quite long and technical, we first give an idea of its structure. We begin the proof by identifying the group W from U and V. We then prove that every element of W decomposes as a product of an element of V and an element of U. The proof of this decomposition is the technical part. It is achieved by a compactness argument taking advantage from the fact that $G_{(i)}^+(Y_0,Y_1)$ is defined by local action conditions.

Let $G= G_{(i)}^+(Y_0,Y_1)$ . As announced at the beginning of the proof, we start by identifying W. Notice that $\mathfrak {T}_0$ is stable under the action of G. Furthermore, for every $g\in \operatorname {\mathrm {Aut}}(T)^+$ and for every subtree $\mathcal {T}$ of T, we have that $g\mathrm {Fix}_G(\mathcal {T})g^{-1}=\mathrm {Fix}_G(g\mathcal {T})$ . In particular, there exist $\mathcal {T}$ , $\mathcal {T}'\in \mathfrak {T}_0$ such that $U=\mathrm { Fix}_G(\mathcal {T})$ and $V=\mathrm {Fix}_G(\mathcal {T}')$ . Since $V\not \subseteq U$ , notice that $\mathcal {T}\not \subseteq \mathcal {T'}$ . If l is even, Lemma 4.5 ensures that $\mathcal {T}=B_T(e,\tfrac {l}{2})$ for some edge $e\in E(T)$ . Furthermore, since $\mathcal {T}\not \subseteq \mathcal {T'}$ and since $l'\geq l$ , there exists a unique vertex $v\in e$ such that $\mathcal {T}'\subseteq T(v,w)\cup B_T(v,\tfrac {l}{2})$ , where w denotes the other vertex of e. In this case, we let $\mathcal {T}_W=B_T(v,\tfrac {l}{2})$ . However, if l is odd, Lemma 4.5 ensures that $\mathcal {T}=B_T(w,({l+1})/{2})$ for some vertex $w\in V(T)$ . Furthermore, since $\mathcal {T}\not \subseteq \mathcal {T'}$ and since $l'\geq l$ , there exists a unique vertex $v\in B_T(w,1)-\{w\}$ such that $\mathcal {T}'\subseteq T(v,w)\cup B_T(\{v,w\},({l-1})/{2})$ . In that case, we let $\mathcal {T}_W=B_T(\{v,w\},({l-1})/{2})$ . In both cases, we set $W=\mathrm {Fix}_G( \mathcal {T}_W)$ . Notice by construction that $W\in \mathcal {S}_0\lbrack l-1\rbrack $ and that $U\subseteq W$ (since $\mathcal {T}_W\subseteq \mathcal {T}$ ). Our purpose is therefore to show that $W\subseteq V U$ . To this end, let $\alpha \in W$ and let us show the existence of an element $\alpha _0\in U$ such that . We start by explaining why the existence of $\alpha _0$ settles the proof. Indeed, if $\alpha _0$ exists, notice that the automorphism $\alpha _1=\alpha _0^{-1}\circ \alpha $ is an element of G for which . In particular, we have that $\alpha _1\in \mathrm {Fix}_G(\mathcal {T}')$ , $\alpha _0\in \mathrm {Fix}_G(\mathcal {T})$ and by construction $\alpha =\alpha _0 \circ \alpha _1$ , which proves that $W\subseteq UV$ . Applying the inverse map on both sides of the inclusion, we obtain that $W\subseteq VU$ which settles the proof.

Now, let us prove the existence of $\alpha _0$ . As announced at the beginning of the proof, we are going to use a compactness argument taking advantage of the fact that $G_{(i)}^+(Y_0,Y_1)$ is defined by local actions conditions. More precisely, we define a descending chain of nonempty compact sets $\Omega _n\subseteq \operatorname {\mathrm {Aut}}(T)^+$ and an increasing chain of finite subtrees $\mathcal {R}_n$ of T such that $T=\bigcup _{n\in \mathbb N} \mathcal {R}_n$ and such that for all $h\in \Omega _n$ :

• • $h\in \mathrm {Fix}_G(\mathcal {T})$ and ;

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h,S_{Y_0}(v))=1$ for all v in $V_{t_0}\cap \mathcal {R}_n$ ;

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h,S_{Y_1}(v))=1$ for all v in $V_{t_1}\cap \mathcal {R}_n$ .

We recall that in the above, $t_0=\max (Y_0)\,\mod 2$ , $t_1=(1+\max (Y_1))\ \pmod{\,2}$ and $\max (\varnothing )=0$ . Let us first show that this settles the existence of $\alpha _0$ . Since the $\Omega _n$ form a descending chain of nonempty compact sets in a Hausdorff space, we obtain $\bigcap _{n\in \mathbb N} \Omega _n\not =\varnothing $ . Let $\alpha _0 \in \bigcap _{n\in \mathbb N} \Omega _n$ . Since $\alpha _0\in \Omega _0$ , notice that , . To see that $\alpha _0$ is as desired, all that remains is to show that $\alpha _0\in G_{(i)}^+(Y_0,Y_1)$ . However, for every $v\in V_{t}(T)$ , there exists a positive integer $n\in \mathbb N$ such that $v\in \mathcal {R}_n$ and since $\alpha _0\in \Omega _{n}$ , we have that $\operatorname {\mathrm {Sgn}}_{(i)}(\alpha _0,S_{Y_t}(v))=1$ . This proves that $\alpha _0$ is as desired.

Figure 2

The tree $\mathcal {T}_W$ and the geodesic $\gamma $ . Colour available online.

What remains is to define the descending chain of nonempty compact sets $\Omega _n\subseteq \operatorname {\mathrm {Aut}}(T)^+$ . Suppose that $\max (Y_0)\leq \max (Y_1)$ (the proof for $\max (Y_1)\leq \max (Y_0)$ is similar). Let $\gamma $ be the smallest geodesic of T that contains both the centre of $\mathcal {T}$ and the centre of $\mathcal {T}'$ , and is oriented from $\mathcal {T}$ to $\mathcal {T}'$ (note that the centre is either a vertex or an edge depending on the values of l and $l'$ ). Since $\mathcal {T}\not \subseteq \mathcal {T}'$ and since $l'\geq l$ , notice that $\gamma $ contains at least two vertices (see Figure 2). The increasing chain of finite subtrees $\mathcal {R}_n$ of T such that $T=\bigcup _{n\in \mathbb N} \mathcal {R}_n$ that we are going to use is $\mathcal {R}_n=B_T(\gamma ,n)$ . We let

Since $\alpha \in \mathrm {Fix}_G(\mathcal {T}_W)$ and since $\mathcal {T}_W$ contains all vertices of $\mathcal {T}\cap \mathcal {T}'$ , notice that $\Omega _{-1}$ is not empty. Now, since $\max (Y_0)\leq \max (Y_1)$ , notice that there exists a unique $r\in \mathbb N$ such that $\max (Y_0)+2r\leq \max (Y_1)\leq \max (Y_0)+2r +1$ (where one of these inequalities is an equality). We let

$$ \begin{align*} \Omega_0=\left\{h\in \Omega_{-1}\ \middle\vert \begin{array}{@{\kern3pt}l@{}} \operatorname{\mathrm{Sgn}}_{(i)}(g,S_{Y_0}(v))=1 \mbox{ for each }v\in B_T(\gamma,2r)\cap V_{t_0}\\ \operatorname{\mathrm{Sgn}}_{(i)}(g,S_{Y_1}(v))=1 \mbox{ for each }v\in B_T(\gamma,0)\cap V_{t_1} \end{array}\right\}. \end{align*} $$

Lemma 4.8 below ensures that this set is not empty. From there, we define the sets $\Omega _n$ by induction on n. For every $n\geq 1$ , let $h_n$ be an element of $\Omega _{n-1}$ and let

For this induction to make sense, it is important for $\Omega _{n}$ to be not empty for all $n\geq 1$ . This is proved by Lemma 4.9 below, which ensures that $\Omega _{n}$ is a nonempty compact set. The result follows.

Proof. We recall that $\max (Y_0)\leq \max (Y_1)$ , that $r\in \mathbb N$ is the unique integer such that $\max (Y_0)+2r\leq \max (Y_1)\leq \max (Y_0)+2r +1$ , that ${t_0=\max (Y_0) \ \pmod{\,2}}$ and that ${t_1=(\max (Y_1)+1) \ \pmod{\,2}}$ . Remember from the proof of Proposition 4.7 that $\Omega _{-1}$ is not empty and let $h_0\in \Omega _{-1}$ . We are going to modify the element $h_0$ with the automorphisms $h_{(v)}$ to get an element of $\Omega _{0}$ . A concrete example of the procedure is given on a $4$ -regular tree with $Y_0=\{0\}$ and $Y_1=\{1,2\}$ by Figures 4, 5 and 6. In these figures:

• • the hollow vertices are those considered by the current and previous steps;

• • the circled vertices are those for which we wish to change the sign $\operatorname {\mathrm {Sgn}}_{(i)}(g,S_{Y_0}(v))$ or $\operatorname {\mathrm {Sgn}}_{(i)}(g,S_{Y_1}(v))$ (depending on the step) without affecting the sign of other hollow vertices;

• • the vertices boxed in squares are the vertices for which a change of the local action is applied to achieve the desired change of sign (note that for our choice $Y_0=\{0\}$ , these vertices are also the circled vertices).

Figure 4

Step I of the proof of Lemma 4.8. Colour available online.

Figure 5

Step II of the proof of Lemma 4.8. Colour available online.

Figure 6

Step III of the proof of Lemma 4.8. Colour available online.

Let $\{w_{0,0},\ldots ,w_{0,m_0}\}$ be the set of vertices $w\in B_T(\gamma ,0)\cap V_{t_0}$ such that ${\operatorname {\mathrm {Sgn}}_{(i)}(h_0,S_{Y_0}(w))=-1}$ . For all $j=0,1,\ldots ,m_0$ , we choose a vertex

$$ \begin{align*}v_{0,j}\in \bigcap_{w\in \gamma-\{w_{0,j}\}}{}T(w_{0,j},w)\end{align*} $$

such that $d(v_{0,j},w_{0,j})=\max (Y_0)$ . In particular, notice that $v_{0,j}\in S_{Y_0}(w_{0,j})$ but that ${v_{0,j}\not \in S_{Y_0}(w)}$ for every $w \in B_T(\gamma ,0)\cap V_{t_0}-\{w_{0,j}\} $ . Furthermore, since $\operatorname {\mathrm {Sgn}}_{(i)}(h_0,S_{Y_0}(w_{0,j}))=-1$ , , , and due to the form of $\mathcal {T}$ and $\mathcal {T}'$ , the vertices $v_{0,j}$ must be such that the automorphisms $h_{(v_{0,0})},\ldots , h_{(v_{0,m_0})}$ fix $\mathcal {T}\cup \mathcal {T}'$ pointwise. Indeed, if $v_{0,j}$ is outside the trees $\mathcal {T}$ and $\mathcal {T}'$ or is a leaf of either of these trees, the conclusion is clear. Furthermore, $v_{0,j}$ can not be an interior vertex of $\mathcal {T}$ or $\mathcal {T}'$ , since $\operatorname {\mathrm {Sgn}}_{(i)}(h_0,S_{Y_0}(w_{0,j})) = -1$ and $h_0\in \Omega _{-1}$ . In particular, if $v_{0,j}$ was an interior vertex of $\mathcal {T}$ , $w_{0,j}$ would be too far inside $\mathcal {T}$ to have $\operatorname {\mathrm {Sgn}}_{(i)}(h_0,S_{Y_0}(w_{0,j})) = -1$ as $h_0$ fixes pointwise a large enough ball around $w_{0,j}$ and if $v_{0,j}$ was an interior vertex of $\mathcal {T}'$ , $w_{0,j}$ would be too far inside $\mathcal {T}'$ to have $\operatorname {\mathrm {Sgn}}_{(i)}(h_0,S_{Y_0}(w_{0,j})) = -1$ as $h_0$ agrees with $\alpha $ on a large enough ball around $w_{0,j}$ . This proves as desired that the automorphisms $h_{(v_{0,0})},\ldots , h_{(v_{0,m_0})}$ fix $\mathcal {T}\cup \mathcal {T}'$ pointwise. In particular, the automorphism

$$ \begin{align*}h_{0,0}= h_0 \circ h_{(v_{0,0})}\circ \cdots\circ h_{(v_{0,m_0})}\end{align*} $$

satisfies , and

$$ \begin{align*}\operatorname{\mathrm{Sgn}}_{(i)}(h_{0,0},S_{Y_0}(w))=1\quad \forall w\in \gamma\cap V_{t_0}.\end{align*} $$

If $r\not =0$ , we iterate this procedure. For every $1\leq \nu \leq 2r$ , let $\{w_{\nu ,0},\ldots ,w_{\nu ,m_{\nu }}\}$ be the set of vertices $w\in B_T(\gamma ,\nu )\cap V_{t_0}$ such that

$$ \begin{align*}\operatorname{\mathrm{Sgn}}_{(i)}(h_{\nu-1,0},S_{Y_0}(w))=-1.\end{align*} $$

For all $j=0,1,\ldots ,m_\nu $ , we choose a vertex

$$ \begin{align*}v_{\nu,j}\in \bigcap_{w\in B_T(\gamma,\nu)\cap V_{t_0}-\{w_{\nu,j}\}}{}T(w_{\nu,j},w)\end{align*} $$

such that $d(v_{\nu ,j},w_{\nu ,j})=\max (Y_0)$ . Hence, notice that $v_{\nu ,j}\in S_{Y_0}(w_{\nu ,j})$ but that $v_{\nu ,j}\not \in S_{Y_0}(w)$ for every $w\in B_T(\gamma ,\nu )\cap V_{t_0}-\{w_{\nu ,j}\}$ . Furthermore, notice that the automorphisms $h_{(v_{\nu ,0})}, \ldots , h_{(v_{\nu ,m_\nu })}$ fix $\mathcal {T}\cup \mathcal {T}'$ pointwise. In particular, the automorphism

$$ \begin{align*}h_{\nu,0}=h_{\nu-1,0} \circ h_{(v_{\nu,0})}\circ \cdots\circ h_{(v_{\nu,m_\nu})}\end{align*} $$

satisfies that , and

$$ \begin{align*}\operatorname{\mathrm{Sgn}}_{(i)}(h_{\nu,0},S_{Y_0}(w))=1\quad \forall w\in B_T(\gamma,\nu)\cap V_{t_0}.\end{align*} $$

Consider the element $h_{2r,0}$ that we have just constructed. This element behaves as desired for the condition given by $Y_0$ on the vertices of $B_T(\gamma ,2r)\cap V_{t_0}$ . However, nothing ensures that the condition given by $Y_1$ on the vertices of $B_T(\gamma ,0)\cap V_{t_1}$ is yet satisfied. We now take care of this task. Let $\{w_0,\ldots ,w_m\}$ be the set of vertices $w\in B_T( \gamma ,0)\cap V_{t_1}$ such that $\operatorname {\mathrm {Sgn}}_{(i)}(h_{2r,0},S_{Y_1}(w))=-1$ . For all $j=0,1,\ldots ,m$ , let ${v_j\in \bigcap _{w\in \gamma -\{w_j\}} T(w_j,w)}$ such that $d(v_j,w_j)=\max (Y_1)$ . In particular, notice that $v_{j}\in S_{Y_1}(w_{j})$ but that $v_{j}\not \in S_{Y_1}(w)$ for every $w\in B_T(\gamma ,0) \cap V_{t_1}-\{w_j\}$ . Furthermore, since $\max (Y_0)+2r\leq \max (Y_1)$ , notice from Remark 3.4 that $v_j\not \in S_{Y_0}(v)$ for every $v\in B_T(\gamma ,2r)\cap V_{t_0}$ . However, just as before, the automorphisms $h_{(v_{0})}, \ldots , h_{(v_{m})}$ fix $\mathcal {T}\cup \mathcal {T}'$ pointwise. In particular, $h_{1}= h_{2r,0}\circ h_{(v_{0})}\circ \cdots \circ h_{(v_{m})}$ satisfies:

• • and ;

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h_{1},S_{Y_0}(w))=1 \text { for all } w\in B_T(\gamma ,2r)\cap V_{t_0};$

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h_{1},S_{Y_1}(w))=1 \text { for all } w\in B_T(\gamma ,0)\cap V_{t_1}.$

This proves that $h_1\in \Omega _0$ and therefore that $\Omega _0$ is not empty.

Proof. We show that $\Omega _n$ is not empty by induction. Lemma 4.8 ensures that $\Omega _0$ is not empty. Suppose that $\Omega _{n-1}$ is not empty and let $h_n\in \Omega _{n-1}$ be the automorphism appearing in the definition of $\Omega _n$ . Just as in the proof of Lemma 4.8, we are going to modify $h_n$ with the automorphisms $h_{(v)}$ to obtain an element of $\Omega _{n}$ . A concrete example of the procedure is given by Figures 7 and 8 on a $4$ -regular tree with $Y_0=\{0\}$ and even $\max {(Y_1)}$ (with the same conventions as before and where the vertices concerned by the current step are highlighted in dotted areas). Let $\{\tilde {w}_0,\ldots ,\tilde {w}_{k}\}$ be the set of vertices $w$ that belong to $ B_T(\gamma ,n+2r)\cap V_{t_0}$ and such that $\operatorname {\mathrm {Sgn}}_{(i)}(h_n,S_{Y_0}(w))=-1$ . For each $j=0,1,\ldots ,k$ , we choose a vertex ${\tilde {v}_{j}\in \bigcap _{w\in B_T(\gamma ,n+2r)\cap V_{t_0}-\{\tilde {w}_j\}}{}T(\tilde {w}_{j},w)}$ such that $d(\tilde {v}_{j},\tilde {w}_{j})=\max (Y_0)$ . Hence, notice that $\tilde {v}_j\not \in S_{Y_0}(w)$ for all ${w\in B_T(\gamma ,n+2r)\cap V_{t_0}-\{\tilde {w}_j\}}$ . Furthermore, since $\max (Y_1)-1\leq 2r+\max (Y_0)$ , notice from Remark 3.4 that $\tilde {v}_j\not \in S_{Y_1}(w)$ for all ${w\in B_T(\gamma ,n-1)\cap V_{t_1}}$ . Furthermore, since ${\operatorname {\mathrm {Sgn}}_{(i)}(h_{n},S_{Y_0}(\tilde {w}_{j}))=-1}$ , , , and due to the form’s of $\mathcal {T}$ and $\mathcal {T}'$ , notice that the automorphisms $h_{(\tilde {v}_{0})}, \ldots , h_{(\tilde {v}_{k})}$ fix $\mathcal {T}\cup \mathcal {T}'$ pointwise. Indeed, if $\tilde {v}_{j}$ is outside the trees $\mathcal {T}$ and $\mathcal {T}'$ or is a leaf of either of these trees, the conclusion is clear. However, $\tilde {v}_{j}$ cannot be an interior vertex of $\mathcal {T}$ or $\mathcal {T}'$ , since $\operatorname {\mathrm {Sgn}}_{(i)}(h_n,S_{Y_0}(\tilde {w}_{j})) = -1$ and $h_n\in \Omega _{n-1}$ . In particular, if $\tilde {v}_{j}$ was an interior vertex of $\mathcal {T}$ , $\tilde {w}_{j}$ would be too far inside $\mathcal {T}$ for $\operatorname {\mathrm {Sgn}}_{(i)}(h_n,S_{Y_0}(\tilde {w}_{j})) = -1$ as $h_n$ fixes pointwise a large enough ball around $\tilde {w}_{j}$ and if $\tilde {v}_{j}$ was an interior vertex of $\mathcal {T}'$ , $\tilde {w}_{j}$ would be too far inside $\mathcal {T}'$ for $\operatorname {\mathrm {Sgn}}_{(i)}(h_n,S_{Y_0}(\tilde {w}_{j})) = -1$ as $h_n$ agrees with $\alpha $ on a large enough ball around $\tilde {w}_{j}$ . This proves as desired that the automorphisms $h_{(\tilde {v}_{0})}, \ldots , h_{(\tilde {v}_{k})}$ fix $\mathcal {T}\cup \mathcal {T}'$ pointwise. In particular, $\tilde {h}_n=h_{n} \circ h_{(\tilde {v}_{0})}\circ \cdots \circ h_{(\tilde {v}_{k})}$ satisfies:

• •

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h_{n},S_{Y_0}(w))=1 \text { for all } w\in B_T(\gamma ,n+2r)\cap V_{t_0};$

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h_{n},S_{Y_1}(w))=1 \text { for all } w\in B_T(\gamma ,n-1)\cap V_{t_1}.$

Now, let $\{w_0,\ldots ,w_m\}$ be the set of vertices $w\in B_T(\gamma ,n)\cap V_{t_1}$ such that $\operatorname {\mathrm {Sgn}}_{(i)}(\tilde {h}_n,S_{Y_1}(w))=-1$ . For each $j=0,1,\ldots ,m$ , choose $v_{j}\in \bigcap _{w\in B_T(\gamma ,n)-\{w_j\}} T(w_{j},w)$ such that $d(v_{j},w_{j})=\max (Y_1)$ . Since $\max (Y_0)+2r\leq \max (Y_1)$ , notice from Remark 3.4 that $v_j\not \in S_{Y_0}(w)$ for every $w\in B_T(\gamma ,n+2r)\cap V_{t_0}$ and $v_j\not \in S_{Y_1}(w)$ for every $w\in B_T(\gamma ,n)\cap V_{t_1}-\{w_j\}$ . Just as before, notice that the automorphisms $h_{(v_{0})}, \ldots ,h_{(v_{m})}$ fix $\mathcal {T}\cup \mathcal {T}'$ pointwise. In particular, $h_{n+1}=\tilde {h}_{n} \circ h_{(v_{0})}\circ \cdots \circ h_{(v_{m})}$ satisfies:

• •

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h_{n+1},S_{Y_0}(w))=1 \text { for all } w\in B(\gamma ,n+2r)\cap V_{t_0};$

• • $\operatorname {\mathrm {Sgn}}_{(i)}(h_{n+1},S_{Y_1}(w))=1 \text { for all } w\in B(\gamma ,n)\cap V_{t_1}.$

This proves that $\Omega _n$ is not empty. We now show that $\Omega _n$ is compact for every integer $n\geq 1$ . To this end, notice that $\Omega _n$ is a closed subset of $\operatorname {\mathrm {Aut}}(T)^+$ and that

$$ \begin{align*} \Omega_n \subseteq h_{n} \mathrm{Fix}_{\operatorname{\mathrm{Aut}}(T)^+}(B_T(\gamma,n-1+\max(Y_1))). \end{align*} $$

Since the right-hand side is a compact subset of $\operatorname {\mathrm {Aut}}(T)^+$ , the results follow.

Proof. Let $G=G^+_{(i)}(Y_0,Y_1)$ . To prove that $\mathcal {S}_0$ factorises $^+$ at depth $l\geq 1$ , we successively check the three conditions of Definition 4.3.

First, we must prove that for every U in the conjugacy class of an element of $\mathcal {S}_0\lbrack l \rbrack $ and every subgroup V in the conjugacy class of an element of $\mathcal {S}_0$ with $V \not \subseteq U$ , there exists a W in the conjugacy class of an element of $\mathcal {S}_0\lbrack l -1\rbrack $ such that

$$ \begin{align*}U\subseteq W \subseteq V U.\end{align*} $$

Let U and V be as above. By the definition of $\mathcal {S}_0$ and since $\mathfrak {T}_0$ is stable under the action of G, there exist two subtrees $\mathcal {T},\mathcal {T}'\in \mathfrak {T}_0$ such that $U=\mathrm {Fix}_G(\mathcal {T})$ and $V=\mathrm {Fix}_G(\mathcal {T}')$ . In particular, $\mathcal {T}'= B_T(o,r)$ , where o is either a vertex or an edge of T and r is a nonnegative integer. Let v be the vertex of $\mathcal {T}$ at maximal distance from o and notice since $V \not \subseteq U$ that $d_T(v,o)> r$ . Now let $\gamma $ be an infinite geodesic ray starting at v and containing o, let $\mathcal {P}$ be the tree centred at $\gamma (d_T(v,o) + l)$ of radius $r+l$ and let $V'=\mathrm {Fix}_G(\mathcal {P})$ . By construction, one has that $\mathcal {T}'\subseteq \mathcal {P}$ so that $V'\subseteq V$ . Furthermore, $V'$ has depth greater than l and since $v\not \in \mathcal {P}$ , Lemma 4.6 ensures that $V'\not \subseteq U$ . Now, Proposition 4.7 ensures the existence of a subgroup $W\in \mathcal {S}_0\lbrack l-1\rbrack $ such that

$$ \begin{align*}U\subseteq W \subseteq V'U\subseteq VU.\end{align*} $$

This proves the first condition.

Next, we must prove that $N_{G}(U, V)= \{g\in {G} \mid g^{-1}Vg\subseteq U\}$ is compact for every V in the conjugacy class of an element of $\mathcal {S}_0$ . Just as before, notice that there exists a $\mathcal {T}'\in \mathfrak {T}_0$ such that $V = \mathrm {Fix}_G(\mathcal {T}')$ . Since G satisfies the hypothesis (H₀), notice that

$$ \begin{align*} N_{G}(U, V)&= \{g\in G \mid g^{-1}Vg\subseteq U\}=\{g\in G \mid g^{-1}\mathrm{Fix}_G(\mathcal{T}')g\subseteq \mathrm{Fix}_G(\mathcal{T})\}\\ &=\{g\in G \mid \mathrm{Fix}_G(g^{-1}\mathcal{T}')\subseteq \mathrm{Fix}_G(\mathcal{T})\}= \{g\in G \mid g\mathcal{T}\subseteq \mathcal{T}'\}. \end{align*} $$

In particular, since both $\mathcal {T}$ and $\mathcal {T}'$ are finite subtrees of T, $N_G(U,V)$ is a compact subset of G which proves the second condition.

Finally, we must prove for every W in the conjugacy class of an element of $\mathcal {S}_0\lbrack l-1\rbrack $ with $U\subseteq W$ that

$$ \begin{align*} W\subseteq N_G(U,U) =\{g\in G \mid g^{-1}Ug\subseteq U\}. \end{align*} $$

For the same reasons as before, there exists $\mathcal {R}\in \mathfrak {T}_0$ such that $W= \mathrm {Fix}_G(\mathcal {R})$ . Furthermore, since $U\subseteq W$ and since G satisfies the hypothesis (H₀), notice that $\mathcal {R}\subseteq \mathcal {T}$ . Moreover, since $\mathrm {Fix}_G(\mathcal {R})$ has depth $l-1$ , notice that $\mathcal {R}$ contains every interior vertex of $\mathcal {T}$ . Since G is unimodular and satisfies hypothesis (H₀), this implies that

$$ \begin{align*} \mathrm{Fix}_G(\mathcal{R})&\subseteq \{h\in G \mid h\mathcal{T}\subseteq \mathcal{T}\}= \{h\in G \mid \mathrm{Fix}_G(\mathcal{T})\subseteq\mathrm{Fix}_G(h\mathcal{T})\}\\ &=\{h\in G \mid h^{-1}\mathrm{Fix}_G(\mathcal{T})h\subseteq\mathrm{Fix}_G(\mathcal{T})\}= N_G(U,U), \end{align*} $$

which proves the third condition.

Proof. Since $l\geq 2$ , Lemma 4.5 ensures the existence of a complete finite subtree $\mathcal {T}$ of T containing a ball of radius $1$ around an edge such that $C=\mathcal {C}(\mathrm { Fix}_{G}(B_T(e,r)))$ . Furthermore, as observed in Section 4.4, we have that $\operatorname {\mathrm {Aut}}_G(C)\cong \operatorname {\mathrm {Stab}}_G(\mathcal {T})/\mathrm {Fix}_G(\mathcal {T})$ and

$$ \begin{align*} \mathfrak{H}_{\mathcal{S}_0}(\mathrm{Fix}_G(\mathcal{T}))&=\{\mathrm{Fix}_G(\mathcal{R})/\mathrm{Fix}_G(\mathcal{T})\mid \mathcal{R}\in \mathfrak{T}_0, \mathcal{R}\subsetneq \mathcal{T} \\ &\qquad\,\mbox{and } \mathcal{R}\mbox{ is maximal for this property}\}. \end{align*} $$

Now, let $\{\mathcal {T}_1,\ldots , \mathcal {T}_s\}$ be the set of maximal complete proper subtrees of $\mathcal {T}$ . These can all be obtained from $\mathcal {T}$ by removing the leaves adjacent to v, where v is a vertex of $\mathcal {T}$ that is not a leaf but such that all but one of its neighbours are leaves of $\mathcal {T}$ . Clearly, $\mathcal {T}_i\cup \mathcal {T}_j=\mathcal {T}$ for every $i\not =j$ and $\operatorname {\mathrm {Stab}}_G(\mathcal {T})$ acts by permutation on the set $\{\mathcal {T}_1,\mathcal {T}_2,\ldots ,\mathcal {T}_s\}$ . Furthermore, for each $i\in \{ 1,\ldots ,s\}$ , notice from Lemma 4.6 that

$$ \begin{align*}\mathrm{ Fix}_G(\mathcal{T})\subsetneq \mathrm{Fix}_G(\mathcal{T}_i).\end{align*} $$

Furthermore, since $\mathcal {T}$ belongs to $\mathfrak {T}_0$ and contains at least a ball of radius $1$ around an edge, notice that $\mathrm {Fix}_G(\mathcal {T}_i)\subseteq \operatorname {\mathrm {Stab}}_G(\mathcal {T})$ . However, this inclusion must be strict since $\operatorname {\mathrm {Stab}}_G(\mathcal {T})$ contains either the fixator of a vertex of $\mathcal {T}_i$ or the fixator of one of its edges. Now, Proposition 4.14 ensures the existence of an irreducible representation of $\operatorname {\mathrm {Stab}}_G(\mathcal {T})/\mathrm {Fix}_G(\mathcal {T})$ without a nonzero $\mathrm {Fix}_G(\mathcal {T}_i)/\mathrm {Fix}_G(\mathcal {T})$ -invariant vector for every $i=1,\ldots ,s$ . Since every proper subtree $\mathcal {R}\in \mathfrak {T}_0$ of $\mathcal {T}$ is contained in one of these $\mathcal {T}_i$ , this representation is an $\mathcal {S}_0$ -standard representation of $\operatorname {\mathrm {Stab}}_G(\mathcal {T})/\mathrm {Fix}_G(\mathcal {T})$ .

Proof. Lemma 4.5 ensures the existence of a vertex $v\in V(T)$ such that ${C=\mathcal {C}(\mathrm {Fix}_{G}(B_T(v,1)))}$ . Let $\mathcal {T}=B_T(v,r)$ . We recall as observed in Section 4.4 that $\operatorname {\mathrm {Aut}}_G(C)\cong \operatorname {\mathrm {Stab}}_G(\mathcal {T})/\mathrm {Fix}_G(\mathcal {T})$ , where

$$ \begin{align*} \operatorname {\mathrm {Stab}}_G(\mathcal {T})=\{g\in G \mid g\mathcal {T}\subseteq \mathcal {T}\}= \{ g\in G \mid gv=v\}= \mathrm { Fix}_G(v).\end{align*} $$

In particular, $\operatorname {\mathrm {Aut}}_G(C)$ can be realised as a the group of automorphisms of $B_T(v,1)$ obtained by restricting the action of $\mathrm {Fix}_G(v)$ . Furthermore,

$$ \begin{align*} \mathfrak{H}_{\mathcal{S}_0}(\mathrm{Fix}_G(\mathcal{T}))&=\{\mathrm{Fix}_G(\mathcal{R})/\mathrm{Fix}_G(\mathcal{T})\mid \mathcal{R}\in \mathfrak{T}_0, \mathcal{R}\subsetneq \mathcal{T} \\ &\qquad\,\mbox{and } \mathcal{R}\mbox{ is maximal for this property}\}\\ &=\{\mathrm{Fix}_G(f)/\mathrm{Fix}_G(B_T(v,1))\mid f\in E(B_T(v,1))\}. \end{align*} $$

Let d be the degree of v in T, let $X=E(B_T(v,1))$ and let $e\in X$ . Since $\operatorname {\mathrm {Alt}}(d)\leq \underline {G}(v)$ and since $d\geq 4$ , notice that $\underline {G}(v)$ is $2$ -transitive. In particular, $\operatorname {\mathrm {Aut}}_{G}(C)$ is $2$ -transitive on X and [Reference SemalSem23, Lemma 2.27] ensures the existence of an irreducible representation $\sigma $ of $\operatorname {\mathrm {Aut}}_{G}(C)$ without a nonzero $\mathrm {Fix}_{\operatorname {\mathrm {Aut}}_{G}(C)}(e)$ -invariant vector. Since $\mathrm {Fix}_G(v)$ is transitive on $E(B_T(v,1))$ , this representation does not admit a nonzero $\mathrm {Fix}_{\operatorname {\mathrm {Aut}}_{G}(C)}(f)$ -invariant vector for any $f\in E(B_T(v,1))$ . The lemma follows from the fact that $\mathrm { Fix}_{\operatorname {\mathrm {Aut}}_{G}(C)}(f)=\mathrm {Fix}_G(f)/\mathrm {Fix}_G(B_T(v,1))$ .