Proof Tsankov [Reference Tsankov18] showed that a topological group is Roelcke precompact if and only if it is the inverse limit of some inverse system of oligomorphic permutation groups. The second part of the statement of this theorem follows from its first part and this result.

Since L is ultrahomogeneous, $\mathrm {Aut}(L)$ is not Roelcke precompact if and only if there are sequences $(a_i)_{i<\omega }, (b_i)_{i<\omega }$ of elements of L such that $\mathrm {tp}^L(a_i / \emptyset ) = \mathrm {tp}^L(b_i / \emptyset ) = \mathrm {tp}^L(a_j / \emptyset ) = \mathrm {tp}^L(b_j / \emptyset )$ for $i, j < \omega $ , and that $\{\mathrm {tp}^L(a_ib_i / \emptyset )\}$ is infinite. Furthermore, since $\mathrm {Th}(L)$ eliminates quantifiers, types realized in L are in one-to-one correspondence with quantifier-free types realized in L. Finally, as L is locally finite, the latter are essentially isomorphism types of subalgebras of L with distinguished generators.

With that in mind, let $F_1$ be the free Heyting algebra whose generator is x. For each term $t(x) \in F_1$ , write $L_t$ for the quotient of $F_1$ by the principal filter $\theta _t$ generated by t. Furthermore, let $L_t^*$ be the Heyting algebra obtained by adding a new minimum element $0^{L_t^*}$ below $0^{L_t}$ . For a term $t(x)$ , we define $t^*(x, y)$ to be the term obtained by replacing every occurrence of $0$ with y. One can check that $(t^*)^{L_t^*}([x]_t, 0^{L_t}) = [t(x)]_t \in L_t^*$ , where $[\cdot ]_t$ denotes the congruence class with respect to $\theta _t$ . Therefore, $L_t^*$ is generated by $0^{L_t}$ and $[x]_t$ . We have obtained 2-generated subalgebras of L of infinitely many isomorphism types. On the other hand, we have $\langle [x]_t \rangle ^{L_t^*} = \langle 0^{L_t} \rangle ^{L_t^*}$ is a 3-chain for all t.

Proof It suffices to show that for every finite subset $S \subseteq L$ there is an infinite orbit in the action of $\mathrm {Aut}(L)_{(S)}$ on $ L$ . Note that for every finite subalgebra $A \subseteq L$ , there exists $a \in L \setminus A$ such that a is join-prime in $\langle Aa \rangle ^L$ . Indeed, consider the dual $\Bbb P$ of A and the disjoint union $\Bbb P' := \Bbb P \sqcup \{w\}$ , where w is a fresh element, and let a be the image of $\{w\}$ under the embedding of the dual of $\Bbb P'$ into L that fixes A pointwise. By repeatedly using this, take an $\omega $ -sequence $(a_i)_{i<\omega }$ of elements of L such that $a_i \in L \setminus \langle Sa_0a_1\dots a_{i-1}\rangle ^L$ is join-prime in $\langle Sa_0a_1\dots a_{i}\rangle ^L$ for $i < \omega $ . By construction, there exists an automorphism $\phi _i : L \to L$ fixing S pointwise such that $\phi _i(a_i) = a_{i+1}$ for $i<\omega $ . Hence, the orbit of $a_0$ under $\mathrm {Aut}(L)_{(S)}$ is infinite.

Proof

1. 1. Let P and $P_1$ be the dual posets of H and $H_1$ , respectively. There is a p-morphism $D(f) : P_1 \to P$ that is the dual of f. $D(f)$ is surjective if f is injective. Let $B(H) = \mathscr {P} (P)$ and $B(H_1) = \mathscr {P}(P_1)$ . $D(f)$ induces a Boolean algebra homomorphism $B(f) : B(H) \to B(H_1)$ . $B(f)$ is injective if $D(f)$ is surjective. Likewise, $B(f)$ is surjective if f is. Let ${}^{\circ }, {}^{\circ _{1}}$ be the operations that take a subset to the maximal up-set contained by that set.

2. 2. Let $(L_i)_{i<\omega }$ be a chain of finite Heyting algebras used in the construction of L; so $\bigcup _i L_i = L$ . Let $B_i = B(L_i)$ as above and ${}^{\circ _{i}}$ be an interior operator such that $B_i{}^{\circ _{i}} \cong L_i$ . We may take $B_i \subseteq B_{i+1}$ for $i < \omega $ . Then ${}^{\circ _{i+1}}$ extends ${}^{\circ _{i}}$ . Let $B = \bigcup _i B_i$ and ${}^{\circ } = \bigcup {}^{\circ _{i}}$ . Then $B{}^{\circ } = \left ( \bigcup _i B_i \right ){}^{\circ } = \bigcup _i B_i{}^{\circ _{i}} = \bigcup _i L_i = L$ . It remains to show that B is atomless. Take an arbitrary $a \in B$ that is nonzero. Take $i < \omega $ such that $a \in B_i$ . Let $P_i$ be the poset dual to $L_i$ ; then a is a nonempty subset of $P_i$ . Take some $w \in a$ . Let $P'$ be the poset obtained from $P_i$ by replacing w with the 2-chain $\{w_1 < w_2\}$ . Let $\pi : P' \twoheadrightarrow P_i$ be the surjection that maps the chain to $\{w\}$ and is the identity elsewhere. This is a p-morphism, and it induces $\iota : L_i \hookrightarrow L'$ , where $L'$ is the dual of $P'$ . Take $k < \omega $ such that there is an embedding $\iota ': L' \hookrightarrow L_k$ such that $\iota ' \circ \iota $ is the identity on $L_i$ . Let $b = (a \setminus \{w \}) \cup \{w_1\}$ . Then $b \in B_k = B(L_k) \subseteq B$ and $0 < b < a$ .

Proof Let $f : L \to L$ be an automorphism. Let $f_k : L_k \to L_k'$ be the restriction of f to $L_k$ where $L_k' = f(L_k)$ . Each $f_k$ is an isomorphism. By the fact above, $f_k$ induces a Boolean algebra isomorphism $B(f_k) : B(L_k) \to B(L_k')$ for each $k < \omega $ ; and by construction $B(f_j)$ extends $B(f_k)$ for each $k < j < \omega $ . Let $\hat f = \bigcup _k B(f_k)$ . Then $\hat f$ is an isomorphism $B \to B$ .

Let $g : L \to L$ be another isomorphism. We have $\hat f \circ \hat g = (f \circ g){\hat {}}$ because each side of the equation extends $f \circ g$ .

Let $\iota : \mathrm {Aut}(L) \to \mathrm {Aut}(B)$ be the map $f \mapsto \hat f$ . The map $\iota $ is a group homomorphism as seen above, and it is clearly injective.

Next, we prove that $\iota $ is continuous. Let $\bar b$ be a tuple in B. It suffices to show that for an automorphism $f : L \to L$ the value of $\hat f(\bar b)$ is determined by the value of $f(\bar a)$ for a tuple $\bar a$ in L. There exists $k < \omega $ such that $\bar b$ is in $B_k = B(L_k)$ . Let $f_k : L_k \to L_k'$ be an isomorphism that is a restriction of f. Then $\hat f(\bar b) = B(f_k)(\bar b)$ . Let $\bar a$ be an enumeration of the finite algebra $L_k$ ; then $\bar a$ is what we needed.

Finally, we show that the image $\iota (U)$ is open in $\operatorname {ran} \iota \subseteq \mathrm {Aut}(B)$ for an arbitrary basic open set U of $\mathrm {Aut}(L)$ . Indeed, let U be the set of $f : L \to L$ fixing the values of f at $\bar a \in L$ ; then $\hat g \in \iota (U)$ if and only if $\hat g \upharpoonright B_0 = \hat f \upharpoonright B_0$ for $g : L \to L$ , where $B_0$ is the Boolean subalgebra of B generated by $\bar a$ .

Proof The set B of fixed points of $\neg \neg$ in L is a Boolean algebra by setting $a \vee ^B b = \neg \neg (a \vee ^L b)$ and the remaining operations of B the restrictions of the corresponding operations of L. (Note that B is not a substructure of L.) By [Reference Ghilardi and Zawadowski9, Proposition 4.28(ii)], B is atomless.

Let $a \in L$ be arbitrary. Take a finite subalgebra $H \subseteq L$ such that $a \in H$ , and let ${\mathbb P}$ be the dual poset of H so we may identify an element of H with an up-set of ${\mathbb P}$ . Possibly by replacing L by another finite Heyting algebra into which L embeds, we may assume that ${\mathbb P}$ is a forest. Furthermore, without loss of generality, we may assume that a is principal as an up-set in ${\mathbb P}$ , generated by $x \in {\mathbb P}$ . If x is a root, then a itself is in B, so there remains nothing to be shown. Suppose not, and let $x^-$ be the predecessor of x. Let ${\mathbb P}_1, {\mathbb P}_2$ be disjoint posets isomorphic to that induced by $a \subseteq {\mathbb P}$ . Let ${\mathbb P}' := ({\mathbb P} \setminus a) \sqcup {\mathbb P}_1 \sqcup {\mathbb P}_2$ whose partial order is the least containing those of the summands and $x^- \le {\mathbb P}_1$ , $x^- \le {\mathbb P}_2$ . Consider the surjective p-morphism ${\mathbb P}' \twoheadrightarrow {\mathbb P}$ that collapses $\{\min {\mathbb P}_1, \min {\mathbb P}_2\}$ to x, and let $i :H \hookrightarrow H'$ be the Heyting algebra embedding it induces. (See also Figure 1.) Note that ${\mathbb P}_i \in H'$ is in B for $i = 1, 2$ and that $i(a) = {\mathbb P}_1 \vee {\mathbb P}_2$ . Let $H_r(a)$ be a subalgebra of L such that there is an isomorphism $\phi : H' \to H_r(a)$ that extends the identity map on H. Let $r_1(a) := \phi ({\mathbb P}_1)$ and $r_2(a) := \phi ({\mathbb P}_2)$ . We have $a = r_1(a) \vee r_2(a)$ and $r_i(a) \in B (i = 1,2)$ as promised.

Proof The first claim is immediate. We show that $h_{\neg \neg }$ is not surjective.

Consider the three-element chain $C_3$ , which can be regarded as a Heyting algebra, and let $a \in C_3$ be such that $0 < a < 1$ . Note that a does not have $\neg \neg a = a$ and a principal up-set in the dual finite poset of $C_3$ . Let D be the diagram $C_3 \hookleftarrow \mathbf 2 \hookrightarrow C_3$ , where $\mathbf 2$ is the two-element Heyting algebra. Let $a_0 = \iota _{\hookleftarrow }^D(a)$ , $a_{1.5} = \iota _{\hookrightarrow }^D(a)$ , and $H = H_r(a_{1.5})$ . Next, let $D'$ be the diagram $H \hookleftarrow \iota _{\hookleftarrow }^D(C_3) \hookrightarrow H$ . Let $a_{1i} = \iota _{\hookleftarrow }^{D'}(r_i(a_{1.5}))$ , $a_{2i} = \iota _{\hookrightarrow }^{D'}(r_i(a_{1.5}))$ , and $a_{0i} = r_i(a_0)$ for $i=1,2$ . Refer to Figure 2 for this construction.

Figure 1 Construction of $H'$ .

Figure 2 Construction by amalgamation.

The Boolean subalgebra $B_6$ generated by $a_{ji}$ ( $0 \le j \le 2, 1 \le i \le 2$ ) in B has six atoms, each permutation of which extends to an automorphism of B. Consider the permutation $a_{ji} \mapsto a_{(j + 1 \bmod 3)i}$ , which extends to an automorphism of $B_6$ , which in turn extends to $\phi \in \mathrm {Aut}(B)$ by ultrahomogeneity of B. By construction,

$$\begin{align*}\bigvee_L \phi(\{a_{11}, a_{12}\}) \neq \bigvee_L \phi(\{a_{21}, a_{22}\}),\end{align*}$$

showing that $\phi $ is not in the image of $h_{\neg \neg }$ .

The last paragraph also shows that the image of $h_{\neg \neg }$ is not dense. To see that $\operatorname {\mathrm {ran}} h_{\neg \neg }$ is not open, let $\overline b$ be an arbitrary tuple in B, and we prove that $\mathrm {Aut}(B)_{(\overline b)} \setminus \operatorname {\mathrm {ran}} h_{\neg \neg } \neq \emptyset $ . Take a finite subalgebra K of L such that K generates $\langle \overline b\rangle ^B$ as a Boolean algebra. Let $D''$ be the diagramFootnote ¹ $K \hookleftarrow \mathbf 2 \hookrightarrow \bigsqcup D^\prime$ such that the image $\operatorname {\mathrm {ran}} \iota _{\hookrightarrow }^{D''}$ generates a copy $B^{\prime }_6$ of $B_6$ . Take an automorphism $\psi _0$ on $\bigsqcup D''$ such that $\psi _0 \upharpoonright B^{\prime }_6$ is as constructed in the preceding paragraph and that $\psi _0 \upharpoonright \operatorname {\mathrm {ran}} \iota _{\hookleftarrow }^{D''}$ is the identity.Footnote ² The automorphism $\psi _0$ extends to another $\phi \in \mathrm {Aut}(B)$ , which is in $\mathrm {Aut}(B)_{(\overline b)} \setminus \operatorname {\mathrm {ran}} h_{\neg \neg }$ .

Proof We show that $\mathcal K^*$ is reasonable and that $\mathcal K^*$ has the amalgamation property. (Other claims are clear.) In what follows, for a totally ordered set $(X, <)$ and $Y, Z \subseteq X$ , we write $Y < Z$ to mean that $y < z$ whenever $y \in Y$ and $z \in Z$ .

Let $H_1 \subseteq H_2$ be finite Heyting algebra, and let $\prec _{1}^{\mathrm {alex}}$ be an arbitrary admissible total order on $H_1$ . We show that there exists an admissible order on $H_2$ extending $\prec _{1}^{\mathrm {alex}}$ . Let $\pi : {\mathbb P}_2 \twoheadrightarrow {\mathbb P}_1$ be the surjective p-morphism dual to the inclusion map $H_1 \hookrightarrow H_2$ . Note that with $I(1_{H_i})$ and ${\mathbb P}_i$ identified as pure sets, an admissible total order of $H_i$ extends the order-theoretic dual of the order of ${\mathbb P}_i$ for $i = 1,2$ .

Suppose that for $p,q \in {\mathbb P}_1$ we have $p \prec _1 q$ . Since $\prec _{1}^{\mathrm {alex}}$ is admissible, $p \not \le q$ . Take arbitrary $p', q' \in {\mathbb P}_2$ such that $\pi (p') = p$ and that $\pi (q') = q$ . Since $\pi $ is order-preserving a fortiori, we have $p' \not \le q'$ .

Let $R = (\mathord {\le } \setminus \Delta ) \cup \{(p',q') \mid \pi (p') \prec _2 \pi (q')\}$ be a binary relation on ${\mathbb P}_2 = I(1_{H_2})$ , where $\Delta $ is the diagonal relation. It can be shown by induction from the fact in the preceding paragraph that R contains no cycle. Therefore, R can be extended to a total order $\prec _2$ . Furthermore, for $p, q \in {\mathbb P}_1$ , we have $\pi ^{-1}(p) \prec _2 \pi ^{-1}(q)$ ; a fortiori, $\pi ^{-1}(p) \prec _{2}^{\mathrm {alex}} \pi ^{-1}(q)$ . This shows that $\prec _{2}^{\mathrm {alex}}$ extends $\prec _{1}^{\mathrm {alex}}$ .

Next, we prove the amalgamation property for $\mathcal K^*$ . Let D be the diagram $H_1 \hookleftarrow H_0 \hookrightarrow H_2$ in $\mathrm {Age}(L)$ and let $\prec _{i}^{\mathrm {alex}}$ be an arbitrary admissible ordering on $H_i$ for $i = 1,2$ . Recall the dual poset ${\mathbb P}$ of $\bigsqcup D$ is a sub-poset of the product order ${\mathbb P}_1 \times {\mathbb P}_2$ , where ${\mathbb P}_i$ is the dual of $H_i$ ( $i = 1, 2$ ) [Reference Maksimova14]. Take a total order $\prec $ on ${\mathbb P}$ so it extends the product order of $\prec _1$ and $\prec _2$ .

We first show that $\prec $ extends the dual of the order of ${\mathbb P}$ . Assume that $(p_1,p_2) \le (q_1, q_2)$ for $(p_i,q_j) \in {\mathbb P}$ and $1 \le i, j \le 2$ . (Recall that $p_i, q_i \in {\mathbb P}_i$ .) Since the order of ${\mathbb P}$ is induced by the product of those of ${\mathbb P}_1$ and ${\mathbb P}_2$ , we have $p_i \le q_i$ for $i = 1,2$ . Because $\prec _i$ extends the dual of the order of ${\mathbb P}_i$ , we have $p_i \succ _i q_i$ ( $i = 1, 2$ ). By the construction of $\prec $ , we have $(p_1,p_2) \succ (q_1, q_2)$ as desired.

We then prove that $(\bigsqcup D, \prec ^{\mathrm {alex}})$ witnesses the amalgamation property. Because of the strong amalgamation property of $\mathrm {Age}(L)$ , it suffices to show that $\prec ^{\mathrm {alex}}$ extends $\iota _{\hookleftarrow }^D(\prec _{1}^{\mathrm {alex}})$ and $\iota _{\hookrightarrow }^D(\prec _{2}^{\mathrm {alex}})$ . Take $p, p' \in {\mathbb P}_1$ , and assume that $p \prec p'$ (the other case can be handled in a similar manner). Since $\iota _{\hookleftarrow }^D$ is induced by the projection $\pi _1 : {\mathbb P} \twoheadrightarrow {\mathbb P}_1$ , it suffices to show that $\pi ^{-1}(p) \prec ^{\mathrm {alex}} \pi ^{-1}(p')$ . Now, it is easy to see that, in fact, $\pi ^{-1}(p) \prec \pi ^{-1}(p')$ by the construction of $\prec $ .

Proof We will make use of the following proposition:

Consider the following construction appearing in [Reference Kechris, Pestov and Todorčević10, Remark 3.1]. Let A be the finite Boolean algebra with the atoms a and b and B with x, y, and z. For the order $<_1$ which extends $a <_1 b$ , define

$$ \begin{align*} \pi_{<_1}(a) := x,&& \pi_{<_2}(b) := y \vee z. \end{align*} $$

Moreover, for the order $<_2$ which extends $b <_2 a$ ,

$$ \begin{align*} \pi_{<_2}(a) := y, && \pi_{<_2}(b) := x \vee z. \end{align*} $$

Let $<'$ be defined as extending $z <' y <' x$ . The objects defined above witness the conditions (i) and (ii). We conclude that $\mathrm {Aut}(L)$ is not amenable.

Proof Suppose that such a class $\mathcal K^*$ exists. Let H be an arbitrary finite Heyting algebra, and consider the action of $\mathrm {Aut}(H)$ on the set of binary relations on H. Since $\mathcal K^*$ is closed under isomorphism types, the set of admissible orderings $A_H$ on H is a union of orbits. Since $\mathcal K^*$ is order-forgetful, $A_H$ consists of a single orbit.

Now consider the poset ${\mathbb P}'$ that is the disjoint union of two 2-chains, with its quotient ${\mathbb P}$ obtained by collapsing one of the 2-chains into a point. The canonical surjection ${\mathbb P}' \twoheadrightarrow {\mathbb P}$ is p-morphic, which induces a Heyting algebra embedding $H \hookrightarrow H'$ . Let $a, b \in H'$ correspond to the two 2-chains. Clearly, H is rigid whereas there is an automorphism $\phi : H' \to H'$ under which a and b are conjugates. Consider an admissible ordering $\prec $ on $H'$ ; without loss of generality, we may assume $a \prec b$ . Writing the action of $\mathrm {Aut}(H')$ by superscripts, we have $b \prec ^\phi a$ . Since $\mathcal K^*$ is a Fraïssé class, the restrictions of $\prec $ and $\prec ^\phi $ to H, respectively, are admissible orderings on H. Now, we have $\mathord {\prec } \cap H^2 \neq \mathord {\prec ^\phi } \cap H^2$ , as witnessed by $(a,b)\in H^2$ . These cannot belong to the same orbit of $A_H$ as H is rigid.

Proof A back-and-forth construction. Enumerate M as $(a_i)_{i<\omega }$ and all the realized 1-types over all finite subsets of M as $(p_i)_{i < \omega }$ . We construct g as the union of the chain $\emptyset = g_0 \subseteq g_1 \subseteq \cdots $ , each of which is a partial isomorphism with a finite domain. Along the way, we construct a chain $\emptyset = S_0 \subseteq S_1 \subseteq \cdots $ of realized 1-types. Suppose that $g_j$ has been constructed. To construct $g_{j+1}$ , one does the following:

If $j = 3i$ . If $a_i$ is in $\operatorname {\mathrm {dom}} g_j$ , then $g_{j+1} := g_j$ . Otherwise, let $g_{j+1}$ extend $g_j$ so $g_{j+1}(a_i)$ may be a realization of $g_j(p)$ outside $\operatorname {\mathrm {ran}} g_j$ , which exists due to the strong amalgamation, where p is the type of $a_i$ over $\operatorname {\mathrm {dom}} g_j$ .

If $j = 3i+1$ . Similar as above, but switch the roles of images and domains.

If $j = 3i+2$ . Let k be the least such that $p_k$ is over X, that $X \subseteq \operatorname {\mathrm {dom}} g_j$ , and that $p_k \not \in S_i$ . (There may not be such k, in which case $g_{j+1} := g_j$ and $S_{i+1} := S_i$ , but there will be such k for infinitely many i because of the other two kinds of stages.) Let $S_{i+1} := S_i \cup \{p_k\}$ . If all realizers of $p_k$ are in $\operatorname {\mathrm {dom}} g_j$ , then $g_{j+1} := g_j$ . If not, apply the strong amalgamation to obtain infinitely many realizers of $p_k$ . Since $\operatorname {\mathrm {dom}} g_j$ is finite, there exists $ a \models p_k$ outside $\operatorname {\mathrm {dom}} g_j$ . Let D be the diagram $\langle aXg_j(X) \rangle \hookleftarrow {\langle Xg_j(X) \rangle } \hookrightarrow \langle aXg_j(X) \rangle $ , and let the diagram $\langle aXg_j(X) \rangle \stackrel {\text {incl.}}{\hookrightarrow } A \stackrel {\iota }{\hookleftarrow } \langle aXg_j(X) \rangle $ , where $A \subseteq M$ , witness the superamalgamation property for D. Now let $g_{j+1} := g_j \cup \{(a, \iota (a))\}$ . (Replace $\iota (a')$ by something else if need be so $\iota (a') \not \in \operatorname {\mathrm {ran}} g_j$ by considering an amalgam with more copies of $\langle aXg_j(X) \rangle $ .) By the superamalgamation property, we have .

Proof By the superamalgamation property of $\mathrm {Age}(M)$ , one can prove that the relation is a stationary independence relation in the sense of Tent and Ziegler [Reference Tent and Ziegler16]. In fact, the invariance of follows from the ultrahomogeneity of M. The monotonicity and the symmetry of are obvious by the shape of the definition of . To show transitivity, assume that and that . To see , take an arbitrary $a \in A$ and $d \in D$ . Suppose $a \le d$ . (The case of $d \le a$ can be handled in a similar way.) Since , there is $b \in BC$ such that $a \le b \le d$ . If $b \in B$ , we are done. Otherwise, $b \in C$ , so by , there is $b' \in B$ such that $a \le b' \le b$ . Now we have $a \le b' \le d$ . To show the existence property of , let p be a realized type over a finite set B and C a finite set. Let $\overline a$ be a tuple realizing p. Now consider the diagram D:

$$\begin{align*}\langle \overline a B\rangle \hookleftarrow \langle B \rangle \hookrightarrow \langle BC \rangle. \end{align*}$$

Let the diagram $\langle \overline a B\rangle \stackrel {\iota }{\hookrightarrow } A \stackrel {\text {incl.}}{\hookleftarrow } \langle BC \rangle $ , where $A \subseteq M$ , witness the superamalgamation property for D. It is clear that . Finally, to show the stationarity of , we may assume the original signature M is $L_0$ without loss of generality. Take two realizations $a, a' \models p$ where p is over a finite set A. We may further assume that A contains all constants in $L_0$ without loss of generality. Consider an arbitrary finite set $A' \supseteq A$ such that . The order types of $aA'$ and $a'A'$ , respectively, are determined by the order types of $aA$ , $a'A$ , and $A'$ . By the hypothesis $\mathrm {Aut}(M) = \mathrm {Aut}(M \upharpoonright L_0)$ , the first-order types of $aA'$ and $a' A'$ are determined by the order types of $aA$ , $a'A$ , and $A'$ . We conclude that $\mathrm {tp}^M(a/A') = \mathrm {tp}^M(a'/A')$ . Therefore, for g constructed in the preceding lemma, every element of $\mathrm {Aut}(M)$ is the product of 16 conjugates of g by Corollary 5.4 of [Reference Tent and Ziegler16].