# Rigidity of the free factor complex

Mladen Bestvina and Martin R. Bridson  Department of Mathematics
University of Utah
Salt Lake City, UT 84103, USA  Mathematical Institute
Andrew Wiles Building
ROQ
Oxford
OX26GG
U.K.

###### Abstract.

We establish the following non-abelian analogue of the Fundamental Theorem of
Projective Geometry: the natural map from ${\rm{Aut}}(F_{n})$ to the
automorphism group of the free-factor complex $\mathcal{AF}_{n}$ is an
isomorphism. We also prove the corresponding theorem for the action of
${\rm{Out}}(F_{n})$ on the complex of conjugacy classes of free factors.

##  1\. Introduction

Our purpose in this article is to describe the symmetries of the complex of
free factors $\mathcal{AF}_{n}$ associated to a finitely generated free group
$F_{n}$. We shall prove that the natural map from ${\rm{Aut}}(F_{n})$ to the
automorphism group of $\mathcal{AF}_{n}$ is an isomorphism. We shall also
prove the corresponding theorem for the action of ${\rm{Out}}(F_{n})$ on the
complex of conjugacy classes of free factors. These results can be viewed as
non-abelian analogues of the Fundamental Theorem of Projective Geometry, as we
shall now explain.

The Fundamental Theorem of Projective Geometry [vS47] establishes that, for
any field $K$, the only bijections of a projective space over $K$ that
preserve incidence relations are the natural ones, i.e. combinations of field
automorphisms and projective-linear maps. This can be rephrased in terms of
the Tits building ${\rm{Tits}}_{n}^{<}(K)$, which is the poset of proper non-
trivial subspaces of $K^{n}$. If $K=\mathbb{Q}$ then there are no field
automorphisms and the theorem tells us that the natural map
${\rm{PGL}}(n,\mathbb{Q})\to{\rm{Aut}}({\rm{Tits}}_{n}^{<}(\mathbb{Q}))$ is an
isomorphism provided $n\geq 3$. The geometric realisation
${\rm{Tits}}_{n}(\mathbb{Q})$ of this poset has an additional symmetry: its
group of simplicial automorphisms is
${\rm{PGL}}(n,\mathbb{Q})\rtimes\mathbb{Z}/2$, with the generator of
$\mathbb{Z}/2$ swapping each vertex $V$ with $V^{\perp}$, where the orthogonal
complement is taken with respect to an inner product on $\mathbb{Q}^{n}$.
(This is an anti-isomorphism of the poset ${\rm{Tits}}_{n}^{<}(\mathbb{Q})$.)

The inclusion $\mathbb{Z}^{n}\hookrightarrow\mathbb{Q}^{n}$ induces an
isomorphism $\mathcal{D}_{n}(\mathbb{Z})\to{\rm{Tits}}_{n}^{<}(\mathbb{Q})$,
where $\mathcal{D}_{n}(\mathbb{Z})$ is the poset of proper direct factors of
$\mathbb{Z}^{n}$, ordered by inclusion. Passing from the free abelian group
$\mathbb{Z}^{n}$ to the non-abelian free group $F_{n}$, the natural analogue
of $\mathcal{D}_{n}(\mathbb{Z})$ is the poset of non-trivial proper free
factors of $F_{n}$, ordered by inclusion. We shall work with the geometric
realisation of this poset, which we denote by $\mathcal{AF}_{n}$. This complex
was introduced by Allen Hatcher and Karen Vogtmann [HV98b, HV98a] who used it
to study the cohomology of ${\rm{Aut}}(F_{n})$; they proved, in analogy with
the Solomon-Tits theorem for ${\rm{Tits}}_{n}(\mathbb{Q})$, that
$\mathcal{AF}_{n}$ has the homotopy type of a wedge of spheres of dimension
$n-2$.

As in the classical case, one has to assume $n\geq 3$ in order to obtain the
desired rigidity for the automorphism group of this complex.

######  Theorem 1.1.

For $n\geq 3$ the natural homomorphism
${\rm{Aut}}(F_{n})\to{\rm{Aut}}(\mathcal{AF}_{n})$ is an isomorphism.

Note in particular that every automorphism of $\mathcal{AF}_{n}$ preserves the
type of each vertex, i.e. the rank of each free factor; there is no equivalent
of the involution $V\leftrightarrow V^{\perp}$ of
${\rm{Tits}}_{n}(\mathbb{Q})$.

We also prove a version of the above theorem for ${\rm{Out}}(F_{n})$. In this
case, the natural analogue of ${\rm{Tits}}_{n}(\mathbb{Q})$ is the geometric
realisation $\mathcal{OF}_{n}$ of the poset of conjugacy classes of non-
trivial proper free factors in $F_{n}$, i.e. the quotient
$\mathcal{AF}_{n}/{\rm{Inn}}(F_{n})$. The large-scale geometry of
$\mathcal{OF}_{n}$ was elucidated by Bestvina and Feighn [BF14a], who proved
that it is a space of infinite diameter that is hyperbolic in the sense of
Gromov.

######  Theorem 1.2.

For $n\geq 3$ the natural homomorphism
${\rm{Out}}(F_{n})\to{\rm{Aut}}(\mathcal{OF}_{n})$ is an isomorphism.

A key similarity between ${\rm{Tits}}_{n}(\mathbb{Q})$, on the one hand, and
$\mathcal{AF}_{n}$ and $\mathcal{OF}_{n}$ on the other, is that each is
composed of standard apartments. In the case of ${\rm{Tits}}_{n}(\mathbb{Q})$,
such an apartment is the full subcomplex whose vertices represent the
subspaces spanned by the proper, non-empty subsets of a basis for
$\mathbb{Q}^{n}$. A standard apartment in $\mathcal{AF}_{n}$ is defined in
much the same way, taking the free factors spanned by the non-empty proper
subsets of a basis. In each case, an apartment is simplicially isomorphic to
the barycentric subdivision of the boundary of an $(n-1)$-simplex.

There are also important differences between ${\rm{Tits}}_{n}(\mathbb{Q})$ and
$\mathcal{AF}_{n}$. The former is a spherical building of diameter 3, while
$\mathcal{AF}_{n}$ has infinite diameter. From a technical point of view, a
major difficulty in understanding the automorphisms of $\mathcal{AF}_{n}$
comes from the fact that, in contrast to ${\rm{Tits}}_{n}(\mathbb{Q})$, there
are many “fake apartments” in $\mathcal{AF}_{n}$, i.e. subcomplexes abstractly
isomorphic to the barycentric subdivision of the boundary of an
$(n-1)$-simplex that are not standard apartments (Section 7).

The first stage in our proof of Theorem 1.1 involves establishing another
difference, to which we have already alluded: every simplicial automorphism of
$\mathcal{AF}_{n}$ preserves the partial ordering on the vertex set, i.e. the
rank of free factors; this is achieved in Section 3.

Our aim in the second stage of the proof (Section 4) is to show that standard
apartments can be recognized intrinsically: they can be distinguished from
fake apartments by metric properties of their neighbourhoods. From this it
follows that the set of standard apartments is preserved by all automorphisms
of $\mathcal{AF}_{n}$. The key technical result in this part of the proof is
the Antipode Lemma (Theorem 4.5), which provides an intrinsic (metric)
characterisation of pairs of vertices $A,L$ such that $A\ast L=F_{n}$.

In the third stage of the proof, working outwards from a fixed standard
apartment, we consider adjacent apartments that have large overlaps. A key
role is played in this part of the argument by sticks – certain rank 1 factors
that, when gathered in appropriate families, provide rigid, highly-symmetric
frames controlling large overlaps between apartments (see Section 5.1).

With these tools in hand, the final step in our proof is straightforward:
${\rm{Aut}}(F_{n})$ acts transitively on the set of standard apartments,
preserving the rank of vertices, so by composing an arbitrary automorphism
$\Phi$ of $\mathcal{AF}_{n}$ with a suitable element of ${\rm{Aut}}(F_{n})$,
we may assume that $\Phi$ fixes a standard apartment; we argue that one can
compose with a further element of ${\rm{Aut}}(F_{n})$ to ensure that $\Phi$
fixes the apartment and all of the adjacent sticks pointwise; this forces
$\Phi$ to fix the neighbouring apartments and their sticks pointwise
(Proposition 5.9), and by propagation $\Phi$ is forced to be the identity
everywhere.

Our proof of Theorem 1.2 follows the same outline but there are some
additional difficulties to address, notably that it is harder to recognise
standard apartments, which are no longer uniquely determined by their rank 1
vertices.

The parallel that we focussed on to motivate Theorem 1.1 compared
$\mathcal{AF}_{n}$ to
$|{\mathcal{D}}_{n}(\mathbb{Z})|\cong{\rm{Tits}}_{n}(\mathbb{Q})$. This is a
facet of the powerful 3-way analogy between automorphism groups of free
groups, lattices such as ${\rm{SL}}(n,\mathbb{Z})$, and mapping class groups
of surfaces of finite type [Bes02, BV06]. In this grand analogy, the object
corresponding to $\mathcal{AF}_{n}$ and $\mathcal{OF}_{n}$ in the setting of
mapping class groups is the curve complex [Har81]. Ivanov [Iva97] proved the
analogue of Theorems 1.1 and 1.2 in this setting: the natural map from the
extended mapping class group of a surface of finite type to the group of
simplicial automorphisms of the corresponding curve complex is an isomorphism
(with some exceptions for small surfaces – cf. [Kor99], [Luo00]).

Ivanov used his theorem to deduce that the extended mapping class group of a
surface of finite type is equal to its own abstract commensurator (with the
same exceptions for small surfaces). In connection with this, we should
comment on the fact that ${\rm{Aut}}(\mathcal{AF}_{n})$ is
${\rm{Aut}}(F_{n})$, whereas ${\rm{Aut}}(\mathcal{D}_{n}(\mathbb{Z}))$ is
${{\rm{PGL}}}(n,\mathbb{Q})$ not ${{\rm{PGL}}}(n,\mathbb{Z})$. This difference
can be interpreted as a manifestation of the fact that
${{\rm{GL}}}(n,\mathbb{Q})$ is the abstract commensurator of
${{\rm{GL}}}(n,\mathbb{Z})$. In contrast, commensurations of
${\rm{Aut}}(F_{n})$ (i.e. isomorphisms between subgroups of finite index) are
as restricted as they are in the mapping class group case: Bridson and Wade
[BW] prove that the action of ${\rm{Aut}}(F_{n})$ on $\mathcal{AF}_{n}$
extends to a faithful action by ${\rm{Comm}}({\rm{Aut}}(F_{n}))$, and it then
follows from Theorem 1.1 that
${\rm{Aut}}(F_{n})={\rm{Comm}}({\rm{Aut}}(F_{n}))$. The corresponding result
for ${\rm{Out}}(F_{n})$ is due to Farb and Handel [FH07] for $n\geq 4$ and to
Horbez and Wade [HW20] for $n\geq 3$ (with proofs that do not follow the
template we have described).

Theorems 1.1 and 1.2 also extend the range of faithful geometric models for
${\rm{Aut}}(F_{n})$ and ${\rm{Out}}(F_{n})$ – by which we mean spaces $X$
where a natural action induces an isomorphism
${\rm{Aut}}(F_{n})\to{\rm{Aut}}(X)$ or ${\rm{Out}}(F_{n})\to{\rm{Aut}}(X)$.
The first such rigidity result was proved by Bridson and Vogtmann, who showed
that ${\rm{Out}}(F_{n})$ is the group of simplicial automorphisms of the spine
of Outer space [BV01]. Other such spaces $X$ include the simplicial closure of
Outer space [AS11], the free and cyclic splitting complexes [AS11, HW14], and
Outer space endowed with the Lipschitz metric [FM09]. This last result, due to
Francaviglia and Martino, is the natural analogue of Royden’s theorem on the
isometries of Teichmüller space [Roy71], which was reproved by Ivanov [Iva97]
using the rigidity of the curve complex (the analogue of Theorem 1.1), with an
argument modelled on the proof of Mostow rigidity in higher rank [Mos73],
which in turn relies on understanding the automorphisms of spherical buildings
such as ${\rm{Tits}}_{n}(\mathbb{R})$, which is where we began.

Acknowledgements. The first author gratefully acknowledges the support by the
National Science Foundation under grant number DMS-1905720.

##  2\. Background and Preliminaries

We shall assume that the reader is familiar with basic algebraic facts about
free groups and their subgroups. For example, if $L<F_{n}$ is a free factor
and $H<F_{n}$ then $H\cap L$ is a free factor of $H$; in particular any
intersection of free factors in $F_{n}$ is a free factor.

Throughout this paper we shall explore subgroups of free groups by working
with labeled graphs that represent them. In this section we gather a range of
facts that we shall need concerning these graphical representations.

###  2.1. Labeled graphs and Stallings folds

We fix a basis $\\{a_{1},\dots,a_{n}\\}$ for $F_{n}$ and identify $F_{n}$ with
the fundamental group of the rose $R_{n}$, which is a graph111we allow graphs
to have multiple edges and loops with one vertex $v$ and $n$ edges, directed
and labeled $a_{1},\dots,a_{n}$. The length of a word $w$ in the letters
$a_{i}^{\pm 1}$ (equivalently, an edge path in $R_{n}$) will be denoted by
$|w|$. A morphism of graphs is a continuous map that sends vertices to
vertices and edges to edges. Formally, a labeled graph is a morphism of graphs
$\lambda:\Gamma\to R_{n}$; in practice, we regard $\Gamma$ as a graph in which
the edges have been oriented and labeled by letters $a_{i}$ so that $\lambda$
preserves the orientation and labeling. Given $H<F_{n}$, the pointed
${\rm{core}}_{*}(H)$ is the labeled graph obtained by restricting the (based)
covering map $(\widetilde{R}_{n},\ast)/H\to(R_{n},v)$ to the minimal connected
subgraph containing all the embedded loops and the basepoint, while the
(unpointed) ${\rm{core}}(H)\subset{\rm{core}}_{*}(H)$ is the minimal connected
subgraph containing all the embedded loops. $H_{1}$ is conjugate to $H_{2}$ if
and only if ${\rm{core}}(H_{1})={\rm{core}}(H_{2})$.

If a pair of directed edges $e,e^{\prime}$ in a labeled graph $\Gamma$ have
the same label and the same initial (resp. terminal) vertex, then the morphism
of labeled graphs $\Gamma\to\Gamma^{\prime}$ that identifies these edges and
their terminal (resp. initial) vertices is called a Stallings fold, [Sta83].
Any morphism of finite graphs can be expressed as a finite sequence of folds
followed by an immersion (locally injective map). There is a unique graph
${\rm{fold}}(\Gamma)$ obtained from $\Gamma$ by a maximal sequence of folds;
such a graph is said to be fully folded; its labeling map
${\rm{fold}}(\Gamma)\to R_{n}$ is an immersion.

We say that a labeled graph with basepoint $(\Gamma,\ast)$ supports a subgroup
$K<F_{n}$ if $K$ is contained in the $\pi_{1}$-image of the labeling map
$\Gamma\to R_{n}$.

For labeled graphs $\Gamma_{1}$ and $\Gamma_{2}$ with basepoints,
$\Gamma_{1}\vee\Gamma_{2}$ will denote the labeled graph obtained from
$\Gamma_{1}\sqcup\Gamma_{2}$ by identifying the basepoints. We refer to
$\Gamma_{1}\vee\Gamma_{2}$ as the wedge of $\Gamma_{1}$ and $\Gamma_{2}$. If
$\Gamma_{1}={\rm{core}}_{*}(H_{1})$ and $\Gamma_{2}={\rm{core}}_{*}(H_{2})$,
then ${\rm{fold}}(\Gamma_{1}\vee\Gamma_{2})={\rm{core}}_{*}\langle
H_{1},H_{2}\rangle$. The following special case of this observation will be
useful.

######  Lemma 2.1.

A subgroup $H<F_{n}$ of rank $k$ is a free factor if and only if there is a
labeled graph $\Gamma$ of rank $(n-k)$ such that
${\rm{core}}_{*}(H)\vee\Gamma$ folds to $R_{n}$.

The following well-known lemma is proved by observing how a graph of rank 1
can fold into ${\rm{core}}(L_{n-1})$.

######  Lemma 2.2.

Let $L_{n-1}=\langle a_{1},\dots,a_{n-1}\rangle$. Then $L_{n-1}*\langle
u\rangle=F_{n}$ if and only if $u=xa_{n}^{\pm 1}y$ for some $x,y\in L_{n-1}$.

The following criterion for recognising factors of corank 1 will also be
useful.

######  Proposition 2.3.

If $H<F_{n}$ is a free factor of rank $n-1$, then either ${\rm{core}}_{*}(H)$
embeds in the rose $R_{n}$ or else the labeled graph obtained by identifying
two of its vertices folds to $R_{n}$.

###### Proof.

Choose $u\in F_{n}$ such that $H\ast\langle u\rangle=F_{n}$. We add a loop
labeled $u$ to $\Gamma={\rm{core}}_{*}(H)$ at $*$ and start folding to obtain
$R_{n}$. Initially, at every step an edge of the $u$-loop folds with an edge
of $\Gamma$. If the process stops before the whole loop is folded in,
${\rm{core}}_{*}(H)$ embeds in $R_{n}$. Otherwise, when the last edge of the
$u$-loop is folded in, two vertices of ${\rm{core}}_{*}(H)$ will be identified
before the folding to $R_{n}$ continues. ∎

###  2.2. Concerning visible factors and powers

The following standard facts will be used without further comment throughout
the paper; the second is used in the proof of the lemma that follows.

  1. •

If $H_{1}<H_{2}$ then there is a unique label-preserving immersion
${\rm{core}}_{*}(H_{1})\to{\rm{core}}_{*}(H_{2})$ restricting to an immersion
${\rm{core}}(H_{1})\to{\rm{core}}(H_{2})$.

  2. •

$u\in F_{n}$ is conjugate into $H<F_{n}$ if and only if there is an oriented
loop in ${\rm{core}}(H)$ whose label is a cyclically reduced word representing
the conjugacy class of $u$; if $H$ is malnormal (e.g. a free factor) then
there is a unique such loop (up to rotation).

We need an elaboration on the second point. To explain this, recall that the
set $L_{H}$ of reduced words representing the elements of a finitely generated
subgroup $H<F_{n}$ consists of the labels on the reduced edge paths in
${\rm{core}}_{*}(H)$ that begin and end at the basepoint. This sits inside
${\rm{sub}}(H)$, the set of labels on all reduced paths in
${\rm{core}}_{*}(H)$, i.e. words $v$ such that some $uvw$ is a reduced word in
$L_{H}$. Define

${\rm{E}}_{a_{i}}(H)=\\{n\mid a_{i}^{n}\in{\rm{sub}}(H)\\}.$

######  Lemma 2.4.

If $H$ is a free factor, then ${\rm{E}}_{a_{i}}(H)$ is infinite if and only if
${\rm{core}}_{*}(H)$ has a loop labeled by the basis element $a_{i}$.

###### Proof.

Any edge path in ${\rm{core}}_{*}(H)$ whose length exceeds the number of
vertices will contain a loop, and a shortest such loop along the path will be
embedded. So if ${\rm{E}}_{a_{i}}$ is infinite then ${\rm{core}}_{*}(H)$
contains an embedded loop labeled $a_{i}^{m}$ for some $m\neq 0$. This
embedded loop represents the conjugacy class of a primitive element, so
$|m|=1$. The converse is obvious. ∎

######  Corollary 2.5.

Let $A<F_{n}$ be a free factor of rank $n-1$. Then, either
${\rm{E}}_{a_{i}}(A)$ is finite for some $i\geq 2$, or else
${\rm{core}}_{*}(A)$ is a tree with $n-1$ loops attached, labeled
$a_{2},\dots,a_{n}$.

###### Proof.

If ${\rm{E}}_{a_{i}}(A)$ is infinite for each $i\geq 2$, then the lemma
provides loops labeled $a_{i}$, and since the rank of ${\rm{core}}_{*}(A)$ is
$n-1$, the remainder of the graph is a tree. ∎

######  Lemma 2.6.

Let $V<F_{n}$ be a free factor of rank $n-1$ and assume that both $\langle
a_{3},\cdots,a_{n}\rangle$ and $\langle a_{2},\cdots,a_{n-1}\rangle$ can be
conjugated into $V$.

  1. (1)

If $n\geq 4$ then $V$ is conjugate to $\langle a_{2},\cdots,a_{n}\rangle$.

  2. (2)

If $n=3$ then $V$ is conjugate to $\langle a_{2}^{\gamma},a_{3}\rangle$ for
some $\gamma\in F_{3}$.

###### Proof.

In this proof factors are considered up to conjugacy so we ignore basepoints
and work with ${\rm{core}}(V)$.

The assumptions imply that the inclusions ${\rm{core}}(\langle
a_{3},\cdots,a_{n}\rangle)\hookrightarrow R_{n}$ and ${\rm{core}}(\langle
a_{2},\cdots,a_{n-1}\rangle)\hookrightarrow R_{n}$ both lift to
${\rm{core}}(V)\to R_{n}$. If $n\geq 4$ these lifts both contain the unique
loop of ${\rm{core}}(V)$ labeled $a_{3}$, so they overlap and their union is a
wedge of $n$ loops labeled $a_{2},\cdots,a_{n}$, thus proving (1).

If $n=3$ we know only that ${\rm{core}}(V)$ contains embedded loops labeled
$a_{2}$ and $a_{3}$. As ${\rm{core}}(V)$ has no vertices of valence $1$ and
${\rm{rank}}(V)=2$, it must be the graph obtained from these two loops by
connecting them with an arc, labeled $\gamma$ say. This proves (2). ∎

The case $n\geq 4$ in the preceding lemma can also be deduced from the
following consequence of the second bullet point above.

######  Lemma 2.7.

If $V<F_{n}$ is a free factor that contains conjugates of $a_{1},a_{2}$ and
$a_{1}a_{2}$, then the loops labeled $a_{1}$ and $a_{2}$ are based at the same
vertex of ${\rm{core}}(V)$.

###### Proof.

The union of the loops in ${\rm{core}}(V)$ labeled $a_{1},a_{2}$ and
$a_{1}a_{2}$ is equal in homology to the union of the loops labeled $a_{1}$
and $a_{2}$, because $H_{1}(V)$ injects into $H_{1}(F)$. It follows that these
subgraphs coincide, and hence the loop labelled $ab$ is based at the same
vertex as either the $a$-loop or the $b$-loop, forcing all three loops to be
based at the same vertex. ∎

###  2.3. Intersections and pullbacks

Given finitely generated $H_{1},H_{2}<F_{n}$ one can compute the intersection
$H_{1}\cap H_{2}$ by constructing the pullback of the labeling maps
${\rm{core}}_{*}(H_{i})\to R_{n}$: the vertex set of the pullback graph $P$
consists of pairs of vertices
$(v,v^{\prime})\in{\rm{core}}_{*}(H_{1})\times{\rm{core}}_{*}(H_{2})$ with the
same image in $R_{n}$, and the directed edges of $P$ are pairs of directed
edges with the same image in $R_{n}$. The component of $P$ that contains the
basepoint $(*,*)$ is ${\rm{core}}_{*}(H_{1}\cap H_{2})$, possibly with trees
attached. Some of the components of $P$ may be trees, while those with non-
trivial fundamental group correspond to the non-trivial intersections of
$H_{1}$ with the conjugates of $H_{2}$.

###  2.4. Free factor graphs, distance in $\mathcal{AF}_{n}$ and
$\mathcal{OF}_{n}$, and links

$\mathcal{AF}_{n}$ is the geometric realisation of the poset of non-trivial
proper free factors of $F_{n}$ ordered by inclusion. For $n\geq 3$ it is a
flag complex, so every automorphism of its 1-skeleton $\mathcal{AF}_{n}^{(1)}$
extends uniquely to a simplicial automorphism of $\mathcal{AF}_{n}$. Thus,
studying the group of simplicial automorphisms of $\mathcal{AF}_{n}$ is
equivalent to studying the group of isometries of the graph
$\mathcal{AF}_{n}^{(1)}$, metrized so that each edge has length $1$. To
lighten the notation, we sometimes write $\mathcal{AF}_{n}$ in place of
$\mathcal{AF}_{n}^{(1)}$, when concentrating on the free factor graph, which
has vertices the non-trivial free factors $A<F_{n}$ and has an edge joining
$A$ to $B$ if $A<B$. Similarly, rather than studying $\mathcal{OF}_{n}$ as a
simplicial complex we shall sometimes concentrate on its 1-skeleton, i.e. the
quotient of the free factor graph by the action of ${\rm{Inn}}(F_{n})$ – so
vertices are conjugacy classes of proper free factors and there is an edge
from $[A]$ to $[B]$ if there are representatives of these conjugacy classes
with $A<B$.

When $n\geq 3$, we write $d_{\mathcal{A}}(A,B)$ for the combinatorial distance
between vertices in (the 1-skeleton of) $\mathcal{AF}_{n}$ and
$d_{\mathcal{O}}([A],[B])$ for the distance in $\mathcal{OF}_{n}$. When there
is no danger of ambiguity, we will simply write $d$. We shall use the terms
“automorphism” and “isometry” interchangeably and supress mention of the
restriction isomorphism from the group of simplicial automorphisms of the full
complex $\mathcal{AF}_{n}$ to the isometry group of its 1-skeleton, writing
both groups as ${\rm{Isom}}(\mathcal{AF}_{n})$ or
${\rm{Aut}}(\mathcal{AF}_{n})$ (and similarly for $\mathcal{OF}_{n}$).

We shall not have to bother much with the case $n=2$, but when we do we must
modify the above definition because $\mathcal{AF}_{2}$ is just a discrete set:
to account for this we regard $\mathcal{AF}_{2}$ as the vertex set of the
graph that has an edge joining $\langle a\rangle$ to $\langle b\rangle$
whenever $\langle a,b\rangle=F_{2}$ and metrize it and $\mathcal{OF}_{2}$
accordingly. (This makes $\mathcal{OF}_{2}$ isometric to the vertex set of the
Farey graph.)

Estimating distances and understanding neighbourhoods in $\mathcal{AF}_{n}$
and $\mathcal{OF}_{n}$ is difficult in general, as we shall see in the proof
of Theorems 1.1 and 1.2, but there are some simple facts relating distance to
the algebra of free factors. For example:

######  Lemma 2.8.

Let $V_{1}$ and $V_{2}$ be vertices in $\mathcal{AF}_{n}$ with
${\rm{rank}}(V_{1})=n-1$. Then $d(V_{1},V_{2})\leq 2$ if and only if
$V_{1}\cap V_{2}\neq 1$.

###### Proof.

$d(V_{1},V_{2})=1$ if and only if $V_{2}<V_{1}$. If $d(V_{1},V_{2})=2$, then
there is a free factor $U$ with $d(V_{1},U)=1=d(U,V_{2})$, whence $U<V_{1}$
and either $V_{2}<U$ or $U<V_{2}$. In the first case $V_{2}<V_{1}$ and in the
second case $U\subset V_{1}\cap V_{2}$. ∎

A similar argument establishes:

######  Lemma 2.9.

Let $[V_{1}]$ and $[V_{2}]$ be vertices in $\mathcal{OF}_{n}$ with
${\rm{rank}}(V_{1})=n-1$. Then $d([V_{1}],[V_{2}])\leq 2$ if and only if
$V_{1}^{w}\cap V_{2}\neq 1$ for some $w\in F_{n}$.

There will be certain points in our argument where it is convenient to work
with the whole complex $\mathcal{AF}_{n}$ rather than just its 1-skeleton.
This is particularly true of arguments that involve links ${\rm{Lk}}(V)$. The
following observations are useful in induction arguments.

######  Lemma 2.10.

If $V\in\mathcal{AF}_{n}$ has rank $n-1$, then there is a rank-preserving
isomorphism ${\rm{Lk}}(V)\cong\mathcal{AF}_{n-1}$. More generally, if
${\rm{rank}}(V)=k$ then the subcomplex ${\rm{Lk}}_{-}(V)\subset{\rm{Lk}}(V)$
spanned by vertices of rank less than $k$ is isomorphic to $\mathcal{AF}_{k}$.
Similarly, if $[V]\in\mathcal{OF}_{n}$ has rank $k$, then
${\rm{Lk}}_{-}[V]\subset{\rm{Lk}}[V]$ is isomorphic to $\mathcal{OF}_{k}$.

###### Proof.

The assertions about $\mathcal{AF}_{n}$ are immediate from the definitions.
For the assertion about $\mathcal{OF}_{n}$ one needs to note that because $V$
is malnormal in $F_{n}$, free factors $A,A^{\prime}<V$ are conjugate in $V$ if
they are conjugate in $F_{n}$. ∎

We shall also need the following observation concerning links.

######  Lemma 2.11.

If $[C]\in\mathcal{OF}_{n}$ is a vertex of rank 1, then
${\rm{Lk}}_{\mathcal{OF}_{n}}([C])\cong{\rm{Lk}}_{\mathcal{AF}_{n}}(C)$.

###### Proof.

Without loss of generality we may assume $C=\langle a_{1}\rangle$. If
$d([C],[L])=1$, then $L$ contains a conjugate of $a_{1}$ and ${\rm{core}}(L)$
contains a unique loop labeled $a_{1}$. We select a conjugate
$L_{a_{1}}\in[L]$ by decreeing the vertex at which this loop is based to be
the basepoint. This choice $[L]\mapsto L_{a_{1}}$ provides an inverse to the
canonical projection
${\rm{Lk}}_{\mathcal{AF}_{n}}(C)\to{\rm{Lk}}_{\mathcal{OF}_{n}}([C])$. ∎

###  2.5. Fully irreducible automorphisms and injectivity radius

Recall that an automorphism $f:F_{n}\to F_{n}$ is called fully irreducible if
for every proper free factor $A<F_{n}$ and every $k>1$, the free factor
$f^{k}(A)$ is not conjugate to $A$. Fully irreducible automorphisms exist in
every rank $n\geq 2$, [GS91]. The results in this section are valid for an
arbitrary fully irreducible automorphism $f$ but when we come to use them in
Section 4.1 we will be free to fix a choice, so it would be enough, for
example, to prove these results for $a_{1}\mapsto a_{2}\mapsto a_{1}a_{2}$ in
the case $n=2$, or

(1) $f_{0}:a_{1}\mapsto a_{2}\mapsto a_{3}\mapsto\dots\mapsto a_{n-1}\mapsto
a_{n}\mapsto a_{1}a_{3}a_{4}\dots a_{n}a_{2},$

in the general case.

The following proposition can be proved using standard facts about stable
laminations [BFH97]. We give an alternative proof suited to the study of free
factor complexes; the general theory is hidden in our appeal to [BF14a].

######  Proposition 2.12.

Let $f\in{\rm{Aut}}(F_{n})$ be a fully irreducible automorphism. For all
$\ell>0,R>0$ and every free factor $A<F_{n}$, there is an integer
$K=K(f,A,\ell,R)$ such that, all $k\geq K$,

$d_{\mathcal{A}}(f^{k}(A),C)\geq d_{\mathcal{O}}([f^{k}(A)],[C])\geq R$

for all rank-1 free factors $C=\langle c\rangle$ with $|c|\leq\ell$.

###### Proof.

The first inequality is obvious. As there are only finitely many rank-1 free
factors with $|c|\leq\ell$, the second inequality is an immediate consequence
of the fact (Theorem 9.3 of [BF14a]) that fully irreducible elements act on
$\mathcal{OF}_{n}$ as isometries with positive translation length, so
$\inf\\{d_{\mathcal{O}}(f^{k}(V),V)\mid V\in\mathcal{OF}_{n}\\}>k\lambda_{f}$
with $\lambda_{f}>0$, hence

$d_{\mathcal{O}}([f^{k}(A)],[C])\geq k\lambda_{f}-d_{\mathcal{O}}([A],[C]).$

∎

In the above proof it was overkill to use the fact that orbits of $f$ grow
linearly: we only needed the orbits to be unbounded.

######  Corollary 2.13.

${\rm{injrad}}({\rm{core}}(f^{k}(A)))\to\infty$ as $k\to\infty$.

###### Proof.

If $c$ is a word of length $\ell$ labeling an embedded loop in
${\rm{core}}(f^{k}(A))$, then $C=\langle c\rangle$ is a rank-1 free factor
conjugate into $f^{k}(A)$, so $d_{\mathcal{O}}([f^{k}(A)],[C])=1$, which
contradicts the proposition unless $k<K(f,A,\ell,2)$. ∎

######  Corollary 2.14.

Let $A<F_{n}$ be a factor of rank $n-1$ and fix $\ell>0$. If $k$ is
sufficiently large, then $\langle f^{k}(A),w\rangle=F_{n}$ implies that every
word conjugate to $w$ has length at least $\ell$.

###### Proof.

Let $C=\langle w\rangle$; it is a free factor. Then $\langle
f^{k}(A),w\rangle=F_{n}$ implies $d(f^{k}(A),C)=3$ when $n\geq 3$ or
$d(f^{k}(A),C)=1$ when $n=2$. The lemma tells us that this cannot happen if
$k$ is sufficiently large and $w$ is conjugate to a word of length less than
$\ell$. ∎

###  2.6. Subfactor Projections

Subfactor projections were introduced by Bestvina and Feighn in [BF14b]. The
definition and use of subfactor projections is motivated by the theory of
subsurface projections introduced by Masur and Minsky [MM00]. For $n\geq 3$,
if $A<F_{n}$ is a free factor of rank $n-1$, then the subfactor projection
$\pi_{A}$ assigns to suitable vertices $[B]\in\mathcal{OF}_{n}$ a subcomplex
$\pi_{A}([B])$ of uniformly bounded diameter in the free factor complex
${\rm{Lk}}[A]\cong\mathcal{OF}_{n-1}$.

In more detail (see [Tay14]), $\pi_{A}$ is defined on $[B]\neq[A]$ provided
that $[B]$ does not contain a conjugate $B^{w}$ antipodal to $A$ and that it
has the following properties:

  * •

the diameter of $\pi_{A}([B])$ is uniformly bounded

  * •

if $B$ is conjugate into $A$ then $\pi_{A}([B])=[B]$

  * •

$\pi_{A}$ is coarsely Lipschitz, i.e. there is a constant $\delta$ such that
if $d_{\mathcal{O}}([B],[C])=1$ and $\pi_{[A]}$ is defined for both $[B]$ and
$[C]$, then the Hausdorff distance between $\pi_{A}([B])$ and $\pi_{A}([C])$
is at most $\delta$.

##  3\. Automorphisms of $\mathcal{AF}_{n}$ preserve the rank of vertices

Let $\mathcal{AF}_{n}(i)\subset\mathcal{AF}_{n}$ be the set of vertices of
rank $i$. The following proposition is the first step in the proof of Theorem
1.1.

######  Proposition 3.1.

Every automorphism of $\mathcal{AF}_{n}$ preserves $\mathcal{AF}_{n}(i)$ for
$i=1,\dots,n-1$.

The proof is broken into several preliminary results.

######  Lemma 3.2.

Every automorphism of $\mathcal{AF}_{n}$ preserves
$\mathcal{AF}_{n}(1)\cup\mathcal{AF}_{n}(n-1)$.

###### Proof.

When $n=3$ there is nothing to prove, so assume that $n>3$. In this proof it
is convenient to work with the whole complex $\mathcal{AF}_{n}$ rather than
just the 1-skeleton. If $A$ is a factor of rank $i$ with $1<i<n-1$ then
${\rm{Lk}}(A)$ can be written as the join ${\rm{Lk}}_{-}(A)*{\rm{Lk}}_{+}(A)$,
where ${\rm{Lk}}_{-}(A)\cong\mathcal{AF}_{i}$ is the full subcomplex spanned
by factors contained in $A$ and ${\rm{Lk}}_{+}(A)$ is the full subcomplex
spanned by factors containing $A$. To finish the proof we need to argue that
links of vertices of rank 1 and $n-1$ are not joins. We will argue that they
have diameter greater than $2$.

In the case of a rank $n-1$ factor $A$, we have
${\rm{Lk}}(A)\cong\mathcal{AF}_{n-1}$. As ${\rm{Aut}}(F_{n})$ acts
transitively on the set of factors of each rank, we may assume $A=\langle
a_{1},\dots,a_{n-1}\rangle$. We could appeal to the non-trivial fact that
$\mathcal{AF}_{n-1}$ has infinite diameter, but it is easy to see that it has
diameter at least $3$, which suffices here: by Lemma 2.8, it is enough to
exhibit a rank $1$ free factor $C<A$ and a rank $n-2$ free factor $B<A$ such
that $C\cap B=1$; let $C=\langle a_{1}\rangle$ and let $B=\langle
a_{2},\dots,a_{n-1}\rangle$.

For the rank 1 case we examine the link of $\langle a_{1}\rangle<F_{n}$,
focusing on $\langle a_{1},a_{2}\rangle$ and $\langle
a_{1},a_{3},\dots,a_{n}\rangle$. The intersection of these factors is $\langle
a_{1}\rangle$ so, arguing as in the proof of Lemma 2.8, we see that their
distance in the link is greater than $2$. ∎

To distinguish rank $1$ vertices from rank $n-1$ vertices, we examine the
geometry of their neighbourhoods in $\mathcal{AF}_{n}$.

######  Lemma 3.3.

Let $A<F_{n}$ be a free factor of rank $n-1$, let $C=\langle u\rangle$ be a
free factor of rank $1$, and suppose $F_{n}=A\ast C$. For any vertex $L$, if
$d(A,L)=1$ then $d(L,C)=2$.

###### Proof.

If $L<A$ then $C$ is not contained in $L$ and $V=\langle L,C\rangle=L\ast C$
is a free factor with $d(L,V)=d(V,C)=1$. ∎

This lemma says that a geodesic from $C$ to $A$ (which has length $3$) cannot
be extended to a geodesic of length $4$; indeed any extension will necessarily
backtrack towards the initial vertex $C$. We shall prove that this metric
property fails if we reverse the roles of rank $1$ and rank $n-1$ vertices,
that is, we find extensions that don’t backtrack.

######  Proposition 3.4.

For every rank $1$ vertex $C$ and every rank $n-1$ vertex $A$, if $d(A,C)>1$
then there exists a vertex $L$ with $d(C,L)=1$ and $d(L,A)>2$.

This proposition is an immediate consequence of Lemma 2.8 and the following
result.

######  Lemma 3.5.

If $A<F_{n}$ is a free factor of rank $n-1$ and $C$ is a free factor of rank
$1$ that is not contained in $A$, then there exists a free factor $L$ of rank
$2$ with $C<L$ and $L\cap A=1$.

###### Proof.

We may assume that $C=\langle a_{1}\rangle$. We analyse $A$ according to the
two cases in Corollary 2.5. Suppose first that ${\rm{E}}_{a_{2}}(A)$ is
finite, fix $M>\max{\rm{E}}_{a_{2}}(A)$ and let $L=\langle a_{1},\
a_{2}^{M}a_{1}a_{3}\rangle$. Note that $L<F_{n}$ is a free factor, since
$\langle L,a_{2}\rangle=\langle a_{1},a_{2},a_{3}\rangle$. A reduced word in
the generators of $L$ either belongs to $C=\langle a_{1}\rangle$ or else
contains $a_{2}^{M}$ as a subword. The intersection of $C$ with $A$ is
trivial, by hypothesis, and reduced words of the latter form do not belong to
$A$, by the definition of $M$, so $L\cap A=1$.

It remains to consider the second case in Corollary 2.5. Thus we assume now
that ${\rm{core}}_{*}(A)$ is a tree with $n-1$ loops attached, labeled
$a_{2},\dots,a_{n}$. Observe that if $p$ is greater than the diameter of
${\rm{core}}_{*}(A)$, then $a_{1}^{p}\not\in{\rm{sub}}(A)$. It follows that no
reduced word in the generators of the rank $2$ free factor $L=\langle a_{1},\
a_{2}a_{1}^{p}a_{3}\rangle$ belongs to $A$. (Again, $L$ is a free factor
because $\langle L,a_{2}\rangle=\langle a_{1},a_{2},a_{3}\rangle$.) ∎

###### Proof of Proposition 3.1.

With Lemma 3.2 in hand, we compare Lemma 3.3 with Proposition 3.4 to deduce
that both $\mathcal{AF}_{n}(1)$ and $\mathcal{AF}_{n}(n-1)$ are preserved by
every isometry of $\mathcal{AF}_{n}$. For $n=3$ there is nothing more to
prove, so we assume $n\geq 4$. Let $A$ be a vertex of rank $i<n-1$ and let $V$
be a rank $(n-1)$ vertex with $A<V$. The action of ${\rm{Aut}}(F_{n})$
preserves the rank of vertices and acts transitively on vertices of each rank,
so by composing an arbitrary automorphism
$\psi\in{\rm{Isom}}(\mathcal{AF}_{n})$ with a suitable element of
${\rm{Aut}}(F_{n})$ we may assume that $\psi(V)=V$. Then $\psi$ restricts to
an isometry of ${\rm{Lk}}(V)\cong\mathcal{AF}_{n-1}$, and by induction on $n$
this restriction preserves the rank of vertices. ∎

##  4\. Recognising Standard Apartments

In the introduction we discussed the significance of standard apartments.

######  Definition 4.1.

A standard apartment in $\mathcal{AF}_{n}$ is the full subcomplex spanned by
the free factors generated by the non-empty proper subsets of a basis for
$F_{n}$.

For the second step in our proof of Theorem 1.1, we must prove that every
isometry of $\mathcal{AF}_{n}$ sends standard apartments to standard
apartments, i.e. the set of standard apartments is characteristic in the
following sense.

######  Definition 4.2.

We say that a collection of subcomplexes of a simplicial complex $X$ is
characteristic (or metrically distinguished) if it is preserved by the
simplicial automorphism group of $X$.

For example, for each $k$ the collection of $k$-simplices of $X$ will be
characteristic. In the previous section we proved that $\mathcal{AF}(i)$ is
characteristic in $\mathcal{AF}_{n}$ for $i=1,\dots,n-1$. Our purpose in this
section is to prove that the set of standard apartments is characteristic, and
a key step in the proof will be to show that the pairs of vertices $\\{A,C\\}$
with ${\rm{rank}}(A)=n-1,\ {\rm{rank}}(C)=1$ and $A\ast C=F_{n}$ is
characteristic (the Antipode Lemma). Along the way, we shall prove that
various other types of subcomplexes are characteristic.

The Antipode Lemma is needed to distinguish standard apartments from fake
apartments (as defined in Definition 4.6). Figure 1 illustrates two of the
concerns that have to be overcome in the case $n=3$ and more elaborate fakes
are discussed in Section 7.

###  4.1. The Antipode Lemma

######  Definition 4.3.

A rank $n-1$ factor $\Lambda$ and a rank 1 factor $\langle u\rangle$ are
algebraically antipodal if $\Lambda*\langle u\rangle=F_{n}$. We write
$\Lambda\perp\langle u\rangle$.

$\Lambda$ and $\langle u\rangle$ are metrically antipodal in
$\mathcal{AF}_{n}$ if $d(\langle u\rangle,L)=2$ for all free factors $L$ with
$d(\Lambda,L)=1$.

######  Remark 4.4.

The condition that $\Lambda$ and $\langle u\rangle$ are metrically antipodal
is equivalent to the following algebraic statement: $u\not\in\Lambda$ and for
all free factors $L\subsetneq\Lambda$ there is a proper free factor of $F_{n}$
that contains both $L$ and $u$. We chose the more concise formulation in the
definition because it makes clear that this property is invariant under
isometries of $\mathcal{AF}_{n}$.

######  Theorem 4.5 (The Antipode Lemma).

Let $\Lambda<F_{n}$ be a free factor of rank $n-1$ and $\langle u\rangle$ a
free factor of rank 1\. Then $\Lambda$ and $\langle u\rangle$ are
algebraically antipodal if and only if they are metrically antipodal.

###### Proof.

It follows easily from the definitions that algebraically antipodal implies
metrically antipodal (Lemma 3.3), so we will assume that $\langle
u\rangle\not\perp\Lambda$ and argue that $\Lambda$ and $\langle u\rangle$ are
not metrically antipodal. The case $u\in\Lambda$ is trivial, so suppose
$u\not\in\Lambda$. By applying a suitable element of ${\rm{Aut}}(F_{n})$ we
may assume $\Lambda=\langle a_{1},\dots,a_{n-1}\rangle$. To complete the
proof, it suffices to exhibit a free factor $L\subset\Lambda$ of rank $n-2$
such that $d(L,\langle u\rangle)>2$. Our proof will show that if
$f:\Lambda\to\Lambda$ is a fully irreducible automorphism and $L_{0}<\Lambda$
is any free factor of rank $n-2$, then $L=f^{k}(L_{0})$ has the desired
property, provided $k>0$ is sufficiently large.

First we consider the case where no conjugate of $u$ is algebraically
antipodal to $\Lambda$. In this case, we argue using the subfactor projection
$\pi_{\Lambda}$ described in section 2.6. Consider $\pi_{\Lambda}([u])$.
Choose $L$ (as above or otherwise) so that the distance between $[L]$ and
$\pi_{\Lambda}([u])$ is large; this is possible because
$Lk([\Lambda])\cong\mathcal{OF}_{n-1}$ has infinite diameter, using the
modified definition of $\mathcal{OF}_{2}$ if $n=3$ (see Proposition 2.12). The
coarse Lipschitz property of $\pi_{\Lambda}$ (section 2.6) tells us any short
path between $[L]$ and $[u]$ in $\mathcal{OF}_{n}$ must pass through a
conjugacy class of factors where $\pi_{\Lambda}$ is not defined. It follows
that there does not exist a free factor $B$ that contains both $L$ and
$\langle u\rangle$, because $\pi_{\Lambda}[B]$ would be well-defined in that
case, and $\pi_{\Lambda}[B]$ would be a distance at most $\delta$ (the
constant of section 2.6) from both $\pi_{\Lambda}([u])$ and
$\pi_{\Lambda}([L])$. ($B$ is not conjugate to $\Lambda$ because $L<B$ and
$u\in B\smallsetminus\Lambda$, whereas distinct conjugates of $\Lambda$
intersect trivially.)

It remains to consider the case where $\langle u\rangle\not\perp\Lambda$ but
some conjugate of $\langle u\rangle$ is antipodal to $\Lambda$. By applying an
automorphism that fixes $\Lambda$ we may assume that, in reduced form,
$u=wa_{n}w^{-1}$ where $w$ is a word whose first letter is $a_{n}^{\pm 1}$.
Let $L=f^{k}(L_{0})$ be as above and assume that $k$ is large enough to ensure
that the injectivity radius of ${\rm{core}}(L)$ is at least $2|u|$ and
Corollary 2.14 holds for $\ell=2|u|$ with $L$ in the role of $A$ and $\Lambda$
in place of $F_{n}$. We will obtain a contradiction from the assumption that
there is a free factor $B$ of rank $n-1$ that contains both $L$ and $u$.

First we observe that if there were such a factor, then $B=\langle L,u\rangle$
and ${\rm{core}}_{*}(B)={\rm{core}}_{*}(L)\vee{\rm{core}}_{*}\langle
u\rangle$. To see this, note that if the canonical map
${\rm{core}}_{*}(L)\to{\rm{core}}_{*}(B)$ were not injective, then the
fundamental group of the image would be a free factor $V\subseteq B$ that
strictly contained $L$. As ${\rm{rank}}(B)={\rm{rank}}(L)+1$, this would imply
$V=B$. But the edges of the graph defining $V$ are labeled by letters from
$\Lambda$, whereas $B$ is not contained in $\Lambda$. Thus
${\rm{core}}_{*}(L)\to{\rm{core}}_{*}(B)$ is injective. As $B$ contains a
conjugate of $a_{n}$ but not $a_{n}$ itself, ${\rm{core}}_{*}(B)$ has a loop
labeled $a_{n}$ based at a vertex $v\neq\ast$. And since $wa_{n}w^{-1}\in B$,
there is path from $*$ to $v$ labeled $w$, which begins with an $a_{n}$-edge.
As $B$ has rank ${\rm{rank}}(L)+1$, this path is disjoint from
${\rm{core}}_{*}(L)$. Thus
${\rm{core}}_{*}(B)={\rm{core}}_{*}(L)\vee{\rm{core}}_{*}\langle u\rangle$.

Proposition 2.3 tells us that if $B=\langle L,u\rangle$ were a free factor,
then by identifying two vertices in ${\rm{core}}_{*}(B)$ we could obtain a
graph $\Gamma$ that folded to the standard rose $R_{n}$. We consider three
cases, depending on the location of the two vertices being identified, and
reach a contradiction in each case.

We shall refer to ${\rm{core}}_{*}\langle u\rangle$ as a lollipop, with stalk
labeled $w$ and loop $a_{n}$.

Case 1: Suppose $v_{0},v_{1}\in{\rm{core}}_{*}(L)$. In this case, the image of
${\rm{core}}_{*}(L)$ in ${\Gamma}$ defines a free factor of rank $n-1$ that
contains $L$ and is contained in $\Lambda$, hence is equal to $\Lambda$. And
$R_{n}$ is obtained by folding this image with ${\rm{core}}_{*}\langle
u\rangle$, so $F_{n}=\Lambda\ast\langle u\rangle$, contrary to the assumption
that $\Lambda\not\perp\langle u\rangle$.

Case 2: Suppose $v_{0}\in{\rm{core}}_{*}(L)$ and
$v_{1}\in{\rm{core}}_{*}\langle u\rangle\smallsetminus\\{*\\}$. In this case,
an arc of the stalk of ${\rm{core}}_{*}\langle u\rangle$ that contains $v_{1}$
but has no edges labeled $a_{n}$ might fold into ${\rm{core}}_{*}(L)$ which,
by construction, has injectivity radius greater than $2|u|$. If $v_{0}$ were a
distance at least $|u|$ from the basepoint, then after this folding we would
have a fully folded graph that still contained ${\rm{core}}_{*}(L)$. If
$v_{0}$ is a distance less than $|u|$ from the basepoint, let $\alpha$ be the
label on the arc from $\ast$ to $v_{0}$, let $\tilde{\beta}$ be the prefix of
$w$ labeling the arc from $\ast$ to $v_{1}$, and let $\beta\in\Lambda$ be the
word obtained from $\tilde{\beta}$ by deleting all occurences of $a_{n}$. Then
$\langle L,\alpha\beta^{-1}\rangle=\Lambda$ if $\Gamma$ folds to $R_{n}$,
because $\langle L,\alpha\beta^{-1}\rangle$ is the fundamental group of the
graph obtained by collapsing the edges of $\Gamma$ labeled $a_{n}$. But this
contradicts Corollary 2.14, because $|\alpha\beta^{-1}|<2|u|$.

Case 3: Suppose $v_{0},v_{1}\in{\rm{core}}_{*}\langle u\rangle$. We fold
$\Gamma_{1}:={\rm{core}}_{*}\langle u\rangle/v_{0}\sim v_{1}$. If the initial
edge on the stalk of the lollipop ${\rm{core}}_{*}\langle u\rangle$ is not
identified with the loop of the lollipop during this folding, then
${\rm{core}}_{*}(L)\vee{\rm{fold}}(\Gamma_{1})$ is fully folded and we are
done. Otherwise, ${\rm{fold}}(\Gamma_{1})$ is the wedge of two loops, one
labeled $a_{n}$ and the other either labeled by a word $c$ in the letters
$a_{i}$ with $i\neq n$, or else labeled $c_{1}c_{2}c_{3}$, where $c_{1}$ and
$c_{3}$ are non-empty words of this form and $c_{2}$ is a non-empty word that
begins and ends $a_{n}^{\pm 1}$. In the former case, we have a contradiction
from Corollary 2.14, because $\langle L,c\rangle\subsetneq\Lambda$. In the
latter case, the arcs labelled $c_{1}$ and $c_{3}$ fold into
${\rm{core}}_{*}(L)$ and the folding stops with ${\rm{core}}_{*}(L)$ still
embedded. ∎

###  4.2. Apartments, fake and standard

The barycentric subdivision of the boundary of the standard $k$-simplex,
$\partial\Delta_{k}$, is the geometric realisation of the poset of nonempty
proper subsets of ${\text{\bf k}}=\\{0,1,\dots,k\\}$ ordered by inclusion. The
barycentre of the face opposite $i\in{\text{\bf k}}$ is ${\text{\bf
k}}\smallsetminus\\{i\\}$.

######  Definition 4.6.

An apartment in $\mathcal{AF}_{n}$ is the image of a simplicial embedding
$\sigma:\partial\Delta_{n-1}\hookrightarrow\mathcal{AF}_{n}$ such that
${\rm{rank}}(\sigma(S))=|S|$ for all $S\subset{\text{\bf n-1}}$. The apartment
is fake if it is not standard.

Note that an apartment is standard if and only if its rank-1 vertices form a
basis for $F_{n}$. Figure 1 illustrates two of the ways in which fake
apartments can arise. There are more examples in Section 7.

$a$$ab^{2}$$c$$\langle a,b\rangle$$\langle ab^{2},c\rangle$$\langle
a,c\rangle$$a$$b$$ac^{2}b$$\langle a,b\rangle$$\langle
b,ac^{2}\rangle$$\langle a,c^{2}b\rangle$ Figure 1. Two fake apartments in
rank 3. In the first, the rank 2 factor $\langle a,b\rangle$ is not generated
by the adjacent rank 1 factors. In the second, the rank 1 factors are not
antipodal to the opposite rank 2 factors.

######  Lemma 4.7.

An apartment in $\mathcal{AF}_{3}$ is standard if and only if each vertex is
antipodal to the barycentre of the opposite face.

###### Proof.

The “if” assertion is the non-trivial one. Suppose that the rank 1 vertices
are $\langle a\rangle,\,\langle b\rangle,\,\langle c\rangle$ and let $V$ be
the barycentre of the face opposite $\langle a\rangle$. Then $V=\langle
b,v\rangle$ for some $v\in V$, and by hypothesis $F_{3}=V\ast\langle
a\rangle$. Thus $\langle a,b\rangle$ is the unique free factor containing $a$
and $b$, and it is therefore the barycentre of the face with vertices $\langle
a\rangle$ and $\langle b\rangle$. This is antipodal to $\langle c\rangle$, so
$\\{a,b,c\\}$ is a basis for $F_{3}$. ∎

######  Proposition 4.8.

For $n\geq 3$, every automorphism of $\mathcal{AF}_{n}$ takes standard
apartments to standard apartments.

###### Proof.

We proceed by induction on $n$. In the light of Lemma 4.7, the Antipode Lemma
(Theorem 4.5) covers the case $n=3$.

Assume now that $n\geq 4$ and consider a rank $n-1$ vertex $V$ of a standard
apartment $\sigma$ and let $\psi$ be an automorphism of $\mathcal{AF}_{n}$. By
composing $\psi$ with an element of ${\rm{Aut}}(F_{n})$, we may assume that
$\psi$ fixes $V$. Then $\psi$ restricts to an automorphism of
${\rm{Lk}}(V)\cong\mathcal{AF}_{n-1}$, where by induction we know that it
takes standard apartments to standard apartments. The intersection
$\sigma\cap{\rm{Lk}}(V)$ is such an apartment, so the image under $\psi$ of
its rank 1 vertices form a basis for $V$. The Antipode Lemma tells us that the
image under $\psi$ of the remaining rank 1 vertex of $\sigma$ is antipodal to
$V$. Thus the image under $\psi$ of the vertex set of $\sigma$ is a basis for
$F_{n}$. ∎

Notation. $\Delta(b_{1},\dots,b_{n})$ will denote the standard apartment
associated to a basis $\\{b_{1},\dots,b_{n}\\}$ of $F_{n}$. A face of rank $k$
is the subcomplex $\Delta[T]$ spanned by a $k$-element subset
$T\subset\\{b_{1},\dots,b_{n}\\}$. The face opposite $\Delta[T]$ is
$\Delta[T^{c}]$, where $T^{c}=\\{b_{1},\dots,b_{n}\\}\smallsetminus T$.

##  5\. Sticks and propagation: the proof of Theorem 1.1

In this section we complete the proof of Theorem 1.1.

Summary of the proof. Given an automorphism $\Phi$ of $\mathcal{AF}_{n}$, with
$n\geq 3$, we now know that $\Phi$ sends standard apartments to standard
apartments. As ${\rm{Aut}}(F_{n})$ acts transitively on the set of standard
apartments, we can compose $\Phi$ with an element of ${\rm{Aut}}(F_{n})$ so as
to assume that $\Phi$ leaves a standard apartment
$\Delta=\Delta(a_{1},\dots,a_{n})$ invariant. The stabilizer of $\Delta$ in
${\rm{Aut}}(F_{n})$ is the group of signed permutations
$W_{n}\cong(\mathbb{Z}/2)^{2}\rtimes{\rm{sym}}(n)$ of the corresponding basis;
its action on $\Delta$ is the full group of rank-preserving symmetries of
$\Delta$. By composing $\Phi$ with an element of $W_{n}<{\rm{Aut}}(F_{n})$ we
may assume that $\Phi$ fixes $\Delta$ pointwise. We would be done if this
modification forced $\Phi$ to be the identity on the whole of
$\mathcal{AF}_{n}$, but it does not. For example, automorphisms of the form
$a_{i}\mapsto a_{i}^{\pm 1}$ fix $\Delta$ but not $\mathcal{AF}_{n}$.

Let $\lambda$ be a Nielsen transformation for the basis
$\\{a_{1},\dots,a_{n}\\}$, that is $[a_{i}\mapsto a_{i}a_{j},\ a_{k}\mapsto
a_{k}\ (k\neq i)]$ or $[a_{i}\mapsto a_{j}a_{i},\ a_{k}\mapsto a_{k}\ (k\neq
i)]$. We say that $\lambda(\Delta)$ is Nielsen adjacent to $\Delta$; it has a
large overlap with $\Delta$.

$\mathcal{AF}_{n}$ is the union of its standard apartments and the index-2
subgroup of ${\rm{Aut}}(F_{n})$ generated by Nielsen transformations acts
transitively on the set of standard apartments. Thus, by propagating to
neighbours throughout $\mathcal{AF}_{n}$, we would be done if any isometry of
$\mathcal{AF}_{n}$ that fixed a standard apartment pointwise had to fix the
Nielsen adjacent apartments pointwise. Although this is not the case, we shall
see that standard apartments have canonical enlargements that make this
argument work: by composing $\Phi$ with a further element of
$W_{n}<{\rm{Aut}}(F_{n})$ we can assume that it fixes the canonical
enlargement of $\Delta$ and this forces $\Phi$ to fix the canonical
enlargement of each Nielsen adjacent apartment.

The vertices of these canonical enlargements are rank-1 vertices adjacent to
$\Delta$ that we call sticks and supersticks.

###  5.1. Sticks and snops

######  Definition 5.1.

The sticks at a face $\Delta[b_{i},b_{j}]$ of rank $2$ in a standard apartment
$\Delta(b_{1},\dots,b_{n})$ are the rank 1 factors of the form $\langle
b_{i}^{\epsilon}b_{j}^{\delta}\rangle,\ \epsilon,\delta=\\{\pm 1\\}$.

Note that this definition depends only on $\langle b_{i},b_{j}\rangle$ and
$\langle b_{i}\rangle,\,\langle b_{j}\rangle$, not on the rest of
$\Delta(b_{1},\dots,b_{n})$. There are 4 sticks at each rank $2$ face, so
$\Delta(b_{1},\dots,b_{n})$ has $4{n\choose 2}$ sticks in total. See Figure 3.

######  Lemma 5.2.

A rank 1 free factor $C<F_{n}$ is a stick of the standard apartment
$\Delta(b_{1},\dots,b_{n})$ if and only if, for some $b_{i}\neq b_{j}$,
$d(C,\,\langle b_{i},b_{j}\rangle)=1$ and $C$ is antipodal to the barycentres
of the rank $n-1$ faces opposite $\langle b_{i}\rangle$ and $\langle
b_{j}\rangle$.

###### Proof.

This follows immediately from Lemma 2.2. ∎

######  Corollary 5.3.

The sets of sticks associated to standard apartments and their faces are
characteristic in $\mathcal{AF}_{n}$.

###### Proof.

Immediate from Proposition 4.8 and the lemma. ∎

######  Remark 5.4.

As an indication of the way in which sticks determine the geometry of
$\mathcal{AF}_{n}$ in a neighbourhood of an apartment, note that in
$\mathcal{AF}_{3}$ each of the $12$ sticks of a standard apartment
$\Delta(a,b,c)$ gives rise to a 2-sphere (after gluing in disks to each
apartment) made from three apartments: for example $bc$ determines the
2-sphere

$\Delta(a,b,c)\cup\Delta(a,bc,b)\cup\Delta(a,bc,c).$

The intersection of each pair of these spheres is $\Delta(a,b,c)$.

######  Remark 5.5 (Sticks and Cubes).

Our formal proofs for $n>3$ do not rely on the following description of sticks
in terms of cubes, but nevertheless we include the general case in our
discussion because it provides useful insight into the local geometry of
$\mathcal{AF}_{n}$.

The $4{n\choose 2}$ sticks associated to a standard apartment parametrize the
codimension-2 faces of an $n$-cube $I^{n}$. The signed permutations of the
basis associated to the apartment form a subgroup
$W_{n}=(\mathbb{Z}/2)^{2}\rtimes{\rm{sym}}(n)<{\rm{Aut}}(F_{n})$ and the
action of this on the sticks is the restriction of the standard representation
of $W_{n}$ as the isometry group of the cube. Figure 2 illustrates the case
$n=3$.

There are 12 sticks associated to a standard apartment (if $n=3$) or rank-3
face (if $n>3$). When three of these sticks lie in a common free factor of
rank $2$ in such a way that any two form a basis of the subgroup they
generate, we say that these sticks form a bonded triple. We also say that two
sticks are bonded to each other if they lie in a common bonded triple. There
are 8 bonded triples associated to each standard apartment (if $n=3$) or
rank-3 face (if $n>3$); they parametrize the vertices of the cube in Figure 2.

The 12 sticks also divide into 3 classes of parallel sticks, such that no pair
of sticks in a given class belong to the same bonded triple; these correspond
to the 3 classes of parallel edges in Figure 2. Each parallelism class divides
into 2 pairs: the opposite of a given stick is the one that labels the edge
that is parallel but has no bonds in common.

With Corollary 5.3 in hand, the following observation is immediate from these
definitions.

######  Lemma 5.6.

Isometries of $\mathcal{AF}_{n}$ preserve bonded triples and parallelism
classes of sticks, as well as pairs of opposite sticks.

$ba$$ab$$a^{-1}b$$ab^{-1}$$ac$$ca$$a^{-1}c$$ac^{-1}$$bc$$cb$$bc^{-1}$$b^{-1}c$
Figure 2. The sticks associated to the standard apartment
$\Delta=\Delta(a,b,c)$ parametrize the edges of the 3-cube. Three sticks form
a bonded triple (snop) when the three edges are adjacent to the same vertex.
The stabilizer of $\\{a,b,c\\}$ in ${\rm{Aut}}(F_{3})$ is the full isometry
group of the cube. $a$$b$$c$$ab$$b^{-1}c$$ac$ Figure 3. The apartment
$\Delta(a,b,c)$ in $\mathcal{AF}_{3}$ with its 12 sticks. The three sticks
that are labeled form a bonded triple (snop).

We have already noted that the sticks associated to a standard apartment
$\Delta$ parametrize the codimension-2 faces of a cube, and that in the case
$n=3$ the vertices of the cube correspond to bonded triples. In the general
case, the vertices of this cube $I^{n}(\Delta)$ correspond to snops, which are
defined as follows. (We shall not rely on this geometric description in our
proofs.)

######  Definition 5.7.

A snop222Croatian for bundle. is a collection $\mathcal{B}$ of sticks
associated to a standard apartment $\Delta(b_{1},\cdots,b_{n})$ with the
following properties:

  1. (1)

Exactly one of the sticks associated to each rank-2 face $\Delta[b_{i},b_{j}]$
belongs to $\mathcal{B}$.

  2. (2)

For every rank-3 face $\Delta[b_{i},b_{j},b_{k}]$, the 3 sticks in
$\mathcal{B}$ form a bonded triple.

The following lemma is an immediate consequence of our previous results.

######  Lemma 5.8.

Snops are characteristic, i.e. every isometry of $\mathcal{AF}_{n}$ takes
snops to snops.

There are $2^{n}$ snops associated to a standard apartment $\Delta$. The
1-skeleton of the cube $I^{n}(\Delta)$ can be constructed by joining two snops
with an edge if they share all but $(n-1)$ of their sticks. (Distinct snops
differ by at least $(n-1)$ sticks.)

The following proposition can be proved by analysing the faithful action of
the stabiliser of $\Delta(b_{1},\dots,b_{n})$ on the cube $I^{n}(\Delta)$,
arguing that if an isometry of the cube fixes sufficiently many codimenson-2
faces then it must be the identity. We leave the details of this proof to the
reader and give a different proof that adapts better to the case of
$\mathcal{OF}_{n}$ considered in the next section.

Notation. The pointwise stabilizer in ${\rm{Aut}}(F_{n})$ of the standard
apartment $\Delta(a_{1},\dots,a_{n})$ is
$(\mathbb{Z}/2)^{n}=\langle\varepsilon_{1},\dots,\varepsilon_{n}\rangle$ where
$\varepsilon_{i}$ is the involution that sends $a_{i}$ to $a_{i}^{-1}$ and
fixes $a_{j}$ if ${j\neq i}$.

To be clear, when we say that an isometry fixes a subcomplex, we mean that it
does so pointwise.

######  Proposition 5.9.

If $\Phi\in{\rm{Isom}}(\mathcal{AF}_{n})$ fixes
$\Delta=\Delta(a_{1},\dots,a_{n})$, then there exists
$\theta\in\langle\varepsilon_{1},\dots,\varepsilon_{n}\rangle$ such that
$\theta\circ\Phi$ fixes $\Delta$ and all of its sticks.

We require a lemma.

######  Lemma 5.10.

If an isometry $\Phi$ of $\mathcal{AF}_{3}$ fixes the standard apartment
$\Delta=\Delta(a,b,c)$ and a stick at $\Delta[a,b]$ then exactly one of
$\\{\Phi,\ \varepsilon_{c}\circ\Phi\\}$ fixes $\Delta$ and all of its sticks.

###### Proof.

The sticks at $\Delta[b,c]$ bonded to $\langle ab\rangle$ are $\langle
b^{-1}c\rangle$ and $\langle cb\rangle$, so if $\Phi$ fixes $\langle
ab\rangle$ then it must either exchange or fix these sticks. Composing with
$\varepsilon_{c}$ if necessary, we may assume that it fixes them. The action
of $\Phi$ as an isometry of the cube in Figure 2 then fixes three edges of the
top face. The only such isometry is the identity. ∎

Proof of Proposition 5.9. We shall proceed by induction. Suppose $n=3$ and
consider a standard apartment $\Delta=\Delta(a,b,c)$ fixed by $\Phi$. If
$\Phi$ does not fix the stick $\langle ab\rangle$ then we can compose $\Phi$
with an element of $\langle\varepsilon_{a},\varepsilon_{b}\rangle$ to arrange
that it does. Then Lemma 5.10 tells us that, composing with $\varepsilon_{c}$
if necessary, we may assume that $\Phi$ fixes all of the sticks of $\Delta$.

We now assume $n>3$ and consider a standard apartment
$\Delta=\Delta(a_{1},\dots,a_{n})$ fixed by $\Phi$. Let
${\rm{Aut}}(F_{n-1})\hookrightarrow{\rm{Aut}}(F_{n})$ be the subgroup fixing
$a_{n}$ and acting in the standard way on $\\{a_{1},\dots,a_{n-1}\\}$.
Consider the barycentre $V=\langle a_{1},\dots,a_{n-1}\rangle$ of the face
opposite $\langle a_{n}\rangle$. We have
${\rm{Lk}}(V)\cong\mathcal{AF}_{n-1}$, where the isomorphism is
${\rm{Aut}}(F_{n-1})$-equivariant. By induction, there exists
$\theta\in\langle\varepsilon_{1},\dots,\varepsilon_{n-1}\rangle$ such that
$\theta\circ\Phi$ fixes $\Delta$ and all of the sticks of
$\Delta[a_{1},\dots,a_{n-1}]$. Applying Lemma 5.10 to
$\Delta[a_{1},a_{2},a_{n}]$, we deduce that by further composing with
$\varepsilon_{n}$ if necessary, we may assume that $\Phi$ fixes $\Delta$, the
sticks of $\Delta[a_{1},\dots,a_{n-1}]$ and the sticks of
$\Delta[a_{1},a_{2},a_{n}]$. The remaining sticks are based at
$\Delta[a_{i},a_{n}]\subset\Delta[a_{1},a_{i},a_{n}]$ with $i>2$. The sticks
at $\Delta[a_{1},a_{i}]\subset\Delta[a_{1},a_{i},a_{n}]$ are fixed by $\Phi$,
as is $\langle a_{1}a_{n}\rangle$. Moreover the latter is not fixed by
$\varepsilon_{n}\circ\Phi$. So Lemma 5.10 tells us that $\Phi$ must fix all
the sticks of $\Delta[a_{1},a_{i},a_{n}]$. This completes the induction. ∎

###  5.2. Supersticks and the end of the proof

We obtain a more rigid framework of rank-1 vertices in the neighbourhood of an
apartment by adding supersticks to sticks. In rank 3, the supersticks
associated to an apartment are at distance 2 from the apartment, but from
$n=4$ onwards they are adjacent to the barycentres of the rank-3 faces of the
apartment.

######  Definition 5.11.

The supersticks associated to a standard apartment $\Delta(a_{1},a_{2},a_{3})$
(if $n=3$) or a rank 3 face $\Delta[a_{1},a_{2},a_{3}]$ (if $n>3$) are the 24
rank 1 factors $\langle
a_{i}^{\delta_{i}}a_{j}^{\delta_{j}}a_{k}^{\delta_{k}}\rangle$ with
$\\{i,j,k\\}=\\{1,2,3\\}$ and $\delta_{i}=\pm 1$.

######  Lemma 5.12.

A rank 1 free factor of $F_{3}$ is a superstick of the standard apartment
$\Delta(a,b,c)$ if and only if it is antipodal to each of the rank 2 vertices
of $\Delta(a,b,c)$.

For $n>3$, a rank 1 free factor $V<F_{n}$ is a superstick of the rank 3 face
$\Delta[a,b,c]$ if and only if $d(V,\langle a,b,c\rangle)=1$ and $V$ is
antipodal in ${\rm{Lk}}_{-}\langle a,b,c\rangle\cong\mathcal{AF}_{3}$ to each
of the rank 2 vertices of $\Delta(a,b,c)$.

###### Proof.

This follows immediately from Lemma 2.2. ∎

######  Corollary 5.13.

The sets of supersticks associated to standard apartments and their rank-3
faces are characteristic in $\mathcal{AF}_{n}$.

We need one last lemma.

######  Lemma 5.14.

If an isometry $\Phi$ of $\mathcal{AF}_{n}$ fixes a standard apartment
$\Delta=\Delta(a,b,c)$ (if $n=3$) or rank-3 face $\Delta=\Delta[a,b,c]$ (if
$n>3$) and it fixes the sticks of $\Delta$, then it also fixes all of the
supersticks of $\Delta$.

###### Proof.

Consider first the superstick $\langle abc\rangle$. As $M_{1}=\langle
a,bc\rangle<F_{n}$ is the unique factor of rank 2 adjacent to both $\langle
a\rangle$ and $\langle bc\rangle$, it must be fixed by $\Phi$. Likewise
$M_{2}=\langle c,ab\rangle$ must be fixed. The unique rank-1 factor adjacent
to $M_{1}$ and $M_{2}$ is $\langle abc\rangle=M_{1}\cap M_{2}$, so it too must
be fixed by $\Phi$. The general case is similar. ∎

End of the Proof of Theorem 1.1. We refer the reader to the summary of the
proof given at the beginning of this section. Given an automorphism $\Phi$ of
$\mathcal{AF}_{n}$, with $n\geq 3$, we compose it with an element of
${\rm{Aut}}(F_{n})$ so as to assume that it leaves a standard apartment
$\Delta=\Delta(a_{1},\dots,a_{n})$ invariant. We use Proposition 5.9 to
compose $\Phi$ with a further element of ${\rm{Aut}}(F_{n})$ so that it fixes
$\Delta$ and all of its sticks. Lemma 5.14 then tells us that $\Phi$ fixes the
supersticks of $\Delta$. We will be done if we can argue that this adjusted
$\Phi$ fixes every standard apartment that is Nielsen adjacent to $\Delta$ and
fixes all the sticks (and hence supersticks) of such an apartment.

Without loss of generality we may assume that the Nielsen transformation is
$\lambda:a_{1}\mapsto a_{1}a_{2}$. Consider
$\Delta_{\lambda}=\Delta(a_{1}a_{2},a_{2},\dots,a_{n})$. The first point to
observe is that every rank 1 vertex of $\Delta_{\lambda}$ is a vertex or stick
of $\Delta$, and hence is fixed by $\Phi$. Each vertex of $\Delta_{\lambda}$
is uniquely determined by its adjacent rank 1 vertices, so $\Phi$ must fix the
whole of $\Delta_{\lambda}$. The second point to observe is that every stick
of $\Delta_{\lambda}$ is a vertex, stick or superstick of $\Delta$, with the
exception of the sticks at $\Delta[a_{1},\,a_{1}a_{2}]$. And since these last
sticks are distinguished from one another by the sticks of
$\Delta[a_{1},\,a_{1}a_{2},a_{3}]$ with which they form bonded triples, they
too must be fixed. ∎

##  6\. $\mathcal{OF}_{n}$ is rigid: Proof of Theorem 1.2

Our proof of Theorem 1.2 follows the same outline of proof as Theorem 1.1 but
there are some additional difficulties to be overcome in the case of
$\mathcal{OF}_{n}$, particularly with regard to the recognition of standard
apartments.

We will typically write $[A]$ for the conjugacy class of a free factor
$A<F_{n}$ but for rank-1 factors abbreviate $[\langle u\rangle]$ to $[u]$, and
often write $[a,b]$ for rank-2 factors.

###  6.1. Step 1: Distinguishing the ranks of vertices.

At various stages in the proof of Theorem 1.1 we used the isomorphism
${\rm{Lk}}(A)\cong\mathcal{AF}_{n-1}$ for vertices of rank $n-1$ to facilitate
induction arguments. Lemma 2.10 assures us that such arguments remain valid in
$\mathcal{OF}_{n}$.

The following lemma can be established by choosing $L$ exactly as in the proof
of Lemma 3.5.

######  Lemma 6.1.

If $A<F_{n}$ is a free factor of rank $n-1$ and $C$ is a free factor of rank
$1$, no conjugate of which is contained in $A$, then there exists a free
factor $L$ of rank $2$ with $C<L$ such that no conjugate of $L$ intersects $A$
non-trivially.

######  Proposition 6.2.

For $n\geq 3$, every isometry of $\mathcal{OF}_{n}$ preserves the set of
vertices of rank $i$, for $i=1,2,\cdots,n-1$.

###### Proof.

The proof is a straightforward adaptation of the proof of Proposition 3.1. To
distinguish vertices of rank $1$ or $n-1$ from those of rank $i$ with
$1<i<n-1$, we prove that the former are not joins, and we do this by showing
that they have diameter greater than $2$. For $n=3$ there is nothing to prove,
so we assume $n\geq 4$ and proceed by induction. The link of a vertex of rank
$n-1$ is isomorphic to $\mathcal{OF}_{n-1}$, which has infinite diameter
(alternatively, as in Lemma 3.2, one can see easily that it has diameter at
least $3$). For vertices of rank 1, Lemma 2.11 tells us that
${\rm{Lk}}_{\mathcal{OF}_{n}}([C])\cong{\rm{Lk}}_{\mathcal{AF}_{n}}(C)$, so
the proof for $\mathcal{AF}_{n}$ applies directly.

The argument for distinguishing vertices of rank $n-1$ from vertices of rank
$1$ also follows the case of $\mathcal{AF}_{n}$: the proof of Lemma 3.3 shows
that for every vertex $[A]\in\mathcal{OF}_{n}$ of rank $n-1$ there exist
vertices $[C]$ of rank $1$ such that $d([A],[L])=1$ implies $d([C],[L])=2$,
and Lemmas 6.1 and Lemma 2.9 tell us this statement becomes false if we
reverse the roles of $A$ and $C$.

The inductive argument in the final paragraph of Section 3 remains valid in
the setting of $\mathcal{OF}_{n}$. ∎

###  6.2. The Antipode Lemma

######  Definition 6.3.

A rank $n-1$ vertex $[\Lambda]\in\mathcal{OF}_{n}$ and a rank 1 vertex
$[u]\in\mathcal{OF}_{n}$ are algebraically antipodal if there are factors
$\Lambda_{0}\in[\Lambda]$ and $\langle u^{\gamma}\rangle\in[u]$ such that
$\Lambda_{0}*\langle u^{\gamma}\rangle=F_{n}$. We write $[\Lambda]\perp[u]$.

$[\Lambda]$ and $[u]$ are metrically antipodal in $\mathcal{OF}_{n}$ if
$d([u],[L])=2$ for all free factors $L$ with $d([\Lambda],[L])=1$.

######  Theorem 6.4 (The Antipode Lemma).

$[\Lambda]\perp[u]$ if and only if $[\Lambda]$ and $[u]$ are metrically
antipodal

###### Proof.

As was the case for $\mathcal{AF}_{n}$, it is easy to see that if
$[\Lambda]\perp[u]$ then $[\Lambda]$ and $[u]$ are metrically antipodal, and
it is obvious that if $u$ is conjugate into $\Lambda$ then $[\Lambda]$ and
$[u]$ are not metrically antipodal. So what we must argue is that if no
conjugate of $u$ is contained in $\Lambda$ and no conjugate of $u$ is
antipodal to $\Lambda$, then there is a free factor $L<\Lambda$ such that
$d([\langle u\rangle],[L])>2$. This is what we proved in the second paragraph
of the proof of Theorem 4.5. ∎

###  6.3. Step 2: Recognising Standard Apartments

The reader should compare the following definition to Definition 4.6. The more
cumbersome definition here reflects the fact that in $\mathcal{OF}_{n}$
apartments are not uniquely determined by their rank 1 vertices. This will
cause us considerable difficulty, as will the fact that standard apartments
are difficult to characterise using the Antipode Lemma alone; see Example 6.9
and Section 7.

######  Definition 6.5.

An apartment in $\mathcal{OF}_{n}$ is the image of a simplicial embedding
$\sigma:\partial\Delta_{n-1}\hookrightarrow\mathcal{OF}_{n}$ such that
${\rm{rank}}(\sigma(S))=|S|$ for all $S\subset{\text{\bf n-1}}$. The apartment
is standard if it is the image under $\mathcal{AF}_{n}\to\mathcal{OF}_{n}$ of
a standard apartment in $\mathcal{AF}_{n}$. We shall maintain the notation
$\Delta(a_{1},\dots,a_{n})$ for the standard apartment associated to the basis
$\\{a_{1},\dots,a_{n}\\}$ and the notation $\Delta[T]$ for its faces; if
$|T|=k+1$ then $\Delta[T]$ is a standard rank-$k$ face.

######  Definition 6.6 (Sticks, supersticks, bonded triples).

We define the sticks, supersticks and bonded triples for standard faces in
$\mathcal{OF}_{n}$ to be the images of the sticks, supersticks and bonded
triples in $\mathcal{AF}_{n}$. For a standard apartment
$\Delta=\Delta(b_{1},\dots,b_{n})$, the sticks of $\Delta$ at the rank-2 face
$\Delta[b_{i},b_{j}]$ are the rank 1 vertices of the form
$[b_{i}^{\epsilon}b_{j}^{\delta}]$ (of which there are only two, because
$b_{i}^{\epsilon}b_{j}^{\delta}$ and $b_{j}^{\delta}b_{i}^{\epsilon}$ are
conjugate and $[x]=[x^{-1}]$).

######  Remark 6.7.

(1) Considerable care is needed with this definition: the “sticks of $\Delta$
at the face $\Delta[b_{i},b_{j}]$”  depend on $\Delta$ and not just
$\Delta[b_{i},b_{j}]$ and its neighbours $[b_{i}],[b_{j}]$. Indeed, if one
drops the reference to $\Delta$ then there are infinitely many sticks at
$\Delta[b_{i},b_{j}]$. To see this note, for example, that for any
$u\in\langle a_{1},a_{2}\rangle$, the triple
$[ua_{1}u^{-1},a_{2}],\,[ua_{1}u^{-1}],\,[a_{2}]$ is identical to
$[a_{1},a_{2}],\,[a_{1}],\,[a_{2}]$, but the sticks of
$\Delta(ua_{1}u^{-1},a_{2},\dots,a_{n})$ at
$[ua_{1}u^{-1},a_{2}]=[a_{1},a_{2}]$ are $[ua_{1}u^{-1}a_{2}]$ and
$[ua_{1}^{-1}u^{-1}a_{2}]$, whereas the sticks of
$\Delta(a_{1},a_{2},\dots,a_{n})$ at $[ua_{1}u^{-1},a_{2}]=[a_{1},a_{2}]$ are
$[a_{1}a_{2}]$ and $[a_{1}^{-1}a_{2}]$.

(2) As $\Delta=\Delta(b_{1},\dots,b_{n})$ has only two sticks at
$\Delta[b_{i},b_{j}]$, it has $2{n\choose 2}$ sticks in total. There are 8
supersticks associated to each standard apartment (if $n=3)$ or rank 3 face
(if $n>3$).

(3) It is no longer useful to discuss which pairs of sticks are bonded,
because any pair of sticks associated to a rank 3 face will be bonded, but it
remains true and useful that any two sticks in a bonded triple uniquely define
the third.

Passing to conjugacy classes $A\mapsto[A]$ preserves the relation of being
algebraically antipodal, so sticks of an apartment remain antipodal to the
barycentres of opposite faces. But at this stage we do not have a metric
characterisation of sticks (as in Lemma 5.2) because we do not yet know that
isometries of $\mathcal{OF}_{n}$ take standard apartments to standard
apartments.

$a$$b$$c$$ab$$bc$$ac$$a^{-1}b$$b^{-1}c$$a^{-1}c$ Figure 4. The apartment
$\Delta(a,b,c)$ in $\mathcal{OF}_{3}$ with its 6 sticks.

######  Example 6.8.

$\Delta(a,b,c)$ has four bonded triples (snops) in $\mathcal{OF}_{3}$

  * •

$ab,b^{-1}c,ac$,

  * •

$ab,bc,a^{-1}c$,

  * •

$a^{-1}b,bc,ac$,

  * •

$a^{-1}b,b^{-1}c,a^{-1}c$.

For any pair of sticks chosen from two rank 2 faces, there is a unique stick
at the third face that forms a bonded triple (snop) with that pair.

The eight supersticks of $\Delta(a,b,c)$ in $\mathcal{OF}_{3}$ are

  * •

$abc,\ abc^{-1},\ ab^{-1}c,\ ab^{-1}c^{-1},\ acb,\ acb^{-1},\ ac^{-1}b,\
ac^{-1}b^{-1}c.$

######  Example 6.9.

We describe an example of a fake (i.e. non-standard) apartment of
$\mathcal{OF}_{3}$ in which all pairs of opposite vertices are antipodal.

Starting with the standard apartment $\Delta(a,b,c)$, we replace $[a,b]$ by
$[a,\gamma b\gamma^{-1}]$ with $\gamma=baca^{-1}$ to obtain the face
$\Delta^{\prime}$. The graph ${\rm{core}}\langle a,\gamma b\gamma^{-1}\rangle$
consists of two loops labeled $a,b$ joined by an arc labeled $baca^{-1}$. To
see that $[a,\gamma b\gamma^{-1}]$ is antipodal to $[c]$, we glue a loop
labeled $c$ to one of the endpoints of the edge of ${\rm{core}}\langle
a,\gamma b\gamma^{-1}\rangle$ labeled $c$ and fold to obtain the rose $R_{3}$.
To see that the apartment $\Delta^{\prime}$ is fake, observe that it has no
sticks at $[a,\gamma b\gamma^{-1}]$: more precisely, there are no rank 1
factors $[u]$ adjacent to $[a,\gamma b\gamma^{-1}]$ that are antipodal to both
$[a,c]$ and $[b,c]$. Indeed, any cyclically reduced word in the conjugacy
class $[u]$ must label a loop in ${\rm{core}}\langle a,\gamma
b\gamma^{-1}\rangle$ that is not $a^{\pm 1}$ or $b^{\pm 1}$, and the label on
any such loop contains more than one occurence of both $a$ and $b$, so is not
antipodal to $[a,c]$ or $[b,c]$.

Fortunately, the problem identified in this example is the only new
obstruction to recognising standard apartments in rank $3$.

######  Proposition 6.10.

Let $\Delta$ be an apartment in $\mathcal{OF}_{3}$ and assume

  1. (1)

opposite vertices of $\Delta$ are antipodal, and

  2. (2)

$\Delta$ has “a potential stick” at each rank 2 vertex, i.e. there is an
adjacent rank 1 vertex that is antipodal to the other two rank 2 vertices of
$\Delta$.

Then $\Delta$ is a standard apartment.

###### Proof.

Let $\\{a,b,c\\}$ be a basis for $F_{3}$. We may assume that $\Delta$ has
opposing vertices $[a]$ and $[b,c]$. By applying an automorphism of $F_{3}$
that fixes $a$ and leaves $\langle b,c\rangle$ invariant, we can assume that
one of the rank 1 vertices adjacent to $[b,c]$ is $[b]$. The rank 2 vertex $V$
between $[a]$ and $[b]$ is then $[b,\gamma a\gamma^{-1}]$ for some $\gamma\in
F_{3}$. If $[u]$ is a potential stick of $\Delta$ at $[b,\gamma
a\gamma^{-1}]$, then it is antipodal to $[b,c]$ and hence the cyclically
reduced form of $u$ contains exactly one occurence of $a$, by Lemma 2.2. This
word labels a tight (i.e. locally-injective) loop in ${\rm{core}}\langle
b,\gamma a\gamma^{-1}\rangle$. The only tight loops with a single occurence of
$a$ in their label, besides $a^{\pm 1}$, are the loops labeled $a^{\pm
1}\gamma^{-1}b^{p}\gamma$ with $p\neq 0$, and these only qualify if there is
no occurence of $a$ in $\gamma$. The loops $a^{\pm 1}$ can be excluded as
potential sticks because they are not antipodal to the rank-2 vertex opposite
$[b]$, since that already contains a conjugate of $a$. Thus the existence of a
potential stick at $V$ forces $\gamma\in\langle b,c\rangle$, and after
applying the automorphism that fixes $b$ and $c$ and sends
$a\mapsto\gamma^{-1}a\gamma$ we may assume $V=[a,b]$.

Consider now the rank 1 vertex of $\Delta$ opposite $V$; call it $[x]$. Since
$[x]$ is antipodal to $[a,b]$, the cyclically reduced word conjugate to $x$
contains exactly one $c^{\pm 1}$, and since $x$ is conjugate into $\langle
b,c\rangle$ we may assume (by conjugating and replacing $x$ with $x^{-1}$)
that $x=b^{m}c$ for some $m$. After applying the automorphism that fixes $a,b$
and sends $c\mapsto b^{-m}c$, we have $x=c$. Then $\Delta$ has 5 of its
vertices in common with the standard apartment $\Delta(a,b,c)$, and the last
one is the conjugacy class of a factor of the form $H=\langle a,\delta
c\delta^{-1}\rangle$. The labeled graph ${\rm{core}}(H)$ has loops $a$ and $c$
connected by an arc $\delta$. Repeating the argument used to analyse $V$, we
see that $H$ can only contain a rank 1 factor antipodal to $[b,c]$ if $\delta$
contains no $c$, and it can only contain a rank 1 factor antipodal to $[a,b]$
if $\delta$ contains no $a$. Thus $\delta=b^{q}$, and the automorphism of
$F_{3}$ that fixes $a$ and $b$ and sends $c\mapsto b^{-q}cb^{q}$ will map
$\Delta$ to the standard apartment $\Delta[a,b,c]$. ∎

######  Corollary 6.11.

Isometries of $\mathcal{OF}_{3}$ take standard apartments to standard
apartments.

###### Proof.

In Step 1 (section 6.1) we proved that isometries of $\mathcal{OF}_{3}$
preserve rank, in the Antipode Lemma we proved that they send antipodal pairs
to antipodal pairs, and in Proposition 6.10 we characterised standard
apartments in terms of these invariants. ∎

The last lemma we need before concluding that isometries preserve standard
apartments is the following. The fake apartments described in Section 7
illustrate the need for condition (3) in this lemma.

######  Lemma 6.12.

Let $n\geq 3$. An apartment $\Delta$ in $\mathcal{OF}_{n}$ is standard if and
only if it satisfies the following conditions:

  1. (1)

Each rank $(n-1)$ face of $\Delta$ is standard.

  2. (2)

Every rank 1 vertex of $\Delta$ is antipodal to the barycentre of the opposite
face.

  3. (3)

Adjacent to each rank $(n-1)$ vertex $V$ of $\Delta$, there is a rank $1$
vertex that is antipodal to every rank $(n-1)$ vertex of $\Delta$ other than
$V$.

###### Proof.

First note that standard apartments satisfy these conditions: for (3), a
suitable rank $1$ factor adjacent to
$[a_{1},\dots,a_{n-1}]\in\Delta(a_{1},\dots,a_{n})$ is $[a_{1}\dots a_{n-1}]$.

For the converse, Proposition 6.10 covers the case $n=3$, so we suppose $n>3$.
Condition (1) lets us assume that there is a basis $\\{a_{1},\dots,a_{n}\\}$
of $F$ such that one of the codimension-1 faces of $\Delta$ is the standard
$\Delta[a_{1},\dots,a_{n-1}]$. Condition (2) says that the rank 1 vertex
opposite this face is $[x]$ where $x$ is antipodal to $\langle
a_{1},\dots,a_{n-1}\rangle$. The action of ${\rm{Aut}}(F_{n})$ (through
${\rm{Out}}(F_{n})$) preserves conditions (1), (2) and (3), so we are free to
apply an automorphism to ensure that $x=a_{n}$.

Consider the codimension 1 face $Y_{1}$ of $\Delta$ opposite $[a_{1}]$. By
condition (1), this is standard, so the barycentre of the face is $[V_{1}]$
where $V_{1}$ is generated by $\langle a_{2},\dots,a_{n-1}\rangle$ and a
conjugate of $a_{n}$, say $a_{n}^{\gamma_{1}}$. We can assume that
$\gamma_{1}\in\langle a_{1},\dots,a_{n}\rangle$ is a word that does not end in
$a_{n}^{\pm 1}$ and (if nontrivial) starts with $a_{n}^{\pm 1}$ – it is the
label on the bridge of ${\rm{core}}(V_{1})$ connecting the rose with petals
$a_{2},\dots,a_{n-1}$ to the loop labeled $a_{n}$. For $j=2,\dots,n-1$, the
barycentre of the edge of $Y_{1}$ joining $[a_{j}]$ to $[a_{n}]$ is
$[a_{j},a_{n}^{\gamma_{1}}]$. Because of our assumptions on $\gamma_{1}$, the
core graph of $[a_{j},a_{n}^{\gamma_{1}}]$ consists of loops labeled $a_{j}$
and $a_{n}$ with the bridge connecting them with the label precisely
$\gamma_{1}$.

Similar considerations apply to face $Y_{i}$ opposite $[a_{i}]$ for
$i=2,\dots,n-1$ and we define $V_{i}$ and $\gamma_{i}$ accordingly. For
example, $V_{2}=\langle a_{1},a_{3},\dots,a_{n-1},a_{n}^{\gamma_{2}}\rangle$
and for $j=1,3,\dots,n-1$, the barycentre edge of $Y_{2}$ joining $[a_{j}]$ to
$[a_{n}]$ is $[a_{j},a_{n}^{\gamma_{2}}]$.

The edge joining $[a_{3}]$ to $[a_{n}]$ in $Y_{1}$ is, of course, the same as
the edge joining them in $Y_{2}$, so
$[a_{3},a_{n}^{\gamma_{1}}]=[a_{3},a_{n}^{\gamma_{2}}]$. Comparing core
graphs, we conclude that $\gamma_{1}=\gamma_{2}$ since both are the label on
the bridge. Proceeding in this manner, we conclude that
$\gamma_{i}=\gamma_{j}$ for all $i,j\in\\{1,\dots,n-1\\}$. If this common
conjugator $\gamma$ lies $\langle a_{1},\dots,a_{n-1}\rangle$, then the
automorphism that fixes $a_{i}$ for $i<n$ and conjugates $a_{n}$ by
$\gamma^{-1}$ will map $\Delta$ to the standard apartment
$\Delta(a_{1},\dots,a_{n})$, so $\Delta$ is standard.

To complete the proof, we argue that if $\gamma\not\in\langle
a_{1},\dots,a_{n-1}\rangle$ then $\Delta$ would not satisfy condition (3). If
a rank 1 vertex $[u]$ is adjacent to $[V_{1}]$, there is a reduced loop in
${\rm{core}}(V_{1})$ labeled $u$. The key point to note is that each reduced
loop in ${\rm{core}}(V_{1})$ either lies in the rose with labels
$a_{2},\dots,a_{n-1}$, or runs only around the loop labeled $a_{n}$, or else
traverses the bridge labeled $\gamma$ twice. In the first case $[u]$ is not
antipodal to $[a_{1},\dots,a_{n-1}]\in\Delta$, in the second case it is not
antipodal to $[V_{2}]$, and in the last case every conjugate of $u$ contains
at least 3 occurrences of the letter $a_{n}$, so $[u]$ is not antipodal to
$[a_{1},\dots,a_{n-1}]$, by Lemma 2.2. ∎

######  Proposition 6.13.

For $n\geq 3$, every isometry of $\mathcal{OF}_{n}$ takes standard apartments
to standard apartments.

###### Proof.

Same as Corollary 6.11. ∎

######  Corollary 6.14.

For $n\geq 3$, every isometry $\Psi$ of $\mathcal{OF}_{n}$ takes the sticks of
a standard apartment $\Delta$ to the sticks of $\Psi(\Delta)$.

###### Proof.

It follows from Lemma 2.2 that the sticks of $\Delta$ at a rank $2$ face
$\Delta[a,b]$ are the unique rank 1 vertices $V$ adjacent to $[a,b]$ with the
property that for every rank $3$ face $\Delta[a,b,c]$, the vertex $V$ is
antipodal to $[a,c]$ and $[b,c]$ in
${\rm{Lk}}_{-}([a,b,c])\cong\mathcal{OF}_{3}$. And $\Psi$ transports this
condition to the sticks of the standard apartment $\Psi(\Delta)$. ∎

Similarly, following Lemma 5.12 we have:

######  Corollary 6.15.

For $n\geq 3$, every isometry $\Psi$ of $\mathcal{OF}_{n}$ takes the
supersticks of a standard apartment $\Delta$ to the supersticks of
$\Psi(\Delta)$.

###  6.4. The endgame

The sum of our previous results tells us that for $n\geq 3$, every isometry of
$\mathcal{OF}_{n}$ maps standard apartments to standard apartments, respecting
their sets of sticks, supersticks and bonded triples (triples of sticks
contained a common factor of rank $2$). We shall deduce Theorem 1.2 by
following the final steps in the proof of Theorem 1.1; only minor adjustments
are needed, except for the issue resolved in Lemma 6.19.

It will be convenient to consider the action of ${\rm{Aut}}(F_{n})$ on
$\mathcal{OF}_{n}$ (with the inner automorphisms acting trivially), as the
subgroups ${\rm{Aut}}(F_{n-1})\hookrightarrow{\rm{Aut}}(F_{n})$ fixing basis
elements appear in the proof. The pointwise stabilizer in ${\rm{Out}}(F_{n})$
of the standard apartment $\Delta(a_{1},\dots,a_{n})$ is
$(\mathbb{Z}/2)^{n}=\langle\varepsilon_{1},\dots,\varepsilon_{n}\rangle$ where
$\varepsilon_{i}$ is the involution that sends $a_{i}$ to $a_{i}^{-1}$ and
fixes $a_{j\neq i}$. The diagonal element
$\iota=\varepsilon_{1}\dots\varepsilon_{n}$ will play a special role, related
to the following observation.

######  Lemma 6.16.

$\iota$ acts trivially on the set of sticks associated to the standard
apartment $\Delta(a_{1},\dots,a_{n})$ in $\mathcal{OF}_{n}$, but it acts
without fixed points on the set of supersticks.

######  Proposition 6.17.

If $\Phi\in{\rm{Isom}}(\mathcal{OF}_{n})$ fixes
$\Delta=\Delta(a_{1},\dots,a_{n})$, then there exists
$\theta\in\langle\varepsilon_{1},\dots,\varepsilon_{n}\rangle$ such that
$\theta\circ\Phi$ fixes $\Delta$ and all of its sticks.

The inductive proof of Proposition 5.9 applies verbatim to this proposition
(replacing $\mathcal{AF}_{n-1}$ with $\mathcal{OF}_{n-1}$) once we have the
following analogue of Lemma 5.10 in hand.

######  Lemma 6.18.

If an isometry $\Phi$ of $\mathcal{OF}_{3}$ fixes the standard apartment
$\Delta=\Delta(a,b,c)$ and the sticks at $\Delta[a,b]$, then one exactly one
of $\\{\Phi,\varepsilon_{c}\circ\Phi\\}$ fixes $\Delta$ and all of its sticks.

###### Proof.

If $\Phi$ exchanges the two sticks at $[b,c]$ then we compose with
$\varepsilon_{c}$ so that it fixes them. It must then fix the sticks at
$[a,c]$, because they are contained in bonded triples where the other two
sticks are fixed, and each pair of sticks in a triple uniquely determines the
third stick (see Example 6.8). ∎

At this stage in the proof of Theorem 1.1 we argued (Lemma 5.14) that if an
isometry $\Phi$ of $\mathcal{AF}_{n}$ fixes a standard apartment $\Delta$ and
its sticks, then it also fixes all of the supersticks of that apartment. This
is not true in the case of $\mathcal{OF}_{n}$; it has to be adjusted as
follows.

Note that since $\iota$ acts freely on the supersticks of
$\Delta(a_{1},\dots,a_{n})$, the word “one” in the following statement means
“exactly one”.

######  Lemma 6.19.

For $n\geq 3$, if an isometry $\Phi$ of $\mathcal{OF}_{n}$ fixes the standard
apartment $\Delta(a_{1},\dots,a_{n})$ and its sticks, then one of $\Phi$ and
$\iota\circ\Phi$ fixes the apartment, its sticks and its supersticks.

###### Proof.

The vertices $[M_{i}]$ and $[M_{i}^{\prime}]$ appearing in this proof should
be regarded as midpoints between the rank 1 vertices of
$\Delta=\Delta(a_{1},\dots,a_{n})$ and the sticks of $\Delta$; these midpoints
come in pairs.

Consider first the superstick $[a_{1}a_{2}a_{3}]$. The rank 2 vertices $[M]$
adjacent to both $[a_{1}a_{2}]$ and $[a_{3}]$ in $\mathcal{OF}_{n}$ are of the
form $[a_{1}a_{2},a_{3}^{\gamma}]$ or $[a_{2}a_{1},a_{3}^{\gamma}]$, where
$\gamma$ is the label on the arc in ${\rm{core}}(M)$ connecting the loop
labeled $a_{1}a_{2}$ or $a_{2}a_{1}$ to the loop labeled $a_{3}$. The key
point to observe is that if $\gamma\neq 1$ then ${\rm{core}}(M)$ does not
contain a loop labeled by a superstick
$a_{i}^{\delta_{1}}a_{j}^{\delta_{2}}a_{k}^{\delta_{3}}$ with
$|\delta_{1}|=|\delta_{2}|=|\delta_{3}|=1$. Thus the only rank 2 vertices
$[M]\in\mathcal{OF}_{n}$ at distance $1$ from $[a_{1}a_{2}]$ and $[a_{3}]$ and
a superstick of $\Delta$ are $[M_{1}]:=[a_{1}a_{2},a_{3}]$ and
$M_{1}^{\prime}:=[a_{2}a_{1},a_{3}]$. Likewise, the only rank 2 vertices
$[M]\in\mathcal{OF}_{n}$ at distance $1$ from $[a_{2}a_{3}]$ and $[a_{1}]$ and
a superstick of $\Delta$ are $[M_{2}]:=[a_{2}a_{3},a_{1}]$ and
$M_{2}^{\prime}:=[a_{3}a_{2},a_{1}]$.

The two supersticks carried by $M_{1}$ are $[a_{1}a_{2}a_{3}]$ and
$[a_{1}a_{2}a_{3}^{-1}]$, while $M_{1}^{\prime}$ carries $[a_{1}a_{3}^{\pm
1}a_{2}]$ and $M_{2}$ carries $[a_{1}(a_{2}a_{3})^{\pm 1}]$ and
$M_{2}^{\prime}$ carries $[a_{1}(a_{3}a_{2})^{\pm 1}]$. Thus $M_{1}$ and
$M_{2}$ have a single superstick in common, as do $M_{1}^{\prime}$ and
$M_{2}^{\prime}$, and no other combination does.

As $\Phi$ fixes $[a_{1}a_{2}]$ and $[a_{3}]$ and takes supersticks to
supersticks, it must fix both of $M_{1}$ and $M_{1}^{\prime}$ or interchange
them. Likewise it must fix both of $M_{2}$ and $M_{2}^{\prime}$ or interchange
them. And if it interchanges $M_{1}$ and $M_{1}^{\prime}$ then it must also
interchange $M_{2}$ and $M_{2}^{\prime}$, since $M_{1}$ and $M_{2}$ have a
superstick in common, whereas $M_{1}$ and $M_{2}^{\prime}$ do not. The action
of $\iota$ fixes $\Delta$ and its sticks while interchanging $M_{1}$ and
$M_{1}^{\prime}$ and interchanging $M_{2}$ and $M_{2}^{\prime}$. So by
composing with $\iota$ if necessary, we may assume that $\Phi$ fixes each of
$M_{1},M_{1}^{\prime},M_{2},M_{2}^{\prime}$. It must then also fix the common
supersticks that pairs of these factors support, and the remaining supersticks
that they carry must then also be fixed. Thus $\Phi$ (possibly adjusted by
$\iota$) must fix all six of the sticks listed above. The remaining
supersticks of $\Delta[a_{1},a_{2},a_{3}]$ are $[a_{1}a_{2}^{-1}a_{3}]$ and
$[a_{1}a_{2}^{-1}a_{3}^{-1}]$. These too must be fixed because the latter is
supported in common with $[a_{1}a_{3}a_{2}]$ on a midpoint graph between
$[a_{1}]$ and $[a_{3}a_{2}]$, whereas the former is not.

At this point we are done in the case $n=3$, but to complete the proof of the
lemma in the general case we must argue that because $\Phi$ fixes the
supersticks associated to one rank-3 face, it fixes the supersticks on all
rank 3 faces. The argument given above shows that $\Phi$ either fixes all or
none of the supersticks at a rank 3 face, so it will be enough to prove that
$\Phi$ fixes one of the supersticks at an adjacent face; we focus on
$[a_{1}a_{2}a_{4}]$.

Observe that $V=[a_{1}a_{2},a_{3},a_{4}]$ is the unique rank-3 vertex adjacent
to $M_{1},\,[a_{3}],\,[a_{4}],\,[a_{3}a_{4}]$, all of which we know to be
fixed by $\Phi$, and ${\rm{core}}(V)$ is the wedge of loops labeled
$a_{3},a_{4},a_{1}a_{2}$. The only superstick of $\Delta[a_{1},a_{2},a_{4}]$
carried by this graph is $[a_{1}a_{2}a_{4}]$; in other words this is the only
such superstick that is a distance $1$ from $V$. Thus the isometry $\Phi$ must
fix $[a_{1}a_{2}a_{4}]$. ∎

The proof of the following observation is contained in the preceding proof.

######  Addendum 6.20.

If $\Phi$ fixes $\Delta(a_{1},\dots,a_{n})$, all of its sticks, and all of its
substicks, then, for all distinct triples $i,j,k\in\\{1,\dots,n\\}$, it also
fixes each of the rank 2 vertices $[M]$ adjacent to both $[a_{i}]$ and
$[a_{j},a_{k}]$

We need one last lemma.

######  Lemma 6.21.

Let $\Delta_{1}=\Delta(b_{1},b_{2},\cdots,b_{n})$ be a standard apartment that
contains all the vertices of $\Delta_{0}=\Delta(a_{1},a_{2},\cdots,a_{n})$
except for $\langle a_{2},a_{3},\cdots,a_{n}\rangle$.

  1. (1)

If $n>3$ then $\Delta_{1}=\Delta_{0}$.

  2. (2)

If $n=3$ then $\Delta_{1}=\Delta_{0}(a_{1},a_{2}^{a_{1}^{k}},a_{3})$ for some
$k\in\mathbb{Z}$.

###### Proof.

The rank $n-1$ factor $V$ that $\Delta_{1}$ has in place of $\langle
a_{2},\cdots,a_{n}\rangle$ contains, up to conjugation, both $\langle
a_{2},\cdots,a_{n-1}\rangle$ and $\langle a_{3},\cdots,a_{n}\rangle$. For
$n>3$ this implies that $V$ is conjugate to $\langle
a_{2},\cdots,a_{n}\rangle$, by Lemma 2.6.

If $n=3$ then $V$ must have the form $V=\langle a_{2}^{\gamma},a_{3}\rangle$,
whose core graph has two loops, labeled $a_{2}$ and $a_{3}$ connected by an
arc labeled $\gamma$. Arguing with the existence of sticks (as in Example 6.9)
we see that $\gamma$ must be a power of $a_{1}$. ∎

End of the Proof of Theorem 1.2. Given an automorphism $\Phi$ of
$\mathcal{OF}_{n}$, with $n\geq 3$, we compose it with an element of
${\rm{Aut}}(F_{n})$ so as to assume that it leaves a standard apartment
$\Delta=\Delta(a_{1},\dots,a_{n})$ invariant. We use Proposition 6.17 to
compose $\Phi$ with a further element of ${\rm{Aut}}(F_{n})$ so that it fixes
$\Delta$ and all of its sticks. Lemma 6.19 then tells us that, after composing
with $\iota$ if necessary, $\Phi$ fixes the supersticks of $\Delta$. We will
be done if we can argue that this adjusted $\Phi$ fixes every standard
apartment that is Nielsen adjacent to $\Delta$ and fixes all the sticks and
supersticks of such an apartment.

Without loss of generality we may assume that the Nielsen transformation is
$a_{1}\mapsto a_{1}a_{2}$. Consider
$\Delta_{\lambda}=\Delta(a_{1}a_{2},a_{2},\dots,a_{n}\\}$. Every rank 1 vertex
of $\Delta_{\lambda}$ is a vertex or stick of $\Delta$, and all of the faces
that do not include the vertex $[a_{1}a_{2}]$ are fixed as they lie in
$\Delta$. Proceeding by induction on the rank we may assume that every vertex
except $V=\langle a_{1}a_{2},a_{3},\cdots,a_{n}\rangle$ is fixed. It then
follows from Lemma 6.21 that $V$ is also fixed. All the sticks of this
apartment except for one (namely $a_{1}a_{2}a_{2}$) are either vertices,
sticks or supersticks of $\Delta$, so all of the sticks are fixed. It follows
from Lemma 6.19 that $\Phi$ fixes all of the supersticks of $\Delta_{\lambda}$
or none of them (because $\iota$ acts without fixed points on the set of
supersticks). But there is one that we know it does fix, namely
$[a_{1}a_{2}^{2}a_{3}]$, because this is the only superstick of
$\Delta_{\lambda}$ that is carried by the rank 2 vertex $[a_{2},a_{1}a_{3}]$,
and this is one of the midpoint vertices $[M]$ that Addendum 6.20 tells us is
fixed by $\Phi$. This completes the proof. ∎

##  7\. Fakery in every rank

In this section we underscore the subtlety of recognising standard apartments
by describing a family of fake apartments in $\mathcal{AF}_{n}$ and
$\mathcal{OF}_{n}$. This family shows that there exist fake apartments in
$\mathcal{OF}_{n}$ with the property that each of their rank $(n-1)$ faces is
standard and each of their rank 1 vertices is antipodal to the barycentre of
the opposite face.

We consider a family of rank $n$ subgroups $H<F_{n}$ for which
${\rm{core}}_{*}(H)$ is obtained from the rose for $\langle
a_{1},\dots,a_{n-1}\rangle$ by connecting it to a loop labelled $a_{n}$ with a
bridge labelled by a word of a particular form. The words that we want are
defined recursively:

$W_{0}:=a_{n}\text{ and }\ W_{k+1}:=W_{k}a_{k+1}W_{k}^{-1}.$

For example, $W_{2}=(a_{n}a_{1}a_{n}^{-1})\,a_{2}(a_{n}a_{1}^{-1}a_{n}^{-1})$.
Define

$H:=\langle a_{1},\dots,a_{n-1},W_{n-1}a_{n}W_{n-1}^{-1}\rangle.$

######  Lemma 7.1.

For $j\leq n-1$, the subgroup $V_{j}<F_{n}$ generated by
$W_{n-1}a_{n}W_{n-1}^{-1}$ and $A_{j}=\\{a_{i}\mid i\leq n-1,\,i\neq j\\}$ is
a rank $(n-1)$ free factor antipodal to $\langle
W_{j-1}a_{j}W_{j-1}^{-1}\rangle$.

###### Proof.

We shall refer to the arc of ${\rm{core}}_{*}(V_{k})$ joining the basepoint
$*$ to the loop labeled $a_{n}$ as the bridge; it is labeled $W_{n-1}$. Let
$p$ be the vertex on the bridge that is the terminus of the path from $*$
labeled by the prefix $W_{j-1}\prec W_{n-1}$. We attach the lollipop
${\rm{core}}_{*}(W_{j-1}a_{j}W_{j-1}^{-1})$ to ${\rm{core}}_{*}(V_{k})$ at $*$
and start folding. The stalk of the lollipop folds entirely into
${\rm{core}}_{*}(V_{k})$, at which point we have the graph obtained from
${\rm{core}}_{*}(V_{j})$ by attaching a loop labeled $a_{j}$ at $p$. The edge
$e$ immediately beyond $p$ then folds around this loop and the folding
continues as the arc beyond $e$ that is labeled $W_{j-1}^{-1}$ folds into the
section of the bridge joining $p$ to $*$ – at this point the folded graph is
the wedge of the rose for $A_{j}$ and two lollipops, one with stalk labeled
$W_{j-1}$ and petal $a_{j}$, and one with stalk
$a_{j+1}W_{j}^{-1}a_{j+2}W_{j}\dots$ and petal $a_{n}$. The initial edge on
the stalk of the second lollipop folds into the rose, then the arc labeled
$W_{j}^{-1}$ traces around the first lollipop, then the edge labeled $a_{j+2}$
folds into the rose, etc.

The folding continues until the entire stalk of the second lollipop has folded
into the wedge of the rose and the first lollipop. At this stage, the loop
labeled $a_{n}$ shares its vertex with the rose for $A_{j}$, and the stalk of
the first lollipop folds into the rank-$(n-1)$ rose that they form. Thus we
obtain the rose $R_{n}$. ∎

######  Proposition 7.2.

The proper subsets of $\\{a_{1},\dots,a_{n-1},\,W_{n-1}a_{n}W_{n-1}^{-1}\\}$
generate free factors of $F_{n}$, and the subcomplex $\Delta<\mathcal{AF}_{n}$
that they span is an apartment with the following properties:

  1. (1)

every codimension-1 face is standard;

  2. (2)

the apartment is fake.

The image of $\Delta$ in $\mathcal{OF}_{n}$ is also fake, and

  1. (3)

the barycentre of each codimension-1 face is antipodal to the rank 1 factor
opposite it.

###### Proof.

Lemma 7.1 assures us that each subset of cardinality $k<n$ generates a free
factor of rank $k$, so $\Delta$ is indeed an apartment and (1) holds. The
lemma also tells us that, in $\mathcal{OF}_{n}$, the codimension-1 face
opposite the vertex $[a_{j}]$ is antipodal to
$[a_{j}]=[W_{j-1}a_{j}W_{j-1}^{-1}]$, so (3) holds.

In a standard apartment of $\mathcal{OF}_{n}$, if $[A],[B]$ are the
barycentres of distinct codimension-1 faces and $A,B$ are representatives with
$A\cap B\neq 1$, then $A\cup B$ will generate $F_{n}$. But in the image of
$\Delta$, such representatives will only generate $H\neq F_{n}$, so the
apartment is fake. ∎

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