1 Introduction
In this paper, we solve Dehn’s conjugacy problem in
$\operatorname {Out}(F_3)$
, the group of outer automorphisms of the free group
$F_3$
of rank 3:
Theorem 1.1. There exists an algorithm which takes two automorphisms
$\phi ,\psi \ \in Aut(F_3)$
and correctly outputs yes or no whether their outer classes are conjugate in
$\operatorname {Out}(F_3)$
.
The conjugacy problem has already been solved within certain classes of outer automorphisms of free groups: the atoroidal fully irreducible ones by Sela [Reference SelaSel95], all the irreducible ones by Los [Reference LosLos96], all the atoroidal ones by [Reference DahmaniDah16], the linearly growing ones by Kristić, Lustig and Vogtmann [Reference Krstí, Lustig and VogtmannKLV01], and the unipotent polynomially growing ones by Feighn and Handel [Reference Feighn and HandelFH19].
In the case of a free group of rank 2,
$\operatorname {Out}(F_2) \simeq \mathrm {GL}(2,\mathbb Z)$
is well-known and virtually free, and there is an algorithm to solve its conjugacy problem.
In the case of a free group of rank 3,
$\operatorname {Out}(F_3)$
is more complicated, and our conjugacy decision algorithm operates by classifying the pair of input automorphisms by means of invariants, in subclasses, in which specific methods can be applied.
Let us mention key conjugacy invariants that are relevant in these classifications and are computable. They give a first frame of reference.
-
• Irreducibility (whether there is an invariant conjugacy class of free factor system, or not), [Reference KapovichKap14], [Reference Francaviglia and MartinoFM22];
-
• exponential growth (whether there is a conjugacy class whose iterated images grow exponentially fast, or not), [Reference Bestvina, Feighn and HandelBFH05];
-
• rank of the maximal polynomially growing subgroups (the maximal subgroups whose elements have iterated images that grow at most polynomially in conjugacy length), and polynomial degree of growth in these subgroups (Proposition 2.7);
-
• arrangement of these subgroups (their orbit under the automorphism group of the ambiant free group) (Gersten’s algorithm);
-
• ranks of the free factors carrying the so-called attracting lamination (Proposition 2.9).
Example 1.2. The map
$a\mapsto a$
,
$b\mapsto ba$
,
$c\mapsto cb$
extends to the free group of basis
$\{a,b,c\}$
as an automorphism that is reducible (
$\langle a, b\rangle $
is an invariant free factor of rank
$2$
), polynomially growing of degree
$2$
(the growth of c is quadratic).
Example 1.3. A pseudo-Anosov mapping class on a twice punctured torus gives an automorphism that is reducible (each puncture corresponds to a free factor of rank one that is preserved), of exponential growth, with two conjugacy classes of maximal polynomially growing subgroups (corresponding to the punctures), both cyclic, on which the automorphism has polynomial degree 0. They are both rank 1 free factors, but not in a same free factor system. The attracting lamination is supported by the entire group.
Example 1.4. Take T a torus with one hole
$\partial T$
, with a base point on its boundary, and a circle C with a base point, and take the wedge of these spaces, identifying the two base points: its fundamental group is
$F_3$
. Consider a pseudo-Anosov mapping class on T, fixing the boundary pointwise, and a map that sends C on the concatenation
$C\cdot \ell $
for a chosen loop
$\ell $
in T. The defined map on
$T\wedge C$
induces an automorphism of
$F_3$
that is reducible, of exponential growth, with an invariant free factor of rank
$2$
(the group of T). The cyclic group of the boundary
$\partial T$
of T is polynomially growing of degree
$0$
. Depending on the choice of
$\ell $
, it might or might not be a maximal polynomially growing subgroup: if
$\ell $
is a power of the loop
$\partial T$
, the maximal polynomially growing subgroup containing
$\pi _1(\partial T)$
is actually of rank 2, on which the automorphism has polynomial degree
$0$
or
$1$
, and it is not a free factor of
$F_3$
but rather a factor of a decomposition of
$F_3$
as some amalgamated free product
$$\begin{align*}\pi_1(T) *_{\pi_1(\partial T)} \pi_1 (C \wedge \partial T).\end{align*}$$
The attracting lamination is carried by the rank 2 free factor
$\pi _1(T)$
. We show in Proposition 6.5 that any exponentially growing automorphism of
$F_3$
with a nontrivial and noncyclic polynomially growing subgroup must be of this form.
It is nevertheless not sufficient to collect these invariants to have characterised the conjugacy class of an outer automorphism. For instance, knowing that the automorphisms are irreducible, with pure exponential growth (i.e., the only polynomially growing subgroup is the trivial one) still requires stamina to decide the conjugation between such outer automorphisms. Actually, all the solved cases listed in the beginning of this introduction all correspond to some particular situation in this frame of reference.
In our study of
$\operatorname {Out}(F_3)$
, we will observe that, given the state of the art, only three configurations in this frame of reference need to be treated. We will cast two of them in the point of view of automorphisms preserving a specific rank 2 free factor. We will cast the last one in the point of view of toral relatively hyperbolic mapping tori. The first configurations is that of the polynomial growth of degree
$2$
on the group
$F_3$
(well illustrated by Example 1.2). The second configuration is the case of exponential growth in which a rank 2 free factor is invariant and ‘attracts’ all the growth. In that case (that is well illustrated by our example 1.4), there is only one class of maximal polynomially growing subgroup, it is not a free factor, and it is either cyclic (generated by the commutator of a basis in the invariant rank 2 free factor), or of rank 2 (attached to this free factor). The last case (illustrated by Example 1.3) is when the growth is exponential, with a rank 1 free factor preserved, and the maximal polynomial growth subgroups are of rank 1 (there might be two of them up to conjugacy).
Our algorithm to solve the conjugacy problem in
$\operatorname {Out}(F_3)$
. If at any time the given automorphisms
$\phi $
and
$\psi $
follow different arrows, they are not conjugate.
Our algorithm is shown in Figure 1, and its correctness is derived from the correctness of every classification step as well as every terminal step. There are several ways to separate cases toward the use of Theorem 3.1 or the use of Theorem 6.4 or of methods in Section 6 since they overlap, but considering complexity, it seems reasonable to leave the later for the smaller number of cases.
In Section 3, we show how to solve the conjugacy problem in
$\operatorname {Out}(F_3)$
among all outer automorphisms with a given invariant rank 2 free factor. This result, in a sense, is a type of induction step from
$\operatorname {Out}(F_2)$
to
$\operatorname {Out}(F_3)$
, but as will be explained in Remark 3.11, it will not generalize similarly to higher ranks. This case is critically important as we will show that most problematic situations (which actually fall in the case where the given automorphism induces a polynomial growth automorphism on a noncyclic subgroup) lead to an invariant free factor of rank 2.
In Section 4, we show that quadratically growing outer automorphisms have an invariant rank 2 free factor. In Section 5, we use train track methods to study the types of attracting laminations that arise. One possibility gives rise (again) to an an invariant rank 2 free factor. In Section 6, we show that the other possibility implies that the outer automorphisms are so-called almost toral relatively hyperbolic and the conjugacy problem for this class of outer automorphisms is established by the main result of [Reference Dahmani and TouikanDT21] in its particular case where the so-called peripheral subgroups are
$\mathbb {Z}\times \mathbb {Z}$
or
$\mathbb Z \rtimes \mathbb Z$
, the fundamental group of the Klein bottle.
Our examples
In order to explain our flowchart, we describe how to distinguish the conjugacy classes of the three automorphisms given at the start of this section.
To start with, each of the automorphisms is reducible. In Example 1.2, there are invariant free factors of rank one and two – namely,
$\langle a \rangle $
and
$\langle a,b \rangle $
. In Example 1.3, each puncture corresponds to an invariant rank one free factor. Finally, in Example 1.4. the torus with one hole is an invariant rank 2 free factor.
(However, in Example 1.3, if the punctures are permuted, then it is irreducible but not fully irreducible, which would entail that it could not be conjugate to the other two).
The next stage of the flowchart would construct a relative train track to determine growth. The automorphisms are essentially already presented in this way with Example 1.2 having quadratic growth, whereas Examples 1.3 and 1.4 have exponential growth. This would already mean that the first is not conjugate to the other two.
One can then find the free factor carrier of the attracting lamination for each of the exponential growth automorphisms, and one finds that in Example 1.3, it has rank three, whereas for Example 1.4, it is carried by the invariant rank two free factor.
Hence, these three automorphisms are not conjugate in
$\operatorname {Out}(F_3)$
.
We note that the flowchart is presented ‘efficiently’ by using a common algorithm for both quadratic polynomial growth automorphisms and for exponential growth automorphisms whose attracting lamination is carried by a rank two free factor. This algorithm – given by Theorem 3.1 – will indeed distinguish the conjugacy classes of those automorphisms, but our global algorithm already determines growth which is a conjugacy invariant.
Remark. We would like to comment that the methods used in this paper are somewhat broad but roughly break down into using mapping torus techniques and relative hyperbolicity versus train track methods and Culler-Vogtmann space.
In choosing which methods to use, we have tried to take the shortest routes that served our proof strategy rather than the most coherent ones. This unfortunately makes the background reading quite extensive. However, in producing this paper, we often produced multiple versions of the argument, and so we are aware that one could rewrite the proofs leaning more towards mapping tori techniques rather than train track techniques and vice versa. For instance, one could solve the conjugacy problem for the exponentially growing case using purely mapping torus techniques or using purely train track techniques. We have opted to mix and match and rely on existing results as far as possible.
The main place where these main techniques are not entirely applicable is in the quadratic growth case, Theorem 3.1, which is somehow a mapping torus situation in the absence of relative hyperbolicity.
2 Computability of a few invariants
Notations
We begin by some general notations and definitions.
Notation 2.1.
-
• In a group G, the conjugation of g by h is
$g^h:=h^{-1} g h$
. -
•
$\operatorname {Aut}(G)$
acts on the right: for
$\phi \in \operatorname {Aut}(G)$
,
$g \in G$
, the image of g under
$\phi $
is
$g \phi $
. -
•
$\operatorname {\mathrm {Ad}}(g)$
denotes the automorphism of G that is conjugation by g:
$x\operatorname {\mathrm {Ad}}(g)=x^g$
. Hence,
$\operatorname {\mathrm {Ad}}(g) \operatorname {\mathrm {Ad}}(h) = \operatorname {\mathrm {Ad}}(gh)$
. -
•
$\operatorname {Out}(G) = \operatorname {Aut}(G)/ \operatorname {Inn} (G)$
, for
$\operatorname {Inn} (G) = \{ \operatorname {\mathrm {Ad}}(g), g\in G\}$
. -
•
$F_3= \langle a,b,c \rangle $
denotes a free group of rank
$3$
. -
• We will prefer the notation K for free factors, and H for subgroups.
-
• We write outer automorphisms with capital greek letters, and automorphisms with small greek letter: if
$\Phi $
is an outer automorphism,
$\phi \in \Phi $
is an automorphism in its class. In this convention, (according to context)
$\mathrm {X}$
reads Chi, and
$\chi \in \mathrm {X}$
is an automorphism.
For
$\Phi \in \operatorname {Out}(F)$
, we say a subgroup
$H\leq F$
is
$\Phi $
-invariant if for any
$\phi \in \Phi $
, we have that H is conjugate to
$H\phi $
.
A free factor system of F is a set of conjugacy classes of subgroups of F,
$\{[K_1], \dots , [K_m]\}$
such that there exists a free subgroup
$F_r<F$
for which
$F= K_1 * \dots *K_m * F_r$
. It is proper if the
$K_i$
are neither
$\{1\}$
nor F. It is
$\Phi $
invariant if all
$K_i$
(not necessarily
$F_r$
) are
$\Phi $
invariants.
A free factor system
$\{[Y_1], \dots , [Y_s]\}$
is smaller than
$\{[K_1], \dots , [K_m]\}$
if for each i, there exists j such that
$Y_i$
has conjugate inside
$K_j$
.
Irreducibility
A first conjugacy invariant, and decidable, property is the irreducibility of outer automorphisms.
Definition 2.2. An outer automorphism
$\Phi $
is called reducible if it admits an invariant proper, nontrivial, free factor system; see [Reference Bestvina, Feighn and HandelBFH00]. Otherwise, it is said to be irreducible.
An automorphism is fully irreducible if every positive power is irreducible.
Remark 2.3. An automorphism is fully irreducible if and only if the only periodic free factors preserved up to conjugacy are given by either the trivial subgroup or the whole group. Every fully irreducible automorphism is irreducible, but not conversely.
An example of an irreducible automorphism which is not fully irreducible is as follows: take a surface with
$p>1$
punctures. Consider a pseudo-Anosov map on the surface which cyclically permutes the punctures. Then the outer automorphism induced on the fundamental group is irreducible but not fully irreducible. In fact, any example of an exponential growth automorphism which is irreducible but not fully irreducible arises in this way; see [Reference MutanguhaMut21] Corollary A.5.
Fully irreducible is often referred to as ‘iwip’ – irreducible with irreducible powers – in the literature.
There is an algorithm to detect whether
$\Phi \in \operatorname {Out}(F_n)$
is fully irreducible given in [Reference KapovichKap14], and whether it is irreducible in [Reference Francaviglia and MartinoFM22]. A more recent algorithm given in [Reference Kapovich and BellKB19] decides whether
$\Phi $
is fully irreducible in polynomial time.
Growth
A second set of conjugacy invariant, and decidable, properties is related to growth.
Definition 2.4. Given
$\phi \in \Phi \in \operatorname {Out}(F_n)$
and
$g \in F_n$
and a fixed basis X of
$F_n$
, we define the growth rate of g as follows:
-
• The element g has polynomial growth of degree at most d under
$\Phi $
if
$\|g\phi ^n\|_X = O(n^d)$
. -
• The element g has exponential growth if there exists
$\lambda>1$
for which
$ \lambda ^n = O(\|g\phi ^n\|_X )$
.Footnote
1
This growth rate is independent of the choice of X and
$\phi \in \Phi $
. Moreover, by [Reference Bestvina and HandelBH92], the growth rate of every
$g\in F_n$
is either at least exponential or at most polynomial.
Definition 2.5. An outer automorphism
$\Phi $
of
$F_n$
is said to have
-
• polynomial growth if all elements
$g\in F_n$
have polynomial growth rate, -
• exponential growth if some element
$g \in F_n$
has exponential growth rate.
In the case of a polynomial growth outer automorphism of
$F_n$
, there exists
$d\geq 0$
for which all elements of F have polynomial growth of degree at most d, and moreover, the smallest such d satisfies
$d \leq n-1$
; see [Reference Bestvina, Feighn and HandelBFH05], [Reference LevittLev09].
Definition 2.6. A polynomial growth outer automorphism of
$F_n$
is unipotent if it induces an automorphism of the abelianisation
$\mathbb {Z}^n$
that is represented by a unipotent matrix.
Such a matrix of the abelianisation of an automorphism of
$F_n$
of polynomial growth only has eigenvalues of modulus
$1$
, and in
$GL(n, \mathbb {Z})$
, so its
$|GL(n, \mathbb {Z}/3)|$
-th power is unipotent; see [Reference Bestvina, Feighn and HandelBFH00], Corollary 5.7.6.
The following is known to specialists and is important to us. We explain a proof, using the theory of CT maps [Reference Feighn and HandelFH18], that are certain homotopy equivalences on graphs, representing outer automorphisms on their fundamental groups; the interested reader is referred to this reference.
Proposition 2.7 [Reference Feighn and HandelFH11, Reference Feighn and HandelFH18].
Given an automorphism,
$\Phi \in \operatorname {Out}(F_n)$
, there is an algorithm to decide
-
• if
$\Phi $
has polynomial or exponential growth and, -
• if
$\Phi $
has polynomial growth, decide the degree of polynomial growth.
Proof. By [Reference Feighn and HandelFH18], there is an algorithm to construct a CT map (which is, in particular, a relative train track map) representing a rotationless power of
$\Phi $
. Since the properties needed here are invariant under taking positive powers, we may as well assume that
$\Phi $
itself is so-called rotationless (the power needed can be determined in advance). In particular, a rotationless automorphism of polynomial growth would be unipotent; see [Reference Feighn and HandelFH11, Lemma 4.2.2].
So take a CT map,
$f: G \to G$
representing
$\Phi $
.
It then follows immediately that if f admits an exponential stratum, then
$\Phi $
will have exponential growth. So we can algorithmically distinguish between exponential and polynomial growth.
So if
$\Phi $
has polynomial growth, then all strata will be NEG (non-exponentially growing); note that by [Reference Feighn and HandelFH11], Theorem 2.19, there will be no zero strata in the absence of exponential strata. Then Lemma 4.2 of [Reference Feighn and HandelFH11] shows that each such NEG stratum consists of a single edge, E, such that
$f(E) = E \cdot u$
, where the dot denotes a splitting. In particular, the growth of
$f^k(E)$
is polynomial of one degree higher than that of u. Hence, by induction, we can determine the growth of every edge.
This almost determines the growth of the automorphism,
$\Phi $
, in the sense that
$\Phi $
has polynomial growth of degree
$d>1$
if and only if some edge grows polynomially with degree d. However, it is possible to represent the identity map as a CT map where some edge grows linearly, and this is the only exception. See [Reference MacuraMac02, Lemma 2.16] and [Reference Andrew, Hughes and KudlinskaAHK22, Lemma 2.3].
However, since we have already passed to a rotationless power (in particular, it will be UPG), our automorphism will have growth of degree 0 if and only if it represents the identity and this is immediately checked.
For an outer automorphism of
$F_n$
of exponential growth, there exists a canonical finite collection of conjugacy classes of finitely generated subgroups of
$F_n$
, such that an element
$g\in F_n$
has polynomial growth rate if and only if it is conjugate in one of these subgroups [Reference LevittLev09]. Those are the maximal polynomially growing subgroups.
Given H such a maximal subgroup, and
$\phi \in \Phi $
, there is an integer
$m> 0$
and
$g\in F_n$
for which
$H( \phi ^m \operatorname {\mathrm {Ad}} (g))=H$
. The automorphism
$\phi ^m \operatorname {\mathrm {Ad}} (g)$
of H is then a polynomial growth automorphism. We again refer to [Reference LevittLev09]. Through standard arguments of quasi-isometry, its polynomial degree of growth rate does not depend on
$\phi \in \Phi $
, m and g satisfying the above. We thus call it the degree of growth of
$\Phi $
on H.
Proposition 2.8. There is an algorithm that, given
$\Phi \in \operatorname {Out}(F_n)$
of exponential growth, computes a basis for each conjugacy-representative of maximal polynomially growing subgroup of
$F_n$
, and the degree of growth of
$\Phi $
on it.
Proposition 2.8 follows from the relative hyperbolicity of
$F_n \rtimes \mathbb {Z}$
given by
$\Phi $
[Reference GhoshGho23, Reference Dahmani and LiDL20], and the computability of its peripheral subgroups [Reference Dahmani and GuirardelDG13]. This detects the maximal polynomially growing subgroups of
$F_n$
. Then, Proposition 2.7 determines the induced growth on each of them.
Lamination carriers are computable
A more involved conjugacy invariant for outer automorphisms of free groups that have exponential growth is that of the carrier of the attracting laminations [Reference Bestvina, Feighn and HandelBFH00, Section 3]. This is a specific conjugacy class of free factor systems of
$F_n$
that in some sense, attracts all the exponential growth of the automorphism.
More concretely, let
$\partial F_n$
, denote Gromov boundary of
$F_n$
, which for a free group is the same as the set of ends. Let
$\widetilde {\mathcal {B}} = (\partial F_n \times \partial F_n - \Delta )/{\mathbb Z}_2$
, where
$\Delta $
is the diagonal subset of
$\partial F_n \times \partial F_n$
, and the
${\mathbb Z}_2$
action is via interchanging the coordinates. That is,
$\widetilde {\mathcal {B}}$
can be thought of as the set of unordered pairs of distinct points on the boundary of
$F_n$
; morally, this is the set of lines in
$F_n$
. The action of
$F_n$
on its boundary extends to an action on
$\widetilde {\mathcal {B}}$
.
We then let
$\mathcal {B} = \widetilde {\mathcal {B}}/F_n$
. We say that
$\beta \in \mathcal {B}$
is carried by (the conjugacy class of) a free factor K if
$\beta $
lies in the image of
$\partial K \times \partial K - \Delta \to \mathcal {B}$
. A subset S of
$\mathcal {B}$
is carried by a free factor system,
$\mathcal {F}$
, if every element of S is carried by some free factor in
$\mathcal {F}$
.
An attracting lamination for
$\Phi $
is then a closed subset of
$\mathcal {B}$
which is the closure of a single point,
$\beta $
, and has some extra properties (
$\beta $
is birecurrent, admits an attracting neighbourhood for some positive iterate of
$\Phi $
, and is not carried by a
$\Phi $
-periodic free factor of rank one; see [Reference Bestvina, Feighn and HandelBFH00, Definition 3.1.5]).
Any automorphism,
$\Phi $
, then admits a finite set of attracting laminations,
$\mathcal {L}(\Phi )$
, which is carried by a
$\Phi $
-invariant free factor system ([Reference Bestvina, Feighn and HandelBFH00, Lemmas 3.1.6 and 3.1.13]). Morally, one has one attracting lamination for every exponential stratum for a relative train track representative for
$\Phi $
(or some power), and moreover,
$\mathcal {L}(\Phi )$
is canonical.
For instance, a fully irreducible automorphism will have a single attracting lamination which is only carried by the whole group. A polyonomially growing automorphism will have no attracting laminations.
Again, the theory of CT maps allows to compute this invariant. We would like to thank Mark Feign and Michael Handel for the proof of the following Proposition.
Proposition 2.9. Given an automorphism,
$\Phi \in \operatorname {Out}(F_n)$
of exponential growth, there is an algorithm to find the smallest free factor system which carries the set of attracting laminations.
Proof. Since the set of attracting laminations is stable under taking positive powers, we are free to replace
$\Phi $
with a (rotationless) power and use [Reference Feighn and HandelFH18] to algorithmically produce a CT map representing this power of
$\Phi $
. Since we are dealing with a rotationless automorphism, each exponential stratum produces exactly one attracting lamination. It is therefore sufficient to produce an algorithm which, given this CT map and a specific exponential stratum,
$H_k$
of it, finds the smallest free factor carrying
$\Lambda $
, the associated attracting lamination.
Let
$f : G \mapsto G$
be the CT map for the power of
$\Phi $
and let
$H_k$
be an exponential stratum with corresponding attracting lamination
$\Lambda $
. Construct a leaf L of
$\Lambda $
by choosing a point fixed by f in the interior of an edge e of
$H_k$
and considering
$$ \begin{align*}e \subset f(e) \subset f^2(e) \subset \ldots \end{align*} $$
Note that L is f-invariant.
Choose a segment of L that starts and ends with a copy of the same oriented edge of
$H_k$
and with e interior to the segment. Glue the initial and terminal edges of the segment to form a loop
$\gamma $
immersing to G. For
$i \ge 0$
, denote by
$\gamma _i$
the immersed loop (or conjugacy class)
$[f^i(\gamma )]$
. Note that
$\gamma _i$
is obtained from L by gluing segments of the form
$[f^i(edge)]$
. In particular, these segments are long if i is big.
Claim: The free factor support
$F(\Lambda )$
of
$\Lambda $
is equal to the free factor support,
$F(\cup _{i\ge 0} \{\gamma _i \})$
of
$$ \begin{align*}\cup_{i\ge 0} \{\gamma_i \}. \end{align*} $$
Comment: The free factor support of a finite collection of elements can be found algorithmically; see [Reference Feighn and HandelFH18, Lemma 4.2]. Moreover, the sequence of free factor supports of
$\{\gamma _1\}, \{\gamma _1, \gamma _2\}, \ldots $
stabilises.
This is because each free factor system has a well-defined complexity (consisting of the ranks of the components of the free factors which comprise it, ordered lexicographically; see [Reference Bestvina, Feighn and HandelBFH00], page 531). There are only finitely many complexities possible, and if one free factor system is strictly contained inside another, then it will have strictly smaller complexity; see [Reference Bestvina, Feighn and HandelBFH00], Lemma 2.6.3. This is exactly our situation since if we label the free factor support of
$\{ \gamma _1, \ldots , \gamma _r\}$
,
$\mathcal {F}_r$
, then we clearly get that
$\mathcal {F}_r \sqsubseteq \mathcal {F}_{r+1}$
. Moreover, we know that
$\mathcal {F}_r \sqsubseteq F(\Lambda )$
for all r and
$\mathcal {F}_r=F(\Lambda )$
as long as
$\mathcal {F}_r$
is
$\Phi $
-invariant, which is algorithmically decidable.
Hence, for an algorithm, it is enough to prove the claim.
Proof of Claim: That
$ F(\Lambda ) \subset F(\cup _{i\ge 0} \{\gamma _i \})$
is clear.
For the other direction, we will show that
$\gamma _i$
is contained in every
$\Phi $
-invariant free factor
$F'$
that carries
$\Lambda $
. It is enough to assume that i is large.
Let
$G'$
be a marked graph with sub-graph
$K'$
representing
$F'$
. Let
$\tilde K' \subset \tilde G'$
and
$\tilde G$
denote universal covers. View L (defined above) as a subset of
$\tilde G$
such that its representation in
$\tilde G'$
is a subset of
$\tilde K'$
. Let T denote the covering translation of G that identifies the segments of L that give
$\gamma _i$
.
If lines in
$\tilde G$
have long (depending on
$G'$
) overlap, then their representations in
$G'$
intersect – this is a consequence of the bounded cancellation lemma. Let
$L'$
denote the representation in
$\tilde G'$
of L and
$T'$
denote the covering translation of
$\tilde G'$
corresponding to T. In particular,
$L'$
and
$T'(L')$
intersect. Since
$L'$
is in
$\tilde K'$
and intersects
$T'(L')$
, the union of
$L'$
and
$T'(L')$
is contained in
$\tilde K'$
– this follows since translates of
$\tilde K'$
are either equal or disjoint.
More generally, the union as j ranges over the set of integers of
$T^{\prime }j(L')$
is connected, contained in
$\tilde K'$
, and
$T'$
-invariant. Hence,
$\gamma _i$
is carried by
$F'$
.
3 The case of a given invariant rank 2 free factor
For a free factor
$K\leq F$
, we define
$$ \begin{align*}\operatorname{Out}(F,K) = \{ \Phi \in \operatorname{Out}(F) : K \text{ is } \Phi\text{-invariant}\}.\end{align*} $$
For any
$\Phi \in \operatorname {Out}(F,K)$
, the restriction to K is not well-defined as an automorphism, but it is well-defined as an outer automorphism because K is its own normaliser in F.
This section is devoted to the proof of the following.
Theorem 3.1. Let
$K\leq F_3$
be a free factor of rank 2. Then the conjugacy problem in
$\operatorname {Out}(F_3,K)$
for pairs elements that induce infinite order outer automorphisms of K is decidable.
It will be enough to consider the case where
$K = \langle a,b\rangle < F_3= \langle a, b, c\rangle $
. Before the proof of Theorem 3.1, we establish some needed results.
Lemma 3.2. Let
$G=\operatorname {Out}(F_3,K)$
, and
$\mathrm {X} \in G$
. Then there exists a unique
$\chi \in \mathrm {X}$
such that there exists
$\epsilon = \pm 1, g \in \langle a,b \rangle $
and
$\chi _0 \in \operatorname {Aut}(\langle a, b \rangle )$
, satisfying
$$ \begin{align} \chi: \left\{ \begin{array}{ccl} c & \mapsto & g c^{\epsilon} \\ b & \mapsto & b\chi_0 \\ a & \mapsto & a \chi_0. \\ \end{array} \right. \end{align} $$
In particular, G is isomorphic to an iterated semi-direct product,
$C_2 \ltimes (\operatorname {Aut}(F_2) \ltimes F_2)$
, where the
$\operatorname {Aut}(F_2)$
action on
$F_2$
is the natural one, and the
$C_2$
action on
$\operatorname {Aut}(F_2) \ltimes F_2$
is given by the involution which maps
$(\chi _0, g)$
to
$(\chi _0 \operatorname {\mathrm {Ad}}(g), g^{-1})$
.
In the light of the decomposition of the lemma, we can represent any
$\operatorname {Out}(F_3,K)$
uniquely as an ordered triple,
$(\epsilon , \chi _0, g)$
.
Proof. The outer automorphism
$\mathrm {X}$
preserves the conjugacy class of
$K = \langle a, b\rangle $
; hence, an element of
$\mathrm {X}$
preserves
$\langle a, b\rangle $
. Note that this element is only well-defined up to the normaliser of
$\langle a, b\rangle $
, which is again
$\langle a, b\rangle $
.
Claim: If
$\chi ' \in \mathrm {X}$
sends
$a,b$
into
$\langle a,b\rangle $
, then it sends c into
$\langle a,b \rangle c^{\pm 1}\langle a,b\rangle $
.
Suppose this was not the case. Then the reduced word for
$c\chi '$
would have more than one occurrence of letters from the set
$\{c,c^{-1}\}$
. Since
$\chi '$
restricts to an automorphism of
$\langle a,b \rangle $
, there is
$\chi _0' \in \operatorname {Aut}(F_3,K)$
that maps
$(a,b,c)$
to
$(a\chi ',b\chi ',c)$
. Then
$\psi =(\chi _0')^{-1}\chi ' \in \operatorname {Aut}(F_3,K)$
maps the triple
$(a,b,c) \mapsto (a,b,c\chi ')$
. We apply Nielsen moves on this triple to cancel out the
$\langle a,b \rangle $
-prefix and suffix of
$c\chi '$
to get the tuple
$\vec t = (a,b,c^{\epsilon _1}\ldots c^{\epsilon _r})$
with
$\epsilon _1,\epsilon _r\in \{\pm 1\}$
. This tuple is Nielsen-reduced and clearly not Nielsen-equivalent to
$(a,b,c)$
, contradicting the fact that
$\vec t$
is a basis of
$F_3$
(see [Reference Lyndon and SchuppLS01, §I.2].) The claim now follows.
Thus,
$\mathrm {X}$
must send c into
$\langle a, b\rangle c^{\pm 1} \langle a, b\rangle $
, and therefore (after composing with another inner automorphism by an element of
$\langle a,b\rangle $
), an element of
$\chi \in \mathrm {X}$
sends c to
$gc^{\epsilon } \in \langle a, b\rangle c^{\epsilon }$
, for
$\epsilon \in \{-1, 1\}$
. An inner automorphism preserving
$\langle a, b\rangle $
and sending c in
$ \langle a, b\rangle c^{\pm 1}$
has to be trivial, and therefore, we obtain the first part of the statement. The set of
$\mathrm {X} \in G$
for which
$\epsilon = 1$
clearly forms a normal subgroup,
$G_1$
. In
$G_1$
, the set of elements for which
$\chi _0= Id$
is again a normal subgroup,
$G_2$
, isomorphic to
$\langle a, b\rangle $
. It is clear that the quotient
$G_1/G_2$
(isomorphic to
$\operatorname {Aut}(F_2)$
) lifts in
$G_1$
, making
$G_1 \simeq (\operatorname {Aut}(F_2) \ltimes F_2)$
. Finally,
$G/G_1 \simeq \mathbb {Z}/2$
also clearly lifts in G, as the automorphism given by
$\chi _0=1, g=1, \epsilon = -1$
, making G an iterated semi-direct product. One verifies that the involution on
$G_1$
given by the later lift is the one given in the statement.
Notation 3.3.
-
• Let G be a group and H a subgroup. For
$x,y \in G$
, we write
$x \sim _H y$
if there exists an
$h \in H$
such that
$x^h=y$
. We do not require
$x,y$
to be in H. -
• We say that
$\sim _H$
is decidable if we can construct an algorithm which decides on inputs
$x,y \in G$
, whether
$x \sim _H y$
. That is,
$\sim _H$
is decidable means that it is a recursive subset of
$G^2$
, and we have an explicit algorithm to decide membership.
Lemma 3.4. Let G be a group, H a subgroup of G and
$H_0$
a finite index subgroup of H. Then if
$\sim _{H_0}$
is decidable and we can compute a complete set
$h_1,\ldots ,h_k$
of coset representatives of
$H/H_0$
, then so is
$\sim _H$
.
Proof. Let
$h_1, \ldots , h_k$
be a computed set of coset representatives of
$H_0$
in H. Then
$x \sim _H y$
if and only if there exists an i such that
$x^{h_i} \sim _{H_0} y$
.
Hence, we can concurrently decide each problem
$x^{h_i} \sim _{H_0} y$
and therefore decide whether
$x \sim _H y$
.
Remark 3.5. Note that the finiteness of the index is crucial here. If one had an infinite recursive set of coset representatives,
$h_i$
, then it would still be true that
$x \sim _H y$
if and only if there exists an i such that
$x^{h_i} \sim _{H_0} y$
. However, we would also require a termination criterion for when to stop checking each conjugacy,
$x^{h_i} \sim _{H_0} y$
, since after a finite time, only finitely many of these checks can have been made.
To put it another way, suppose for some N that
$x^{h_i} \not \sim _{H_0} y$
, for
$1 \leq i \leq N$
. Then is that because
$x \not \sim _H y$
, or is it because
$x^{h_{N+1}} \sim _{H_0} y$
?
Remark 3.6. Some care needs to be taken here. It is false – there are counterexamples – that if the conjugacy problem is solvable in a finite index subgroup, then it is solvable in the group. That is, there are examples of groups G with finite index subgroups H such that
-
•
$\sim _H \cap H^2$
is recursive, -
•
$\sim _G$
is not recursive.
Hence, conjugacy is solvable in H but not in G. This does not contradict the Lemma above since we have the stronger hypothesis that
$\sim _H$
is decidable.
The main tool we will use to solve conjugacy in G is the twisted conjugacy problem.
Definition 3.7. Let
$\phi \in \operatorname {Aut}(F_n)$
. Then
$x, y \in F_n$
are said to be twisted
$\phi $
conjugate, written
$x \sim _{\phi } y$
if there exists a
$w \in F_n$
such that
$$ \begin{align*} x = (w \phi) y w^{-1}. \end{align*} $$
We need the following later:
Lemma 3.8. Let
$\phi \in \operatorname {Aut}(F_n)$
and
$x \in F_n$
. Then for any
$k \in {\mathbb Z}$
,
$x \sim _{\phi } x \phi ^k$
.
Proof. Simply notice that
$x\phi = (x \phi ) x x^{-1}$
and
$x \phi ^{-1} = ((x^{-1} \phi ^{-1}) \phi ) x (x \phi ^{-1})$
and that
$\sim _{\phi }$
is an equivalence relation.
Theorem 3.9 (Theorem 1.5 of [Reference Bogopolski, Martino, Maslakova and VenturaBMMV06]).
Let
$\phi \in \operatorname {Aut}(F_n)$
. Then the twisted conjugacy problem for
$\phi $
is solvable. That is,
$\sim _{\phi }$
is a recursive subset of
$F_n^2$
.
We will also need the following:
Theorem 3.10. The conjugacy problem is solvable in
$\operatorname {Out}(F_2)$
and
$\operatorname {Aut}(F_2)$
. Moreover, for any
$\Phi \in Out(F_2)$
of infinite order and for any
$\phi \in \Phi $
,
-
• the centraliser of
$\Phi $
in
$\operatorname {Out}(F_2)$
is virtually cyclic; -
• there is an algorithm that computes the coset representatives of
$\langle \Phi \rangle $
in its centraliser; -
• the centraliser of
$\phi $
in
$\operatorname {Aut}(F_2)$
has a finite index subgroup
$$ \begin{align*}C_0=\{ \phi^k \operatorname{\mathrm{Ad}}(g) \ : \ k \in {\mathbb Z}, \ g \in Fix(\phi)\} = \langle \phi \rangle \times \langle \operatorname{\mathrm{Ad}}(g) \ : \ g \in Fix(\phi) \rangle;\end{align*} $$
-
• the group
$D=\{ \chi \in \operatorname {Aut}(F_2) \ : \ \phi ^\chi = \phi \operatorname {\mathrm {Ad}}(h), h \in F_2\}$
has a finite index subgroup
$D_0 = \{ \phi ^k \operatorname {\mathrm {Ad}}(x) \ : \ k \in {\mathbb Z}, x \in F_2 \}$
.
Proof. We note that
$\operatorname {Out}(F_2) \cong GL_2({\mathbb Z})$
is virtually a free group of rank 2, and therefore word hyperbolic. Therefore, the conjugacy problem is solvable, and centralisers of infinite order elements are virtually cyclic (see [Reference Bridson and HaefligerBH99,
$\varGamma$
-2.12, and
$\varGamma$
-3 Coro. 3.10]). For any element of infinite order, g, in a hyperbolic group, it is known to be algorithmic to find the coset representatives of
$\langle g \rangle $
in its centraliser. This is folklore, but a detailed proof appears in Proposition 4.11 of [Reference Bogopolski, Martino and VenturaBMV10].
The solution for the conjugacy problem for
$\operatorname {Aut}(F_2)$
appears in [Reference BogopolskiBog00] and also as Corollary 5.2 of [Reference Bogopolski, Martino and VenturaBMV10].
The rest of the statements about
$\operatorname {Aut}(F_2)$
are then just translations of the corresponding statements about
$\operatorname {Out}(F_2)$
.
We now address the conjugacy problem in
$\operatorname {Out}(F_3,K)$
for elements that induce infinite order outer automorphisms on K.
Proof of Theorem 3.1.
Let
$G=\operatorname {Out}(F_3,K)$
, and
$\Phi , \Psi \in G$
, given by data, as in Lemma 3.2, as
$$\begin{align*}\Phi=(\epsilon_1, \phi_0, u)\text{ ~and~ }\Psi= (\epsilon_2, \psi_0, v).\end{align*}$$
We assume that
$\phi _0, \psi _0$
define infinite order outer automorphisms of K. We must decide whether there exists a conjugator,
$\chi = (\epsilon _3, \chi _0, h) \in G$
, such that
$\Phi ^{\chi } = \Psi $
. By Lemma 3.4, we may assume that
$\epsilon _3 =1$
; namely, we take
$H=G$
and
$H_0$
to be the index two subgroup of G where
$\epsilon = 1$
. We calculate the possibilities for
$\Phi ^\chi $
, listed as a triple:
$$ \begin{align} \begin{array}{ll} \text{If } \epsilon_1=1 & \Phi^\chi = (1, \phi_0^{\chi_0}, (h^{-1} \phi_0^{\chi_0}) (u \chi_0) h) \\ \\ \text{If } \epsilon_1=-1 & \Phi^\chi = (-1, \phi_0^{\chi_0} \operatorname{\mathrm{Ad}}(h), h^{-1} (h^{-1} \phi_0^{\chi_0}) (u \chi_0) ). \\ \end{array} \end{align} $$
Hence, if
$\epsilon _1 =1$
, then for
$\Phi $
and
$\Psi $
to be conjugate, we would require that
$\phi _0$
and
$\psi _0$
are conjugate. If
$\epsilon _1 =-1$
, they would need to be conjugate as outer automorphisms.
We next observe that if
$\epsilon _1=\epsilon _2 =1$
and
$\phi _0$
and
$\psi _0$
are conjugate in
$\operatorname {Aut}(F_2)$
, then there is a conjugator
$\chi \in \operatorname {Out}(F_3)$
so that the restriction of
$\Phi ^{\chi }$
to
$F_2$
equals the restriction of
$\Psi $
to
$F_2$
. This is clear from the equation above.
We also want to say that same in the case when
$\epsilon _1=\epsilon _2 = -1$
, and we argue as follows. Suppose
$\phi _0, \psi _0$
are conjugate in
$\operatorname {Out}(F_2)$
(more correctly, the outer automorphisms they represent are conjugate). Then there is a
$\chi _0 \in \operatorname {Aut}(F_2)$
and
$h \in F_2$
such that
${\phi _0}^{\chi _0} = \psi _0 \operatorname {\mathrm {Ad}}(h^{-1})$
. But then if we set
$\chi = (1, \chi _0, h)$
, we get that
$$ \begin{align*}\begin{array}{rcl} \Phi^{\chi} & = & (-1, {\phi_0}^{\chi_0} \operatorname{\mathrm{Ad}}(h), h^{-1} (h^{-1} \phi_0^{\chi_0}) (u \chi_0) ) \\ \\ & = & (-1, \psi_0, h^{-1} (h^{-1} \phi_0^{\chi_0}) (u \chi_0) ). \end{array} \end{align*} $$
Hence, if
$\Phi , \Psi $
are to be conjugate, we would require
$\epsilon _1 = \epsilon _2$
(otherwise, we know that they are not conjugate). Moreover, by solving conjugacy in
$\operatorname {Aut}(F_2)$
or
$\operatorname {Out}(F_2)$
, we may assume that
$\phi _0=\psi _0$
as automorphisms. To summarise, we have reduced the problem to the situation where
-
•
$\epsilon _1 = \epsilon _2$
, -
•
$\phi _0 = \psi _0$
in
$\operatorname {Aut}(F_2)$
.
Case 1: ϵ 1 = ϵ 2 = 1.:
In this case, we must have that
$\phi _0^{\chi _0} = \psi _0 = \phi _0$
. Hence,
$\chi _0$
centralises
$\phi _0$
. Let
$C=C_{\operatorname {Aut}(F_2)}(\phi _0)$
be the centraliser of
$\phi _0$
in
$\operatorname {Aut}(F_2)$
. Then
$\chi $
lies in the subgroup,
$C \ltimes F_2$
of G. By Lemma 3.4, it will be enough to solve the problem where we look at conjugators that lie in the subgroup,
$C_0 \ltimes F_2$
, where
$C_0$
is the finite index subgroup of C given by Theorem 3.10. Hence, we may assume that
$\chi _0 = \phi _0^k \operatorname {\mathrm {Ad}}(x)$
, where
$k \in {\mathbb Z}$
and
$x \in Fix(\phi _0)$
.
But then we get that
$$ \begin{align*}\Phi^\chi = (1, \phi_0, (h^{-1} \phi_0) (u \phi_0^k \operatorname{\mathrm{Ad}}(x)) h) = (1, \phi_0, ((h^{-1} x^{-1}) \phi_0) (u \phi_0^k) xh), \end{align*} $$
using the fact that
$x \phi _0 = x$
. Since
$\Psi = (1, \phi _0, v)$
, that means we are trying to decide whether there exist
$k \in {\mathbb Z}$
,
$h \in F_2$
,
$x \in Fix(\phi _0)$
such that
$$ \begin{align*}(h^{-1} x^{-1}) \phi_0 (u \phi_0^k) xh = v. \end{align*} $$
Putting
$h'=xh$
, this is equivalent to deciding
$$ \begin{align*}({h'}^{-1}) \phi_0 (u \phi_0^k) h' = v. \end{align*} $$
But by Lemma 3.8, this is equivalent to saying that
$u \sim _{\phi _0} v$
, which is decidable by Theorem 3.9.
Case 2: ϵ 1 = ϵ 2 = −1.:
As before, we have that
$\phi _0 = \psi _0$
. This time, by Equation 2, we get that
$\chi _0$
belongs to the subgroup D given by Theorem 3.10. Hence, by Lemma 3.4, we may assume that
$\chi _0 \in D_0$
. Hence,
$\chi _0 = \phi _0^k \operatorname {\mathrm {Ad}}(x)$
for some
$x \in F_2$
. Therefore, using Equation 2 and simplifying,
$$ \begin{align*}\Phi^\chi = (-1, \phi_0 \operatorname{\mathrm{Ad}}(w), h^{-1} (h^{-1} \phi_0 \operatorname{\mathrm{Ad}}(w h^{-1})) (u \chi_0) ), \end{align*} $$
where
$w = (x^{-1} \phi _0)xh$
. However, since we have reduced to the case where
$\phi _0= \psi _0$
, we get the extra restriction that
$w=1$
is equivalent to
$h=x^{-1} (x \phi _0)$
. And hence for
$\Phi $
and
$\Psi $
to be conjugate, we would need to simply equate the third entry of the triple:
$$ \begin{align*}(h^{-1} \phi_0) h^{-1} x^{-1} (u \phi_0^k) x = v. \end{align*} $$
Putting
$h=x^{-1} (x \phi _0)$
, this gives
$$ \begin{align*}(x^{-1} \phi_0^2) (x \phi_0) (x^{-1} \phi_0) (u \phi_0^k) x = (x^{-1} \phi_0^2) (u \phi_0^k) x = v. \end{align*} $$
Hence, this problem is equivalent by Lemma 3.8 to
$u \sim _{\phi _0^2} v$
or
$u \phi _0 \sim _{\phi _0^2} v$
, both of which are solvable by Theorem 3.9.
Remark 3.11. A natural question to ask is whether the strategy above can be generalized to higher rank free groups. We note that ingredients used in the proof above are
-
• the conjugacy problem in
$\operatorname {Aut}(F_2)$
. The conjugacy problem in
$\operatorname {Aut}(F_n)$
appears to be harder than the conjugacy problem in
$\operatorname {Out}(F_n)$
. -
• explicit computations of centralizers in
$\operatorname {Out}(F_2)$
. In our case, we were able to exploit the fact that
$\operatorname {Out}(F_2)$
is virtually free. It was shown in [Reference Bestvina, Feighn and HandelBFH97] that centralizers of irreducible automorphisms are virtually cyclic. As for more general automorphisms, the current state of the art for polynomially growing automorphisms [Reference RodenhausenRod13, Reference Rodenhausen and WadeRW15, Reference Andrew and MartinoAM22a] amounts to establishing finiteness properties.
For these reasons, the argument above does not obviously generalize.
4 Polynomially growing automorphisms
Before turning our attention to the quadratic growth case in
$F_3$
(i.e., the largest polynomial growth permitted there), let us record the combination of classical results on the conjugacy problem for low degree polynomial growth outer automorphisms of
$F_n$
.
Theorem 4.1 [Reference KrstićKrs89, Reference Krstí, Lustig and VogtmannKLV01].
The conjugacy problem in
$\operatorname {Out}(F_n)$
is decidable among outer automorphisms with growth of polynomial degree 0 or 1.
Recall also that the conjugacy problem among unipotent polynomially growing outer automorphisms of
$F_n$
has been solved by [Reference Feighn and HandelFH19].
Quadratic polynomial growth in
$F_3$
.
Every
$\Phi \in \operatorname {Out}(F_n)$
of polynomial growth has a power
$\Phi ^n$
that is unipotent (see [Reference Feighn and HandelFH18, Coro. 3.14]). In fact, one can take this power to be uniform for every n; the exponent (or order) of
$GL_n({\mathbb Z}_3)$
suffices, as one can define a UPG automorphism as one of polynomial growth which induces the trivial map in
${\mathbb Z}_3$
homology (see [Reference Bestvina, Feighn and HandelBFH05, Prop. 3.5]).
Definition 4.2. Two automorphisms
$\phi , \psi \in \operatorname {Aut}(F_n)$
are said to be isogredientif they are conjugate by an inner automorphism.
Recall that any outer automorphism of
$F_n$
is a coset of
$\operatorname {Inn}(F_n)$
in
$\operatorname {Aut}(F_n)$
. On each such outer automorphism, the relation of isogredience is an equivalence relation.
Theorem 4.3 (Bestvina-Handel Theorem, [Reference Bestvina and HandelBH92]).
Let
$\Phi \in \operatorname {Out}(F_n)$
. Then,
$$\begin{align*}\sum \max\{ \operatorname{rank}(\mathrm{Fix } \phi) - 1, 0 \} \leq n-1, \end{align*}$$
where the sum is taken over representatives,
$\phi $
, of isogredience classes in
$\Phi $
.
Theorem 4.4. Let
$\Phi \in \operatorname {Out}(F_n)$
have quadratic growth. Then,
$$\begin{align*}1 \leq \sum \max\{ \operatorname{rank}(\mathrm{Fix } \phi) - 1, 0 \} \leq n-2. \end{align*}$$
If
$n=3$
, this sum is exactly equal to 1 and exactly one isogredience class has a nonzero contribution to this sum.
Proof. This follows from results in either [Reference LevittLev09] or [Reference MartinoMar02]. A full discussion of this is also given in [Reference Andrew and MartinoAM22b].
More specifically, as argued in [Reference Andrew and MartinoAM22b, Theorems 2.4.6 and 2.4.8], any outer automorphism,
$\Phi $
of
$F_n$
, which satisfies the equality from 4.3, namely,
$$\begin{align*}\sum \max\{ \operatorname{rank}(\mathrm{Fix } \phi) - 1, 0 \} = n-1, \end{align*}$$
must have linear growth.
Lemma 4.5. Let
$\Phi \in \operatorname {Out}(F_2)$
be UPG. Then some automorphism in
$\Phi $
has fixed subgroup of rank 2.
Proof. It is easy to show that there is a basis of
$F_2$
,
$\langle a,b \rangle $
where (some automorphism of)
$\Phi $
acts as:
$a \mapsto a, b \mapsto b a^k$
for some integer k. Hence, the fixed subgroup is
$\langle a, bab^{-1} \rangle $
or
$\langle a, b \rangle $
when
$k=0$
.
Theorem 4.6. Let
$\Phi \in \operatorname {Out}(F_3)$
be a UPG automorphism of quadratic growth. Then
$\Phi $
admits a rank 2 invariant free factor, K which is algorithmically computable. Moreover, K is unique in the sense that if
$K_1$
is any other invariant rank 2 free factor, then
$K_1$
is conjugate to K.
Proof. By [Reference Bestvina, Feighn and HandelBFH05], there is a basis,
$a,b,c$
for
$\Phi $
such that some
$\phi \in \Phi $
has the following representation. In fact, by [Reference Feighn and HandelFH18], there is an algorithm to produce the following basis:
$$ \begin{align*} \begin{array}{rcl} & \phi & \\ a & \mapsto & a \\ b & \mapsto & ba^k \\ c & \mapsto & u c v, \end{array} \end{align*} $$
where
$k \in {\mathbb Z}$
and
$u, v \in \langle a,b \rangle $
and
$\phi \in \Phi $
.
Notice that
$k=0$
implies that
$\Phi $
has linear growth, and hence, we deduce that
$k \neq 0$
. In particular,
$\mathrm {Fix } \phi = \langle a, bab^{-1} \rangle $
(it cannot have higher rank due to Theorems 4.3 and 4.4). Hence,
$\langle a,b \rangle $
is the smallest free factor containing
$\mathrm {Fix } \phi $
.
Thus, we have algorithmically produced a
$\Phi $
-invariant rank 2 free factor,
$\langle a,b \rangle $
, and all that remains is to show is that it is unique.
If
$K_1$
were another
$\Phi $
-invariant rank 2 free factor, then the restriction of
$\Phi $
to
$K_1$
would again be UPG, and Lemma 4.5 would imply that some
$\psi \in \Phi $
would have a fixed subgroup of rank 2, contained in
$K_1$
. But Theorem 4.4 then implies that
$\phi $
and
$\psi $
are isogredient and hence have conjugate fixed subgroups implying that
$\langle a,b \rangle $
and
$K_1$
are also conjugate.
Corollary 4.7. Let
$\Phi \in \operatorname {Out}(F_3)$
be an automorphism of quadratic growth. Then
$\Phi $
admits a unique invariant rank 2 free factor. Moreover, there is an algorithm to determine this free factor.
Proof. Some positive power,
$\Phi ^k$
of
$\Phi $
, is UPG. By Theorem 4.6,
$\Phi ^k$
admits a unique invariant rank 2 free factor, K. But
$K \Phi $
is also
$\Phi ^k$
invariant, and hence, the uniqueness of K implies that K is conjugate to
$K \Phi $
.
Finally, if
$K_1$
is any
$\Phi $
-invariant rank 2 free factor, it must also be
$\Phi ^k$
-invariant, and hence, by Theorem 4.6 again,
$K_1$
is conjugate to K.
This free factor is produced algorithmically as in Theorem 4.6.
Conjugacy problem in polynomial growth for
$\operatorname {Out}{F_3}$
.
We may conclude for this section.
Recall (Proposition 2.7) that given
$\Phi $
and
$\Psi $
two outer automorphisms of
$F_3$
, we may decide whether they are of polynomial growth, and we may compute their degree.
Assume first that both are polynomial growth of degree
$2$
. We may compute by 4.7 the unique conjugacy classes
$[K_\Phi ], [K_\Psi ]$
of invariant free factor of rank 2 of
$F_3$
. After conjugating
$\Psi $
by a (computable) automorphism that sends
$K_\Phi $
to
$K_\Psi $
, we may assume that these two groups are equal. We may then apply Theorem 3.1 in order to decide whether
$\Phi $
and
$\Psi $
are conjugate in
$\operatorname {Out}{(F_3,K)}$
.
Since by Corollary 4.7, the rank 2 invariant free factor of
$\Phi $
is unique, the two outer automorphisms are conjugate in
$\operatorname {Out}{(F_3,K)}$
if and only if they are conjugate in
$\operatorname {Out}{(F_3)}$
.
This, together with Theorem 4.1, solves the conjugacy problem in
$\operatorname {Out}{F_3}$
for all polynomially growing outer automorphisms.
5 Exponential growth: laminations
Recall (references in Section 2, Proposition 2.7) that one can decide whether an outer automorphism is of exponential growth and whether it is irreducible. In that case, we recall the following.
Theorem 5.1 [Reference LosLos96, Reference LustigLus07, Reference KapovichKap14, Reference Francaviglia and MartinoFM22].
The conjugacy problem is decidable among irreducible elements of
$\operatorname {Out}(F_n)$
.
We will now focus on the reducible case.
Invariant free factor systems in
$F_3$
We will now work towards describing reducible exponentially growing automorphisms. Following [Reference Bestvina, Feighn and HandelBFH00], we have the following.
Proposition 5.2. Let
$\Phi \in \operatorname {Out}(F_3)$
. Then any relative train track representative of
$\Phi $
has at most one exponential stratum.
Proof. Suppose that
$f: \Gamma \to \Gamma $
is a relative train track representative of
$\Phi $
. Suppose that
$H_r$
is an exponential stratum and that
$G_r$
is the corresponding f-invariant subgraph (the union of all the strata, up to and including
$H_r$
).
Similarly, let
$G_{r-1}$
be the f-invariant subgraph consisting of the union of the strata up to, but not including,
$H_r$
.
Let
$C_r$
be a connected component of
$H_r$
and let
$C_{r-1}$
be a connected component of
$C_r \cap G_{r-1}$
(which we allow to be empty). Since the transition matrix of
$H_r$
is irreducible, some power of f must leave both
$C_r$
and
$C_{r-1}$
invariant. However, since
$H_r$
is an exponential stratum, the rank of
$\pi _1(C_r)$
is at least 2 more than the rank of
$\pi _1(C_{r-1})$
. (For instance, take an edge of
$H_r$
incident to
$C_{r-1}$
and take a power, k, of f such that
$f^k(e)$
crosses e at least three times). This immediately gives the result: either the rank of
$\pi _1(C_{r-1})$
is zero and the rank of
$\pi _1(C_{r})$
is at least two, in which case there can only be zero or polynomial strata above
$H_r$
, or the rank of
$\pi _1(C_{r-1})$
is one, and there can be no strata above
$H_r$
.
Corollary 5.3. Let
$\Phi \in \operatorname {Out}(F_3)$
have exponential growth. Then
$\Phi $
has exactly one attracting lamination. This lamination is carried by a
$\Phi $
-invariant conjugacy class of a free factor, K of
$F_3$
.
The rank of K is either 2 or 3. If the rank of K is 2, then it is unique in the following sense: if
$K_1$
is another free factor of rank 2 which is
$\Phi $
-invariant (up to conjugacy), then
$K_1$
is conjugate to K.
Proof. Some positive power of
$\Phi $
has an improved relative train track representative by [Reference Bestvina, Feighn and HandelBFH00, Theorem 5.1.5].
By the proof of Lemma 3.1.13 of [Reference Bestvina, Feighn and HandelBFH00], there is a one-to-one correspondence between exponential strata of an eg-aperiodic relative train track map (such as an improved relative train track map) and the attracting laminations of
$\Phi $
.
However, by Proposition 5.2, our improved relative train track for
$\Phi $
only admits one exponential stratum, and therefore,
$\Phi $
only has one attracting lamination which is therefore fixed by
$\Phi $
(and not just periodic for
$\Phi $
).
As in [Reference Bestvina, Feighn and HandelBFH00, Corollary 2.6.5 and Definition 3.2.3], there is a unique free factor whose conjugacy class carries the lamination of
$\Phi $
. Since the lamination is fixed by
$\Phi $
, this free factor is
$\Phi $
-invariant up to conjugacy.
For the final part, we note that an exponential stratum cannot have a component which has rank 1 as a graph (is homotopic to a circle), so K must have rank 2 or 3.
Furthermore, if K has rank 2, and if
$K_1$
is a
$\Phi $
-invariant free factor (up to conjugacy) also of rank 2, then we can form – via [Reference Bestvina, Feighn and HandelBFH00, Theorem 5.1.5] – an improved relative train track representative for some power of
$\Phi $
, where
$K_1$
is the fundamental group of some invariant subgraph,
$G_r$
. If the restriction of the relative train track map to
$G_r$
is polynomial, then the whole automorphism will have polynomial growth; this is excluded by hypothesis. Hence, by [Reference Bestvina, Feighn and HandelBFH00, Lemma 3.1.9], there is a lamination carried by (the conjugacy class of)
$K_1$
. But there is only one lamination for
$\Phi $
; hence, K and
$K_1$
are conjugate.
Conjugacy problem for exponential growth with rank
$2$
lamination
Consider
$\Phi $
and
$\Psi $
two outer automorphisms of
$F_3$
that are of exponential growth, and whose attracting lamination is carried by a free factor of rank
$2$
, respectively
$K_\Phi $
and
$K_\Psi $
.
By Proposition 2.9 (and the unicity of Corollary 5.3), the groups
$K_\Phi $
and
$K_\Psi $
can be computed.
Since both
$K_\Phi $
and
$K_\Psi $
are free factors of same rank, there exists (and one can compute) an automorphism
$\chi $
of
$F_3$
sending
$K_\Psi $
to
$K_\Phi $
, and after conjugating
$\Psi $
by the outer-class
$\mathrm {X}$
of
$\chi $
, we may assume that
$K_\Phi = K_\Psi $
, and we denote it K.
Since by Corollary 5.3, this invariant free factor is unique, the two automorphisms
$\Phi $
and
$\Psi $
are conjugated in
$\operatorname {Out}(F_3)$
if and only if they are conjugated in
$\operatorname {Out}(F_3, K)$
. This can be decided by Theorem 3.1.
6 Exponential growth: mapping tori
In this section, we take the point of view of analysing the semi-direct products of
$F_3$
that are associated to automorphisms (see also [Reference SelaSel95, Reference DahmaniDah16, Reference DahmaniDah17, Reference Dahmani and TouikanDT21]). Although this point of view allows to treat the conjugacy problem for all the exponentially growing outer automorphisms of
$F_3$
, we restrict our presentation to the remaining case in Flowchart 1 – namely, the case of automorphisms whose attracting lamination carrier is the entire group
$F_3$
.
Given
$\phi \in \operatorname {Aut}(F_n)$
, the associated mapping torus is
$F_n\rtimes _\phi \langle t \rangle $
. The normal subgroup
$F_n < F_n\rtimes _\phi \langle t \rangle $
is called a fiber. If
$\phi = \psi \operatorname {\mathrm {Ad}}(g)$
, then we have
$F_n\rtimes _{\psi }\langle t \rangle = F_n\rtimes _{\phi }\langle tg \rangle $
; in particular,
$\Phi =[\phi ]\in \operatorname {Out}(F_n)$
has a well-defined mapping torus. The following proposition describes how it relates to conjugacy in
$\operatorname {Out}(F_n)$
. We refer the reader to [Reference Dahmani and TouikanDT21] for precise definititions.
Proposition 6.1 (Standard mapping torus criterion for conjugacy).
Let G be a group,
$\Phi ,\Psi \in \operatorname {Out}(G)$
and
$\phi \in \Phi , \psi \in \Psi $
. Then
$\Phi $
is conjugate to
$\Psi $
in
$\operatorname {Out}(G)$
if and only if there is an isomorphism
$$\begin{align*}f: G \rtimes_\phi \langle t \rangle \to G \rtimes_\psi \langle t \rangle \end{align*}$$
such that
$f(G) = G$
and
$f(t) = tw$
for some
$w \in G$
.
The following theorem relates dynamical characteristics of outer automorphisms to the structure of their mapping tori.
Theorem 6.2 (Relative hyperbolicity in exponential growth).
For any
$\Phi \in \operatorname {Out}(F_n)$
, its mapping torus admits a properly relatively hyperbolic (possibly word-hyperbolic) metric if, and only if,
$\Phi $
has exponential growth. Peripheral subgroups of the relatively hyperbolic structure can be taken to be the mapping tori of
$\Phi $
restricted to maximal polynomially growing subgroups.
Proof. The converse implication was obtained in [Reference GhoshGho23, Theorem 3.1] and [Reference Dahmani and LiDL20, Theorem 4]. The direct implication is found in [Reference MacuraMac02], [Reference DahmaniDah17, Prop 1.3] see also [Reference HagenHag19], [Reference BrinkmannBri00, Theorem 1.2].
Using this strategy, in [Reference Dahmani and TouikanDT21], the conjugacy problem in
$\operatorname {Out}(F_n)$
is completely reduced to specific algorithmic problems in the peripheral subgroups. These problems are the algorithmic tractability for their subgroups (effective coherence, conjugacy problem, generation problem), the Minkowski property for certain subgroups, the mixed Whitehead problem, and the conjugacy problem for the induced automorphisms on maximal polynomially growing subgroups. In this reduction, the exponential growth part of the outer automorphisms is completely evacuated from the discussion (it is treated during the reduction).
In the case of exponentially growing automorphisms of
$F_3$
, the polynomially growing subgroups are sufficiently small that we can complete a solution of the conjugacy problem in their case. We will explain this in the remaining case of Flowchart 1, in which the polynomially growing subgroups are even simpler.
This will require two steps. The first step is giving the solution to the criterion of Proposition 6.1 in the case the mapping tori of the given autmorphisms are so-called almost toral. The second step is proving that if the mapping torus of an automorphism is not almost toral, then there is a rank 2 free factor carrying the attracting lamination. This allows to conclude since this later case was already treated. Treating it alternatively through the criterion of Proposition 6.1 is still possible and involves cases covered by the larger study [Reference Dahmani and TouikanDT23], but it is not done here.
The almost toral case
We say
$\Phi \in \operatorname {Out}(F_n)$
or its mapping torus
$F_n \rtimes _\phi \mathbb Z, \phi \in \Phi $
is almost toral relatively hyperbolic if
$F_n \rtimes _\phi \mathbb Z$
is hyperbolic relative to a collection of subgroups isomorphic to
$\mathbb Z \times \mathbb Z$
or to
$\mathbb Z \rtimes \mathbb Z$
, the fundamental group of a Klein bottle. By extension, in that case, we say that the automorphism is almost toral relatively hyperbolic.
In the relatively hyperbolic structure of Theorem 6.2, the peripheral subgroups are semi-direct products of free groups; hence, the groups
$\mathbb Z \times \mathbb Z$
and
$\mathbb Z \rtimes \mathbb Z$
are the only possible virtually abelian peripheral subgroups. Actually, the following is immediate from Theorem 6.2 and Nielsen-Schreier theorem.
Proposition 6.3. The mapping torus of an outer automorphism of
$F_n$
is almost toral if and only if its maximal polynomially growing subgroups have rank one.
It is furthermore decidable whether the mapping torus of a given automorphism of
$F_n$
is almost toral, by [Reference Dahmani and GuirardelDG13] (this is decidable in the following sense: there is an algorithm that will terminate if the automorphism is exponentially growing, and provide presentations for each conjugacy representative of peripheral subgroup and indicate whether or not they are abelian or isomorphic to
$\mathbb {Z}\rtimes \mathbb {Z}$
).
Theorem 6.4. The conjugacy problem in
$\operatorname {Out}(F_n)$
is decidable among the almost toral relatively hyperbolic automorphisms.
Proof. We are given
$\phi , \psi $
two automorphisms of
$F_n$
, such that
$F_n\rtimes _\phi \langle t\rangle $
and
$F_n\rtimes _\phi \langle t\rangle $
are relatively hyperbolic groups with peripheral subgroups either
$\mathbb {Z}^2$
or
$K= \mathbb {Z} \rtimes \mathbb {Z}$
.
By [Reference Dahmani and GuirardelDG13], we may compute the peripheral subgroups of both groups. More specifically, we may get a list of conjugacy representatives of these subgroups given by generating pairs.
If the numbers of conjugacy classes of peripheral subgroups in
$F_n\rtimes _\phi \langle t\rangle $
and in
$F_n\rtimes _\phi \langle t\rangle $
are different, then the two groups cannot be isomorphic, and
$\phi $
and
$ \psi $
cannot be conjugate in
$\operatorname {Out}(F_n)$
. We thus assume these numbers are equal.
If there are no peripheral subgroups in
$F_n\rtimes _\phi \langle t\rangle $
and in
$F_n\rtimes _\phi \langle t\rangle $
, these groups are hyperbolic, and we may use [Reference DahmaniDah16] to determine whether
$\phi $
and
$ \psi $
are conjugate in
$\operatorname {Out}{F_n}$
. We now assume that there are peripheral subgroups in both mapping tori.
Using [Reference Dahmani and TouikanDT21], in order to determine whether
$\phi $
and
$ \psi $
are conjugate in
$\operatorname {Out}(F_n)$
, it suffices to establish that these possible peripheral subgroups (i.e.,
$\mathbb {Z}^2$
and
$K= \mathbb {Z} \rtimes \mathbb {Z}$
) and their subgroups form an algorithmically tractable class of groups, with the (effective) Minkowski property, and with solvable fiber-and-orientation preserving mixed Whitehead problem.
First, notice that all the subgroups of
$\mathbb {Z}^2$
and K are themselves isomorphic to either
$\mathbb {Z}^2, K, \mathbb {Z}$
, or
$\{1\}$
, so we only discuss the properties for
$\mathbb {Z}^2$
and K. Algorithmic tractability is actually immediate (despite its definition).
We now address the Minkowski property for
$\mathbb {Z}^2$
and K. Recall that it means finding a characteristic finite quotient of them such that every finite order outer automorphism of
$\mathbb {Z}^2$
or K induces a non-inner automorphism in it.
For
$\mathbb {Z}^2$
, the quotient
$\mathbb {Z}^2 \to (\mathbb {Z}/3)^2$
is sufficient because (and this is the classic observation of Minkowski giving the name of the property)
$GL_2(\mathbb {Z}) \to GL_2(\mathbb {Z}/3)$
has a torsion-free kernel.
For K, let us write it as
$K= \langle a\rangle \rtimes \langle t \rangle $
. Recall that the derived subgroup is
$\langle a^2\rangle $
, and its abelianisation is
$K/\langle a^2\rangle \simeq (\mathbb {Z}/2) \times \mathbb {Z}$
. It follows that
$K\to (\mathbb {Z}/2) \times (\mathbb {Z}/3)$
(sending a to
$(1,0)$
and t to
$(0,1)$
) is a characteristic quotient. We will show that it is a suitable quotient for the Minkowski property.
Any automorphisms of K preserve
$\langle t^2\rangle $
(the center) and
$\langle a \rangle $
(the preimage of the torsion of the abelianisation). Since the set of square roots of
$t^2$
is
$\{a^k t, k\in \mathbb {Z} \}$
, the automorphisms of K are of the form
$( a\mapsto a^{\epsilon _1}, t\mapsto a^kt^{\epsilon _2})$
for
$\epsilon _1; \epsilon _2 \in \{-1, 1\}$
and
$k\in \mathbb {Z}$
. Since the automorphisms
$(a\mapsto a^{-1}, t\mapsto t )$
and
$(a\mapsto a, t\mapsto a^2 t)$
are both inner, it follows that
$$\begin{align*}\operatorname{Out}{K}\simeq (\mathbb{Z}/2) \times (\mathbb{Z}/2),\end{align*}$$
with representatives being
$(a\mapsto a, t\mapsto a^ut^{\epsilon })$
, for
$u \in \{0,1\}$
and
$ \epsilon \in \{-1,1\}$
. This shows that the proposed quotient is suitable for establishing the Minkowski property of K.
It remains to solve the fiber-and-orientation preserving mixed Whitehead problem for
$\mathbb {Z}^2 = \langle a\rangle \times \langle t \rangle $
, and for
$K = \langle a\rangle \rtimes \langle t \rangle $
(the fiber being
$\langle a\rangle $
). It is the problem of determining whether two tuples of conjugacy classes of tuples are in the same orbit for the group of automorphisms preserving
$\langle a\rangle $
and
$t\langle a \rangle $
.
In the case of
$\mathbb {Z}^2 = \langle a\rangle \times \langle t \rangle $
, this subgroup of automorphisms is
$\{ (a\mapsto a^{\epsilon }, t\mapsto a^kt), \epsilon \in \{-1,1\}, k\in \mathbb {Z} \}$
. The tuples of conjugacy classes of tuples are just tuples of elements in an abelianisation group. This immediately translates as a trivial divisibility problem for integers.
In the final case of
$K = \langle a\rangle \rtimes \langle t \rangle $
, the subgroup of fiber and orientation preserving automorphisms consists of only two outer classes (since
$\operatorname {Out}{K} \simeq (\mathbb {Z}/2) \times (\mathbb {Z}/2)$
). The orbit problem is then easily solved by applying the two automorphisms and using a solution to the conjugacy problem.
The polynomially growing subgroups in
$F_3$
By Levitt’s theorem [Reference LevittLev09, Theorem 4.1], given an outer automorphism of
$F_n$
, its maximal polynomially growing subgroups have rank
$\leq n$
, they have rank
$<n$
if the automorphism has an exponentially growing conjugacy class, and in the later case, if one such group has rank
$n-1$
, it is unique up to conjugacy.
In the case of
$n=3$
, the possible polynomially growing subgroups for an exponentially growing outer automorphism of
$F_3$
are then either trivial, or cyclic, or of rank
$2$
. Given the previous discussion, we consider the case of a single conjugacy class of polynomially growing subgroup of rank
$2$
.
It turns out, as we will show, that in this case, the polynomially growing subgroup must be placed very specifically in the group
$F_3$
, revealing a lamination carrier of rank 2. The next proposition shows in particular that all such automorphisms must be conjugated to automorphisms given in Example 1.4. Although this particular detail is not used in the solution to the conjugacy problem, it gives insights into how to represent the conjugacy classes of outer automorphisms that fall into this case.
Recall that if D is a
$\Phi $
-invariant subgroup, we can find
$\phi \in \Phi $
such that
$\phi (D)=D$
.
Proposition 6.5. If
$\phi \in \operatorname {Aut} (F_3)$
is of exponential growth and has an invariant subgroup D of rank at least
$2$
on which it induces a polynomially growing automorphism, then it preserves the conjugacy class of a rank
$2$
free factor Q on which it induces an exponentially growing outer automorphism.
Moreover, the conjugacy class of such a free factor is unique, and there exists a conjugate
$D'$
of D, containing a subgroup
$C'$
of
$Q \cap D'$
, such that
$C'$
is generated by the commutator of a basis in Q, and
$F_3= Q*_{C'}D'$
.
Proof. We assume without loss of generality that D is a maximal
$\phi $
-invariant subgroup on which
$\phi |_D$
is polynomially growing. By maximality, we cannot have
$D <H <F_3$
with H of rank 2 and
$\phi $
-invariant. Indeed, if this were the case, either
$\phi |_H$
is either polynomially growing, which contradicts maximality, or exponentially growing with a polynomially growing invariant subgroup of rank 2, which is impossible in
$\operatorname {Out}(F_2)$
. Furthermore, D cannot be a free factor of
$F_3$
, as the expression (1) from Lemma 3.2 would imply that
$\phi $
is also polynomially growing. It follows that
$(F_3,D)$
is relatively one-ended.
By [Reference Gaboriau, Jaeger, Levitt and LustigGJLL98, Theorem II.2],
$F_3$
acts faithfully on an
$\mathbb R$
-tree
$T_\infty $
with trivial arc stabilizers, and there is a homotethy
$H:T_\infty \to T_\infty $
such that for all
$x\in T_\infty $
and
$f \in F_3$
,
$f\cdot H(x) = H((f\phi ) x)$
(here, a homotethy is a map satisfying
$d(H(x),H(y)) = \lambda d(x,y)$
for some fixed stretch factor
$\lambda \geq 1$
). Note that from [Reference Gaboriau, Jaeger, Levitt and LustigGJLL98, §B - §E], the action of
$F_3$
on
$T_\infty $
is obtained as a limit of rescaled actions on a fixed free cocompact action of
$F_3$
on a simplicial metric tree
$\tau $
. Exponential growth of
$\phi $
implies that D fixes a point in
$T_\infty $
.
Trivial arc stabilizers and the nontrivial subgroup D (which cannot be in a proper free factor) that fixes a point (so the action is not free) imply that we may apply [Reference HorbezHor14, Lemma 4.6] to the action of
$F_3$
on
$T_\infty $
with the empty free factor system. This lemma gives three possibilities. Two of these imply that point stabilizers must all either be cyclic or contained in proper free factors of
$F_3$
, which is impossible because of the properties of D. The remaining possibility is that the action of
$F_3$
on
$T_\infty $
has a so-called dynamical proper free factor. This implies that
$T_\infty $
is not a simplicial
$\mathbb R$
-tree, since by definition, a dynamical proper free factor must act on its minimal invariant tree with dense orbits.
We now apply [Reference GuirardelGui08, Theorem 5.1]. The triviality of arc stabilizers, one-endeness of
$F_3$
relative to D, and the faithfullness of the action of
$F_3$
on
$T_\infty $
imply that the action of
$F_3$
on
$T_\infty $
decomposes into a graph of actions where each vertex action is either simpicial, Seifert type or axial. Free groups cannot admit faithful axial actions, and since
$T_\infty $
is not simplicial, the graph of actions must contain a Seifert type vertex. Orbifolds with free fundamental groups are surfaces with boundary. Therefore,
$F_3$
decomposes as a graph of groups Y, and one of the vertex groups
$Y_q$
must be isomorphic to the fundamental group of a surface
$\Sigma $
with boundary, equipped with a measured foliation
$\mathcal {F}$
with dense leaves, for which the action of
$Y_q$
on the minimal invariant subtree
$(T_\infty )_{Y_q}$
is equivariantly isomorphic to the action of
$\pi _1(\Sigma )$
on the
$\mathbb R$
-tree dual to lifted measured foliation on the universal cover
$(\tilde \Sigma ,\tilde {\mathcal {F}})$
. The subgroup D, elliptic in
$T_\infty $
, must also lie in some vertex group
$Y_D$
. Since D is a free group of rank 2 that fixes a point, it cannot be conjugate into the subgroup
$Y_q$
, so
$Y_q\neq Y_D$
.
By the definition of a graph of actions (see [Reference GuirardelGui08, §1.3]), the edge group incident to
$Y_q$
must be a point stabilizer, which in turn must either be trivial or conjugate to the
$\pi _1$
-image of
$\partial \Sigma .$
The former case gives a free decomposition of
$F_3$
relative to D which contradicts earlier considerations. The latter case implies that all the edges adjacent to q have cyclic edge groups. It also follows that if we collapse all edges of Y that have noncyclic edge group to get a new graph of groups
$\bar Y$
,
$Y_q$
is still a vertex group of this collapsed splitting and
$\bar Y$
has another vertex group containing D. It follows that
$F_3$
admits a nontrivial cyclic splitting relative to D containing
$Y_q$
as a quadratically hanging (QH) subgroup – that is to say, it is isomorphic to
$\pi _1(\Sigma )$
, where
$\Sigma $
is a compact surface with boundary and the (conjugacy classes) of the incident edge groups coincide with the
$\pi _1$
-image of
$\partial \Sigma $
.
Because
$F_3$
is one ended relative to D, by [Reference Guirardel and LevittGL17, Theorem 9.5],
$F_3$
admits a canonical cyclic JSJ decomposition J relative to D. By our description of
$\bar Y$
above, we know that J has at least two vertex groups and that one of them is a maximal QH subgroup. Arc stabilizers of the Bass-Serre tree dual to J are cyclic, and there are at least two vertex orbits so we can apply – for example, [Reference Gaboriau and LevittGL95] – to conclude that J has exactly two noncyclic vertex groups and that these vertex groups have rank exactly 2. One of these vertex groups contains (a conjugate of)
$Y_q$
and is a maximal QH subgroup. The other vertex group contains D. The automorphism invariance of JSJ decompositions, [Reference Guirardel and LevittGL17, Corollary 7.4], implies that
$\phi $
must preserve J (i.e.,
$\phi $
maps vertex groups and edge groups to conjugates of vertex and edge groups (respectively)). Since Q is the unique flexible subgroup of J, we must have that
$\phi (Q)$
is mapped to a conjugate of
$Q.$
Now the vertex group
$Y_q$
from the graph of actions Y sits inside the QH subgroup Q as the
$\pi _1$
-image of some subsurface since both groups have the same rank,
$Y_q=Q$
(up to conjugacy). By [Reference Culler, Vogtmann, Chern, Kaplansky, Moore, Singer and AlperinCV91], the only Seifert-type action of
$F_2$
is dual to an irrational foliation on a once punctured torus. It follows that the QH strucutre on Q is that of an orientable surface of genus 1 with one boundary component. Therefore, there is exactly one edge group incident to Q.
Let
$\bar J$
be the graph of groups obtained by contracting all edges except the unique edge adjacent to the unique QH vertex group Q. Then
$\bar J$
is the amalgamated product
$F_3=D'*_C Q$
.
$D'$
must contain a conjugate of D, and by construction, this splitting is
$\phi $
-invariant, so
$D'=D$
. Since C is conjugate to the
$\pi _1$
-image of
$\partial \Sigma $
in
$\pi _1(\Sigma )=Q$
, we have that Q is one-ended relative to C. Since
$F_3$
is many-ended, by [Reference ShenitzerShe55, Reference SwarupSwa86, Reference TouikanTou15], D is forced to admit a free decomposition
$D=\langle d \rangle * \langle c\rangle $
with
$C\leq \langle c \rangle $
. This gives
$$\begin{align*}F_3 = \underbrace{\langle d \rangle *(\langle c \rangle}_{D}*_C Q). \end{align*}$$
Since Q must be the fundamental group of a torus
$\Sigma $
with a boundary component, we have that C, the
$\pi _1$
-image of
$\partial \Sigma $
, must be a commutator and therefore cannot be a proper power; thus,
$C=\langle c \rangle $
, so Q is a free factor of
$F_3$
, as required. Its uniqueness follows from the canonicity of the JSJ decomposition.
By [Reference Gaboriau, Jaeger, Levitt and LustigGJLL98, Theorem II.1], if the stretch factor of the homotethy H is
$\lambda =1$
, then
$T_\infty $
is simplicial. Since that Q acts on
$(T_\infty )_Q$
with dense orbits, we have
$\lambda>1$
. Since up to composition with an inner automorphism we have
$\phi (Q)=Q$
, we have that
$\phi $
induces an exponentially growing automorphism on
$Q.$
In particular, we get the following.
Corollary 6.6. Let
$\Phi \in \operatorname {Out}(F_3)$
be exponentially growing, not fully irreducible, and such that no power has an invariant proper free factor of rank 2. Then the mapping torus
$ F_3 \rtimes _\phi \mathbb Z$
is almost toral relatively hyperbolic.
This concludes all cases of Flowchart 1, and thus proves Theorem 1.1.
Acknowledgements
We would like to thank Vincent Guirardel, Mark Feighn and Michael Handel for valuable discussions as well as the anonymous referee for their valuable input.
Competing interest
The authors have no competing interest to declare.
Funding statement
F.D. is supported by ANR-22-CE40-0004 GoFR. S.F. is supported by INdAM group GNSAGA and the European Union - NextGenerationEU under the National Recovery and Resilience Plan (PNRR) – Mission 4 Education and Research – Component 2 From Research to Business - Investment 1.1, Notice Prin 2022 issued with Decree No. 104 on 2/2/2022, with the title ‘Geometry and Topology of Manifolds’, proposal code 2022NMPLT8 - CUP J53D23003820001. N.T. is supported by an NSERC Discovery Grant.
Open access
