JSJ III: Flexible vertices are QH

June 07, 2025

JSJ III: Flexible vertices are QH

In this post we finish our discussion of the basic properties of JSJ decompositions following Guirardel-Levitt. This is based on a talk given by Hannah Boẞ.

Previously we discussed the elliptic vertices in a JSJ decomposition in some detail. Today we give an overview of the properties of the flexible vertices and discuss how, for JSJ decompositions over slender groups, they are all "surface-like" in a sense to be made precise below.

Orbifolds

In order to deal with groups with torsion, our notion of surface-like will have to include orbifolds. The reader who, like me, doesn't find orbifolds intuitive is welcome to assume the group is torsion-free and skip this section, replacing further instances of orbifold by surface.

Most compact 2-dimensional orbifolds $\Sigma$, including all those that will concern us, are Euclidean or hyperbolic.
Euclidean orbifolds whose fundamental group is not virtually cyclic have empty boundary, so they can only arise from flexible vertices in a trivial way. For instance, in the case of cyclic splittings of torsion-free groups, $\mathbb{Z}^2=\pi_1(T^2)$ and the Klein bottle group may appear as flexible vertex groups, but only as free factors (all incident edge groups must be trivial).

We therefore restrict to hyperbolic orbifolds: $\Sigma$ is a compact 2-dimensional orbifold, equipped with a hyperbolic metric with totally geodesic boundary. It is the quotient of a convex subset $\tilde\Sigma\subset \mathbb{H}^2$ by a proper discontinuous group of isometries
$G_\Sigma\subset \text{Isom}(\mathbb{H}^2)$ (isometries may reverse orientation); we denote by $p:\tilde\Sigma\to\Sigma$ the quotient map. By definition, the (orbifold) fundamental group of $\Sigma$ is $\pi_1(\Sigma)=G_\Sigma$.
We may also view $\Sigma$ as the quotient of a compact orientable hyperbolic surface $ \Sigma_0$ with geodesic boundary by a finite group of isometries $\Lambda$. A point of $\Sigma$ is singular if its preimages in $\tilde\Sigma$ (or in $\Sigma_0$) have non-trivial stabilizer.

If we forget the orbifold structure, $\Sigma$ is homeomorphic to a surface   $\Sigma_{top}$. The boundary of $\Sigma_{top}$ comes from the boundary $\partial \tilde\Sigma$, and from {mirrors} corresponding to reflections in $G_\Sigma$ (see below).
We define the boundary (in an orbifold) $\partial\Sigma$ of $\Sigma$ as the image of $\partial \tilde\Sigma$ in $\Sigma$ (thus excluding mirrors). Equivalently, it is the image of $\partial\Sigma_0$.
Each component $C$ of $\partial\Sigma$ is either a component of $\partial\Sigma _{top}$ (a circle) or an arc contained in $\partial\Sigma _{top}$.
The (orbifold) fundamental group of $C$ is $\mathbb{Z}$ or an infinite dihedral group $D_\infty$   accordingly.
A boundary subgroup is a subgroup $B\subset \pi_1(\Sigma)$ which is conjugate to the fundamental group of a component $C$ of $\partial \Sigma $.
Equivalently, it is the setwise stabilizer of a connected component of $\partial \tilde\Sigma$.

The closure of the complement of $\partial\Sigma$ in $\partial\Sigma_{top}$ is a union of mirrors: a mirror is the image of a component of the fixed point set of an orientation-reversing element of $\pi_1(\Sigma)$ in $\tilde\Sigma$.
Equivalently, a mirror is the image of a component of the fixed point set of an orientation-reversing element of $\Lambda$ in $\Sigma_0$.
Each mirror is itself a circle or an arc contained in $\partial\Sigma _{top}$. Mirrors may be adjacent in $\partial\Sigma_{top}$, whereas boundary components of $\Sigma$ are disjoint.

Singular points not contained in mirrors are conical points the stabilizer of their preimages in $\mathbb{H}^2$   is a finite cyclic group consisting of orientation-preserving maps (rotations). Points belonging to two mirrors are corner reflectors;the associated stabilizer is a finite dihedral group $D_{2r}$ of order $2r$.

As in the case of surfaces, compact hyperbolic 2-orbifolds may be characterized as those with negative Euler characteristic, defined as below:

Definition: The Euler characteristic $\chi(\Sigma)$ is defined as the Euler characteristic of the underlying topological surface $\Sigma_{top}$, minus contributions coming from the singularities:
a conical point of order $q$ (with isotropy group $\mathbb{Z}/q\mathbb{Z}$) contributes $1-\frac1q$,   a corner reflector with isotropy group the dihedral group $D_{2r}$ of order $2r$ contributes $\frac12(1-\frac1r)$, and a point adjacent to a mirror and a component of $\partial\Sigma$ contributes $\frac14$.

Much like the surface case, we also have splittings dual to closed geodesics. More precisely:

Proposition 1: Let $\Sigma$ be a compact hyperbolic 2-orbifold. Assume that $\pi_1(\Sigma)$ acts on a tree $T$ non-trivially, without inversions, minimally, with small edge stabilizers, and with all boundary subgroups elliptic.

Then $T$ is equivariantly isomorphic to the Bass-Serre tree of the splitting dual to a family $\mathcal{L}$ of disjoint essential simple closed geodesics of $\Sigma$.

If edge stabilizers are not assumed to be small, $T$ is still dominated by a tree dual to a family of geodesics.

Definition: Let $\Sigma$ be a compact hyperbolic $2$-orbifold, and let $\mathcal{C}$ be a non-empty collection of (non-disjoint) essential simple closed geodesics in $\Sigma$.
We say that $\mathcal{C}$ fills $\Sigma$ if the following equivalent conditions hold:

For every essential simple closed geodesic $\alpha$ in $\Sigma$, there exists $\gamma\in \mathcal{C}$ that intersects $\alpha$ non-trivially (with $\gamma\neq \alpha$).
For every element $g\in\pi_1(\Sigma)$ of infinite order that is not conjugate into a boundary subgroup, there exists $\gamma\in \mathcal{C}$ such that $g$ acts hyperbolically in the splitting of $\pi_1(\Sigma)$ dual to $\gamma$.
The full preimage $\tilde{\mathcal{C}}$ of $\mathcal{C}$ in the universal covering $\tilde{\Sigma}$ is connected.

Proof of equivalence: $(2)\Rightarrow(1)$ is clear, using a $g$ representing $\alpha$.

To prove $(1)\Rightarrow (3)$, consider a connected component $\mathcal{C}_0$ of $\tilde{\mathcal{C}}$, and its convex hull $A$ in $\tilde\Sigma\subset \mathbb{H}^2$.
If $\tilde{\mathcal{C}}$ is not connected, then $A\neq\tilde\Sigma$; indeed, $A$ is contained in a half space bounded by a geodesic in $\tilde{\mathcal{C}} \setminus \mathcal{C}_0$.
Let $\alpha$ be a connected component of the boundary of $A$ in $\tilde\Sigma$, a bi-infinite geodesic.
We note that $\alpha$ cuts no geodesic of $\tilde{\mathcal{C}}$. Indeed, if $\gamma\in\tilde{\mathcal{C}}$ cuts $\alpha$, then $\gamma\notin \mathcal{C}_0$, so all elements of $\mathcal{C}_0$ are disjoint from $\gamma$, and $A$ is contained in a half-space bounded by $\gamma$, contradicting $\alpha\subset \bar A$.
Now $\alpha$ projects to a simple closed geodesic in $\Sigma$ (if $g\alpha$ did intersect $\alpha$ transversely, then $A$ would be contained in the intersection of the half spaces bounded by $\alpha$ and $g\alpha$, a contradiction).
Assumption $(1)$ ensures that $\alpha$ is a boundary component of $\tilde\Sigma$. This implies that $A=\tilde\Sigma$, and that $\tilde{\mathcal{C}}$ is connected.

To prove $(3)\Rightarrow(2)$, consider $g\in\pi_1(\Sigma)$ of infinite order, and let $A(g)$ be its axis in $\tilde\Sigma$. If (2) does not hold, it intersects no geodesic in $\tilde{\mathcal{C}}$. By connectedness, $\tilde{\mathcal{C}}$ is contained in one of the half-spaces bounded by $A(g)$. It follows that the convex hull $A$ of $\tilde{\mathcal{C}}$ is properly contained in $\tilde\Sigma$.
Since $A$ is $\pi_1(\Sigma)$-invariant, this is a contradiction. $\blacksquare$

Corollary: If $\Sigma$ contains at least one essential simple closed geodesic, then the set of simple closed geodesics fills $\Sigma$.

Quadratically hanging subgroups

As usual, we fix $G$, $\mathcal{A}$, and $\mathcal{H}$. Let $Q$ be a subgroup of $G$.

Definition: We say that $Q$ is a QH subgroup (over $\mathcal{A}$, relative to $\mathcal{H}$) if:

$Q=G_v$ is the stabilizer of a vertex $v$ of an $(\mathcal{A},\mathcal{H})$-tree $T$;
$Q$ is an extension $1\to F\to Q \to \pi_1(\Sigma )\to 1$, with $\Sigma$ a compact hyperbolic 2-orbifold; we call $F$ the fibre, and $\Sigma$ the underlying orbifold;
each incident edge stabilizer, and each intersection $Q\cap gHg$ for $H\in \mathcal{H}$, is an extended boundary subgroup: by definition, this means that its image in $\pi_1(\Sigma)$ is finite or contained in a boundary subgroup $B$ of $\pi_1(\Sigma)$.

Condition (3) may be rephrased as saying that all groups in $\mathrm{Inc}^{\mathcal{H}}_v$ are extended boundary subgroups.

In general, the isomorphism type of $Q$ does not necessarily determine $F$ and $\Sigma$; when we refer to a QH subgroup, we always consider $F$ and $\Sigma$ as part of the structure.

A vertex $v$ as above, as well as its image in $\Gamma=T/G$, is called a QH vertex. It is the only point of $T$ fixed by $Q$ because extended boundary subgroups are proper subgroups of $Q$. Also note that the preimage in $Q$ of the finite   group carried by a conical point or a corner reflector of $\Sigma$ is an extended boundary subgroup.

Any incident edge stabilizer $G_e$ is contained in an extension of $F$ by a virtually cyclic subgroup of $\pi_1(\Sigma)$. But, even if $G$ is one-ended, $G_e$ may meet $F$ trivially, or have trivial image in $\pi_1(\Sigma)$. In particular, in full generality, $F$ does not have to belong to $\mathcal{A}$, or be universally elliptic if $T$ is a JSJ tree.

Definition: A splitting of $\pi_1(\Sigma)$ dual to a family of geodesics $\mathcal{L}$ induces a splitting of $Q$ relative to $F$. By the third condition in the definition of quadratically hanging, this splitting is also relative to $\mathrm{Inc}^{\mathcal{H}}_{|Q}$, so extends to a splitting of $G$ relative to $\mathcal{H}$ (this is a fact that we quote). We say that this splitting of $G$ is dual to $\mathcal{L}$(or determined by $\mathcal{L}$). The edge group associated to $\gamma\in\mathcal{L}$ is denoted $Q_\gamma$.

The edge groups $Q_\gamma$ are extensions of the fiber $F$ by $\mathbb{Z}$ or $D_\infty$, and in general they do not have to be in $\mathcal{A}$.
For many natural classes of groups $\mathcal{A}$, though, such extensions are still in $\mathcal{A}$, and we will see an example of this below.

Conversely, we have seen by Proposition 1 that small splittings of an orbifold group relative to the boundary subgroups are dual to families of geodesics. We will prove a similar statement for splittings of QH groups relative to incident edge stabilizers and groups in $\mathcal{H}_{ | Q}$ (i.e.\ to $\mathrm{Inc}^{\mathcal{H}}_{|Q}$), but one has to make   additional assumptions.

First, the fiber has to be elliptic in the splitting; this is not automatic in full generality, but this holds as long as $F$ and groups in $\mathcal{A}$ are slender. Next, we need to ensure that boundary subgroups are elliptic; this motivates the following definition.

Definition: A boundary component $C$ of $\Sigma$ is used if the group $B=\pi_1(C)$ (isomorphic to $\mathbb{Z}$ or $D_\infty$) contains with finite index the image of   an incident edge stabilizer or of a subgroup of $Q$ conjugate (in $G$) to a group in $\mathcal{H}$.

Equivalently, $C$ is used if there exists some subgroup $H\in \mathrm{Inc}^{\mathcal{H}}_v$ whose image in $\pi_1(\Sigma)$ is infinite and contained in $\pi_1(C)$ up to conjugacy.

Lemma 2: Let $C$ be a boundary component of a compact hyperbolic 2-orbifold $\Sigma$. There exists a non-trivial splitting of $\pi_1(\Sigma)$ over $\{1\}$ or $\mathbb{Z}/2\mathbb{Z}$ relative to the fundamental groups $B_k$ of all boundary components distinct from $C$.

Proof: Any arc $\gamma$ properly embedded in $\Sigma_{top}$ and with endpoints on $C$ defines a free splitting of $\pi_1(\Sigma)$ relative to the groups $B_k$. In most cases one can choose $\gamma$ so that this splitting is non-trivial. We study the exceptional cases:
$\Sigma_{top}$ is   a disc or an annulus, and $\Sigma$ has no conical point.

If $\Sigma_{top}$ is   a disc, its boundary circle consists of components of $\partial\Sigma$ and mirrors. Since $\Sigma$ is hyperbolic, there must be a mirror $M$ not adjacent to $C$ (otherwise $\partial\Sigma_{top}$ would consist of $C$ and one or two mirrors, or two boundary components and two mirrors, and $\chi(\Sigma)$ would not be negative).
An arc $\gamma$ with one endpoint on $C$ and the other on $M$ defines a splitting over $\mathbb{Z}/2\mathbb{Z}$, which is non-trivial because $M$ is not adjacent to $C$.

If $\Sigma_{top}$ is   an annulus, there are two cases. If $C$ is an arc, one can find an arc $\gamma$ from $C$ to $C$ as in the general case. If $C$ is a circle in $\partial\Sigma_{top}$, the other circle contains a mirror $M$ (otherwise $\Sigma$ would be a regular annulus) and an arc $\gamma$ from $C$ to $M$ yields a splitting over $\mathbb{Z}/2\mathbb{Z}$.

Remark: If the splitting constructed is over $K=\mathbb{Z}/2\mathbb{Z}$, then $\Sigma$ contains a mirror and $K$ is contained in an infinite dihedral subgroup (generated by $K$ and a conjugate).

Lemma 3: Let $Q=G_v$ be a QH vertex group of a tree $T$, with fiber $F$. If some boundary component is not used, then $G$ splits relative to $\mathcal{H}$ over a group $F'$ containing $F$ with index at most 2. Moreover, $T$ can be refined at $v$ using this splitting.

Proof sketch: Let $C$ be a boundary component of $\Sigma$. Lemma 2 yields a non-trivial splitting of $Q$ over a group $F'$
containing F with index    $\le2$ (with $F'=F$ if there is no mirror).
If $C$ is not used, this splitting is relative to $\mathrm{Inc}^{\mathcal{H}}_{ | Q}$. One may use it to refine $T$ at $v$ (this is where we are being vague). One obtains a splitting of $G$ relative to $\mathcal{H}$, which may be collapsed to a one-edge splitting over $F' \blacksquare$.

Proposition 1 implies:

Lemma 4: Let $Q $ be a QH vertex group, with fiber $F$.

Any non-trivial (minimal) splitting of $Q$ relative to $F$ factors through a splitting of $\pi_1(\Sigma)=Q/F$.
If the splitting is also relative to $\mathrm{Inc}^{\mathcal{H}}_{|Q}$, and every boundary component of $\Sigma$ is used, then the splitting of $\pi_1(\Sigma)$ is dominated by a splitting dual to a family of geodesics. In particular, $\Sigma$ contains an essential simple closed geodesic.
If, moreover, the splitting of $Q$ has small edge groups, the splitting of $\pi_1(\Sigma)$ is dual to a family of geodesics.

Remark: The induced splitting of $\pi_1(\Sigma)$ is relative to the boundary subgroups if (2) holds, but not necessarily if only (1) holds.

Proof: Let $S$ be the Bass-Serre tree of the splitting of $Q$. The group $F$ acts as the identity on $S$: its fixed point set is nonempty, and $Q$-invariant because $F$ is normal in $Q$. We deduce that the action of $Q$ on $S$ factors through an action of $\pi_1(\Sigma)$.

Under the assumptions of (2), this action is relative to all boundary subgroups because all boundary components of $\Sigma$ are used (by an incident edge stabilizer or a conjugate of a group in $\mathcal{H}$). We apply Proposition 1. $\blacksquare$

Flexible QH vertex groups $G_v$ of JSJ decompositions have non-trivial splittings relative to $\mathrm{Inc}^{\mathcal{H}}_v$. We wish to deduce that $\Sigma$ contains an essential simple closed geodesic, so as to rule out the small orbifolds which don't split

Proposition 5: Let $Q=G_v$ be a QH vertex stabilizer of a JSJ tree over $\mathcal{A}$ relative to $\mathcal{H}$. Assume that $F$, and any subgroup of $G_v$ containing $F$ with index 2, belongs to $\mathcal{A}$, and that $F$ is universally elliptic. Then:

every boundary component of $\Sigma$ is used;
if $Q$ is flexible, then $\Sigma$ contains an essential simple closed geodesic;
let $T$ be an $(\mathcal{A},\mathcal{H})$-tree such that $Q$ acts on $T$ with small edge stabilizers; if $Q$ is not elliptic in $T$, then the action of $Q$ on its minimal subtree is dual, up to subdivision, to a family of essential simple closed geodesics of $\Sigma$.

Proof: If some boundary component of $\Sigma$ is not used, Lemma 3 yields a refinement of the JSJ tree. The new edge stabilizers contain $F$ with index at most 2, so belong to $\mathcal{A}$ and are universally elliptic. This contradicts the maximality of the JSJ tree.

If $Q=G_v$ is flexible, it acts non-trivially on an $(\mathcal{A},\mathcal{H})$-tree. This tree is also relative to $F$ because $F$ is universally elliptic. This yields a splitting of $G_v$ relative to $F$, which is relative to $\mathcal{H}_{ | G_v}$ and to incident edge stabilizers because they are universally elliptic. Since every boundary component is used, we obtain a geodesic by Lemma 4.
Applying the previous argument to $T$ as in (3) shows that the splitting of $Q$ is dual to a family of geodesics by the third assertion of Lemma 4. $\blacksquare$

The first part of the following proposition shows that, conversely, the existence of an essential geodesic implies flexibility.

Proposition 6: Let $Q$ be a QH vertex group. Assume that $\Sigma$ contains an essential simple closed geodesic, and that, for all essential simple closed geodesics $\gamma$, the group $Q_\gamma$ belongs to $\mathcal{A}$.

Any universally elliptic element (resp.\ subgroup) of $Q$ is contained in an extended boundary subgroup. In particular, $Q$ is not universally elliptic.
If $J<Q$ is small in $(\mathcal{A},\mathcal{H})$-trees, its image in $\pi_1(\Sigma)$ is virtually cyclic.

Proof: Any subgroup of $Q$ contained in the union of all extended boundary subgroups is contained in a single extended boundary subgroup, so to prove (1) it suffices to show that, if an element $g$   does not lie in an extended boundary subgroup, then $g$ is not   universally elliptic.

The image of $g$ in $\pi_1(\Sigma)$ has infinite order, so acts non-trivially in a splitting of $\pi_1(\Sigma)$ dual to a geodesic $\gamma$. The splitting of $G$ dual to $\gamma$ is relative to $\mathcal{H}$, and the edge group $Q_\gamma$ is assumed to be in $\mathcal{A}$, so $g$ is not universally elliptic.

    If $J$ as in (2) has infinite image   in $\pi_1(\Sigma)$, it acts non-trivially in a splitting of $G$ dual to $\gamma$ as above. The action preserves a line or an end by smallness of $J$, so the image of $J$ in $\pi_1(\Sigma)$ is virtually cyclic (this might require a moment's thought to realise). $\blacksquare$

Proposition 5 requires $F$ to be universally elliptic. We show that this is automatic for splittings over slender groups.

Lemma 7: Let $Q$ be a QH vertex group. If $F$ and all groups in $\mathcal{A}$ are slender, then $F$ is universally elliptic.

Proof: Suppose that $F$ is slender but not universally elliptic. It acts non-trivially on some tree, and there is a unique $F$-invariant line $\ell$. This line is $Q$-invariant because   $F$ is normal in $Q$. It follows that $Q$ is an extension of a group in $\mathcal{A}$ by its image in $\mathrm{Isom}(\ell)$, a   virtually cyclic group. If groups in $\mathcal{A}$ are slender, we deduce that $Q$ is slender, a contradiction because $Q$ maps onto $\pi_1(\Sigma)$ with $\Sigma$ hyperbolic $\blacksquare$.

Corollary 8: Let $Q$ be a QH vertex group $G_v$ of a JSJ tree $T$. Assume that all groups in $\mathcal{A}$ are slender, and that every extension of the fiber $F$ by a virtually cyclic group belongs to $\mathcal{A}$.
Then:

$F$ is universally elliptic; it is the largest slender normal subgroup of $Q$;
all boundary components of the underlying orbifold $\Sigma$ are used;
if $G$ acts on a tree $S$, and $Q$ does not fix a point, then the action of $Q$ on its minimal subtree is dual to a family of essential simple closed geodesics of $\Sigma$;
$Q$ is flexible if and only if $\Sigma$ contains an essential simple closed geodesic;
if $Q$ is flexible, than any universally elliptic subgroup of $Q$ is an extended boundary subgroup.

Quadratically hanging subgroups are elliptic in the JSJ

The goal of this subsection is to prove that, under suitable hypotheses, any QH vertex group is elliptic in the JSJ deformation space (if we do not assume existence of the JSJ deformation space, we obtain ellipticity in every universally elliptic tree).

We start with the following fact:

Lemma 9: If $G$ splits over a group $K\in\mathcal{A}$, but does not split over any infinite index subgroup of $K$,
then $K$ is elliptic in the JSJ deformation space.

Proof: Apply Assertion 3 of Lemma 2 of the previous post with $T_1$ a JSJ tree and $T_2$ a one-edge splitting over $K$. $\blacksquare$

Remark 10: If $K$ is not universally elliptic, it fixes a unique point in any JSJ tree. Also note that being elliptic or universally elliptic is a commensurability invariant, so the same conclusions hold for groups commensurable with $K$.

It has been proved that, if $Q$ is a QH vertex group in some splitting (in the class considered), then $Q$ is elliptic in the JSJ deformation space. This is not true in general, even if $\mathcal{A}$ is the class of cyclic groups: $\F_n$ contains many
QH subgroups, none of them elliptic in the JSJ deformation space.
This happens because $G=\F_n$ splits over groups in $\mathcal{A}$ having infinite index in each other, something which is prohibited by the hypotheses of the papers which prove this result, $G$ is allowed to split over a subgroup of infinite index in a group in $\mathcal{A}$, but $Q$ has to be the enclosing group of minimal splittings.

There are also counterexamples with $\mathcal{A}$ the family of abelian groups and the QH subgroup $Q$ only has one abelian splitting, which is universally elliptic, so $Q$ is not elliptic in the JSJ space.

These examples explain the hypotheses in the following result.

Theorem 11: Let $Q$ be a QH vertex group. Assume that, if $\hat J\subset Q$ is the preimage of a virtually cyclic subgroup $J\subset \pi_1(\Sigma)$, then $\hat J$ belongs to $\mathcal{A}$ and $G$ does not split over a subgroup of infinite index of $\hat J$. If one of the following conditions holds, then $Q$ is elliptic in the JSJ deformation space:

$\Sigma$ contains an essential simple closed geodesic $\gamma$;
$\mathcal{H}=\emptyset$, and the boundary of $\Sigma$ is nonempty;
the fiber $F$ is elliptic in the JSJ deformation space;
all groups in $\mathcal{A}$ are slender.

Note that $ \hat J$ contains $F$, so $F\in\mathcal{A}$. The point of the proof is to use Lemma 4 to reduce to case 1, and then apply the following fact: Let $\tilde\gamma,\tilde\gamma'\subset \tilde\Sigma\setminus\partial\tilde\Sigma$ be lifts of simple geodesics. If $\tilde\gamma$ and $\tilde\gamma'$ intersect, then $c_{\tilde \gamma}=c_{{\tilde \gamma}'}$.

JSJ over slender groups

The main result of this section is the description of JSJ decompositions over slender groups. Recall that a subgroup $A\subset G$ is slender if $A$ and all its subgroups are finitely generated. Whenever $G$ acts on a tree, $A$ fixes a point or leaves a line invariant.

Our approach essentially follows Fujiwara and Paposoglu but with simplifications. In particular, we do not have to ``deal with a third splitting'' We now state the results.

Let $G$ be a finitely generated group, $\mathcal{A}$ a family of subgroups stable under conjugation and taking subgroups, and $\mathcal{H}$ a finite set of finitely generated subgroups of $G$ such that $G$ is finitely presented relative to $\mathcal{H}$.

The goal of this section is to show that non-slender flexible vertex groups $Q$ of JSJ decompositions over $\mathcal{A}$ relative to $\mathcal{H}$ are QH}. We need two assumptions on $\mathcal{A}$. First, groups in $\mathcal{A}$ should be slender. The second is a stability condition involving a subfamily $\mathcal{F}\subset \mathcal{A}$ (we will show that fibers of QH flexible vertex groups belong to $\mathcal{F}$).

Definition: We say that $\mathcal{A}$ satisfies the stability condition ($SC$), with fibres in a family of subgroups $\mathcal{F}$, if the following hold for every short exact sequence $$1\to F\to A\to K\to1$$ with $A< G$:

if $A\in \mathcal{A}$, and $K$ is isomorphic to $\mathbb{Z}$ or $D_\infty$, then $F\in\mathcal{F}$;
if $F\in \mathcal{F}$ and $K$ is isomorphic to a quotient of $\mathbb{Z}$ or $D_\infty$, then $A\in\mathcal{A}$.

The group $\mathbb{Z}$ acts on the line in an orientation-preserving way. The infinite dihedral group $D_\infty$ also acts on the line, but orientation is not preserved. In (2), the group $K$ may be finite (cyclic or dihedral) or infinite (isomorphic to $\mathbb{Z}$ or $D_\infty$).

If $Q$ is a QH subgroup with fiber $F$, recall that any simple geodesic $\gamma$ on $\Sigma$ defines a splitting of $G$ over a group $Q_\gamma$ which is an extension of $F$ by $\mathbb{Z}$ or $D_\infty$. If there is a geodesic $\gamma_0$ such that $Q_{\gamma_0}\in\mathcal{A}$, the stability condition ensures first that $F\in\mathcal{F}$, and then that $Q_\gamma\in\mathcal{A}$ for every $\gamma$, Corollary 8 and Theorem 11). Failure of the stability condition explains why Theorem 12below does not apply to abelian JSJ splittings.

On the other hand, one easily checks that the stability condition holds in the following cases:

$\mathcal{A}$=\{slender\} (i.e.\ $\mathcal{A}$ consists of all slender subgroups of $G$), with $\mathcal{F}=\mathcal{A}$;
$\mathcal{A}$=\{virtually cyclic\}, with $\mathcal{F}$=\{finite\};
$\mathcal{A}$=\{virtually polycyclic\}, with $\mathcal{F}=\mathcal{A}$;
$\mathcal{A}=\{VPC_{\leq n}\}$,the virtually polycyclic subgroups of Hirsch length at most $n$, with $\mathcal{F}$=\{VPC$_{\leq n-1}$\};
$G$ is a torsion-free CSA group, $\mathcal{A}$=\{finitely generated abelian\}, with $\mathcal{F}=\mathcal{A}$ (recall that a group is CSA if all maximal abelian subgroups are malnormal)

Theorem 12: Suppose that all groups in $\mathcal{A}$ are slender and $\mathcal{A}$ satisfies the stability condition ($SC$), with fibers in a family $\mathcal{F}$. Let $\mathcal{H}$ be a finite family of finitely generated subgroups. Let $G$ be finitely presented (or only finitely presented relative to $\mathcal{H}$).

If $Q$ is a non-slender flexible vertex group of a JSJ decomposition of $G$ over $\mathcal{A}$ relative to $\mathcal{H}$, then:

$Q$ is QH with fiber in $\mathcal{F}$ (it maps onto $\pi_1(\Sigma)$, where $\Sigma$ is a compact hyperbolic 2-orbifold, with kernel $F\in \mathcal{F}$; the image of an incident edge group in $\pi_1(\Sigma)$ is finite or contained in a boundary subgroup);
if $G$ acts on a tree and $Q$ does not fix a point, the action of $Q$ on its minimal subtree is dual to a family of geodesics of $\Sigma$;
$F$ and extended boundary subgroups are universally elliptic;
every boundary component of $\Sigma$ is used;
$\Sigma$ contains an essential closed geodesic;
every universally elliptic subgroup of $Q$ is an extended boundary subgroup.

Once (1) is known, Assertions (2)-(6) are consequences of the results proved above.

Corollary 13: Let $\mathcal{A}$ be the class of all slender subgroups of $G$. Let $\mathcal{H}$ be a finite family of finitely generated subgroups. Let $G$ be finitely presented (or only finitely presented relative to $\mathcal{H}$). Then every flexible vertex of any JSJ decomposition over $\mathcal{A}$ relative to $\mathcal{H}$ is either slender or QH with slender fiber.

The result by Fujiwara-Papasoglu is the case when $\mathcal{H}=\emptyset$ and $\mathcal{A}$ is the class of all slender groups. Dunwoody-Sageev consider (a generalisation of) the case when $\mathcal{H}=\emptyset$ and $\mathcal{A}=\{VPC_{\leq n}\}$; if $G$ does not split over a $VPC_{\leq n-1}$-subgroup, flexible vertex groups are QH with $VPC_{ n-1}$ fiber.

The class of all (finite or infinite) cyclic groups does not satisfy the stability condition ($SC$) if $G$ contains dihedral subgroups. To recover Rips and Sela's description of JSJ decompositions over cyclic groups, we introduce a modified stability condition $(SC_\mathcal{Z}$) preventing groups in $\mathcal{A}$ from acting dihedrally on a line.

We say that $\mathcal{A}$ satisfies the stability condition ($SC_\mathcal{Z}$), with fibers in $\mathcal{F}$, if the following hold:

no group of $\mathcal{A}$ maps onto $D_\infty$; given a short exact sequence $1\to F\to A\to \mathbb{Z}\to1$ with $A\in \mathcal{A}$, we have $F\in\mathcal{F}$;
given a short exact sequence $1\to F\to A\to K\to1$ with $F\in \mathcal{F}$ and $K$ isomorphic to a quotient of $\mathbb{Z}$, we have $A\in\mathcal{A}$.

This condition is satisfied by the classes consisting of

all cyclic subgroups,with $\mathcal{F}=\{1\}$,
all subgroups which are finite or cyclic, with $\mathcal{F}=\{1\}$,
all virtually cyclic subgroups which do not map onto $D_\infty$ (i.e.\ are finite or have infinite center), with $\mathcal{F}=$\{finite\},

and any class satisfying ($SC$) and consisting of groups which do not map onto $D_\infty$.

Theorem 14: Theorem 12 holds if $\mathcal{A}$ satisfies the stability condition $(SC_\mathcal{Z})$ rather than ($SC$). In this case the underlying orbifold $\Sigma$ has no mirror (all singular points are conical points).

In the next subsection we shall reduce Theorems 12 and 14 to Theorem 15.

Reduction to totally flexible groups

Unlike Fujiwara-Papasoglu, we know in advance that JSJ decompositions exist, and we only need to show that flexible vertex groups $G_v$ are QH. This allows us to forget $G$ and concentrate on $G_v$, but we have to remember the incident edge groups and we therefore consider splittings of $G_v$ that are relative to $\mathrm{Inc}^{\mathcal{H}}_v$: since incident edge groups are universally elliptic, these are exactly the splittings that extend to splittings of $G$ relative to $\mathcal{H}$. Even if we are interested only in non-relative JSJ decompositions of $G$ ($\mathcal{H}=\emptyset$), it is important here that we work in a relative context.

The fact that $G_v$ is a flexible vertex of a JSJ decomposition says that $G_v$ splits relative to $ \mathrm{Inc}^{\mathcal{H}}_v$, but not over a universally elliptic subgroup. This motivates the following general definition.

Definition: $G$ is totally flexible (over $\mathcal{A}$ relative to $\mathcal{H}$) if it admits a non-trivial splitting, but none over a universally elliptic subgroup. Equivalently, the JSJ decomposition of $G$ is trivial, and $G$ is flexible.

The example to have in mind is the fundamental group of a compact hyperbolic surface other than a pair of pants, with $\mathcal{A}$ the class of cyclic groups and $\mathcal{H}$ consisting of the fundamental groups of the boundary components.

Theorems 12 and 14 follow from the following result, which says that totally flexible groups are QH.

Theorem 15 Let $G$ be finitely presented relative to a finite family $\mathcal{H}$ of finitely generated subgroups.
Let $\mathcal{A}$ be a class of slender groups satisfying the stability condition $(SC)$ or ($SC_\mathcal{Z}$) with fibers in $\mathcal{F}$.

Assume that $G$ is totally flexible over $\mathcal{A}$ relative to $\mathcal{H}$, and not slender. Then $G$ is QH with fiber in $\mathcal{F}$, and $\Sigma$ has no mirror in the $SC_\mathcal{Z}$ case.

Since there are no incident edge groups, being QH means that $G$ is an extension of a group $F\in \mathcal{F}$ by the fundamental group of a hyperbolic orbifold $\Sigma$, and the image of each group $H\in\mathcal{H}$ in $\pi_1(\Sigma)$ is either finite or contained in a boundary subgroup. By Corollary 8, every component of $\partial \Sigma$ is used by a group of $\mathcal{H}$ (if $\mathcal{H}=\emptyset$, then $\Sigma$ is a closed orbifold).

This theorem implies Theorems 12 and 14: we apply it to $G_v$, with $\mathcal{A}=\mathcal{A}_v$ (the family of subgroups of $G_v$ belonging to $\mathcal{A}$) and $\mathcal{H}= \mathrm{Inc}^{\mathcal{H}}_{v}$. $ \mathrm{Inc}^{\mathcal{H}}_v$ turns out to be a finite family of finitely generated subgroups and $G_v$ finitely presented relative to $\mathrm{Inc}^{\mathcal{H}}_v$, and that $\mathcal{A}_{ v}$ satisfies $(SC)$ or $(SC_\mathcal{Z})$ with fibers in $\mathcal{F}_{ v}$.

There are three main steps in the proof of Theorem 15.

Given two trees $T_1,T_2$ such than no edge stabilizer of one tree is elliptic in the other, we follow Fujiwara-Papasoglu's construction of a core $\mathcal{C}\subset T_1\times T_2$. This core happens to be a surface away from its vertices. More precisely, if one removes from $\mathcal{C}$ its cut vertices and the vertices whose link is homeomorphic to a line, one gets a surface whose connected components are simply connected. The decomposition of $\mathcal{C}$ dual to its cut points yields a tree $R$ having QH vertices coming from the surface components. We call this tree $R$ the regular neighborhood of $T_1$ and $T_2$. The stability condition is used to ensure that edge stabilizers of $R$ are in $\mathcal{A}$.
Given a totally flexible $G$, we construct two splittings $U$ and $V$ of $G$ which ``fill'' $G$, and we show that their regular neighborhood is the trivial splitting of $G$. We deduce that $G$ itself is QH, as required (in the case of cyclic splittings of a torsion-free group, $G$ is the fundamental group of a compact surface and one should think of $U$ and $V$ as dual to transverse pair of pants decompositions).
The previous steps require that splittings of $Q$ be minuscule. This is a condition which controls the way in which $Q$ may split over subgroups $A,A'$ with $A'\subset A$, and we check that it is satisfied.

We will content ourselves with a description of the core.

As mentioned earlier, we let $\mathcal{H}$ be arbitrary and we only assume that $G$ is finitely generated. Groups in $\mathcal{A}$ are slender, and one of the stability conditions $(SC)$ or ($SC_\mathcal{z}$) is satisfied.

Minuscule splittings

Definition: Given $A,A'\in\mathcal{A}$, we say that $A'\ll A$ if $A'\subset A$ and there exists a tree $S$ such that $A'$ is elliptic in $S$ but $A$ is not. We also write $A\gg A'$. We say that $A\in\mathcal{A}$ is minuscule if, whenever a subgroup $A'\ll A$ fixes an edge $e$ of a tree, $A'$ has infinite index in the stabilizer $G_e$. Equivalently, $A$ is minuscule if and only if no $A'\ll A$ is commensurable with an edge stabilizer. A tree is minuscule if its edge stabilizers are minuscule. We say that all trees (or all splittings) are minuscule if all $(\mathcal{A},\mathcal{H})$-trees are minuscule.

If $A'\ll A$, then $A'$ has infinite index in $A$. If $A'\subset A\ll B\subset B'$, then $A'\ll B'$. In particular, the relation $\ll $ is transitive.

Being minuscule is a commensurability invariant. It is often used in the following way: if an edge stabilizer $G_e$ of a tree $T$ is elliptic in another tree $S$, then any minuscule $A$ containing $G_e$ is also elliptic in $S$.

Remark: If $G$ does not split (relative to $\mathcal{H}$) over a subgroup commensurable with a subgroup of infinite index of a group in $\mathcal{A}$, then every $A\in\mathcal{A}$ is minuscule (because, if $A$ is not minuscule, some $G_e$ has a finite index subgroup $A'$ contained in $A$ with infinite index).

The following lemma says that ellipticity is a symmetric relation among minuscule trees

Lemma 16 If $T_1$ is elliptic with respect to $T_2$, and $T_2$ is minuscule, then $T_2$ is elliptic with respect to $T_1$.

Proof: Let $\Hat T_1$ be a standard refinement of $T_1$ dominating $T_2$. Let $e_2$ be an edge of $T_2$, and $e$ an edge of $\Hat T_1$ with $G_e\subset G_{e_2}$. The group $G_e$ is elliptic in $T_1$, and since $G_{e_2}$ is minuscule $G_{e_2}$ itself is elliptic in $T_1$. This argument applies to any edge of $T_2$, so $T_2$ is elliptic with respect to $T_1 \blacksquare$.

Lemma 17: Assume that $\mathcal{H}$ is a finite family of finitely generated subgroups, $G$ is finitely presented relative to $\mathcal{H}$, and all trees are minuscule. There exists a tree $T$ which is maximal for domination: if $T'$ dominates $T$, then $T$ dominates $T'$ (so $T$ and $T'$ are in the same deformation space).

Finite presentation is necessary as evidenced by Dunwoody's inaccessible group with $\mathcal{H}=\emptyset$ and $\mathcal{A}$ the family of finite subgroups.

Example: If we consider cyclic splittings of the fundamental group of a closed orientable surface $\Sigma$, then $T$ is maximal if and only if it is dual to a pair of pants decomposition of $\Sigma$.

Proof: Let $\mathcal{T}$ be the set of all trees with finitely generated edge and vertex stabilizers, up to equivariant isomorphism. By Dunwoody's accessibility every tree is dominated by a tree in $\mathcal{T}$, so it suffices to find a maximal element in $\mathcal{T}$.
The set $\mathcal{T}$ is countable, so it suffices to show that, given any sequence $T_k $ such that $T_{k+1}$ dominates $T_k$, there exists a tree $T$ dominating every $T_k$. We produce inductively a tree $S_k$ in the same deformation space as $T_k$ which refines $S_{k-1}$. Start with $S_0=T_0$, and assume that $S_{k-1}$ is already defined.
Since $T_{k}$ dominates $T_{k-1}$, hence $S_{k-1}$, it is elliptic with respect to $S_{k-1}$. All trees being assumed to be minuscule, $S_{k-1}$ is elliptic with respect to $T_{k}$ by Lemma 16. We may therefore define $S_k$ as a standard refinement of $S_{k-1}$ dominating $T_k$. It belongs to the same deformation space as $T_k$ by Assertion 4 of Proposition 1 from the previous post. By Dunwoody's accessibility there exists a tree $T\in\mathcal{T}$ dominating every $S_k$, hence every $T_k \blacksquare$.

Definition of the core

Let $T_1$ and $T_2$ be trees. Recall that $T_1$ is elliptic with respect to $T_2$ if every edge stabilizer of $T_1$ is elliptic in $T_2$. If $T_1$ has several orbits of edges, it may happen that certain edge stabilizers are elliptic in $T_2$ and others are not (being slender, they act on $T_2$ hyperbolically, leaving a line invariant).
This motivates the following definition.

Definition: $T_1$ is fully hyperbolic with respect to $T_2$ if every edge stabilizer of $T_1$ acts hyperbolically on $T_2$.
This implies that edge stabilizers of $T_1$ are infinite, and no vertex stabilizer of $T_1$ is elliptic in $T_2$ (except if $T_1$ and $T_2$ are both trivial).

Example: Suppose that $G$ is a surface group and $T_1$, $T_2$ are dual to families of disjoint geodesics $\mathcal{L}_1$, $\mathcal{L}_2$. Then $T_1$ is elliptic with respect to $T_2$ if each curve in $\mathcal{L}_1$ is either contained in $\mathcal{L}_2$ or disjoint from $\mathcal{L}_2$.
It is fully hyperbolic with respect to $T_2$ if each curve in $\mathcal{L}_1$ meets $\mathcal{L}_2$ and every intersection is transverse.

Let $T_1$ be fully hyperbolic with respect to $T_2$. We consider the product $T_1\times T_2$. We view it as a complex made of squares $e_1\times e_2$, with the diagonal action of $G$. An edge of the form $e_1\times \{v_2\}$ is horizontal, and an edge $\{v_1\}\times e_2$ is vertical.

We define the asymmetric core $\mathcal{C}(T_1,T_2)$ as follows. For each edge $e$ and each vertex $v$ of $T_1$, let $\mu_{T_2}(G_e)$ and $\mu_{T_2}(G_v)$ be the minimal subtrees of $G_e$ and $G_v$ respectively in $T_2$ (with $\mu_{T_2}(G_e)$ a line since $G_e$ is slender).

Definition: Let $T_1$ be fully hyperbolic with respect to $T_2$. The asymmetric core $\mathcal{C}(T_1,T_2)\subset T_1\times T_2$ is
$$\mathcal{C}(T_1,T_2)=\left( \bigcup_{v\in V(T_1)} \{v\}\times \mu_{T_2}(G_v) \right) \cup \left(\bigcup_{e\in E(T_1)}e\times \mu_{T_2}(G_e)\right).$$

If we also assume that $T_2$ is fully hyperbolic with respect to $T_1$, we denote by $\check{\mathcal{C}}(T_1,T_2)\subset T_1\times T_2$ the opposite
construction: $$\check{\mathcal{C}}(T_1,T_2)=\left(\bigcup_{v\in V(T_2)}\mu_{T_1}(G_v)\times \{v\}\right) \cup
\left(\bigcup_{e\in E(T_2)} \mu_{T_1}(G_e)\times e\right).$$

When no confusion is possible, we use the notations $\mathcal{C}$, $\check{\mathcal{C}}$ instead of $\mathcal{C}(T_1,T_2)$, $\check{\mathcal{C}}(T_1,T_2)$. Note that $\mathcal{C}$ consists of all $(x,y)\in T_1\times T_2$ such that $y$ belongs to the minimal subtree of $G_x$. It is an important fact that when two miniscule trees are fully hyperbolic with respect to each other, $\mathcal{C}$ = $\check{\mathcal{C}}$

Every $\mu_{T_2}(G_v) $, $ \mu_{T_2}(G_e)$ being a non-empty subtree, with $ \mu_{T_2}(G_e)\subset \mu_{T_2}(G_v) $ if $v$ is an endpoint of $e$, a standard argument shows that $\mathcal{C}(T_1,T_2)$ is simply connected.

The group $G$ acts diagonally on $T_1\times T_2$, and $\mathcal{C}$ is $G$-invariant. Since $G_e$ acts cocompactly on the line $\mu_{T_2}(G_e)$ for $e\in E(T_1)$, there are finitely many $G$-orbits of squares in $\mathcal{C}$.

Remark: If $H\subset G$ is elliptic in $T_1$ and $T_2$, it fixes a vertex in $\mathcal{C}$ (because $H$ fixes some $v_1\in T_1$, and being elliptic in $T_2$ it fixes a point in the $G_{v_1}$-invariant subtree $\mu_{T_2}(G_{v_1})\subset T_2$).

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On the shoulders of giants