JSJ II: the JSJ deformation space

Today we will see that although there is no canonical splitting of a group, one can meaningfully talk about the set of all splittings, which is organised in a sensible way into an object known as the JSJ deformation space. This is based on a talk by Achille de Ridder, following Guirardel-Levitt.

We will see a lot of groups acting on trees, so let's say once and for all that we consider groups acting on simplicial/combinatorial trees without edge inversions, i.e. if the group action on the geometric realisation fixes an edge setwise it fixes the edge pointwise.

We say that the action of a group $G$ on a tree $T$ is trivial if there is a global fixed point for the action and minimal if there is no proper $G-$invariant subtree. Unless otherwise indicated, all trees which are not a single point will be endowed with a minimal action of $G$ without inversion.

To motivate the definition of the JSJ deformation space, let us first consider free decompositions of a group $G$,
\ie decompositions of $G$ as the fundamental group of a graph of groups with trivial edge groups,
or equivalently   actions of $G$ on a simplicial  tree $T$ with trivial edge stabilizers.

Let $G=G_1*\dots * G_p *F_q$ be a Grushko decomposition, as defined above ($G_i$ is non-trivial, not $\mathbb{Z}$, freely indecomposable). One may view $G$ as the fundamental group of one of the graphs of groups The corresponding Bass-Serre trees have trivial edge stabilizers, and the vertex stabilizers are precisely the conjugates of the $G_i$'s; we call a tree with these properties a Grushko tree (if $G$ is freely indecomposable, Grushko trees are points).

Since the $G_i$'s are freely indecomposable, Grushko trees $T_0$ have the following maximality property: if $T$ is any tree on which $G$ acts with trivial edge stabilizers, $G_i$ fixes a point in $T$, and therefore $T_0$ dominates $T$ in the sense that there is a $G$-equivariant map $T_0\to T$. In other words,
among free decompositions of $G$, a Grushko tree $T_0$ is as  far  as possible from the trivial tree (a point): its vertex stabilizers are as small as possible (they are conjugates of the $G_i$'s). This maximality property does not determine $T_0$ uniquely, as it is shared by all Grushko trees; hence even in this simple case we need to consider the set of all splittings if we want a canonical object.

Groups acting on trees

In general, we want to be able to control the way certain classes of subgroups act on trees: e.g. in the 3-manifold case we wanted to split over tori and by Bass-Serre theory this is equivalent to asking for the tori to fix an edge when acting on the tree. We make these notions precise:

An element $g$ or a subgroup $H$ of $G$ is elliptic in $T$ if it fixes a point. This is equivalent to $g$ or $H$ being contained in a conjugate of a vertex group of $\Gamma$. We    denote by $Fix (g)$ or $Fix (H)$ its fixed point set in $T$. If $H$ is elliptic, any $H$-invariant subtree meets $Fix (H)$.
 If $H_1\leq H_2\leq G$ with $H_1$ of finite index in $H_2$, then $H_1$ is elliptic if and only if $H_2$ is.

Besides $G$, we usually also fix a (nonempty)  family $\mathcal{A}$of subgroups of $G$ which is stable under conjugation and under taking subgroups. An $\mathcal{A}$-tree is a tree $T$ whose edge stabilizers belong to $\mathcal{A}$. We often say that $T$, or the corresponding splitting of $G$, is over $\mathcal{A}$, or over groups in $\mathcal{A}$. We say cyclic tree (abelian tree, slender tree, \dots) when $\mathcal{A}$ is the family of cyclic (abelian,   \dots) subgroups.

We also fix an arbitrary set $\mathcal{H}$ of subgroups of $G$, and we restrict to $\mathcal{A}$-trees $T$ such that each $H\in\mathcal{H}$ is elliptic in $T$  (in terms of graphs of groups, $H$ is contained in a conjugate of a vertex group;  if $H$ is not finitely generated, this is stronger than requiring that every $h\in H$ be elliptic). We call such a tree an $(\mathcal{A}, \mathcal{H})$-tree, or a tree over $\mathcal{A}$ relative to $\mathcal{H}$.

The set of $(\mathcal{A}, \mathcal{H})$-trees does not change if we replace a group of $\mathcal{H}$ by a conjugate, or if we enlarge $\mathcal{H} $ by making it invariant under conjugation.
 

If $G$ acts non-trivially on an $(\mathcal{A},\mathcal{H})$-tree, we say that $G$splits over $\mathcal{A}$ (or over a group of $\mathcal{A}$) relative to $\mathcal{H}$}.
The group $G$ is freely indecomposable relative to $\mathcal{H}$ if it does not split over the trivial group relative to $\mathcal{H}$. Equivalently (unless $G=\mathbb{Z}$ and $\mathcal{H}$ is trivial), one cannot write $G=G_1*G_2$ with $G_1,G_2$ non-trivial, and every group in $\mathcal{H}$ contained in a conjugate of $G_1$ or $G_2$.
 
 One says that $G$ is one-ended relative to $\mathcal{H}$ if $G$ does not split over a finite group relative to $\mathcal{H}$

Maps between trees

We now define how to compare trees (with their $G-$actions). Maps between trees will always be $G$-equivariant, send vertices to vertices and edges to edge paths (maybe a point). By minimality of the actions, they are always surjective;
each edge of $T'$ is contained in the image of an edge of $T$.
Any edge stabilizer $G_{e'}$ of $T'$ contains an edge stabilizer $G_e$ of $T$.
Also note that any edge or vertex stabilizer of $T$ is contained in a vertex stabilizer of $T'$.

We mention two particular classes of maps.
 
Definition: A map $f:T\rightarrow T'$ between two trees is a morphism if and only if one may subdivide $T$ so that $f$ maps each edge onto an edge; equivalently,
no edge of $T$ is collapsed to a point. If $f$ is a morphism, any edge stabilizer of $T$ is contained in an edge stabilizer of $T'$. 

If you know what this is, Stallings foldings are a good example of morphisms.

A collapse map $f: T\to T'$ is a map obtained by collapsing certain edges to points, followed by an isomorphism (by equivariance, the set of collapsed edges is $G$-invariant).
Equivalently, $f$ preserves alignment: the image of any arc $[a,b]$ is a point or  the arc $[f(a),f(b)]$. Another characterization is that the preimage of every subtree is a subtree. In terms of graphs of groups, one obtains $T'/G$ by collapsing edges in $T/G$. If $T'$ is irreducible, so is $T$. If $T$ is irreducible, one easily checks that  $T'$ is  trivial (a point) or irreducible.  A tree $T'$ is a collapse of $T$ if there is a collapse map  $T\rightarrow T'$; conversely, we say that $T$ refines $T'$.
In terms of graphs of groups, one passes from $\Gamma=T/G$ to $\Gamma'=T'/G$ by collapsing edges; for each vertex $v$ of $\Gamma'$,
the vertex group $G_{v}$ is the fundamental group of the graph of groups $\Gamma_{v}$ occurring as the preimage of $v$ in $\Gamma$.

Conversely, suppose that $v$ is a vertex of a splitting $\Gamma'$, and $\Gamma_{v}$ is  a splitting of $G_{v}$ in which incident edge groups are elliptic. One may then refine $\Gamma'$ at $v$ using $\Gamma_{v}$, so as to obtain a splitting $\Gamma$ whose edges are those of $\Gamma'$ together with those of $\Gamma_{v}$ . Note that $\Gamma$ is not uniquely defined because there is flexibility in the way edges of $\Gamma'$ are attached to vertices of $\Gamma_{v}$.
Two trees $T_1,T_2$ are compatible if they have a common refinement: i.e. there exists a tree $\hat {T}$ with collapse maps $g_i:\hat {T}\to T_i$. There is such a $\hat{T}$ with the additional property that no edge of $\hat{ T}$ gets collapsed in both $T_1$ and $T_2$.

Domination

A tree $T_1$ dominates another tree $T_2$ if there is an equivariant map $f:T_1\rightarrow T_2$ from $T_1$ to $T_2$. We call $f$ a domination map. Equivalently, $T_1$ dominates $T_2$ if every vertex stabilizer of $T_1$ fixes a point in $T_2$: every subgroup which is elliptic in $T_1$ is also elliptic in $T_2$. In particular, every refinement of $T_1$ dominates $T_1$. Beware that domination is defined by considering ellipticity of subgroups, not just of elements (this may make a difference if vertex stabilizers are not finitely generated).

 Deformation spaces are defined by saying that two  trees   belong to the same deformation space $\mathcal{D}$ if they have the same elliptic subgroups (ie each one dominates the other).  When we restrict to $\mathcal{A}$-trees, we say that $\mathcal{D}$ is a deformation space over $\mathcal{A}$. If a tree in $\mathcal{D}$ is irreducible, so are all others, and we say that $\mathcal{D}$ is irreducible.

A deformation space $\mathcal{D}$  {dominates} a space $\mathcal{D}'$ if trees in $\mathcal{D}$ dominate those of $\mathcal{D}'$.  Every deformation space dominates the deformation space of the trivial tree, which  is called the trivial deformation space. It is the only deformation space in which $G$ is elliptic. 

A tree $T$ is reduced if no proper collapse of $T$ lies in the same deformation space as $T$.. Observing that the inclusion from $G_v$ into $G_u*_{G_e}G_v$ is onto if and only if the inclusion $G_e\to G_u$ is, one sees that $T$ is reduced if and only if, whenever
 $e=uv$ is an edge with $G_e=G_u$, then $u$ and $v$ belong to the same orbit (i.e.\ $e$ projects to a loop in $\Gamma=T/G$). Another characterization is that, for any edge $uv$ such that $<G_u,G_v>$ is elliptic, there exists a hyperbolic element $g\in G$ sending $u$ to $v$
(in particular the edge maps to a loop in $\Gamma$).

If $T$ is not reduced, one obtains a reduced tree $T'$ in the same deformation space by  collapsing  certain  orbits of edges ($T'$ is not uniquely defined in general).

When more general decompositions are allowed, for instance when one considers  splittings over $\mathbb{Z}$, there may not exist a tree with the same maximality property. The  fundamental example is the following.
Consider an   orientable  closed surface $\Sigma$, and  two simple closed curves $c_1,c_2$ in $\Sigma$ with non-zero intersection number.
Let $T_i$ be the Bass-Serre tree of the associated splitting of $G=\pi_1(\Sigma)$ over $\mathbb{Z}\simeq\pi_1(c_i)$.
Since $c_1$ and $c_2$ have positive intersection number, $\pi_1(c_1)$ is hyperbolic in $T_2$ (it does  not fix a point) and vice-versa.
Using  the fact that $\pi_1(\Sigma)$ is freely indecomposable, it is an easy exercise to check that there is no splitting of $\pi_1(\Sigma)$
which dominates both $T_1$ and $T_2$. In this case there is no hope of having a splitting over cyclic groups  similar to $T_0$ above.

To overcome this difficulty, one restricts to universally elliptic splittings, which will be defined in the following sections

Definition Let $T_1,T_2$ be trees. $T_1$ is elliptic with respect to $T_2$ if every edge stabilizer of $T_1$ fixes a point in $T_2$.

Note that $T_1$ is elliptic with respect to $T_2$ whenever there is a refinement $\hat {T_1}$  of $T_1$ that dominates $T_2$: edge stabilizers of $T_1$ are elliptic in $\hat {T_1}$, hence in $T_2$. We show a converse statement.

Proposition 1 If $T_1$ is elliptic with respect to $T_2$, there is  a tree $\hat {T_1}$ with maps $p:\hat {T_1}\rightarrow T_1$ and $f:\hat {T_1}\rightarrow T_2$ such that:

• $p$ is a collapse map;
•  for each $v\in T_1$, the restriction of $f$ to the subtree $Y_v=p^{-1}(v)$ is injective.

In particular:

1.   $\hat {T_1}$ is a refinement of $T_1$ that dominates $T_2$;
2.  the stabilizer of any edge of $\hat {T_1}$ fixes an edge in $T_1$ or in $T_2$;
3.  every edge stabilizer of $T_2$ contains an edge stabilizer of $\hat {T_1}$;
4.  a subgroup of $G$ is elliptic in $\hat {T_1}$ if and only if it is elliptic in both $T_1$ and $T_2$.

Assertions 2 and 4 guarantee that $\hat {T_1}$ is an $(\mathcal{A},\mathcal{H})$-tree since $T_1$ and $T_2$ are.

Proof: We construct $\hat{T_1}$ as follows.
For each vertex $v\in V(T_1)$, with stabilizer $G_v$, choose any
$G_v$-invariant subtree $\tilde{ Y_v}$ of $T_2$ (for instance, $\tilde{ Y_v}$ can be a minimal $G_v$-invariant subtree, or the whole of  $ T_2$). For each edge $e=vw\in E(T_1)$, choose vertices $p_v\in \tilde{ Y_{v}}$ and $p_w\in \tilde{ Y_w}$ fixed by $G_e$; this is possible because $G_e$ is elliptic in $T_2$ by assumption, so has a fixed point in any  $G_v$-invariant subtree. We make these choices $G$-equivariantly.

We can now define a  tree $\tilde{ T_1}$
by blowing up each vertex $v$ of $T_1$ into $\tilde{ Y_v}$, and attaching     edges of $T_1$ using the points $p_v$.   Formally, we consider the disjoint union $( \bigsqcup_{v\in V(T_1)}\tilde{ Y_v}) \cup (\bigsqcup _{e\in E(T_1)}e)$, and for
each edge $ e=vw$ of
$T_1$ we identify $v$ with $p_v\in \tilde{ Y_v}$ and $w$ with $p_w\in\tilde{  Y_w}$.
We define $\tilde{ p}:\tilde{ T_1}\rightarrow T_1$ by sending $Y_v$ to $v$, and sending $e\in E(T_1)$ to itself.  We also define a map
 $\tilde{ f}:\tilde{ T_1}\rightarrow T_2$  as equal to the inclusion $\tilde{ Y_v}\hookrightarrow T_2$   on $\tilde{ Y_v}$, and   sending the edge $e=vw$ to the segment $[p_v,p_w]\subset T_2$.

In general, $\tilde{ T_1}$ may fail to be minimal, so we define $\hat {T_1}\subset \tilde{ T_1}$ as the unique minimal $G$-invariant subtree $\mu_{ \tilde{ T_1}}(G)$ (the action of $G$ on $\hat {T_1}$ is non-trivial unless $T_1$ and $T_2$ are both points).
We then define $p$ and $f$ as the restrictions of $\tilde{ p}$ and $\tilde{ f}$ to $\hat {T_1}$.
These maps clearly satisfy the first two requirements.
For reasons of time and space we sketch 2 and 4.

If $e$ is an edge of $\hat{ T_1}$ that is not collapsed by $p$, then $G_e$ fixes an edge of $T_1$.
Otherwise, $f$ maps $e$ injectively to a non-degenerate segment of $T_2$, so $G_e$ fixes an edge in $T_2$.

To prove the non-trivial direction of 4, assume that $H$ is elliptic in $T_1$ and $T_2$. Then $H\subset G_v$ for some $v\in T_1$,
so $H$ preserves the subtree $Y_v\subset \hat {T_1}$.
Since $f$ is injective in restriction to $Y_v$, it is enough to prove that $H$ fixes a point in $f(Y_v)$. This holds because $H$ is elliptic in $T_2$. $\blacksquare$

Definition Any tree $\hat {T_1}$ as in Proposition 1 will be called a standard refinement of $T_1$ dominating $T_2$.
 
In general, there is no uniqueness of standard refinements. However, by Assertion 4 of Proposition 1, all standard refinements belong to the same deformation space, which is the lowest deformation space dominating the deformation spaces containing $T_1$ and $T_2$ respectively. If $T_1$ dominates $T_2$ (resp.\ $T_2$ dominates $T_1$),
then $\hat {T_1}$ is in the same deformation space as $T_1$ (resp.\ $T_2$).
Moreover,  there is some symmetry: if $T_2$ also happens to be elliptic with respect to $T_1$,  then   any standard refinement $\hat{ T_2}$ of $T_2$ dominating $T_1$ is in the same deformation space as $\hat {T_1}$.

Lemma 2

1. If $T_1$ refines $T_2$ and does not belong to the same deformation space, some $g\in G$ is hyperbolic in $T_1$ and elliptic in $T_2$.
2. If  $T_1$ is elliptic with respect to  $T_2$, and every $g\in G$ which is elliptic in $T_1$ is also elliptic in $T_2$, then $T_1$ dominates $T_2$.  
3. If $T_1$ is elliptic with respect to  $T_2$, but $T_2$ is not elliptic with respect to $T_1$, then $G$ splits over a group    which has infinite index in an edge stabilizer of $T_2$.

Recall that all trees are assumed to be $(\mathcal{A},\mathcal{H})$-trees; the splitting obtained in (3) is relative to $\mathcal{H}$. The proof is mostly definition chasing and omitted.

Universal ellipticity

Definition A subgroup $H\subset G$ is universally elliptic if it is elliptic in every  \AH-tree. A tree $T$ is universally elliptic if its edge stabilizers are universally elliptic, i.e. if $T$ is elliptic with respect to every \AH-tree. 

When we need to be specific, we say universally elliptic over $\mathcal{A}$ relative to $\mathcal{H}$, or $(\mathcal{A},\mathcal{H})$-universally elliptic. Otherwise we just say universally elliptic, recalling that all trees are assumed to be $(\mathcal{A},\mathcal{H})$-trees.

Groups with Serre's property (FA), in particular finite groups,  
are universally elliptic. If $H$ is universally elliptic and $H'$ contains $H$ with finite index, then $H'$ is universally elliptic.

Lemma 3 Consider  two trees $T_1,T_2$.

• If $T_1$ is universally elliptic, then some refinement  of $T_1$ dominates $T_2$.
• If $T_1$ and $T_2 $ are universally elliptic, any standard refinement  $\hat {T_1}$ of $T_1$ dominating $T_2$ is universally elliptic. In particular, there is a universally elliptic tree $\hat {T_1}$ dominating both $T_1$ and $T_2$.
• If $T_1$ and $T_2 $ are universally elliptic and have the same elliptic elements, they belong to the same deformation space.

The proof is left as an exercise to the reader.

Having fixed $\mathcal{A}$, we define $\mathcal{A}_\ell \subset \mathcal{A}$ as the set of groups in $\mathcal{A}$ which are universally elliptic (over $\mathcal{A}$ relative to $\mathcal{H}$); $\mathcal{A}_\ell$ is stable under conjugating and taking subgroups.

A tree is universally elliptic if and only if it is an $\mathcal{A}_\ell$-tree.

Definition If there exists a deformation space $\mathcal{D}_{JSJ} of $(\mathcal{A}_{\ell},\mathcal{H})$-trees which is maximal for domination, it is unique by the second assertion of Lemma 3.
It is called the JSJ deformation space of $G$ over $\mathcal{A}$ relative to $\mathcal{H}$.

Trees in $\mathcal{D}_{JSJ}$ are called JSJ trees (of $G$ over $\mathcal{A}$ relative to $\mathcal{H}$).
They are precisely those trees $T $ which are universally elliptic, and which dominate every universally elliptic tree. We also say that trees $T\in \mathcal{D}_{JSJ}$, and  the associated graphs of groups $\Gamma=T/G$, are JSJ decompositions.

We will show that the JSJ deformation space exists if $G$ is finitely presented. In general there are
many JSJ trees, but they  all belong to the same deformation space so can be studied all at once. In
particular, it turns out that they have the same vertex stabilizers, except possibly for
vertex stabilizers in $\mathcal{A}_{\ell}$.

Definition [Rigid and flexible vertices] Let $H=G_v$ be a vertex stabilizer   of  a JSJ  tree $T$ (or a vertex group  of the graph of groups $\Gamma=T/G$).
We say that $H$ is rigid if it is universally elliptic, flexible if it is not. We also say that the vertex $v$ is rigid (flexible). If $H$ is flexible, we say that it  is a  flexible  subgroup of $G$ (over $\mathcal{A}$ relative to $\mathcal{H}$).

The definition of  flexible subgroups of $G$ does not
depend on the choice of the  JSJ tree $T$. Understanding flexible groups is the hard part of JSJ theory.

Lemma 4 Let $T$ be a JSJ tree, and $S$ any tree.

•   There is a tree $\hat {T}$ which refines $T$ and dominates $S$.
•  If $S$ is universally elliptic, it may be refined to a JSJ tree.

Proof Since $T$ is elliptic with respect to $S$, one can construct a standard refinement $\hat {T}$ of $T$ dominating $S$. It satisfies the first assertion.

For the second assertion, since $S$ is elliptic with respect to $T$, we can consider a standard refinement $\hat {S}$ of $S$ dominating $T$.
It is universally elliptic  by the second assertion of  Lemma 3,
and dominates $T$, so it is a JSJ tree. $\blacksquare$

Existence of the JSJ deformation space: the non-relative case

We prove the existence of JSJ decompositions assuming $\mathcal{H}=\emptyset$.   

Theorem 5 If $G$ is finitely presented, then the JSJ deformation space $\mathcal{D}_{JSJ}$ of $G$ over $\mathcal{A}$ exists. It contains a tree whose edge and vertex stabilizers are finitely generated.
 
There is no hypothesis, such as smallness or finite generation, on
the elements of $\mathcal{A}$.
Recall that, $G$ being finitely generated, finite generation of edge stabilizers implies  finite generation of vertex stabilizers.


The existence of $\mathcal{D}_{JSJ}$ will be deduced from the
following version of Dunwoody's accessibility.

Dunwoody accessibility: Let $G$ be finitely presented. Assume that $T_1\leftarrow\dots\leftarrow T_k \leftarrow T_{k+1}\leftarrow \dots$
is a sequence of refinements of trees.
There exists a tree $S$ such that:

• for $k$ large enough, there is a morphism $S\rightarrow T_k$ (in particular, $S$ dominates $T_k$);
• each edge and vertex  stabilizer of $S$ is finitely generated.

Note that the maps $T_{k+1}\rightarrow T_k$ are required to be collapse maps. We will defer the proof of this to a later time.
 
Recall that $f:S\rightarrow T_k$ is  a morphism if $S$ may be subdivided so that $f$ maps edges to edges (a collapse map is not a morphism). In particular, edge stabilizers of $S $ fix an edge in $T_k$, so $S$ is an $\mathcal{A}$-tree since  every $T_k$ is. It is universally elliptic if every $T_k$ is.

Remark Unfortunately, it is not true that the deformation space of $T_k$ must stabilize as $k$ increases, even if all edge stabilizers are cyclic.
For example, let $A$ be  a group with a sequence of nested infinite cyclic groups $A\supsetneqq  C_1\supsetneqq C_2\supsetneqq\dots$, let $G=A*B$ with $B$ non-trivial, and let $T_k$ be the Bass-Serre tree of the iterated amalgam
$$G=A*_{C_1} C_1 *_{C_{2}} C_{2} *_{C_{3}} \cdots *_{C_{k-1}} C_{k-1} *_{C_k} <{C_k,B}>.$$
These trees refine each other, but are not in the same deformation space since ${C_k,B}$ is not elliptic in $T_{k+1}$.
They are dominated by the tree $S$ dual to the free decomposition $G=A*B$, in accordance with the proposition.


Applying Dunwoody accessibility to a constant sequence yields the following standard result:
Corollary 6 If $G$ is finitely presented, and $T$ is a tree, there exists a morphism $f:S\rightarrow T$ where $S$ is a tree with finitely generated edge and vertex stabilizers.

If $T$ is a universally elliptic $\mathcal{A}$-tree, so is $S$. 

Proof of the existence of a JSJ deformation space: Let $\mathcal{U} $ be the set of universally elliptic trees with finitely generated edge and vertex stabilizers, up to equivariant isomorphism. It is non-empty since it contains the trivial tree. An element of $\mathcal{U}$ is described by a finite graph of groups with finitely generated edge and vertex groups. Since $G$ only has countably many finitely generated subgroups, and there are countably many homomorphisms from a given finitely generated group to another, the set $\mathcal{U}$ is countable.  

By Corollary 6, every universally elliptic tree is dominated by one in $\mathcal{U}$, so it suffices to produce a universally elliptic tree dominating every $U\in\mathcal{U}$. Choose an enumeration $\mathcal{U}=\{U_1,U_2,\dots,U_k,\dots\}$. We define inductively a universally elliptic tree $T_k$ which refines $T_{k-1}$ and dominates $U_1,\dots, U_k$  (it may have infinitely generated edge or vertex stabilizers).

We start with $T_1=U_1$. Given $T_{k-1}$ which dominates $U_1,\dots,U_{k-1}$,
we  let $T_k$ be a  standard refinement of $T_{k-1}$ which dominates $U_k$ (it exists by Proposition 1 because $T_{k-1}$ is universally elliptic). Then $T_k$ is universally elliptic by the second assertion of Lemma 3, and it dominates $U_1,\dots, U_{k-1}$ because $T_{k-1}$ does.

Apply Dunwoody accessibility to the sequence $T_k$. The tree $S$ is universally
elliptic and it dominates every $T_k$, hence every $U_k$. It follows that $S$ is a JSJ tree over $\mathcal{A}$. $\blacksquare$