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# A survey of relative Dunwoody’s accessibility theorem

## Motivation

This is not (even remotely) an original work. For example it contains large excerpts from a variety of papers (often without reference).

More importantly beware! What follows may contain outrageously false statements. This was created for an in-class presentation while the author was exploring these ideas for the first time..

Consider a group G (finitely generated). We are interested to ‘split’ the group into simpler pieces. In this direction we have the following result:

### Grushko Decomposition Theorem

Any non trivial finitely generated group G can be decomposed as a free product $$G = G_1 * G_2 * \cdots * G_r * F_s, r, s \geq 0$$ where each of the groups $$G_i$$ is non trivial, freely indecomposable (that is it cannot be decomposed as a free product) and not infinite cyclic, and where $$F_s$$ is a free group of rank s.

For a given G, the groups G1, …, Gr are unique up to a permutation of their conjugacy classes in G (and, in particular, the sequence of isomorphism types of these groups is unique up to a permutation) and the numbers s and r are unique as well.

More precisely, if G = B1∗…∗BkFt is another such decomposition then k = rs = t, and there exists a permutation σ∈Sr such that for each i=1,…,r the subgroups Gi and Bσ(iare conjugate in G.

This gives a decomposition of the group G as the fundamental group of a graph of groups,  or equivalently actions of G on a simplicial  tree T (Grushko Tree) with trivial edge  stabilizers.

#### Maximality of Grushko Trees

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.

### Next level of generalization

We want to consider more general decompositions. In Grushko decomposition, the edge stabilizers (of corresponding Grushko Trees) are trivial. What if we wanted the edge stabilizers to be cyclic? What if they were finite?

Toward this end, we first choose our favorite class of edge stabilizers and call it $$\mathcal{A}$$. For example $$\mathcal{A}$$ could be the class of cyclic subgroups or the class of virtually cyclic subgroups or the class of finite subgroups and so on.

Suppose $$\mathcal{A}$$ is closed under taking subgroups and conjugation. We investigate the action of G on all trees whose edge stabilizers are from $$\mathcal {A}$$.

These trees are known as $$\mathcal{A}$$ – trees.

#### Question

Is it possible to find a tree with a maximality property as in the case of Grushko trees? That is, is it possible to find an $$\mathcal{A}$$ tree on which G acts such that the vertex stabilizers are as small as possible? More explicitly speaking, is it possible to split the vertex groups finitely many times and terminate the process?

### Dunwoody’s inaccessible group

The answer to the above question, for finitely generated groups, is negative. Even if we demand the edge stabilizers to be finite, there exists groups which do not yield to such maximal decompositions.

Suppose G be a finitely generated group. $$\mathcal{A}$$ = {finite subgroups of G}.

A finitely generated group G is said to be accessible if it is the fundamental group of a graph of groups in which all edge groups are finite (from $$\mathcal{A}$$ ) and every vertex group has at most one end.

Stallings Theorem showed that a group G has more than one end if and only if G ≍ A*FB, where F is finite, A ≠ F ≠ B, or G is an HNN-extension with finite edge group F. Hence if we can find a graph of groups in which every vertex group has at most one end then, these vertex groups won’t split any more over finite groups.

We say that G is inaccessible if it is not accessible.

M.J. Dunwoody (1993) constructed a finitely generated group that is inaccessible

.

### Next best hope: finitely presented groups

Suppose G is a finitely presented group. $$\mathcal {A}$$ be the class of cyclic groups (closed under taking subgroups and conjugation). We may hope that such a group has a maximal splitting over cyclic subgroups.

Consider the following example: $$G = \mathbb{Z} * B$$ where B is non-empty, indecomposable, finitely presented group. Then we have

$$G = \mathbb{Z} * B = \mathbb{Z} *_{2\mathbb{Z}} <2\mathbb{Z}, B>$$

But this can be split further. For example, we have

$$G = \mathbb{Z} *_{2\mathbb{Z}} {2\mathbb{Z}}*_{4\mathbb{Z}} <4\mathbb{Z}, B>$$

Hence we have found a finitely presented group, which has a graph of groups decomposition over cyclic groups that does not terminate.

But all hope is not lost. In a sense $$G = \mathbb{Z} * B$$ dominates all other decompositions that we have created. That is vertex stabilizers of $$G = \mathbb{Z} * B$$ are stabilize vertices in the subsequent splittings

In fact, what actually happens is this: for finitely presented groups, we will be able to find a universally elliptic $$\mathcal {A}$$ -tree that will dominate all other universally elliptic $$\mathcal {A}$$ -trees. That is, though we may get non-terminating splittings, there will exist one that will dominate all of them eventually.

## Motivation from manifold topology

William Jaco, Peter Shalen, and Klaus Johannson (1979) proved the following:

Irreducible orientable closed (i.e., compact and without boundary) 3-manifolds have a unique (up to isotopy) minimal collection of disjointly embedded incompressible tori such that each component of the 3-manifold obtained by cutting along the tori is either atoroidal or Seifert-fibered.

“Differentiable manifolds can always be given the structure of PL manifolds, which can be triangulated into simplicial complexes. By shrinking a spanning tree of the 1-skeleton of this simplicial complex, we can obtain a CW complex 𝑋 with a single 0-cell. This complex is no longer a manifold, but has the same fundamental group as the original manifold, since quotienting out by a contractible subspace is a homotopy equivalence.

If the manifold is compact, it has a simplicial decomposition with a finite number of cells. This carries over to 𝑋. But the fundamental group of a 𝐶𝑊 complex with a single 0-cell has a presentation with a generator for each 1-cell and a relation for each 2-cell. Thus 𝑋, and therefore the original manifold, has a finitely presented fundamental group.”

This idea motivates search for ‘a’ maximal splitting of the (finitely) group over favorite subgroups.

### JSJ Decomposition

A JSJ decomposition (or JSJ tree) of G over $$\mathcal {A}$$ is an $$\mathcal{A}$$ – tree such that:

1. T is universally elliptic (its edge stabilizers are elliptic in every $$\mathcal{A}$$ – tree or fixes a point in every $$\mathcal {A}$$ – tree \)
2. T dominates any other universally elliptic tree T’ (the vertex stabilizers are as small as possible; they are elliptic in every universally elliptic tree).

If $$\mathcal {A}$$ only contains the trivial group, JSJ trees are same as Grushko trees.

What if $$\mathcal {A}$$ is something else? Cyclic, Abelian, Slender etc. Dunwoody’s accessibility theorem concludes that a finitely presented group has JSJ decompositions over any class $$\mathcal{A}$$ of subgroups.

Theorem: Let $$\mathcal {A}$$ be an arbitrary family of subgroups of G, stable under taking subgroups and under conjugation. If G is finitely presented, it has a JSJ decomposition over $$\mathcal {A}$$. In fact there exists a JSJ tree whose edge and vertex stabilizers are finitely generated.

Dunwoody’s accessibility theorem is a consequence of this (general) existence theorem.

### Dunwoody’s Accessibility Theorem

G is finitely presented. There exist a graph of groups decomposition of G such that all edge groups are finite and all vertex groups are finite or one-ended.

## Lemma

Let G be finitely presented relative to $$\mathcal{H} = \{ H_1, H_2 , \cdots , H_p \}$$. Assume that $$T_1 \leftarrow \cdots \leftarrow T_k \leftarrow T_{k+1} \leftarrow \cdots$$ is a sequence of refinements of $$(\mathcal {A, H} )$$ trees. There exists an $$(\mathcal {A, H} )$$ – tree S such that:

1. for k large enough, there is a morphism $$S \to T_k$$;
2. each edge stabilizer of S is finitely generated.

#### Morsels of the statement

• Relatively finitely presented: G is relatively finitely presented relative to its subgroups $$H_1 , \cdots , H_p$$ if there exists a finite subset $$\Omega \subset G$$ such that the natural morphism $$F ( \Omega ) * H_1 * \cdots * H_p \to G$$ is onto and its kernel is normally generated by a finite subset $$\mathcal{R}$$
• $$(\mathcal {A, H} )$$ trees
• Equivariant Maps: A map $$f: T \to T’$$ is said to be G-equivariant if $$f (g \cdot x ) = g \cdot f(x)$$.
• Maps between trees will always be G – equivariant, send vertices to vertices, and edges to edge-paths (may be a point)
• Morphism: A map $$f : T \to T’$$ between two trees is a morphism if an only if one can subdivide T so that f maps each edge onto an edge; equivalently, no edge of T is collapsed to a point. Folds are examples of morphism. (Fig. 1)
• Collapse Map: A collapse map $$f : T \to T’$$ between two trees 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)]. (Fig. 2)
• Refinements (of trees): A tree T’ is a collapse of T if there is a collapse map $$T \to T’$$; conversely, we say that T refines T’.

#### And the big picture

Why do we care about refinements?

Action of a group on a tree is a description of the group in simpler terms or a splitting of the group. In figure 2, $$G_1, L_1, H_1, I_1$$ and edges connecting them maps to the vertex $$N_1$$ in tree T’.

Since the collapse map is G-equivariant this implies if g stabilizes $$G_1$$ then $$N_1 = f(G_1) = f( g \cdot G_1 ) = g \cdot f(G_1) = g \cdot N_1$$ or g stabilizes $$N_1$$. In other words, stabilizers of $$G_1, L_1, H_1, I_1$$ and the edges connecting them are contained in stabilizer of $$N_1$$.

The vertex group $$G_{N_1}$$ is the fundamental group of the graph of groups $$\Gamma_{N_1}$$ in red, occurring as the preimage of $$N_1$$.

Why do we want the edge stabilizers of S to be finitely generated?

## Proof

Make the 2-complex, universal cover, and fix the lifts of elliptic portions, and vertices of remaining parts.

1. Construct a connected simplicial 2-complex X such that $$\pi_1 (X) = G$$
1. $$(Y_i, u_i)$$ be a pointed 2-complex with $$\pi_1 (Y_i, u_i) = H_i$$
2. Starting from the disjoint union of the $$Y_i’s$$, add p edges joining the $$u_i$$ ‘s to an additional vertex u.
3. Add # $$\Omega$$ additional edges joining u to itself.
4. Represent each element of $$\mathcal {R}$$ by a loop in this space, and glue a disc along this loop to obtain the desired space X.
5. I wonder if it helps to build the Eilenberg Maclane complex
2. X has a universal cover $$\pi : X \to \tilde {X}$$
3. G acts on $$\tilde {X}$$ by deck transformation.
4. For $$i \in \{1, \cdots , p \}$$ consider a connected component $$\tilde {Y}_i$$ of $$\pi^{-1} (Y_i)$$, whose stabilizer is $$H_i$$.
5. Fix lifts $$v_1 , \cdots , v_q \in \tilde {X}$$ of all the vertices in X \ $$Y_1 \cup \cdots \cup Y_p$$ . (It is a finite simplicial complex. Hence it has finitely many vertices)

Construct equivariant maps $$f_k : \tilde {X} \to T_k$$ inductively.

1. $$p_k : T_{k+1} \to T_k$$ be the collapse map. By definition of collapse map the pre-image of the midpoint of an edge of $$T_k$$ is a single point, namely the midpoint of the edge of $$T_{k+1}$$ mapping onto $$e_k$$
2. We demand $$f_k : \tilde {X} \to T_k$$ to be G – equivariant, mapping $$\tilde {Y}_i$$ to a vertex fixed by $$H_i$$, sends each $$v_j$$ to a vertex and sends each edge of $$\tilde {X}$$ either to a point or injectively onto a segment in $$T_k$$. We further require that $$p_k (f_{k+1} (x) ) = f_k (x)$$ if x is a vertex of $$\tilde {X}$$ or if $$f_k (x)$$ is a midpoint of an edge in $$T_k$$ Fig 3: Construction of $$f_k$$

#### Construction

We start with $$T_0$$ a point, and $$f_0, p_0$$ the constant maps. We then assume that $$f_k : \tilde{X} \to T_k$$ has been constructed and we construct $$f_{k+1}$$.

To define $$f_{k+1}$$ in $$\tilde {Y}_i$$ we note that $$f_k (\tilde {Y}_i)$$ is a vertex of $$T_k$$ fixed by $$H_i$$. Since $$p_k$$ preserves alignment, $$p_k^{-1} (f_k (\tilde {Y}_i ))$$ is an $$H_i$$ invariant subtree. Since $$H_i$$ is elliptic in $$T_{k+1}$$, it fixes some vertex 0f this subtree, and we map $$\tilde {Y}_i$$ to such a vertex.

(confusion: why should that vertex be contained in that subtree)

We then define $$f_{k+1} (v_j )$$ as any vertex in $$p_k^{-1} (f_k (v_j) )$$, and we extend by equivariance.

(Recall that $$v_j$$ were the chosen lifts of vertices of X. For example a vertex u in X may be lifted to $$v_{u_1} , v_{u_2}, \cdots$$. If we chose $$v_{u_1}$$ as the favorite lift then $$v_{u_i} = g_i \cdot v_{u_1}$$. Since we fixed $$f_{k+1} (v_{u_1} )$$ then we define $$f_{k+1} (v_{u_i} ) = g_i \cdot f_{k+1} (v_{u_1} )$$ )

Notice that we finished defining $$f_{k+1}$$ on all vertices of $$\tilde {X}$$ .

Now consider an edge e of X not contained in any $$Y_i$$, and a lift $$\tilde {e} \subset \tilde{X}$$. The map $$f_{k+1}$$ is already defined on the endpoints of $$\tilde {e}$$.

The restriction of $$p_k$$ to the segment of $$T_{k+1}$$ joining the images of the endpoints of $$\tilde {e}$$ is a collapse map. In particular, the preimage of the midpoint of an edge $$e_k$$ of $$T_k$$ is the midpoint of the edge of $$T_{k+1}$$ mapping onto $$e_k$$

$$f_k$$ is constant or injective on $$\tilde {e}$$, this allows us to define $$f_{k+1}$$ on $$\tilde {e}$$ as a map which is either constant or injective and satisfies $$p_k ( f_{k+1} (x) = f_k (x)$$ if $$f_k(x)$$ is the midpoint of an edge of $$T_k$$

Doing this equivariantly we have now defined $$f_{k+1}$$ on the 1- skeleton of $$\tilde {X}$$

We then extend $$f_{k+1}$$ in a standard way to every triangle abc not contained in $$\pi ^{-1} (Y_i )$$; in particular, if $$f_{k+1}$$ is not constant on abc, preimages of the midpoints of the edges in $$T_{k+1}$$ are straight arcs joining two distinct sides.

Pattern of Dunwoody, its dual tree

1. Define $$\tilde {\tau}_k \subset \tilde{X}$$ as the preimage (under $$f_k$$ of the midpoints of all edges of $$T_k$$. This is a pattern in the sense of Dunwoody
2. $$\tilde {\tau}_k$$ does not intersect any $$\tilde {Y}_i$$
3. $$\tilde{\tau}_k \subset \tilde {\tau}_{k+1}$$
4. $$\tau_k = \pi ( \tilde{\tau}_k )$$ be the projection of in X. It is a finite graph as it is contained in the complement of $$Y_1 \cup \cdots \cup Y_p$$.
5. $$S_k$$ be the tree dual to the pattern $$\tilde{\tau}_k$$

Claim 1: $$S_k$$ is an $$(\mathcal {A, H} )$$ tree.

Proof: Each edge in $$S_k$$ is ‘cut’ by a unique track which corresponds to a midpoint of an edge in $$T_k$$. This defines a map from $$S_k \to T_k$$. Hence edge stabilizers of $$S_k$$ are in edge stabilizers of $$T_k$$ implying $$S_k$$ is $$\mathcal {A}$$ tree.

Every $$H_i$$ is elliptic in $$S_k$$ because $$\tilde {Y}_i$$ does not intersect $$\tilde{\tau}_k$$.

Claim 2: $$S_k$$ has finitely generated edge stabilizers.

Proof: (Forwarded by Professor Hruska, explained by Arka Banerjee)

Let A be a subcomplex of B. Consider the universal cover $$\tilde{B}$$. Each component of the preimage of A in the universal cover is a cover of A corresponding to the subgroup $$N = ker (\pi_1(A) \to \pi_1(B) )$$

Why? Some loops in A may shrink to the basepoint in B but not in A. These loops are precisely the members of N (apart from the trivial loop that shrinks both in A and in B). Pre-images of the homotopy classes of these loops (under the covering projection in the cover) are distinct homotopy classes of loops in a connected component of the pre-image of A. Hence N is the fundamental group of each connected component.

Such a component is called an “elevation” of A to the cover. (It’s not the same thing as a “lift” of A.)

Why? After all, a lift will include A in the universal cover of its super space. These components are covers of A and is possibly larger than A

The deck transformations that stabilize the elevation of A would be
$$im ( \pi_1(A) \to \pi_1(B) ) = \pi_1(A) / N$$

Why? The elements in $$im ( \pi_1(A) \to \pi_1(B) )$$ are the homotopy classes of loops in A which do  not shrink to basepoint in A and in B (except the trivial one). These are precisely all those group elements that keep members of elevations of A in elevations of A.

(The fundamental group of the base space acts on the universal cover by deck transformations. The action is determined as follows. Take a base point $$x_0 \in X$$ and a pre-image $$\tilde{x}_0$$ of $$x_0$$ in the universal cover $$\tilde {X}$$. Then each element of $$\pi_1 (X, x_0)$$ is represented by a loop f : I → X based at $$x_0$$. There is a unique lift $$\tilde{f} : I → \tilde{X}$$ starting at $$\tilde{x}_0$$. Then we define the action of the homotopy class [f] on $$\tilde{x}_0$$ by $$[f] (\tilde{x}_0 = \tilde {f}(1)$$. The quotient under this group action is the base space.)

If A is a finite graph, it is clear that any quotient of $$\pi_1(A)$$ is finitely generated.

Why? Fundamental group of a finite graph is generated by the finite number of edges (of that graph sans the maximal tree in it). Hence it is finitely generated. Finally quotient of finitely generated group is finitely generated. The quotient group is generated by the images of the generators of G under the canonical projection.

Number of non-parallel tracks are bounded

Let $$X’ \subset X$$ be the closure of the complement of $$Y_1 \cup \cdots \cup Y_p$$. By construction this is a finite complex.

#### Theorem [Dun85, Theorem 2.2]

There is a bound on the number of non-parallel tracks in X’.

This implies that there exists $$k_0$$ such that for all $$k \geq k_0$$, for every connected component $$\sigma$$ of $$\tau_k$$ \ $$\tau_{k_0}$$, there exists a connected component $$\sigma’$$ of $$\tau_{k_0}$$ such that $$\sigma \cup \sigma’$$ bounds a product region containing no vertex of X’.

It follows that, for $$k \geq k_0$$, one can obtain $$S_k$$ from $$S_{k_0}$$ by subdividing edges. We then take $$S = S_{k_0}$$

Proof of Dunwoody bounded track theorem:

#### Morsels of the Statement

Tracks: Let L be a connected 2-dimensional complex. A track is a subset S of |L| with the following properties.

• S is connected
• For each two simplex $$\sigma$$ of |L|, $$S \cap | \sigma |$$ is a union of finitely many disjoint straight lines
• If $$\gamma$$ is a 1-complex of L and $$\gamma$$ is not a face of a 2-complex, then either $$S \cap | \gamma | = \phi$$ or S consists of a single point in the interior of $$| \gamma |$$ . This implies S is 1 point set.

If |L| is a 2-manifold then S is a connected 1 dimensional submanifold.

Band: A band is a subset of B of |L| with the following properties.

• B is connected
• For each 2-complex $$\sigma$$ 0f L, $$B \cap | \sigma |$$ is a union of finitely many components each of which is bounded by two closed intervals in distinct faces of $$\sigma$$ and the disjoint lines joining the endpoints of these intervals.
• If $$\gamma$$ is a 1-complex of L, which is not a face of 2-complex, then either $$B \cap | \gamma | = \phi$$ or B consists of a subset of $$| \gamma |$$ bounded by two points in the interior of $$| \gamma |$$

If B is a band then we get a track $$t(B)$$ by choosing the midpoint of each component of $$| \gamma | \cap B$$ for every 1-simplex $$\gamma$$ of L and joining these points in the appropriate components of $$| \sigma | \cap B$$ where $$\sigma$$ is a 2-complex of L.

Untwisted Band: B is untwisted if B is homeomorphic to $$t(B) \times [0, 1]$$ in which case $$\partial B$$ has two components each homeomorphic to t(B).

Twisted Band: If B is not untwisted then it is twisted. In this case $$\partial B$$ is a track which double covers t(B).

Twisted and Untwisted tracks: If S is a track then S is called twisted (untwisted) if there is a twisted (untwisted) band B such that t(B) = S.

Parallel Tracks: Two tracks $$S_1$$ and $$S_2$$ are parallel if there is an untwisted band B such that $$\partial B = S_1 \cup S_2$$

1-cochains with $$\mathbb{Z}_2$$ coefficients: Let S be a track in the connected 2-dimensional complex L. If $$\gamma$$ is a 1-simplex of L, 1-cochain $$z(S) (\gamma$$ = k \mod 2 \) where k = # $$(|\gamma| \cap S)$$. That is the number of times S intersects $$\gamma$$  modulo 2. (Is it hitting $$\gamma$$  even or odd times ?)

If $$\sigma$$ is a 2-simplex then $$\partial \sigma \cap \S$$ has an even number of points.

If AB, BC and CA are the sides of a triangle and S hits AB and BC, then every time it hits AB, it will also hit BC. In this particular case z(S) (AB) – z(S) (BC) + z(CA) is even or 0 mod 2. Hence it z(S) is a map which goes to 0 under $$\delta$$ or difference function.

Hence z(S) is a 1-cocycle.

Claim 1: z(S) is a coboundary (an image of the differential) if and only if S separates |L|, i.e. |L| – S has two components.

Notice that z(S) is a 1-cochain. Can it be regarded as a map obtained by taking difference of the value assigned to the vertices? That is, if $$\phi : \Delta^0 \to \mathbb{Z}_2$$, then is $$\delta \phi = z(S)$$ ?

This is possible if and only if S splits L into two components.

Claim 2: A twisted track cannot separate L.

Proof: Follow the track along from side of the track long enough to reach the other side as there is only one side per se.

Hence if S is twisted, z(S) is not a coboundary. Hence it is a non-trivial element of $$H^1 (L; \mathbb{Z}_2 )$$. Why? Notice that z(S) is a cocycle (as shown earlier). Hence it is in kernel. $$H^1 (L; \mathbb{Z}_2 )$$ is formed by quotienting out the coboundaries from the kernel. Is S is twisted, z(S) is not coboundary (but nevertheless a cocycle). Hence it is not quotiented out. Hence it represents a non trivial element of $$H^1 (L; \mathbb{Z}_2 )$$ .

In fact no non empty union of disjoint twisted tracks separates |L|. Hence the corresponding elements of $$H^1 (L; \mathbb{Z}_2 )$$ are linearly independent.

### Theorem

Suppose $$\beta = \textrm{rank} H^1 (L; \mathbb{Z}_2 )$$ is finite. Let $$T = \{ t_1, \cdots , t_n \}$$ be a set of disjoint tracks. Then $$|L| – \cup t_i$$ has at least $$n – \beta$$ components, The set T contains at most $$\beta$$ twisted tracks.

Proof: Let M be the subgroup of $$H^1 (L; \mathbb{Z}_2 )$$ generated by the elements corresponding to $$z(t_1), z(t_2), \cdots , z(t_n)$$. Consider the epimorphism $$\theta : \underbrace{ \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \cdots \oplus \mathbb{Z}_2}_\text{n copies } \to M$$. We can think of this epimorphism as a ‘string picker’. For example (1, 0, 0 , … 0) is mapped to $$z(t_1)$$ etc.

Kernel of this epimorphism are elements mapping to the identity element of $$H^1 (L; \mathbb{Z}_2 )$$; that is the coboundary maps. Basis elements of this kernel corresponds to the number of untwisted tracks. Since untwisted (disjoint) tracks splits $$|L| – \cup t_i$$ into components, hence the basis elements of the kernel corresponds to the components of $$|L| – \cup t_i$$

The proposition follows. There are atmost $$\beta$$ twisted, disjoint, tracks. In T, therefore there are at least n – $$\beta$$ untwisted disjoint tracks. Which leads to $$n – \beta$$ components of $$L – \cup t_i$$

Suppose now that L is finite. Let

$$v_L$$ = number of vertices of L

$$f_L$$ = number of 2 simplexes of L

and n(L) = $$2 \beta + v_L + f_L$$

### Theorem

Suppose $$t_1 , t_2 , \cdots , t_k$$ are disjoint tracks in |L|. If $$k > n(L)$$, then there exists $$i \neq j$$ such that $$t_i$$ and $$t_j$$ are parallel.

Proof: If $$\sigma$$ is a 2-simplex of L and D is the closure of a component of $$|\sigma| – \cup t_i$$ then D is a disc. We say that D is good if $$\partial D \cap \partial |\sigma|$$ consists of two components in distinct faces of $$\sigma$$

For any $$\sigma$$ there are at most three D’s which contain a vertex of $$\sigma$$ and at most one other component which is not good.

If $$k > n(L)$$ then $$|L| – \cup t_i$$ has more than $$2 \beta + v_L + f_L – \beta$$ = $$\beta + v_L + f_L$$ components.

Confusion: It follows that there are at least $$\beta + 1$$ components whose closures are bands.

Since |L| contains at mosts $$\beta$$ disjoint twisted bands, there is at least one component whose closure is an untwisted band. The theorem follows immediately.

#### Lemma (3.2 – Levitt)

Assume that all groups in $$\mathcal {A}$$ are universally elliptic. If T is a JSJ tree, then the vertex stabilizers are universally elliptic.

Morsels of the proof

As mentioned above, Grushko trees have a strong maximality property: their vertex stabilizers are elliptic in any free splitting of G. This does not hold any longer when one considers JSJ decompositions over infinite groups, in particular cyclic groups: a vertex stabilizer $$G_v$$ of a JSJ tree may fail to be elliptic in some splitting (over the chosen family A).
If this happens, we say that the vertex stabilizer $$G_v$$ (or the corresponding vertex v, or the vertex group of the quotient graph of groups) is flexible. The other stabilizers (which are elliptic in every splitting over A) are called rigid. In particular, all vertices of Grushko trees are rigid (because their stabilizers are freely indecomposable). On the other hand, in the example of G = π1(Σ), the unique vertex stabilizer $$G_v = G$$ is flexible.

Proof:

Suppose vertex stabilizer $$G_v$$ is flexible (fails to be elliptic in some splitting). Consider T’ such that $$G_v$$ is not elliptic in T.

Since T is a JSJ decomposition, it is universally elliptic (its edges stabilizers fixes a point in every $$\mathcal{A}$$ tree). So one can consider a standard refinement $$\hat{T}$$ of T dominating T’.

By our assumption on $$\mathcal{A}$$, the tree $$\hat {T}$$ is universally elliptic. So by definition of the JSJ deformation space T dominates $$\hat {T}$$. This implies $$G_v$$ is elliptic in $$\hat{T}$$, hence in T’, a contradiction.

### Corollary (4.14 – Levitt)

Let $$G_v$$ be a vertex stabilizer of a JSJ tree $$T_J$$.

1. $$G_v$$ does not split over a universally elliptic subgroup relative to $$Inc^{\mathcal{H}}_v$$
2. $$G_v$$ is flexible if it splits relative to $$Inc^{\mathcal{H}}_v$$, rigid other wise.

#### Morsels of Theorem

Let T be a tree (minimal, relative to $$\mathcal {H}$$ with edge stabilizers in $$\mathcal{A}$$. Let $$v$$ be a vertex with stabilizer $$G_v$$.

Incident edge groups $$Inc_v$$

Given a vertex v of a tree T, there are finitely many $$G_v$$ orbits of edges with origin v. We choose representatives $$e_i$$ and we define $$Inc_v$$ (or $$Inc_{G_v}$$ as the family of stabilizers $$G_{e_i}$$. We call $$Inc_v$$, the set of incident edge groups. It is a family of subgroups of $$G_v$$, each well defined up to conjugacy.

Proof:

If there is a splitting as in (1), we may use it to refine $$T_J$$ to a universally elliptic tree. (see Lemma 4.12 Levitt). This tree must be in the same deformation space as $$T_J$$, so the splitting of $$G_v$$ must be trivial.

(2) follows from Lemma 4.13 Levitt applied with $$H = G_v$$

(Lemma 4.12 Levitt)

Let $$G_v$$ be a vertex stabilizer of an $$( \mathcal {A, H} )$$ tree T. Any splitting of $$G_v$$ relative to $$Inc^{\mathcal{H}}_v$$ extends (non-uniquely) t0 a splitting of G relative to $$\mathcal {H}$$.

More precisely, given an $$(\mathcal{A}_v Inc^{\mathcal{H}}_v )$$ tree $$S_v$$, there exist an $$( \mathcal {A, H} )$$ tree $$\hat {T}$$ and a collapse map $$p: \hat{T} \to T$$ such that $$p^{-1} (v)$$ is $$G_v$$ – equivariantly isomorphic to $$S_v$$

We say that $$\hat{T}$$ is obtainedby refining T at v using $$S_v$$. More generally, one may choose a splitting for each orbit of vertices of T, and refine T using them. Any refinement of T may be obtained by this construction (possibly with non minimal trees $$S_v$$ ).

Proof: ## By Ashani Dasgupta

Pursuing Ph.D. in Geometric Group Theory at University of Wisconsin, Milwaukee