## From K to L

• $$\mathcal{K} = \{\phi_i :A_i \to B_i \}_{i=1, … , k}$$ be a non-nesting closed system on a finite tree with at least one infinite orbit.
• Non nesting forces a special structure for the set of finite regular $$\mathcal{K}$$ – orbits
1. Union of all finite regular orbits = a finite union of open intervals
2. Its complement $$K_1$$ is a finite union of closed subtrees, not all of them points.
• Disregard isolated points of $$K_1$$
• D = disjoint union of all closed edges of $$K_1$$
• D is a multi – interval (a finite disjoint union of compact intervals)
• $$\delta D$$ = set of endpoints of compact intervals in D.
• int D = $$D \backslash \delta D$$
• The system of maps $$\mathcal{K}$$ naturally induces a system $$\mathcal{L}$$ on D
• Replace each $$\phi_i$$ by the collection of its restrictions to edges of $$K_i$$, keeping only those maps whose domains contains more than a single point.
• We may split domains at preimages of vertices of $$B_i$$
• The system $$\mathcal{L}$$ is closed and non-nesting
• Every regular orbit of $$\mathcal{L}$$ is infinite
• Every $$\mathcal{L}$$ invariant probability measure without atoms corresponds to a $$\mathcal{K}$$ invariant probability measure (supported on $$K_1$$ )
• Hence it is sufficient to find an $$\mathcal{L}$$ invariant measure

## From L to M

• Let $$\overset{o}{\mathcal{L}}$$ be the open system obtained by replacing each element $$\phi : A \to B$$ of $$\mathcal{L}$$ to the open interval A \ {endpoints}.
• Hardest case: every minimal closed(?) $$\overset{o}{\mathcal{L}}$$ invariant set is a finite singular orbit (we need dense orbits).
• Why is this hardest? What is singular orbit?
• We want every orbit to be dense (and the system to be open) to apply Sacksteders theorem.
• In the worst case, none-of the orbits are dense. Since they are infinite, there must be accumulation points (every infinite sequence has a convergent subsequence). These accumulation points could be inside or at the end point
• If we look at the accumulation points they form a minimal invariant set. If they are finite and singular, it is hardest to get and infinite dense orbit out of them.
• There are finitely many minimal sets, among which are the endpoints of $$d \in \delta D$$
• By splitting D (thus changing $$\mathcal{L}$$ and $$\overset{o}{\mathcal{L}}$$, we may assume that the points of $$\delta D$$ are the only minimal sets of $$\overset{o}{\mathcal{L}}$$
• Note that the closure of every $$\overset{o}{\mathcal{L}}$$-orbit contains an end point of D.
• For every $$d \in \delta D$$, we choose an $$\mathcal{L}$$ word $$w_d$$ defined on a non – degenerate closed interval $$I_d$$ containing d and sending d to a point $$\tilde {d} \in int D$$
• Such a word exists because otherwise non-nesting of $$\mathcal{L}$$ would imply finiteness of orbits near d.
• $$E = \{ p \in \delta D\}$$ such that some $$\overset{o}{\mathcal {L}}$$ – orbit accumulates on p.
• This implies we can get points of some orbit arbitrarily close to p.
• Choose $$x_p , y_p$$ from these points (see next).
• For each $$p \in E$$ choose $$x_p, y_p \in I_p$$ such that some $$\overset{o}{\mathcal {L}}$$ – word $$\tau_p$$ sends $$x_p$$ to $$y_p$$ in an orientation preserving way.
• Since every $$\overset{o}{\mathcal {L}}$$ – orbit accumulates on E, we may require, furthermore, that for every $$d \in \delta D$$ the orbit of $$\tilde{d}$$ meets the union of all intervals $$(x_p , y_p ), p \in E$$.
• Notice that previously we found every end point d went to some interior point $$\tilde {d}$$
• Look at the orbit of $$\tilde {d}$$. Clearly it accumulates at some end point $$p \in E$$
• There are finitely many d (hence $$\tilde {d}$$ ). Thus for these we may arrange $$(x_p , y_p ), p \in E$$ such that orbit of d intersects the union of $$(x_p , y_p ), p \in E$$
• After changing $$w_d$$, we may assume that every $$\tilde {d}$$ belongs to some $$(x_p , y_p ), p \in E$$.
• Note that every $$\overset{o}{\mathcal {L}}$$ orbit in int D meets some $$(x_p , y_p )$$ infinitely often (why?)
• For $$p \in E$$, $$C_p$$ = circle obtained by identifying the endpoints of $$[x_p , y_p]$$.
• C = disjoint union of these circles.

Goal: Construct a non-nesting system of maps $$\mathcal{M}$$ on C with every orbit infinite, in such a way that an $$\mathcal{M}$$ -invariant measure on C provides an $$\mathcal{L}$$ invariant measure on D

• Construct a finite family $$\mathcal {P}$$ of homeomorphisms $$\lambda : U \to V$$ between open intervals $$U \subset int D$$ and intervals $$V \subset C$$. It will have three types of maps:
1. For each $$p \in E$$, include the natural map $$\pi_p : (x_p , y_p) \to C_p$$
2. Then for each p, we consider the points $$(x_p , y_p )$$ and their common image $$z_p \in C$$. Restrict $$\tau_p$$ to a homeomorphism between a small neighborhood $$X_p$$ of $$x_p$$ and a neighborhood $$Y_p$$ of $$y_p$$. Since $$\tau_p$$ preserves orientation, both these neighborhoods are naturally homeomorphic to a neighborhood $$Z_p$$ of $$z_p$$ in $$C_p$$. We include the maps $$X_p \to Z_p$$ and $$Y_p \to Z_p$$
3. The ranges of the maps constructed so far cover C. If domains cover int D then we may stop. Otherwise, we need a third type of maps. Let $$x \in int D$$. Some $$\overset{o}{\mathcal {L}}$$ word $$\alpha_x$$ sends a neighborhood $$U_x$$ of x to an open interval $$V_x$$ contained in some $$(x_p , y_p)$$. The word $$\alpha_x$$ may be chosen to be constant near the end point of D. But compactness (?), we deduce that int D may be covered by finitely many $$U_x$$. We include the corresponding finite set of maps from $$U_x \to \pi_p (V_x) \subset C_p$$
• This defines the family $$\mathcal {P}$$. We now use it to carry $$\overset{o}{\mathcal {L}}$$ over to a system $$\mathcal {M}$$ on C.
• Let $$\gamma_1 : U_1 \to V_1$$ and $$\gamma_2 : U_2 \to V_2$$ be two elements of $$\mathcal{P}$$. If $$U_1 \cap U_2 \neq \phi$$, then $$\gamma_2 \gamma_1^{-1}$$ is a homeomorphism between two subintervals of C. We include it in $$\mathcal{M}$$.
• Similarly, for $$\theta \in \overset{o}{\mathcal {L}}$$ , we include $$\gamma_2 \theta \gamma_1^{-1}$$ is its domain in non-empty.
• $$\mathcal {M}$$ = set of all maps thus obtained, for all possible choices of $$\gamma_1, \gamma_2 \in \mathcal {P}$$ and $$\theta \in \overset{o}{\mathcal {L}}$$ .
• It is an open system of maps in C.
• The construction of $$\mathcal {P}$$ and $$\mathcal{M}$$ was done in such a way that the following properties hold:
• Given $$\gamma_1, \gamma_2 \in \mathcal {P}$$ and an $$\overset{o}{\mathcal {L}}$$ word w, given x in the fomain of $$\gamma_2 w \gamma_1^-1$$, there is an $$\mathcal {M}$$ word equal to $$\gamma_2 w \gamma_1^{-1}$$ on the neighborhood of x.
• Given an $$\mathcal {M}$$ word w, and y in the domain of $$\gamma_2^{-1} w \gamma_1$$ some $$\overset{o}{\mathcal {L}}$$ word coincides with $$\gamma_2^{-1} w \gamma_1$$ near y.
• Every $$\mathcal {M}$$ orbit is infinite, because every $$\overset{o}{\mathcal {L}}$$ orbit in int D meets some $$(x_p , y_p)$$ infinitely often.
• Also note that $$\mathcal{M}$$ is non nesting because every orientation preserving $$\mathcal{M}$$ word that has a fixed point is a restriction of the identity (why?).
• Furthermore, any $$\mathcal {M}$$ invariant probability measure with no atom lifts to an $$\overset{o}{\mathcal {L}}$$ invariant measure on D.
• The existence of the words $$w_d$$ (which depends on the non nesting of $$\mathcal {L}$$ implies that this measure has finite total mass, hence extends to $$\mathcal{L}$$ invariant measure on D.

We have now reduced our problem to finding an $$\mathcal{M}$$ invariant measure with no atom (assuming that every minimal set of $$\overset{o}{\mathcal {L}}$$ is infinite )

## From M to N

• Let $$F \subset C$$ be a minimal set of $$\mathcal {M}$$.
• It is infinite and has no isolated point.
• For each component $$C_p$$ of C that meets F, we choose a collapsing map $$\rho_p : C_p \to C_p’$$ where $$C_p’$$ is another circle and $$\rho_p$$ sends each component of $$C_p \backslash (F \cap C_p )$$ to a point.
• Let C’ be the union of circles $$C_p’$$
• Given an element $$\phi : A \to B$$ of $$\mathcal {M}$$ whose domain meets F, we consider images A’, B’ of A, B in C’ and the natural homeomorphism $$\phi’$$ between interiors of A’ and B’ (note that A’ , B’ are non degenerate intervals, but they need not be open).
• $$\mathcal{N}$$ = collection of these $$\phi’$$ . It is an open system of maps on C’.
• $$\mathcal {N}$$ is non -nesting and every regular orbit is dense, because every $$\mathcal{M}$$ orbit contained in F is dense in F (why?) but there may be finite singular orbits.

## From N to O

• Choose an interval [x, y] disjoint from all finite singular orbits such that some orientation preserving $$\mathcal {N}$$ word sends x to y.
• Perform the same operation as in (L to M) so as to obtain a system $$\mathcal{O}$$ on the circle obtained by identifying the endpoints of [x, y]
• This last system is open, non-nesting with every orbit dense.
• By Sacksteder’s theorem it admits an invariant measure.
• This measure first lifts to an $$\mathcal {N}$$ invariant measure, then to the required $$\mathcal {M}$$ invariant measure.
• This completes the proof when every minimal set of the original system $$\overset{o}{\mathcal {L}}$$ is finite.
• If $$\overset{o}{\mathcal {L}}$$ has an infinite minimal set F, we first collapse to a point every component of D \F (as in M to N) .
• We obtain a system $$\mathcal {N}$$ on a multi-interval with every regular orbit dense and we deal with it as before.

## Dunwoody – Day 6

A JSJ decomposition (or JSJ tree) of G over A is an A-tree T such that:

• T is universally elliptic;
• T dominates any other universally elliptic tree T’.

An A-tree is universally elliptic if its edge stabilizers are elliptic in every A-tree.

Recall that H is elliptic in T if it fixes a point in T (in terms of graphs of groups, H is contained in a conjugate of a vertex group).

Definition 2.1 (Ellipticity of trees). T1 is elliptic with respect to T2 if every edge stabilizer of T1 fixes a point in T2.

Note that T1 is elliptic with respect to T2 whenever there is a refinement $$\hat{T}1$$ of T1 that dominates T2. Edge stabilizers of T1 are elliptic in $$\hat{T}1$$, hence in T2. We show a converse statement.

Theorem: If G is finitely presented, then the JSJ deformation space D(JSJ) of G over A exists. It contains a tree whose edge and vertex stabilizers are finitely generated.

## Rips Machine

Let G be a finitely presented group acting minimally, stably and non trivially by isometries on an $$\mathbb{R}$$ tree S. If G does not split over an arc stabilizer of S, then one of the following is true:

1. There is a line $$L \subset S$$ acted on by a subgroup H < G and $$N \triangleleft H$$ the kernel of the action of H on S, so that H / N is virtually $$\mathbb{Z}^n$$ for some n > 1.
2. There is a closed hyperbolic cone 2-orbifold F and a normal subgroup $$N \triangleleft G$$ with $$\pi_1 (F) \cong G/N$$. Furthermore the action of G on S factors through $$\pi_1 (F)$$.
3. There is a finite graph of groups decomposition, $$\Gamma_1$$ of G with H < G a vertex group having the following properties.
1. There is F, a cone 2-orbifold with boundary, and a normal subgroup $$N \triangleleft H$$ with $$H/N \cong \pi_1 (F)$$, so that the action of H on S factors through $$\pi_1(F)$$
2. The edge groups of H in $$\Gamma$$ correspond to the peripheral subgroups of F.
3. All edge groups act trivially.

## Dunwoody’s accessibility theorem – Talk Day 4

This is a personal musing. Possible errors, uncredited excerpts lie ahead.

We constructed sequence of equivariant maps $$f_k$$ from the universal cover $$\tilde {X}$$ to the sequence of refinements $$T_k$$.

The construction was complete up to the 1-skeleton. We want to extend the maps to 2-skeleton in a certain way. To motivate the extension, let us consider the following example.

### Motivating example

Suppose S be a four holed torus. We will cut it along curves that are homotopically non-trivial, and non parallel.

1. Each such cut corresponds to an action on a tree T (by Van Kampen theorem).
2. Adding more curves is like refining the tree.

This process must terminate and eventually we will end up drawing curves that are parallel to the previously drawn curves. This is because, each time we draw a curve, euler characteristic of the parts would add up to give the euler euler characteristic of sum (circle’s euler characteristic is 0). Hence we have a reduction in the complexity of euler characteristic.

(This was presented by Professor Jonah Gastor in the colloquium in fall )

### Back to $$\tilde {X}$$

We wish to reverse this process.

Given a tree we wish to ‘cut’ the simplicial complex using 1-dimensional structure called tracks.

### Tracks:

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.

Refinement of trees will add more tracks.

We will show that eventually we will end up having parallel tracks (by a complexity reduction argument of the ran of first cohomology group).

### Extension to 2-simplices

Consider the map $$f_1$$ from $$\tilde {X}$$ to $$T_1$$.

We have already defined the map on the 1-skeleton. Suppose abc is a 2-simplex in $$\tilde {X}$$ such that $$f_1(a), f_1(b), f_1(c)$$ are not pairwise equal.

As $$T_1$$ is a tree, $$f_1(a), f_1(b), f_1(c)$$ is a tripod. Suppose M is the median of the tripod. It will have unique pre-image on each edge of abc. Also midpoint of each edge of the tripod will have exactly one pre-image on two of the three edges of the 2-simplex [abc].

We join the pair by a straight arc and map that arc to the corresponding midpoint.

We may extend the $$f_1$$ to the remainder of 2-simplex linearly. In the next step we will throw away whatever is happening in the remaining space and recreate the map.

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. $$\tilde{\tau}_k \subset \tilde {\tau}_{k+1}$$
3. $$\tau_k = \pi ( \tilde{\tau}_k )$$ be the projection of in X. It is a finite graph.
4. $$S_k$$ be the tree dual to the pattern $$\tilde{\tau}_k$$

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

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 B 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

#### 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:

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.

## Cut points in Bowditch Boundary of Relatively hyperbolic groups 2

Understanding Swenson (large excerpt.. some diagrams .. some remarks). Please be cautious. Potentially wrong remarks lie ahead.

### Continuum

A continuum is a compact connected Hausdorff space.

### Cut Point

In a continuum Z, $$c \in Z$$ is a cut point if $$Z = A \cup B$$ where A and B are non-singleton continua and $$A \cap B = \{c\}$$. If in addition $$D \subset A – \{c\}$$ and $$E \subset B – \{ c \}$$, we say that c separates D from E.

For the remainder of this section, Z will be a metric continuum, G will be a group (possibly trivial) of homeomorphisms of Z, and $$C \subset Z$$ will be a G-equivariant (GC = C) set of cut points of Z.

### Is a member of (interval)

For $$a, b \in Z$$ and $$c \in C$$, we define $$c \in (a, b)$$ if there exist non-singleton continua A containing a and B containing b with $$A \cup B = Z$$ and $$A \cap B = \{ c \}$$.

Any point in an interval is a cut point by definition of interval.

We define the closed and half open intervals in the obvious way i.e., $$[a, b] = \{a, b \} \cup (a, b)$$, and $$[a, b) = \{a\} \cup (a, b)$$ for $$a \neq b$$ ( $$[a, a) = \phi$$ )

Notice that if $$c \in (a, b)$$ then for any subcontinuum $$Y \subset Z$$ from a to b ($$a, b \in Y$$ ), $$c \in Y$$. Why? Suppose Y is a subcontinuum containing a and b. $$c \in (a, b)$$ implies there are continua A and B such that $$a \in A, b \in B, A \cup B = Z, A \cap B = \{ c \}$$.

$$A \cap Y$$ and $$B \cap Y$$ compact and Hausdorff (hence are subcontinua) containing a and b respectively. Their union is Y (as all y in Y are either in A or B as A union B is Z and Y is contained in Z). Their intersection is either $$\phi$$ or {c}.

confusion

### Equivalence of points

For $$a, b \in Z – C$$ we say that a is equivalent to b, a ~ b, if $$(a, b) = \phi$$. That is there is no cut point between a and b.For $$c \in C$$, c is equivalent only to itself. This is clearly an equivalence relation, so let P bet the set of equivalence classes of Z. We will abuse notation and say $$C \subset P$$ since each element of C is its own equivalence class.

Observe that for $$a, b, d\in Z$$ if a ~ b, then (a, d) = (b, d). We can therefore translate the interval relation on Z to P and we also enlarge it as follows.

### Membership in P (interval in P)

For $$x, y, z \in P$$, we say $$y \in (x, z)$$ if either

1. $$y \in C$$ where $$y \in (a, b)$$ for some $$a, b \in Z$$ with $$a \in x$$ and $$b \in z$$ (note that x, and z are equivalence classes ) or
2. $$y \notin C and if \( a, b, d \in Z$$ with $$a\in x , b \in Y$$ and $$d \in z$$, then $$[a, b) \cup (b, d] = \phi$$

Since C was chosen to be G invariant, the action of G on Z gives an action of G on P which preserves the interval structure ( we have not given P a topology so it doesn’t make sense to ask if the action is by homeomorphism).

## Dunwoody’s Accessibility Theorem – Day 2

Suppose G is a finitely presented group. Let us fix $$\mathcal{A}$$ – a favorite class of subgroups of G (closed under taking subgroups and conjugation).

If G acts on an $$\mathcal{A}$$ – tree T, we have a graph of groups decomposition for G.

If H < G then H acts on T and we have graph of groups decomposition for H.

## Definitions

1. H acts elliptically on T means the splitting of G does not split H (or H is contained in a vertex stabilizer).
2. H is universally elliptic if it acts elliptically on all $$\mathcal{A}$$ – trees. Example: Groups with property FA (Finite Groups).
3. An $$\mathcal {A}$$ – tree is universally elliptic, if its edges stabilizers are elliptic in every $$\mathcal {A}$$ – tree
4. T dominates T’ means there exists a G-equivariant map from T to T’. This further implies that if some subgroup H is not split by T then H is not split by T’.
5. A JSJ decomposition (or JSJ tree) of G over  is an  – 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).
3. Example: If  $$\mathcal {A}$$ only contains the trivial group, JSJ trees are same as Grushko trees.

## Motivation from JSJ Decomposition of compact 3 manifold

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.

### Comparing JSJ decomposition of groups and manifold

Graph of groups decomposition of G over $$\mathcal{A}$$.

Graph of groups for H < G

H acts elliptically on T

H is universally elliptic

JSJ decomposition of a 3 manifold by cutting along embedded family of tori

H ‘cut’ by the family of tori.

Tori can be pushed off of H

H is not cut by any embedded torus

## Goal

We wish to show that for every sequence of graph of groups decomposition of a finitely presented group G, over a favored class of subgroups $$\mathcal {A}$$, there exists a JSJ decomposition that dominates all decompositions eventually.

## Cut points in Bowditch Boundary of Relatively hyperbolic groups 1

This document is a personal musing. It has many excerpts without credit, potentially false claims, and misquotes. If some cosmic accident has lead you to this page, then take a deep breath and assume caution. If you are worried about copyright infringement, kindly let me know. I will modify the document.

B.H. Bowditch thought about cut points in the boundary of relatively hyperbolic groups. His comments in the page 64 of the following monologue is particularly interesting.

Bowditch says that if the boundary of a relatively hyperbolic group is connected then each of its global cut points is a parabolic fixed point.

“For this one need to place certain mild restrictions on the class of groups that can occur as maximal parabolic groups. (It is sufficient to assume that they are one or two-ended, finitely presented, and not infinite torsion groups. Probably only the last of these assumptions is really important.)”

Do we really need all three of these conditions? That is, do the maximal parabolic subgroups need to be

• one or two ended
• finitely presented
• not infinite torsion group (Bowditch thinks only this one is really important)

In the same monologue he refers to Swenson’s paper which has an ‘alternative route’.

Bestvina’s paper hints at a route that may get rid of these extra conditions.

There are three components of his thought:

1. relatively hyperbolic group
2. boundary of a group
3. cut points

We will try to explore each of these ideas.

## Day 3 (notes from Craig’s Lecture)

#### Definition

A subset N of a space X is a neighborhood of infinity if $$\bar{X /N }$$ is compact.

We say that X has k ends ( $$k \in \mathbb{N} \cup \{ \infty \}$$ ) if $$k = sup \{ j | X \textrm{has a nbd of} \infty \textrm{with j unbounded components} \}$$

#### Example

• X has zero ends iff X is compact (for nice spaces)
• $$\mathbb{R}$$ has at least two ends. (Ex. exactly two)
• $$\mathbb{R}^n, n \geq 2$$ has at least one end. (Ex. exactly one)
• 3 ends
• Infinitely many ends
• Infinitely many ends

Remark: The above does not work well with bad spaces.

Goal: Develop a better theory of ends which will allow us to distinguish the last two examples.

Assume from now on that X is connected, locally connected, locally compact and Hausdorff.

### Lemma A

Let $$C \subseteq X$$, compact. Then X\C has finitely many unbounded components.

Proof: Let $$\{ U_{\alpha} \}_{\alpha \in A}$$ be the set of unbounded components of X\C.

By local connectedness each $$U_{\alpha}$$ is open in X \ C, and in X.

Also each $$U_{\alpha}$$ is closed in X\C.

By connectedness each $$U_{\alpha}$$ has limit points in C.

By local compactness there exists a compact set $$D \subseteq X$$ such that $$C \subseteq int D$$

No $$U_{\alpha}$$ is contained in D, otherwise it would be bounded.

Claim: No $$U_{\alpha}$$ is contained in X \ D.

Otherwise $$\bar {U_{\alpha}} \subseteq$$ X \int D which implies $$U_ {\alpha }$$ has no limit point in C.

Since $$U_{\alpha}$$ is connected and contains points of both D and X\D then $$U_{\alpha} \cup Fr D \neq \phi$$

For each $$\alpha \in A$$, choose $$x_{\alpha} \in U_{\alpha} \cap Fr D$$.

By compactness, $$\{y_{\alpha} \}$$ has a limit point x in Fr D. Then $$x \in$$ X\C. Choose a connected nbd V of xlying in X \C

Choose $$x_{\alpha_1 } \neq x_{\alpha_2} \in V$$ (uses Hausdorff). Then $$U_{\alpha_1} \cup V \cup U_{\alpha_2 }$$ is connected and lies in X \ C.

Remark

It is possible for X \ C to have infinitely many bounded components.

### Lemma B

If $$C \subseteq X$$ compact, and $$C_0$$ is the union of C with all of the bounded components of X \ C, then $$C_0$$ is compact.

(and its complementary components are precisely the unbounded components of C).

Proof: Choose compact set D, such that $$C \subseteq int D$$.

Let $$\mathcal{E}$$ be the set of bounded components of X \ C.

& C’ = $$C \cup ( \cup \{ E \in \mathcal{E} | E \subseteq D \}$$

Then $$C’ \subseteq D$$ and X \ C’ is open so C’ is closed and hence compact.

claim: All but finitely many elements of $$\mathcal {E}$$ lie in D.

Proof: Each element of $$\mathcal{E}$$ has its frontier in C. so if $$E \in \mathcal{E}$$ does not lie in D it contains points of Fr D.

Now use an argument like before to finish the claim.

Now $$C_0 = C’ \cup \{\cup_{i=1}^n E_i \}$$ whose each $$E_i$$ is bounded and has frontier in C.

So $$C_0$$ is compact.

Recall: A partially ordered set (poset) is a set A together with a relation $$\leq$$ satisfying:

• $$a\leq a \forall a \in A$$
• if $$a \leq b$$ and $$b \leq a$$ then a = b
• if $$a \leq b$$ and $$b \leq c$$ then $$a \leq c$$

Ex. For any set X, $$(P(X), \subseteq )$$ is a poset.

Ex. Every subset of a poset is a poset.

Hence if $$\mathcal{W} \subseteq {W}$$ then $$\mathcal{W}$$ is a poset under $$\subseteq$$ relation.

A poset is a pair, $$(A, \leq )$$, is directed, if for any pair of elements if $$\forall a, b, \in A$$ there exist $$c \in A$$ such that $$a \leq c$$ and $$b \leq c$$

Let $$\mathcal{C}$$ be a category, for example: Sets, Topological Spaces, Simplicial maps, CW complexes, Groups, Rings, Modules

#### Inverse System

An inverse system $$\{X_{\alpha} , f_{\alpha}^{\beta}, A \}$$ in $$\mathcal{C}$$ is

• a directed set $$(A, \leq)$$
• An object $$X_{\alpha} \in \mathcal{C}$$ for each $$\alpha \in A$$
• For any $$\alpha \leq \beta$$ in A, a morphism $$f_{\alpha}^{\beta} : X_{\beta} \to X_{\alpha}$$ satisfying
• $$f_{\alpha}^{\alpha} = id_{X_{\alpha}} \forall \alpha \in A$$
• if $$\alpha \leq \beta \leq \gamma$$ then $$f_{\alpha}^{\gamma} = f_{\alpha}^{\beta} \cdot f_{\beta}^{\gamma}$$
• We call the $$f_{\alpha}^{\beta}$$’s bonds (or bonding maps).

An inverse limit of $$\{X_{\alpha} , f_{\alpha}^{\beta}, A \}$$ is an object X of $$\mathcal{C}$$ satisfying $$\forall \alpha \in A$$ a morphism $$p_{\alpha} : X \to X_{\alpha}$$ such that

(i) if $$\alpha \leq \beta$$ then $$p_{\alpha} = f_{\alpha}^{\beta} \cdot p_{\beta}$$

(ii) If Z an object of $$\mathcal {C}$$ satisfies: $$\forall \alpha \in A$$ there exists $$q_ {\alpha} : Z \to X_{\alpha}$$ such that $$q_{\alpha} = f_{\alpha}^{\beta} \cdot q_{\beta} \forall \alpha \leq \beta$$ then there exists unique morphism $$q : Z \to X$$ such that $$p_{\alpha} \cdot q = q_{\alpha} \forall \alpha \in A$$

Ex. If X as above exists, then it is unique up to isomorphism in $$\mathcal{C}$$

## 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 $$T_k$$

#### 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: