In this note we discuss two notions of connectedness in the context of a metric space. These are Linear Connectedness and Local Connectedness. We wish to prove that linear connectedness implies local connectedness (however the converse is not true).
Linear Connectedness or bounded turning is a geometric condition that is defined as follows:
Suppose $(X, d)$ is a metric space. We say it is $L$-linearly connected for some $L \geq 1$ if for all $x, y \in X$ there exists a compact, connected set $J \ni x, y$ of diameter less than or equal to $Ld(x, y)$.
In order to understand local connectedness, let us first define the notion of a neighbourhood.
If $X$ is a topological space and $p$ is a point in $X$, then a neighbourhood of $p$ is a subset $V$ of $X$ that includes an open set $U$ containing $p$,
$$ p \in U \subseteq V \subseteq X $$
Local connectedness in a metric space is defined as follows.
Suppose $(X, d)$ is a metric space. Let $x \in X$ and $V \ni x$ a neighbourhood of $x$. Then there exists a connected neighbourhood $U$ such that $x \in U \subseteq V$.
We wish to show that if a metric space $X$ is linearly connected then it is locally connected. In particular, suppose $x \in X$ is any point and $V$ is any neighbourhood of $x$ in $X$. Then there exists a connected neighbourhood $U$ of $x$ that is contained in $V$.
If $V$ is a neighbourhood for $x$, then $x \in \operatorname{int}(V)$ and then there exist $\epsilon>0$ such that $B(x, \epsilon) \subseteq V$.
Recall that, by assumption, $(X, d)$ is $L$-linearly connected. Choose a point $y \in B(x, \epsilon)$ such that $d(x, y) < \frac{\epsilon}{L}$. If there is no such point then $B(x,\frac{\epsilon}{L}) = \{x\}$ is a connected open set containing $x$ and we are done.
Otherwise proceed as follows.
Since $(X, d)$ is $L$-linearly connected, there exists a compact, connected set $J \ni x, y$ of diameter less than or equal to $L \times d(x, y)$. Hence $diam(J) < L \times \frac{\epsilon}{L} = \epsilon$. This implies $ J \subseteq B(x, \epsilon)$. Therefore we have a connected compact set $J$ that contains $x$ and is contained in $V$.
We essentially proved that every neighbourhood of $x \in X$ contains a connected set $J$ that also contains $x$. This property is known as weakly locally connected or connected im-kleinan.
The precise definition goes as follows:
$X$ is connected im kleinen at $x$ iff each open neighbourhood $U$ of $x$ contains an open neighbourhood $V$ of $x$ such that any pair of points in $V$ lie in some connected subset of $U$.
or equivalently $X$ is connected im kleinen at $x$ iff each open neighbourhood $U$ of $x$ contains a connected neighbourhood $V$ of $x$ (not necessarily open).
It is a theorem that:
If $X$ is connected im kleinen at each point, then $X$ is locally connected.
Hence we are done!
If $X$ is connected im kleinen at all points $x \in X$ then $X$ is locally connected.
Suppose $x \in X$ be any point. Let $U$ be any neighbourhood of $x$ in $X$. We wish to show that there is a open set $V$ contained in $U$ that contains $x$.
Sur les Groupes Hyperboliques d’après Mikhael Gromov is a celebrated work in elementary geometric theory. In the geometric group theory research group at Cheenta Academy, we translated it from French to English.
The final copy is retracted due to a note from springer. If need more information drop an email at ashani.dasgupta@cheenta.com
Yesterday we were discussing the Heisenberg group in our weekly Geometric group theory workshop. The game of Mario came up! If you have not played Mario, then here is a the only thing you need to know. Mario can jump and hit floating bricks. These bricks may pop open to produce gold coins.
The method of simplifying products of elements in Heisenberg group is similar. First a few definitions.
Let \( \Gamma \) be a group generated by \( \{a, b, c\} \). Moreover set of relators contain \( ac = ca, bc = cb, ba = cab \). We say \(\Gamma\) is the Heisenberg group.
One way to think about this is: \( c\) commutes with everything, \(a\) and \( b\) do not commute but they are \( c\) away from actually commuting. Hence we do not have \( ba = ab \) but we have \( ba = c \cdot ab \).
We claim that every element of Heisenberg group looks like \( a^m b^n c^p \) where \( (m, n, p) \in \mathbb{Z}^3 \). We play a game of Mario with the elements of Heisenberg group to illustrate this.
Choose two elements \( a^m b^n c^p \) and \( a^r b^s c^t \) in \( \Gamma \). Lets multiply them and see what happens.
\[ a^m b^n c^p \cdot a^r b^s c^t \]
Since \( c \) commutes with everything, hence we can pull the \( c \)’s to the right and get the first simplification.
Now we focus on the \( b^n \cdot a^r \) in the middle of the expression. Think of each \( b\) as Mario and each of the \(a\)’s in \(a^r \) as a gold coin bearing brick. We will push the \( b\) to the right replacing \( b \cdot a \) by \(c \cdot a \cdot b\). The gold coin is \(c\). That is if we push the \(b\) to the right through one \(a\) then one \(c \) is produced and \(ba\) becomes \(ab\).
Now we can push that \( b \) through all the \( r \) copies of \(a\) so that it moves to the right and sits with \( b^s \). In this process \( r \) gold coins (\(c\)’s) will be produced. Hence we have \( c^r \) produced.
We will need to do this \(n \) times (for each of the Mario’s or \(b\)’s in \(b^n \)). Hence \( n \times r \) gold coins or \(c\)’s are produced in the process. After this is done we have the following expression.
We use \(NFNN\) as an abbreviation for ‘not far not near’.
Annulus \(A(w, r, R)\) in a metric space \((X, d)\), for \(w \in X, 0 < r < R < \infty \) is defined as the set \[A =\{ x \in X | r \leq d(w, x) \leq R \} \].
Definition
Suppose \(P\) is a finitely generated, one-ended group. Let \(\Gamma(P, S)\) be its Cayley graph with respect to some finite, symmetric generating set \(S\). We say \(P\) is \(M-NFNN\) if there exists an integer \(M\) such that given any two points \(x,y \in \Gamma(P)\) with \(2M \leq r_x \leq r_y\), where \(r_x = d(e,x)\) and \(r_y = d(e,y)\), there exists an arc in \(A(e, \frac{r_x}{3}, 2 r_y) \subset \Gamma(P)\) that connects \(x\) and \(y\).
Question
Characterise NFNN groups.
Remark
Finitely presented groups are \(M-NFNN\) for \(M\) equal to the largest length of a relator.
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.
Each such cut corresponds to an action on a tree T (by Van Kampen theorem).
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
Fig 6: Finitely many (2) straight lines in MOP
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.
straight arcs in 2-simplex map to edge midpoints
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.
extension to \(T_2 \)
Pattern of Dunwoody, its dual tree
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
\( \tilde{\tau}_k \subset \tilde {\tau}_{k+1} \)
\( \tau_k = \pi ( \tilde{\tau}_k ) \) be the projection of in X. It is a finite graph.
\( S_k \) 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.
Tree dual to the pattern
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.
Fig 7: Twisted tracks and bands
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.
Fig 8: Track leading to co-boundary.
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 \( \sigma \)
Fig 9: only one ‘not good’ disc D without a vertex
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.
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.
continuum
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.
cut point
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.
not an 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
\( y \in C\) 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
\( y \notin C and if \( a, b, d \in Z \) with \( a\in x , b \in Y \) and \( d \in z \), then \( [a, b) \cup (b, d] = \phi \)
interval in P – first kind
interval in P – second kind
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).
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
H acts elliptically on T means the splitting of G does not split H (or H is contained in a vertex stabilizer).
H is universally elliptic if it acts elliptically on all \( \mathcal{A} \) – trees. Example: Groups with property FA (Finite Groups).
An \( \mathcal {A} \) – tree is universally elliptic, if its edges stabilizers are elliptic in every \(\mathcal {A} \) – tree
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’.
A JSJ decomposition (or JSJ tree) of G over is an – tree such that:
T is universally elliptic (its edge stabilizers are elliptic in every \(\mathcal {A} \) – tree or fixes a point in every \(\mathcal {A} \) – tree \)
T dominates any other universally elliptic tree T’ (the vertex stabilizers are as small as possible; they are elliptic in every universally elliptic tree).
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.
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’.
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∗…∗Bk∗Ft is another such decomposition then k = r, s = t, and there exists a permutation σ∈Sr such that for each i=1,…,r the subgroups Gi and Bσ(i) are 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
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 ? 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 ?. But the fundamental group of a ?? complex with a single 0-cell has a presentation with a generator for each 1-cell and a relation for each 2-cell. Thus ?, 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:
T is universally elliptic (its edge stabilizers are elliptic in every \( \mathcal{A} \) – tree or fixes a point in every \( \mathcal {A} \) – tree \)
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:
for k large enough, there is a morphism \( S \to T_k \);
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)
Fig 1: Example of Morphism
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’.
Fig 2: Example of Collapse map, refinement
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.
Construct a connected simplicial 2-complex X such that \( \pi_1 (X) = G \)
\( (Y_i, u_i) \) be a pointed 2-complex with \( \pi_1 (Y_i, u_i) = H_i \)
Starting from the disjoint union of the \( Y_i’s \), add p edges joining the \( u_i \) ‘s to an additional vertex u.
Add # \( \Omega \) additional edges joining u to itself.
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.
I wonder if it helps to build the Eilenberg Maclane complex
X has a universal cover \(\pi : X \to \tilde {X} \)
G acts on \( \tilde {X} \) by deck transformation.
For \( i \in \{1, \cdots , p \} \) consider a connected component \( \tilde {Y}_i \) of \( \pi^{-1} (Y_i) \), whose stabilizer is \( H_i \).
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)
\( p_k : T_{k+1} \to T_k \) 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 \( e_k \)
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.
Fig 4: Extension to 2-skeleton
Pattern of Dunwoody, its dual tree
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
\( \tilde {\tau}_k \) does not intersect any \( \tilde {Y}_i \)
\( \tilde{\tau}_k \subset \tilde {\tau}_{k+1} \)
\( \tau_k = \pi ( \tilde{\tau}_k ) \) be the projection of in X. It is a finite graph as it is contained in the complement of \( Y_1 \cup \cdots \cup Y_p \).
\( S_k \) 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 \( \tilde{\tau}_k \).
Fig 5: Tree dual to the pattern
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
Fig 6: Finitely many (2) tracks in MOP
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.
Fig 7: Twisted tracks and bands
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.
Fig 8: Track leading to co-boundary.
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 \( \sigma \)
Fig 9: only one ‘not good’ disc D without a vertex
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 \( T_J \).
\( G_v \) does not split over a universally elliptic subgroup relative to \( Inc^{\mathcal{H}}_v \)
\( G_v \) 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 \( S_v \) ).
