# 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.