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**Normal form theorem for graphs of groups.** Let be a graph of groups and .

- Any can be written as

as before. - If , this expression includes `backtracking’, meaning that for some , with , and furthermore that if , then .

Similar to the case of the free group, the proof boils down to the fact that the Bass–Serre tree is a tree.

**Proof.** To simplify notation, set , so

.

Fix base points in the vertex spaces , which are chosen to coincide when the vertices do. Then is a loop in based at , and is a path, crossing the corresponding edge space, from to . This allows us to consider as a loop in based at . (We may assume by adding letters from a maximal tree.)

Consider the universal covering and fix a base point over in . Let be the lift of based at and its image in the Bass–Serre tree . We now analyze and closely.

Choose adjoining and so that the edge traversed by when lifted at corresponds to the coset .

Then lifts to a path in which terminates at . Similarly, lifts at to a path across the edge to the vertex space terminating at . Therefore, lifts at to a path which crosses the edge space and ends at .

Then, lifts at to a path in ending at , and lifts at to a path across the edge into the vertex space , and terminating at . Thus lifts at to a path which crosses , through , across , and ending at

.

We continue this process until we have explicitly constructed . By hypothesis, , so and are both loops in and , respectively. Since is a tree, must backtrack.

This implies that and that . That is, by Lemma 18,

.

Therefore, we have found a backtracking, and can accordingly shorten . This proves the theorem.

Last time we proved that if is a graph of groups, then acts on a tree , namely the underlying graph of the universal cover of . Lemmas 16, 17, and 18 describe the vertex and edge sets of and the action of . The following is immediate.

**Theorem 14 **(Serre)**:** Every graph of groups is developable.

**Corollary:** If is a graph of groups and , then the map is injective.

The tree is called the *Bass-Serre Tree* of . We will usually equip with a length metric in which each edge has length 1. The approach we’ve taken is due to Scott-Wall. Here’s a sample application.

**Lemma 19:** Let . If is torsion, then is conjugate into or .

**Proof.** Let be the Bass-Serre Tree. Fix a vertex amd consider . This is a finite set.

**Exercise 19:** There is an that minimizes (when is torsion).

Such an is fixed by . Therefore, stabilizes a vertex. The stabilizers of the vertices are precisely the conjugates of and .

Our next goal is to understand the elements of — which are trivial and which not? We’ll start with a presentation. If is a tree, then can be thought of as a sequence of amalgamated free products. If is a rose, then can be thought of as a sequence of HNN-extensions. In general, fix a maximal tree . This determines a way of decomposing as a sequence of amalgamated products followed by a sequence of HNN-extensions. This process is sometimes called “Excision.” It follows that has a “presentation” like this:

We can write any in the following form:

where , is a loop in , and where is the terminus of and the origin of . A priori, we just know that . Suppose, say, that doesn’t originate from . Then insert stable letters corresponding to edges in that form a path from to the origin of . Now repeat.

Last time we saw that if acts on a tree, then has the structure of a graph of groups (remeber stabilizers). Such a is called *developable.*

**Exercise 18:** Show that .

**Theorem 13 (Scott-Wall): **Let be a graph of spaces. The universal cover of is itself a graph of spaces. Also, each vertex space (resp. edge space) is the universal cover of a vertex space of (.edge space).

**Sketch of Proof:** For any , let , where is the set of edges that are the edges incident to . It should be noted that is a deformation retract of . Also, recall that the edge maps are injective. From covering space theory we know that given a map and a covering space of that lifts to a map if and only if . It therefore follows that . So is built from by attaching covering spaces of edge spaces. Because is injective, these covering spaces of edge spaces really are universal covers. By iteratively gluing together copies of the we can construct a simply connected cover of .

Given we have constructed the universal cover . The underlying graph, of is a tree because is a surjection. We now need to check that the action of on is interesting.

Let be a graph of groups and let . Fix a base point and a choice of lift . Let . The space is a universal cover of by Theorem 13, and so the lift of to the universal cover of at is contained in . Therefore the preimages of that are contained in correspond to the elements of .

Now consider . If is lifted at then the terminus of this lift is not in , but in some other component of the preimage of . Call the component where the lift terminates . If are such that both have lifts that terminate in then . We have just proved the following lemma.

**Lemma 16:** Let be a graph of groups and let be the underlying graph of the universal cover . For any vertex the set of vertices of lying above is in bijection with and acts by left translation.

We can also prove the following two lemmas in a similar fashion.

**Lemma 17: **For any the set of edges of that lie above is in bijection with and acts by left translation.

**Lemma 18:** If adjoins a vertex then for any lying above the set of edges of adjoining lying above is in bijection wiht and acts by left translation.

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