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Normal form theorem for graphs of groups. Let $\mathcal{G}$ be a graph of groups and $G=\pi_1(G)$.

1. Any $g \in G$ can be written as
$g = g_0 t_{e_1}^{\epsilon_1} g_1 \cdots t_{e_n}^{\epsilon_n} g_n\quad$
as before.
2. If $g = 1$, this expression includes `backtracking’, meaning that for some $i$, $e_i = e_{i + 1}$ with $\epsilon_i = - \epsilon_{i + 1}$, and furthermore that if $\epsilon_i = \pm 1$, then $g_i \in \partial_\pm ( G_{e_i} )$.

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

Proof. To simplify notation, set $t_i = t_{e_i}^{\epsilon_i}$, so

$g = g_0 t_1 g_1 \cdots t_n g_n$.

Fix base points $*_i$ in the vertex spaces $X_{v_i}$, which are chosen to coincide when the vertices do. Then $g_i$ is a loop in $X_{v_i}$ based at $*_i$, and $t_i$ is a path, crossing the corresponding edge space, from $*_i$ to $*_{i + 1}$. This allows us to consider $g$ as a loop in $X_{\mathcal{G}}$ based at $*_0$. (We may assume $v_0 = v_n$ by adding letters from a maximal tree.)

Consider the universal covering $\widetilde{X}_{\mathcal{G}}$ and fix a base point $\widetilde{*}_0$ over $*_0$ in $\widetilde{X}_{\widetilde{v}_0}$. Let $\widetilde{g}$ be the lift of $g$ based at $\widetilde{*}_0$ and $\gamma$ its image in the Bass–Serre tree $T$. We now analyze $\widetilde{g}$ and $\gamma$ closely.

Choose $\widetilde{e}_i$ adjoining $\widetilde{X}_{\widetilde{v}_0}$ and $\widetilde{X}_{\widetilde{v}_1}$ so that the edge traversed by $t_i$ when lifted at $\widetilde{*}_{i - 1}$ corresponds to the coset $1 \cdot G_{e_i} \subseteq G_{v_i} / G_{e_i}$.

Then $g_0$ lifts to a path in $\widetilde{X}_{\widetilde{v}_0}$ which terminates at $g_0 \widetilde{*}_0$. Similarly, $t_1$ lifts at $*_0$ to a path across the edge $\widetilde{e}_1$ to the vertex space $t_1\widetilde{X}_{\widetilde{v}_1}$ terminating at $t_1 \widetilde{*}_1$. Therefore, $g_0 t_1$ lifts at $\widetilde{*}_0$ to a path which crosses the edge space $g_0 \widetilde{e}_1$ and ends at $g_0 t_1 \widetilde{*}_1$.

Then, $g_1$ lifts at $\widetilde{*}_1$ to a path in $\widetilde{X}_{\widetilde{v}_1}$ ending at $\widetilde{*}_1$, and $t_2$ lifts at $\widetilde{*}_1$ to a path across the edge $\widetilde{e}_2$ into the vertex space $\widetilde{X}_{\widetilde{v}_2}$, and terminating at $t_2 \widetilde{*}_2$. Thus $g_0 t_1 g_1 t_2$ lifts at $\widetilde{*}_0$ to a path which crosses $g_0 \widetilde{e}_1$, through $g_0 \widetilde{X}_{\widetilde{v}_1}$, across $g_0 t_1 g_1 \widetilde{e}_2$, and ending at

$g_o t_1 g_1 t_2 \widetilde{*}_2 \in g_0 t_1 g_1 t_2 \widetilde{X}_{\widetilde{v}_2}$.

We continue this process until we have explicitly constructed $\widetilde{g}$. By hypothesis, $g = 1$, so $\widetilde{g}$ and $\gamma$ are both loops in $\widetilde{X}_{\mathcal{G}}$ and $T$, respectively. Since $T$ is a tree, $\gamma$ must backtrack.

This implies that $\widetilde{e}_i=\widetilde{e}_{i + 1}$ and that $\epsilon_i = -\epsilon_{i + 1}$. That is, by Lemma 18,

$g_{i + 1} \in \partial_\pm^{e_i} ( G_{e_i} )$.

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

Last time we proved that if $\mathcal{G}$ is a graph of groups, then $G = \pi_1 \mathcal{G}$ acts on a tree $T$, namely the underlying graph of the universal cover of $X_\mathcal{G}$.  Lemmas 16, 17, and 18 describe the vertex and edge sets of $T$ and the action of $G$.  The following is immediate.

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

Corollary: If $\mathcal{G}$ is a graph of groups and $v \in V(\Gamma)$, then the map $G_v \rightarrow G = \pi_1 \mathcal{G}$ is injective.

The tree $T$ is called the Bass-Serre Tree of $\mathcal{G}$.  We will usually equip $T$ 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 $G = A \ast_C B$.  If $g \in G$ is torsion, then $g$ is conjugate into $A$ or $B$.

Proof. Let $T$ be the Bass-Serre Tree.  Fix a vertex $v \in T$ amd consider $\{g^n v \mid n \in \mathbb{Z} \}$.  This is a finite set.

Exercise 19: There is an $x \in T$ that minimizes $\mathrm{max}_{n \in \mathbb{Z}} d(x, g^n v)$ (when $g$ is torsion).

Such an $x$ is fixed by $g$.  Therefore, $g$ stabilizes a vertex.  The stabilizers of the vertices are precisely the conjugates of $A$ and $B$$\Box$

Our next goal is to understand the elements of $G = \pi_1 \mathcal{G}$ — which are trivial and which not?  We’ll start with a presentation.  If $\Gamma$ is a tree, then $G$ can be thought of as a sequence of amalgamated free products.  If $\Gamma$ is a rose, then $G$ can be thought of as a sequence of HNN-extensions.  In general, fix a maximal tree $\Lambda \subseteq \Gamma$.  This determines a way of decomposing $G$ as a sequence of amalgamated products followed by a sequence of HNN-extensions.  This process is sometimes called “Excision.”  It follows that $G = \pi_1 \mathcal{G}$ has a “presentation” like this:

$G = \langle \{ G_v \mid v \in V(\Gamma) \}, \{t_e \mid e \in E(\Gamma)\} \mid \\ \{t_e \partial_+^e(g)t^{-1}_e = \partial_-^e(g) \mid e \overline{t}(\Gamma), g \in G_e\}, \{t_e = 1 \mid e \in E(\Lambda)\} \rangle$

We can write any $g \in G$ in the following form:

$g = g_0 t^{\epsilon_1}_{e_1} g_1 t^{\epsilon_2}_{e_2} \cdots t^{\epsilon_{n-1}}_{e_{n-1}} g_{n-1}t^{\epsilon_n}_{e_n} g_n$

where $\epsilon_i = \pm 1$, $e_1^{\epsilon_1} \cdots e_n^{\epsilon_n}$ is a loop in $\Gamma$, and $g_i \in G_{v_i}$ where $v_i$ is the terminus of $e_i^{\epsilon_i}$ and the origin of $e_{i+1}^{\epsilon_{i+1}}$.  A priori, we just know that $g = g_0 t^{\epsilon_1}_{e_1} g_1 \cdots t^{\epsilon_n}_{e_n} g_n$.  Suppose, say, that $e_1$ doesn’t originate from $v_0$.  Then insert stable letters corresponding to edges in $\Lambda$ that form a path from $v_0$ to the origin of $e_1$.  Now repeat.

Last time we saw that if $G$ acts on a tree, $T$ then $G/T$ has the structure of a graph of groups (remeber stabilizers). Such a $\mathcal{G}=G/\Gamma$ is called developable.

Exercise 18: Show that $SL_2\mathbb{Z}\cong(\mathbb{Z}/4\mathbb{Z})\ast_{\mathbb{Z}/2\mathbb{Z}}(\mathbb{Z}/6\mathbb{Z})$.

Theorem 13 (Scott-Wall): Let $\mathscr{X}$ be a graph of spaces. The universal cover of $X_{\mathscr{X}}$ is itself a graph of spaces. Also, each vertex space (resp. edge space) is the universal cover of a vertex space of $\mathscr{X}$ (.edge space).

Sketch of Proof: For any $v\in V(\Xi)$, let $L_v=X_v\cup(\cup_{e\in E}X_e\times [\pm 1,0])$, where $E$ is the set of edges that are the edges incident to $v$. It should be noted that $X_v$ is a deformation retract of $L_v$. Also, recall that the edge maps are $\pi_1$ injective. From covering space theory we know that given a map $f:Y\to X$ and a covering space $\hat X$ of $X$ that $f$ lifts to a map $\hat f:Y\to \hat X$ if and only if $f_\ast (\pi_1(Y))\subset \pi_1(\hat X)$. It therefore follows that $\tilde X_v \hookrightarrow \tilde L_v$. So $\tilde L_v$ is built from $\tilde X_v$ by attaching covering spaces of edge spaces. Because $\partial_{\pm}$ is $\pi_1$ injective, these covering spaces of edge spaces really are universal covers. By iteratively gluing together copies of the $\tilde L_v$ we can construct a simply connected cover of $X_{\mathscr{X}}$.

Given $X_{\mathscr{X}}$ we have constructed the universal cover $\tilde X_{\mathscr{X}}$. The underlying graph, $T$ of $\tilde X_{\mathscr{X}}$ is a tree because $\pi_1(\tilde X_{\mathscr{X}})\to\pi_1(T)$ is a surjection. We now need to check that the action of $G=\pi_1(X_{\mathscr{X}})$ on $T$ is interesting.

Let $\mathcal{G}$ be a graph of groups and let $G=\pi_1(\mathcal{G})$. Fix a base point $\ast\in X_v \subset X_{\mathcal{G}}$  and a choice of lift $\tilde\ast\in \tilde X_{\tilde v}\subset \tilde X_{\mathcal{G}}$. Let $g\in G_v=\pi_1(X_v)$.  The space $\tilde X_{\tilde v}$ is a universal cover of $X_v$ by Theorem 13, and so the lift of $g$ to the universal cover of $X_{\mathcal{G}}$ at $\tilde\ast$ is contained in $\tilde X_{\tilde v}$. Therefore the preimages of $\ast$ that are contained in $\tilde X_{\tilde v}$ correspond to the elements of $G_v$.

Now consider $g\in G\smallsetminus G_v$.  If $g$ is lifted at $\tilde \ast$ then the terminus of this lift is not in $\tilde X_{\tilde v}$, but in some other component of the preimage of $X_v$.  Call the component where the lift terminates $\tilde X_{\tilde v_1}$. If $g,h\in G$ are such that both have lifts that terminate in $\tilde X_{\tilde v_1}$ then $h^{-1}g\in G_v$. We have just proved the following lemma.

Lemma 16: Let $\mathcal{G}$ be a graph of groups and let $T$ be the underlying graph of the universal cover $\tilde X_{\mathcal{G}}$. For any vertex $v\in V(\Gamma)$ the set of vertices of $T$ lying above $v$ is in bijection with $G/G_v$ and $G$ acts by left translation.

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

Lemma 17: For any $e\in E(\Gamma)$ the set of edges of $T$ that lie above $e$ is in bijection with $G/G_e$ and $G$ acts by left translation.

Lemma 18: If $e\in E(\Gamma)$ adjoins a vertex $v\in V(\Gamma)$ then for any $\tilde v\in V(T)$ lying above $v$ the set of edges of $T$ adjoining $\tilde v$ lying above $e$ is in bijection wiht $G_v/G_e$ and $G_v$ acts by left translation.