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As before, are $\chi'$ and $\chi$ are graphs of spaces equipped with maps $\Phi \colon X_{\chi'} \to X_{\chi}$, $\varXi' \to \varXi$, $\phi_{v'} \colon X_{v'} \to X_v$, and $\phi_{e'} \colon X_{e'} \to X_e$ such that

commutes.

Lemma 28: Suppose that every edge map of $\chi'$ is an elevation.  Then the map $\Phi$ is $\pi_1$-injective.

Proof: The idea is to add extra vertex spaces to $\chi'$ so that $\chi'$ satisfies Stalling’s condition.  As before, we have inclusions:

If $\chi'$ does not satisfy Stalling’s condition then one of these maps is not surjective.  Without loss of generality, suppose that

does not arise as an edge map of $\chi'$.  Suppose $\partial_{e}^{-} \colon X_e \to X_u$.  Then $(\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'}))$ is a subgroup of $\pi_1(X_u)$.  Let $X_{u'}$ be the corresponding covering space of $X_u$.  Since $\pi_1(X_{u'})$ contains $(\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'}))$, there is a lift $\partial_{e'} \colon X_{e'} \to X_{u'}$ of $\partial_{e'} \circ \phi_{e'}$.  Now replace $\chi'$ by $\chi' \cup X_{u'}$ and repeat.  After infinitely many repetitions, the result $\hat{\chi}$ satisfies the hypothesis of Stalling’s condition, and $\chi'$ is contained in a subgraph of spaces.

Theorem 18: If $G_{+}$ and $G_{-}$ are residually finite groups then $G_{+} * G_{-}$ is also residually finite.

Proof: Let $X_{\pm} = K(G_{\pm}, 1)$, let $e = *$ be a point, and fix maps $\partial^{\pm} \colon e \to X_{\pm}$.  This defines a graph of spaces $\chi$.  By the Seifert-van Kampen Theorem, $\pi_1(X_{\chi}) \cong G_{+} * G_{-}$.  Let $\tilde{\chi}$ be the graph of spaces structure on the universal cover of $X_{\chi}$, and let $\Delta \subseteq X_{\tilde{\chi}}$ be a compact subset.  We may assume that $\Delta$ is connected; we may also assume that if $\tilde{e} \in \tilde{\varXi}$ (Bass-Serre tree) and $\Delta \cap \tilde{e} \ne \emptyset$, then $\tilde{e} \subseteq \Delta$.  Let $\eta \colon X_{\tilde{\chi}} \to \tilde{\varXi}$ be the map to the underlying map of the Bass-Serre tree.  Then $\eta(\Delta)$ is a finite connected subgraph, $\varXi'$.  For each $v' \subseteq v(\varXi')$, let $\Delta_{v'} = \Delta \cap X_{v'}$, a compact subspace of $X_{v'}$.  Because $G_{\pm}$ are residually finite, we have a diagram

where $X_{\hat{v}} \to X_{\pm}$ is a finite-sheeted covering map and $\Delta_{v'}$ embeds in $X_{\hat{v}}$.  Let $\hat{\chi}$ be defined as follows.  Set $\hat{\varXi} = \varXi'$; for a vertex $\hat{v} \in \hat{\varXi}$, the vertex space is the $X_{\hat{v}}$ corresponding to $v'$.  For each $\hat{e} \in E(\hat{\varXi})$ corresponding to $e' \in E(\varXi')$, define $\partial_{\hat{e}}^{\pm}$ so that the diagram

commutes.  Now sum.

Exercise 26: If $G_1$ and $G_2$ are residually finite and $H$ is finite, prove that $G_1 *_H G_2$ is residually finite.

Proof of Theorem 6: Let $c \colon [a,b] \to X$ be a quasi-geodesic in a $\delta$-hyperbolic space $X$.  We can replace $c$ by $c' \colon [a,b] \to X$ as in Lemma 8.  Let $p = c'(a)$, $q = c'(b)$, and $x \in [p,q]$ such that $D = d(x, \textnormal{im}(c'))$ is maximal.  We need to bound $D$.  Let $y \in [p,x]$ be such that $d(x,y) = \min \lbrace d(x,p), \: 2D \rbrace$, and $z \in [x, q]$ such that $d(x,z) = \min \lbrace d(x,q), \: 2D \rbrace$.  Let $y' \in \textnormal{im}(c')$ such that $d(y,y') \le D$, and $y' \in \textnormal{im}(c')$ such that $d(z,z') \le D$.  Finally, let $\gamma$ be a path in $X$ obtained by concatenating $[y,y']$, the section of the image of $c'$ with endpoints $y'$ and $z'$, and $[z,z']$.

By construction, $\gamma$ does not intersect $B(x,D)$, and part 3 of Lemma 8 shows that

$l(\gamma) \le k_1 d(y', z') + k_2 + 2D \le k_1 (6D) + k_2 + 2D$.

Lemma 7 gives a bound on the length of paths based on the distance between their endpoints:

$D \le \delta \log_2 (l(\gamma)) + 1 \le \delta \log_2 (k_1 (6D) + k_2 + 2D) + 1$.

The left hand side of this inequality increases linearly in $D$, while the right increases logarithmically, so that $D$ must be bounded above for the equality to hold, and clearly this upper bounded depends only on the constants $k_1$, $k_2$ and $\delta$QED.

It now makes sense to make the following definition.

Definition: A finitely generated group $\Gamma$ is called (word-) hyperbolic if some (any) Cayley graph for $\Gamma$ is Gromov-hyperbolic.  Equivalently, a group $\Gamma$ is hyperbolic if it acts properly discontinuously and cocompactly by isometries on a proper Gromov-hyperbolic metric space.

Examples:

a) Free groups
b) $\mathbb{Z}^n$ is not hyperbolic for $n > 1$.
c) Let $M$ be any closed hyperbolic manifold.  Then $\pi_1(M)$ is word-hyperbolic.
d)  More generally, $\pi_1(M)$ is word-hyperbolic for any $M$ with negative sectional curvature bounded away from $0$.

Without getting into too much detail, we briefly mention the following theorem of Gromov as an indication of how general the class of hyperbolic groups really is.

Theorem (Gromov): A “randomly chosen” finitely presented group is “almost surely” word-hyperbolic.

Definition: A subspace $Y$ of a geodesic metric space $X$ is quasiconvex if there exists a $K \ge 0$ such that, for all $y_1, y_2 \in Y$ and for all $x \in [y_1, y_2]$, $d(x, Y) \le K$.

Example: Consider $\mathbb{Z}^2$ with the $\ell^1$-metric.  Then the diagonal subgroup $\Delta \subset \mathbb{Z}^2$ is not quasiconvex (though it is quasi-embedded).

Theorem 6 implies that this kind of poor behavior does not occur in hyperbolic space.

Corollary: Suppose that $\Gamma$ is a word-hyperbolic group and $H$ is a subgroup.  Then $H$ is quasiconvex in some (any) Cayley graph of $\Gamma$ if and only if $H$ is finitely generated and $H \hookrightarrow \Gamma$ is a quasi-embedding.

Proof:  $(\Leftarrow)$ is immediate from Theorem 6.  For the other direction, fix a generating set $S$ for $\Gamma$, assume $H$ is quasiconvex in the Cayley graph of $\Gamma$ with constant $K$, and let $h$ be in $H$.  Consider a geodesic in the Cayley path of $\Gamma$ from $1$ to $h$, which we can take to be of the form $s_1...s_n$ for $s_i$ in $S$.  Let $v_i$ be the vertices of this geodesic, so $v_i= s_1...s_i$.

By quasiconvexity, for each $v_i$ there exists $g_i$ in $H$ such that $d(v_i,g_i) \leq K$.  Take $g_0=1$ and $g_n=h$.  Let $h_i=g_{i-1}^{-1}g_i$, so $h=h_1... h_n$.  Let $u_i=v_ig_i^{-1}$.  Note that $l_S(u_i)\leq K$.  For each $i$, we have that $h_i=u_is_iu_{i+1}^{-1}$ and so $l_S(h_i)\leq 2K+1$.  Therefore, $H$ is generated by $T=B(1,2K+1)\cap H$, a finite set.  Furthermore, we have shown that $l_T(h)\leq n=l_S(h)$.

But it’s clear that $l_S(h)\leq (2K+1)l_T(h)$ (as each element of $T$ has $S$-length at most $2K+1$) so the inclusion of $H$ into $\Gamma$ is a quasi-isometric embedding.  QED.

In light of this, the following definition makes sense

Definition: A subgroup $H$ of a hyperbolic group $\Gamma$ is called quasiconvex if it is a quasiconvex space of some (any) Cayley graph of $\Gamma$.

Exercise 14: If $G \stackrel{\rho}{\longrightarrow} H$ is a retraction and $G$ is finitely generated, then the inclusion $H \hookrightarrow G$ is a quasi-isometric embedding.  (Hint: you can choose a generating set $S$ for $G$ such that $\rho(s) = s$ or $\rho(s) = 1$ for all $s \in S$.

Example: Marshall Hall’s Theorem implies that every finitely generated subgroup of a free group is a retract of a finite-index subgroup, so every finitely generated subgroup of a free group is quasiconvex.