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Our goal is to understand the abelian subgroups of a hyperbolic group; e.g., can a hyperbolic group contain $\mathbb{Z}^2$ as a subgroup?  We already know that it cannot be a quasiconvex subgroup, but it may be possible for $\mathbb{Z}^2$ to be “twisted” in some manner.

Theorem 10. The intersection of two quasiconvex subgroups is quasiconvex.

Pf. As usual, we work in a $\delta$-hyperbolic $\text{Cay}_S ( \Gamma )$.  Let $H , K \subset \Gamma$ be quasiconvex subgroups each with the same corresponding constant $\kappa$.  Let $\gamma \in H \cap K$, and let $g_0 \in [ 1 , \gamma ] \cap \Gamma$.  Our goal is therefore to show that $g_0$ is in a bounded neighborhood of $H \cap K$.  Let $g_D$ be the (or more precisely, a particular) closest element of $H \cap K$ to $g_0$, and suppose that $d ( g_0 , g_D ) = D$.  Let $h_0 \in H$ be such that $d ( g_0 , h_0 ) \leq \kappa$ and let $k_0 \in K$ be such that $d ( g_0 , k_0 ) \leq \kappa$; such elements exist since $H$ and $K$ are quasiconvex.  Let $g_t \in [ g_0 , g_D ]$ be such that $d ( g_0 , g_t ) = t$.  We sketch the situation below.

Consider the geodesic triangle with vertices $g_0$, $h_0$, and $g_D$.  Because this triangle is $\delta$-slim, for each $t$ there exists some $h_t \in H$ such that $d ( h_t , g_t ) \leq \delta + \kappa$.

Likewise, for each $t$ there exists $k_t \in K$ such that $d ( g_t , k_t ) \leq \delta + \kappa$.  Next, let $u_t = g_t^{-1} h_t$ and $v_t = g_t^{-1} k_t$.  Then $\ell ( u_t ) , \ell ( v_t ) \leq \delta + \kappa$.

Suppose $D > ( \# B ( 1 , \delta + \kappa ) )^2$.  Then by the Pigeonhole Principle there are integers $s$ and $t$ with $s > t$ such that $u_s = u_t$ and $v_s = v_t$.  Now, we can use this information to find a closer element of $H \cap K$.

Consider $g_t g_s^{-1} g_D$.  By the triangle inequality,

$d ( g_0 , g_t g_s^{-1} g_D ) \leq d ( g_0 , g_t ) + d ( g_t , g_t g_s^{-1} g_D ) = t + D - s < D$.

All that remains is to prove that $g_t g_s^{-1} g_D \in H \cap K$.  But since $u_t = u_s$,

$g_t g_s^{-1} g_D = g_t u_t u_t^{-1} g_s^{-1} g_D = ( g_t u_t ) ( g_s u_s )^{-1} g_D = h_t h_s^{-1} g_D \in H$.

Playing this same game with $v_t = v_s$, we get $g_t g_s^{-1} g_D = k_t k_s^{-1} g_D \in K$, and hence we have found our contradiction.$\square$

For a group $G$, recall that $Z ( G ) = \{ g \in G : g g' = g' g \text{ for all } g' \in G \}$ is the center of $G$.

Corollary. For any $\gamma \in \Gamma$ (where $\Gamma$ is $\delta$-hyperbolic) of infinite order, the subgroup generated by $\gamma$ is quasiconvex.  Equivalently, the map $c : \mathbb{Z} \to \text{Cay}_S ( \Gamma )$ sending $n \mapsto \gamma^n$ is a quasigeodesic (which is a sensible statement to make given that $\mathbb{Z}$ is quasi-isometric to $\mathbb{R}$).

Pf. By Theorem 9, we deduce that $C ( \gamma )$ is quasiconvex, and so is finitely generated.  Let $T$ be a finite generating set for $C ( \gamma )$.  Notice

$Z ( C ( \gamma ) ) = \displaystyle \bigcap_{t\in T} C_{C(\gamma)} ( t )$

where $C_{C(\gamma)} ( t )$ is the centralizer in $C ( \gamma )$ of $t$, so $Z ( C ( \gamma ) )$ is quasiconvex by Theorem 10 and thus $Z ( C ( \gamma ) )$ is a finitely generated abelian group containing $\langle \gamma \rangle$.  By Exercise 16 (below), we deduce that $\langle \gamma \rangle$ is quasi-isometrically embedded in $Z ( C ( \gamma ) )$, and hence is quasiconvex in $\Gamma$. $\square$

Exercise 16. Cyclic subgroups of finitely generated abelian groups are quasi-isometrically embedded.

Lemma 10. Suppose the order of $\gamma \in \Gamma$ is infinite.  If $\gamma^p$ is conjugate to $\gamma^q$, then $| p | = | q |$.

Pf. Suppose $t \gamma^p t^{-1} = \gamma^q$.  An easy induction on $n$ shows $t^n \gamma^{(p^n)} t^{-n} = \gamma^{(q^n)}$.  Therefore, applying the triangle inequality,

$\ell ( \gamma^{(q^n)} ) \leq \ell ( \gamma^{(p^n)} ) + 2 \ell ( t^n ) \leq | p |^n \ell ( \gamma ) + 2 n \ell ( t )$.

But $\langle \gamma \rangle \hookrightarrow \Gamma$ is a $( \lambda , \epsilon )$-quasi-isometric embedding, so

$| p |^n \ell ( \gamma ) + 2 n \ell ( t ) \geq \ell ( \gamma^{(q^n)} ) \geq \lambda^{-1} | q |^n - \epsilon$.

This is impossible unless $| q | \leq | p |$ (eventually the exponential growth dominates).  Similarly, reversing the roles of $p$ and $q$ in the above argument implies that $| p | \leq | q |$, so $| p | = | q |$. $\square$

Theorem 8: Let $\Gamma$ be a $\delta$-hyperbolic group with respect to $S$. If $u,v \in \Gamma$ are conjugate then there exists $\gamma\in\Gamma$ such that

$\gamma u\gamma^{-1}=v\\l(\gamma)\leq M(l(u),l(v))$

where $M$ depends only on $\Gamma$.

Proof: We work in $Cay_S(\Gamma)$. Let $\gamma\in \Gamma$ be such that $\gamma u\gamma^{-1}=v$.  Let $\gamma_t \in [1,\gamma]$ be such that $d(1,\gamma_t)=t$. We want to find a bound on $d(\gamma_t, v\gamma_t)$.

Let $c=[1,\gamma u]$. By Lemma 9,

$d(\gamma_t,c(t))\leq 2(\delta+l(v))\\ d(v\gamma_t,c(l(\gamma u)-(l(\gamma)-t)))\leq 2(\delta+l(v))$

Also

$d(c(t),c(l(\gamma u)-(l(\gamma)-t)))=l(\gamma u)-l(\gamma)+t-t\\=l(\gamma u)-l(\gamma)\leq l(u)$

So $d(\gamma_t,v\gamma_t)\leq 4(\delta +l(u)+l(v))= R(l(u),l(v))$. Thus $l(\gamma_t^{-1}v\gamma_t)\leq R$. Suppose that $l(\gamma)> \#B(1,R)$. By the Pigeonhole Principle there exist integers $s>t$ such that $\gamma_t^{-1}v\gamma_t=\gamma_s^{-1}v\gamma_s$. It follows that one can find a shorter conjugating element by cutting out the section of $\gamma$ between $\gamma_t$ and $\gamma_s$.

Recall, for $\gamma \in \Gamma$, $C(\gamma)=\{g\in\Gamma: g\gamma=\gamma g\}$ is the centralizer of $\gamma$.

Theorem 9: If $\Gamma$ is $\delta$-hyperbolic with respect to $S$ and $\gamma\in\Gamma$, then $C(\gamma)$ is quasi-convex in $\Gamma$.

Proof: Again we work in $Cay_S(\Gamma)$. Let $g\in C(\gamma)$, $h\in [1,g]$. We need to prove that $H$ is in a bounded neighborhood $C(\gamma)$.

Just as in the proof of Theorem 8,

$l(h^{-1}\gamma h)=d(h, \gamma h)\leq 4(\delta+2l(\gamma))$

Well, $g$ and $h^{-1}\gamma h$ are conjugate. By Theorem 8 there exists $k\in \Gamma$ such that

$k^{-1}\gamma k=h^{-1}\gamma h \\ l(k)\leq M(l(\gamma), l(h^{-1}\gamma h))\\ \leq M(l(\gamma),4(\delta+2l(\gamma)))$

But $h^{-1}k\gamma=\gamma hk^{-1}$ so that $hk^{-1}\in C(\gamma)$ and $d(h,hk^{-1})=d(1,k^{-1})=l(k^{-1})=l(k)\leq M$.

Exercise 15: Prove that

$\Gamma_A=\mathbb{Z}^2\rtimes_A\mathbb{Z}$

is not hyperbolic for any Anosov $A$.

We will see two examples of non-quasiconvex subgroups in this section. The first one is NOT a hyperbolic group, while the second one is.

Example: For the first example, let

$A=\left(\begin{array}{cc}2&1\\1&1\end{array}\right)$,

with one eigenvalue (the larger one) $\lambda>1$. Notice that $A$ does not fix any non-zero vectors in $\mathbf{Z}^2$ (such a map $A$ is called Anosov).

Now let $\Gamma_A=\mathbf{Z}^2\rtimes_A\mathbf{Z}=(\langle a\rangle\oplus\langle b\rangle)\rtimes_A\langle t\rangle$. This is a group.  The group law works like this: for any $g\in\mathbf{Z}^2$, $tgt^{-1}=Ag$. Pick $S=\{a,b\}$, $T=\{a,b,t\}$. The map $\mathbf{Z}^2\hookrightarrow\Gamma_A$ is, by the following analysis,  NOT a quasi-embedding:

Choose $g\in\mathbf{Z}^2$ such that $\lim_{n\rightarrow\infty}\displaystyle\frac{||A^ng||_2}{\lambda^n||g||_2}=1$. All norms on $\mathbf{R}^2$ are bilipschitz, so there exists $k\geq 1$ such that $k^{-1}||g||_2\leq||g||_1\leq k||g||_2$. Therefore, for sufficiently large $n$$||A^ng||_1\geq k^{-1}||A^ng||_2\geq k^{-2}\lambda^n||g||_2\geq k^{-3}\lambda^n||g||_1$, and so $l_S(A^ng)\geq k^{-3}\lambda^n l_S(g)$. On the other side, we have $l_T(t^ngt^{-n})\leq l_T(g)+2n$. It follows that $\mathbf{Z}^2\hookrightarrow\Gamma_A$ is not a quasi-embedding.

Example: For the second example, let $\Sigma$ be a hyperbolic surface. An automorphism $\psi$ of $\Sigma$ is called pseudo-Anosov if for any smooth closed curve $\gamma$ on $\Sigma$ and any $n\in \mathbf{Z}\smallsetminus\{0\}$, $\psi^n(\gamma)$ is not homotopic to $\gamma$. Let $M_{\psi}$ be the mapping torus of $\psi$, i.e., $M_{\psi}:=\Gamma\times [0,1]/\sim$, with the relation $\sim$ generated by $(x,0)\sim (\psi(x),1)$.

Under these assumptions, we are able to use a theorem of Thurston asserting that, $M$ must be a hyperbolic 3-manifold. (W. Thurston, “On the geometry and dynamics of diffeomorphisms of surfaces,” Bull. Amer. Math. Soc. vol 19 (1988), 417-431)

Hence, if $\Gamma$ is closed, then $M_{\psi}$ is also closed. So $\pi_1(M_{\psi})$ acts nicely on $\mathbf{H}^3$ (actually $\pi_1(M_{\psi})=\pi_1(\Sigma)\rtimes_{\psi_*}\mathbf{Z}$), and so is word-hyperbolic by the Švarc-Milnor Lemma. Then a similar argument to the previous shows the natural map $\pi_1(\Sigma)\hookrightarrow\pi_1(M_{\psi})$ is NOT a quasi-embedding.

For concrete examples, see A. Casson & S. Bleiler, “Automorphisms of Surfaces After Nielsen and Thurston”.

After the two examples, let us switch to a property for all hyperbolic groups:

Theorem 7: Hyperbolic groups are finitely presented.

In order to prove this theorem, we need the following lemma:

Lemma 9: Let $c,c': [0,T]\rightarrow X$ be two geodesics in a $\delta$-hyperbolic metric space $X$, $c(0)=c'(0)$.  (If $c$ is longer than $c'$, say, then extend $c'$ by the constant map). Then for any $t\in [0,T]$, $d(c(t),c'(t))\leq 2(\delta+d(c(T),c'(T)))$.

Proof: Case 1: there is $t'\in [0,T]$ such that $d(c(t),c'(t'))\leq\delta$. Without loss of generality, assume $t'>t$, then $|t'-t|=d(c'(t'),c'(0))-d(c'(t),c'(0))\leq$ $d(c'(t'),c(t))+d(c(t),c(0))-d(c'(t),c'(0))=d(c'(t'),c(t))+t-t=$ $d(c'(t'),c(t))\leq\delta$. So, $d(c(t),c'(t))\leq d(c(t),c'(t'))+d(c'(t'),c'(t))\leq$ $\delta+|t-t'|\leq 2\delta$.

Case 2: there is no $t'\in [0,T]$ such that $d(c(t),c'(t'))\leq\delta$. Then $c(t)$ must be within distance $\delta$ of $[c(T),c'(T)]$. Apply a similar argument to the previous, we see $d(c(t),c'(t))\leq 2(\delta+d(c(T),c'(T)))$.

Proof of Theorem 7: Let $\Gamma$ be $\delta$-hyperbolic, with the generating set $S$. Let $w$ be any relation, which corresponds to a loop in the Cayley graph $Cay_s(\Gamma)$. We can always take $w=\gamma s\delta^{-1}$ with $\gamma$ and $\delta$ geodesics in $Cay_s(\Gamma)$ and $s\in S$, by “triangulating”.

Write $\gamma_t=\gamma(t)$$\delta_t=\delta(t)$. Denote $u_t=\gamma_t\cdot\gamma_{t-1}^{-1}$$v_t=\delta_t\cdot\delta_{t-1}^{-1}$$\alpha_t=\delta_t\cdot\gamma_t^{-1}$.  An easy induction shows that

$w=\prod_{t=1}^T\delta_{t-1}\alpha_{t-1}^{-1}u_t\alpha_t v_t^{-1}\delta_{t-1}^{-1}$.

But Lemma 9 implies that $l(\alpha_t)\leq 2(\delta+1)$ for all $t$, so we have written the loop $w$ as a product of conjugates of loops of length at most $4\delta+6$.  Therefore, the set of all loops of length at most $4\delta+6$ is a finite set of relations for $\Gamma$.