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14. Centralizers (continued)

23 February 2009 in Course notes | Tags: Hyperbolic groups, Quasiconvex subgroups | by smhart | 1 comment

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 $d ( h_t , g_t ) \leq \delta + \kappa$ .

fig2

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 $H \cap K$ .

fig3

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 $| p | = | q |$ . $\square$

13. Conjugacy & Centralizers

22 February 2009 in Course notes | Tags: Hyperbolic groups, Quasiconvex subgroups | by dgirao | Leave a comment

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 $A$ .

12. Hyperbolic groups and quasiconvex subgroups (continued)

18 February 2009 in Course notes | Tags: Hyperbolic groups, Quasiconvex subgroups | by yyao392c | 1 comment

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 $\Gamma$ .

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14. Centralizers (continued)

13. Conjugacy & Centralizers

12. Hyperbolic groups and quasiconvex subgroups (continued)

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