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31. Doubles & virtual retractions

15 April 2009 in Course notes | Tags: Doubles of free groups, Virtual retractions | by smhart | Leave a comment

Last time, we used the following lemma without justification, so let’s prove it now.

Lemma 30. Let $\mathcal{G}$ be a graph of groups with $\Gamma$ finite and $G = \pi_1 \mathcal{G}$ finitely generated. If $G_e$ is finitely generated for every edge $e \in E ( \Gamma )$ , then $G_v$ is finitely generated for every $v \in V ( \Gamma )$ .

This is not completely trivial: it is certainly possible for finitely generated group to have subgroups that are not finitely generated. Indeed, recall the proof that every countable group embeds in the free group on three generators.

Pf. Let $S$ be a finite generating set for $G$ , and for each $e \in E ( \Gamma )$ let $S_e$ be a finite generating set for the edge group $G_e$ . By the Normal Form Theorem, every $g \in S$ can be written in the form

$g = g_0 t_1^{\pm 1} g_1 \cdots t_n^{\pm 1} g_n$

where each $t_i$ is a stable letter and each $g_i \in G_{v_i}$ for some $v_i \in V ( \Gamma )$ . For a fixed $v \in V ( \Gamma )$ , let

$\displaystyle S_v = \bigcup_{g\in S} \{ g_i : g_i \in G_v \} \cup \bigcup_{e \text{ adjoining } v} \partial_e^{\pm} ( S_e )$ ,

where for each $e \in E ( \Gamma )$ adjoining $v$ the plus or minus is chosen so that $\partial_e^\pm : G_e to G_v$ . It is clear then that $S_v$ is contained in $G_v$ . To see that $S_v$ is finite, note that since $S$ is finite, the first union is finite; and since $\Gamma$ is finite there can be only finitely many edges adjoining a given vertex, so the second union is finite. Hence it remains to prove that $S_v$ generates $G_v$ .

Let $\gamma \in G_v$ . Because $S$ generates $G$ , we have

$\gamma = \gamma_1 \cdots \gamma_m$

where $\gamma_j \in S$ . Each $\gamma_j$ has a normal form as above, so we get an expression of the form

$\gamma = g_0 t_1^{\pm 1} g_1 t_n^{\pm 1} g_n$ , or $1 = \gamma^{-1} g_0 t_1^{\pm 1} g_1 t_n^{\pm 1} g_n$ .

By the Normal Form Theorem, the expression on the right can then be simplified. After the simplification process, we have no stable letters left, and every $g_i$ is either contained in $G_v$ or is a product of elements of the incident edge groups, and in both cases lie in $S_v$ . $\square$

Remember that Theorem 19 said $D$ is LERF, answering our question (b). But in fact, we get more from the proof of Theorem 19.

Definition. Recall that $H \subset G$ is a retract if the inclusion $H \hookrightarrow G$ has a left inverse $\rho : G \to H$ . Similarly, we call $H$ a virtual retract if $H$ is a retract of a finite index subgroup of $G$ .

For instance, Marshall Hall’s Theorem implies that every finitely generated subgroup of a free group is a virtual retract.

Theorem 20. Every finitely generated subgroup of $D$ is a virtual retract.

Pf. Consider the setup of the proof of Theorem 19. We start with a subgroup $H = \pi_1 X_{\mathfrak{X}'}$ and end up with a finitely sheeted covering space $X_{\widehat{\mathfrak{X}}} \to X_{\mathfrak{X}'}$ . The graph of spaces $\widehat{\mathfrak{X}}$ is built using the “obvious” bijection between elevations to $\mathfrak{X}^+$ and elevations to $\mathfrak{X}^-$ . Thus the identification $X_{\mathfrak{X}^+} \to X_{\mathfrak{X}^-}$ extends to a topological retraction $X_{\widehat{\mathfrak{X}}} \to X_{\mathfrak{X}^+}$ . Now, we shall build a map $\pi_1 X_{\mathfrak{X}^+} \to \pi_1 X_{\mathfrak{X}'} = H$ . We build it at each vertex space, one at a time. From the proof of Lemma 29, we see that the core of each $X_{v'}$ is a topological retract of the corresponding $X_{v^+}$ . Furthermore, we can choose the retraction so that for each long loop of degree $d$ that we added is mapped to a null-homotopic loop in $X_{v'}$ . This allows you to piece together the map $X_{v^+} \to X_{v'}$ into a retraction $\pi_1 X_{\mathfrak{X}^+} \to \pi_1 X_{\mathfrak{X}'} = H$ . $\square$

Exercise 14 asserted that a (virtual) retract is quasi-isometrically embedded, and so Theorem 20 has the following corollary:

Corollary. Every finitely generated subgroup of $D$ is quasi-convex.

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

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31. Doubles & virtual retractions

14. Centralizers (continued)

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