You are currently browsing the tag archive for the ‘Quasiconvex subgroups’ tag.

Last time: Theorem 21 (Groves–Manning–Osin): If G is hyperbolic rel \mathcal P then there exists a finite subset A \subseteq G\setminus 1 such that if \bigcup_i N_i \cap A = \emptyset then
(a) P_i/N_i \to G/\mathcal N is injective;
(b) G is hyperbolic rel P_i/N_i.

Theorem 22 (Gromov, Olshanshkii, Delzant): If G is hyperbolic relative to the infinite cyclic \{\langle g_1\rangle,\dots,\langle g_n \rangle\} then there is a K>0 such that for all l_1,\dots,l_n>0 there exists a \phi : G \to G' hyperbolic such that o(\phi(g_i))=Kl_i for each i.

The proof is an easy application of Groves–Manning–Osin.

Definition: If \{\langle g_1 \rangle,\dots,\langle g_n\rangle\} (infinite cyclic) is malnormal then we say g_1,\dots,g_n are independent. A group G is omnipotent if for every independent g_1,\dots,g_n there exists a K>0 such that for all l_1,\dots,l_n>0 there exists a homomorphism $\phi$ from G to a finite group such that o(\phi(g_i)) = Kl_i for all i.

Omnipotence strengthens residual finiteness for torsionfree groups.

Exercise 29: If every hyperbolic group is residually finite then every hyperbolic group is omnipotent.

We’ll finish off by talking about a similar theorem of Agol–Groves–Manning. I’m going to seem a little cavalier about torsion. This is OK. In fact, if every hyperbolic group is residually finite then every hyperbolic group is virtually torsionfree.

Theorem 22 (Agol–Groves–Manning): If every hyperbolic group is residually finite then every quasi-convex subgroup H of any hyperbolic group G is separable.

Let g \in G \setminus H. The idea is to Dehn fill H to get a new hyperbolic group \bar G in which the image \bar H is finite and \bar g \not\in\bar H. If we could do this, we would be done by residual finiteness. This works if H is malnormal. But it probably isn’t. Fortunately, we can quantify how far H is from being malnormal:

Definition: The height of H is the maximal n \in \mathbb N such that there are distinct cosets g_1 H,\dots,g_n H \in G/H such that the intersection
g_1 H g_1^{-1} \cap \dots \cap g_n H g_n^{-1}
is infinite.

H is height 0 iff H is finite. In a torsionfree group, H is height 1 iff H is malnormal.

Theorem 23 (Gitik, Mitra, Rips, Sageev): A quasiconvex subgroup of a hyperbolic group has finite height.

Agol, Groves and Manning are able to prove:

Theorem 24: Let G be a (torsionfree) residually finite hyperbolic group, and H a quasiconvex subgroup of height k. Let g \in G\setminus H. Then is an epimorphism \eta: G \to \bar G to a hyperbolic group such that
(i) \eta(H) is quasiconvex in \bar G;
(ii) \eta(g) \not\in\eta(H);
(iii) \eta(H) has height \leq k-1.

The idea of the proof of Theorem 24 is to Dehn fill a finite index subgroup of a maximal infinite intersection of conjugates of H. Theorem 22 is an easy consequence.

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.

Figure 1

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

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.

Follow

Get every new post delivered to your Inbox.

Join 36 other followers