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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.

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 .


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))


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


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


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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