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Lemma 29: Suppose \{f_i':C_i\longrightarrow\Gamma^{H}\} is a finite set of infinite degree elvations and \Delta \subseteq \Gamma^{H} is compact. Then for all sufficiently large d>0, there exists an intermediate covering \Gamma_d such that

(a) \Delta embeds in \Gamma_d

(b) every f'_i descends to an elevation \hat{f_i}:\hat{C_i}\longrightarrow \Gamma_d of degree d

(c) the \hat{f_i} are pairwise distinct

Proof: We claim that the images of f_i' never share an infinite ray (a ray is an isometric embedding of [0,\infty)). Neither do two ends of the same elevation f_i'. Let’s claim by passing to the universal cover of \Gamma, a tree T.

For each i, lift f_i' to a map \tilde{f_i}:\mathbb{R}\longrightarrow T. If f'_i and f'_j share an infinite ray then there exists h\in H such that \tilde{f_i} and h\tilde{f_j} overlay in an infinite ray. The point is that \tilde{f_i}, \tilde{f_j} correspond to cosets g_if_{\ast}(\pi_1(C)) and g_jf_{\ast}(\pi_1(C)). But this implies that


This implies that Hg_if_{\ast}(\pi_1(C))=Hg_jf_{\ast}(\pi_1(C)). So f'_i=f'_j. A similar argument implies that the two ends of f'_i do not overlap in an infinite ray. This proves the claim.

Let \Gamma' be the core of \Gamma^{H}. Enlarging \Delta if necessary, we can assume that

(i) \Gamma'\subseteq\Delta;

(ii) \Delta is a connected subgraph;

(iii) for each i, for some x_i\in C'_i, f'_i(x_i)\in\Delta;

(iv) for each i, |im(f'_i)\cap \delta\Delta|=2.

For each i identifying C'_i with \mathbb{R} so that C is identified with \mathbb{R}/\mathbb{Z} and x_i is identified with 0. Let

\Delta_d=\Delta\cup(\bigcup_i f'_i([-d/2,d/2]))

For all sufficiently large d,

f'_i(\pm d/2)\notin\Delta

Now, the restriction of \Delta_d \longrightarrow \Delta factors through \Delta_d/\sim\longrightarrow\Gamma, where f'_i(d/2)\sim f'_i(-d/2). This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required. \Box

Theorem 19: D is LERF.

Recall the set-up from the previous lecture. We built a graph of spaces \mathscr{X} for D.

Proof: Let H\subset D be finitely generated. Let X_H be the corresponding covering space of X_{\mathscr{X}} and let \Delta\subseteq X_H be compact. Because H is finitely generated, there exists a subgraph of spaces X' such that \pi_1(X') =H. We can take X' large enough so that \Delta \subseteq X'. We can enlarge \Delta so that it contains every finite-degree edge space of X'. Also enlarge \Delta so that

\partial^{\pm}_{e'}(\Delta\times (\mathrm{interval}))\subseteq\Delta

for any e'\in E(\Xi'). For each v'\in V(\Xi') let \Delta_{v'}=\Delta\cap X_{v'} and let {f'_i}=\{ incident edge map of infinite degree \}.

Applying lemma 29 to \Gamma^H=X_{v'}, for some large d, set X_{\hat{v}}=\Gamma_d. (Here we use the fact that vertex groups of \mathscr{X}' are finitely generated)

Define \mathscr{X}^+ as follows:

\bullet \Xi^+=\Xi^-

\bullet For each v^+\in V(\Xi^+), the edge space is the X_{v^+} that the lemma produced from the corresponding v'.

Now, by construction, \bigcup_{v^+}X_{v^+} can be completed to a graph of spaces \mathscr{X}^+ so that the map

X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}}

factors through X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}} and \Delta embeds. Let \mathscr{X}^- be identical to \mathscr{X}^+ except with +’s and -‘s exchanged. Clearly \mathscr{X}^+\cup\mathscr{X}^- satisfies Stallings condition, as required. \Box

Agol-Groves-Manning’s Theorem predicts that, for every word-hyperbolic group we can easily construct, every quasiconvex subgroup is separable (otherwise, we would find a non-residually finite hyperbolic group!).

In this section, we use graphs of groups to build new hyperbolic groups:

Combination Theorem (Bestvina & Feighn): If H is a quasiconvex malnormal subgroup of hyperbolic groups G_1, G_2, then G_1\ast_H G_2 is hyperbolic.

Recall: H is called a malnormal subgroup of G if it satisfies: if gHg^{-1}\cap H\neq 1, then g\in H.

For a proof, see M. Bestvina and M. Feighn, “A combination theorem for negatively curved groups”, J. Differential Geom., 35 (1992), 85–101.

Example: Let F be free, w\in F not a proper power. By Lemma 11, \langle w\rangle\leq F is malnormal, so D:=F\ast_{\langle w\rangle} F is hyperbolic. As a special case, if \Sigma is closed surface of even genus n=2k, considered as the connected sum of two copies of the closed surface of genus k, then by Seifert-van Kampen Theorem, \pi_1(\Sigma)=F_{2k}\ast_{\langle w\rangle} F_{2k} for some w\in F_{2k}.

Question: (a) Which subgroups of D are quasiconvex? (b) Which subgroups of D are separable?

We will start by trying to answer (b). The following is an outline of the argument: Let \Gamma be a finite graph so that \pi_1(\Gamma)=F, let \Gamma_{\pm} be two copies of \Gamma. Realize w\in F as  maps \partial^{\pm}: C\rightarrow \Gamma_{\pm}, where C\simeq S^1. Let X be the graph of spaces with vertex spaces \Gamma_{\pm}, edge space C, and edge maps \partial^{\pm}. Then clearly, D\simeq \pi_1(X), and finitely generated subgroups H\leq D are in correspondence with covering spaces X^H\rightarrow X. We can then use similar technique to sections 27 and 28.

Let us now make a few remarks about  elevations of loops. Let f: C\rightarrow X be a loop in some space X, i.e., C\simeq S^1 and \pi_1(C)\simeq\mathbf{Z}. Consider an elevation of f:


The conjugacy classes of subgroups of \mathbf{Z} are naturally in bijection with \mathbf{N}\cup\{\infty\}. The degree of the elevation is equal to the degree of the covering map C'\rightarrow C.

Definition: Suppose X'\rightarrow X is a covering map and \widehat{X} is an intermediate covering space, i.e., X'\rightarrow X factors through \widehat{X}\rightarrow X, and we have a diagram


If f' and \widehat{f} are elevations of f and the diagram commutes, then we say that f' descends to \widehat{f}.

Let \Gamma  be a finite graph, H\leq \pi_1(\Gamma) a finitely generated subgroup and f: C\rightarrow\Gamma a loop. Let \Gamma^H\rightarrow\Gamma be a covering space corresponding to H.

Lemma 29: Consider a finite collection of elevations \{f'_i: C'_i\rightarrow\Gamma^H\} of f to \Gamma^H, each of infinite degree. Let \Delta\leq\Gamma^H be compact. Then for all sufficiently large d>0, there exists an intermediate, finite-sheeted covering space \Gamma_d\rightarrow\Gamma satisfying: (a) \Delta embeds in \Gamma_d; (b) every f'_i descends to an elevation \widehat{f}_i: \widehat{C}_i\rightarrow\Gamma_d of degree exactly d; (c) these \widehat{f}_i are pairwise distinct.

Recall that our goal is to determine, for a given map of graphs of spaces such as the one shown below, whether the map \Phi can be extended to a covering map \widehat{\Phi}.

Figure 1

Figure 1

Let \mathcal{X}, \mathcal{X}' be graphs of spaces equipped with maps \Xi' \to \Xi, \phi_{v'} : X_{v'} \to X_v and \phi_{e'} : X_{e'} \to X_e as before.  Recall that in Stallings’ proof of Hall’s Theorem, we completed an immersion to a covering map by gluing in edges.  We will aim to do the same thing with graphs of spaces.

Definition: Let \mathcal{X} be a graph of spaces, and let \eta : X_{\mathcal{X}} \to \Xi be the map to the underlying graph.  If \Delta \subseteq \Xi is a subgraph, then \eta^{-1}(\Delta) \subseteq X_{\mathcal{X}} has a graph-of-spaces structure \mathcal{Y} with underlying graph \Delta.  Call \mathcal{Y} a subgraph of spaces of \mathcal{X}.

We’re seeking a condition on \mathcal{X}' such that \mathcal{X}' is realized as a subgraph of spaces of some \widehat{\mathcal{X}} with a covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}} such that the following diagram commutes:

diagram1Definition: For each edge map \partial_e^{\pm} : X_e \to X_v of \mathcal{X}, and each v' \mapsto v a vertex of \mathcal{X}', let

\mathcal{E}^{\pm}(e) = \bigcup_{v' \mapsto v} \mathcal{E}^{\pm}(e, v').

For each possible degree \mathcal{D}, let \mathcal{E}_{\mathcal{D}}^{\pm}(e) \subseteq \mathcal{E}^{\pm}(e) be the set of elevations of degree \mathcal{D}.  We will say \mathcal{X}' satisfies Stallings’ condition if and only if the following two things hold:

(a) Every edge map of \mathcal{X}' is an elevation of the appropriate edge map of \mathcal{X}.
(b) For each e \in E(\Xi) and \mathcal{D}, there is a bijection \mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e).

So in Figure 1, the graph of spaces \mathcal{X}' is something you might be able to turn into a covering.  In the picture, \mathcal{E}_{\mathcal{D}}^+(e) is represented by the blue circles, and \mathcal{E}_{\mathcal{D}}^-(e) is represented by the green circles.  Observe that the blue circles are in bijection with the green circles.

Corollary: \mathcal{X}' satisfies Stallings’ condition if and only if \mathcal{X}' can be realized as a subgraph of spaces of some \widehat{\mathcal{X}} such that

(a) V(\Xi ') = V(\widehat{\Xi}), and
(b) there is a covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}} such that the following diagram commutes:


Proof of Corollary. First we’ll show that if \Phi can be extended to a covering map as described above, then \mathcal{X}' satisfies Stallings’ condition.  By Theorem 17, every edge map of \widehat{\mathcal{X}} is an elevation.  So there are inclusions


Furthermore, these maps are surjective, and clearly degree-preserving.

Now assume that \mathcal{X}' satisfies Stallings’ condition.  Then we build \widehat{\mathcal{X}} as follows.  Let V(\widehat{\Xi}) = V(\Xi ').  As above, we have degree-preserving inclusions (this time, not surjections)


Extend these inclusions to bijections \mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e).  Now we set E(\widehat{\Xi}) = \bigcup_{e \in E(\Xi)} \mathcal{E}^+(e).  Each of these \widehat{e} is an elevation


This defines an edge space X_{\widehat{e}} and an edge map \partial_{\widehat{e}}^+.  Consider the corresponding elevation in \bigcup_{e \in E(\Xi)} \mathcal{E}^-(e):


Because \partial_{\widehat{e}}^+ and \partial_{\widehat{e}}^- are of the same degree, we have a covering transformation X_{\widehat{e}} ' \to X_{\widehat{e}}.  So we can identify them, and use \partial_e^- as the other edge map.  By construction, \widehat{\mathcal{X}} satisfies the conditions of Theorem 17, so there is a suitable covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}}.

Exercise 25: (This will be easier later, but we have the tools necessary to do this now.)  Prove that if G_1 and G_2 are LERF groups, then so is G_1 * G_2.