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Lemma 27 Revisited. Suppose \tau : X' \to X is a covering map. Then there is a covering map \sigma: Y'\to Y such that X' is the fibre product of \sigma and d.

Proof. Let Y' = \left\{ (\xi',\eta) \in X' \times Y : \tau (\xi) = i (\eta) \right\} be the fibre product of \tau and i. There is a map d' : X' \to Y' given by \xi' \mapsto (\xi' , d \circ \tau (\xi')).  Let \hat{X} be the fibre product of \sigma and d; i.e.

\hat{X} = \left\{ (\xi,\eta') \in X \times Y' : d(\xi) = \sigma(\eta') \right\}.

fig1note04011

There is a map X' \to \hat{X} given by \xi' \mapsto (\tau(\xi') , d'(\xi')).  This is a covering map and injective, so it is a homeomorphism.

Let f: X \to Y be continuous, Y' \to Y be a covering map and x \in X, y= f(x) \in Y choices of basepoint. We have already seen that a choice of y' \in Y' such that y' \mapsto y determines an elevation of f to Y' at y'.  Fix such a y'. The pre-image of y in Y' is in bijection with the set of cosets

\pi_1(Y',y') \backslash \pi_1(Y,y)

This raises the question, when do two cosets determine the same elevation?

Exercise 24. \pi_1(Y',y')g_1 and \pi_1(Y',y')g_2 determine the same elevation if and only if

\pi_1(Y',y') g_1 f_* \pi_1(X,x) = \pi_1(Y',y') g_2 f_* \pi_1(X,x);

that is, the set of elevations of f to Y' is in bijection with \pi_1(Y') \backslash \pi_1(Y) / f_*(\pi_1(X)).

Let \mathfrak{X} and \mathfrak{X}' be graphs of spaces and suppose we have the following data.

(a) A combinatorial map \Xi' \to \Xi given by e' \mapsto e and v' \mapsto v.

(b) Covering maps \varphi_{v'} : X_{v'} \to X_v for each v' \in V(\Xi').

(c) Covering maps \varphi_{e'} : X_{e'} \to X_{e} for each e' \in E (\Xi'), such that \varphi_{v'}\circ\partial_{e'} = \partial_e\varphi_{e'} whenever e' adjoins v'.

This determines a continuous map \Phi : X_{\mathfrak{X}'} \to X_{\mathfrak{X}}.  When is \Phi really a covering map?

Theorem 17. \Phi is a covering map if

(i) for all e' \in E(\Xi') adjoining v' \in V(\Xi'), the edge map \partial_{e'}^{\pm} : X_{e'} \to X_{v'} is an elevation of \partial_{e}^{\pm} : X_{e} \to X_{v}; and

(ii) wherever e \in E(\Xi) adjoining v \in V(\Xi) and v' \in V(\Xi'), every elevation \partial_{e}^{\pm} : X_{e} \to X_{v} to X_{v'} arises as an edge map of \mathfrak{X}'.

Proof (sketch). It’s enough to consider our local model : X_{\mathfrak{X}} = X and Y = X_v. \varphi_{v'} : X_{v'} \to X_v and \varphi_{e'} : X_{e'} \to X_{e} be covering maps defining \mathfrak{X}' and a map \Phi : X_{\mathfrak{X}'} \to X_{\mathfrak{X}}. By Lemma 27, \Phi is a covering map if and only if X_{\mathfrak{X}'} is a fibre product with respect to some covering:

fig2note0401Every map in the diagram is \pi_1 injective, so (for each component)

\pi_1(Y') \cong \pi_1(X_{\mathfrak{X}'}) \cong \pi_1(X_{v'})

and it follows that X_{\mathfrak{X}'} is the fibre product of d and \varphi_{v'}. The result follows.

Question (Gromov). Classify groups up to quasi-isometry.

1) Ends. Roughly, if X is a metric space, Ends(X) is the number of components of the boundary at \infty of X. If X = (\Gamma, d_s), then Ends captures algebraic information.

Definition. For functions f_1, f_2 : \mathbb{N} \longrightarrow \mathbb{ N}, we say f_1 \preceq f_2 if there exists C such that f_1(n) \leq C f_2 (Cn + C) + Cn + C.  If f_1 \preceq f_2 and f_1 \succeq f_2 then f_1 \simeq f_2.

2) Growth. If \Gamma is a group and S is a finite generating set.

f_\Gamma (n) = \# B(1, n),

where B(1, n) is the set of elements \gamma \in \Gamma such that l_S(\gamma) \leq n. This is a quasi-isometric invariant of \Gamma.

Example. f_{\mathbb{Z}^k}(n) \simeq n^k

Example. f_{F_2}(n) is exponential.

3) If \Gamma is finitely presented and \Gamma' is quasi-isometric to \Gamma then \Gamma' is also finitely presented.

4) Let \Gamma = \langle S | R \rangle is finitely presented; so \Gamma = F_S / \langle \langle R \rangle \rangle. Let r \in \langle \langle R \rangle \rangle. Then

(1)               http://www.codecogs.com/eq.latex?r=\prod_{i=1}^{n}g_{i}r_{i}^{{\varepsilon}_i}g_{i}^{-1}

where r_i \in R, \varepsilon_i \in \left\{ \pm 1 \right\}, and g_i \in \Gamma.  The question: how hard is it to write r in such a product? Define Area (r) to be minimum n in any such expression of r in (1). Let

\delta_\Gamma(n) = \max \left\{ Area(r) | r \in \langle \langle R \rangle \rangle, l_S(r) \leq n \right\}

This function \delta_\Gamma is the Dehn function of \Gamma, which measures how hard the word problem is to solve in \Gamma.  The \simeq-class of \delta_\Gamma is a quasi-isometric invariant.

Remark. Having a solvable word problem is equivalent to having a computable Dehn function.

Hyperbolic Metric Spaces

We want a notion of metric spaces (and hence for groups) that captures hyperbolicity (that is, for one, that triangles are thin).

In what follows, X is always a geodesic metric space.  We’ll write {[x,y]} for a geodesic between x and y (not necessarily unique).

Definition. Let x,y,z \in X, and let \bigtriangleup = [x,y] \cup [y,z] \cup [z,x].  We say that \bigtriangleup is \delta-slim if

{ [y,z] \subseteq B ( [x,y] \cup [z,x], \delta)},

where B(A, \delta) = \bigcup_{a \in A} B(a, \delta), and the same for both {[x,y]} and {[z,x]} (that is, for each geodesic “side” of the triangle, it is contained in a \delta neighborhood of the other two geodesic sides of the triangle).

Definition. X is Gromov hyperbolic (or \delta-hyperbolic, or just hyperbolic) if every geodesic triangle, \bigtriangleup, is uniformly \delta-slim; that is, there exists \delta such that every \bigtriangleup is \delta-slim.

Example (a). Any tree is {0}-hyperbolic.  Every geodesic triangle is a “tripod”.

Example (b). \mathbb{R}^2 is not \delta-hyperbolic for any \delta.

Example (c). \mathbb{H}^2 (and hence \mathbb{H}^n) is hyperbolic (and indeed, any space of principal negative sectional curvature bounded away from zero).

Given a geodesic triangle \bigtriangleup \subseteq \mathbb{H}^2 and let a \in \bigtriangleup.  We ask how far from the other sides is a? Well, inscribe a semi-circle centered at a inside of \bigtriangleup; pick the largest such inscribed semi-circle, and call its radius \delta_a.  So \bigtriangleup is \delta-slim, where \delta is the largest \delta_a; that is, \delta is the radius of the largest semi-circle that can be inscribed in \bigtriangleup.

So to find \delta, we look at semi-circles; for this, we need a fact about \mathbb{H}^2.

Fact. For any \bigtriangleup \subseteq \mathbb{H}^2, Area(\bigtriangleup) = \pi - \alpha - \beta - \gamma < \pi, where \alpha, \beta, \gamma are  angles of the triangle.

This leads to a uniform bound on the area, and hence the radius of semi-circles inscribed in \bigtriangleup.

To define hyperbolic groups, we want to prove hyperbolicity is a quasi-isometric invariant of geodesic metric spaces.  We need to “quasi-fy” the definition of \delta-hyperbolic.

Definition. A quasi-geodesic is a quasi-isometric embedding of a closed interval.

Exercise 13. Let c : [1,\infty) \longrightarrow \mathbb{R}^2 by c(t) = (t, \log t) in polar coordinates. Show that c is a quasi-isometric embedding.

We will prove this behavior does not happen in hyperbolic metric spaces.

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