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26. Building Covering Spaces

6 April 2009 in Course notes | by krammai | Leave a comment

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:

fig2note0401 Every 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.

9. More Quasi-isometry and Hyperbolic Metric Spaces

12 February 2009 in Course notes | Tags: Hyperbolic metric spaces, Quasi-isometry | by krammai | 1 comment

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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26. Building Covering Spaces

9. More Quasi-isometry and Hyperbolic Metric Spaces

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