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We want to show that the notion of hyperbolicity is preserved under quasi-isometry. One problem is that geodesics are not preserved under quasi-isometry; one can show that there are quasi-geodesics in the plane that look nothing like geodesics.

(In the following, X is always a geodesic metric space.)

Definition: If A, B \subseteq X are compact, then

d_{Haus}(A, B) = \max \lbrace \sup_{a\in A} \ \inf_{b \in B} \ d(a, b), \sup_{b \in B} \ \inf_{a \in A} \ d(a, b) \rbrace

is the Hausdorff distance between A and B.

Recall: A quasi-geodesic is a quasi-isometrically embedded interval.

Theorem 6: For all \delta \geq 0, \lambda \geq 1, \epsilon \geq 0, there exists R = R(\delta, \lambda, \epsilon) with the following property: If X is a \delta-hyperbolic metric space, c : [a, b] \to X is a (\lambda, \epsilon)-quasi-geodesic, and \ [c(a), c(b)] is any geodesic from c(a) to c(b), then d_{Haus} (im \ c, [c(a), c(b)]) \leq R(\delta, \lambda, \epsilon).

Corollary: A geodesic metric space X is hyperbolic if and only if for every \lambda \geq 1, \epsilon \geq 0, there exists an M such that every (\lambda, \epsilon)-quasi-geodesic triangle is M-slim.

Corollary: If X is \delta-hyperbolic, Y is geodesic, and f : Y \to X is a quasi-isometric embedding, then Y is hyperbolic.

Corollary: Hyperbolicity is a quasi-isometry invariant of geodesic metric spaces.

(Note: Gromov provides a definition of hyperbolicity that works for non-geodesic metric spaces, but this notion of hyperbolicity is not quasi-isometry invariant.)

To prove Theorem 6, we must first think about how to find the length of a curve in a metric space. The idea is to choose several points on the curve, and draw geodesic segments between pairs of consecutive points. The length of the curve should then be greater than or equal to the total of the lengths of these segments. We then define the length of the curve to be the supremum of this sum over all possible choices of points on the curve.

rectifyDefinition: A continuous path c : [a, b] \to X has length

l(c) = \sup_{a = t_0 < t_1 < \cdots < t_n = b} \sum_{i=1}^n d(c(t_{i-1}), c(t_i)).

If l(c) < \infty, then we say that c is rectifiable.

Now we’ll show that a path in a hyperbolic metric space can’t go very far from a geodesic between its endpoints unless the path is very long.

Lemma 7: Let X be \delta-hyperbolic. Let c be a continuous, rectifiable path from p to q. Then for any x \in [p, q],

d(x, im \ c) \leq \delta | \log_2 l(c) | + 1.

quasigeodesic1Proof. Without loss of generality, we may assume that c : [0, 1] \to X is parametrized proportionally to length. Let N \in \mathbb{N} such that

\frac{l(c)}{2^N} < 1 \leq \frac{l(c)}{2^{N-1}}.

It’s enough to prove that d(x, im \ c) \leq \delta N + 1. The proof is by induction on N. If N = 0, then l(c) < 1, so a point on \ [p, q] can’t be more than one unit away from the image of c. So in this case the inequality follows immediately.

For the inductive step, consider the triangle \Delta = \Delta(p, c(\frac{1}{2}), q):

quasigeodesic2Now d(x, x') \leq \delta for some x' on one of the edges of the triangle other than \ [p, q]; without loss of generality we’ll say \ [p, c(\frac{1}{2})]. By induction, we have d(x', im \ c) \leq \delta(N - 1) + 1. So d(x, im \ c) \leq d(x, x') + d(x', im \ c) \leq \delta + \delta(N - 1) + 1 = \delta N + 1, as desired.

In the next lemma, we show that given an arbitrary quasi-geodesic, we can find a “nicer” quasi-geodesic that is close to the given quasi-geodesic.

Lemma 8: Let X be a geodesic metric space. Given any (\lambda, \epsilon)-quasi-geodesic c : [a, b] \to X, there exists a continuous (\lambda, \epsilon')-quasi-geodesic c' : [a, b] \to X such that

(i) c'(a) = c(a) and c'(b) = c(b).
(ii) \epsilon' = 2(\lambda + \epsilon).
(iii) l(c'|_{[s, t]}) \leq k_1 d(c'(s), c'(t)) + k_2 for all s and t in \ [a, b], where k_1 = \lambda(\lambda + \epsilon) and k_2 = (\lambda \epsilon' + 3)(\lambda + \epsilon).
(iv) d_{Haus} (im \ c, im \ c') \leq \lambda + \epsilon.

Proof. First we’ll choose some points where c' will coincide with c. These will be

\Sigma = \{ a, b \} \cup ([a, b] \cap \mathbb{Z}).

So we set c'(t) = c(t) for all t \in \Sigma (and thus (i) immediately follows). Choose geodesic segments joining these points, and parametrize c' linearly along these segments.

So each segment is of length at most \lambda + \epsilon, since c is a (\lambda, \epsilon)-quasi-geodesic. Every point of im \ c \cup im \ c' is at most \frac{\lambda + \epsilon}{2} from c(\Sigma) = c'(\Sigma), and so (iv) follows.

Now we prove (ii). For t \in [a, b], let \ [t] be a choice of nearest element of \Sigma. Then for s, t \in [a, b], we have

d(c'(s), c'(t)) \leq d(c([s]), c([t])) + (\lambda + \epsilon)
\quad \leq \lambda | [s] - [t] | + \epsilon + (\lambda + \epsilon)
\quad \leq \lambda (|s - t| + 1) + (\lambda + 2\epsilon)
\quad = \lambda |s - t| + 2(\lambda + \epsilon).

The other inequality is similar, and (ii) follows.

Finally, to prove (iii), we’ll start by looking at integer subintervals. For all integers n, m \in [a, b] with n < m, we have

l(c'|_{[n, m]}) = \sum_{i=n}^{m-1} d(c(i), c(i+1))
\quad \leq (\lambda + \epsilon) |m - n|.

Similarly, l(c'|_{[a, m]}) \leq (\lambda + \epsilon)(m - a + 1) and l(c'|_{[n, b]}) \leq (\lambda + \epsilon)(b - n + 1). Therefore, for all s, t \in [a, b] we have

l(c'|_{[s, t]}) \leq (\lambda + \epsilon)(|[s] - [t]| + 2).

Combine this with

d(c'(s), c'(t)) \geq \frac{1}{\lambda} |s - t| - \epsilon' \geq \frac{1}{\lambda}(|[s] - [t]| - 1) - \epsilon'.

Then (iii) 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 \deltaslim 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 \deltahyperbolic, 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.

Example: G is quasi-isometric to 1 if and only if G is finite.

Definition: A metric space X is proper if closed balls of finite radius in X are compact.  The action of a group \Gamma on a metric space X is cocompact if X/\Gamma is compact in the quotient topology.

The Švarc-Milnor Lemma: Let X be a proper geodesic metric space.  Let \Gamma act cocompactly and properly discontinuously on X.  (Properly discontinuously means that for all compact K\subseteq X, |\{\gamma\in \Gamma | \gamma K\cap K \neq \emptyset \}| < \infty.)  Then \Gamma is finitely generated and, for any x_0\in X, the map

\Gamma \rightarrow X
\gamma \mapsto \gamma.\ x_0

is a quasi-isometry (where \Gamma is equipped with the word metric).

Proof: We may assume that \Gamma is infinite and X is non-compact.  Let R be large enough that the \Gamma-translates of B=B(X, R) cover X.  Set

S=\{s\in \Gamma\setminus 1 | s\bar{B}\cap \bar{B} \neq \emptyset \}

Let r=\inf\{d(\bar{B},\gamma\bar{B}) | \gamma\in\Gamma , \gamma \not\in S \cup \{1\} \}.  Let \lambda=\max_{s\in S} d(x_0, sx_0).   We want to prove that:
(a) S generates,
(b) \forall \gamma\in\Gamma,

\underbrace{ \lambda^{-1}d(x_0,\gamma x_0)}_{i} \leq \ell_S(\gamma)\leq \underbrace{ \frac{1}{r}d(x_0,\gamma x_0) + 1}_{ii}
(c) \forall x\in X, there exists \gamma\in\Gamma such that

d(x,\gamma x_0) \leq R

Note: d_S(1,\gamma)=\ell_S(\gamma) and d_S(\gamma, \delta)=\ell_S(\gamma^{-1}\delta).

(c) is obvious.
(b-i) is also obvious.
To complete the proof we need to show (a) and (b-ii).

Assume \gamma\not\in S\cup\{1\}.  Let k be such that

R + (k-1)r \leq d(x_0, \gamma x_0) < R + kr

As \gamma\not\in S\cup\{1\}, k>1.  Choose x_1, \cdots, x_{k+1} =\gamma x_0, such that d(x_0, x_1)<R and d(x_i, x_{i+1})<r for each i>0.   Choose 1=\gamma_0,\gamma_1,\cdots, \gamma_{k-1},\gamma_k=\gamma such that x_{i+1}\in\gamma_i\bar{B} for each i.  Let s_i=\gamma_{i-1}^{-1}\gamma_i, so \gamma=s_1\cdots s_k.  Now

d(\bar{B}, s_i\bar{B}) \leq d(\gamma_{i-1}^{-1} x_i, s_i\gamma_i^{-1}x_{i+1}) = d(x_i, x_{i+1}) <r

So, s_i\in S\cup\{1\}.  Therefore S generates \Gamma.

Also,

\ell_S(\gamma) \leq k \leq \frac{1}{r}d(x_0, \gamma x_0) + \frac{r-R}{r} < \frac{1}{r}d(x_0,\gamma x_0)+1

as required.

Corollary: If K\subseteq \Gamma is a finite index subgroup of a finitely generated group then K is quasi-isometric to \Gamma.

Two groups G_1 and G_2 are commensurable if they have isomorphic subgroups of finite index.  Clearly, if G_1 and G_2 are commensurable then they are quasi-isometric.

Example: Sol = \mathbb{R}^2 \rtimes_E \mathbb{R}.
Semidirect product is taken over the matrix eqlatex This means that Sol=\{(x,y,t) | x,y,t\in\mathbb{R} \}, but

(x,y,t)(x',y',t')=(x+e^tx', y+e^{-1}y', t+t')

Let A\in SL_2(\mathbb{Z}) with eigenvalues \lambda, \lambda^{-1} with \lambda >1.  Let \Gamma=\mathbb{Z}^2 \rtimes_A \mathbb{Z}.

\Gamma_A sits inside Sol as a uniform lattice, meaning Sol/\Gamma_A is a compact space.

Exercise 11: What is this quotient?

So, \Gamma_A is a quasi-isomorphic to Sol  But, Bridson-Gersten showed that \Gamma_A and \Gamma_{A'} are commensurable if and only if the corresponding eigenvalues \lambda, \lambda' have a common power.

Exercise 12: Let T_k be the infinite regular 2k valent tree.  Prove that for all k, k' \geq 2, T_k is quasi-isometric to T_{k'}.