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25. Elevations and Fibre Products

1 April 2009 in Course notes | Tags: Elevations, fibre products | by AHM | 2 comments

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Lemma 24: Fix basepoints $x,y,y' \in X,Y,Y'$ as usual. There is a lift $f':X'\rightarrow Y'$ of $f:X\rightarrow Y$ such that $f(x)=y'$ if and only if

$f_*\pi_1(X,x) \subseteq \pi_1(Y',y')$ .

Furthermore, if the lift exists, it is unique.

Lemma 25: Given a choice of $y'$ , there exists an elevation $f':X'\rightarrow Y'$ of $f:X\rightarrow Y$ at $y'$ . Furthermore, $f'$ is unique in the sense that if $\bar{f}:\bar{X}\rightarrow Y'$ is another elevation of $f$ and $y'$ then there is a homeomorphism $X'\rightarrow \bar{X}$ and the diagram

commutes.

Proof:

Let $(X',x')\rightarrow (X,x)$ be defined by $\pi_1(X',x')=f^{-1}_*\pi_1(Y',y')$ . By Lemma 24, the composition

$(X',x')\rightarrow (\bar{X},\bar{x})\rightarrow (Y,y)$

lifts at $y'$ , call this lift $f'$ . Suppose

Then $\pi_1(X',x')\subseteq \pi_1(\bar{X}\bar{x})$ . But by Lemma 24, $f_*\pi_1(\bar{X},\bar{x})\subseteq \pi_1(Y',y')$ . This implies $\pi_1(X',x') =\pi_1(\bar{X},\bar{x})$ .

Another, more categorical construction uses the fibre product.

The fibre product $\hat{X}$ is defined by

$\hat{X} =\{(\xi,\eta)\in X\times Y'|f(\xi)=\sigma(\eta) \}$ .

There are obvious maps

$\hat{X}\stackrel{\hat{f}}{\rightarrow} Y'$
$\hat{X}\stackrel{f}{\rightarrow} X$

given by forgetting factors.

Exercise: If $\sigma$ is a covering map then $\rho$ is a covering map.

Lemma 26: Fix $x\in X$ . Let $y=f(x)\in Y$ and let $\sigma(y')=y$ for $y'\in Y$ . Let $x'=(x,y')\in \hat{X}$ , and let $X\subseteq\hat{X}$ be the connected component containing $x'$ . Then $f'=\hat{f}|_{x'}:X'\rightarrow Y'$ is an elevation of $f$ at $y'$ , and every elevation of $f$ arises in this way.

Proof: To prove that $f'$ is an elevation we just observe that $p_*\pi_1(X',x')=f_*^{-1}\sigma_*\pi_1(Y,'y')$ . Now suppose

is an elevation. Then $\bar{X}\rightarrow \hat{X}=\{(\xi,\eta)|f(\xi)=\sigma(\eta) \}$ , with $\xi\mapsto(\tau(\xi),\bar{f}(\xi))$ .

The covering map $\tau$ factors trhough $\bar{X}\rightarrow\hat{X}$ , and so $\bar{X}\mapsto\hat{X}$ is a covering map. Because $\bar{X}$ is an elevation, $\bar{X}\mapsto\hat{X}$ is a homeomorphism onto its image, a connected component of $\hat{X}$ .

What has this got to do with graphs of spaces/groups?

Let $Y$ be a vector space, and let $\partial: E\rightarrow Y$ be an edge map. Define $X$ to be the mapping cylinder

$X =(E\times[0,1]) \sqcup Y/\sim ;\;\; (x,1)\sim\partial(x)$

$X$ comes with a map $d:X\rightarrow Y$ such that $d|_Y=id_Y$ and $d(x,t)=\partial(x)$ . This is an inclusion $\iota:Y\rightarrow X$ , and $d\circ\iota =id_Y$ . Let $\sigma:Y'\rightarrow Y$ be a covering map.

Let $\hat{X}$ be the fibre product. There’s a map $\iota':Y'\rightarrow \hat{X}$ ; $\eta\mapsto (\iota\circ\sigma (\eta),\eta)$ . Clearly, $d'\circ\iota'=id_{Y'}$ .

Therefore, $\iota'$ is an injection. It’s easy to see that $\iota'$ induces a bijection at the level $\pi_0$ .

Lemma 27: Any covering space $X'\rightarrow X$ arises as the fibre product of a covering map $Y'\rightarrow Y$ .

Proof: Let $\tau:X'\rightarrow X$ be a covering map

let $Y'$ be the fibre product of $\tau$ and $\iota$ .

$Y'=\{(\xi,\eta)\in X'\times Y |\tau(\xi)=\iota(\eta)\}$

Define $d':X'\rightarrow Y'$ by $\xi\mapsto (\xi,d\circ\tau(\xi))$ . As before, $\sigma$ is a covering map.

8. The Švarc-Milnor Lemma

9 February 2009 in Course notes | Tags: Quasi-isometry, Švarc-Milnor Lemma | by AHM | 3 comments

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 $d_S(\gamma, \delta)=\ell_S(\gamma^{-1}\delta)$ .

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.