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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

Image 1



Image 2

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

Image 3

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.

Image 4

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

Image 5

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.

Image 6

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

Image 7

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.

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.


\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'}.