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Section 2.4 What is an end, anyway?

In the previous lecture we presented the notion of a many ended graph, but didn't quite define what an end was. We will do this in this lecture and apply the concept to quasi-isometric rigidity.

Subsection 2.4.1 Ends are like essential components at infinity.

Previously we defined a locally finite connected graph \(X\) to be many ended if there was some point \(x_0\) such that the deletion of some ball \(B(x_0,r)\) in \(X\) produced multiple infinite components.

Consider some sequence \(0\lt r_1 \lt r_2 \lt \ldots\) of radii growing to infinity and denote by

\begin{equation*} \mathcal{K^i}=\{K^i_{j} \subset X; 1\leq j \leq N_i\} \end{equation*}

the infinite connected components of \(X \setminus B(x_0,r_i)\text{.}\) Common sense about sets implies that since we have \(B(x_0,r_1)\subset B(x_0,r_2)\subset \cdots\) then if \(n>m\) then each element of \(\mathcal{K}^n\) is contained in a unique element of \(\mathcal{K}^m\) which gives a well defined surjective function \(f_{nm}:\mathcal K^n \to \mathcal K ^m\text{.}\)

If you want you can make a rooted tree where level \(i\) is a component in \(\mathcal K^i\) and the ancestor of each component is the element of \(\mathcal K^{i-1}\) which contains it.

Figure 2.4.1. Essential components

The ends of the graph \(X\) correspond to the infinite sequences \(K^1_1 \supset K^2_{n_2} \supset K^3_{n_3} \supset \cdots\) of nested components. Formally the set of ends is the defined as the set of inverse limits of components

\begin{equation*} \mathrm{Ends}(X) = \lim_{\leftarrow} \mathcal K^n \end{equation*}

with respect to the system of functions \(f_{nm}, n>m\text{.}\) The elements of the inverse limit are formally sequences of infinite connected components, and so can be seen as "components" at infinity and they are in bijective correspondence with the infinite branches of the tree shown in Figure 2.4.1.

We suppress the concern that ends seem to be highly dependent on the choice of growing balls (we will confront this momentarily) and will consider two examples.

Let \(L\) be the infinite line (i.e. the standard Cayley graph for \(\ZZ\)) and let \(T_4\) be the infinite regular tree of valence 4 (i.e. \(T_4 = \cay{\{a,b\}}{F(a,b)}\text{.}\)) In both metric spaces consider the balls \(B(x_0,1)\subset B(x_0,2) \subset B(x_0,3)\subset \cdots\text{.}\)

  • For \(L\) we always have that \(L\setminus B(x_0,n)\) has exactly two components, so as the balls grow we get the following containements of components
    \begin{equation*} \cdots \subset K^2_1 \subset K^1_1 \subset L \supset K^1_2 \supset K^2_2 \supset \cdots \end{equation*}
    so the diagram given in figure Figure 2.4.1 has precisely two infinite branches and, so \(L\) has two ends, which we should think of as being \(\pm \infty\text{.}\)
  • In \(T_4\text{,}\) the complement \(T_4 \setminus B(x_0,n)\) has precisely \(4 \cdot 3^n\) components. In particular each \(K^n_j \in \mathcal K^n = T_4 \setminus B(x_0,n)\) contains exactly 3 components in \(\mathcal K^{n+1}\text{.}\) It follows that the ends of \(T_4\) are in bijective correspondence with infinite strings of the form:
    \begin{equation*} ab_1b_2b_3\cdots, a\in\{1,2,3,4\}, b\in\{1,2,3\} \end{equation*}
    and therefore form an infinite set of continuum size.

Subsection 2.4.2 Ends are robust

A compact (or finite) exhaustion of a set \(X\) is a sequence of compact (or finite) subsets \(K_1 \subset K_2 \subset \cdots\) such that:

\begin{equation*} X = \bigcup_{i=1}^\infty K_i. \end{equation*}

A sequence of balls with a common center whose radii grow to infinity is an example of a compact exhaustion.

The main issue about our definition of ends is that it seems to depend on our choice of compact exhaustion. We will state our main robustness result, but only sketch a proof. The first item is a straightforward, yet non trivial exercise in inverse limits, which are beyond the scope of the prerequisites, and the second item is obvious given the material that was previously covered, but also ultimately depends on inverse limits.

We first sketch the proof of the first point. Let \(A_1 \subset A_2 \subset \cdots \) and \(B_1 \subset B_2 \subset \cdots\) be two different compact exhaustions of \(X\) and denote by \(\mathcal A^i\) and \(\mathcal B^j\) the sets of infinite components of \(X\setminus A_i\) and \(X \setminus B_i\) we want to show an identification between the inverse limits

\begin{equation*} \lim_{\leftarrow} \mathcal A^n, \lim_{\leftarrow} \mathcal B^n \end{equation*}

but the difficulty is that we may not have a well defined map \(\mathcal A^n \to \mathcal B^m\) where we want to send \(Z \in \mathcal A^n\) to the unique component \(W \in \mathcal B^m\) such that \(Z \subset W\) even if \(n>m\text{.}\)

The issue is resolved by observing that for some sufficiently large \(N(m)\) such maps \(\mathcal A^k \to \mathcal B^m\) will exist provided \(k> N(m)\) and then applying the abstract nonsense of inverse limits which enables us to consider the inverse limits of the union \(\{\mathcal A^n\}\cup\{\mathcal B^n\}\) equipped with a compatible system of surjective functions.

We now sketch a proof of the second point. If \(K \subset X\) is a finite essential disconnecting set, then it's quasi isometric image \(f(K)\subset Y\) can be enlarged (by a uniformly bounded amount) to a finite essential disconnecting set. Furthermore the quasi isometric image of a compact \(A_1 \subset A_2 \subset \cdots \) exhaustion of \(X\) gives rise (perhaps after bounded enlargement) to a compact exhaustion \(\widehat{f(A_1)} \subset \widehat{f(A_2)} \subset \cdots \)of \(Y\text{.}\) Again, the image \(f(K^i_j)\subset Y\) of an infinite component of \(K^i_j \subset X\setminus A_i\) may not quite lie in a component of the counterpart \(Y \setminus \widehat{f(A_i)}\text{,}\) but if \(n \gt\gt m\) then we will have \(f(K^n_j)\subset Y\setminus \widehat{f(A_m)}\text{.}\) The bijection then follows from the universal property of inverse limits. !

Here's an even more informal explanation:

  • given different compact exhaustions of a space, we can always find a way to eventually match things up so that we have the same ends, and
  • Quasi isometries sent compact exhaustions to compact exhaustions and finite essential disconnecting sets to finite essential disconnecting sets, and everything works out great.

The main point of all this is that we get a quasi-isometric invariant of groups.

Subsection 2.4.3 2 endedness and the quasi isometric rigidity of \(\ZZ\)

We have now developed just enough machinery to show that \(\ZZ\) is quasi isometrically rigid, what we will prove is precisely the following:

Here's the first step in the proof. Note that in this proof, quasi isometry is not used, but rather action by isometries.

Since \(G\) acts on \(X\) by isometries, then in particular it acts on \(X\) by quasi isometries, and every compact exhaustion is sent to another compact exhaustion. By Theorem 2.4.3, this implies that \(G\) acts on \(\mathrm{Ends}(G)\text{.}\) Now if \(G \onto F\) is a homomorphism to a finite group then the kernel has index \(|F|\text{,}\) an the first point follows from the action of \(G\) on the two element set \(\{\pm \infty\}\text{.}\)

Suppose now that we passed to \(H\) (note that we allow \(H=G\)) and let

\begin{equation*} \cdots \subset K^2_1 \subset K^1_1 \subset L \supset K^1_2 \supset K^2_2 \supset \cdots \end{equation*}

be some nested sequence of components of complements of sets in a compact exhaustion starting with \(C\text{.}\) Without loss of generality (reversing \(E^+\) and \(E^-\) if necessary) we may assume that \(K^n_1 \subset E^-\) and \(K^n_2 \subset E^+\text{,}\) for all \(n\text{.}\) Then in particular the point \(\infty\) corresponds to the chain of inclusions \(L \supset K^1_2 \supset K^2_2 \supset \cdots\text{.}\) Now if for some \(h \in H\) we have \(h \cdot E^+ \subset E^-\) then this means that the translated chain \(L \supset h\cdot K^1_2 \supset h \cdot K^2_2 \supset \cdots\) is contained in \(E^-\) and its terms converge to \(-\infty\) so we get \(h \cdot \infty = -\infty\) contradicting our assumption that \(H \leq G\) didn't permute the ends of \(G\text{.}\)

We also have the following useful fact.

Let \(F_1,\ldots, F_m\) be the the finite components of \(X \setminus K\) and let \(I_1,\ldots,I_n\) be the infinite components of \(X \setminus K\text{.}\) Then every component \(F_i\) touches \(K\) (actually, to be accurate the closure of each component intersects \(K\) in a non-empty set.) Take the set

\begin{equation*} K' = K \cup F_1 \cup \cdots \sup F_m. \end{equation*}

It is an exercise in point-set topology to see that \(K'\) is also compact. Graph theoretically, finiteness of \(K'\) s obvious. Either the complement \(X\setminus K'\) consists of a finite union of infintie components.

Exercises 2.4.4 Exercises

1.

Draw the complements of the first three balls in \(T_4\) as described in Example 2.4.2.

2.

Prove Theorem 2.4.5

Hint: Follow these steps.

  1. Start with \(G\) and take \(X = \cay A G\) for some finite generating set.
  2. Take \(H\leq G\) as in Lemma 2.4.6.
  3. Let \(C \subset X\) be a finite essential cutset and prove that there exists some \(h \in H\) such that \(h \cdot C \cap C = \emptyset\text{.}\)
  4. By Lemma Lemma 2.4.7 we can assume that \(X \setminus C\) consists of precisely two infinite components.
  5. If \(K_1,K_2\) are the infinite components of \(X \setminus C\text{,}\) argue that \(g\cdot C\) must lie in one of these components an by the assumption that \(g\) doesn't permute ends that (w.l.og. )\(g\cdot K_1 \subset K_1\text{.}\)
  6. Iterate this nesting, i.e \(K_1 \supset g\cdot K_1 \supset g^2\cdot K_1 \supset \cdots\text{,}\) to argue that \(g\) has infinite order so that \(\gen g \approx \ZZ.\)
  7. Argue that \(\gen h\) is finite index in \(G\) by showing that the action of \(\gen h\) on \(X\) is co-compact. This can be done by showing that every component of
    \begin{equation*} X \setminus\left(\bigcup_{n\in \ZZ} h^n\cdot C\right) = Y \end{equation*}
    is finite, so that every \(g \in G\) can be joined to an element of \(\gen h\) by a finite path. Show that if this is not the case, then it is possible to find an essential finite disconnecting set whose complement has at least 3 infinite components (one of these components must lie in a connected component of \(Y\) ), contradicting 2-endedness.