1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

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5.2. Groups acting on trees 179

= μs.x,s.y


for every s E E and ( x, y) E X x Y. 0
Remark 5.2.2. This result really does generalize the fact that extensions
of exact groups are exact. Let r be a group and A<1r be a normal subgroup
such that A and r /A are exact. The hypotheses of Proposition 5.2.1 are
satisfied with X = ,B(r /A) and K = r /A.
The hard work is essentially over. We now recall lots of definitions
and prove a few well-known facts about trees, compactifications and groups
acting on these objects.
Let T be a tree, which we identify (as a metric space) with its vertex
set (see Appendix E). A finite or infinite sequence x(O), x(l), ... in T is
called a geodesic path if d(x(n), x(m)) = In - ml for every n and m.^7 For
convenience, if (x(n))f:=o is a finite geodesic path, then we extend it to
an infinite sequence by setting x(m) = x(N) for every m ~ N; we still
call this sequence a (finite) geodesic, even though it isn't, strictly speaking.
Two geodesic paths x and x' are equivalent if they eventually flow together,
i.e., if there exist mo, no E N such that x(mo + n) = x'(no + n) for every
n ~ 0. We can (and will) identify T with a subset of the equivalence classes
of geodesics: every point x in T is identified with the equivalence class
of geodesic paths ending at x. The ideal boundary 8T of T is defined as
the set of all equivalence classes of infinite geodesic paths. We define the
compactification of the tree T to be T = TLJ8T (a topology will be described
shortly). If (x(n))n is a geodesic path with equivalence class x ET, then
we say the geodesic path (x(n))n connects x(O) with x. For a bi-infinite
geodesic path (x(n))~=-oo' we let x(oo) E 8T (resp. x(-oo) E 8T) be the
equivalence class of the geodesic path (x(n))n2':0 (resp. (x(-n))n2':o), and we
say (x(n))n connects x(-oo) with x(oo).
Lemma 5.2.3 (See also Lemma E.2). Let x ET and y E 8T. Then, there
exists a unique geodesic connecting x with y.

Proof. Pictorially, the proof is totally transparent. Here's the recipe in
words: Let (y(n)) be a representative of y and let ( w(j) )f= 1 be a finite
geodesic connecting x with y(O) (which exists, since T is connected). Let
No :::::; N be the first integer such that there exists no with w(No) = y(no)


  • i.e., find the first point of intersection of the two geodesics. Define a new
    geodesic (z(k)) by z(k) = w(k) for 1:::::; k:::::; No and z(k) = y(no + (k - No))
    fork> N 0. Evidently (z(k)) is a geodesic connecting x with y.
    Uniqueness of (z(k)) follows from the fact that Tis a tree - any other
    geodesic connecting x with y would yield a loop in T. 0


7rn a tree, a path is geodesic if and only if it never backtracks.
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