186 5. Exact Groups and Related Topicsand o; is a (C, r)-quasigeodesic path, the length of a' is at most 6CDo + r.
By joining /o, a' and 11, we obtain a sequence/ connecting bo to bi. For
the reader's convenience, we list the properties of 1: it connects bo to bi;
d(po, 1) 2:: Do; the length 111 of the sequence / is at most (6C + 2)Do + r;
and d(!'(k ), 1(k + 1)) :S C(l + r) for every k.
Now we apply the Weierstrass bisection process. Set bZ = bk, pg =Po
and 1° =I· Let c^0 be (one of) the midpoint(s) of 1° and consider a geodesic
triangle [b8, c^0 ] U [ c^0 , b~] U [b~, b8J. Since K is hyperbolic, there exists PB in
[b8, c^0 ] U [c^0 , b~] such that d(pg, PB) :S 8. If PB is on [b8, c^0 ], then we set b6 = b8
and bi = c^0 - otherwise let b6 = c^0 and bi = b~. Let 11 be the subsequence
of I connecting b6 to bi. We note that 1111 :S (2/3)111- Now, we continue
this process by letting c^1 be (one of) the midpoint(s) of 11 , and so on. This
process terminates in l steps, with l :Slog Iii/ log(3/2), and gives rise to Pb
on [bb, bi] such that bb and bi are consecutive points on I· It follows that
Since linear functions grow faster than logarithms, it follows that for each
fixed C, rand 8, the numbers Do must be uniformly bounded (independent
of o; and /3). D
Having established geodesic stability, the following definition is indepen-
dent of the choice of generating set.
Definition 5.3.6 (Hyperbolic group). Let r be a finitely generated group.
We say that r is hyperbolic if its Cayley graph is hyperbolic.
Remark 5.3.'T. Since the Cayley graph of lFn is a tree, free groups are hy-
perbolic. Other examples include co-compact lattices in simple Lie groups
of real rank one and the fundamental groups of compact Riemannian man-
ifolds of negative sectional curvature (cf. [70]). Some of these groups have
Kazhdan's property (T) (see [15, 84]).
Our next goal is to define the Gromov boundary; drawing lots of pictures
will help.
Let K be a hyperbolic graph. We say that two infinite geodesic paths o;
and f3 in K are equivalent if
m,n--)-oo liminf (a(m),/3(n))o = oo
for some o E K. Geometrically, this means o; and f3 are pointing in the
same direction. It is clear that the definition is independent of the choice
of o E K. It's not so clear that we have an equivalence relation, hence a
lemma.