Margulis Lemma

STRUCTURE OF FUNDAMENTAL GROUPS 17
where we used ε ≤ 1
2C
in the inequality. This easily implies that vol(B r/10(c(t)))
are comparable for all t. More precisely, for any t 1 , t 2 ∈ [0, 1] we have that
(8)
1
C 0
vol B r/10 (c(t 1 )) ≤ vol B r/10 (c(t 2 )) ≤ C 0 vol B r/10 (c(t 1 ))
with a computable universal C 0 = C 0 (n). Put
∫
h(s) = −
dt r (s)(p, q) dµ(p) dµ(q),
B r/10 (p 0) ′ ×B r(c(0))
U := {(p, q) ∈ B r/10 (p 0 ) ′ × B r (c(0)) | dt r (s)(p, q) < r},
φ s (U) := {(φ s (p), φ s (q)) | (p, q) ∈ U},
and dt ′ dt
r(s)(p, q) := lim sup r(s+h)(p,q)
h↘0 h
. Since dt r is monotonously increasing
and bounded by r, we deduce that if dt r (s)(p, q) = r, then dt ′ r(s)(p, q) = 0.
Since φ s is measure preserving, it follows
∫
h ′ d
(s) ≤ −
dt |t=0 dt′ r(0)(p, q)
φ s(U)
∫
≤
4 vol(B3r(c(s))2
vol(B r/10 (c(0)))
− dt ′ r(0)(p, q),
2
B 3r(c(s)) 2
where we used that φ s (B r/10 (p 0 ) ′ ) 2 ⊂ φ s (U) ⊂ B 3r (c(s)) 2 . If p is not in the cut
locus of q and γ pq : [0, 1] → M is a minimal geodesic between p and q then
dt ′ r(0)(p, q) ≤ d(p, q)
∫ 1
0
‖∇·X‖(γ pq (t)) dt.
Combining the last two inequalities with the segment inequality we deduce
h ′ (s) ≤
∫
C 1 (n)r− ‖∇·X‖
B 6r(c(s))
≤ C 1 (n)r Mx ‖∇·X‖(c(s))
and thus h(1) ≤ C 1 (n)rε. Note that the choice of the constant C 1 (n) can be made
explicit and independent of the induction assumption.
This in turn implies that that the subset
{
∫
}
B r (p 0 ) ′ := p ∈ B r (p 0 ) ∣ − dt r (1)(p, q) dµ(q) ≤ r/2
B r/10 (p 0) ′
satisfies
(9)
It is elementary to check that
vol(B r (p 0 ) ′ ) ≥ (1 − 2C 1 (n)ε) vol(B r (p 0 )).
φ t (B r (p 0 ) ′ ) ⊂ B 2r (c(t)) for all t ∈ [0, 1].
Then arguing as before we estimate that
∫
−
dt r (s)(p, q)dµ(p)dµ(q) ≤ C 2 (n) · r · ε.
B r(c(t)) ′ ×B r(c(t))