- INCOMPRESSIBILITY OF BOUNDARY TORI 249
Since the area of V(t) is less than or equal to t he area of D(t),(33.73) 1
15
L (p) dp:::; A (p) :::; A (p) = 115
L15 (p) dp.In view of (33.59) and since we are in an almost hyperbolic cusp, we have t hatgiven any TJ > 0 and 0 < B < A :::; A, there exists t,,., < oo such that for t ~ t,,.,.
:P ( eP-PL 15 (P)) :::; 7JePL 15 (p) for p E [p, p] C [O, PA,B(t)).
Integrating this inequality yieldsIE 15 (p) - e15-Pf: 15 (p)I = 1115d~ ( eil-Pf: 15 (p)) dpl
:::; T/ 1 15 ePL15 (p) dp= T/ ( eP - eP) L15 (p) '
which implies that
L 15 (p) :::; e^15 -P (1 + rJe^2 P-i5 ( e^15 -P - 1)) L 15 (p).Thus for any c: > 0, by choosing T/ = 2c:e-PA,B so that 7Je^2 P- 15 :::; 2c:, we have
(33.74)
Motivated by (33.60), we have the following property for t he almost hyperbolic
cusp CA,B CM.
LEMMA 33 .3 4 (Monotonicity-type formula for length). Let 0 < B < A :::; A,
where A is as in (h5) in Subsection 1.1 of this chapter. Then for any c: > 0
sufficiently small, there exists tc: < oo such that fort ~ tc: and p E (0, PA,B(t)] we
have that the function L (p) in (33.66) satisfies(33.75) d ( ec:p^115 )
d- _ - L(p)dp ~O.
p 1 -eP 0
PROOF. In view of the calculation
(33. 76)d ( eEP r p ) L (p) e- 15
dp ln 1 - e-P Jo L (p) dp = It L (p) dp + c: - 1 - e-15'we desire to bound L (p) from below. As a consequence of (33.74) and (33.71), we
have that for 0:::; p:::; p < PA,B,
L 15 (p):::; e^15 ((1 -2c:) e-P + 2c:ei5-^2 P) L (p).
Integrating this, we obtain It L 15 (p) dp :::; (e15 - 1) (1 + c: (eP - 1)) L (p). Hence, it
follows from (33.73) that
( 15 e15 - 1j o L (p) dp :::; 1 - c( eP - 1) L (p).
This implies that the RHS of (33.76) is nonnegative. Observe that c: > 0 was
introduced to enable this nonnegativity. D