- THE ~-GAP THEOREM FOR 3-DIMENSIONAL ~-SOLUTIONS 119
since o E (0, 1).^32 From (19.70) and
Vol (Bsz(p) (xoo, JU) x BIR (iJ 00 , 7J2))
2: 14v'20VolB 82 (v'2) (*,v'2)= 28v'20 (1 - cos 1)
::::: 188,
we may estimate crudely that in M 00 (taking into account a possible Z 2 -
quotient)
Vol9 00 (1) B9 00 (1)(q 00 , 10) 2: 90.
Hence, for i sufficiently large, we have
VolfJ,,.i(l) B_g,,.i(l)(qi, 10) 2: 80.We now may estimate the reduced volume Vat Ti from below:
v h) ::::: r (47rTi)-3/2 e-£(q,Ti)dμgh) (q)
j B9h) (qi,lOy'Ti)
:;:: (47rTi)-^3 l^2 e_^8 -1
·Vol9( 7 i)B.9h)(qi,10.JTi)= (47r)-^3 l^2 e-^8 -
1
· Vol 9 ,,.i(l) B_g,,.i(l)(qi, 10)(19.71) :;:: (47r)-^3 /^2 e-^8 -^1. 80.
STEP 3. Finishing the proof of Theorem 19. 56.
Let c1 (n) be the constant in Theorem 19.51. Suppose that (M^3 ,g (t))is ~1-collapsed at the scale ro > 0 at some point (xo, 0) E M x (-oo, O] for
some ~1 E (~, ci (3)). It then follows from the trace Harnack estimate (which
bounds the curvature backward in time) that g (t) is strongly ~1-collapsed
at (xo, 0) at scale ro (see Definition 19.50).
Choosing the basepoint p used in the above discussion to be xo and
applying Theorem 19.51 below, we obtain(19. 72) V ~ ( ~ 1 1/3 r 0 2) ~ e 312 ~1/2 1 +w2e -1/2 exp ( ---m 1 ) ,
(47r) 2~1
where w2 = 7r is the volume of the unit Euclidean 2-ball. For i sufficiently
large we have Ti 2: ~i/nrfi, so that by the monotonicity of the reduced volume
we have
(19.73)for such i. Combining (19.72), (19.73), and (19.71), we conclude that
0 < ( 47r )-3/2 e -8-1 80 ~ e 312 ~1/2 1 + 7re -1/2 exp ( ---m 1 ).
(47r) 2~1(^32) We still denote the lifted metric by fj 00 (1). Note also that (J2)^2 +(7J2)^2 =10^2.