1547845440-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_III__Chow_

4. NECKS IN MANIFOLDS WITH POSITIVE SECTIONAL CURVATURE 67

If c: ( n) > 0 is small enough, since the geometry of the neck 91 is close to that

of the standard cylinder, then we have^20

diam(b~^1 (bo)) :s;diam('lf;(Sn-l x [-57r,57rl))

11 _ I.

< /1l\1cnR (qo)^1 2 diam (sn-l x [-57r, 57rl)

• v101

< ~cnR (qo)-1/2. / 1f2 + (l07r)2

- v101 V

::; ll1fcnR (qo)-^1 /^2.

Finally we prove the first inequality in (18.51), i.e., diam (b~^1 (bo)) 2:

g1rcnR(qo)-^1 l^2. Note that for fixed z E sn-1, the function b'Y('lf;(z,u)) is

continuous in u. Inequalities (18.55) and (18.58) imply that for any z E sn-l

we have

b'Y ('lj; (z, 57r)) > bo and b'Y (1/; (z, -57r)) < bo.

Hence there exists u (z) E [-57r, 57r] such that

b'Y (1/; (z, u (z))) = bo.

If we choose z E sn-l to be the antipodal point of zo (recall 'lj; (zo, 0) = qo),
then since the geometry of the neck 91 is close to that of the standard
cylinder, it is clear that
diam (b~^1 (bo)) 2: d ( 1f; (z, u (z)), 1f; (zo, 0))

2:^11 ( )-1/2

12

cnR qo dsn-lx[-57r,57r] ((z,u(z)), (zo,O))

2:^11 -1/2

12

7rcnR (go).

D

The existence of a 'sufficiently good' neck implies the following.

LEMMA 18.31 (Manifold containing an c:(n)-neck has b'Y attaining its

minimum). There exists c: (n) > 0 such that if (Nn, h) is a complete non-

compact Riemannian manifold with positive sectional curvature containing
an embedded c:( n )-neck, then for any ray / the corresponding Busemann
function b'Y is bounded from below and attains its minimum.

PROOF. By the proof of Lemma 18.30, if E (n) > 0 is sufficiently small

and if 91 C N is an embedded c:(n)-neck, then Ninn U 91 is compact a11d for

x E N - (Ninn U 91) , we have b'Y ( x) > bo. Hence inf N b'Y = minMnn us.n b'Y is

attained at some point in Ninn U 91 C N. D

REMARK 18.32. Without the existence of a sufficiently good neck, it is
easy to come up with a counterexample to b'Y being bounded from below.
See Exercise I.14 in Appendix I.

(^20) In particular we choose s(n) > 0 small enough so that the second inequality holds
simply because ffor > 1.