- CHARACTERIZING ROUND SOLUTIONS
c:-neck, we have f3 n C 0 = 0. Let u E (3. Since f3 c 5; o, +' we have
d(u,a)::::; d(u,p) + d(p, y)
::::; sup d(z,p)+lOO
zEs; 0 ,+::::; sup d(z,p)+202,
zES5,+93where we used !sol ::::; 101 and provided c: is small enough. By our hypothesis we
then obtain
3
d (u , a)::::; 2,d (x , p) + 202::::; 1.6ry.It follows that there exist points a E a and b E f3 such that
d (a, b) = d (a, (3) ~ u::::; 1.6ry.
Consider a geodesic triangle in ( 52 , g) with vertices x, b, a and minimal geodesic
sides xb, xa, ba having corresponding side lengths ry, ry , u. Let x, b~ be aEuclidean triangle with the same side lengths L(xb) = ry , L(xa) = ry, L(ba) = u.
Then the angle at a satisfies Lxab:::;:: cos-^1 (0.8) since u ::::; 1.6ry. By the triangle
version of the Toponogov comparison theorem , we conclude that
L xab ::::: Lxab ::::: cos-^1 (0.8).
However, since a is near the center of an c:-neck and since x and b are outside the
c:-neck, the angle Lxab must be small and we obtain a contradiction. DIn view of Lemma 29.24, the following result is useful.LEMMA 29.26 (Distance circles in the vicinity of the center of an c:-neck). Letg be a Riemannian m etric on 52 with scalar curvature R > 0 and let p E 52.
For any O > 0 th ere exists c: E (0 , 0.01) such that if Ne: is a normalized c:-neck
centered at p, then for any x rj. Bp ( c^1 ), y E Bp (100), and for almost every s E
(d(y, x) - c:, d(y, x) + c:), we have the following. Let a 8 be the connected component
of 8Bx(s) with d (a 8 , y) < c:. Then the loop a 5 is smooth except at finitely many
points and <:Xs is o-close to fSy ~ cp-^1 ( {Sy} x 51 ) in c^0 ' where (Sy, By) = cp(y).
Furthermore, the values L(as), L(rsy), and 27r are all o-close to each other.PROOF. STEP l. c^0 closeness. By [376], for any c: > 0 and for a.e. s E
( d(y, x) - c:, d(y, x) + c:) we have that [)Bx ( s) is a disjoint union of piecewise smooth
embedded loops. By choosing c: sufficiently small, we guarantee that there exists aunique connected component a 5 of 8Bx (s) with d(as,Y) < €.
Let g ~ (cp-^1 )*(g). Then g is c:-close in the cf^0 -
1l+^1 -topology to gcyl on
( -c^1 + 4, c^1 - 4) x 51 ; compare with (28.26). Since (29. 77) holds approximately
with gcyl replaced by g, we have that for s E (d(y, x) - c:, d(y, x) + c:) that cp(as)
is lOc:-close to {Sy} x 51 in c^0 with respect to g. That is , a 5 is lOc:-close to
rsy = cp-^1 ({sy} x 51 ) in c^0 with respect tog. So, first of all , we require that
€ < b/10.
STEP 2. Closeness of the lengths. Parametrize a 5 by arc length u and choose
any point z ~ a 5 (uz) a t which a 5 is smooth. Join z to x by a minimal unit speedgeodesic f3z : [O, s] --+ 52 , so that f3z (0) = z and f3z ( s) = x. Let [O, L1) be the largest
subinterval for which fJz([O, L 1 )) C Bp (c:-^1 ). Without loss of generality, we may