60 2. SPECIAL AND LIMIT SOLUTIONS
such that each metric in the family has positive scalar curvature, satisfies
I ls 1JiA,B I :S 1, has a neck of radius rmin (0) = JA, and has a bump of height
no less that A VB. Moreover, the scale-invariant measure a of its curvature
pinching is bounded uniformly for all A E ( 0, A).
Lemma 2 .32 implies that the solution of the Ricci flow starting from
gA,B must lose its neck before time
T ....:... rmin (0)2
A
A"7" n-l n - l
On the other hand, the solution will have a bump at some x* (t); and since
lal is uniformly bounded for all solutions under consideration, the height of
this bump will be bounded from below by
2 - r;:; - r;:; CA
1jJ (x* (t), t) ?: Av B - Ct?: Av B - n _
1.If A E (0, A) is small enough, the neck must disappear before the bump can
vanish. This concludes our proof of Proposition 2.42
5.6. Single point pinching. Finally, we justify Remark 2.17. We
consider the special case of a reflection-invariant metric on sn+l with a single
symmetric neck at x = 0 and two bumps. (See Figure 2.) For solutions
of this type whose diameter remains bounded as the singularity time is
approached, the neckpinch singularity will occur only on the totally-geodesic
hypersurface {O} x sn. We shall prove this claim by constructing a family
of subsolutions Q for v = 1/i 8.
Recall that x(t) denotes the location of the right-hand bump. (For
reflection symmetric data, this is the unique point in (0, 1) where 1/imax(t) is
attained). Recall too that D = limvr1Ji(x(t),t) denotes the final height
of that bump. By Proposition 2.42, we may assume that D > 0. Define the
function
p(t) = n lat foD ( ~)2
dsdt,noting that p is monotone increasing in time, so that p(T) = limvr p(t)
exists.
Now let s(t) = s(x(t), t) denote the distance from the equator to the
right-hand bump. By Lemma 2.21, one has 11/isl :S 1; and because Ot1/imax :S
-(n - 1)/1/imax, one also has 1/i(s(t), t) > D for all t E [O, T). Together,
these results imply that s(t) > D for all t E [O, T). Let XD(t) denote the
unique point in (O,x*(t)) such that s(xD(t),t) = D. Notice that for any
s, the hypothesis of reflection symmetry about x = 0 lets one integrate by
parts to obtain the identityOS = n rx 1/iss OS dx = n { 1/is + r(x) (1/is) 2 ds}.