1547845440-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_III__Chow_

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  1. SETTING UP THE PROOF BY CONTRADICTION AND POINT PICKING 167


THEOREM 21.9 (ro = 1 case of pseudolocality). For every a > 0 and


n 2: 2 there exist 6 > 0 and co > 0 depending only on a and n such that if


(Mn,g (t)), t E [O,c^2 J, where c E (O,co], is a complete solution of the Ricci

flow with bounded curvature and if xo E M is a point such that


(21.9) R (x, 0) 2: -1 in B 9 (o) (xo, 1)


and

(21.10)

for any regular domain n c Bg(o)(xo, 1), where Cn = nnwn is the Euclidean
isoperimetric constant, then^5
a 1
IRml(x,t):::; -+ 2
t co

(21.11)

for x EM such that dg(t) (x, xo) <co and t E (0, c^2 ].^6


REMARK 21.10 (Monotonicity of 6, co, and a). Note that, in the above
theorem, the dependence of 6 and co on a are monotone in the following
sense. If 6 > 0 and co > 0 work for some a > 0, then for any 0 < f/ :::; 8,
0 < c~ :::; co, and a :::; a' < oo, we have that 6' and c~ work for a'.

Since the proof of Theorem 21.9 is somewhat long, we divide it into
steps which we present in the remaining sections. The proofs of some of the
technical results are deferred to the next chapter.

2.1. If the pseudolocality Theorem 21.9 is false - a sequence


of counterexamples.


We begin the proof of the pseudolocality Theorem 21.9. From Remark
21.10, clearly we only need to prove the theorem for a sufficiently small, say
0 <a< 13 (n!l)fo (this is used in the proof of Claim 2 in §1 of Chapter 22).
STEP 1. If the theorem is false, then there exists a bad sequence of solu-
tions, points, times, and scales (by 'bad' we mean having 'large curvature').
Suppose that the theorem is false. We then have the following.

Counterstatement A. There exist a E ( 0, 13 (n!l)fo) and sequences of
positive numbers Ji ---+ o+, cOi ---+ o+, and smooth complete pointed solutions

(^5) We leave it as an exercise to show that one can replace the RHB of (21.11) by
max{%,~}·
(^6) 0bserve the following elementary fact: For any f (t) which is o (1) near t = 0 and
for any a > 0 there exists c: 0 > 0 such that
fort E (0, c:5J.
f (t) < ~
t - t

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