1547845440-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_III__Chow_

APPENDIX J

Solutions to Selected Exercises

It's not what you know; it's what you can prove.

• From the movie "Training Day".

SOLUTION TO EXERCISE 22.2. We prove this by contradiction. If the

result is not. true, .then for all j EN there exists

· (xj, tj) E P(xj-i, tj-i, (c:Qj-i)-i/^2 , -(c:Qj-i)-i)

with

and

... T :.
t J ·· > - t J ·-1 - HQ-:-i J-1 > - to - HQ-i (1 + 2-i + · · · + 2-(j-i)) > -- 4 (1 + 2-j)

si~ce to 2:'.: ~ and HQ-^1 = ~· This contradicts SUPMx[O,T] I Rm l(x, t) < oo.

Now assume by induction that I Rm l(Xj-1' tj-1) > ta J-1 (we verified this

for j = 2, 3). We· then have ·

2a a
I Rm l(xj, tj) > 21 Rm l(xj-1, tj-i) > - 2:'.: -
tj-i tj
(since tj 2:'.: tj-i - HQ-i2-(j-i) = tj-i - ~2-(j-i) 2:'.: ~tj-1).
SOLUTION TO EXERCISE 22.5. Suppose that, inductively, for some m E
N the points (xk, tk) E S with Xk E Bg(tk) (xo, (2A + l)c:) and tk > 0 have
been defined for 1:::::; k:::::; m. Moreover, suppose that the point (x, f) in the
claim cannot be taken to be (xm, tm) (otherwise we are done). Then we may

define (xm+l, tm+i) ES to be a point .such that

(J.1) I Rm l(xm+l, tm+i) > 41 Rm l(xm, tm),

(J.2)

1

0 < tm+l :::S tm and dg(trn+i)(Xm+i,xo) ::'.S dg(trn)(Xm,Xo)+AIRml-2(xm,tm)·

By induction we also have that (J.l) and (J.2) hold with m replaced by
k E [1, m - 1]. Thus, for all 1 :::::; k :::::; m + 1 we have 0 < tk :::::; ti :::::; c:^2 and
the curvatures at (xk, tk) are increasing at least geometric;:ally:

(J.3) I Rm l(xk, tk) > 4k-il Rm l(xi, ti),

so that

(J.4)

1 1. 1

I Rm 1-2 (xk, tk) <

2

k_ 1 IRm1-2 (xi, ti).

497