1547845440-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_III__Chow_

500 J. SOLUTIONS TO SELECTED EXERCISES

where r, (), z are cylindrical coordinates. That is, X is obtained by revolving
Y about the z-axis. Endow X with the intrinsic metric induced from the
ambient space ffi.^3. Note that X is compact. (Why does X have finite
diameter?)^1 Let p = (0, 0) be the origin and take for example

for i EN. Then

lYi ~ J(2i + 1)!. (2i)!,

f3i ~ J(2i)!. (2i -1)1,

. ,lim (X, aid,p) = ffi.^2 x {O},
i->oo
,lim (X, f3id, p) = { (0, O)} x [O, oo ).
i-too

We may also consider the sequences defined by /i ~ (2i)! and Oi ~ (2i -1)!.
What are the corresponding limits for these sequences?
SOLUTION TO EXERCISE H.44. Go parallel to the x-or y-axis to the
closest of the two diagonals {y = x} and {y = -x} and then go along this
diagonal to the origin (0, 0).

SOLUTION TO EXERCISE I.14. We first consider a 2-:dimensional flat

cone and then approximate it by smooth complete Riemannian surfaces

with positive curvature. Let s; denote the circle of length a E (0, 21f) and

consider the Euclidean metric cone Cone ( s;). If d ( e1, e2) = ~, then by

(G.12) we have

dcone(S,1;) ([(()1, r1)], [(()2, r2)]) = Jrr + r~ -2/3r1r2,.

where f3 ~ cos ( % ) E ( -1, 1). Now consider the 2-dimensional positively

curved rotaticmally symmetric expanding gradient solitons discussed in §4

of Chapter · 2 ~f Volume One; we do not need that they satisfy the Ricci

flow, just that they approximate the cone well. For any a E (0, 21f) there

is such a solution, call it (R^2 , 9a ( t)), which is asymptotic to Cone ( s;). In

fact, (ffi.^2 , 9a (t)) converges to Cone ( s;) in the Gromov-Hausdorff distance

as t. -+ o+, while the incomplete solutions (R^2 --:-{ 0} , 9a ( t)) converge to

Cone ( s;) -{origin} in the C^00 Cheeger-Gromov sense as t -+ o+ .. For the

Riemannian surface (R^2 , 9a ( t)), where a E ( 1r, 21r) and t > 0 is sufficiently

small depending on a, consider the ray

I : [O, oo) -+ ffi.2

emanating from the origin defined by 'Y (s)-.;... (() 2 , s). On the cone Cone (s;)

consider the Busemann function for the corresponding ray I ( s) = [ ( () 2 , s)],

(^1) Answer: Essentially.because :z:=:=o ~ = e < oo. ·