1547845447-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_IV__Chow_

110 29. COMPACT 2 -DIMENSIONAL ANCIENT SOLUTIONS

The 3-t ensor H is fully symmetric and trace-free and is easily seen to b e H =

TF(B). By (29.135), (29.136), and (H , X®gs2) = (H , Y ® gs2) = 0, we see that

(29.133) also follows from

(29.137) ITF(B) l^2 = IHl^2 = JV^3 v J

2

• 21x1

2

• 6 IYl

2

• 4 (X , Y).

11.2. The vanishing of Q on the King-Rosenau solution.

Consider t he King- Rosenau ancient solution (S^2 , 9KR(t)). For simplicity, we

takeμ= 1 in (29.17). Differentiating (29.8) then yields dvKR = -/3 sin 2 '!/Jd'lj;, where

'ljJ is the latitude. In this discussion we sh all treat f3 as a positive constant, a lthough

we may assume that it satisfies (29.17).

Using formula (29.18) for the Christoffel symbols, we compute that the Hessian

and Laplacian of v are given by

(29.138a)

1 1

~ \7^2 vKR = -2 cos 2 '1/Jd'l/J ® d'lj; +

2

sin^2 2 '1/JdB ®dB,

(29.138b)

1

~6. 8 2vKR = -2 cos^2 'ljJ + 4sin^2 'lj;,

where e is the longitude. Let (x^1 , x^2 ) = ('l/J, e). Since 'VfjW = -r}j and 'VfjB =

• f fj, the Hessian of the latitude and the longitude are

\7^2 '!/J = -sin

2

2

W dB® dB, \7^2 B =tan 'ljJ (d'lj; ®dB+ dB® d'lj;).

Hence, covariantly differentiating (29.138a), we obtain

(29.139)

1

~ \7^3 vKR = 4 sin 2 '1/Jd'lj; Q9 d'lj; Q9 d'lj; + 2 sin 2 '!/J cos^2 'lj;d'lj; ®dB® dB

+ sin2'1j;cos^2 'ljJ (de® dB® d'l/J +dB® d'l/J ®dB).

Equivalently, we may compute directly that the components of \7^3 vKR are given by

\7 1 \71 \71 VKR = 4 /3 sin 2 '1j;,

\7 1 \7 2 \7 2VKR = 2 /3 sin 2 '!/J cos^2 'ljJ = 2\7 2 \7 1\72v ,

\71\71\72vKR = \7 2\71\71v = \72\7 2 \72v = 0.

From this we can derive the following identities for the King-Rosen au solut ion:

132 sin2 2 '!/J = ~ IV3vKRl2 = ~IV 6. 2vKRl2

22 36 s

= IVvKRl2 = -~ ( \76.52vKR, 'VvKR).

Substituting these formulas into (29.133) yields that

QKR ~ VKR ITF(BKR)J2 = 0.

Alternat ively, from ~dvKR = -sin 2 '1j;d'lj; and ~ \7 6. 8 2vKR = 6 sin 2 '!/Jd'lj;, we find

this vanishing of QKR directly from (29.139), (29.135), and H = TF(B).
Let TFS ~ TF o S be the trace-free part of the symmetrization.