12 The heat-type equation satisfied by Q
EXERCISE 29.43.
(1) Show t hat
S (V^3 vKR) = 4/3 sin 2'lj; S (d'l/; 0 g5 2 ).From this we again see that TF(BKR) =TFS (V^3 vKR) = O.
111(2) Using (29.9) and that \7^2 z + z g 52 = 0 on 52 , where {x,y ,z} are t he
Euclidean coordinates on JR^3 , show that
n 3 v v KR = -d v KR ® g 1 d KR 1 d KR
52 - 2.g 52 0 v - 2 g52c 13 ) 0 v( 2 ),where (952(13) 0 dvc2J) (X1, X2, X 3) ~ g52 (X1, X3) dv (X2).12. The heat-type equation satisfied by Q
Let g(t), t E (-oo, 0) , be a maximal ancient solution to the Ricci fl.ow on 52 and
let v(t) b e its pressure function. Throughout this section, the covariant derivative
\7, Laplacian 6. 5 2, and norms of tensors are all defined with respect to g 5 2.
We have t he following nice evolution equation for Q = v ITF(B) 12 , as defined
in (29.134).PROPOSITION 29.44 ( Q is a subsolut ion of the heat equation). W e have
a 2
(29.140) at Q = v6.52Q - 4RQ - 2 ITF (v'VTF(B) + 2dv 0 TF(B))I- 2 ltr^1 '^2 (v'VTF(B) - dv 0 TF(B))i
2
.
In particular, since R > 0, we have that gt Q :::; 6. 9 Q.
The rest of this section is devoted to the proof of this proposition. We begin
by calculating the evolut ion of B = S (\7^3 v). Using (29.7), we compute that
:t ('Vrjkv) = 'Vrjk ( v 6. 52v - 1Vvl
2
- 2v^2 )
= v'Vrjk6.52V + 6.52v'Vrjkv + 'Vrjk (-1Vvl
2
+ 2v^2 )
+'Vi (\7kv\7j6.52v) + 'Vj (\7iv\7k6.52v) + 'Vk ('Vjv\7i6.52v).By symmetrizing both sides of this equation, we obtain
a
(29.141) at S(\7^3 v) = v S(\7^3 6. 52v) + 6.52v S(\7^3 v) + 3 S(\7 (dv 0 \7 6.52v))+ S(\7^3 ( - 1Vvl
2
+ 2v^2 )).On the other hand, for a function f on a Riemannian manifold (Mn,g), we
h ave (see (2.34) in [77] for example)
v;j6. 9 f = 6. 9 v;jJ + (-'VjRie + \7 eRij - 'ViRje)\7 d- 2Rikjp v;d -R u 'VJef - Rje v;ef.
Hence, for v on ( 52 , g 5 2), we h ave
(29.142) 'VJk6.52V = 6.5 2 'VJkV + 26.52v(g52)jk - 4'VJkV.
Taking another covariant derivative, we have
(29.143) \7rjk6.52V = \7i 6.5 2 'VJ kV+ 2\7i6.52v(g52 )jk - 4\7rjkV.