- THE HEAT-TYPE EQUATION SATISFIED BY Q 115
where Z = ~ \7 6. 5 2v + idv. Now
1 2.
3tr' (\7 S(Z 0 g52))jk = \7' (Zi(gs2)jk + Zj(gs2)ik + Zk(gs2)ij)
= div(Z)(gs2)jk + \7kZj + 'VJZk.
Therefore
(29.159)
tr^1 '^2 (\7TF(B)) = \7^2 6. 5 2v + l
3
0\7^2 v - ~6. 5 2vg 52 - div(Z)g52 - 2S(\7Z)
= 2 1( \7^2 6.52v -^1 2 ) (^2 1 )
2
- 5 2vg 52 + 3 \7 v -
2
6. 5 2vg 52.
We conclude that a in (29.155) satisfies
(29.160) tr^12 ' (a)= 2 v ( \7^2 6.52v - 26.^1 52 2vg52 ) + 3v ( \7^2 v -^1 )
2
6.s2vg52
- 2tr^1 '^2 (dv 0 TF(B)).
Second, for the second line of (29.154), we shall derive
(29.161) JVvJ^2 JTF(B)J^2 = 2 ftr^1 '^2 (dv 0 TF(B))f
2
.
Given any point p E S^2 , we shall compute in local coordinates where 9iJ = Oij at
p. We have
(29.162a)
(29.162b)
This implies that
2
(29.163) JTF(B)J^2 = L (TF(B)iJk)^2 = 4((TF(B)m)^2 + (TF(Bb2)^2 ).
i ,j,k=l
Using (29.162) and (29.163), we calculate
ftr^1 '^2 (dv 0 TF(B))f
2
= (\71vTF(B)m)
2
- (\71vTF(B)122)
2
+ 2 (\71 v TF(B)112)^2 + (\72v TF(B)211)^2
+ (\72v TF(B)222)^2 + 2 (\72v TF(B)212)
2
= 2 (\71 v)
2
(TF(B)m)
2
- 2 (\71 v)
2
(TF(Bb2)
2
- 2 (\72v)^2 (TF(Bh22)^2 + 2 (\72v)^2 (TF(B)m)^2
= ~ J'Vv J
2
JTF(B)J
2
.
2
This establishes (29.161).