112 29. COMPACT^2 -DIMENSIONAL ANCIENT SOLUTIONS
The first term on the RHS is equal to
'Vi.6. 52 v;kv = v rnjkv
= \7 Wj^5 k V - R^5 uem^2 \7^3 mjk V - R iijm^52 \7 imk^3 V - R5iikm^2 \7^3 ijm V
3 52 3 52 3 52 3
= .6. 52 \lijkv - Rim 'Vmjkv - 2Riejm 'Vemkv - 2Riekm 'Vejmv
= .6.5 2 \l[jkv - v ;jkv + 2(-V'Jikv - \l~ijv)
+ 2 ((g52)ij (\7k.6.52v + \lkv) + (g52)ik (\7j.6.52v + 'Vjv)).Symmetrizing this formula , we obtain
S(\7 .6. 52 \7^2 v) = .6.52 S(\7^3 v) - 5 S(\7^3 v) + 4 S( (\7 .6.52v + dv) 0 g52)
since S and .6. 52 commute. Therefore, by (29.143), we have
(29.144) S(\7^3 .6.52v) = .6.52 S(\7^3 v) - 9 S(\7^3 v) + 24 S(Z 0 g52 ) ,
where Z = i(V .6. 5 2v + ~dv) as in (29.131b).
Now substituting (29.144) into (29.141) yieldsLEMMA 29.45. The 3-tensor B = S(\7^3 v) satisfies the following heat-type equa-
tion(29.145)
8
ot B = v.6.52B + (.6.52v - 9v) B + 24v S(Z 0 g 5 2)- 3 S(\7 (dv 0 \7.6.52v)) + S(\7^3 (-l'Vvl
2 - 2v^2 )).
We now proceed to prove t he proposition by carrying out a long derivation
of the formula for Q = v ITF(B)l^2. We are guided by co llecting terms to form
complete squares.
STEP l. Taking the trace-free part of (29.145) yields
(29.146)
8
ot TF(B) = v.6.5 2 TF(B) + (.6.52v - 9v) TF(B) + 24vTFS(Z 0 g 5 2)+ TFS( 3 \7^2 v 0 \7.6.52V + 3dv 0 \7^2 .6.52v - \7^3 l'Vv l
2
+ 2 \7^3 ( v^2 ) ) ,where TFS = TF o S. On the other hand,
(:t - v.6.52) Q = 2vTF(B) · (:t -v.6.52) (TF(B)) - 2v
2
l'VTF(B)l2- 2v\7v · \7 ITF(B)l
2
+ ITF(B)l
2
( :t -v.6.52) v.By combining the above two displays, while applying (29.7) and the fact that
TFS( Z 0 g 5 2) = 0, we obtain
(29.147) (:t -v.6.52) Q = -2v^2 l'VTF(B)l^2 -4v(dv®TF(B)) · 'VTF(B)
+ (2v.6.52v - l'Vv l^2 - 16v^2 ) ITF(B)l^2
+ 4vTF(B) ·TFS(\7^3 (v^2 ))
+ 6v TF(B) · TFS(\7^2 v 0 \7.6. 5 2v)- 2v TF(B) · TFS(\7^3 l'Vvl^2 )
- 6v TF(B) · TFS(dv 0 \7^2 .6. 5 2v).