- THE HEAT-TYPE EQUATION SATISFIED BY Q 113
Toward the goal of obtaining nonpositive terms on the RI-IS of (29.147), we first
rewrite the last four lines therein.
Third line. Since S(\7^3 (v^2 )) = 2vB + 6 S ( dv ® \7^2 v) by a direct calculation, we
have(29.148) TF(B) · TFS(\7^3 (v^2 )) = 2v ITF(B)l^2 + 6TF(B) ·TFS (dv ® \7^2 v).
Fourth line. Since TF(B) is totally trace-free and symmetric, we have(29.149)Fifth line. We compute that(29.150)
where U 8 p,q V denotes a single tensor contraction of t he p-th component of U with
the q-th component of V. A computation yields the commutator formula (we use
\7 Rm 52 = 0 in the first line)
(29.151) n^4 n^4 R5(^2) n2
vijkev = vifjkv - j fkm vimv
4 52 2 52 2 52 2
= \7 fijk V - Rifjm \7 mk V - Rifkm \7 jm V - RJfkm \7 im V
= 'Vjijk v - 2(g52 )ek vtv - (g52 )ej v;k v + (g52 )ij v ;k v
+ (g52 )ik 'VJeV + (g52 )jk \7 ;ev,and consequently,
'Vev'V [jkev = 'Vev'Viijkv - 2\7kvv;jv - 'Vjv'V;kv
+ (g52 )ij \7 ev'V;k v + (g52 )ik \7 ev'V]ev + (g52 )jk \7 ev'V;ev.Thus, by taking trace-free symmetric parts of this, we have
TFS(dv 81 ,4 \7^4 v) = dv 8 1, 1 'VTF(B) - 3TFS (dv ® \7^2 v).Substituting this into (29.150) and taking the inner product with -4vTF(B) yields
(29.152) -2vTF(B) · TFS(\7^3 l'Vvl
2
) = -4v(dv ® TF(B)) · 'VTF(B)
- 12 vTF(B) ·TFS (dv ® \7^2 v)
- 12 v TF(B) ·TFS (\7^3 v 8 3, 2 \7^2 v).
Sixth line. Since TF(B) is totally trace-free and symmetric, we have(29.153)
TF(B) · TFS(dv ® \7^265 2v) = TF(B) · ( dv ® ( \7^265 2v - ~D.i2vg52))
=tr^1 •^2 (dv ® TF(B)) · ( \7^2 65 2v - ~D.i2vg52).