116 29. COMPACT 2-DIMENSIONAL ANCIENT SOLUTIONS
Third, regarding the third line of equation (29.154), observe that by using that
TF(B) is totally trace-free and symmetric, we have(29. 164 ) TF(B) ·TFS (dv ® \7^2 v) = TF(B) · ( dv ® ( \7^2 v - ~6 5 2vg 52 ))= tr^1 •^2 (dv ® TF(B)) · ( \7^2 v - ~652vg52).Now, by applying to (29.154) the formulas (29.160), (29.161), and (29.164), we
obtain(29.165)( :t -v6s2 + 4R) Q + 2 JTF (v\7TF(B) + 2dv ® TF(B))J
2
=-21~ (v^2 t:. 5 2v-
6
1
2
;
952
) +3v(v^2 v-652
;
952
) +2tr^1 ·^2 (dv®TF(B))l2
+ 6v6s2V JTF(B)J
2
+ 6 ltr^1 •^2 (dv ® TF(B)) 12
+ 36vtr^1 ·^2 (dv ® TF(B)) · ( \7^2 v -652
;
952
) + 6vTF(B) · (\76 5 2v ® \7^2 v)- 12 v TF(B) ·TFS (\7^3 v 8 3,2 \7^2 v)
- 6vtr^1 •^2 (dv ® TF(B)) · ( \7^2 6s2V - ~2'12vg52).
STEP 3. We can improve this by completing the square on t he second line in a
slightly different way, which reduces the number of terms, as follows:(29.166)( :t -v6s2 + 4R) Q + 2 JTF (v\7 TF(B) + 2dv ® TF(B))J
2
= -21~ ( \7^265 2v -6
1
2
;
952
) + 3v ( \7^2 v -652
;
952
) - tr^1 •^2 (dv ® TF(B))l2
+ 6v6s2V JTF(B)J^2 + 6vTF(B) · (\76 5 2v ® \7^2 v)- 12v TF(B) ·TFS (\7^3 v 03 , 2 \7^2 v).
We claim that
(29.167)
2's2v ITF(B) 12 + TF(B) · (V 65 2v ® \7^2 v) - 2 TF(B) ·TFS (\7^3 v 03 , 2 \7^2 v) = 0.
By (29.159), this completes the proof of P roposit ion 29.44.
To see t he claim, we compute in local coordinates that
(29.168) TF(B) · (\7 2's2v ® \7^2 v)
= TF(B)rn((Vi 1 v - \7~ 2 v)\7 1 2's2v - 2Vi 2 v\722's2v)
+ TF(Bb2((\7~ 2 v - V i 1 v)\7 2 2's2v - 2Vi 2 v\712's2v).