13. THAT Q = 0 IMPLIES THE SOLUTION IS THE KING-ROSENAU SOLUTION 117
We also computeTF(B) ·TFS (V^3 v 83,2 V^2 v)= TF(B) · V^3 v 83,2 V^2 v
= TF(B)rn (ViievViev - ViuvV~ev - V~ievV~ev - V~2evViev)
- TF(Bh22 (-Vi1evV~ev - V i2evV iev - V~ievViev + V~2evV~ev)
= TF(B)rn (Vi 11 v - V~ 21 v) V i 1 v
+ TF(B)rn (Vr12v - Vi21V - V~11V - V~22v) Vi2v
+ TF(B)rn (-V i 22 v - V~ 12 v) V~ 2 v + TF(B)222 (V~ 22 v - Vii 2 v) V~ 2 v
+ TF(B)222 (V~21V - V~12v - Vi22V - V i 11 v) Vi2v
+ TF(B)22 2 (-v~ll v - V i2 1 v ) V i i v.Hence, by this and (29.168), while using V i6.s2v = V y 11 v + V y 22 v and 6.s2v =
V t 1 v + V~ 2 v, we obtainTF(B) · (V 6. 5 2v ® V^2 v) - 2 TF(B) ·TFS (V^3 v 8 3 , 2 V^2 v)= -TF(B)rn (Vi 11 v - Vi 22 v - 2V~ 21 v ) 6.s2V
- TF(B)222 (V~22V - v~ll v - 2Vi12V) 6.52V.
Now, commuting covariant derivatives and using (29.162), we h ave(29.169a) V i11v- Vi22V- 2V~21V = V i11v- 3Vi22V- 2V1v = 4TF(B)rn,
(29.169b) V~ 22 v - V~ 11 v - 2Vi 12 v = V~ 22 v - 3V~ 11 v - 2V2v = 4 TF(B)222.From the a bove formulas and (29.163), we obtain (29.167). This finishes the proof
of Proposition 29.44. D13 That Q = 0 implies the solution is t he King- Rosenau solution
Let Q be the quantity defined by (29.134) for a maximal ancient solution
(S^2 ,g(t)), t E (-oo,O). In this section we prove the following characterization
of the King- Rosen au solution.
PROPOSITION 29.46 (Q = 0 implies being the King- Rosenau solution). W e
have that Q = 0 if and only if g (t) is either round or the King- Rose nau solution.
Throughout this section, let x = x^1 and y = x^2 denote the standard Euclidean
coordinat es , let V = 8, and let I · I denote the Euclidean norm.13.1. Plane version of Q.
To prove the main proposition of this section, we shall find it convenient to
consider the plane version of Q. Computations in the plane, as compared to com-
putations on the round 2-sphere, a re slightly sitnplified since partial derivatives
commute.
Recall by (29.7) that the pressure function v satisfies
[)-