118 29. COMPACT 2-DIMENSIONAL ANCIENT SOLUTIONSTo define the plane version Q of Q, we first consider the irreducible orthogonal
decomposition for the 3-tensor cJ3v ~ L;,j,k=l Vijkdxi 0 dxj 0 dxk on IR^2. The
decomposition is given via defining
1
(29.171) aijk ~ Vijk - 4 (6.eucViOJk+t.eucVjOik+t.eucVkOij).Th e 3 t -ensor -a ~"^2 i,j,k=l aijk - d x i '<Y '°' d x j '<Y '°' d k x. is t t o a (^11) y t race-f ree an d t t o a (^11) y
symmetric. Define on IR^2 the quantity
(29.172) Q ~ v lal^2.
Observe that Q is scale-invariant in the following sense. Given K > 0 and
to < 0, define
u(x, y , t) = K^2 u(Kx, Ky, t 0 + t), ii(x, y , t) = K-^2 v(Kx, Ky, t 0 + t).
Then(29.173) Q(x, y, t) ~ii lal^2 (x, y, t) = Q(Kx, Ky, to+ t),
where
1
aiJk ~ iiiJk - 4 (6.euciiiOJk + 6.euciiJoik + 6.euciikoij).EXAMPLE 29.47.
(1) On the King- Rosenau solution we compute, using (29.11), that v 11112ax = 3v221 and that ii222 = 12ay = 3fJ 11 2, so that Q = 0.
(2) The cigar soliton metric on IR^2 is 9cig = v-^1 9euc, where fJ = 1 +r^2. Thus,
for the cigar soliton we have that Vijk = 0 and hence Q = 0.
(3) Q = 0 for a round 2-sphere.
Next, we observe formulas for a and lal^2. We compute that
1 1
(29.174) am=
4
(vm - 3v122) = -a122 and a222 =
4(v222 - 3v112) = -a112are all of the components of a up to symmetry. Since the decomposition (29.171)
is orthogonal, we find that
2 2
l8^3 vl2
= L (viJk)
2
and lal2
= L (aiJk)
2
i,j,k=l i,j,k=lsatisfy
(29.175) l8^3 vl^2 - ~l86.euciil^2 = lal
2
= ~ ( (vm - 3ii221)^2 + (ii222 - 3fJ112)^2 ).
13.2. Proof of Proposition 29.46.By Lemma 29.50 below, our hypothesis is equivalent to Q = 0.
STEP 1. For each t E (-oo,O) there exist (x 0 (t), y 0 (t)) E JR^2 , A(t) > 0, and a
quadratic polynomial q(x, y, t) such that
(29.176) v(x, y, t) = A(t)((x - x 0 (t))^2 + (y -y 0 (t))^2 )^2 + q(x, y, t).Fix a time t. Since Q = 0, the components of a in (29.174) vanish and we have
(29.177) ii111 - 3ii122 = 0 and ii222 - 3ii112 = 0.
Differentiating this implies that
(29.178) fJ1111 = 3fJ1221 = 3ii1122 = fJ2222.