1547671870-The_Ricci_Flow__Chow

15. AN ALTERNATIVE STRATEGY FOR THE CASE x (M^2 > 0) 169

To prove the claim we consider the trace Harnack quantity for the un-
normalized Ricci flow on a surface with bounded positive scalar curvature,
namely
{) 2

Q ~ otlogR- IV'logRI = 6.logR+R.

This is the quantity defined in (5.35) without the -r term. Following the
proof of Lemma 5.55, one computes that it evolves by

(5.56) ot {) Q = 6.Q + 2 (V'logR, V'Q) + 2 I V'V' logR +^1 12

2

Rg ,

hence estimates that

(5.57)

{)

ot Q 2 6.Q + 2 (V'log R, V'Q) + Q^2.

(The reader should compare the equality and inequality above to (5.37) and

(5.38), respectively.) If Q ever becomes nonnegative, then it remains so by

the maximum principle. On the other hand, if Qmin (t 1 ) < 0 for some t 1 ,

then Qmin (to) < 0 for all to ::; t1, and one has

1 1

Qmin (t)^2 (Q (to))-^1 - (t - to) > -t - to

for all t >to. Since (N^2 , g (t)) is an eternal solution, one may let to ~ -oo,

obtaining

. 1
Qmin (t) 2 hm --= 0.
to--+-oo to - t
Hence on (N^2 , g (t)), we have
6. log R + R = Q 2 0.
This is the ancient version of Hamilton's trace Harnack inequality. (Note
the similarity between this argument and the proof of Lemma 9.15.)
Now by hypothesis, there exists (po, to) E N^2 x IR such that
R (po, to) = sup R.
N^2 xlR

Thus we have 8R/8t = 0 and IV' R I = 0 at (po, to), and hence Q (po, to)= 0.

Since Q 2 0, applying the strong maximum principle to equation (5.56) then

shows that Q must vanish identically on N^2 x R By examining its evolution

equation (5.56), we see that this is possible only if the tensor

1

V'V' log R +

2

Rg

vanishes on N^2 x R In other words

{)

{)tg = -Rg = .C(VlogR)u g.

Hence g(t) is a gradient Ricci soliton fl.owing along grad (log R). D