1547845440-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_III__Chow_

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  1. 2-AND 3-DIMENSIONAL w-SOLUTIONS 105


Let
9i (t) ~ R (xi, 0) g ( R (xi, 0)-^1 t)

be the rescaled solutions. Since ~~ ;:::: 0 and since g (t) is 11:-noncollapsed
at all scales, with the above choices of {xi} and {ri} and by Hamilton's
Cheeger-Gromov compactness theorem, a subsequence of (incomplete) so-
lutions (B 9 (o) (xi,ri) ,gi (t) ,xi) will converge to a complete limit ancient
solution
(M~,9 00 (t) ,x 00 ), t E (-oo,O],
with
(19.41)
and sup Moo Rgoo (·, 0):::; 4.^23
On the other hand, by dimension reduction, this limit solution must
contain a line (see Theorem 18.19) and it splits as

(M~,goo (t)) = (ffi.xW1,du^2 + 9W (t)).


Since (W,gw (t)) is I-dimensional, we have that (M 00 ,g 00 (t)) is flat, which


contradicts (19.41). D

In contrast, recall that the cigar soliton solution has ASCR (g ( 0)) =


0, dimension reduces to a flat cylinder, and the cigar soliton is not 11:-

noncollapsed at all scales for any 11: > 0.


4.2. The equivalence of the notions of 11:-solution and 11:-solution


with Harnack in dimension 3.
In Corollary 19.43 we proved that a 2-dimensional 11:-solution with Har-
nack is a 11:-solution (i.e., has bounded curvature). This result extends to
dimension 3 using the same idea of proof, namely dimension reduction, to-
gether with the fact that 3-dimensional 11:-solutions with Harnack dimension
reduce to round cylinders.
We now prove the following.

PROPOSITION 19.44 (3-dimensional 11:-solutions with or without Harnack
are equivalent). Any 3-dimensional 11:-solution with Harnack (M^3 , g (t)), t :S
0, has bounded curvature on M x (-oo, OJ.

PROOF. Case (1). We first assume that (M, g (0)) does not have positive
curvature operator. Then since Rm ::'.". 0, it follows from Hamilton's strong
maximum principle that the universal cover of (M, g (0)) contains a line and
splits as the product of a line with a 2-dimensional 11:-solution with Harnack.
It then follows from Corollary 19.43 that g (0) has bounded curvature.


Case (2). Now suppose that (M, g (0)) has positive curvature operator.
If (M, g (t)) does not have bounded curvature, then supxEM R (x, 0) = oo.


(^23) Note that since R (xi, 0) rr --+ =,the radii of the balls, that are defined with respect
to gi ( 0), limit to infinity.

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