1547845447-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_IV__Chow_

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8 27. NONCOMPACT GRADIENT RICCI SOLITONS


Combining Theorem 27.4 with Lemma 27.6 yields the following result of Z.-H.
Zhang.


COROLLARY 27 .7 (Completeness of g implies completeness of V' f). If (Mn, g ,
f , t:) is a complete GRS, then V' f is a complete vector field.

1.4. Equality case of the scalar curvature lower bound.


We now return to Theorem 27.2 and consider the equ ality case. We have the
following result of Pigola, Rimoldi, and Setti.


PROPOSITION 27.8. Let (Mn, g , f ,E) be a complete GRS.
(1) If E = 1 and if there exists xo EM such that R(xo) = -~, then Re=
1.

-29·


(2) If E = 0 and if there exists x o E M such that R (xo) = 0, then Re= 0.


(3) If E = -1 and if there exists xo E M such that R (xo) = 0, then
(M,g,f,-l) is a Gaussian so liton, so that (M,g) is isometric to Eu-
clidean space.

PROOF. The idea is to apply the strong maximum principle. By Corollary

27 .7 and by Theorem 4.1 in [77], we may extend the GRS structure (M, g, f , c)


canonically in time so that g (t) is a complete solution of the Ricci flow with g (0) = g


and such that g ( t) and f ( t) satisfy


2 E

Rcg(t) +V'g(t)f (t) + 2 (ct+ l)g (t) = 0.


Recall the standard equation


(27.35)

8R 2


8t = D.R + 2 IRcl.


We thus have that R ~ R + 2(c~~l) satisfies


(27.36) f)R - I R^1


2


  • = D.R + 2 Re --g + -^2 ( -R - --nt: ) R , -
    ot n n ct+ 1


where we used


(^2) nt:^2 I R 1
2
21Rcl - = 2 Rc--g + -^2 (-R---nt: ) R. -


2 (ct + 1)^2 n n Et + 1


Case i: E = 0 or l. By Theorem 27.2 we have


R+ nt: :'.::: 0.


2 (ct+ 1)


Since there exists xo E M su ch that R (x 0 , 0) = 0, by applying the parabolic strong
maximum principle (which is a local result) to (27.36), we have R (x, t) = 0 for
x EM and t :S: 0 such that ct+ 1 > 0. Thus, by (27.36), we have Re= ~g = -~g.
This proves parts (1) and (2).


Case ii: E = -1. By Theorem 27.2 we have R :'.::: 0. Applying the parabolic


strong maximum principle to (27.35), we have R = 0, which in turn implies Re= 0.


Since g is a shrinker , we then obtain


(27.37)

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