- VOLUME GROWTH OF SHRINKING GRADIENT RICCI SOLITONS 21
Since the !-Ricci tensor is nonnegative, i.e., Re / = Re+ \7^2 f = ~ g 2 0, the Bakry-
Emery volume comparison theorem (see (27.104) below) implies that
(27.88) Vol1 Ba(an) = r e-f dμ:::; Wne^3.^4 nana~.
}Bo(an)
Finally,
Vol Bo(an):::; emaxao(an) f Vol/ Bo(an) :::; e!Can+V2ri")
2
Wne^3 ·^4 nana~ ~ C(n)
since f:::; t(an + ffn)^2 in Bo(an)· D
PROBLEM 27.34. For shrinkers, must we have AVR (g) :::; 1?
4.3. Characterizing when AVR > 0 for noncompact shrinkers.
Let V(e) and R(e) be defined as in (27.72). For shrinkers, we have the following
characterization for AVR > 0. This is due to B. Yang and two of the authors, based
on earlier works of others.
PROPOSITION 27 .35 (Necessary and sufficient condition for AVR > 0).
(Mn, g , f , -1) be a complete normalized noncompact shrinking GRS. Then
(27.89)
1
(^00) R(e)
AVR(g) > 0 if and only if -V( ) de< oo,
n+2 e e
or, equivalently,
(27.90)
J
oo 1Bo(r) Rdμ dr
-"------- < 00
1 Vol B 0 (r) r ·
That is, AVR (g) = 0 if and only if 1; 2 er;J(l) de = oo, or, equivalently,
J
oo 1B -(r) Rdμ dr
1 V~lB 0 (r) r = oo.
PROOF. Integrating (27.83) yields
(27.91)
(1-n +^2 ) N(c )
P(e)=P(n+2)e-Jn+2 c 1-N(2C c ) de
Let
fore > n + 2. Regarding the integral on the RHS, from (27.79) it is easy to see that
for any e E [n + 2, oo) we have
(27.92) -^11 e N (e) de:::; 1e ( 1 - --n+2) N(e) N ( ) de:::; 2 1e N (e) de.
2 n+2 n+2 2e 1 - e n+2
If 1: 2 N (e) de= oo, then by (27.91) we have AVR(g) = 2 n~n lime-too P(e) = 0.
On the other hand, if 1: 2 N (e) de < oo, then it follows from (27.91) and
(27.92) that
P(e) 2". P(n+ 2) e-^2 f~2N(e)de > 0
for e 2 n + 2; hence AVR(g) > 0. We have shown that AVR (g) = 0 if and only if
1:2 er;J(l) de =^00 ·
Finally, we observe that by inequalities (27.42) and (27.46) we h ave that
roo R(e)d 'f d 1 'f Joo f B a(r) Rdμ, dr D
Jn+2 CV(C} e = oo 1 an on y 1 1 Vol Bo(r) r = oo.
Using Proposition 27.35, one may obtain the following qualitatively sharp result
due to S.-J. Zhang.