1547845439-The_Ricci_Flow_-_Techniques_and_Applications_-_Part_I__Chow_

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  1. REDUCED VOLUME FOR RICCI FLOW 389


See Lemma 8.16(ii) below for why V (7) is well defined even when Mis
noncom pact.

In the case of a Ricci flat solution g (7) = g, we have


V (T) = f)4orr)-nfZ exp {-d~~q)'} dμg.


(Compare with (8.1).)
We now give heuristic (e.g., unjustified) proofs of the reduced volume
monotonicity. Provided we can differentiate (8.16) under the integral sign,
we obtain


dV (7) = .!!:___ ( { (47r7)-n/^2 e-£(.,T)dμ ( 7 ))
d7 d7 JM g

(8.17) ~JM :7 ( (47r7)-n/2 e-e(.,T)dμg(T))


1 (


= --n - -8£ + R ) ( 41T7 )-n/2 e -£ dμ,


M 27 fh

where ~ denotes an unjustified inequality.^2 By applying inequalities (7.91)
and (7.148), which results in (7.150), we obtain


~~ (7):::; JM (1\7£12 - .6.e) (47r7)-n/2 e-edμ:::; 0.

Recall that for any vector field X, we have .Cxdμ =div (X) dμ. In partic-

ular, £\Jhdμ = .6.h dμ for any C^2 function h. Define the first-order differential
operator acting on time-dependent tensors and forms:


D 8
d7 ~ f)T +£x.

EXERCISE 8.15. Show that for any C^1 -vector field X and C^1 func-
tion ¢, one of them with compact support, under the backward Ricci flow
(Mn,g (7)), we have


d~ JM <f>dμ =JM~ (<f>dμ)

=JM (:7</>+X·\7</>+</>R+</>div(x))dμ.


SOLUTION TO EXERCISE 8.15. The result follows from fTdμ = Rdμ
and the integration by parts identity


JM (X · \7¢ +¢div (X)) dμ = 0.

Now we consider again the time-derivative of the reduced volume of
(Mn, g ( 7)) under the backward Ricci flow and we heuristically discuss some

(^2) For the justification, see the proof of Theorem 8.20 below.

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