Advances in Pure Mathematics
Vol.07 No.11(2017), Article ID:80554,2 pages
10.4236/apm.2017.711039

Erratum to “Manifolds with Bakry-Emery Ricci Curvature Bounded Below”, Advances in Pure Mathematics, Vol. 6 (2016), 754-764

Issa Allassane Kaboye1, Bazanfaré Mahaman2

1Faculté de Sciences et Techniques, Université de Zinder, Zinder, Niger

2Département de Mathématiques et Informatique, Université Abdou Moumouni, Niamey, Niger

Copyright © 2017 by authors and Scientific Research Publishing Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

Received: August 24, 2016; Accepted: October 14, 2016; Published: October 17, 2016


The original online version of this article (Issa Allassane Kaboye, Bazanfaré Mahaman (2016) Manifolds with Bakry-Emery Ricci Curvature bounded below 6, 754-764. http://dx.doi.org/10.4236/apm.2016.611061 unfortunately contains a mistake. The author wishes to correct the errors.

Lemma 3.5. Let (M, g, e−fdvolg) be a complete smooth metric measure space with R i c f 0 ; fix p M ; if there exists c so that | f ( x ) | c then for R r > 0

V o l f ( B ( p , R ) ) V o l f ( B ( p , r ) ) e 4 c ( R r ) n

Proof

Let x be a point in M and let γ : [ 0 , r ] M be a minimal geodesic joining p to x and ( e i ( t ) ) 1 i n 1 be a parallel orthonormal vector fields along γ . Set Y i ( t ) = t r e i ( t ) .

By the second variation formula we have:

m ( r ) = Δ r i = 1 n 1 I ( Y i , Y i ) = 0 r ( i = 1 n 1 Y i ( t ) 2 R ( Y i ( t ) , γ ( t ) ) γ ( t ) , Y i ( t ) ) d t 1 r 2 0 r ( n 1 t 2 R i c ( γ ' ( t ) ) ) d t

n 1 r + 0 r t 2 r 2 H e s s ( f ) ( γ , γ ) d t

n 1 r + 0 r t 2 r 2 ( f γ ) d t = n 1 r + 1 r 2 0 r d d t ( t 2 ( f γ ) ( t ) ) d t 2 r 2 0 r t ( f γ ) ( t ) d t = n 1 r + r f 2 r f ( x ) + 2 r 2 0 r ( f γ ) ( t ) d t

Hence

m f ( r ) = Δ r r f = r ( ln ( A f ( r , θ ) ) ) n 1 r + r f 2 r f ( x ) + 2 r 2 0 r ( f γ ) ( t ) d t

For all positive reals r and s, integrating this relation we have:

r s m f ( t ) d t = ln ( A f ( s , θ ) A f ( r , θ ) ) ( s r ) n 1 + 2 r s ( 1 t 2 0 t f ( u ) d u 1 t f ( t ) ) d t = ln ( s r ) n 1 ( 2 t 0 t f d t ) | r s + 2 r s 1 t f d t 2 r s 1 t f d t = ln ( s r ) n 1 ( 2 t 0 t f d t ) | r s ln ( s r ) n 1 + 4 c

Therefore we have r n 1 A f ( s , θ ) e 4 c A f ( r , θ ) s n 1 . Hence

0 R S n 1 r n 1 A f ( s , θ ) d θ d r e 4 c 0 R S n 1 A f ( r , θ ) d θ d r

which implies

R n n S n 1 A f ( s , θ ) d θ e 4 c s n 1 0 R S n 1 A f ( r , θ ) d θ d r = e 4 c s n 1 v o l f ( B ( p , R ) )

and integrating from 0 to R' with respect to s we obtain the conclusion.