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

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.   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 $Ri{c}_{f}\ge 0$ ; fix $p\in M$ ; if there exists c so that $|f\left(x\right)|\le c$ then for $R\ge r>0$

$\frac{Vo{l}_{f}\left(B\left(p,R\right)\right)}{Vo{l}_{f}\left(B\left(p,r\right)\right)}\le {e}^{4c}{\left(\frac{R}{r}\right)}^{n}$

Proof

Let x be a point in M and let $\gamma :\left[0,r\right]\to M$ be a minimal geodesic joining p to x and ${\left({e}_{i}\left(t\right)\right)}_{1\le i\le n-1}$ be a parallel orthonormal vector fields along $\gamma$ . Set ${Y}_{i}\left(t\right)=\frac{t}{r}{e}_{i}\left(t\right)$ .

By the second variation formula we have:

$\begin{array}{c}m\left(r\right)=\Delta r\le \sum _{i=1}^{n-1}I\left({Y}_{i},{Y}_{i}\right)\\ ={\int }_{0}^{r}\left(\sum _{i=1}^{n-1}{‖{{Y}^{\prime }}_{i}\left(t\right)‖}^{2}-〈R\left({Y}_{i}\left(t\right),{\gamma }^{\prime }\left(t\right)\right){\gamma }^{\prime }\left(t\right),{Y}_{i}\left(t\right)〉\right)\text{d}t\\ \le \frac{1}{{r}^{2}}{\int }_{0}^{r}\left(n-1-{t}^{2}Ric\left(\gamma \text{'}\left(t\right)\right)\right)dt\end{array}$

$\frac{n-1}{r}+{\int }_{0}^{r}\frac{{t}^{2}}{{r}^{2}}Hess\left(f\right)\left({\gamma }^{\prime },{\gamma }^{\prime }\right)\text{d}t$

$\begin{array}{l}\frac{n-1}{r}+{\int }_{0}^{r}\frac{{t}^{2}}{{r}^{2}}{\left(f\circ \gamma \right)}^{\prime \text{​}\prime }\text{d}t\\ =\frac{n-1}{r}+\frac{1}{{r}^{2}}{\int }_{0}^{r}\frac{\text{d}}{\text{d}t}\left({t}^{2}{\left(f\circ \gamma \right)}^{\prime }\left(t\right)\right)\text{d}t-\frac{2}{{r}^{2}}{\int }_{0}^{r}t{\left(f\circ \gamma \right)}^{\prime }\left(t\right)\text{d}t\\ =\frac{n-1}{r}+{\partial }_{r}f-\frac{2}{r}f\left(x\right)+\frac{2}{{r}^{2}}{\int }_{0}^{r}\left(f\circ \gamma \right)\left(t\right)\text{d}t\end{array}$

Hence

$\begin{array}{c}{m}_{f}\left(r\right)=\Delta r-{\partial }_{r}f=\frac{\partial }{\partial r}\left(\mathrm{ln}\left({A}_{f}\left(r,\theta \right)\right)\right)\\ \le \frac{n-1}{r}+{\partial }_{r}f-\frac{2}{r}f\left(x\right)+\frac{2}{{r}^{2}}{\int }_{0}^{r}\left(f\circ \gamma \right)\left(t\right)\text{d}t\end{array}$

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

$\begin{array}{c}{\int }_{r}^{s}{m}_{f}\left(t\right)\text{d}t=\mathrm{ln}\left(\frac{{A}_{f}\left(s,\theta \right)}{{A}_{f}\left(r,\theta \right)}\right)\le {\left(\frac{s}{r}\right)}^{n-1}+2{\int }_{r}^{s}\left(\frac{1}{{t}^{2}}{\int }_{0}^{t}f\left(u\right)\text{d}u-\frac{1}{t}f\left(t\right)\right)\text{d}t\\ =\mathrm{ln}{\left(\frac{s}{r}\right)}^{n-1}-{\left(\frac{2}{t}{\int }_{0}^{t}f\text{d}t\right)|}_{r}^{s}+2{\int }_{r}^{s}\frac{1}{t}f\text{d}t-2{\int }_{r}^{s}\frac{1}{t}f\text{d}t\\ =\mathrm{ln}{\left(\frac{s}{r}\right)}^{n-1}-{\left(\frac{2}{t}{\int }_{0}^{t}f\text{d}t\right)|}_{r}^{s}\le \mathrm{ln}{\left(\frac{s}{r}\right)}^{n-1}+4c\end{array}$

Therefore we have ${r}^{n-1}{A}_{f}\left(s,\theta \right)\le {e}^{4c}{A}_{f}\left(r,\theta \right){s}^{n-1}$ . Hence

${\int }_{0}^{R}{\int }_{{S}^{n-1}}{r}^{n-1}{A}_{f}\left(s,\theta \right)\text{d}\theta \text{d}r\le {e}^{4c}{\int }_{0}^{R}{\int }_{{S}^{n-1}}{A}_{f}\left(r,\theta \right)\text{d}\theta \text{d}r$

which implies

$\frac{{R}^{n}}{n}{\int }_{{S}^{n-1}}{A}_{f}\left(s,\theta \right)\text{d}\theta \le {e}^{4c}{s}^{n-1}{\int }_{0}^{R}{\int }_{{S}^{n-1}}{A}_{f}\left(r,\theta \right)\text{d}\theta \text{d}r={e}^{4c}{s}^{n-1}vo{l}_{f}\left(B\left(p,R\right)\right)$

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