Rigidity for the isoperimetric inequality of negative effective dimension on weighted Riemannian manifolds

Abstract

We study, on a weighted Riemannian manifold of $\hbox {Ric}_{N} \ge K > 0$ for $N < -1$, when equality holds in the isoperimetric inequality. Our main theorem asserts that such a manifold is necessarily isometric to the warped product ${\mathbb {R}} \times _{\cosh (\sqrt{K/(1-N)}t)} \Sigma ^{n-1}$ of hyperbolic nature, where $\Sigma ^{n-1}$ is an $(n-1)$-dimensional manifold with lower weighted Ricci curvature bound and ${\mathbb {R}}$ is equipped with a hyperbolic cosine measure. This is a similar phenomenon to the equality condition of Poincaré inequality. Moreover, every isoperimetric minimizer set is isometric to a half-space in an appropriate sense.

Appendix A

In the appendix, we will prove that $I_{K,N,D}(\theta ) > I_{K,N,\infty }(\theta )$ for all $\theta \in (0,1)$, $N < 0$ and $D < \infty$. We abbreviate in this appendix $I(\theta ) := I_{K,N,\infty }(\theta )$.

Lemma A.1

$K_{3,D}(\theta ) > I_{K,N,\infty }(\theta )$ for all $\theta \in (0,1)$, $N < 0$ and $D < \infty$.

Proof

By the definitions of $c(\theta )$,$d_{3}(\theta )$ in $\S$2.2, we have $1 = I(\theta )c'(\theta ) = K_{3,D}(\theta )d_{3}'(\theta )$. Therefore

$$\begin{aligned} I'(\theta )= & {} (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }c(\theta )),\\ K_{3,D}'(\theta )= & {} (N-1)\sqrt{\sigma }. \end{aligned}$$

Put $h(\theta ) = K_{3,D}(\theta )-I(\theta )$. Since $|\tanh x| < 1$, we have $h'(\theta ) < 0$. Hence $h(\theta ) \ge h(1) = \frac{e^{(N-1)\sqrt{\sigma }D}}{\int _{0}^{D} e^{(N-1)\sqrt{\sigma }s}ds} > 0$. $\square$

Lemma A.2

$K_{2,D}(\theta ) > I_{K,N,\infty }(\theta )$ for all $\theta \in (0,1)$, $N < 0$ and $D < \infty$.

Proof

Put $\varphi (t) := \sinh ^{N-1}(\sqrt{\sigma }t)$. For $t>0$, we have

$$\begin{aligned} \varphi '(t) = (N-1)\sqrt{\sigma }\varphi (t)\coth (\sqrt{\sigma }t) < 0. \end{aligned}$$

Fix $\xi >0$ and set $g_{\xi }(\theta ) := \frac{\varphi (d_{2,\xi }(\theta ))}{\int _{\xi }^{\xi + D} \varphi (s)ds} > 0$. By the definitions of $c(\theta )$, $d_{2,\xi }(\theta )$, we have $1 = I(\theta )c'(\theta ) = g_{\xi }(\theta )d_{2,\xi }'(\theta )$. Therefore

$$\begin{aligned} I'(\theta )= & {} (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }c(\theta )),\\ g_{\xi }'(\theta )= & {} (N-1)\sqrt{\sigma }\coth (\sqrt{\sigma }d_{2,\xi }(\theta )). \end{aligned}$$

Put $h_{\xi }(\theta ) = g_{\xi }(\theta )-I(\theta )$. Since $|\tanh x| < 1$ and $\coth y > 1$ for $y>0$, $h_{\xi }(\theta ) \ge h_{\xi }(1)$. Note that $I(1) = 0$, hence $h_{\xi }(\theta ) > 0$ for each $\xi > 0$. Fix $\theta$ and put $m(\xi ) = \frac{\int _{\xi }^{\xi + D} \varphi (s)ds}{\varphi (\xi +D)} > 0$. Denote $\cosh (\sqrt{\sigma } D)$, $\sinh (\sqrt{\sigma } D)$ by $c_{D}$ and $s_{D}$. We have

$$\begin{aligned} \lim _{\xi \rightarrow \infty } m(\xi )&= \lim _{\xi \rightarrow \infty } \frac{\varphi (\xi +D)-\varphi (\xi )}{\varphi '(\xi +D)} = \lim _{\xi \rightarrow \infty }\frac{\varphi (\xi +D)}{\varphi '(\xi +D)} - \lim _{\xi \rightarrow \infty } \frac{\varphi (\xi )}{\varphi (\xi +D)} \frac{\varphi (\xi +D)}{\varphi '(\xi +D)}\\&= \frac{1}{(N-1)\sqrt{\sigma }}-\frac{1}{(N-1)\sqrt{\sigma }}\lim _{\xi \rightarrow \infty } \frac{\varphi (\xi )}{\varphi (\xi +D)}\\&= \frac{1}{(N-1)\sqrt{\sigma }} - \frac{1}{(N-1)\sqrt{\sigma }}\lim _{\xi \rightarrow \infty } \big ( c_{D} + s_{D}\coth (\sqrt{\sigma }\xi )\big )^{1-N}\\&= \frac{1}{(N-1)\sqrt{\sigma }} \big (1- (c_{D}+s_{D})^{1-N}\big ). \end{aligned}$$

Note that $\varphi (d_{2,\xi }(\theta )) \ge \varphi (\xi + D)$, hence

$$\begin{aligned} \lim _{\xi \rightarrow \infty } h_{\xi }(\theta ) \ge \lim _{\xi \rightarrow \infty } h_{\xi }(1) \ge \lim _{\xi \rightarrow \infty } (m(\xi ))^{-1} > 0. \end{aligned}$$

Now we consider the other direction $\lim _{\xi \rightarrow 0} h_{\xi }(\theta )$. We also define ${\bar{d}}(\xi )$ as

$$\begin{aligned} \theta = \frac{\int _{\xi }^{{\bar{d}}(\xi )} (\sqrt{\sigma }s)^{N-1}ds}{\int _{\xi }^{\xi +D} (\sqrt{\sigma }s)^{N-1}ds} = \frac{{\bar{d}}^{N}_{\xi }-\xi ^{N}}{(\xi +D)^N-\xi ^{N}}. \end{aligned}$$

Put $f(\Delta ) := \frac{\int _{\xi }^{\xi +\Delta } \varphi (s)ds}{\int _{\xi }^{\xi +\Delta } (\sqrt{\sigma }s)^{N-1}ds} \ge 0$. We have

$$\begin{aligned} f'(\Delta )&= f(\Delta )\bigg ( \frac{\varphi (\xi +\Delta )}{\int _{\xi }^{\xi +\Delta } \varphi (s)ds} - \frac{(\sqrt{\sigma }(\xi +\Delta ))^{N-1}}{\int _{\xi }^{\xi +\Delta } (\sqrt{\sigma }s)^{N-1}ds}\bigg )\\&= \frac{f(\Delta )(\sqrt{\sigma }(\xi +\Delta ))^{N-1}}{\int _{\xi }^{\xi +\Delta } \varphi (s)ds}\bigg ( \frac{\varphi (\xi +\Delta )}{(\sqrt{\sigma }(\xi +\Delta ))^{N-1}} - \frac{\int _{\xi }^{\xi +\Delta } \varphi (s)ds}{\int _{\xi }^{\xi +\Delta } (\sqrt{\sigma }s)^{N-1}ds} \bigg ). \end{aligned}$$

By Cauchy’s mean value theorem, there exists $c\in (\xi ,\xi +\Delta )$ such that $\frac{\int _{\xi }^{\xi +\Delta } \varphi (s)ds}{\int _{\xi }^{\xi +\Delta } (\sqrt{\sigma }s)^{N-1}ds} = \frac{\varphi (c)}{(\sqrt{\sigma }c)^{N-1}}$. Note that $\frac{\sinh x}{x}$ is a monotone increasing function when $x > 0$, we have $\frac{\varphi (x)}{(\sqrt{\sigma }x)^{N-1}}$ is monotone decreasing. Hence $f'(\Delta ) < 0$ and f is monotone decreasing. Therefore

$$\begin{aligned} \frac{\int _{\xi }^{{\bar{d}}(\xi )} (\sqrt{\sigma }s)^{N-1}ds}{\int _{\xi }^{d_{2,\xi }(\theta )} \varphi (s)ds} = \frac{\int _{\xi }^{\xi + D} (\sqrt{\sigma }s)^{N-1}ds}{\int _{\xi }^{\xi +D} \varphi (s)ds} \ge \frac{\int _{\xi }^{d_{2,\xi }(\theta )} (\sqrt{\sigma }s)^{N-1}ds}{\int _{\xi }^{d_{2,\xi }(\theta )} \varphi (s)ds}. \end{aligned}$$

Thus we can deduce that $d_{2,\xi }(\theta ) \le {\bar{d}}(\xi ) = \big ( \theta (\xi +D)^{N} + (1-\theta )\xi ^{N}\big )^{1/N}$. Hence

$$\begin{aligned} \lim _{\xi \rightarrow 0_{+}} h_{\xi }(\theta )&\ge \lim _{\xi \rightarrow 0_{+}} \frac{\varphi ({\bar{d}}(\xi ))}{\int _{\xi }^{\xi + D} \varphi (s)ds} - I(\theta ) \\ {}&= \lim _{\xi \rightarrow 0_{+}} \frac{(N-1)\sqrt{\sigma }\varphi ({\bar{d}}(\xi ))\coth ({\sqrt{\sigma }} {\bar{d}}(\xi )){\bar{d}}'(\xi )}{\varphi (\xi +D)-\varphi (\xi )} - I(\theta )\\&= \lim _{\xi \rightarrow 0_{+}} \frac{(N-1)\sqrt{\sigma }^{N}{\bar{d}}^{N-1}(\xi ){\bar{d}}'(\xi )}{\sinh ({\sqrt{\sigma }}{\bar{d}}(\xi ))\big (\varphi (D)-\varphi (\xi )\big )} - I(\theta )\\&= \lim _{\xi \rightarrow 0_{+}} \frac{1}{\sinh ({\sqrt{\sigma }}{\bar{d}}(\xi ))}\cdot \frac{(N-1)\sqrt{\sigma }^{N}\big (\theta (\xi +D)^{N-1} + (1-\theta )\xi ^{N-1}\big )}{\varphi (D)-\varphi (\xi )} - I(\theta )\\ {}&= \lim _{\xi \rightarrow 0_{+}} \frac{1}{\sinh ({\sqrt{\sigma }}{\bar{d}}(\xi ))}\cdot \frac{(1-N) \sqrt{\sigma }\big (\theta (\xi +D)^{N-1} + (1-\theta )\xi ^{N- 1}\big )}{\xi ^{N-1}}- I(\theta )\\&= \infty . \end{aligned}$$

Therefore $K_{2,D}(\theta ) - I(\theta ) = \inf _{\xi>0} h_{\xi }(\theta ) > 0$. $\square$

Lemma A.3

$K_{1,D}(\theta ) > I_{K,N,\infty }(\theta )$ for all $\theta \in (0,1)$, $N < 0$ and $D < \infty$.

Proof

Put $\varphi (t) := \cosh ^{N-1}(\sqrt{\sigma }t)$. We have

$$\begin{aligned} \varphi '(t) = (N-1)\sqrt{\sigma }\varphi (t)\tanh (\sqrt{\sigma }t). \end{aligned}$$

Put $g_{\xi }(\theta ) := \frac{\varphi (d_{1,\xi }(\theta ))}{\int _{\xi }^{\xi + D} \varphi (s)ds}$. By the definitions of $c(\theta )$,$d_{1.\xi }(\theta )$ in $\S$2.2, we have $1 = I(\theta )c'(\theta ) = g_{\xi }(\theta )d_{1,\xi }'(\theta )$. Therefore

$$\begin{aligned} I'(\theta )= & {} (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }c(\theta )),\\ g_{\xi }'(\theta )= & {} (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }d_{1,\xi }(\theta )). \end{aligned}$$

Put $h_{\xi }(\theta ) = g_{\xi }(\theta )-I(\theta )$. Since $\tanh (x)$ is monotone increasing we have $h_{\xi }'(\theta _{0}) = 0$ if and only if $c(\theta _{0}) = d_{1,\xi }(\theta _{0})$. Since $I(0) = I(1) = 0$, we can deduce $h_{\xi }(\theta ) \ge \min \{g_{\xi }(0),g_{\xi }(1), h_{\xi }(\theta _{0})$ where $c(\theta _{0}) = d_{1,\xi }(\theta _{0})\}$.

Put $m_{0}(\xi ) := (g_{\xi }(0))^{-1} = \frac{\int _{\xi }^{\xi + D} \varphi (s)ds}{\varphi (\xi )}$. We have

$$\begin{aligned} m'_{0}(\xi )&= \frac{\varphi (\xi +D)-\varphi (\xi )}{\varphi (\xi )} - m_{0}(\xi )\frac{\varphi '(\xi )}{\varphi (\xi )} \\&= m_{0}(\xi ) \bigg ( \frac{\varphi (\xi +D)-\varphi (\xi )}{\int _{\xi }^{\xi + D} \varphi (s)ds} - (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }\xi )\bigg ). \end{aligned}$$

By Cauchy’s mean value theorem, there exists $c\in (\xi ,\xi +D)$ such that $\frac{\varphi (\xi +D)-\varphi (\xi )}{\int _{\xi }^{\xi + D} \varphi (s)ds} = \frac{\varphi '(c)}{\varphi (c)} = (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }c)$. Note that $\tanh x$ is a monotone increasing function, we have $m_{0}(\xi )$ is a monotone decreasing function. Hence $g_{\xi }(0) \ge \lim _{\xi \rightarrow -\infty } \frac{\varphi (\xi )}{\int _{\xi }^{\xi + D} \varphi (s)ds}$. Denote $\cosh (\sqrt{\sigma } D)$, $\sinh (\sqrt{\sigma } D)$ by $c_{D}$ and $s_{D}$. We have

$$\begin{aligned} \lim _{\xi \rightarrow -\infty } m_{0}(\xi )&= \lim _{\xi \rightarrow -\infty } \frac{\varphi (\xi +D)-\varphi (\xi )}{\varphi '(\xi )} = \lim _{\xi \rightarrow -\infty } \frac{\varphi (\xi + D)}{\varphi (\xi )} \frac{\varphi (\xi )}{\varphi '(\xi )} - \lim _{\xi \rightarrow -\infty }\frac{\varphi (\xi )}{\varphi '(\xi )}\\&= \frac{-1}{(N-1)\sqrt{\sigma }}\lim _{\xi \rightarrow -\infty } \frac{\varphi (\xi + D)}{\varphi (\xi )} - \frac{-1}{(N-1)\sqrt{\sigma }}\\&= \frac{-1}{(N-1)\sqrt{\sigma }}\lim _{\xi \rightarrow -\infty } \big ( c_{D} + s_{D}\tanh (\sqrt{\sigma }\xi )\big )^{N-1} - \frac{-1}{(N-1)\sqrt{\sigma }}\\&= \frac{-1}{(N-1)\sqrt{\sigma }} \big ( (c_{D}-s_{D})^{N-1} - 1 \big ). \end{aligned}$$

Hence $g_{\xi }(0) \ge \displaystyle \frac{(1-N)(\sqrt{\sigma })}{( (c_{D}-s_{D})^{N-1} - 1 )} > 0$.

On the other hand, put $m_{1}(\xi ) := (g(1))^{-1} = \frac{\int _{\xi }^{\xi + D} \varphi (s)ds}{\varphi (\xi +D)}$. We have

$$\begin{aligned} m'_{1}(\xi )&= \frac{\varphi (\xi +D)-\varphi (\xi )}{\varphi (\xi +D)} - m_{1}(\xi )\frac{\varphi '(\xi +D)}{\varphi (\xi +D)} \\&= m_{1}(\xi ) \bigg ( \frac{\varphi (\xi +D)-\varphi (\xi )}{\int _{\xi }^{\xi + D} \varphi (s)ds} - (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }(\xi +D)\bigg ). \end{aligned}$$

By Cauchy’s mean value theorem, there exists $c\in (\xi ,\xi +D)$ such that $\frac{\varphi (\xi +D)-\varphi (\xi )}{\int _{\xi }^{\xi + D} \varphi (s)ds} = \frac{\varphi '(c)}{\varphi (c)} = (N-1)\sqrt{\sigma }\tanh (\sqrt{\sigma }c)$. Note that $\tanh x$ is a monotone increasing function, we have $m_{1}(\xi )$ is a monotone increasing function. Hence $g_{\xi }(1) \ge \lim _{\xi \rightarrow \infty } \frac{\int _{\xi }^{\xi + D} \varphi (s)ds}{\varphi (\xi +D)}.$ We have

$$\begin{aligned} \lim _{\xi \rightarrow \infty } m_{1}(\xi )&= \lim _{\xi \rightarrow \infty } \frac{\varphi (\xi +D)-\varphi (\xi )}{\varphi '(\xi +D)} = \lim _{\xi \rightarrow \infty }\frac{\varphi (\xi +D)}{\varphi '(\xi +D)} - \lim _{\xi \rightarrow \infty } \frac{\varphi (\xi )}{\varphi (\xi +D)} \frac{\varphi (\xi +D)}{\varphi '(\xi +D)}\\&=\frac{1}{(N-1)\sqrt{\sigma }}-\frac{1}{(N-1)\sqrt{\sigma }}\lim _{\xi \rightarrow \infty } \frac{\varphi (\xi )}{\varphi (\xi +D)}\\&= \frac{1}{(N-1)\sqrt{\sigma }} - \frac{1}{(N-1)\sqrt{\sigma }}\lim _{\xi \rightarrow \infty } \big ( c_{D} + s_{D}\tanh (\sqrt{\sigma }\xi )\big )^{1-N}\\&= \frac{1}{(N-1)\sqrt{\sigma }} \big (1- (c_{D}+s_{D})^{1-N}\big ). \end{aligned}$$

Hence $g_{\xi }(1) \ge \displaystyle \frac{(N-1)(\sqrt{\sigma })}{( 1-(c_{D}+s_{D})^{1-N} )} > 0$.

Now we consider the case where there exists $\theta _{0}\in (0,1)$ such that $c(\theta _{0}) = d_{1,\xi }(\theta _{0})\in (\xi ,\xi +D)$. Then

$$\begin{aligned} h_{\xi }(\theta _{0}) = \varphi (c(\theta _{0}))\bigg ( \frac{1}{\int _{\xi }^{\xi +D}\varphi (s)ds} - \frac{1}{\int _{-\infty }^{\infty }\varphi (s)ds}\bigg ) > 0. \end{aligned}$$

Note that $\varphi (c(\theta )) \ge \min \{ \varphi (\xi ),\varphi (\xi +D)\}$. By a similar calculation as above, we have

$$\begin{aligned} \lim _{\xi \rightarrow -\infty } \varphi (\xi +D)\bigg ( \frac{1}{\int _{\xi }^{\xi +D}\varphi (s)ds} - \frac{1}{\int _{-\infty }^{\infty }\varphi (s)ds}\bigg )&= \lim _{\xi \rightarrow -\infty } \varphi (\xi )\bigg ( \frac{1}{\int _{\xi }^{\xi +D}\varphi (s)ds} - \frac{1}{\int _{-\infty }^{\infty }\varphi (s)ds}\bigg ) \\ {}&= \frac{(1-N)(\sqrt{\sigma })}{( (c_{D}-s_{D})^{N-1} - 1 )} > 0 \end{aligned}$$

and

$$\begin{aligned} \lim _{\xi \rightarrow \infty } \varphi (\xi )\bigg ( \frac{1}{\int _{\xi }^{\xi +D}\varphi (s)ds} - \frac{1}{\int _{-\infty }^{\infty }\varphi (s)ds}\bigg )&= \lim _{\xi \rightarrow \infty } \varphi (\xi +D)\bigg ( \frac{1}{\int _{\xi }^{\xi +D}\varphi (s)ds} - \frac{1}{\int _{-\infty }^{\infty }\varphi (s)ds}\bigg ) \\ {}&= \frac{(N-1)(\sqrt{\sigma })}{( 1-(c_{D}+s_{D})^{1-N} )} > 0. \end{aligned}$$

Taking the infimum over $\xi \in (-\infty ,\infty )$ shows

$$\begin{aligned} K_{1,D}(\theta ) - I_{K,N,\infty }(\theta ) \ge \inf _{\xi \in {\mathbb {R}}} \{ g_{\xi }(0), g_{\xi }(1), h_{\xi }(\theta _{0}) \text { with } c(\theta _{0}) = d_{1,\xi }(\theta _{0}) \}> 0. \end{aligned}$$

$\square$

Proposition A.4

$I_{K,N,D}(\theta ) > I_{K,N,\infty }(\theta )$ for all $\theta \in (0,1)$, $N < 0$ and $D < \infty$.

Proof

This is just a corollary of the lemmas above. $\square$