Prove $\int_{\Sigma_r} |\nabla\varphi|^2 d\sigma_r \ge \frac{2}{u(r)^2} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r $

Question

Reference: this paper

Given the deSitter-Schwarzschild metric with mass $m > 0$ and scalar curvature equal to $2$ is the metric

$$\bigg( 1 -\frac{r^2}{3}-\frac{2m}{r} \bigg)^{-1} dr^2 + r^2 dg_{\mathbb S^2} \tag{1}$$

defined on $(a_0 ,b_0) \times \mathbb S^2$, where $(a_0,b_0)=\{r>0:1 -\frac{r^2}{3}-\frac{2m}{r}>0 \}$ and $g_{\mathbb S^2}$ is the standard metric on $\mathbb S^2$ with constant Gauss curvature equal to $1$.

In order to deal with the metric in $(1)$, we use the warped product metric $g=dr^2 + u(r)^2 dg_{\mathbb S^2}$ on $\mathbb R \times \mathbb S^2$, where $u(r)$ is a positive real function. If we assume that $g$ has constant scalar curvature equal to $2$, then $u$ solves the following second-order differential equation

$$u''(r)=\frac{1}{2}\bigg( \frac{1-u'(r)^2}{u(r)} - \frac{u(r)}{2} \bigg) \tag{2}$$

Considering only positive solutions $u(r)$ to $(2)$ which are defined for all $r \in \mathbb R$, we get a one-parameter family of periodic rotationally symmetric metrics $g_a = dr^2 + u_a(r)^2 g_{\mathbb S^2}$ with constant scalar curvature equal to $2$, where $a \in (0,1)$ and $u_a(r)$ satisfies $u_a(0)=a= \text{min} ~u$ and $u'_a(0) =0$. These metrics are precisely the deSitter-Schwarzschild metrics on $\mathbb R \times \mathbb S^2$ defined in $(1)$.

Next, let $(M,g)$ be a three-manifold and consider a two-sided compact surface $\Sigma \subset M$.

The mass of $\Sigma \subset M$ is defined by

$$m(\Sigma) =\ \bigg( \frac{|\Sigma|}{16\pi} \bigg)^{1/2} \bigg( 1 - \frac{1}{ 16\pi } \int_{\Sigma} H^2 d\sigma - \frac{ \Lambda}{24\pi } |\Sigma| \bigg) \tag{3}$$

where $\Lambda = \text{inf}_M ~ R$, $R$ is the scalar curvature of $M$, $H$ is the mean curvature of $\Sigma$, $K_\Sigma$ is the Gauss curvature of $\Sigma$, and $|\Sigma| = \int_\Sigma d\sigma$.

The first variation of $m$:

$$\frac{d}{dt}m(\Sigma(t))\bigg|_{t=0} = - \frac{2|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \varphi \Delta_\Sigma H d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_{\Sigma} \bigg[ 2K_\Sigma - \frac{8\pi}{|\Sigma| }+ \bigg( \frac{1}{2|\Sigma|}\int_\Sigma H^2 d\sigma - |A|^2 \bigg) \bigg] H \phi d\sigma \\ + \frac{|\Sigma|^{1/2}}{(16\pi)^{3/2}} \int_\Sigma (\Lambda - R) H\varphi d\sigma \tag{4}$$

Remark $1$. It follows from $(4)$ that if a two-sphere $\Sigma \subset M$ is umbilic where $|A|^2 = \frac{H^2}{2}$ and has constant Gauss curvature and $M$ has constant scalar curvature equal to $2$ along $\Sigma$, then $\Sigma$ is a critical point of the mass in $(3)$.

Denote by $\Sigma_r$ the slice $\{r\} \times \mathbb S^2$. By Remark $1$, $\Sigma_r$ is a critical point for the mass in $(\mathbb R \times \mathbb S^2 , g_a)$, for all $r \in \mathbb R$ and $a \in (0,1)$. Moreover the mass of $\Sigma_r \subset (\mathbb R \times \mathbb S^2 , g_a)$ is constant for all $r \in \mathbb R$. It follows by a straightforward computation:

$$\frac{d}{dr} m(\Sigma_r) = \frac{1}{2} u'(r)(1-u'(r)-u(r)^2-2u(r)u''(r)) \tag{5}$$

which is zero once $u(r)$ solves $(2)$, we obtain therefore that $m(\Sigma_r)$ is constant equal to $m(\Sigma_0)$.

Now, since $g_{\Sigma_r} = u(r)^2 g_{\mathbb S^2}$, by the Poincare inequality we have $$ \int_{\Sigma_r} |\nabla\varphi|^2 d\sigma_r \ge \frac{2}{u(r)^2} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r = \frac{8\pi}{|\Sigma_r|}\int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r \tag{6}$$

where $\varphi \in C^\infty(\Sigma_r)$.

Question:

Where does $(6)$ come from?

Thank you.

Answer 1

Although I am not really sure with my calculation, but here is my progress.

Any suggestions or corrections are very welcome.

Given the conformal metric:

\begin{align} g|\Sigma_r = u(r)^2 g_{\Bbb S^2} ~, \tag{7} \end{align}

where $u(r)>0$ is some positive function and solves the second-order differential equation below

\begin{align} - 6 \Delta u + R u = \tilde{R}_{g|\Sigma_r} u^3 \\ \int_{\Sigma_r} |\nabla u|^2 d\sigma_r + \int_{\Sigma_r} \frac{1}{6}2 u^2 d\sigma_r = \int_{\Sigma_r}\frac{1}{6}\frac{8\pi}{|\Sigma|} u^4 d\sigma_r \tag{8} \end{align}

where $R =2 >0$ is the scalar curvature of the metric $g_{S^n}$ and $\tilde{R}_{g|\Sigma_r} = 2K_{\Sigma_r} = 2 . \frac{4\pi}{|\Sigma|} = \frac{8\pi}{|\Sigma|} \in \mathbb R$ is the scalar curvature of $g|\Sigma_r$.

The Yamabe invariant of a compact manifold $(M^3,g)$ :

\begin{align} Y(g) = \text{inf }\frac{\int_{\Sigma_r} \left( |\nabla u|^2 + \frac{1}{6}R u^2 \right) d\sigma_r} { \bigg( \int_{\Sigma_r} |u|^4 d\sigma_r \bigg)^{1/2} } ~ , \tag{9} \end{align}

where $Y(g)$ is the Yamabe constant.

From $(9)$ and $(8)$, it shows that

\begin{align} Y(g) \bigg( \int_{\Sigma_r} |u|^4 d\sigma_r \bigg)^{1/2} = \frac{1}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} u^4 d\sigma_r ~ . \tag{10} \end{align}

Substitute (10) to (9) we get

\begin{align} \int_{\Sigma_r} |\nabla u|^2 d\sigma_r \ge\ & \frac{u^2}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} u^2 d\sigma_r \nonumber\\ % % & - \frac{u^2}{6} 2{u^2} \int_{\Sigma_r} u^2 d\sigma_r ~ . \tag{11} \end{align}

$\int_{\Sigma_r} u^2 d\sigma_r$ could also be written as

\begin{align} \int_{\Sigma_r} u^2 d\sigma_r =\ & u \int_{\Sigma_r} u\ d\sigma_r \nonumber\\ %% % =\ & \frac{1}{|\Sigma|} u |\Sigma| \int_{\Sigma_r} u\ d\sigma_r \nonumber\\ % % =\ & \frac{1}{|\Sigma|} u \int_{\Sigma_r} d\sigma_r \int_{\Sigma_r} u\ d\sigma_r \nonumber\\ % % =\ & \frac{1}{|\Sigma|} \int_{\Sigma_r} u\ d\sigma_r \int_{\Sigma_r} u\ d\sigma_r \nonumber\\ % % =\ & \frac{1}{|\Sigma|} \bigg( \int_{\Sigma_r} u\ d\sigma_r \bigg)^2 ~ , \tag{12} \end{align}

Plugging $(12)$ to the second term on RHS of the inequality $(11)$ we get

\begin{align} \int_{\Sigma_r} |\nabla u|^2 d\sigma_r \ge\ & \frac{u^2}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} u^2 d\sigma_r \nonumber\\ % % & - \frac{u^{2}}{6} \frac{2}{u^2} \frac{1}{|\Sigma|} \bigg( \int_{\Sigma_r} u\ d\sigma_r \bigg)^2 ~ . \tag{13} \end{align}

Since $\frac{8\pi}{|\Sigma|} = \frac{2}{u^2}$ and $\int_{\Sigma_r} u^2 d\sigma_r - \frac{1}{|\Sigma|} \bigg( \int_{\Sigma_r} u\ d\sigma_r \bigg)^2 = \int_{\Sigma_r} (u - \bar{u})^2 d\sigma_r$, we could write $(13)$ as

\begin{align} \int_{\Sigma_r} |\nabla u|^2 d\sigma_r \ge\ & \frac{u^2}{6} \frac{2}{u^2} \int_{\Sigma_r} (u - \bar{u})^2 d\sigma_r \\ = & \frac{u^2}{6} \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} (u - \bar{u})^2 d\sigma_r \tag{14} ~ , \end{align}

which is kind of similar to $(6)$.

My questions now are:

1. Where does $\frac{u^2}{6}$ go?

2. Despite of the vanishing $\frac{u^2}{6}$ in $(6)$, if we use notation $u = \varphi$, why isn't it just written as

\begin{align} \int_{\Sigma_r} |\nabla \varphi|^2 d\sigma_r \ge\ & \frac{2}{\varphi^2} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r \\ = & \frac{8\pi}{|\Sigma|} \int_{\Sigma_r} (\varphi - \bar{\varphi})^2 d\sigma_r \tag{15} \end{align}

Why $\frac{2}{u^2}$ and not $\frac{2}{\varphi^2}$ ?

3. Is it legal if I take $\bar{\varphi}=0$ and why?

Thank you.

Answer 2

As mentioned by the @user99914, we just need to know the best constant in the Poincare inequality, which I will simply call here as the "Poincare constant". To be precise, here it means the largest constant $C>0$ such that \begin{align} C\int_M(f-\bar{f})d\mu\leq\int_M|\nabla f|^2d\mu \end{align}

It turns out that the Poincare constant $C$ is related to the first eigenvalue $\lambda_1$ of the Laplacian on the manifold (in this case, the $2$-sphere); by this I mean $-g^{ij}\nabla_i\nabla_j f=\lambda_1f$; that is, there is a minus sign in front. In fact, we have $C=\lambda_1$. This fact probably can be found in many standard Riemannian geometry or geometric analysis texts, and I'll just quote one which I encountered: Jurgen Jost's book. In short, it follows from the following: \begin{align} \lambda_1=\inf\left\{\frac{\int_{M}|\nabla f|^2d\mu}{\int_{M}f^2d\mu}:f\in W^{1,2}(M), \int_{M}fd\mu=0\right\} \end{align} (For simplicity, I only consider the case where $M$ is compact.)

Now, it is sort of regarded as a standard fact in geometric analysis that the first eigenvalue of an $n$-sphere with radius $1/\sqrt{K}$ is equal to \begin{align} \lambda_1=nK \end{align} For a more general result which includes this as a special case, see Lichnerowicz and Obata's work. Now, the radius of our $\Sigma_r$ is $u(r)$, while its dimension is $2$. Hence \begin{align} C=\lambda_1=2\cdot\left(\frac{1}{u(r)}\right)^2 \end{align} which gives you the desired result.

P/S: I haven't checked out your new questions in your answer. Maybe I'll also try to answer them when I'm free.