## Tag Archives: yamabe

### Invariance under the conformal mapping

Consider the 2-sphere ${\mathbb{S}^2=\{x\in \mathbb{R}^3||x|^2=1\}}$ with the standard metric ${g_0}$. All conformal diffeomorphism of ${\mathbb{S}^2}$ are composing a suitable isometries of ${\mathbb{S}^2}$ and some ${\pi^{-1}\circ M_t\circ \pi}$, where ${\pi}$ is the stereographic projection and ${M_t:x\rightarrow tx}$ on ${\mathbb{R}^2}$. For any conformal metric ${g=e^{2u}g_0}$, define

$\displaystyle S[u]= \int_{\mathbb{S}^2} |\nabla u|^2+2u\,d\mu_0$

If ${\phi}$ is a conformal transformation, then we can find ${u_\phi}$ such that ${\phi^*g=e^{2u_\phi}g_0}$, where

$\displaystyle u_\phi=u\circ\phi+\frac{1}{2}\log \det|d\phi|$

here we use the notation ${\phi^*g_0=\det|d\phi|g_0}$.

Fact: for any conformal map ${\phi}$ of ${S^2}$, ${u=\frac{1}{2}\log\det|d\phi|}$ satisfies the following identity

$\displaystyle \frac 12\Delta \log\det|d\phi|+\det|d\phi|=1$

Actually, this identity means ${(\mathbb{S}^2,\phi^*g_0)}$ has Gaussian curvature ${1}$, the same as ${(\mathbb{S}^2, g_0)}$.${\hfill\square}$

Upon this fact, we have the invariance of ${S[u]}$.

Proposition: ${S[u]=S[u_\phi]}$.

Proof:

$\displaystyle S[u_\phi]=\int |\nabla (u\circ \phi)+\frac{1}{2}\nabla\log \det|d\phi||^2+ 2u\circ \phi+\log \det|d\phi|$

$\displaystyle =\int |\nabla(u\circ\phi)|^2+\nabla (u\circ \phi)\cdot\nabla\log \det|d\phi|+2u\circ\phi+S[\frac{1}{2}\log \det|d\phi|]$

$\displaystyle =\int |\nabla(u\circ\phi)|^2+2u\circ\phi\det|d\phi|+S[\frac{1}{2}\log \det|d\phi|]$

from integration by parts of the middle term. Suppose ${\nabla u= g^{ij}\frac{\partial u}{\partial x^i}\frac{\partial }{\partial x^j}}$, then

$\displaystyle \nabla(u\circ\phi)=g^{ik}\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial\phi^\alpha}{\partial x^i}\frac{\partial}{\partial x^k}$

So

$\displaystyle |\nabla(u\circ\phi)|^2=g^{ik}\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial\phi^\alpha}{\partial x^i}g^{jl}\frac{\partial u}{\partial x^\beta}\circ\phi\cdot\frac{\partial\phi^\beta}{\partial x^j}g_{kl}$

$\displaystyle =\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial u}{\partial x^\beta}\circ\phi\cdot g^{ij}\frac{\partial\phi^\alpha}{\partial x^i}\frac{\partial\phi^\beta}{\partial x^j}$

$\displaystyle =\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial u}{\partial x^\beta}\circ\phi\cdot g^{\alpha\beta}\circ\phi\cdot\det|d\phi|=|(\nabla u)|^2\circ\phi\det|d\phi|$

where we have used ${\phi^*g_0=\det|d\phi|g_0}$. Continuing our simplication of ${S[u_\phi]}$,

$\displaystyle S[u_\phi]=\int |(\nabla u)|^2\circ\phi\det|d\phi|+2u\circ\phi\det|d\phi|+S[\frac{1}{2}\log \det|d\phi|]$

$\displaystyle =S[u]+S[\frac{1}{2}\log \det|d\phi|]$

by changing variables. So we only need to prove the last term is ${0}$, which is

$\displaystyle \frac{1}{4}\int |\nabla \log\det|d\phi||^2+\log\det|d\phi|=0$

integration by parts, this is equivalent to

$\displaystyle \int \log\det|d\phi|=-\int\det|d\phi|\log\det|d\phi| \ \ \ \ \ (1)$

From ${g_0=\phi^*(\phi^{-1})^*g_0}$, we get

$\displaystyle \det|d\phi|\circ\phi^{-1}\cdot\det|d\phi^{-1}|=1$

Changing variable by ${x=\phi^{-1}(y)}$,

$\displaystyle -\int\det|d\phi|(x)\log\det|d\phi|(x)d\mu_0(x)$

$\displaystyle =-\int\det|d\phi|\circ\phi^{-1}(y)\log\det|d\phi|\circ\phi^{-1}(y)\det|d\phi^{-1}|d\mu_0(y)$

$\displaystyle =\int \log\det|d\phi^{-1}|(y)d\mu_0(y)$

So we only need to justify ${\int \log\det|d\phi^{-1}|d\mu_0=\int \log\det|d\phi|d\mu_0}$. As mentioned at the begining, up to some isometry, ${\phi=\pi^{-1}\circ M_t\circ\pi}$ for some ${t}$, where ${\pi}$ is the stereographic projection of north pole. Then ${\phi^{-1}=\tilde{\pi}^{-1}\circ M_{1/t}\circ\tilde{\pi}}$, where ${\tilde{\pi}}$ is the stereographic projection of south pole. Note that

$\displaystyle \det|d\phi|(x)=\det|d\phi^{-1}|(\tilde{x})$

where ${\tilde{x}=(x_1,x_2,-x_3)}$ if ${x=(x_1,x_2,x_3)}$. By changing variables

$\displaystyle \int \log\det|d\phi^{-1}|(x)d\mu_0(x)=\int \log\det|d\phi^{-1}|(\tilde{x})d\mu_0(\tilde{x})$

$\displaystyle =\int \log\det|d\phi^{-1}|(\tilde{x})d\mu_0(x)=\int \log\det|d\phi|(x)d\mu_0(x)$

$\Box$

Remark1: Under sterographic projection, ${(\mathbb{S}^2\backslash\{P\},g_0)}$ is isometric to ${(\mathbb{R}^2, \frac{4}{(1+|x|^2)^2}dx^2)}$. Then for ${\phi:x\rightarrow tx}$ on ${\mathbb{R}^2}$

$\displaystyle \det|d\phi|=\frac{t^2(1+|x|^2)}{(1+t^2|x|^2)}$

Another point of view is thinking ${\phi:\mathbb{R}^2\rightarrow \mathbb{R}^2}$ as a diffeomorphism, then ${d\phi}$ is the transformation of corresponding tangent space. One can also get ${\det|d\phi|}$ is the above expression.

Remark2: Alice Chang, Paul Yang, prescribing curvature on ${\mathbb{S}^2}$, 1987.

### Yamabe flow as gradient flow

Case 1: Suppose we have a compact manifold ${(M,g_0)}$ without boundary. For any metric ${g= u^{\frac{4}{n-2}}g_0}$, define

$\displaystyle I(g)=\int_M \left(|\nabla u|^2+c(n)R_0u^2\right) d\mu_0=c(n)\int_M R_gd\mu_g$

where ${c(n)=\frac{n-2}{4(n-1)}}$, ${R_0}$ and ${R_g}$ are the scalar curvatures under metric ${g_0}$ and ${g}$. If ${\tilde{g}=v^{\frac{4}{n-2}}g=(uv)^{\frac{4}{n-2}}g_0}$, using the conformal invariance, one can verify that

$\displaystyle I(\tilde{g})=\int_M \left(|\nabla v|^2+c(n)R_gv^2\right) d\mu_g=c(n)\int_M R_{\tilde{g}}d\mu_{\tilde{g}}$

Consider a special set of metrics which preserve the volume,

$\displaystyle \mathcal{N}=\left\{g\bigg|g=u^{\frac{4}{n-2}}g_0,\int_M u^{\frac{2n}{n-2}}d\mu_0=1, u>0\right\}$

If we view all conformal metrics form a Banach manifold, then ${\mathcal{N}}$ is a hypersurface. The tangent space at ${g\in \mathcal{N}}$ is

$\displaystyle T_g\mathcal{N}=\left\{wg\bigg|\int_M wd\mu_g=0\right\}$

Choose the inner product on tangent space as

$\displaystyle (wg,\tilde{w}g)_g=\int_{M} wg^{ij}\tilde{w}g_{ij}d\mu_g=n\int_M w\tilde{w}d\mu_g$

We will restrict functional ${I}$ on ${\mathcal{N}}$ and still use ${I}$ to denote ${I|_{\mathcal{N}}}$. We want to find a flow ${g(t)\in\mathcal{N}}$ for every time ${t>0}$ which converges to some special metric. Then necessarily ${\partial_t g}$ must belong to ${T_{g(t)}\mathcal{N}}$ for ${t>0}$.

Fix any ${g\in \mathcal{N}}$. Suppose ${w\in C^\infty(M)}$, ${Vol(t)=\int_M (1+tw)^{2^*}d\mu_g}$, ${2^*=\frac{2n}{n-2}}$, ${t}$ small enough. Then

$\displaystyle g(t)=Vol(t)^{-\frac 2n}(1+tw)^{\frac{4}{n-2}}g=v(t)^{\frac{4}{n-2}}g\in \mathcal{N}$

where ${v(t)=Vol(t)^{-1/2^*}(1+tw)}$. We will get

$\displaystyle I(g(t))=\frac{\int_M t^2|\nabla w|^2+c(n)R_g(1+tw)^2d\mu_g}{Vol(t)^{(n-2)/n}}$

Note that

$\displaystyle \frac{\partial}{\partial t}\bigg|_{t=0}Vol(t)=2^*\int_M wd\mu_g$

One can calculate

$\frac{\partial}{\partial t}\bigg|_{t=0}I(g(t))=2\int_M c(n)R_gwd\mu_g-\frac{n-2}{n}\cdot2^*\int_M wd\mu_g\int_M c(n)R_gwd\mu_g$
$=2c(n)\int_M(R_g-r_g)wd\mu_g=\frac{2c(n)}{n}\left((R_g-r_g)g,wg\right)_g$

Here ${r_g}$ is the average scalar curvature. Note that ${(R_g-r_g)g\in T_g\mathcal{N}}$. On the other hand, for any ${wg\in T_g\mathcal{N}}$,

$\displaystyle \dot{g}(0)=\frac{\partial}{\partial t}\bigg|_{t=0}g(t)=\frac{4}{n-2}wg$

$\displaystyle \frac{\partial}{\partial t}\bigg|_{t=0}I(g(t))=(\nabla I, \dot{g}(0))_g=(\nabla I,\frac{4}{n-2}wg)_g$

So

$\displaystyle \nabla I=\frac{c(n)}{2^*}(R_g-r_g)g$

Then we can construct a negative gradient flow, which is

$\displaystyle \partial_tg=-\frac{c(n)}{2^*}(R_g-r_g)g$

By scaling on time variable, one immediately have the yamabe flow

$\displaystyle \partial_tg=-(R_g-r_g)g$

Case 2: ${(M,g_0)}$ without boundary. Consider the following functional

$\displaystyle I(u)=\frac{1}{2}\int_M|\nabla u|^2+c(n)R_0u^2d\mu_0-K\int_M u^{2^*}d\mu_0$

equivalently

$\displaystyle I(g)=\frac{1}{2}c(n)\int_M R_gd\mu_g-K\int_Md\mu_g$

where ${K}$ is some fixed constant. One can verify that for any ${0, there exists a unique ${\lambda=\lambda(u)}$ such that

$\displaystyle I(\lambda(u)u)=\max\limits_{t>0}\,I(tu).$

So there exists a notion of Nehari manifold, ${\{u|\langle I'(u),u\rangle=0\}}$, converting to metric sense, we get a hypersurface

$\mathcal{N}=\left\{u^{\frac{4}{n-2}}g_0\bigg|\int_M|\nabla u|^2+c(n)R_0u^2d\mu_0=K2^*\int_M u^{2^*}d\mu_0\right\}$
$=\left\{g:c(n)\int_M R_gd\mu_g=K2^*\int_Md\mu_g\right\}=\{g:r_g=K2^*/c(n)\}.$

The tangent space at ${g}$ is

$\displaystyle T_g\mathcal{N}=\left\{wg:2c(n)\int_MwR_gd\mu_g=K(2^*)^2\int_M wd\mu_g\right\}$

We also use the inner product as before

$\displaystyle (wg,\tilde{w}g)_g=\int_{M} wg^{ij}\tilde{w}g_{ij}d\mu_g=n\int_M w\tilde{w}d\mu_g$

Fix ${g\in\mathcal{N}}$, ${g(t)=(1+tw)^{\frac{4}{n-2}}g}$,

$\displaystyle r(t)=\frac{\int_M R_{g(t)}d\mu_{g(t)}}{\int_Md\mu_{g(t)}}$

Fact: for ${M}$, if metric was changed to ${\tilde{g}=\lambda g}$, then ${\tilde{R}=\frac{1}{\lambda}R}$, ${\tilde{r}=\frac{1}{\lambda}r}$.

Using the above fact, ${\bar{g}(t)=\frac{c(n)r(t)}{K2^*}g(t)\in \mathcal{N}}$, ${\forall\, t}$ small enough. Suppose

$\displaystyle \bar{g}(t)=u_1^{\frac{4}{n-2}}g,\quad u_1(t)=\left(\frac{c(n)}{K2^*}\right)^{\frac{n-2}{4}}r(t)^{\frac{n-2}{4}}(1+tw)$

For simplicity on notations, let us use ${L=c(n)/K2^*}$, recall that ${Lr(0)=1}$. Plugging in back to ${I}$,

$I(\bar{g}(t))=\frac{1}{2}\int_M |\nabla u_1(t)|^2+c(n)R_gu_1(t)^2d\mu_g-K\int_Mu_1^{2^*}(t)d\mu_g$

$=\frac{1}{2} (Lr(t))^{\frac{n-2}{2}}\int_M t^2|\nabla w^2|+c(n)R_g(1+tw)^2d\mu_g-K(Lr(t))^{\frac{n}{2}}\int_M(1+tw)^{2^*}d\mu_g$

where

$\displaystyle r(t)=\frac{\int_M R_{g(t)}d\mu_{g(t)}}{\int_Md\mu_{g(t)}}=\frac{c(n)^{-1}\int_M t^2|\nabla w|^2+c(n)R_g(1+tw)^2d\mu_g}{\int_M(1+tw)^{2^*}d\mu_g}$

Differentiating this, we get

$\displaystyle \dot{r}(0)=\frac{2\int_MR_gwd\mu_g}{\int_Md\mu_g}-\frac{2^*\int_Mwd\mu_g\int_MR_gd\mu_g}{\left(\int_Md\mu_g\right)^2} \ \ \ \ \ (1)$

$\frac{\partial}{\partial t}\bigg|_{t=0}I(\bar{g}(t))=\frac{1}{2}(Lr(0))^{\frac{n-2}{2}}\int_M2c(n)R_gd\mu_g+\frac{1}{2}L^{\frac{n-2}{2}}\frac{n-2}{2}r(0)^{\frac{n-2}{2}-1}\int_Mc(n)R_gd\mu_g\dot{r}(0) -KL^{\frac{n}{2}}\frac{n}{2}r(0)^{\frac{n}{2}-1}\dot{r}(0)\int_Md\mu_g-K(Lr(0))^{\frac{n}{2}}2^*\int_Mwd\mu_g$
$=c(n)\int_MR_gd\mu_g+\frac{n-2}{4}r(0)^{-1}\dot{r}(0)c(n)\int_MR_gd\mu_g-K\frac{n}{2}r(0)^{-1}\dot{r}(0)\int_Md\mu_g-K2^*\int_Md\mu_g$

Considering the middle two terms

$\frac{n-2}{4}r(0)^{-1}\dot{r}(0)c(n)\int_MR_gd\mu_g-K\frac{n}{2}r(0)^{-1}\dot{r}(0)\int_Md\mu_g$
$=\dot{r}(0)r(0)^{-1}\left[\frac{n-2}{4}c(n)\int_MR_gd\mu_g-K\frac{n}{2}\int_Md\mu_g\right]=0$

from the defnition of ${\mathcal{N}}$. This implies

$\frac{\partial}{\partial t}\bigg|_{t=0}I(\bar{g}(t))=c(n)\int_M R_gwd\mu_g-K2^*\int_Mwd\mu_g$

On the other hand we have

$\displaystyle \frac{\partial}{\partial t}\bigg|_{t=0}I(\bar{g}(t))=(\nabla I, \dot{\bar{g}}(0))_g=(\nabla I,\frac{4}{n-2}wg+\frac{\dot{r}(0)}{r(0)}g)_g$

So we want ${\nabla I\in T_g\mathcal{N}}$ and also $c(n)\int_M R_gwd\mu_g-K2^*\int_Mwd\mu_g=(\nabla I,\frac{4}{n-2}wg+\frac{\dot{r}(0)}{r(0)}g)_g$ holds for every ${wg\in T_g\mathcal{N}}$.

It is easy to verify that ${\dot{r}(0)=0}$ if ${w\in T_g\mathcal{N}}$. So we have

$\displaystyle c(n)\int_M R_gwd\mu_g-K2^*\int_Mwd\mu_g=(\nabla I,\frac{4}{n-2}wg)_g \ \ \ \ \ (2)$

Let us assume ${\nabla I}$ has the form ${(aR_g+b)g}$, then the above equation is equivalent to

$\displaystyle c(n)\int_M R_gwd\mu_g-K2^*\int_Mwd\mu_g=\frac{4n}{n-2}\left(a\int_MR_gwd\mu_g+b\int_M wd\mu_g\right) \ \ \ \ \ (3)$

Recall ${wg\in T_g\mathcal{N}}$, which means

$\displaystyle 2c(n)\int_MR_gwd\mu_g=K(2^*)^2\int_M wd\mu_g$

using this relation one can simplify $(3)$ to be $c(n)(1-2/2^*) \int_M R_gwd\mu_g=\frac{4n}{n-2}(a+b\frac{2c(n)}{K(2^*)^2}) \int_M R_gwd\mu_g$

$\displaystyle a+b\frac{2c(n)}{K(2^*)^2}=\frac{n-2}{4n}c(n)(1-2/2^*) \ \ \ \ \ (4)$

The restirction ${\nabla I\in T_g\mathcal{N}}$ will give us

$\displaystyle 2c(n)\int_MR_g(aR_g+b)d\mu_g=K(2^*)^2\int_M (aR_g+b) d\mu_g \ \ \ \ \ (5)$

This is equivalent to

$\displaystyle a\left(\int_M R_g^2d\mu_g-\frac{K(2^*)^2}{2c(n)}\int_M R_gd\mu_g\right)+b\left(1-\frac{2^*}{2}\right)\int_MR_gd\mu_g=0 \ \ \ \ \ (6)$

combining $(4)$ and $(6)$, we get

$\displaystyle a=\frac{\frac{n-2}{4n}c(n)(1-2/2^*)(1-2^*/2)\int_M R_gd\mu_g}{\left(2-2^*/2\right)\int_M R_gd\mu_g-\frac{2c(n)}{K(2^*)^2}\int_M R^2_gd\mu_g}=-\frac{K}{(n-2)^2}\frac{\int_M R_gd\mu_g}{\int_M (R_g-K(2^*)^2/2c(n))^2d\mu_g}$
$\displaystyle b=-\frac{\frac{2c(n)}{K(2^*)^2}\int_M R^2_gd\mu_g-\int_M R_gd\mu_g}{\left(2-2^*/2\right)\int_M R_gd\mu_g-\frac{2c(n)}{K(2^*)^2}\int_M R^2_gd\mu_g}=-\frac{\int_M{R_g(R_g-\frac{K(2^*)^2}{2c(n)})d\mu_g}}{\int_{M}(R_g-K(2^*)^2/2c(n))^2d\mu_g}$

Remark: For the second case, it is firstly Professor Yan Yan Li told me the idea to construct flow on Nehari manifold.

### Approriate scalling in Yamabe equation

Suppose ${(M,g)}$ is a Riemannian manifold, and ${L_g=\Delta_g -\frac{n-2}{4(n-1)}R_g}$ is the conformal Laplacian. Assume ${u>0}$ satisfies

$\displaystyle L_gu+Ku^p=0$

where ${K}$ is some fixed constant, ${1. Suppose near a point ${x_0\in M}$, there is a coordinates ${x^1,x^2,\cdots, x^n}$. We want to scale the coordinates to ${x^i=\lambda y^i}$,

$\displaystyle g(x)=g_{ij}(x)dx^idx^j=\lambda^2 g_{ij}(\lambda y)dy^idy^j=\lambda^2 \hat{g}(y)$

By the conformal invariance of ${L}$, for any ${\phi}$, we get

$\displaystyle L_{g}(\lambda^{-\frac{n-2}{2}}\phi)=\lambda^{-\frac{n+2}{2}}L_{\hat{g}}(\phi)$

We want to choose ${\phi(y)=\lambda^{\alpha}u(\lambda y)}$ such that

$\displaystyle L_{\hat{g}}(\phi)+K\phi^p=0$

which means

$\displaystyle L_{\hat{g}}(\lambda^{\alpha}u(\lambda y))=\lambda^{\frac{n+2}{2}}L_g (\lambda^{\alpha-\frac{n-2}{2}}u(\lambda y))=-K\lambda^{\alpha+2} (u(\lambda y))^p$

Letting

$\displaystyle \alpha+2=\alpha p$

we get ${\alpha=\frac{2}{p-1}}$.

The above proof may not be right.

Or we should look it more directly

$\displaystyle L_g(u(x))=\lambda^2 L_{\hat{g}}(u(\lambda y))=\lambda^2 Ku(\lambda y)^p$

then

$\displaystyle L_{\hat{g}}(\lambda^\alpha u(\lambda y))=K(\lambda^\alpha u(\lambda y))^p$

with ${\alpha=\frac{2}{p-1}}$.