## 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.