## Category Archives: Yamabe flow

### Conformal killing operator and divergence transformation under conformal change

Suppose ${(M,g)}$ is a Riemannian manifold. For each vector field ${V}$, we can define the conformal killing operator ${\mathcal{D}}$ to be the trace free part of Lie derivative ${\mathcal{L}_Vg}$, more precisely

$\displaystyle \mathcal{D}V=\mathcal{L}_Vg-\frac{2}{n}(div_g V)g$

Obviously ${\mathcal{D}}$ maps vector field to trace free symmetric two tensors. Now suppose we have a conformal transformation ${\tilde{g}=e^{2f}g}$, then what happen to the conformal killing operator ${\tilde{\mathcal{D}}}$? Notice that

$\displaystyle \mathcal{L}_V\tilde g=\mathcal{L}_V(e^{2f}g)=2e^{2f}V(f)g+e^{2f}\mathcal{L}_v g$

By using the identity ${\mathcal{L}_V d\mu_g=(div_g V)d\mu_g}$ for any vector field ${V}$, here ${d\mu_g}$ is the volume element, one can get the transfromation of divergence under confromal change

$\displaystyle div_{\tilde g}V=div_g V+nV(f)$

therefore

$\displaystyle \tilde{\mathcal{D}}V=\mathcal{L}_V\tilde g-\frac{2}{n}(div_{\tilde g} V)\tilde g=e^{2f}\mathcal{D}V.$

${\mathcal{D}}$ induces a formal adjoint ${\mathcal{D}^*}$ on trace free 2-tensors. Suppose we have a symmetric 2-tensor ${h=h_{ij}dx^i\otimes dx^j}$, where ${x^i}$ are coordinates on ${M}$. If one has

$\displaystyle \tilde{\mathcal{D}}^*(h-\tilde{\mathcal{D}}V)=0$

for some vector field ${V}$ and trace free 2-tensor ${h}$. What does this coorespond to under metric ${g}$? To see that, we first need a formula about symmtric 2-tensors,

$\displaystyle \langle h,w\rangle_{\tilde g}=\int_M h_{ij}w_{kl}\tilde g^{ik}\tilde g^{jl}d\mu_{\tilde {g}}=\langle e^{(n-4)f}h,w\rangle_g$

Now choose any vector field ${W}$, then

$\displaystyle 0=\langle h-\tilde{\mathcal{D}}V,\tilde{\mathcal{D}}W\rangle_{\tilde g}=\langle h-e^{2f}\mathcal{D}V, e^{2f}\mathcal{D}w \rangle_{\tilde g}=\langle e^{nf}(e^{-2f}h-\mathcal{D} V),\mathcal{D} W\rangle_{g}$

This is equivalent to

$\displaystyle \mathcal D^*(e^{nf}(e^{-2f}h-\mathcal DV))=0$

Next consider the divergence operator ${\delta:\mathscr{S}^{p+1}M\rightarrow \mathscr{S}^p M}$ and its adjoints ${\delta^*}$.

$\displaystyle \delta T=-g^{ik}\nabla_iT_{k....}$

It is well know that on 1-forms

$\displaystyle \delta^*\alpha(X,Y)=\frac{1}{2}{\nabla_X\alpha(Y)+\nabla_Y\alpha(X)}=\frac 12 (L_{\alpha^\sharp}g)(X,Y).$

where ${\sharp}$ operator turns 1-form to a vector field by using metric ${g}$. What is the relation of ${\delta h}$ and ${\tilde \delta h}$? To find that, choose any 1-form ${\alpha}$,

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle \tilde \delta h,e^{-(n-2)f}\alpha\rangle_{\tilde g}=\langle h,\tilde\delta^*(e^{-(n-2)f}\alpha)\rangle_{\tilde g} \ \ \ \ \ (1)$

using the formula about ${\delta^*}$, one gets

$\displaystyle \tilde \delta^*(e^{-(n-2)f}\alpha)=e^{-(n-2)f}\tilde \delta^*\alpha-(n-2)e^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

$\displaystyle =e^{-(n-2)f}\delta^*\alpha+e^{-(n-2)f}\alpha(f)g-ne^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

then using ${h}$ is symmetric, continue from (1)

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle e^{(n-4)f},\tilde \delta^*(e^{-(n-2)f}\alpha)\rangle_{g}=\langle e^{-2f}h,\delta^*\alpha+\alpha(f)g-ndf\otimes\alpha\rangle_{g}$

$\displaystyle =\langle\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f,\alpha\rangle_g$

In other words,

$\displaystyle \tilde \delta h=\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f$

$\displaystyle =\delta h-(n-2)e^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f \ \ \ \ \ (2)$

In the other way, we can calculate more directly

$\displaystyle \nabla_k h_{ij}=\frac{\partial}{\partial x^k}h_{ij}-\Gamma^p_{ki}h_{pj}-\Gamma^p_{kj}h_{ip}$

$\displaystyle \tilde \Gamma^k_{ij} = \Gamma^k_{ij}+ \delta^k_i\partial_j f + \delta^k_j\partial_i f -g_{ij}\nabla^k f$

We get

$\displaystyle \tilde \delta h=-\tilde g^{ki}\tilde \nabla_kh_{ij}=-e^{-2f}g^{ki}\nabla_kh_{ij}+e^{-2f}g^{ki}h_{pj}(\delta_k^p\partial_if+\delta^p_i\partial_kf-g_{ki}\nabla^pf)$

$\displaystyle +e^{-2f}g^{ki}h_{ip}(\delta^p_k\partial_j f+\delta^p_j\partial_kf-g_{kj}\nabla^pf)$

$\displaystyle \tilde \delta h=e^{-2f}\delta h+e^{-2f}(g^{ki}h_{kj}\partial_if+g^{ki}h_{ij}\partial_kf-nh_{pj}\nabla^p f+g^{ki}h_{ik}\partial_j f+g^{ki}h_{ij}\partial_k f- h_{jp}\nabla^p f)$

therefore

$\displaystyle \tilde \delta h=e^{-2f}\delta h-(n-2)e^{-2f}g^{ki}h_{kj}\partial_if+e^{-2f}(tr_gh)\nabla f$

One can compare this with (2).

### The perturbation of and metric and standard bubble

Suppose ${\bar u_\varepsilon=\left(\frac{\varepsilon}{\varepsilon^2+|x|^2}\right)^{\frac{n-2}{2}}}$, then it is well know that it satisfies

$\displaystyle -\Delta \bar u_\varepsilon=n(n-2)\bar u^{\frac{n+2}{n-2}}_\varepsilon$

on ${\mathbb{R}^n}$. Now we want to perturb the Euclidean metric a little bit and an approximate solution of above equation. Suppose ${g(t)=\delta+tS+O(t^2)}$ where ${S=\frac{d}{dt}g(t)|_{t=0}}$ is a symmetric 2-tensor and ${\delta}$ is the Euclidean metric. The reasonable equation of ${u_\varepsilon=u_{\varepsilon}(t)}$ must satisfy should be

$\displaystyle -\Delta_{g}u_{\varepsilon}+\frac{n-2}{4(n-1)}R_gu_\varepsilon=n(n-2)u^{\frac{n+2}{n-2}}_\varepsilon$

Suppose ${u_\varepsilon(t)=\bar u_\varepsilon+tw+O(t^2)}$. Taking derivative on both sides of the above equation and restrict to ${t=0}$, one can get the equation ${w}$ must satisfy

$\displaystyle \Delta w+n(n+2)\bar u_\varepsilon^{\frac{4}{n-2}}w=\frac{n-2}{4(n-1)}\bar u_\varepsilon(\partial_i\partial_j S_{ij}-\Delta tr S)+\partial_{i}(\partial_j\bar u_\varepsilon S_{ij})-\partial_i(tr S)\partial_i\bar u_\varepsilon$

where we have use the fact that

$\displaystyle \frac{d}{dt}\big|_{t=0}\Delta_{g(t)}u_\varepsilon(t)=\Delta w-\partial_i(S_{ij}\partial_j\bar u_\varepsilon)+\frac 12\partial_i(tr S)\partial_i \bar u_\varepsilon$

$\displaystyle \frac{d}{dt}\big|_{t=0}R_{g(t)}u_\varepsilon(t)=(\partial_{i}\partial_jS_{ij}-\Delta tr S)\bar u_\varepsilon$

### Functional of surface with boundary

Suppose ${M}$ is a compact surface with boundary ${\partial M}$ and metric ${g_0}$. Denote by ${K_0}$ the Gauss curvature of ${M}$ and by ${k_0}$ the geodesic curvature of ${\partial M}$. Define a functional

$\displaystyle E_0[u]=\frac{1}{2}\int_M |\nabla_0u|^2d\mu_0+\int_M K_0ud\mu_0+\int_{\partial M}k_0uds_0$

Lemma: The first variation of the functional ${E[u]}$

$\displaystyle \langle E'_0[u],v\rangle=\int_M Kvd\mu+\int_{\partial M}kvds$

where ${K}$, ${k}$ correspond to the curvature ${g=e^{2u}g_0}$. Suppose there is another metric ${g_1=e^{2u_1}g_0}$, similarly we can define ${E_1[u]}$ replacing by ${K_1}$ and ${k_1}$. Then

Proposition: ${E'_0[u]=E_1'[u-u_1]}$; ${E_0[u]=E_1[u-u_1]+E_0[u_1]}$

Proof: The first conclusion is easy to see from the lemma.

$\displaystyle E_0[u]=E_0[u-u_1+u_1]=E_0[u_1]+\frac{1}{2}\int_M |\nabla_0(u-u_1)|^2d\mu_0$

$\displaystyle +\int_M K_0(u-u_1)d\mu_0+\int_{\partial M}k_0(u-u_1)ds_0+\int_M\langle \nabla_0 u_1,\nabla_0(u-u_1)\rangle_0 d\mu_0$

$\displaystyle =E_0[u_1]+\langle E'_0[u],u-u_1\rangle-\frac{1}{2}\int_{M}|\nabla_0(u-u_1)|^2_0d\mu_0$

$\displaystyle =E_0[u_1]+\langle E'_1[u-u_1],u-u_1\rangle-\frac{1}{2}\int_{M}|\nabla_0(u-u_1)|^2_0d\mu_0$

$\displaystyle =E_0[u_1]+E_1[u-u_1]+\frac{1}{2}\int_{M}|\nabla_1(u-u_1)|^2_1d\mu_1-\frac{1}{2}\int_{M}|\nabla_0(u-u_1)|^2_0d\mu_0$

So we need to prove

$\displaystyle |\nabla_1(u-u_1)|^2_1d\mu_1=|\nabla_0(u-u_1)|^2_0d\mu_0$

By using ${\langle \nabla_0 u,X\rangle_0=X(u)}$, we get

$\displaystyle \nabla_1(u-u_1)=e^{-2u_1}\nabla_0(u-u_1)$

$\displaystyle |\nabla_1(u-u_1)|^2_1=e^{2u_1}|\nabla_0(u-u_1)|^2_0$

which exactly means the former identity. $\Box$

### 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.