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.

 

 

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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<u\in C^\infty(M)}, 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<p\leq \frac{n+2}{n-2}}. 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}}.