## Category Archives: Riemannian Geo

### Bubble functions under different setting

Bubble function can be defined either on $\mathbb{R}^n$, $\mathbb{S}^n$ or $\mathbb{B}^n$. For the following notations, $c_n$ will denote suitable constants which may be different from line to line.

• For every $\epsilon>0$ and  $\xi\in\mathbb{R}^n$, define

$\displaystyle u_{\epsilon,\xi}=c_n\left(\frac{\epsilon}{\epsilon^2+|x-\xi|^2}\right)^{\frac{n-2}{2}}$

It is well know that $-\Delta u= c_nu^{\frac{n+2}{n-2}}$. Moreover $(\mathbb{R}^n,u^{\frac{4}{n-2}}_{\epsilon,\xi}g_E)$ is isometric to the standard sphere minus one point.

• For any $a\in \mathbb{S}^n$ and $\lambda>0$ define

$\displaystyle\delta(a,\lambda)=c_n\left(\frac{\lambda}{\lambda^2+1+(\lambda^2-1)\cos d(a,x)}\right)^{\frac{n-2}{2}}$

where $d(a,x)$ is the geodesic distance of $a$ and $x$ on $\mathbb{S}^n$. Actually $\cos d(a,x)=a\cdot x$

• For each $p\in \mathbb{B}^{n+1}$, define $\delta_p(x):\mathbb{S}^n\to \mathbb{R}$ by

$\displaystyle\delta_p(x)=c_n\left(\frac{1-|p|^2}{|x+p|^2}\right)^{\frac{n-2}{2}}$

Both the second and third one satisfy

$\displaystyle \frac{4(n-1)}{n-2}\Delta_{\mathbb{S}^n}\delta-n(n-1)\delta+c_n\delta^{\frac{n-2}{n+2}}=0$

If we make $p=\frac{\lambda-1}{\lambda+1}a$, the third one will be changed to the second one.

To get the second one from the first one, let us deonte $\Phi_a:\mathbb{S}^n\to \mathbb{R}^n$ be the stereographic projection from point $a$. Then

$\displaystyle \delta(a,\lambda)\circ \Phi^{-1}_a=c_n\left(\frac{\lambda(1+|y|^2)}{\lambda^2|y|^2+1}\right)^{\frac{n-2}{2}}=c_nu_{\lambda,0}u_{1,0}^{-1}$

It should be able to see the third one from hyperbolic translation directly.

### Parallel surfaces and Minkowski formula

Suppose ${X:M^n\rightarrow \mathbb{R}^{n+1}}$ is an immersed orientable closed hypersurface. ${N}$ is the inner unit normal for ${X(M^n)}$ and denote by ${\sigma}$ the second fundamental form of the immersion and by ${\kappa_i}$, ${i=1,\cdots,n}$ the principle curvatures at an arbitrary point of ${M}$. The ${r-}$th mean curvature of ${H_r}$ is obtained by applying ${r-}$elementary symmetric function to ${\kappa_i}$. Equivalently, ${H_r}$ can be defined through the identity

$\displaystyle P_n(t)=(1+t\kappa_1)\cdots(1+\kappa_n)=1+\binom{n}{1}H_1 t+\cdots+\binom{n}{n}H_n t^n$

for all real number ${t}$. One can see that ${H_1}$ represents the mean curvature of ${X}$, ${H_n}$ is the gauss-Kronecker curvature. ${H_2}$ can reflect the scalar curvature of ${M}$ on the condition that the ambient manifold is a space form.

We want to study the consequence of moving the hypersurface parallel. Namely, define ${X_t}$ to be

$\displaystyle X_t= X-tN.$

When ${t}$ is small enough, ${X_t}$ is well defined immersed hypersurface. Suppose ${e_1,\cdots, e_n}$ are principle directions at a point ${p}$ of ${M}$, then

$\displaystyle \quad(X_t)_*(e_i)=(1+\kappa_it)e_i$

here we identify ${X_*(e_i)=e_i}$ as abbreviation. This implies that ${N_t= N\circ X_t^{-1}}$ is also an unit normal field of ${X_t}$. The area element ${dA_t}$ will be

$\displaystyle dA_t=(1+t\kappa_1)\cdots(1+t\kappa_n)dA=P_n(t)dA.$

The second fundamental form of ${X_t}$ with respect to ${N}$ will be

$\displaystyle \sigma_t(v,w)=\langle N_t,\nabla^{\mathbb{R}^{n+1}}_vw\rangle=-\langle \nabla^{\mathbb{R}^{n+1}}_vN_t,w\rangle$

for all ${v,w}$ tangent vector fields on ${X_t(M)}$. Plugging in ${v=(X_t)_*(e_i)}$ and ${w=(X_t)_*(e_j)}$, we get

$\displaystyle (\nabla^{\mathbb{R}^{n+1}}_vw)(X_t(p))=(\nabla^{\mathbb{R}^{n+1}}_{e_i}e_j)(X(p))$

$\displaystyle \nabla_{v}^{\mathbb{R}^{n+1}}N_t=-\frac{\kappa_i}{1+t\kappa_i}v$

So ${e_1,\cdots, e_n}$ are also principle directions for ${X_t}$ and principle curvatures are

$\displaystyle \frac{\kappa_i}{1+t\kappa_i}$

Another way to see this is by choosing a geodesic local coordinates such that ${\partial_iX}$ are the principle directions of ${X}$ at ${p}$. Then

$\displaystyle \partial_j\partial_iX=\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij}$

$\displaystyle \partial_iN=-\kappa_i\partial_iX$

$\displaystyle \partial_i X_t=\partial_i X-t\partial_i N=\partial_i X+t\kappa_i\partial_iX$

$\displaystyle \partial_j\partial_iX_t=(1+t\kappa_i)\partial_j\partial_iX=(1+t\kappa_i)(\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij})$

Since ${g^{ij}_t=(1+\kappa_it)^{-2}\delta_{ij}}$ at ${p}$. Therefore we get the principle curvature are ${\frac{\kappa_i}{1+t\kappa_i}}$.

Therefore the mean curvature ${H(t)}$ for ${X_t}$ is

$\displaystyle H(t)=\frac{1}{n}\frac{P_n'(t)}{P_n(t)}$

Since we have identity

$\displaystyle \Delta|X_t|^2=2n(1+H\langle X_t,N\rangle)$

which implies

$\displaystyle \int_M\left(1+H(t)\langle X_t,N\rangle\right)dA_t=0$

Plugging in all the information,

$\displaystyle \int_M\left(nP_n(t)+P_n'(t)\langle X,N\rangle-tP_n'(t)\right)dA=0$

Reorder the terms in the above identity by the order of ${t}$, we get

$\displaystyle \int_M (H_{r-1}+H_r\langle X,N\rangle )dA=0$

One can use this to prove Heintze-Karcher inequality. There are Minkowski formula in Hyperbolic space and $\mathbb{S}^n$ also.

Remark: S. Montiel and Anotnio Ros, compact hypersurfaces: the alexandrov theorem for higher order mean curvatures. Differential Geometry, 52, 279-296

### f-extremal disk

In the last nonlinear analysis seminar, Professor Espinar talked about the overdetermined elliptic problem(OEP) which looks like the following

$\Delta u+f(u)=0\quad\text{ in }\Omega$

$u>0\quad \text{ in }\Omega$

$u=0 \quad \text{on }\partial \Omega$

$\frac{\partial u}{\partial\eta}=cst\quad\text{on }\partial \Omega$

There is a BCN conjecture related to this

BCN: If $f$ is Lipschitz, $\Omega\subset \mathbb{R}^n$ is a smooth(in fact, Lipschitz) connected domain with $\mathbb{R}^n\backslash\Omega$ connected where OEP admits a bounded solution, then $\Omega$ must be either a ball, a half space, a generalized cylinder or the complement of one of them.

BCN is false in $n\geq 3$. Epsinar wih Mazet proved BCN when $n=2$. This implies the Shiffer conjecture in dimension 2. In higher dimension of Shiffer conjecture, if we know the domain is contained in one hemisphere of $\mathbb{S}^n$, then one can use the equator or the great circle to perform the moving plane.

### 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).

### Product metric on product manifold

Suppose we have two Riemannian manifolds $(M^m,g)$ and $(N^n,\tilde g)$, what happened to their product manifold with product metric?
$(M\times N, g\times\tilde g)$.

We will use $a,b,c,..=1..m$ and $A,B,C,..=1..n$. Then by definition $g_{aA}=0$. Therefore

$\Gamma_{aA}^b=\frac 12g^{bc}(\partial_Ag_{ac}+\partial_ag_{Ac}-\partial_cg_{aA})=0$

Similarly for all mixed indices on $\Gamma$. This implies

$\nabla_{\partial_a}\partial_A=\Gamma_{aA}^b\partial_b+\Gamma_{aA}^B\partial_B=0.$

Therefore the Riemannian curvature operator is

$R(\partial_a,\partial_A)\partial_B=R(\partial_a,\partial_A)\partial_b=0.$

and Ricci tensor

$R_{aA}=0.$

When you have the warped product metric like $\mathbb{R}\times N$ with metric $dr^2+\phi^2(r)\tilde g$, the curvature is given like the following

$R(\partial _r,X,\partial_r,Y)=-{\phi''}\phi\tilde g(X,Y)$

$R(\partial_r,X,Y,Z)=0$

$R(X,Y,Z,W)=\phi^2\tilde R(X,Y,Z,W)-(\phi'\phi)^2(\tilde g(X,Z)\tilde g(Y,W)-\tilde g(X,W)\tilde g(Y,Z))$

### Geodesic Normal Coordinates, Gauss lemma and Identity

Suppose ${p}$ is a point on ${n-}$dim manifold ${M}$. Consider the exponential map near ${p}$, at suitable neighborhood any point within it can be expressed uniquely as

$\displaystyle \text{exp}(x^1e_1+x^2e_2+\cdots+x^ne_n)$

where ${\{e_i\}}$ is an orthonormal basis. Then ${\{x^i\}}$ can be coordinate around ${p}$. Let us deduce useful identities using this coordinate. Obviously, the geodesic in this coordinate will be

$\displaystyle \gamma(t)=tv, \quad v\in T_pM$

In particular ${v=e_i}$,

$\displaystyle g_{ij}(p)=\langle \frac{\partial }{\partial x^i}(0),\frac{\partial }{\partial x^j}(0)\rangle=\langle e_i,e_j\rangle=\delta _{ij}$

The geodesic equation will be

$\displaystyle \Gamma^k_{ij}(\gamma(t))v^iv^j=0$

Multiplying ${t^2}$, we get

$\displaystyle \Gamma^k_{ij}(\gamma(t))x^ix^j=0$

Since the tangent vector has constant length along ${\gamma(t)}$, i.e.

$\displaystyle g_{ij}(\gamma(t))v^iv^j=g_{ij}(p)v^iv^j=v^iv^i$

Multiplying ${t^2}$, it is equivalent to say

$\displaystyle g_{ij}x^ix^j=x^ix^i$

Furthermore, recalling the definition of Christoffel symbol, one can get

$\displaystyle \frac{1}{2}(\partial_j g_{ik}+\partial_i g_{jk}-\partial_k g_{ij})x^ix^j=0$

Or equivalently

$\displaystyle \partial_j g_{ik}x^ix^j=\frac{1}{2}\partial_kg_{ij}x^ix^j=\frac{1}{2}\partial_k(g_{ij}x^ix^j)-g_{kj}x^j=x^k-g_{kj}x^j$

While on the left hand side

$\displaystyle \partial_j g_{ik}x^ix^j=\partial_j(g_{ik}x^i)x^j-g_{ik}x^i$

Combining this two facts,

$\displaystyle \partial_j(g_{ik}x^i)x^j=x^k$

$\displaystyle \partial_j(g_{ik}x^i-x^k)x^j=0$

This means along ${\gamma(t)}$,

$\displaystyle \frac{d}{dt}(g_{ik}x^i-x^k)=0$

Since at ${p}$, ${g_{ik}x^i-x^k=0}$, one get

$\displaystyle g_{ik}x^i=x^k$

on any point in the neighborhood. This identity is called Gauss Lemma.

Suppose ${g(x)=\text{exp}(h(x))}$, ${h_{ij}(x)}$ is a symmetric 2-tensor. Then the above identity means ${h_{ij}x^j=0}$.

### Torus and Flat Torus

Torus

$\displaystyle T(u,v)=((R+r\cos u)\cos v,(R+r\cos u)\sin v,r\sin u)$

One can calculate the Gauss curvature by the following formula

$\displaystyle E=T_u\cdot T_u, F=T_u\cdot T_v=0, G=T_v\cdot T_v$

$\displaystyle K=-\frac{1}{2\sqrt{EG}}\left(\frac{\partial }{\partial v}\left(\frac{E_v}{\sqrt{EG}}\right)+\frac{\partial }{\partial u}\left(\frac{G_u}{\sqrt{EG}}\right)\right)$

In fact

$\displaystyle K=\frac{\cos u}{r(R+r\cos u)}$

So one can see that some places ${K}$ is positive and some places ${K}$ is negative. There is another torus homeomorphic to this one, which is also ${S^1\times S^1}$ with different metric,

$\displaystyle T=(\cos u,\sin u,\cos v,\sin v)$

lies in the space ${\mathbb{R}^4}$. Using the formula above, one can get

$\displaystyle E=1,F=0,G=1,\text{ then }K=0$

So this torus has a name as flat torus. Flat torus can not be put in ${R^3}$ because the following therem:

Theorem: On every compact surface ${M\subset\mathbb{R}^3}$ there is some point ${p}$ with ${K(p)>0}$.

Remark:Oprea, John. Differential geometry and its applications. p129

### 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.

### Mean curvature of sphere cap

Suppose one has the disc ${B_1=\{x\in \mathbb{R}^n||x|\leq 1\}}$, ${n\geq 3}$, prescribe the ball with metric

$\displaystyle g_{ij}=4u^{\frac{4}{n-2}}\delta_{ij}, \quad u=\left(\frac{\epsilon}{\epsilon^2+|x|^2}\right)^{(n-2)/2}$

What is the mean curvature of the boundary? As we all know that under the Euclidean metric, the boundary of unit ball has mean curvature ${h=1}$. We want to use the formula of mean curvature under the conformal transmformation. Namely, suppose the ${(M,g_0)}$ has mean curvature ${h_0}$, then under metric ${g=v^{\frac{4}{n-2}}g_0}$, the mean curvature of ${(M,g)}$ will be

$\displaystyle h_g=\frac{2}{n-2}v^{-\frac{n}{n-2}}\left(\frac{\partial v}{\partial \eta}+\frac{n-2}{2}h_0 v\right)$

where ${\eta}$ is the normal outer unit vector under ${g_0}$. Using the above principle, let ${v=2^{\frac{n-2}{2}}u}$, ${g_0}$ be the Euclidean flat metric, then ${h_0=1}$.

$\displaystyle \frac{\partial v}{\partial \eta}=(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)$

$\displaystyle \frac{n-2}{2}h_0 v=\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}$

$\displaystyle h_g=\frac{2}{n-2}\frac{(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)+\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}}{(2\epsilon)^{-\frac{n}{2}}(\epsilon^2+1)^{-\frac{n}{2}}}=\frac{\epsilon^2-1}{2\epsilon}$

Remark: Escobar. Conformal Defromation of a Riemannnian metric to a constant scalar curvature metric with constant mean curvature on the boundary. Indiana University Mathematics Journal 1996.

### Conformally invariant Laplacian

On a compact manifold ${(M^n,g)}$, ${n\geq 3}$,

$\displaystyle L_g=-\Delta_g+\frac{n-2}{4(n-1)}R_g$

is called conformal laplacian operator. This follows from the following fact. Suppose ${\tilde{g}=\phi^{\frac{4}{n-2}}g}$, then

$\displaystyle L_{\tilde{g}}(f)=\phi^{-\frac{n+2}{n-2}}L_g(\phi f), \quad\forall\,f\in C^\infty(M)$

Proof: Firstly suppose ${f>0}$, define ${\bar{g}=(\phi f)^{\frac{4}{n-2}}g}$, then

$\displaystyle L_{\bar{g}}(\phi f)=\frac{n-2}{4(n-1)}R_{\bar{g}}(\phi f)^{\frac{n+2}{n-2}}$

on the other hand ${\bar{g}=f^{\frac{4}{n-2}}\tilde{g}}$

$\displaystyle L_{\tilde{g}}(f)=\frac{n-2}{4(n-1)}R_{\bar{g}}(f)^{\frac{n+2}{n-2}}$

So we proved the equality. For general ${f\in C^\infty(M)}$, ${\exists\, C>0}$ such that ${f+C>0}$. By the special case

$\displaystyle L_{\tilde{g}}(f+C)=\phi^{-\frac{n+2}{n-2}}L_g(\phi (f+C))$

$\displaystyle L_{\tilde{g}}(C)=\phi^{-\frac{n+2}{n-2}}L_g(\phi (C))$

Thus the general case is also true. $\Box$