## 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.