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]}.


\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}


\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)


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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