## Tag Archives: conformally invariant

### Upper half plane and disc

Suppose ${\mathbb{R}^{n+1}_+=\{(y,t)|y\in \mathbb{R}^n,t>0\}}$ is the upper half plane. Define the map ${\pi: \mathbb{R}^{n+1}_+=\{(y,t)\}\rightarrow \mathbb{R}^{n+1}=\{(x,s)\}}$ by the following

$\displaystyle x=\frac{y}{|y|^2+(t+1)^2}$

$\displaystyle s=\frac{t+1}{|y|^2+(t+1)^2}-1$

One can see that ${\pi}$ maps ${\mathbb{R}_+^n}$ onto the open ball ${B=B_{\frac12}((0,-\frac12))\subset \mathbb{R}^{n+1}}$. It is easy to verify that ${\pi^{-1}}$ looks like

$\displaystyle y=\frac{x}{|x|^2+(s+1)^2}$

$\displaystyle t=\frac{s+1}{|x|^2+(s+1)^2}-1$

We want to pull the metric of ${\mathbb{R}^{n+1}_+}$ to ${B}$, that is ${(\pi^{-1})^*(dy^2+dt^2)}$. Denote ${A=|x|^2+(s+1)^2}$. We have

$\displaystyle (\pi^{-1})^*dy_i=A^{-1}dx_i-A^{-2}x_idA$

$\displaystyle (\pi^{-1})^*dt=A^{-1}ds-A^{-2}(s+1)dA$

$\displaystyle (\pi^{-1})^*(dy^2+dt^2)=A^{-2}(dx^2+ds^2)$

Therefore, ${(\pi^{-1})^*(dy^2+dt^2)}$ is conformal to ${dx^2+ds^2}$

Next, for some ${\alpha\geq 0}$, we want to pull the solution ${u}$ of

$\displaystyle -\text{div}(t^\alpha \nabla u)=\alpha(n+\alpha-1)t^{\alpha-1}u^{\frac{n+\alpha+1}{n+\alpha-1}}$

to ${B}$. That is defining ${\psi(x,s)=A^{-\frac{n+\alpha-1}{2}}u}$ on ${B}$. We shall derive the equation ${\psi}$ satisfy on ${B}$. Note that the equation means

$\displaystyle \Delta u+\frac{\alpha}{t}\partial_t u=-\alpha(n+\alpha-1)\frac{1}{t}u^{\frac{n+\alpha+1}{n+\alpha-1}}.$

Let us use the following notations

$\displaystyle \beta=n+\alpha-1.$

It follows from the covariant property of conformal laplacian that for any ${u=u(y,t)}$

$\displaystyle \Delta_{(y,t)}u=A^{\frac{n+3}{2}}\Delta_{(x,s)}(A^{-\frac{n-1}{2}}u).$

Note that ${A^{-\frac{n-1}{2}}u=A^{\frac{\alpha}{2}}\psi}$. We turn to calculate

$\displaystyle \Delta_{(x,t)}(A^{\frac{\alpha}{2}}\psi)=(\Delta A^{\frac{\alpha}{2}})\psi+A^{\frac{\alpha}{2}}\Delta\psi+2\nabla A^{\frac{\alpha}{2}}\nabla \psi.$

It is not hard to see

$\displaystyle \Delta A^{\frac{\alpha}{2}}=\alpha\beta A^{\frac{\alpha}{2}-1},\quad 2\nabla A^{\frac{\alpha}{2}}\nabla \psi=2\alpha A^{\frac{\alpha}{2}-1}\nabla\psi\cdot(x,s+1).$

Therefore

$\displaystyle \Delta_{(y,t)}u=A^{\frac{n+3}{2}}\Delta_{(x,s)}(A^{\frac{\alpha}{2}}\psi)=A^{\frac{\beta+4}{2}}\Delta\psi+2\alpha A^{\frac{\beta+2}{2}}\nabla \psi\cdot(x,s+1)+\alpha\beta A^{\frac{\beta+2}{2}}\psi.$

Next, to handle the term ${\partial_t u}$, applying ${\partial_t A=-2(1+t)A^{2}=-2(1+s)A^{3}}$

$\displaystyle \partial_t u=\partial_t(A^{\frac{\beta}{2}}\psi)=-\beta(1+t)A^{\frac{\beta+2}{2}}\psi+A^{\frac{\beta}{2}}\partial_x \psi[-2y(1+t)A^2]+$

$\displaystyle A^{\frac{\beta}{2}}\partial_s\psi[A-2(1+t)^2A^2].$

That is

$\displaystyle \partial_t u=A^{\frac{\beta}{2}}(1+s)[-\beta\psi-2x\partial_x\psi-2(1+s)\partial_s\psi]+A^{\frac{\beta+2}{2}}\partial_s\psi$

$\displaystyle =-A^{\frac{\beta}{2}}(1+s)[\beta\psi+2\nabla\psi\cdot(x,s+\frac12)]+A^{\frac{\beta}{2}}[A-(1+s)]\partial_s\psi.$

It follows from the transformation formula that

$\displaystyle \frac{\alpha}{t}\partial_t u=\frac{\alpha A}{s+1-A}\partial_t u$

To summarize the above calculation, one the one hand,

$\displaystyle \Delta u+\frac{\alpha}{t}\partial_t u=A^{\frac{\beta+4}{2}}\Delta\psi-\frac{2\alpha}{s+1-A}A^{\frac{\beta+4}{2}}\nabla\psi\cdot(x,s+\frac12)-\frac{\alpha\beta}{s+1-A}A^\frac{\beta+4}{2}\psi$

on the other hand

$\displaystyle -\alpha(n+\alpha-1)\frac{1}{t}u^{\frac{n+\alpha+1}{n+\alpha-1}}=-\alpha\beta\frac{A}{s+1-A}A^{\frac{n+\alpha+1}{2}}\psi^{\frac{n+\alpha+1}{n+\alpha-1}}.$

Then we get the equation of ${\psi}$

$\displaystyle \Delta \psi-\frac{2\alpha \nabla\psi\cdot(x,s+\frac12)}{s+1-A}=\alpha\beta\frac{\psi-\psi^{\frac{n+\alpha+1}{n+\alpha-1}}}{s+1-A}.$

Notice that ${s+1-A=\frac14-r^2}$ where ${r=\sqrt{|x|^2+(s+\frac12)^2}}$ is the distance of ${(x,s)}$ to the center ${(0,-\frac12)}$ of the the ball ${B}$. The equation that ${\psi}$ satisfies is rotationally symmetric with respect to the center of ${B}$

### 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. $\Box$

### Fully nonlinear ellipticity and conformal invariancy

Let $\mathcal{S}^{n\times n}$ be the space of symmetric $n\times n$ matrices and $\mathcal{O}(n)$ is the orthogonal matrices. Suppose $U\subset \mathcal{S}^{n\times n}$ satisfies

$\displaystyle O^{-1}UO=U$, $\forall\, O\in \mathcal{O}(n)$

Let $F\in C^1(U)$ satisfy

$\displaystyle F(O^{-1}MO)=F(M)$, $\forall\, M\in U$, and $O\in \mathcal{O}(n)\quad$      (1)

The fully nonlinear equation $F(D^2u)=\psi$ is called elliptic in $U$ if

$(F_{ij})(D^2u)>0$ for any $D^2u\in U\quad$      (2).

From (1), we know there exists $f\in C^1$ such that $F(M)=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$, where $\lambda_1,\lambda_2,\cdots,\lambda_n$ are the eigenvalues of $M$. And $f$ must be symmetric.

$\mathbf{Problem:}$ $F$ is elliptic is equivalent to $\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i$

$\mathbf{Proof:}$ Firstly, let us assume $F$ is elliptic, i.e. $F_{ij}(M)>0$, $\forall M\in U$.

Then in particular choose $M=\Lambda=diag\{\lambda_1,\lambda_2,\cdots,\lambda_n\}\subset U$,

$f(\lambda_1,\lambda_2,\cdots,\lambda_n)=F(\Lambda)$

Then                                             $\displaystyle \frac{\partial f}{\partial \lambda_i}=F_{ii}(\Lambda)>0$

Secondly, assume $\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i$.

$\forall\, M\in U$, there exists $O\in \mathcal{O}(n)$ such that $M=O\Lambda O^{-1}$, here we denote $O=(O_{ij}), O^{-1}=(O^{ij})$.

Then $F(M)=F(O\Lambda O^{-1})=F((O_{ik}\lambda_{k}O^{kj}))=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$

Differentiating (1), we get       $\displaystyle \sum_{i,j}O^{ik}F_{ij}(O^{-1}MO)O_{lj}=F_{kl}(M)$

That is                       $\displaystyle \sum_{i,j}O^{ik}F_{ij}(\Lambda)O_{lj}=F_{kl}(M)\quad (3)$

Since $F(\Lambda)=f(\lambda_1,\lambda_2,\cdots,\lambda_n)$, then $\displaystyle F_{ii}(\Lambda)=\frac{\partial f}{\partial \lambda_1}$, $\forall\, i>0$

For $F_{ij}(\Lambda)$, when $i\neq j$, by definition and use the fact that $\boldsymbol{F_{ij}=F_{ji}}$(possible to remove?)

$\displaystyle 2F_{ij}(\Lambda)=\lim\limits_{a \to 0}\frac{F(\Lambda+aE_{ij})-F(\Lambda)}{a}$

here $E_{ij}$ is symmertric and has entries which are 0 except at $(i,j)$-th and $(j,i)$-th positions are 1.

$\displaystyle \Lambda+tE_{ij}$ has eigenvalue $\lambda_k$ with $k\neq i,j$ and $\displaystyle \frac{\lambda_i+\lambda_j+\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2}$, $\displaystyle \frac{\lambda_i+\lambda_j-\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2}$.

So $\displaystyle F_{ij}(\Lambda)=\lim\limits_{a \to 0}\frac{F(\Lambda+aE_{ij})-F(\Lambda)}{2a}=\lim\limits_{a \to 0}\frac{f(\lambda_1,\lambda_2,\cdots,\lambda_n)-f(\lambda_1,\lambda_2,\cdots,\lambda_n)}{2a}$

$\displaystyle =\lim\limits_{a \to 0}\frac{f(\lambda_1,\cdots, \frac{\lambda_i+\lambda_j+\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2},\cdots,\frac{\lambda_i+\lambda_j-\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2},\cdots,\lambda_n)-f(\lambda_1,\lambda_2,\cdots,\lambda_n)}{2a}\quad (4)$

• If $\lambda_i=\lambda_j=\mu$, then (4) becomes

$\displaystyle F_{ij}(\Lambda)=\lim\limits_{a \to 0}\frac{f(\lambda_1,\cdots, \mu+a,\cdots,\mu-a,\cdots,\lambda_n)-f(\lambda_1,\lambda_2,\cdots,\lambda_n)}{2a}$

$\displaystyle =\frac{1}{2}\left(\frac{\partial f}{\partial \lambda_i}(\lambda_1,\lambda_2,\cdots,\lambda_n)-\frac{\partial f}{\partial \lambda_j}(\lambda_1,\lambda_2,\cdots,\lambda_n)\right)=0,$

because $f$ is symmetric and $\lambda_i=\lambda_j$.

• If $\lambda _i\neq \lambda_j$, (4) becomes

$\displaystyle F_{ij}(\Lambda)=\frac{\partial f}{\partial \lambda_i}(\lambda_1,\lambda_2,\cdots,\lambda_n)\lim\limits_{a\to 0}\frac{\lambda_j-\lambda_i+\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2a}+\frac{\partial f}{\partial \lambda_j}(\lambda_1,\lambda_2,\cdots,\lambda_n)\lim\limits_{a\to 0}\frac{\lambda_i-\lambda_j-\sqrt{(\lambda_i-\lambda_j)^2+4a^2}}{2a}$

$\quad =0$

Combing the above results, we get $\displaystyle (F_{ij}(\Lambda))=diag\left\{\frac{\partial f}{\partial \lambda_1},\frac{\partial f}{\partial \lambda_2}\cdots,\frac{\partial f}{\partial \lambda_n}\right\}$.

Applying $(F_{ij}(\Lambda))$ to (3), it becomes $\displaystyle \sum_{i}O^{ik}\frac{\partial f}{\partial \lambda_i}(\lambda_1,\lambda_2,\cdots,\lambda_n)O_{li}=F_{kl}(M)$

This means $\displaystyle O\left(diag\left\{\frac{\partial f}{\partial \lambda_1},\frac{\partial f}{\partial \lambda_2}\cdots,\frac{\partial f}{\partial \lambda_n}\right\}\right)O^{-1}=(F_{kl}(M))^T$.

$(\displaystyle F_{kl}(M))$ is a positive definite matrix because  $\displaystyle \frac{\partial f}{\partial \lambda_i}>0$, $\forall\, i$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$