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

Upper half plane and disc

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:}