Tag Archives: conformally invariant

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

 

Advertisements

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:}