## Two ways of extension

Let ${(X^{n+1},M^n,g_+)}$ be a Poincaé-Einstein manifold and let ${\rho}$ be any boundary defining function such that ${ g=\rho^2 g_+}$ is a metric of the compact manifold ${\bar X}$. For ${\gamma\in (0,1)}$ and ${f\in C^\infty(M)}$, there are different ways to define the fractional derivative of ${f}$. Denote ${m=1-2\gamma}$.

In Jeffrey and Alice’s paper, they use the notion of smooth metric measure space ${(X^{n+1}, g:=\rho^2g_+,\rho^m dvol_g,m-1)}$

$\displaystyle L_{2,\phi}^m=-\Delta_{\phi}+\frac{m+n-1}{2}J_{\phi}^m=-\Delta_g -m\rho^{-1}\partial_\rho+\frac{m+n-1}{2}J_{\phi}^m$

$\displaystyle =-\rho^{-m}\text{div}_g(\rho^m \nabla \cdot)+\frac{m+n-1}{2}J_{\phi}^m$

here ${J_\phi^m}$ is the weighted Schouten scalar defined by

$\displaystyle J_\phi^m=\frac{1}{2(m+n)}\left[R_g-2m\rho^{-1}\Delta_g \rho-m(m-1)\rho^{-2}(|\nabla \rho|^2-1)\right].$

Suppose ${U}$ is the solution of the extension problem

$\displaystyle \begin{cases}L_{2,\phi}^mU=0&\text{ in }(X, g)\\U=f&\text{ on }M\end{cases}$

then

$\displaystyle P_{\gamma}f-\frac{n-2\gamma}{2}d_\gamma\Phi f=\frac{d_\gamma}{2\gamma}\lim_{\rho\rightarrow 0}\frac{\partial U}{\partial \rho}$

here ${\rho=r+\Phi r^{1+2\gamma}+o(r^{1+2\gamma})}$ and ${r}$ is the geodesic defining function.

In Maria and Alice’s paper, one can define ${U}$ as the solution of the extension problem

$\displaystyle \begin{cases}-\text{div}_g(\rho^{m}\nabla U)+E(\rho)U=0&\text{ in }(X, g)\\U=f&\text{ on }M\end{cases}$

here the derivative are taken with respect to the metric ${g}$

$\displaystyle E(\rho)=\rho^{-1-s}(-\Delta_{g_+}-s(n-s))\rho^{n-s}$

or equivalently

$\displaystyle E(\rho)=[-\Delta_g(\rho^{\frac{m}{2}})\rho^{-\frac{m}{2}}+\frac14m(m-1)\rho^{-2}+\frac{n-1}{4n}R_g]\rho^{{m}}$

$\displaystyle =[-\frac{m}{2}\rho^{-1}\Delta_g\rho-\frac14m(m-1)\rho^{-2}(|\nabla\rho|^2-1)+\frac{n-1}{2}J_g]\rho^m$

One may wonder what the difference of the two ways of extension is? The leading operator are both second order. It attempts to compare the error term ${E(\rho)\rho^{-m}}$ and ${\frac{m+n-1}{2}J_\phi^m}$. Since ${g_+}$ has constant scalar curvature ${-n(n+1)}$, one has the following identity

$\displaystyle J_g+\rho^{-1}\Delta_g \rho=\frac{n+1}{2}\rho^{-2}(|\nabla\rho|_g^2-1).$

If ${\rho}$ is the geodesic defining function, that is ${|\nabla \rho|_g=1}$, then

$\displaystyle E(\rho)=\frac{m+n-1}{2}J_g\rho^m,\quad J_{\phi}^m=J_g.$

Therefore, in this case the two error term are the same. Consequently the equations of extension are the same.

It seems for general defining function, the two ways of extension are different.

## Convex function, Legendre transform and support function

Suppose ${u}$ is a (smooth) locally strictly convex function on ${\mathbb{R}^n}$, ${\phi}$ is the Legendre transform of ${u}$. Namely, suppose ${u=u(x)}$ and ${y=Du(x)}$, then

$\displaystyle \phi(y)=x\cdot Du(x)-u(x),\quad x=D\phi(y)$

It follows that the Hessian ${D^2u=(D^2\phi)^{-1}}$. One can verify that the Legendre transformation of ${\phi}$ is ${u}$

Now consider the graph of ${u}$ by ${\Sigma=(x,u(x))}$. The downward unit normal of ${\Sigma}$ is

$\displaystyle \nu(x)=\frac{(Du,-1)}{\sqrt{1+|Du|^2}}=\frac{(y,-1)}{\sqrt{1+|y|^2}}.$

One can think of ${\nu}$ as the Gauss map of ${\Sigma}$ to ${\mathbb{S}^n}$. Denote ${z=\nu(x)}$ the coordinates in ${\mathbb{S}^n}$ and ${X=(x,u(x))}$ is the position vector of ${\Sigma}$. Then the support function of Gaussian mapping is

$\displaystyle w(z)=X\cdot \nu=\frac{\phi(y)}{\sqrt{1+|y|^2}}.$

One can also interpret this relation as the following: treat ${\phi}$ is defined over the hyperplane ${(y,-1)\subset \mathbb{R}^{n+1}}$. Then the homogeneous degree one extension of ${w(z)}$ restricted to this plane is just ${\phi(y)=\sqrt{1+|y|^2}w(z)}$.

One also recover ${\Sigma}$ by just the support function ${w(z)}$. Since ${\Sigma}$ has nonvanishing curvature, the Gauss map is invertible. Let ${X=\nu^{-1}(z)}$ be its inverse. Then

$\displaystyle w(z)=z\cdot \nu^{-1}(z).$

It is easy to show that

$\displaystyle \overline{\nabla}_e w(z)=e\cdot \nu^{-1}(z)$

for any tangent vector of ${\mathbb{S}^n}$. Here ${\overline{\nabla}= \nabla^{\mathbb{S}^n}}$. This tells us that ${\overline{\nabla}w(z)}$ is the projection of ${\nu^{-1}(z)}$ onto the tangent plane

$\displaystyle X=\nu^{-1}(z)=w(z)z+\overline{\nabla}w(z).$

Then ${d\nu^{-1}(z)=d\overline{\nabla} w+wI}$. Therefore, the Weingarten map of ${d\nu=(d\overline{\nabla} w+wI)^{-1}}$. The eigenvalues of ${d\overline{\nabla} w+wI}$ are the reciprocals of the principle curvatures of ${\Sigma}$.

There is another way to verify this fact. Suppose the metric on ${\Sigma}$ induced from ${\mathbb{R}^{n+1}}$ is

$\displaystyle g=g_{ij}dx^idx^j,\quad g_{ij}=(\delta_{ij}+u_iu_j)$

where ${u_i=\partial_{x_i}u}$. Denote ${\mu=\sqrt{1+|Du|^2}}$. Then ${g^{ij}=\delta_{ij}-\mu^{-2}u_iu_j}$. The second fundamental form of ${\Sigma}$ is ${A_{ij}=\mu^{-1}u_{ij}}$. Then the principle curvature ${\kappa_1,\cdots,\kappa_n}$ of ${\Sigma}$ are the eigenvalues of the matrix

$\displaystyle g^{ik}A_{kj}=\frac{g^{ik}u_{kj}}{\mu}$It is equivalent to say ${\frac{1}{\kappa_1},\cdots, \frac{1}{\kappa_n}}$ are the eigenvalues of ${\mu \phi_{ik}g_{ij}=\mu\phi_{ik}(1+y_ky_j)}$

Consider the Stereographic projection from origin (see here), one can prove that the metric of ${\mathbb{S}^n}$ reads

$\displaystyle \sigma^{ij}={\mu^2}(\delta_{ij}+y_iy_j)$

and more importantly

$\displaystyle \phi_{y_iy_j}=\mu(\overline{\nabla}_{ij}w+w\sigma_{ij})$

then ${\mu \phi_{ik}g_{ij}=\sigma^{ik}\phi_{kj}\mu^{-1}=\overline{\nabla}^i_jw+w\delta^i_j}$. This just means ${\frac{1}{\kappa_1},\cdots, \frac{1}{\kappa_n}}$ are the eigenvalues of ${d\overline{\nabla}w+wI}$.

## Conformal change of divergence

Suppose $M$ is $n-$dimensional, $\hat g=e^{2f}g$ and $\hat V=e^{-f}V$, then

$\displaystyle \text{div}_{\hat g}\hat V=e^{-f}[\text{div}_{g}V+(n-1)\langle V,f\rangle]$

## One identity in asymptotically hyperbolic manifold

Suppose ${(X^{n+1},\bar g)}$ is a smooth Riemannian manifold with boundary ${M^n}$ with ${\hat h=\bar g|_M}$. Let us denote ${\rho}$ as the geodesic defining function for ${(M,\hat h)}$. In Fermi coordinates on a neighborhood ${M\times [0,\varepsilon)}$, we have the expansion of ${\bar g}$ as

$\displaystyle \bar g=\hat h+h^{(1)}\rho+h^{(2)}\rho^2+o(\rho^2).$

For the following calculation, we shall use the indices ${1\leq i,j,k\leq n}$ and ${1\leq a,b,c\leq n+1 }$

Firstly of all, ${h^{(1)}}$ is related to the second fundamental form of ${M}$. Indeed,

$\displaystyle h_{ij}^{(1)}=\partial_\rho \bar g_{ij}=\partial_\rho\langle e_i,e_j\rangle_{\bar g}=2\langle \nabla_{\partial_\rho}e_i,e_j\rangle_{\bar g}=-2\pi_{ij}.$

Secondly, one can see ${h^{(2)}}$ through

$\displaystyle h^{(2)}_{ij}=\frac{1}{2}\partial_{\rho\rho}^2\bar g_{ij}=\langle \nabla_{\partial\rho}\nabla_{\partial_\rho}e_i,e_j\rangle+\langle\nabla_{\partial_\rho}e_i,\nabla_{\partial\rho}e_j\rangle$

$\displaystyle =\langle R(\partial_\rho,e_i)\partial_\rho,e_j\rangle+\langle\nabla_{e_i}\partial_\rho,\nabla_{\partial\rho}e_j\rangle=-R_{i\rho j\rho}[\bar g]+\pi_{ik}\pi^k_j.$

here we have use ${\nabla^2_{\partial\rho}e_i=R(\partial_\rho,e_i)\partial_\rho}$. Therefore,

$\displaystyle \bar g^{ij}h_{ij}^{(2)}=-Ric_{\rho\rho}[\bar g]+||\pi||^2$

## Non-Holder continuous solution

Consider the following elliptic equation on $\mathbb{R}^2$

$\displaystyle \partial_{x_i}\left(\frac{1}{r^2(\log r)^2}\partial_{x_j}u\right)=0,\quad r^2=x_1^2+x_2^2<1$

It has a solution $u=(\log r^{-1})^{-1}$ which is not holder continuous around origin. Note that the above equation is not uniformly elliptic.

See the paper, Trudinger, on the regularity of generalized solutions of linear, non-unformly elliptic equations, 1970.

Another example is $w(x_1,\cdots, x_n)=|x_1|\log^2\frac{1}{|x_1|}$ and the function $u(x_1,\cdots, x_n)=\frac{\text{sign } x_1}{\log (1/|x_1|)}$ satisfy the elliptic equation $\partial_i(w(x)\partial_j u)=0$ in $\mathbb{R}^n$. This equation is degenerate elliptic. But $w$ is not an $A_2$ weight.

Check the paper Fabes Kenig and Serapioni, the local regularity of solutions of degenerate elliptic equations. 1982.

## 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 ${B}$

## Green’s function and parametrix

Suppose ${(M^n,g)}$ is a Riemannian manifold without boundary with ${n\geq 2}$. Consider the Beltrami-Laplacian on ${M}$

$\displaystyle \Delta_g f= |g|^{-\frac{1}{2}}\partial_i[|g|^{\frac12}g^{ij}\partial_jf]$

We want to find the Green’s function ${G}$ for ${\Delta}$, namely

$\displaystyle \Delta^{dist}_{g,y} G(x,y)=\delta_x(y)$

in distribution sense, or equivalently for any ${\psi,\phi\in C^2(M)}$

$\displaystyle \phi(x)=\int_{M}G(x,y)\Delta_g \phi(y)dv_y.$

$\displaystyle \psi(y)=\Delta_{g,y}\int_M G(x,y)\psi(x)dv_x$

Moreover ${\psi\rightarrow \int_M G\psi dv}$ looks like the inverse operator of ${\Delta_g}$. Of course such ${G}$ would only be unique modulo some constants.

To do that, we know on ${(\mathbb{R}^n,|dx|^2)}$, the Green function is

$\displaystyle K(x,y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}& n>2\\\frac{1}{2\pi}\log|x-y|,&n=2.\end{cases}$

In a general Riemannian manifold, ${G}$ should look like ${K}$ more or less.

Suppose ${M}$ has injectivity radius ${\delta}$ and ${\eta}$ is a smooth radial cut off function on ${\mathbb{R}^n}$ whose support lies in ${B_\delta(0)}$ and ${\eta=1}$ in ${B_{\delta/2}(0)}$ . Then ${K(x,y)\eta(|x-y|)}$ is well defined on ${M}$ through local coordinate patches, which we still deonte as ${K}$. Easy computation shows

$\displaystyle |\Delta_{g,y}K(x,y)|\leq C|x-y|^{2-n}\in L^1(M).$

$\displaystyle \psi(y)=\int_M K(x,y)\Delta \psi(y)dv_y-\int_M\Delta_yK(x,y)\psi(y)dv_y$

for all ${\psi\in C^2(M)}$. This actually means

$\displaystyle \Delta_{g,y}^{dist} K(x,y)=\Delta_{g,y}K(x,y)+\delta_x(y)$

and

$\displaystyle \int_M \Delta_{g,y}K(x,y)dv_y=-1.$

After changing the order of integration in the Green’s formula, we could establish that

$\displaystyle \phi(y)=\Delta_{g,y}\int_M K(x,y)\phi(x)dv_x-\int_M \Delta_{g,y}K(x,y)\phi(x)dv_x$

for any ${\phi\in C^2(M)}$. One can compare this with the equation ${G}$ satisfy. Denote ${\Gamma_1=\Delta_{g,y}K(x,y)\in L^1}$. Define two operators

$\displaystyle Q\phi(y)=\int_M K(x,y)\phi(x)dv_x$

$\displaystyle R\phi(y)=\int_{M}\Gamma_1(x,y)\phi(x)dv_x$

Also denote ${\Gamma_{k+1}=R\Gamma_k}$. Then Green’s formula means

$\displaystyle Q\Delta_g=\Delta_g Q=I-R$

To find the Green’s function, it is equivalent to find the inverse operator of ${\Delta_g}$. One can apply the idea of parametrix, which says the inverse operator can be constructed by

$\displaystyle (I-R)^{-1}Q=Q+RQ+R^2Q+\cdots$

Therefore if we define ${\Gamma_k=}$

$\displaystyle G(x,y)=K(x,y)+\sum \int_{M}\Gamma_k(x,z) K(z,y)dv_z+remainder$

## Some calculation on asymptotic analysis of Poincare Einstein manifold

Suppose ${\gamma\in (1,2)}$ and ${m=3-2\gamma}$. Let ${(X^{n+1}, M^n,g_+)}$ be a Poincare-Einstein manifold and ${r}$ be the geodesic defining function. Suppose ${\rho}$ be another defining function for ${M}$ such that, asymptotically near ${M}$

$\displaystyle \rho=r+\rho_2r^3+\Phi r^{1+2\gamma}+o(r^{1+2\gamma}).$

Denote ${g_0=r^2g_+}$ and ${g=\rho^2g_+}$. Then ${g=(\frac{\rho}{r})^2g_0}$. It is easy to get

$\displaystyle |\nabla^0\rho|_0^2=1+6\rho_2r^2+2(1+2\gamma) \Phi r^{2\gamma}+o(r^{2\gamma})$

then

$\displaystyle |\nabla^g\rho|_g^2=(\frac{r}{\rho})^2|\nabla^0\rho|_0^2=1+4\rho_2r^2+4\gamma\Phi r^{2\gamma}+o(r^{2\gamma}).$

$\displaystyle \rho^{-2}(|\nabla^g\rho|_g^2-1)=4\rho_2+4\gamma\Phi r^{2\gamma-2}+o(r^{2\gamma-2}).$

Consequently

$J+\rho^{-1}\Delta_g\rho=2(n+1)(\rho_2+\gamma\Phi r^{2\gamma-2})+o(r^{2\gamma-2})$

Now let us see the expansion of ${\Delta_{g_0}r}$ and ${\Delta_g \rho}$. To do that, we need to formula of laplacian

$\displaystyle \Delta_g f=\frac{1}{\sqrt{|g|}}\partial_i(\sqrt{|g|}g^{ij}f_j).$

which works for any local coordinates. Since ${g_0=dr^2+h_r}$ near ${M}$ and ${h_r=h+h_2r^2+o(r^2)}$, then ${\sqrt{|g_0|}=\sqrt{|h_r|}}$ and

$\displaystyle \Delta_0 r=\frac{1}{\sqrt{|h_r|}}\partial_r(\sqrt{|h_r|})=\frac12tr_h(\partial_r h_r)=(tr_h h_2)r+o(r).$

Remark 1 We have ${tr_hh_2=-\bar J}$

On the other hand, we have ${g=(\frac{\rho}{r})^2g_0}$. Consequently ${\sqrt{|g|}=(\frac{\rho}{r})^{n+1}\sqrt{|h_r|}}$ and

$\Delta_g\rho=\frac{1}{\sqrt{|g|}}\partial_r(\sqrt{|g|}g^{rr}\partial_r \rho)=(\frac{r}{\rho})^{n+1}\frac{1}{\sqrt{|h_r|}}\partial_r(\sqrt{|h_r|}(\frac{\rho}{r})^{n-1}\partial_r\rho)$
$=(\frac{r}{\rho})^2\partial_r\rho\frac{1}{\sqrt{|h_r|}}\partial_r(\sqrt{|h_r|})+(\frac{r}{\rho})^{n+1}\partial_r((\frac{\rho}{r})^{n-1}\partial_r\rho)$
$=(\frac{r}{\rho})^2\partial_r\rho [(tr_hh_2) r+o(r)]+(\frac{r}{\rho})^{n+1}\partial_r((\frac{\rho}{r})^{n-1}\partial_r\rho)$

Then

$\rho^{-1}\Delta_{g}\rho=(\frac{r}{\rho})^3\partial_r\rho [(tr_hh_2) +o(1)]+(\frac{r}{\rho})^{n+2}\frac{1}{r}\partial_r((\frac{\rho}{r})^{n-1}\partial_r\rho)$

It follows from the expansion of ${\rho}$ that

$\displaystyle \partial_r\rho=1+3\rho_2r^2+(2\gamma+1)\Phi r^{2\gamma}+o(r^{2\gamma})$

$\displaystyle \frac{r}{\rho}=1-\rho_2r^2-\Phi r^{2\gamma}+o(r^{2\gamma})$

$\displaystyle \frac{1}{r}\partial_r[(\frac{\rho}{r})^{n-1}\partial_r\rho]=2(n+2)\rho_2+2\gamma(n+2\gamma)\Phi r^{2\gamma-2}$

Putting all these back to $\Delta_g\rho$, it yields

$\rho^{-1}\Delta_g\rho=tr_hh_2+2(n+2)\rho_2+2\gamma(n+2\gamma)\Phi r^{2\gamma-2}+o(r^{2\gamma-2})$

Recall the equation of $J+\Delta_g\rho$, it leads to

$\displaystyle J=-tr_hh_2-2\rho_2+2\gamma(1-2\gamma)\Phi r^{2\gamma-2}+o(r^{2\gamma-2}).$

Since ${J_\phi^m=J-\frac{m}{n+1}(J+\rho^{-1}\Delta_g\rho)}$, then

$\displaystyle J_\phi^m=-tr_hh_2-2(m+1)\rho_2-4\gamma\Phi r^{2\gamma-2}+o(r^{2\gamma-2}).$

Moreover, since ${\eta=-\frac{\rho}{r}\partial_r}$, because ${|\eta|_g^2=(\frac{\rho}{r})^2|\partial_r|_g^2=|\partial_r|^2_0=1}$, then

$\displaystyle \rho^{m}\eta J_\phi^m=-\rho^{m}\frac{\rho}{r}\partial_rJ_\phi^m=8\gamma(\gamma-1)\Phi +o(1)$

One can use the (3-5) in \cite{Case2017} and ${P(\eta,\eta)=-2\rho_2}$ on ${M}$ to get

$\displaystyle T_2^{2\gamma}=\frac{2-\gamma}{\gamma-1}[\bar J+4(\gamma-1)\rho_2]+o(1).$

Remark:

Jeffrey S Case. Some energy inequalities involving fractional GJMS operators. Analysis and PDE , 10(2):253–280, 2017.

Jeffrey S Case and Sun-Yung Alice Chang. On Fractional GJMS Operators. Communications on Pure and Applied Mathematics , 69(6):1017–1061, 2016

Many thanks to the help of Jeffrey S Case. The dimension of $X$ is $n+1$ and $P(\eta,\eta)=-2\rho_2$.

## A negative gradient flow under constraint

Suppose we have two $C^1$ functionals $J$ and $L$ on a Hilbert space $H$. Suppose we want to have a flow $u(t)$ such that along the flow $J(u(t))$ decrease and $L(u(t))$ remains the same.

Suppose $\nabla L(u)\neq 0$, for any $u\in H$, which means that $\mathcal{C}=\{u:L(u)=C\}$ is a regular hypersurface of $H$. Define the unit normal vector field

$\displaystyle N(u)=\frac{\nabla u}{||\nabla u||_H}$

and

$\displaystyle \nabla^{\mathcal{C}}J=\nabla J-\langle \nabla J,N \rangle N$

Then it is easy to the flow $\partial_t u=-\nabla^{\mathcal{C}}J$ will satisfy our request.

For example let $P_{g_0}$ be the Paneitz operator of fourth order. Assume it is a positive operator. Define

$H=W^{2,2}\text{ with norm }||u||_H^2=\int uP_{g_0}ud\mu_0$

Choose our $L(u)=||u||_H$ and $J(u)=-\frac{n-4}{2n}\int |u|^{\frac{2n}{n-4}}d\mu_0$

then $\nabla L(u)=u$ and $\nabla J(u)=-P_{g_0}^{-1}|u|^{\frac{n+4}{n-4}}$. Consequently

$\nabla^{\mathcal{C}}J=-P_{g_0}^{-1}|u|^{\frac{n+4}{n-4}}+\frac{\int |u|^{\frac{2n}{n-4}}d\mu_0}{||u||_H^2}u$

Then one may have the following flow

$\displaystyle \partial_t u=-u+\mu P_{g_0}^{-1}(|u|^{\frac{n+4}{n-4}})$

where $\mu=||u||_H^2/\int |u|^{\frac{2n}{n-4}}d\mu_0$.

P. Baird, A. Fardoun, R. Regbaoui, Calculus of Variations Q-curvature flow on 4-manifolds, 27 (2006) 75–104.

M. Gursky, A. Malchiodi, A strong maximum principle for the Paneitz operator and a non-local flow for the $Q$-curvature, J. Eur. Math. Soc. 17 (2015) 2137–2173.

## Some calculation of sigma2 II

We want find the critical metric of the following functional restricted to the space of conformal metrics of unit volume

$\displaystyle I(g)=\frac{n-4}{2}t\int_M J^2_gd\mu_g+4\int_M\sigma_2(A_g)d\mu_g$

Here ${(M^n,g)}$ is a Riemannian metric with ${n\geq 5}$, ${A}$ is the Schouten tensor and ${J}$ is ${tr_gA}$.

Now let us differentiate the two terms respectively, suppose ${g(s)=(1+s\psi)^{\frac{4}{n-4}}g=u(s)^{\frac{4}{n-2}}g}$. From conformal change property of scalar curvature, we get

$\displaystyle J_{g(s)}=u(s)^{-\frac{n+2}{n-2}}\left(-\frac{2}{n-2}\Delta_{g}u(s)+Ju(s)\right)$

$\displaystyle \dot J=-\frac{2}{n-2}\Delta_g\dot u-\frac{4}{n-2}J\dot u$

Then

$\displaystyle \frac{d}{ds}|_{t=0}\int_{M}J_{g(s)}^2d\mu_{g(s)}=\int_{M}2J\dot{J}+\frac{2n}{n-2}J^2\dot ud\mu$

$\displaystyle =\int_M -\frac{4}{n-2}J\Delta \dot u+\frac{2(n-4)}{n-2}J^2\dot ud\mu$

$\displaystyle =\int_M\left(-\frac{4}{n-4}\Delta J+2J^2\right)\psi d\mu$

where we have used the fact ${\dot u(0)=\frac{n-2}{n-4}\psi}$. Next, it is easy to know

$\displaystyle \frac{d}{ds}|_{t=0}\int_M \sigma_2(A_{g(s)})d\mu_{g(s)}=2\int_{M}\sigma_2(A)\psi d\mu$So

$\displaystyle I'(g)\psi=2t\int_{M}\left(-\Delta J+\frac{n-4}{2}J\right)\psi d\mu_g+8\int_{M}\sigma_2(A)\psi d\mu_g$

Now assume ${g(s)}$ keep the unit volume infinitesimally, that is ${\int_M \psi d\mu=0}$, then ${I'(g)=0}$ under this constraint means

$\displaystyle t\left(-\Delta J+\frac{n-4}{2}J\right)+4\sigma_2(A)=const.$

Remark: M.J. Gursky, F. Hang, Y.-J. Lin, Riemannian Manifolds with Positive Yamabe Invariant and Paneitz Operator, Int. Math. Res. Not.