## Category Archives: Riemannian Geo

### 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 ${B}$

### 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 $P(\eta,\eta)=-2\rho_2$.

### Surface intersection with a ball

Suppose $M$ is 2-dimensional surface in $\mathbb{R}^3$. If $M$ is simply connected, then $B\cap M$ may not be simply connected, where $B$ is ball in $\mathbb{R}^3$. See the following figure.

But if $M$ is minimal surface, then $M\cap B$ must be simply connected. The reason is $M$ has convex haul property.

Learn this example from Jacob Bernstein.

Suppose ${u=u(r)}$ is a radial function on ${\mathbb{R}^n}$, here ${r=|x|}$.

$\displaystyle u_{x_i}=u'\frac{x_i}{r}$

$\displaystyle u_{x_ix_j}=u''\frac{x_ix_j}{r^2}+u'(\frac{\delta_{ij}}{r}-\frac{x_ix_j}{r^3})=\frac{u'}{r}\delta_{ij}+(\frac{u''}{r^2}-\frac{u'}{r^3})x_ix_j$

therefore

$\displaystyle \det D^2u = \left(\frac{u'}{r}\right)^{n}\det[ \delta_{ij}+\frac{r}{u'}(u''-\frac{u'}{r})\frac{x_ix_j}{r^2}]$

$\displaystyle =\left(\frac{u'}{r}\right)^{n}[1+\frac{r}{u'}(u''-\frac{u'}{r})]=\left(\frac{u'}{r}\right)^{n-1}u''$

If we use the polar coordinates ${(r,\theta_1,\cdots, \theta_{n-1})}$, and ${g=dr^2+r^2\sum_{i=1}^{n-1}d\theta_i^2}$ and the following fact

$\displaystyle \nabla_X\partial_r=\begin{cases}\frac{1}{r}X&\text{if } X\text{ is tangent to }\mathbb{S}^{n-1}\\0 \quad &\text{if } X=\partial_r\end{cases}$

then one can calculate the Hessian of ${u}$ under this coordinates

$\displaystyle Hess (u)(\partial_r,\partial_r)=u''$

$\displaystyle Hess (u)(\partial_{\theta_i},\partial_r)=0$

$\displaystyle Hess (u)(\partial_{\theta_i},\partial_{\theta_j})=ru'\delta_{ij}.$

Then

$\displaystyle \frac{\det Hess (u)}{{\det g}}=\left(\frac{u'}{r}\right)^{n-1}u''$

If the metric is $g=dr^2+\phi^2ds_{n-1}^2$, then we will have

$\displaystyle \frac{\det Hess (u)}{{\det g}}=\left(\frac{u'\phi'}{\phi}\right)^{n-1}u''$

### Unit normal to a radial graph over sphere

Consider $\Omega\subset \mathbb{S}^n$ is a domain in the sphere. $S$ is a radial graph over $\Omega$.

$\boldsymbol{F}(x)=\{v(x)x:x\in \Omega\}$

What is the unit normal to this radial graph?

Suppose $\{e_1,\cdots,e_n\}$ is a smooth local frame on $\Omega$. Let $\nabla$ be the covariant derivative on $\mathbb{S}^n$. Tangent space of $S$ consists of $\{\nabla_{e_i}\boldsymbol{F}\}_{i=1}^n$ which are

$\nabla_{e_i}\boldsymbol{F}=v(x)e_i+e_i(v)\cdot x$

In order to get the unit normal, we need some simplification. Let us assume $\{e_i\}$ are orthonormal basis of the tangent space of $\Omega$ and $\nabla v=e_1(v)e_1$. Then

$\nabla_{e_1}\boldsymbol{F}=v(x)e_1+e_1(v)x, \quad \nabla_{e_i}\boldsymbol{F}=v(x)e_i, \quad i\geq 2$

Then we obtain an orthonormal basis of the tangent of $S$

$\{\frac{1}{\sqrt{v^2+|\nabla v|^2}}\nabla_{e_1}\boldsymbol{F},\nabla_{e_2}\boldsymbol{F},\cdots,\nabla_{e_n}\boldsymbol{F}\}$

We are able to get the normal by projecting $x$ to this subspace

$\nu=x-\frac{1}{v^2+|\nabla v|^2}\langle x,\nabla_{e_1}\boldsymbol{F}\rangle\nabla_{e_1}\boldsymbol{F}=\frac{v^2x-v\nabla v}{v^2+|\nabla v|^2}.$

After normalization, the (outer)unit normal can be written

$\frac{vx-\nabla v}{\sqrt{v^2+|\nabla v|^2}}$

Remark: Guan, Bo and  Spruck, Joel. Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.

### Laplacian on graph

Suppose ${\Omega\subset \mathbb{R}^n}$ is a domain and ${\Sigma=graph(F)\subset \mathbb{R}^{n+1}}$ is a hypersurface, where ${F=F(u_1,\cdots,u_n)}$ is a function on ${\Omega}$. Define ${f=f(u_1,\cdots,u_n)}$ on ${\Omega}$. Then ${f}$ also can be considered as a function on ${\Sigma}$. How do we understand ${\Delta_\Sigma f}$?

Denote ${\partial_i=\frac{\partial}{\partial{u_i}}}$ for short. If we pull the metric of ${\mathbb{R}^{n+1}}$ back to ${\Omega}$, denote as ${g}$, then

$\displaystyle g_{ij}=g(\partial_i,\partial_j)=\delta_{ij}+F_{u_i}F_{u_j},\quad g^{ij}=\delta_{ij}-\frac{F_{u_i}F_{u_j}}{W^2},\quad \det g=W^2$

where ${W=\sqrt{1+|\nabla F|^2}}$ and ${F_{u_i}=\frac{\partial F}{\partial u_i}}$. Then one can use the local coordinate to calculate ${\Delta_\Sigma f}$

$\displaystyle \Delta_\Sigma f=\frac{1}{\sqrt{\det g}}\partial_{u_i}\left(\sqrt{\det g}\,g^{ij}f_{u_j}\right)$

$\displaystyle =g^{ij}f_{u_iu_j}+f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}$

also one can see from another definition of Laplacian

$\displaystyle \Delta_\Sigma f=g^{ij}Hess(f)(\partial_i,\partial_j)=g^{ij}[\partial_j(\partial_i(f))-(\nabla_{\partial_j}\partial_i)f]$

$\displaystyle =g^{ij}f_{u_iu_j}-g^{ij}(\nabla_{\partial_j}\partial_i)f$

By using the expression of ${g^{ij}}$ stated above, we can calculate

$\displaystyle \frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=\frac{-F_{u_j}F_{u_{i}u_{k}}g^{ik}}{W^2}$

It follows from the definition of tangential derivative on ${\Sigma}$, see, that

$\displaystyle \langle\nabla^\Sigma f, e_{n+1}\rangle=g^{ij}f_{u_{i}}F_{u_{j}}=f_{u_{i}}F_{u_{i}}-\frac{f_{u_i}F_{u_i}|\nabla F|^2}{W^2}=\frac{f_{u_{i}}F_{u_{i}}}{W^2}$

then

$\displaystyle f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=-\langle \nabla^\Sigma f,e_{n+1}\rangle F_{u_iu_k}g^{ik}=-\langle \nabla^\Sigma f,e_{n+1}\rangle HW$

where ${H}$ is the mean curvature of the ${\Sigma}$. Combining all the above calculations,

$\displaystyle \Delta_\Sigma f=g^{ij}f_{u_iu_j}-\langle \nabla^\Sigma f,e_{n+1}\rangle HW$

### Stereographic projection from center

Suppose we are doing stereographic projection at the center. Namely consider the following map

$\displaystyle \phi:\mathbb{S}^2\rightarrow \mathbb{R}^2$

$\displaystyle (x,y,z)\mapsto\frac{(u_1,u_2,-1)}{\lambda}$

where ${\lambda=\sqrt{u_1^2+u_2^2+1}}$. Then one can see ${u_1=\frac{x}{z}}$, ${u_2=\frac{y}{z}}$. Under this stereographic projection, a strip will be equivalent to some lens domain on the sphere.

Let us pull the standard metric of ${\mathbb{S}^2}$ to the ${\mathbb{R}^2}$. For the following statement, we will always omit ${\phi^*}$ and ${\phi_*}$. Calculation shows,

$\displaystyle dx=\left(\frac{1}{\lambda}-\frac{u_1^2}{\lambda^3}\right)du_1-\frac{u_1u_2}{\lambda^3}du_2=z(-1+x^2)du_1+xyzdu_2$

$\displaystyle dy=-\frac{u_1u_2}{\lambda^3}du_1+\left(\frac{1}{\lambda}-\frac{u_2^2}{\lambda^3}\right)du_2=xyzdu_1+z(-1+y^2)du_2$

$\displaystyle dz=\frac{1}{\lambda^3}(u_1du_1+u_2du_2)=z^2(xdu_1+ydu_2)$

therefore

$\displaystyle dx^2+dy^2+dz^2=z^2(1-x^2)du_1^2-2xyz^2du_1du_2+z^2(1-y^2)du^2_2$

$\displaystyle =\frac{1}{\lambda^2}(\delta_{ij}-\frac{u_iu_j}{\lambda^2})du_idu_j$

here we used the fact that ${(x,y,z)\in \mathbb{S}^2}$. Suppose ${\bar\nabla}$ is the connection on ${\mathbb{S}^2}$ equipped with the standard metric. We want to calculate ${\bar \nabla_{\partial_{u_i}}\partial_{u_j}}$. To that end, it is better to use the ${x,y,z}$ coordinates in ${\mathbb{R}^3}$

$\displaystyle \partial_{u_1}=z(x^2-1)\partial_x+xyz\partial_y+xz^2\partial_z$

$\displaystyle \partial_{u_2}=xyz\partial_x+z(y^2-1)\partial_y+yz^2\partial_z$

Since we know

$\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=\left(\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}\right)^T=\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}-\langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle\partial_r$

where ${T}$ means the tangential part to ${\mathbb{S}^2}$ and ${\partial_r=x\partial_x+y\partial_y+z\partial_z}$ is the unit normal to ${\mathbb{S}^2}$. Using the connection in ${\mathbb{R}^3}$, we get

$\displaystyle \nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}=z^2\left[3x(x^2-1)\partial_x+y(3x^2-1)\partial_y+z(3x^2-1)\partial_z\right]$

$\displaystyle \langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle=z^2(x^2-1)$

One can verify from the above equalities that

$\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=2xz\partial_{u_1}$

Similarly

$\displaystyle \bar \nabla_{\partial_{u_1}}\partial_{u_2}=yz\partial_{u_1}+xz\partial_{u_2}$

$\displaystyle \bar\nabla_{\partial u_2}\partial_{u_2}=2yz\partial_{u_2}$

### Scalar curvature of cylinder

Consider the scalar curvature ${R_g}$ of a cylinder ${\mathbb{R}\times \mathbb{S}^{n-1}}$ with the standard product metric ${g_{prod}}$, where ${n\geq 2}$.

First approach, the cylinder is diffeomorphic to the puncture plane,

$\displaystyle \phi:\mathbb{R}^n\backslash\{0\}\longmapsto \mathbb{R}\times \mathbb{S}^{n-1}$

$\displaystyle x\rightarrow (\log|x|,\frac{x}{|x|})$

It is easy to verify that ${(\phi^{-1})^*g_{prod}=|x|^{-2}|dx|^2}$. One can calculate ${R_g}$ from the conformal change formula

$\displaystyle R_g=-\frac{4(n-1)}{n-2}|x|^{-\frac{n+2}{2}}\Delta |x|^{-\frac{n-2}{2}}=(n-2)(n-1)$

Second approach, remember the Ricci curvature splits on product manifold with product metric. Therefore

$\displaystyle R_g=R_{g_{S^n}}=(n-2)(n-1)$

From the above, we know that if ${n=2}$, then ${R_g=0}$ while ${R_g>0}$ if ${n\geq 3}$. I always get confused with this two cases. There is a nice way to see this result for ${n=2}$. ${\mathbb{S}^1\times \mathbb{R}}$ can be covered locally isometrically by ${\mathbb{R}^2}$ with Euclidean metric. Therefore ${R_g=0}$ in this case. However, ${\mathbb{R}^n}$ should cover ${\mathbb{S}^1\times \mathbb{R}^{n-1}}$ instead of ${\mathbb{R}\times \mathbb{S}^{n-1}}$.

### Cylinder, Puncture plane and Cone

Suppose $\phi: \mathbb{R}^n-\{0\}\longmapsto \mathbb{R}\times S^{n-1}$ defined by $\phi(x)=(\log|x|,\frac{x}{|x|})$. Use $(t,\xi)$ denote the points on $\mathbb{R}\times S^{n-1}$, then

$\displaystyle dt=\frac{x_i}{|x|^2}dx^i,\quad d\xi^i=\frac{dx^i}{|x|}-\frac{x_ix_j}{|x|^3}dx^j$

This means

$\displaystyle dt^2+d\xi^2=\frac{1}{|x|^2}|dx|^2$

then $\phi$ is a conformal diffeomorphism from $(\mathbb{R}^n-\{0\},|dx|^2)$ to $(\mathbb{R}\times S^{n-1}, dt^2+d\xi^2)$. Suppose $g$ is a conic metric such that $g(x)=|x|^\beta|dx|^2$ on $\mathbb{R}^n-\{0\}$. Thus $(\mathbb{R}^n-\{0\},g)$ is isometric to $(\mathbb{R}\times S^{n-1},e^{t(2+\beta)}(dt^2+d\xi^2))$ through $\phi$. Therefore $Y(\mathbb{R}^n-\{0\},[g])=Y(\mathbb{R}\times S^{n-1},[dt^2+d\xi^2])$. It is easy to see that $Y(\mathbb{R}\times S^{n-1},[dt^2+d\xi^2])=Y(S^{n})$ because

$\displaystyle g_{S^n}=\frac{4}{(1+|x|^2)}|dx|^2=(\cosh t)^{-2}(dt^2+d\xi^2)$

### Bach flat four dimensional manifold and sigma2 functional

We want to find the necessary condition of being the critical points of ${\int\sigma_2}$ on four dimensional manifold.

1. Preliminary

Suppose ${(M^n,g)}$ is a Riemannian manifold with ${n=4}$. ${P_g}$ is the Schouten tensor

$\displaystyle P_g=\frac{1}{n-2}\left(Ric-\frac{R}{2(n-1)}g\right)$

and denote ${J=\text{\,Tr\,} P_g}$. Define

$\displaystyle \sigma_2(g)=\frac{1}{2}[(\text{\,Tr\,} P_g)^2-|P_g|_g^2]$

$\displaystyle I_2(g)=\int_M \sigma_2(g)d\mu_g$

where ${|P|_g^2=\langle P,P\rangle_g}$. It is well known that ${I_2(g)}$ is conformally invariant.

Suppose ${g(t)=g+th}$ where ${h}$ is a symmetric 2-tensor. We want to calculate the first derivative of ${I_2(g(t))}$ at ${t=0}$. To that end, let us list some basic facts (see the book of Toppings). Firstly denote ${(\delta h)_j=-\nabla^i{h_{ij}}}$ the divergence operator and

$\displaystyle G(h)=h-\frac{1}{2}(\text{\,Tr\,} h) g$

$\displaystyle (\Delta_L h)_{ij}=(\Delta h)_{ij}-h_{ik}Ric_{jl}g^{kl}-h_{jk}Ric_{il}g^{kl}+2R_{ikjl}h^{kl}$

where ${\Delta_L}$ is the Lichnerowicz Laplacian. Then the first variation of Ricci curvature and scalar curvature are

$\displaystyle \dot{R}=\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle \ \ \ \ \ (1)$

$\displaystyle \dot{Ric}=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta G(h))^\sharp}g=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta h)^\sharp}g-\frac{1}{2}Hess(\text{\,Tr\,} h)$

$\displaystyle =-\frac{1}{2}\Delta_Lh-d(\delta h)-\frac{1}{2}Hess(\text{\,Tr\,} h)$

where we were using upper dot to denote the derivative with respect to ${t}$.

2. First variation of the sigma2 functional

Lemma 1 ${(M^4,g)}$ is a critical point of ${I_2(g)}$ if and only if

$\displaystyle \Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g=0 \ \ \ \ \ (2)$

where ${(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}$.

Proof:

$\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\int_M J\dot J-\langle\dot P,P\rangle+\langle h,P\wedge P\rangle+\frac{1}{2}\sigma_2\text{\,Tr\,} h \,d\mu_g$

where ${(P\wedge P)_{ij}=P_{ik}P_{jl}g^{kl}}$. Since we have

$\displaystyle \int_M\langle P,\dot P\rangle\\ =\frac{1}{n-2}\int_M\langle P, \dot Ric-\dot Jg-Jh\rangle =\frac{1}{n-2}[\langle P, \dot Ric\rangle-\dot J J-J\langle h,P\rangle]$

$\displaystyle =\frac{1}{n-2}[-\frac{1}{2}\langle h,\Delta_L P\rangle+\langle h,Hess(J)\rangle-\frac{1}{2}\Delta J \text{\,Tr\,} h-\dot J J-J\langle h,P\rangle]$

Plugging this into the derivative of ${I_2}$ to get

$\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t))\\ =\int_M\frac{1}{2}\langle h,\Delta_L P\rangle-\langle h,Hess(J)\rangle+\frac{1}{2}\Delta J \text{\,Tr\,} h\\$

$\displaystyle \quad +(n-1)\dot J J+J\langle h,P\rangle+(n-2)\langle h,P\wedge P\rangle+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g$

In order to simplify the above equation, we recall the definition of Lichnerowicz Laplacian ${\Delta_L}$

$\displaystyle (\Delta_LP)_{ij}=(\Delta P)_{ij}-2P_{ik}Ric_{jl}g^{kl}+2R_{ikjl}P^{kl}$

$\displaystyle =(\Delta P)_{ij}-2(n-2)P_{ik}P_{jl}g^{kl}-2JP_{ij}+2R_{ikjl}P^{kl}$

Apply (1) to get

$(n-1)\dot J J=\frac12J\dot R=\frac12J[\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle]$
$=\frac{1}{2}[\langle h, Hess(J)\rangle-\text{\,Tr\,} h\Delta J-(n-2)J\langle h, P\rangle-J^2\text{\,Tr\,} h]$

Therefore we can simplify it to be

$\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t)) =\int_M\frac{1}{2}\langle h,\Delta P\rangle-\frac{1}{2}\langle h, Hess(J)\rangle+h^{ij}R_{ikjl}P^{kl}$

$\displaystyle -\frac{n-2}{2}J\langle h,P\rangle-\frac{1}{2} J^2 \text{\,Tr\,} h+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g$

Let us denote ${(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}$. Using the fact ${n=4}$ and the definition of ${\sigma_2}$,

$\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g$

where

$\displaystyle Q=\Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g$

$\Box$

Remark 1 It is easy to verify ${\text{\,Tr\,} Q=0}$, this is equivalent to say ${I_2}$ is invariant under conformal change. More precisely, letting ${h=2ug}$, then

$\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g=\frac{1}{2}\int_M u\text{\,Tr\,} Q d\mu_g=0.$

Remark 2 If ${g}$ is an Einstein metric with ${Ric=2(n-1)\lambda g}$, then ${P=\lambda g}$, ${J=n\lambda}$ and

$\displaystyle \mathring{Rm}(P)=\lambda Ric=2(n-1)\lambda^2 g$

It is easy to verify that ${Q=0}$. In other words, Einstein metrics are critical points of ${I_2}$.

Are there any non Einstein metric which are critical points of ${I_2}$?

Here is one example. Suppose ${M=\mathbb{S}^2\times N}$, where ${\mathbb{S}^2}$ is the sphere with standard round metric and ${(N,g_N)}$ is a two dimensional compact manifolds with sectional curvature ${-1}$. ${M}$ is endowed with the product metric. We can prove ${Ric=g_{S^2}-g_N}$, ${P=\frac{1}{2}g_{S^2}-\frac{1}{2}g_N}$, ${J=0}$, ${\mathring{Rm}(P)=g_{prod}}$ and consequently ${Q=0}$.

Note that the above example is a locally conformally flat manifold. For this type of manifold, we have the following lemma which can say

Lemma 2 Suppose ${g}$ is locally conformally flat and ${Q=0}$, then

Proof: When ${g}$ is locally conformally flat,

$\displaystyle \mathring{Rm}(g)=JP+|P|_g^2g-2P\wedge P$

${Q=0}$ is equivalent to

$\displaystyle \Delta P-Hess(J)+|P|_g^2g-4P\wedge P=0$

Actually this is equivalent to the Bach tensor ${B}$ is zero. $\Box$

3. Another point of view

We have the Euler Characteristic formula for four dimensional manifolds

$\displaystyle 8\pi^2\chi(M)=\int_M (|W|_g^2+\sigma_2) d\mu_g$

therefore the critical points for ${\int_M \sigma_2d\mu_g}$ will be the same as the critical points of ${\int_M |W|_g^2d\mu_g}$. However, the functional

$\displaystyle g\rightarrow \int_M |W|_g^2d\mu_g$

is well studied by Bach. The critical points of this functional satisfy Bach tensor equal to 0.

$\displaystyle B_{ij}=\nabla^k\nabla^l W_{likj}+\frac{1}{2}Ric^{kl}W_{likj}$

Obviously, ${B=0}$ for Einstein metric, but not all Bach flat metrics are Einstein. For example ${B=0}$ for any locally conformally flat manifolds.