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

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}

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.

nose

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.

 

 

Hessian of radial functions

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.

new_lens

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.