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### 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.

### Principle curvature of translator

Suppose ${\Sigma\subset\mathbb{R}^{3}}$ is a translator in ${e_{3}}$ direction. Denote ${V=e_{3}^T}$ and ${A}$ is the second fundamental form of ${\Sigma}$. Then it is well know that

$\displaystyle \Delta A+\nabla_VA+|A|^2A=0 \ \ \ \ \ (1)$

Choose a local orthonormal frame ${\{\tau_1,\tau_2\}}$ such that ${A(\tau_1,\tau_1)=\lambda}$, ${A(\tau_2,\tau_2)=\mu}$ and ${A(\tau_1,\tau_2)=0}$ in the neighborhood of some point ${p}$. We want to change (1) to some expression on ${\lambda}$ or ${\mu}$. To do that, we need to apply both sides of (1) to ${\tau_1,\tau_1}$, getting

$\displaystyle (\nabla_VA)(\tau_1,\tau_1)=\nabla_V \lambda-2A(\nabla_V \tau_1,\tau_1)=\nabla_V\lambda \ \ \ \ \ (2)$

where we have used ${\nabla_V\tau_1\perp \tau_1}$. Similarly

$\displaystyle (\nabla_{\tau_k} A)(\tau_1,\tau_1)=\nabla_{\tau_k}\lambda$

$\displaystyle (\nabla_{\tau_k}A)(\tau_1,\tau_2)=(\lambda-\mu)\langle \nabla_{\tau_k}\tau_1,\tau_2\rangle$

Now let us calculate the Laplacian of second fundamental form

$\displaystyle (\nabla_{\tau_k}^2A)(\tau_k,\tau_1)=\tau_k(\nabla_{\tau_k}\lambda)-2(\nabla_{\tau_k}A)(\nabla_{\tau_k}\tau_1,\tau_1)-(\nabla_{\tau_k}\tau_k)\lambda$

$\displaystyle =\tau_1(\nabla_{\tau_1}\lambda)-2\frac{(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}{\lambda-\mu}.$

Then

$\displaystyle (\Delta A)(\tau_1,\tau_1)=(\nabla_{\tau_k}^2A)(\tau_k,\tau_1)= \Delta \lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}.$

Combining all the above estimates to (1), we get

$\displaystyle \Delta \lambda+\nabla_V\lambda+|A|^2\lambda-2\sum_{k=1}^2\frac{|A_{12,k}|^2}{\lambda-\mu}=0.$

where we write ${A_{12,k}=(\nabla_{\tau_k}A)(\tau_1,\tau_2)^2}$. Using this equation and the other one on ${\mu}$, one can derive that if ${\Sigma}$ is mean convex then it is actually convex.

### 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,

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

### One example of blowing up corner

Consider ${u(x,y)=\sqrt{x^2+y^4}}$ on ${\mathbb{R}^2_+=\{(x,y):x\geq 0, y\geq 0\}}$. Notify ${\mathbb{R}^2_+}$ has a corner at the origin and ${u}$ is not smooth at the origin. ${u\approx x}$ when ${x\geq y^2}$ and ${u\approx y^2}$ when ${x\leq y^2}$. We want to resolve ${u}$ by blowing up the origin through a map ${\beta}$. After blowing up, ${W}$ looks like the following picture.

Denote ${W=[\mathbb{R}^2_+,(0,0)]}$ and ${\beta:W\rightarrow \mathbb{R}^2_+}$ is the blow down map. On ${W\backslash lb}$(near A), ${\beta}$ takes the form

$\displaystyle \beta_1(\xi_1,\eta_1)=(\xi_1^2,{\xi_1\eta_1})$

where ${\xi_1}$ is the boundary defining function for ff and ${\eta_1}$ is boundary defining function for rb. Similarly on ${W\backslash rb}$(near B), ${\beta}$ takes the form

$\displaystyle \beta_2(\xi_2,\eta_2)=(\xi_2\eta^2_2,\eta_2)$

where ${\xi_2}$ is a bdf for lb and ${\eta_2}$ is a bdf for ff. One can verify that ${\beta}$ is a diffeomorphism ${\mathring{W}\rightarrow \mathring{\mathbb{R}}^2_+}$. Let ${w=\beta^* u}$. Then ${w}$ is a polyhomogeneous conormal function on ${W}$. Its index can be denoted ${(E,F,H)}$ correspond to lb, ff and rb.

$\displaystyle E=\{(n,0)\}, F=\{(2n,0)\}, H=\{(2n,0)\}$

Suppose ${\pi_1}$ is the projection to ${x}$ coordinate. Consider ${f=\pi_1\circ \beta:W\rightarrow \mathbb{R}_+}$, ${f}$ is actually a ${b-}$fibration. Then the push forward map ${f_*}$ maps ${w}$ to a polyhomogeneous function on ${\mathbb{R}^+}$.

$\displaystyle f_*w=\pi_* u=\int_0^\infty u(x,y)dy$

In order to make ${u}$ is integrable, let us assume ${u}$ support ${x\leq 1}$ and ${y\leq 1}$. What is the index for ${f_* w}$ on ${\mathbb{R}^+}$?

$\displaystyle \int_0^1\sqrt{x^2+y^4}dy=\int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy+\int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy$

For the first integral, letting ${y^2/x=t}$

$\displaystyle \int_0^{\sqrt{x}}\sqrt{x^2+y^4}dy=x\sqrt{x}\int_0^1\sqrt{1+t^4}dt=c_0x\sqrt{x}$

For the second integral, letting ${x/y^2=t}$

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=\frac{1}{2}x\sqrt{x}\int_x^1t^{-\frac{5}{2}}\sqrt{t^2+1}dt$

Since the Taylor series

$\displaystyle \sqrt{1+t^2}=1+\frac{1}{2}t^2-\frac{1}{8}t^4+\cdots,\quad \text{for }|t|<1$

Consequently

$\displaystyle \int_{\sqrt{x}}^1\sqrt{x^2+y^4}dy=a_0+a_1x+a_2x\sqrt{x}+\cdots$

Combining all the above analysis, the index for ${f_*w}$ is ${\{(n,0)\}\cup \{\frac{n}{2},0\}}$

From another point of view, the vanishing order of ${f}$ on each boundary hypersurface of ${W}$ are ${e_f(lb)=1}$, ${e_f(\text{ff})=2}$ and ${e_f(rb)=0}$. ${f}$ maps lb and ff to the boundary of ${\mathbb{R}^+}$. Therefore the index of ${f_*w}$ is contained in

$\displaystyle \frac{1}{e_f(lb)}E\overline{\cup}\frac{1}{e_f(\text{ff})}F=E\overline{\cup}\frac{1}{2}F$

Remark: Daniel Grieser, Basics of ${b-}$Calculus.

### 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

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

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

### Frist and second variation of translating soliton

Suppose ${F(x,t):\Sigma\times(-\epsilon,\epsilon)\rightarrow \mathbb{R}^{n+1}}$ be a variation of ${\Sigma}$ with compact support and fixed boundary. Consider the weighted area functional

$\displaystyle \mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}dv$

What is the critical point of this area functional? To see that,

$\displaystyle \frac{\partial}{\partial t}\mathcal{A}(\Sigma)=\int_\Sigma F_t(x_{n+1})e^{x_{n+1}}dv+e^{x_{n+1}}\partial_t dv$

From some basic calculation in minimal surface(see colding minicozzi’s book)

$\displaystyle \partial_t dv=-\langle F_t,\mathbf{H}\rangle+div_{\Sigma}F_t^T$

Stokes’ theorem implies

$\displaystyle \int_\Sigma div_\Sigma F_t^Te^{x_{n+1}}dv=-\int_{\Sigma}\langle F_t^T,\nabla_\Sigma x_{n+1}\rangle e^{x_{n+1}}dv=-\int_\Sigma F_t^T(x_{n+1})e^{x_{n+1}}dv$

Combining these above fact, we get

$\displaystyle \partial_t\mathcal{A}(\Sigma)=\int_\Sigma F_t^{\perp}(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma \langle F_t^{\perp},N\rangle N(x_{n+1})-\langle F_t,\mathbf{H}\rangle dv$

$\displaystyle =\int_\Sigma \langle F_t,N\rangle \langle N,e_{n+1}\rangle-\langle F_t,\mathbf{H}\rangle dv=\int_\Sigma\langle F_t,\langle N,e_{n+1}\rangle N-\mathbf{H}\rangle dv$

where ${N}$ is the unit normal of ${\Sigma}$. Therefore the critical point of ${\mathcal{A}}$ will satisfy

$\displaystyle \mathbf{H}=\langle N, e_{n+1}\rangle N=e_{n+1}^{\perp}$

This is so called translating soliton.

Now consider the second variation at a translation soliton,

$\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+2F_t(x_{n+1})^2dv+2F_t(x_{n+1})\partial_tdv+\partial_{tt}dv\right]$

$\displaystyle =\int_\Sigma e^{x_{n+1}}\left[F_{tt}(x_{n+1})dv+\partial_{tt}dv\right]$

While

$\displaystyle \partial_{tt}dv=-|\langle A(\cdot,\cdot),F_t\rangle|^2+|\nabla_\Sigma^NF_t|^2+div_\Sigma(F_{tt})$

Suppose ${F_t=\eta N}$, where ${\eta\in C_c^\infty(\Sigma)}$, then ${|\nabla_\Sigma^NF_t|^2=|\nabla \eta|^2}$

$\displaystyle \partial_t^2\mathcal{A}(\Sigma)=\int_\Sigma [|\nabla\eta|^2-|A|^2\eta^2]e^{x_{n+1}}dv$

### 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.

### Bubble functions under different setting

Bubble function can be defined either on $\mathbb{R}^n$, $\mathbb{S}^n$ or $\mathbb{B}^n$. For the following notations, $c_n$ will denote suitable constants which may be different from line to line.

• For every $\epsilon>0$ and  $\xi\in\mathbb{R}^n$, define

$\displaystyle u_{\epsilon,\xi}=c_n\left(\frac{\epsilon}{\epsilon^2+|x-\xi|^2}\right)^{\frac{n-2}{2}}$

It is well know that $-\Delta u= c_nu^{\frac{n+2}{n-2}}$. Moreover $(\mathbb{R}^n,u^{\frac{4}{n-2}}_{\epsilon,\xi}g_E)$ is isometric to the standard sphere minus one point.

• For any $a\in \mathbb{S}^n$ and $\lambda>0$ define

$\displaystyle\delta(a,\lambda)=c_n\left(\frac{\lambda}{\lambda^2+1+(\lambda^2-1)\cos d(a,x)}\right)^{\frac{n-2}{2}}$

where $d(a,x)$ is the geodesic distance of $a$ and $x$ on $\mathbb{S}^n$. Actually $\cos d(a,x)=a\cdot x$

• For each $p\in \mathbb{B}^{n+1}$, define $\delta_p(x):\mathbb{S}^n\to \mathbb{R}$ by

$\displaystyle\delta_p(x)=c_n\left(\frac{1-|p|^2}{|x+p|^2}\right)^{\frac{n-2}{2}}$

Both the second and third one satisfy

$\displaystyle \frac{4(n-1)}{n-2}\Delta_{\mathbb{S}^n}\delta-n(n-1)\delta+c_n\delta^{\frac{n-2}{n+2}}=0$

If we make $p=\frac{\lambda-1}{\lambda+1}a$, the third one will be changed to the second one.

To get the second one from the first one, let us deonte $\Phi_a:\mathbb{S}^n\to \mathbb{R}^n$ be the stereographic projection from point $a$. Then

$\displaystyle \delta(a,\lambda)\circ \Phi^{-1}_a=c_n\left(\frac{\lambda(1+|y|^2)}{\lambda^2|y|^2+1}\right)^{\frac{n-2}{2}}=c_nu_{\lambda,0}u_{1,0}^{-1}$

It should be able to see the third one from hyperbolic translation directly.