## Category Archives: Geometry

63411

### 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.

### Parallel surfaces and Minkowski formula

Suppose ${X:M^n\rightarrow \mathbb{R}^{n+1}}$ is an immersed orientable closed hypersurface. ${N}$ is the inner unit normal for ${X(M^n)}$ and denote by ${\sigma}$ the second fundamental form of the immersion and by ${\kappa_i}$, ${i=1,\cdots,n}$ the principle curvatures at an arbitrary point of ${M}$. The ${r-}$th mean curvature of ${H_r}$ is obtained by applying ${r-}$elementary symmetric function to ${\kappa_i}$. Equivalently, ${H_r}$ can be defined through the identity

$\displaystyle P_n(t)=(1+t\kappa_1)\cdots(1+\kappa_n)=1+\binom{n}{1}H_1 t+\cdots+\binom{n}{n}H_n t^n$

for all real number ${t}$. One can see that ${H_1}$ represents the mean curvature of ${X}$, ${H_n}$ is the gauss-Kronecker curvature. ${H_2}$ can reflect the scalar curvature of ${M}$ on the condition that the ambient manifold is a space form.

We want to study the consequence of moving the hypersurface parallel. Namely, define ${X_t}$ to be

$\displaystyle X_t= X-tN.$

When ${t}$ is small enough, ${X_t}$ is well defined immersed hypersurface. Suppose ${e_1,\cdots, e_n}$ are principle directions at a point ${p}$ of ${M}$, then

$\displaystyle \quad(X_t)_*(e_i)=(1+\kappa_it)e_i$

here we identify ${X_*(e_i)=e_i}$ as abbreviation. This implies that ${N_t= N\circ X_t^{-1}}$ is also an unit normal field of ${X_t}$. The area element ${dA_t}$ will be

$\displaystyle dA_t=(1+t\kappa_1)\cdots(1+t\kappa_n)dA=P_n(t)dA.$

The second fundamental form of ${X_t}$ with respect to ${N}$ will be

$\displaystyle \sigma_t(v,w)=\langle N_t,\nabla^{\mathbb{R}^{n+1}}_vw\rangle=-\langle \nabla^{\mathbb{R}^{n+1}}_vN_t,w\rangle$

for all ${v,w}$ tangent vector fields on ${X_t(M)}$. Plugging in ${v=(X_t)_*(e_i)}$ and ${w=(X_t)_*(e_j)}$, we get

$\displaystyle (\nabla^{\mathbb{R}^{n+1}}_vw)(X_t(p))=(\nabla^{\mathbb{R}^{n+1}}_{e_i}e_j)(X(p))$

$\displaystyle \nabla_{v}^{\mathbb{R}^{n+1}}N_t=-\frac{\kappa_i}{1+t\kappa_i}v$

So ${e_1,\cdots, e_n}$ are also principle directions for ${X_t}$ and principle curvatures are

$\displaystyle \frac{\kappa_i}{1+t\kappa_i}$

Another way to see this is by choosing a geodesic local coordinates such that ${\partial_iX}$ are the principle directions of ${X}$ at ${p}$. Then

$\displaystyle \partial_j\partial_iX=\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij}$

$\displaystyle \partial_iN=-\kappa_i\partial_iX$

$\displaystyle \partial_i X_t=\partial_i X-t\partial_i N=\partial_i X+t\kappa_i\partial_iX$

$\displaystyle \partial_j\partial_iX_t=(1+t\kappa_i)\partial_j\partial_iX=(1+t\kappa_i)(\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij})$

Since ${g^{ij}_t=(1+\kappa_it)^{-2}\delta_{ij}}$ at ${p}$. Therefore we get the principle curvature are ${\frac{\kappa_i}{1+t\kappa_i}}$.

Therefore the mean curvature ${H(t)}$ for ${X_t}$ is

$\displaystyle H(t)=\frac{1}{n}\frac{P_n'(t)}{P_n(t)}$

Since we have identity

$\displaystyle \Delta|X_t|^2=2n(1+H\langle X_t,N\rangle)$

which implies

$\displaystyle \int_M\left(1+H(t)\langle X_t,N\rangle\right)dA_t=0$

Plugging in all the information,

$\displaystyle \int_M\left(nP_n(t)+P_n'(t)\langle X,N\rangle-tP_n'(t)\right)dA=0$

Reorder the terms in the above identity by the order of ${t}$, we get

$\displaystyle \int_M (H_{r-1}+H_r\langle X,N\rangle )dA=0$

One can use this to prove Heintze-Karcher inequality. There are Minkowski formula in Hyperbolic space and $\mathbb{S}^n$ also.

Remark: S. Montiel and Anotnio Ros, compact hypersurfaces: the alexandrov theorem for higher order mean curvatures. Differential Geometry, 52, 279-296

### f-extremal disk

In the last nonlinear analysis seminar, Professor Espinar talked about the overdetermined elliptic problem(OEP) which looks like the following

$\Delta u+f(u)=0\quad\text{ in }\Omega$

$u>0\quad \text{ in }\Omega$

$u=0 \quad \text{on }\partial \Omega$

$\frac{\partial u}{\partial\eta}=cst\quad\text{on }\partial \Omega$

There is a BCN conjecture related to this

BCN: If $f$ is Lipschitz, $\Omega\subset \mathbb{R}^n$ is a smooth(in fact, Lipschitz) connected domain with $\mathbb{R}^n\backslash\Omega$ connected where OEP admits a bounded solution, then $\Omega$ must be either a ball, a half space, a generalized cylinder or the complement of one of them.

BCN is false in $n\geq 3$. Epsinar wih Mazet proved BCN when $n=2$. This implies the Shiffer conjecture in dimension 2. In higher dimension of Shiffer conjecture, if we know the domain is contained in one hemisphere of $\mathbb{S}^n$, then one can use the equator or the great circle to perform the moving plane.

### Hypersurface in Hyperbolic space and its tangent horospheres

Hypersurface in hyperbolic space and its tangent horospheres

### Hypersurface in hyperbolic space and conformal metric on Sphere

Suppose $\Omega$ is a domain in $\mathbb{S}^n$, $g=e^{2\rho}g_0$, where $g_0$ is the standard metric on $\mathbb{S}^n$. One can construct a hypersurface in hyperboloid $\mathbb{H}^{n+1}\subset \mathbb{L}^{n+2}$ by

$\phi(x)=\frac{1}{2}(1+\sigma^2+|\nabla^0\sigma|^2)\psi(x)-\sigma(0,x)-(0,\nabla^0 \sigma)$

where $\sigma=e^{-\rho}$ and $\psi=e^{\rho}(1,x)=\sigma^{-1}(1,x)$. What are the induced metric on $\phi$? One can calculate as the following,

Suppose $u\in T_x\Omega$ is a unit eigenvector associated to the eigenvalue $s$ of $Hess(\sigma)_x$. We need to calculate $d\psi_x(u)$. Firstly it is easy to see

$d\psi_x(u)=-\sigma^{-1}u(\sigma)\psi+\sigma^{-1}(0,u)$

and

$d(\nabla^0 \sigma)_x(u)=\ \nabla_u^0\nabla^0 \sigma+(d \nabla^0\sigma_{x}(u))^{\perp}=Hess(\sigma)_x(u)+\langle d \nabla^0\sigma_{x}(u),x\rangle x$
$= \ su-\langle\nabla^0\sigma_{x},u\rangle x= \ su-u(\sigma)x$

Then

$d\phi_x(u)=u(\sigma)\sigma \psi+\frac{1}{2}u(|\nabla \sigma|^2)\psi+\frac{1}{2}(1+\sigma^2+|\nabla \sigma|^2)d\psi_x(u)-\sigma(0,u)-(0,su)$

So this gives the following formula

$d\phi_x(u)=\ \left(\frac{1}{2}-\lambda\right)d\psi_x(u).$

where $\lambda=s\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2)$. Suppose $u_i$ is the eigenvector corresponding to $s_i$ and $u_i,i=1,\cdots, n$ are orthonormal basis, then

$\ll d\phi_x(u_i),d\phi_x(u_j)\gg=\left(\frac{1}{2}-\lambda_{i}\right)\left(\frac{1}{2}-\lambda_{j}\right)\frac{1}{\sigma^{2}}\delta_{ij}$

where $\lambda_i=s_i\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2)$. Recall that Schouten tensor has the following formula under conformal change,

$Sch_g=Sch_0+d\rho\otimes d\rho-\nabla^{2}_0\rho-\frac{1}{2}|\nabla^0\rho|^2g_0$

$=\frac{1}{2}g_0+\frac{1}{\sigma}\nabla_0^2\sigma-\frac{1}{\sigma^2}|\nabla_0\sigma|^2$

for $g=e^{2\rho}g_0$ and $Sch_0=\frac{1}{2}g_0$. Therefore the eigenvalues of $Sch_g$ are $\lambda_i$.

$\ll d\phi,d\phi\gg=\left(\frac{1}{2}g-Sch_g\right)^2{\sigma^{2}}\delta_{ij}$

$\textbf{Remark:}$ I am using the thesis of Dimas Percy Abanto Silva and some communication with him.

### Hyperbolic translation in Poincare disc model

Hyperboloid model ${\mathbb{H}^{n+1}=(x_0,x_1,\cdots,x_{n+1})}$ is hyperquadric in ${\mathbb{L}^{n+2}}$, with

$\displaystyle -x_0^2+\sum_{i=1}^{n+1} x_i^2=-1$

Half-space model ${\mathbb{R}^{n+1}_+=\{(y_1,y_2,\cdots,y_{n+1}),y_{n+1}>0\}}$. Denote ${(y_1,\cdots,y_{n},y_{n+1})=(y,y_{n+1})}$,

Poincaré ball model ${\mathbb{B}^{n+1}=(u_1,\cdots,u_{n+1})\subset\mathbb{R}^{n+1}}$ There are some transformation formula between these hyperbolic models, say

$\Psi_{12}:\mathbb{H}^{n+1}\longmapsto \mathbb{R}^{n+1}_+$

$(x_0,\cdots,x_{n+1})\longmapsto \frac{1}{x_0+x_{n+1}}\left(x_1,\cdots,x_{n},1\right)$

$\Psi_{21}:\mathbb{R}^{n+1}_+\longmapsto \mathbb{H}^{n+1}$

$(y_1,\cdots,y_{n+1})\longmapsto \left(\frac{1+|y|^2+y_{n+1}^2}{2y_{n+1}},\frac{y}{y_{n+1}},\frac{1-|y|^2-y_{n+1}^2}{2y_{n+1}}\right)$

$\Psi_{23}:\mathbb{R}^{n+1}_+\longmapsto \mathbb{B}^{n+1}$

$(y_1,\cdots,y_{n+1})\longmapsto\frac{1}{|y|^2+(y_{n+1}+1)^2}(2y,|y|^2+y_{n+1}^2-1)$

$\Psi_{32}:\mathbb{B}^{n+1}\longmapsto\mathbb{R}^{n+1}_+$
$(u_1,\cdots, u_{n+1})\longmapsto \frac{1}{|u|^2+(u_{n+1}-1)^2}\left(2u,2(1-u_{n+1})\right)-(0,\cdots,0,1)$

On ${\mathbb{R}^{n+1}_+}$, there are vertical scaling transformations which are isometries of ${\mathbb{R}^n_+}$

$\displaystyle \tau(y_1,\cdots,y_{n+1})= (ty_1,\cdots, ty_{n+1}),\quad t>0$

On ${\mathbb{B}^{n+1}}$, they are called translation in ${e_{n+1}}$ direction through the map

$\displaystyle (u,u_{n+1})\mapsto\Psi_{23}\circ\tau\circ\Psi_{32}(u,u_{n+1})$

In component,

$\displaystyle u_i\mapsto \frac{4tu_i}{(1-t)^2|u|^2+[u_{n+1}(t-1)+t+1]^2}$
$\displaystyle u_{n+1}\mapsto \frac{(t^2-1)|u|^2+4t^2u_{n+1}+(t^2-1)(u_{n+1}-1)^2}{(1-t)^2|u|^2+[u_{n+1}(t-1)+t+1]^2}$

On $\mathbb{H}^{n+1}$,

$(x_0,x,x_{n+1})\longmapsto \Psi_{21}\circ \tau\circ\Psi_{12}(x_0,x,x_{n+1})$

In component,

$\displaystyle x_0=\frac{(1+t^2)x_0^2+(1-t^2)x_{n+1}^2}{2t}$

$x_i=x_i,i=1,\cdots,n$

$\displaystyle x_{n+1}=\frac{(1-t^2)x_0^2+(1+t^2)x_{n+1}^2}{2t}$

### Conformal killing operator and divergence transformation under conformal change

Suppose ${(M,g)}$ is a Riemannian manifold. For each vector field ${V}$, we can define the conformal killing operator ${\mathcal{D}}$ to be the trace free part of Lie derivative ${\mathcal{L}_Vg}$, more precisely

$\displaystyle \mathcal{D}V=\mathcal{L}_Vg-\frac{2}{n}(div_g V)g$

Obviously ${\mathcal{D}}$ maps vector field to trace free symmetric two tensors. Now suppose we have a conformal transformation ${\tilde{g}=e^{2f}g}$, then what happen to the conformal killing operator ${\tilde{\mathcal{D}}}$? Notice that

$\displaystyle \mathcal{L}_V\tilde g=\mathcal{L}_V(e^{2f}g)=2e^{2f}V(f)g+e^{2f}\mathcal{L}_v g$

By using the identity ${\mathcal{L}_V d\mu_g=(div_g V)d\mu_g}$ for any vector field ${V}$, here ${d\mu_g}$ is the volume element, one can get the transfromation of divergence under confromal change

$\displaystyle div_{\tilde g}V=div_g V+nV(f)$

therefore

$\displaystyle \tilde{\mathcal{D}}V=\mathcal{L}_V\tilde g-\frac{2}{n}(div_{\tilde g} V)\tilde g=e^{2f}\mathcal{D}V.$

${\mathcal{D}}$ induces a formal adjoint ${\mathcal{D}^*}$ on trace free 2-tensors. Suppose we have a symmetric 2-tensor ${h=h_{ij}dx^i\otimes dx^j}$, where ${x^i}$ are coordinates on ${M}$. If one has

$\displaystyle \tilde{\mathcal{D}}^*(h-\tilde{\mathcal{D}}V)=0$

for some vector field ${V}$ and trace free 2-tensor ${h}$. What does this coorespond to under metric ${g}$? To see that, we first need a formula about symmtric 2-tensors,

$\displaystyle \langle h,w\rangle_{\tilde g}=\int_M h_{ij}w_{kl}\tilde g^{ik}\tilde g^{jl}d\mu_{\tilde {g}}=\langle e^{(n-4)f}h,w\rangle_g$

Now choose any vector field ${W}$, then

$\displaystyle 0=\langle h-\tilde{\mathcal{D}}V,\tilde{\mathcal{D}}W\rangle_{\tilde g}=\langle h-e^{2f}\mathcal{D}V, e^{2f}\mathcal{D}w \rangle_{\tilde g}=\langle e^{nf}(e^{-2f}h-\mathcal{D} V),\mathcal{D} W\rangle_{g}$

This is equivalent to

$\displaystyle \mathcal D^*(e^{nf}(e^{-2f}h-\mathcal DV))=0$

Next consider the divergence operator ${\delta:\mathscr{S}^{p+1}M\rightarrow \mathscr{S}^p M}$ and its adjoints ${\delta^*}$.

$\displaystyle \delta T=-g^{ik}\nabla_iT_{k....}$

It is well know that on 1-forms

$\displaystyle \delta^*\alpha(X,Y)=\frac{1}{2}{\nabla_X\alpha(Y)+\nabla_Y\alpha(X)}=\frac 12 (L_{\alpha^\sharp}g)(X,Y).$

where ${\sharp}$ operator turns 1-form to a vector field by using metric ${g}$. What is the relation of ${\delta h}$ and ${\tilde \delta h}$? To find that, choose any 1-form ${\alpha}$,

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle \tilde \delta h,e^{-(n-2)f}\alpha\rangle_{\tilde g}=\langle h,\tilde\delta^*(e^{-(n-2)f}\alpha)\rangle_{\tilde g} \ \ \ \ \ (1)$

using the formula about ${\delta^*}$, one gets

$\displaystyle \tilde \delta^*(e^{-(n-2)f}\alpha)=e^{-(n-2)f}\tilde \delta^*\alpha-(n-2)e^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

$\displaystyle =e^{-(n-2)f}\delta^*\alpha+e^{-(n-2)f}\alpha(f)g-ne^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

then using ${h}$ is symmetric, continue from (1)

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle e^{(n-4)f},\tilde \delta^*(e^{-(n-2)f}\alpha)\rangle_{g}=\langle e^{-2f}h,\delta^*\alpha+\alpha(f)g-ndf\otimes\alpha\rangle_{g}$

$\displaystyle =\langle\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f,\alpha\rangle_g$

In other words,

$\displaystyle \tilde \delta h=\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f$

$\displaystyle =\delta h-(n-2)e^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f \ \ \ \ \ (2)$

In the other way, we can calculate more directly

$\displaystyle \nabla_k h_{ij}=\frac{\partial}{\partial x^k}h_{ij}-\Gamma^p_{ki}h_{pj}-\Gamma^p_{kj}h_{ip}$

$\displaystyle \tilde \Gamma^k_{ij} = \Gamma^k_{ij}+ \delta^k_i\partial_j f + \delta^k_j\partial_i f -g_{ij}\nabla^k f$

We get

$\displaystyle \tilde \delta h=-\tilde g^{ki}\tilde \nabla_kh_{ij}=-e^{-2f}g^{ki}\nabla_kh_{ij}+e^{-2f}g^{ki}h_{pj}(\delta_k^p\partial_if+\delta^p_i\partial_kf-g_{ki}\nabla^pf)$

$\displaystyle +e^{-2f}g^{ki}h_{ip}(\delta^p_k\partial_j f+\delta^p_j\partial_kf-g_{kj}\nabla^pf)$

$\displaystyle \tilde \delta h=e^{-2f}\delta h+e^{-2f}(g^{ki}h_{kj}\partial_if+g^{ki}h_{ij}\partial_kf-nh_{pj}\nabla^p f+g^{ki}h_{ik}\partial_j f+g^{ki}h_{ij}\partial_k f- h_{jp}\nabla^p f)$

therefore

$\displaystyle \tilde \delta h=e^{-2f}\delta h-(n-2)e^{-2f}g^{ki}h_{kj}\partial_if+e^{-2f}(tr_gh)\nabla f$

One can compare this with (2).

### Product metric on product manifold

Suppose we have two Riemannian manifolds $(M^m,g)$ and $(N^n,\tilde g)$, what happened to their product manifold with product metric?
$(M\times N, g\times\tilde g)$.

We will use $a,b,c,..=1..m$ and $A,B,C,..=1..n$. Then by definition $g_{aA}=0$. Therefore

$\Gamma_{aA}^b=\frac 12g^{bc}(\partial_Ag_{ac}+\partial_ag_{Ac}-\partial_cg_{aA})=0$

Similarly for all mixed indices on $\Gamma$. This implies

$\nabla_{\partial_a}\partial_A=\Gamma_{aA}^b\partial_b+\Gamma_{aA}^B\partial_B=0.$

Therefore the Riemannian curvature operator is

$R(\partial_a,\partial_A)\partial_B=R(\partial_a,\partial_A)\partial_b=0.$

and Ricci tensor

$R_{aA}=0.$

When you have the warped product metric like $\mathbb{R}\times N$ with metric $dr^2+\phi^2(r)\tilde g$, the curvature is given like the following

$R(\partial _r,X,\partial_r,Y)=-{\phi''}\phi\tilde g(X,Y)$

$R(\partial_r,X,Y,Z)=0$

$R(X,Y,Z,W)=\phi^2\tilde R(X,Y,Z,W)-(\phi'\phi)^2(\tilde g(X,Z)\tilde g(Y,W)-\tilde g(X,W)\tilde g(Y,Z))$

### Geodesic Normal Coordinates, Gauss lemma and Identity

Suppose ${p}$ is a point on ${n-}$dim manifold ${M}$. Consider the exponential map near ${p}$, at suitable neighborhood any point within it can be expressed uniquely as

$\displaystyle \text{exp}(x^1e_1+x^2e_2+\cdots+x^ne_n)$

where ${\{e_i\}}$ is an orthonormal basis. Then ${\{x^i\}}$ can be coordinate around ${p}$. Let us deduce useful identities using this coordinate. Obviously, the geodesic in this coordinate will be

$\displaystyle \gamma(t)=tv, \quad v\in T_pM$

In particular ${v=e_i}$,

$\displaystyle g_{ij}(p)=\langle \frac{\partial }{\partial x^i}(0),\frac{\partial }{\partial x^j}(0)\rangle=\langle e_i,e_j\rangle=\delta _{ij}$

The geodesic equation will be

$\displaystyle \Gamma^k_{ij}(\gamma(t))v^iv^j=0$

Multiplying ${t^2}$, we get

$\displaystyle \Gamma^k_{ij}(\gamma(t))x^ix^j=0$

Since the tangent vector has constant length along ${\gamma(t)}$, i.e.

$\displaystyle g_{ij}(\gamma(t))v^iv^j=g_{ij}(p)v^iv^j=v^iv^i$

Multiplying ${t^2}$, it is equivalent to say

$\displaystyle g_{ij}x^ix^j=x^ix^i$

Furthermore, recalling the definition of Christoffel symbol, one can get

$\displaystyle \frac{1}{2}(\partial_j g_{ik}+\partial_i g_{jk}-\partial_k g_{ij})x^ix^j=0$

Or equivalently

$\displaystyle \partial_j g_{ik}x^ix^j=\frac{1}{2}\partial_kg_{ij}x^ix^j=\frac{1}{2}\partial_k(g_{ij}x^ix^j)-g_{kj}x^j=x^k-g_{kj}x^j$

While on the left hand side

$\displaystyle \partial_j g_{ik}x^ix^j=\partial_j(g_{ik}x^i)x^j-g_{ik}x^i$

Combining this two facts,

$\displaystyle \partial_j(g_{ik}x^i)x^j=x^k$

$\displaystyle \partial_j(g_{ik}x^i-x^k)x^j=0$

This means along ${\gamma(t)}$,

$\displaystyle \frac{d}{dt}(g_{ik}x^i-x^k)=0$

Since at ${p}$, ${g_{ik}x^i-x^k=0}$, one get

$\displaystyle g_{ik}x^i=x^k$

on any point in the neighborhood. This identity is called Gauss Lemma.

Suppose ${g(x)=\text{exp}(h(x))}$, ${h_{ij}(x)}$ is a symmetric 2-tensor. Then the above identity means ${h_{ij}x^j=0}$.