Category Archives: Geometry


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}}.


\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


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


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


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 horosphers

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(\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


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


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}


(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}


\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)


\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)


\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


 Therefore the Riemannian curvature operator is


 and Ricci tensor


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(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}.