Category Archives: Geometry


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}


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

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

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

Stereographic projection from center

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

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

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

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


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

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

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

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


\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}


\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}}.


\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