## Mean curvature of a graph on cylinder

Let ${\Sigma}$ is the cylinder ${\{x: x_1^2+x_2^2=1\}}$ in ${\mathbb{R}^3}$. Suppose ${M}$ is a graph on the cylinder ${\Sigma}$ can be written as

$\displaystyle M=\{u(x)\nu_{\Sigma}(x):x\in \Sigma\},$

where ${\nu_{\Sigma}}$ is the normal vector of ${\Sigma}$, that is ${\nu=(0,0,1)}$. What is the mean curvature of ${M\subset \mathbb{R}^3}$?

We can parametrize the cylinder by ${(\theta,z)}$ where ${x_1=\cos\theta}$, ${x_2=\sin\theta}$, ${x_3=z}$, ${0\leq \theta\leq 2\pi}$. One can write ${M}$ as

$\displaystyle F(\theta,z)=(u(\theta,z)\cos\theta,u(\theta, z)\sin\theta,z)$

Then the induced metric on ${M}$ is

$\displaystyle g=|dx|^2=[u^2+u_\theta^2]d\theta^2+2u_\theta u_zd\theta dz+(u_z^2+1)dz^2$

$\displaystyle g^{-1}=\frac{1}{u^2(1+u_z^2)+u_\theta^2}\begin{bmatrix} 1+u_z^2 & -u_\theta u_z \\ -u_\theta u_z& u^2+u_\theta^2 \end{bmatrix}$

Since

$\displaystyle F_\theta=(-u\sin\theta+u_\theta \cos\theta, u\cos\theta+u_\theta\sin \theta,0)$

$\displaystyle F_z=(u_z\cos\theta,u_z\sin\theta,1)$

The normal vector of ${M}$ is

$\displaystyle \nu_M=\frac{1}{\sqrt{u^2(1+u_z^2)+u_\theta^2}}(u\cos\theta+u_\theta\sin\theta,u\sin \theta-u_\theta\cos \theta,-u\,u_z)$

The second fundamental form is

$\displaystyle A_{\theta\theta}=-F_{\theta\theta}\cdot \nu_{M}=u^2-uu_{\theta\theta}+2u_\theta^2$

$\displaystyle A_{\theta z}=-F_{\theta z}\cdot \nu_{M}=-uu_{\theta z}+u_\theta u_z$

$\displaystyle A_{zz}=-F_{zz}\cdot \nu_{M}=-uu_{zz}$

Then the mean curvature is

$\displaystyle H=\text{tr}A=g^{\theta\theta}A_{\theta\theta}+2g^{\theta z}A_{\theta z}+g^{zz}A_{zz}$

$\displaystyle =\frac{-[(1+u_z^2)u_{\theta\theta}+(u^2+u_\theta^2)u_{zz}]u+(1+u_z^2)u^2+2u_\theta^2+2uu_\theta u_zu_{\theta z}}{[u^2(1+u_z^2)+u_\theta^2]^{3/2}}$

If ${v=u-1}$ and ${|v|_{C^4}\ll 1}$, then

$\displaystyle H\sim 1-v_{\theta\theta}-v_{zz}-v$

## Troyanov’s work on prescrbing curvature on compact surfaces with conical singuarities

I will describe some parts of Troynov’s work on conical surface. For details, check his paper. For simplicity, we consider a closed Riemann surface ${S}$ with a real divisor ${\boldsymbol{\beta}=\sum_{i}\beta_ip_i}$. A conformal metric ${ds^2}$ on ${S}$ is said to represent the divisor ${\boldsymbol{\beta}}$ if ${ds^2}$ is smooth Riemannian metric on ${S\backslash supp(\boldsymbol{\beta})}$ and near each ${p_i}$, there exists a neighborhood ${\mathcal{O}_i}$ of ${p_i}$ and coordinate function ${z_i\mathcal{O}_i:\rightarrow \mathbb{R}^2}$ and ${u:\mathcal{O}_i\rightarrow\mathbb{R}}$ such that ${u\in C^2(\mathcal{O}_i-\{p_i\})}$ and

$\displaystyle ds^2=e^{2u}|z_i-a_i|^{2\beta_i}|dz_i|^2\quad \text{ on }\mathcal{O}_i$

where ${a_i=z_i(p_i)}$. ${\theta_i=2\pi(\beta_i+1)}$ is called the angle of the conical singularity. We will always assume ${\beta_i>-1}$.

For example ${\mathbb{C}}$ equipped with the metric ${|z|^{2\beta}|dz|^2}$ is isometric to an Euclidean cone of total angle ${\theta=2\pi(\beta+1)}$.

Suppose now ${(S,\boldsymbol{\beta},ds_0^2)}$ is a closed Riemann surface with conical metric ${ds_0^2}$. Assume that the Gauss curvature ${K_0}$ extends on ${S}$ as a H\”{o}lder continuous function.

Suppose ${ds^2=e^{2u}ds_0^2}$ is a conformal change of the conical metric. Then necessarily we have ${K=e^{-2u}(\Delta_0u+K_0)}$. In order to prescribe the Gauss curvature ${K}$, we need to solve the equation

$\Delta_0u+K_0=Ke^{2u}$.

Any reasonable solution of (0) will satisfy

$\displaystyle \int K e^{2u}d\mu_0=\int K_0d\mu_0:=\gamma$

where ${\gamma=2\pi (\chi(S)+\sum_i\beta_i)}$ follows from the Gauss-Bonnet formula for conical surface. We will use variational method to attack it. To that end, define ${H(ds_0^2)}$ to be Sobolev space of functions with

$\displaystyle \int_S|\nabla u|_0^2+u^2d\mu_0<\infty$

and functionals

$\displaystyle \mathscr{F}(u):=\int_{S}|\nabla u|_0^2+2K_0u^2d\mu_0,\quad\mathscr{G}(u):=\int Ke^{2u}d\mu_0$

It is easy to verify that the minimizer of the following variation problem will be a weak solution of (0)

$\displaystyle m=\inf_{u\in H(ds_0^2)}\{\mathscr{F}(u):\mathscr{G}(u)=\gamma\}$

As one can see, there are two immediate questions we need to answer,

(a) ${m>-\infty}$

(b) minimizer exists

The first question follows from the Trudinger inequality. Namely, define ${\tau(S,\boldsymbol{\beta})=4\pi+4\pi\min\{0,\min_i\beta_i\}}$. Then

Lemma: For any fixed ${b<\tau(S,\beta)}$, there exist constant ${C>0}$ such that,

$\displaystyle \int_S e^{bu^2}d\mu_0\leq C$

for any ${u\in H(ds_0^2)}$ and ${\int_S ud\mu_0=0}$ and ${\int_S |\nabla u|_0^2d\mu_0\leq 1}$.

Lemma: Suppose ${\psi\in L^2(ds_0^2)}$ and ${\int_S\psi d\mu_0=1}$. Let ${p>1}$. Then for any fixed ${b<\tau(S,\boldsymbol{\beta})}$, there exists constants ${C=C(b, p,\psi,ds_0^2)}$ such that

$\displaystyle |\int ve^{2u}d\mu_0|\leq C||v||_{L^p(ds_0^2)}\exp\left\{\frac{q}{b}\int_S |\nabla u|_0^2d\mu_0+\int_S2\psi ud\mu_0\right\}$

for any ${v\in L^p(ds_0^2)}$ and ${u\in H(ds_0^2)}$. Here ${q=p/(p-1)}$.

From this key lemma, one can derive that ${\mathscr{F}(u)}$ is lower bounded on ${\mathscr{G}(u)=\gamma}$ provided ${\gamma<\tau}$ and ${\sup K>0}$. To do that, choose ${p}$ and ${b}$ such that ${\gamma< \frac{b}{q}<\tau}$, then it follows from the above lemma that

$\displaystyle \gamma=\int_S Ke^{2u}d\mu_0\leq C||K||_{L^p(ds_0^2)}\exp\left\{\frac{q}{b}\int_S |\nabla u|_0^2+2\int_S\frac{K_0 u}{\gamma}d\mu_0\right\}$

$\displaystyle \leq C|K|_\infty\exp\left\{\frac{1}{\gamma}\mathscr{F}(u)+(-\frac{1}{\gamma}+\frac{q}{b})\int|\nabla u|_0^2\right\}\leq C|K|_\infty\exp\{\frac{1}{\gamma}\mathscr{F}(u)\}.$

For the second question, we need to prove the embedding ${H(ds_0^2)\rightarrow L^2(ds_0^2)}$ is compact(actually it is true for any ${L^p(ds_0^2)}$ for any ${p<\infty}$). Note that this is true if ${ds_0^2}$ is some smooth metric, however ${ds_0^2}$ is conical one. Therefore, we want to compare ${H(ds_0^2)}$ with ${H(ds_1^2)}$ of some smooth metric.

Suppose ${S}$ is equipped with smooth metric ${ds_1^2}$ such that ${ds_0^2=\rho ds_1^2}$. Here ${\rho}$ is smooth and positive outside the support of ${\boldsymbol{\beta}}$ and ${\rho(z)=O(|z|^{2\beta_i})}$ near ${p_i}$. If we use the ${^i\nabla u}$ denote the gradient of ${u}$ with respect to ${ds_i^2}$, then ${^0\nabla=(1/\rho)^1\nabla}$ and

$\displaystyle \int_{S}|\nabla u|_0^2d\mu_0=\int_{S}|\nabla u|_1^2d\mu_1$

This is only true in two dimension. Now ${H(ds_1^2)}$ and ${H(ds_0^2)}$ have inner product

$\displaystyle (u,v)_1=\int_S \langle\nabla u,\nabla v\rangle_1+uv d\mu_1,\quad (u,v)_0=\int_S \langle\nabla u,\nabla v\rangle_0+uv d\mu_0$

From the above analysis, we need to examine the difference of ${L^2(ds^2_1)}$ and ${L^2(ds_0^2)}$. It follows from the singular behavior of ${\rho }$ that

$\displaystyle L^p(ds_0^2)\subset L^q(ds_1^2),\quad L^p(ds_1^2)\subset L^q(ds_0^2)$

for any ${p\geq 1}$ and some ${q\gg p}$. Then for any ${u\in H(ds_0^2)}$,

$\displaystyle |u|_{L^2(ds_1^2)}\leq C|u|_{L^p(ds_0^2)} \leq C|u|_{H(ds_0^2)}$

Then ${H(ds_0^2)\subset H(ds_1^2)}$. On the contrary, we need the following inequality

Lemma: For conical metric ${ds_0^2}$

$\displaystyle |u|_{L^p(ds_0^2)}\leq |u|_{H(ds_0^2)}$

Then we have ${H(ds_0^2)= H(ds_1^2)}$. Therefore ${H(ds_0^2)\subset L^p(ds_0^2)}$ is compact for ${p<\infty}$. Furthermore, after some effort, one can prove

$\displaystyle H(ds_0^2)\rightarrow L^p(ds_0^2)$

$\displaystyle u\mapsto e^{2u}$

is also compact for ${p<\infty}$.

Remark:

[1] M. Troyanov, Prescribing curvature on compact surfaces with conical singularities, Trans. Am. Math. Soc. 324 (1991) 793–821. doi:10.1090/S0002-9947-1991-1005085-9.

## 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.

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.

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

$\displaystyle \frac{\det Hess (u)}{{\det g}}=\left(\frac{u'\phi'}{\phi}\right)^{n-1}u''$

## An identity related to generalized divergence theorem

I am trying to verify one proposition proved in Reilly’s paper. For the notation of this, please consult the paper.

Propsotion 2.4 Let ${{D}}$ be a domain in ${\mathbb{R}^n}$. ${f}$ is a smooth function on ${{D}}$. Then

$\displaystyle \int_{{D}}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\left(\tilde S_q(z)z_n-\sum_{\alpha\beta}\tilde T_{q-1}(z)^{\alpha\beta}z_{\alpha}z_{n,\beta}\right)dA$

Proof: Take an orthonormal frame field ${\{e_\alpha,e_n\}}$ such that ${\{e_\alpha\}}$ is tangent to ${\partial D}$. Notice

$\displaystyle D^2(f)(X,Y)=X(Yf)-(\nabla_{X}Y)f$

$\displaystyle D^2(f)=\left(\begin{matrix} z_{,\alpha\beta}-z_nA_{\alpha\beta} & z_{n,\alpha}\\ z_{n,\beta} & f_{nn} \end{matrix} \right)$

where ${f_{nn}=D^2(f)(e_n,e_n)}$. It follows from Remark 2.3 that

$\displaystyle \int_{D}(q+1)S_{q+1}(f)dx_1\cdots dx_m=\int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA$

where ${t=(t_1,\cdots,t_n)}$ is the outward unit normal to ${\partial D}$. Changing the coordinates to ${e_\alpha}$ and ${e_n}$, we can get

$\displaystyle \int_{\partial D}\sum_{i,j}T_q(f)^{ij}f_jt_idA=\int_{\partial D}T_q(f)^{\alpha n}z_{\alpha}+T_{q}(f)^{nn}z_ndA$

It is easy to see

$\displaystyle T_q(f)^{nn}=\tilde S_{q}(z)$

and

$\displaystyle T_q(f)^{\alpha n}z_\alpha=\sum\delta\binom{i_1,i_2\cdots,n,\alpha}{j_1,j_2\cdots,\beta, n}z_\alpha$

$\displaystyle =-\sum\delta\binom{\alpha_1,\alpha_2\cdots,\alpha}{\beta_1,\beta_2\cdots,\beta}f_{n\beta}z_\alpha=-\tilde T_{q-1}(z)^{\alpha\beta}z_\alpha z_{n,\beta}$

Therefore the proposition is established.

Remark: Robert Reilly, On the hessian of a function and the curvature of its graph

## Some identities related to mean curvature of order m

For any ${n\times n}$ (not necessarily symmetric) matrix ${\mathcal{A}}$, we let ${[\mathcal{A}]_m}$ denote the sum of its ${m\times m}$ principle minors. For any hypersurface which is a graph of ${u\in C^2(\Omega)}$, where ${\Omega\subset \mathbb{R}^n}$. We have its downward unit normal is

$\displaystyle (\nu,\nu_{n+1})=\left(\frac{Du}{\sqrt{1+|Du|^2}},\frac{-1}{\sqrt{1+|Du|^2}}\right).$

The principle curvatures are taken from the eigenvalues of the Jacobian matrix ${[D\nu]}$. One can define its ${m}$ mean curvature using the notation of above

$\displaystyle H_m=\sum_{i_1<\cdots

Now let us consider general matrix ${\mathcal{A}}$,

$\displaystyle A_m=[\mathcal{A}]_m=\frac{1}{m!}\sum \delta\binom{i_1,\cdots,i_m}{j_1,\cdots,j_m}a_{i_1j_1}\cdots a_{i_mj_m}$

where ${\delta}$ is the generalized Kronecker delta.

$\displaystyle \delta\binom{i_1 \dots i_p }{j_1 \dots j_p} = \begin{cases} +1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an even permutation of } j_1 \dots j_p \\ -1 & \quad \text{if } i_1 \dots i_p \text{ are distinct and an odd permutation of } j_1 \dots j_p \\ \;\;0 & \quad \text{in all other cases}.\end{cases}$

Then we define the Newton tensor

$\displaystyle T^{ij}_m=\frac{\partial A_m}{\partial a_{ij}}=\frac{1}{(m-1)!}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}a_{i_2j_2}\cdots a_{i_m j_m}.$

For any vector field ${X}$ on ${\mathbb{R}^n}$, ${DX}$ is a matrix, where ${D=(D_{1},\cdots,D_n)}$ and ${|X|\neq 0}$, denote ${\tilde X=X/|X|}$, we have

$\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}X_{j_2}]\cdots [D_{i_m}X_{j_m}].$

Since for any ${1\leq p,k,l\leq n}$

$\displaystyle D_k\tilde X_l=\frac{D_kX_{l}}{|X|}-\frac{\sum_{p=1}^nX_pD_kX_pX_l}{|X|^3}$

$\displaystyle \sum_{i,j,i_2,j_2}\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_jX_pX_{j_2}D_{i_2}X_p=0$

because $\delta$ is skew-symmetric in $j,j_2$. Then

$\displaystyle (m-1)!T^{ij}_m(DX)X_iX_j=|X|^{m-1}\sum\delta\binom{i,i_2 \dots i_m }{j,j_2 \dots j_m}X_iX_j[D_{i_2}\tilde X_{j_2}]\cdots [D_{i_m}\tilde X_{j_m}].$

$=(m-1)!|X|^{m-1}T_m^{ij}(D\tilde X)X_iX_j=(m-1)!|X|^{m+1}T_m^{ij}(D\tilde X)\tilde X_i\tilde X_j.$

Applying the formula $(T^{ij}_m(\mathcal{A}))=[\mathcal{A}]_{m-1}I-T_{m-1}(\mathcal{A})\cdot \mathcal{A}$(Check [1] Propsition 1.2) and $(D\tilde X)\tilde X=0$, we get

$T^{ij}_m(DX)X_iX_j=|X|^{m+1}[D\tilde X]_{m-1}$

It follows from the result of Reilly, Remark 2.3(a), that

$\displaystyle mA_m[DX]=D_i[T^{ij}_mX_j]$

Suppose ${X}$ is a vector field normal to ${\partial \Omega}$, then

$\displaystyle m\int_\Omega [DX]_m=\int_{\partial \Omega} T^{ij}_m X_j\gamma_i=\int_{\partial \Omega}(X\cdot \gamma)^m[D\gamma]_{m-1}=\int_{\partial \Omega} (X\cdot \gamma)^m H_{m-1}(\partial \Omega)$

where ${\gamma}$ is the outer ward unit normal to ${\partial \Omega}$.

Remark: [1] R.C. Reilly, On the Hessian of a function and the curvatures of its graph., Michigan Math. J. 20 (1974) 373â€“383. doi:10.1307/mmj/1029001155.

[2] N. Trudinger, Apriori bounds for graphs with prescribed curvature.

We probably have to assume $DX$ is a symmetric matrix in order to use the formula of Reilly. Not sure about this.

## Sherman-Morrison Formula

Suppose $\eta\in \mathbb{R}^n$ is a column vector and $M_{n\times n}$ is an invertible matrix.  Set $A=M+\eta \eta^T$, then

$A^{-1}=M^{-1}-\frac{M^{-1}\eta\eta^TM^{-1}}{1+\eta^T M^{-1}\eta}$

This formula has more general forms.

## 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 known 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

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

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 ${A_{12,k}=(\nabla_{\tau_k}A)(\tau_1,\tau_2)}$. 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,

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