## Category Archives: Uncategorized

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### 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.

### 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

### 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,

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

### 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

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

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

### 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 ${\mathbb{R}\times \mathbb{S}^{n-1}}$.

### Unique continuation property on the boundary

I am writing a theorem proved by Jin Zhiren in his thesis.

Suppose $\Omega$ is a smooth domain in $\mathbb{R}^n$, $x_0\in \partial \Omega$ and $u$ is a harmonic function in $\Omega$. If there exists $A, b>0$ such that

$\displaystyle |u(x)|\leq Ae^{-\frac{b}{|x-x_0|}}\quad x\in \Omega$

for $|x-x_0|$ small, then $u=0$. If $n=2$, the same conclusion holds for the solutions of a general second order linear elliptic equation.

A borderline example for this theorem is $u$ be the real part of $e^{-1/z^\alpha}$, $\alpha\in (0,1)$. $u$ is harmonic in the right half plane and $u\leq Ae^{-1/|x|^\alpha}$ and consequently $D^\beta u(0)=0$.