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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.

 

b_map

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.

new_lens

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.