Category Archives: Uncategorized


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 know 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}.


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


\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


\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)


\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}}.

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.





Newton tensor

Suppose {A:V\rightarrow V} is a symmetric endomorphism of vector space {V}, {\sigma_k} is the {k-}th elementary symmetric function of the eigenvalue of {A}. Then

\displaystyle \det(A+tI)=\sum_{k=0}^n \sigma_k t^{n-k}

One can define the {k-}th Newton transformation as the following

\displaystyle \det(A+tI)(A+tI)^{-1}=\sum_{k=0}^{n-1}T_k(A)t^{n-k-1}

This means

\displaystyle \det(A+tI)=\sum_{k=0}^{n-1}T_k(A+tI)t^{n-k-1}

\displaystyle =T_0 t^n+\sum_{k=0}^{n-2}(A\cdot T_k(A)+T_{k+1}(A))t^{n-k-1}+T_{n-1}(A)

By comparing coefficients of {t}, we get the relations of {T_k}

\displaystyle T_0=1,\quad A\cdot T_k(A)+T_{k+1}(A)=\sigma_{k+1}I,\, 0\leq k\leq n-2\quad T_{n-1}(A)=\sigma_n

Induction shows

\displaystyle T_{k}(A)=\sigma_kI-\sigma_{k-1}A+\cdots+(-1)^kA^k

For example

\displaystyle T_1(A)=\sigma_1I-A

\displaystyle T_2(A)=\sigma_2-\sigma_1A+A^2

One of the important property of Newton transformation is that: Suppose {F(A)=\sigma_k(A)}, then

\displaystyle F^{ij}=\frac{\partial F}{\partial A_{ij}}=T_{k-1}^{ij}(A)

The is because

\displaystyle \frac{\partial }{\partial A_{ij}}\det(A+tI)=\det(A+tI)((A+tI)^{-1})_{ij}.

If {A\in \Gamma_k}, then {T_{k-1}(A)} is positive definite and therefore {F} is elliptic.

Remark: Hu, Z., Li, H. and Simon, U. . Schouten curvature functions on locally conformally flat Riemannian manifolds. Journal of Geometry, 88(1{-}2), (2008), 75{-}100.

Self-shrinker and polynomial volume growth

Proposition: If M is an entire graph of at most polynomial volume growth and H=\langle X,\nu\rangle, namely M is a self-shrinker. Then M is a plane.

Proof: Suppose

\displaystyle v=\frac{1}{\langle \nu, w\rangle}

Then one can derive the following equation

\displaystyle \Delta v=\langle\nabla v,X\rangle+|A|^2v+2v^{-1}|\nabla v|^2

Multiplying both sides by e^{-|X|^2/2} and integration on M, which makes sense because of the polynomial volume growth, we get

\int_M (\Delta v-\langle\nabla v,X\rangle)e^{-\frac{|X|^2}{2}}d\mu=\int_M (|A|^2v+2v^{-1}|\nabla v|^2)e^{-\frac{|X|^2}{2}}d\mu

However, integration by parts shows the LHS is zero. Thus A\equiv v\equiv 0, M must be a plane.

Some calculations of sigma_2

On four-manifold {(M^4,g_0)}, we define Shouten tensor

\displaystyle A = Ric-\frac 16 Rg

and Einstein tensor and gravitational tensor

\displaystyle E=Ric - \frac 14 Rg\quad S=-Ric+\frac{1}{2}Rg

Suppose {\sigma_2} is the elemantary symmetric function

\displaystyle \sigma_2(\lambda)=\sum_{i<j}\lambda_i\lambda_j

Thinking of {A} as a tensor of type {(1,1)}. {\sigma_2(A)} is defined as {\sigma_2} applied to eigenvalues of {A}. Then

\displaystyle \sigma_2(A)= \frac{1}{2}[(tr_g A)^2-\langle A, A\rangle_g] \ \ \ \ \ (1)

Notice {A=E+\frac{1}{12}Rg}. Easy calculation reveals that

\displaystyle \sigma_2(A)=-\frac{1}{2}|E|^2+\frac{1}{24}R^2 \ \ \ \ \ (2)

Under conformal change of metric {g=e^{2w}g_0}, we have

\displaystyle R= e^{-2w}(R_0-6\Delta_0 w-6|\nabla_0 w|^2) \ \ \ \ \ (3)

\displaystyle A=A_0-2\nabla^2_0 w+2dw\otimes dw-|\nabla_0w|^2g_0 \ \ \ \ \ (4)

\displaystyle S=S_0+2\nabla_0^2w-2\Delta_0wg_0-2dw\otimes dw-|\nabla_0 w|^2g_0 \ \ \ \ \ (5)

We want to solve the equation {\sigma_2(A)=f>0}, which is equivalent to solve

\displaystyle \sigma_2(A_0-2\nabla^2_0w+2dw\otimes dw-|\nabla_0w|^2g_0)=f

This is an fully nonlinear equation of Monge-Ampere type. Under local coordinates, the above equation can be treated as

\displaystyle F(\partial_i\partial_j w,\partial_kw,w,x)=f

where {F(p_{ij},v_k,s,x):\mathbb{R}^{n\times n}\times\mathbb{R}^n\times\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}}. This equation is elliptic if the matrix {\left(\frac{\partial F}{\partial p_{ij}}\right)} is positive definite. In order to find that matrix, we need the linearized operator

\displaystyle L[\phi]=\frac{\partial F}{\partial p_{ij}}(\nabla_0^2\phi)_{ij}=\frac{d}{dt}|_{t=0}F(\partial_i\partial_j w+t\partial_i\partial_j\phi,\partial_kw,w,x) \ \ \ \ \ (6)

Using the elementary identity

\displaystyle \frac{d}{dt}\rvert_{t=0}\sigma_2(H+tG)=tr_gH\cdot tr_gG-\langle H, G\rangle_g. \ \ \ \ \ (7)

for any fixed matrix {H} and {G}. Now plug in {H=A} is Schouten tensor and {B=-2\nabla_0^2\phi}. One can calculate them as

\displaystyle tr_g H\cdot tr_g G=\langle \frac{1}{3}Rg, G\rangle_g \ \ \ \ \ (8)

Then we get

\displaystyle L[\phi]=\langle S,G\rangle_g=-2\langle S,\nabla^2_0\phi\rangle_g