Category Archives: Geometry

63411

Parallel surfaces and Minkowski formula

March 20, 2017 – 2:09 pm

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

By Sun's World | Also posted in Mean curvature flow, Riemannian Geo | Comments (0)

f-extremal disk

March 2, 2017 – 8:06 pm

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.

By Sun's World | Also posted in Elliptic PDE, PDE, Riemannian Geo | Tagged | Comments (0)

Hypersurface in Hyperbolic space and its tangent horospheres

December 7, 2016 – 4:02 pm
Hypersurface in hyperbolic space and horosphers

Hypersurface in hyperbolic space and its tangent horospheres

By Sun's World | Also posted in Hyperbolic Geo | Comments (0)

Hypersurface in hyperbolic space and conformal metric on Sphere

November 30, 2016 – 4:09 pm

Suppose \Omega is a domain in \mathbb{S}^n, g=e^{2\rho}g_0, where g_0 is the standard metric on \mathbb{S}^n. One can construct a hypersurface in hyperboloid \mathbb{H}^{n+1}\subset \mathbb{L}^{n+2} by

\phi(x)=\frac{1}{2}(1+\sigma^2+|\nabla^0\sigma|^2)\psi(x)-\sigma(0,x)-(0,\nabla^0 \sigma)

where \sigma=e^{-\rho} and \psi=e^{\rho}(1,x)=\sigma^{-1}(1,x). What are the induced metric on \phi? One can calculate as the following,

Suppose u\in T_x\Omega is a unit eigenvector associated to the eigenvalue s of Hess(\sigma)_x. We need to calculate d\psi_x(u). Firstly it is easy to see

d\psi_x(u)=-\sigma^{-1}u(\sigma)\psi+\sigma^{-1}(0,u)

and

d(\nabla^0 \sigma)_x(u)=\ \nabla_u^0\nabla^0 \sigma+(d \nabla^0\sigma_{x}(u))^{\perp}=Hess(\sigma)_x(u)+\langle d \nabla^0\sigma_{x}(u),x\rangle x
= \ su-\langle\nabla^0\sigma_{x},u\rangle x= \ su-u(\sigma)x

Then

d\phi_x(u)=u(\sigma)\sigma \psi+\frac{1}{2}u(|\nabla \sigma|^2)\psi+\frac{1}{2}(1+\sigma^2+|\nabla \sigma|^2)d\psi_x(u)-\sigma(0,u)-(0,su)

So this gives the following formula

d\phi_x(u)=\ \left(\frac{1}{2}-\lambda\right)d\psi_x(u).

where \lambda=s\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2). Suppose u_i is the eigenvector corresponding to s_i and u_i,i=1,\cdots, n are orthonormal basis, then

\ll d\phi_x(u_i),d\phi_x(u_j)\gg=\left(\frac{1}{2}-\lambda_{i}\right)\left(\frac{1}{2}-\lambda_{j}\right)\frac{1}{\sigma^{2}}\delta_{ij}

where \lambda_i=s_i\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2). Recall that Schouten tensor has the following formula under conformal change,

Sch_g=Sch_0+d\rho\otimes d\rho-\nabla^{2}_0\rho-\frac{1}{2}|\nabla^0\rho|^2g_0

=\frac{1}{2}g_0+\frac{1}{\sigma}\nabla_0^2\sigma-\frac{1}{\sigma^2}|\nabla_0\sigma|^2

for g=e^{2\rho}g_0 and Sch_0=\frac{1}{2}g_0. Therefore the eigenvalues of Sch_g are \lambda_i.

\ll d\phi,d\phi\gg=\left(\frac{1}{2}g-Sch_g\right)^2{\sigma^{2}}\delta_{ij}

\textbf{Remark:} I am using the thesis of Dimas Percy Abanto Silva and some communication with him.

By Sun's World | Also posted in Hyperbolic Geo | Comments (0)

Hyperbolic translation in Poincare disc model

November 19, 2016 – 8:15 pm

Hyperboloid model {\mathbb{H}^{n+1}=(x_0,x_1,\cdots,x_{n+1})} is hyperquadric in {\mathbb{L}^{n+2}}, with

\displaystyle -x_0^2+\sum_{i=1}^{n+1} x_i^2=-1

Half-space model {\mathbb{R}^{n+1}_+=\{(y_1,y_2,\cdots,y_{n+1}),y_{n+1}>0\}}. Denote {(y_1,\cdots,y_{n},y_{n+1})=(y,y_{n+1})},

Poincaré ball model {\mathbb{B}^{n+1}=(u_1,\cdots,u_{n+1})\subset\mathbb{R}^{n+1}} There are some transformation formula between these hyperbolic models, say

\Psi_{12}:\mathbb{H}^{n+1}\longmapsto \mathbb{R}^{n+1}_+

(x_0,\cdots,x_{n+1})\longmapsto \frac{1}{x_0+x_{n+1}}\left(x_1,\cdots,x_{n},1\right)

\Psi_{21}:\mathbb{R}^{n+1}_+\longmapsto \mathbb{H}^{n+1}

(y_1,\cdots,y_{n+1})\longmapsto \left(\frac{1+|y|^2+y_{n+1}^2}{2y_{n+1}},\frac{y}{y_{n+1}},\frac{1-|y|^2-y_{n+1}^2}{2y_{n+1}}\right)

\Psi_{23}:\mathbb{R}^{n+1}_+\longmapsto \mathbb{B}^{n+1}

(y_1,\cdots,y_{n+1})\longmapsto\frac{1}{|y|^2+(y_{n+1}+1)^2}(2y,|y|^2+y_{n+1}^2-1)

\Psi_{32}:\mathbb{B}^{n+1}\longmapsto\mathbb{R}^{n+1}_+
(u_1,\cdots, u_{n+1})\longmapsto \frac{1}{|u|^2+(u_{n+1}-1)^2}\left(2u,2(1-u_{n+1})\right)-(0,\cdots,0,1)

 On {\mathbb{R}^{n+1}_+}, there are vertical scaling transformations which are isometries of {\mathbb{R}^n_+}

\displaystyle \tau(y_1,\cdots,y_{n+1})= (ty_1,\cdots, ty_{n+1}),\quad t>0

On {\mathbb{B}^{n+1}}, they are called translation in {e_{n+1}} direction through the map

\displaystyle (u,u_{n+1})\mapsto\Psi_{23}\circ\tau\circ\Psi_{32}(u,u_{n+1})

In component,

\displaystyle u_i\mapsto \frac{4tu_i}{(1-t)^2|u|^2+[u_{n+1}(t-1)+t+1]^2}
\displaystyle u_{n+1}\mapsto \frac{(t^2-1)|u|^2+4t^2u_{n+1}+(t^2-1)(u_{n+1}-1)^2}{(1-t)^2|u|^2+[u_{n+1}(t-1)+t+1]^2}

On \mathbb{H}^{n+1},

(x_0,x,x_{n+1})\longmapsto \Psi_{21}\circ \tau\circ\Psi_{12}(x_0,x,x_{n+1})

In component,

\displaystyle x_0=\frac{(1+t^2)x_0^2+(1-t^2)x_{n+1}^2}{2t}

x_i=x_i,i=1,\cdots,n

\displaystyle x_{n+1}=\frac{(1-t^2)x_0^2+(1+t^2)x_{n+1}^2}{2t}

By Sun's World | Also posted in Hyperbolic Geo | Comments (0)

Conformal killing operator and divergence transformation under conformal change

May 13, 2016 – 1:00 am

Suppose {(M,g)} is a Riemannian manifold. For each vector field {V}, we can define the conformal killing operator {\mathcal{D}} to be the trace free part of Lie derivative {\mathcal{L}_Vg}, more precisely

\displaystyle \mathcal{D}V=\mathcal{L}_Vg-\frac{2}{n}(div_g V)g

Obviously {\mathcal{D}} maps vector field to trace free symmetric two tensors. Now suppose we have a conformal transformation {\tilde{g}=e^{2f}g}, then what happen to the conformal killing operator {\tilde{\mathcal{D}}}? Notice that

\displaystyle \mathcal{L}_V\tilde g=\mathcal{L}_V(e^{2f}g)=2e^{2f}V(f)g+e^{2f}\mathcal{L}_v g

By using the identity {\mathcal{L}_V d\mu_g=(div_g V)d\mu_g} for any vector field {V}, here {d\mu_g} is the volume element, one can get the transfromation of divergence under confromal change

\displaystyle div_{\tilde g}V=div_g V+nV(f)

therefore

\displaystyle \tilde{\mathcal{D}}V=\mathcal{L}_V\tilde g-\frac{2}{n}(div_{\tilde g} V)\tilde g=e^{2f}\mathcal{D}V.

{\mathcal{D}} induces a formal adjoint {\mathcal{D}^*} on trace free 2-tensors. Suppose we have a symmetric 2-tensor {h=h_{ij}dx^i\otimes dx^j}, where {x^i} are coordinates on {M}. If one has

\displaystyle \tilde{\mathcal{D}}^*(h-\tilde{\mathcal{D}}V)=0

for some vector field {V} and trace free 2-tensor {h}. What does this coorespond to under metric {g}? To see that, we first need a formula about symmtric 2-tensors,

\displaystyle \langle h,w\rangle_{\tilde g}=\int_M h_{ij}w_{kl}\tilde g^{ik}\tilde g^{jl}d\mu_{\tilde {g}}=\langle e^{(n-4)f}h,w\rangle_g

Now choose any vector field {W}, then

\displaystyle 0=\langle h-\tilde{\mathcal{D}}V,\tilde{\mathcal{D}}W\rangle_{\tilde g}=\langle h-e^{2f}\mathcal{D}V, e^{2f}\mathcal{D}w \rangle_{\tilde g}=\langle e^{nf}(e^{-2f}h-\mathcal{D} V),\mathcal{D} W\rangle_{g}

This is equivalent to

\displaystyle \mathcal D^*(e^{nf}(e^{-2f}h-\mathcal DV))=0

Next consider the divergence operator {\delta:\mathscr{S}^{p+1}M\rightarrow \mathscr{S}^p M} and its adjoints {\delta^*}.

\displaystyle \delta T=-g^{ik}\nabla_iT_{k....}

It is well know that on 1-forms

\displaystyle \delta^*\alpha(X,Y)=\frac{1}{2}{\nabla_X\alpha(Y)+\nabla_Y\alpha(X)}=\frac 12 (L_{\alpha^\sharp}g)(X,Y).

where {\sharp} operator turns 1-form to a vector field by using metric {g}. What is the relation of {\delta h} and {\tilde \delta h}? To find that, choose any 1-form {\alpha},

\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle \tilde \delta h,e^{-(n-2)f}\alpha\rangle_{\tilde g}=\langle h,\tilde\delta^*(e^{-(n-2)f}\alpha)\rangle_{\tilde g} \ \ \ \ \ (1)

using the formula about {\delta^*}, one gets

\displaystyle \tilde \delta^*(e^{-(n-2)f}\alpha)=e^{-(n-2)f}\tilde \delta^*\alpha-(n-2)e^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)

\displaystyle =e^{-(n-2)f}\delta^*\alpha+e^{-(n-2)f}\alpha(f)g-ne^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)

then using {h} is symmetric, continue from (1)

\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle e^{(n-4)f},\tilde \delta^*(e^{-(n-2)f}\alpha)\rangle_{g}=\langle e^{-2f}h,\delta^*\alpha+\alpha(f)g-ndf\otimes\alpha\rangle_{g}

\displaystyle =\langle\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f,\alpha\rangle_g

In other words,

\displaystyle \tilde \delta h=\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f

\displaystyle =\delta h-(n-2)e^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f \ \ \ \ \ (2)

In the other way, we can calculate more directly

\displaystyle \nabla_k h_{ij}=\frac{\partial}{\partial x^k}h_{ij}-\Gamma^p_{ki}h_{pj}-\Gamma^p_{kj}h_{ip}

\displaystyle \tilde \Gamma^k_{ij} = \Gamma^k_{ij}+ \delta^k_i\partial_j f + \delta^k_j\partial_i f -g_{ij}\nabla^k f

We get

\displaystyle \tilde \delta h=-\tilde g^{ki}\tilde \nabla_kh_{ij}=-e^{-2f}g^{ki}\nabla_kh_{ij}+e^{-2f}g^{ki}h_{pj}(\delta_k^p\partial_if+\delta^p_i\partial_kf-g_{ki}\nabla^pf)

\displaystyle +e^{-2f}g^{ki}h_{ip}(\delta^p_k\partial_j f+\delta^p_j\partial_kf-g_{kj}\nabla^pf)

\displaystyle \tilde \delta h=e^{-2f}\delta h+e^{-2f}(g^{ki}h_{kj}\partial_if+g^{ki}h_{ij}\partial_kf-nh_{pj}\nabla^p f+g^{ki}h_{ik}\partial_j f+g^{ki}h_{ij}\partial_k f- h_{jp}\nabla^p f)

therefore

\displaystyle \tilde \delta h=e^{-2f}\delta h-(n-2)e^{-2f}g^{ki}h_{kj}\partial_if+e^{-2f}(tr_gh)\nabla f

One can compare this with (2).

By Sun's World | Also posted in PDE, Riemannian Geo, Yamabe flow | Tagged | Comments (1)

Product metric on product manifold

May 11, 2016 – 5:24 pm

Suppose we have two Riemannian manifolds (M^m,g) and (N^n,\tilde g), what happened to their product manifold with product metric?
(M\times N, g\times\tilde g).

We will use a,b,c,..=1..m and A,B,C,..=1..n. Then by definition g_{aA}=0. Therefore

\Gamma_{aA}^b=\frac 12g^{bc}(\partial_Ag_{ac}+\partial_ag_{Ac}-\partial_cg_{aA})=0

Similarly for all mixed indices on \Gamma. This implies

 \nabla_{\partial_a}\partial_A=\Gamma_{aA}^b\partial_b+\Gamma_{aA}^B\partial_B=0.

 Therefore the Riemannian curvature operator is

 R(\partial_a,\partial_A)\partial_B=R(\partial_a,\partial_A)\partial_b=0.

 and Ricci tensor

 R_{aA}=0.

When you have the warped product metric like \mathbb{R}\times N with metric dr^2+\phi^2(r)\tilde g, the curvature is given like the following

R(\partial _r,X,\partial_r,Y)=-{\phi''}\phi\tilde g(X,Y)

R(\partial_r,X,Y,Z)=0

R(X,Y,Z,W)=\phi^2\tilde R(X,Y,Z,W)-(\phi'\phi)^2(\tilde g(X,Z)\tilde g(Y,W)-\tilde g(X,W)\tilde g(Y,Z))

By Sun's World | Also posted in Riemannian Geo | Comments (0)

Geodesic Normal Coordinates, Gauss lemma and Identity

January 4, 2015 – 1:27 am

Suppose {p} is a point on {n-}dim manifold {M}. Consider the exponential map near {p}, at suitable neighborhood any point within it can be expressed uniquely as

\displaystyle \text{exp}(x^1e_1+x^2e_2+\cdots+x^ne_n)

where {\{e_i\}} is an orthonormal basis. Then {\{x^i\}} can be coordinate around {p}. Let us deduce useful identities using this coordinate. Obviously, the geodesic in this coordinate will be

\displaystyle \gamma(t)=tv, \quad v\in T_pM

In particular {v=e_i},

\displaystyle g_{ij}(p)=\langle \frac{\partial }{\partial x^i}(0),\frac{\partial }{\partial x^j}(0)\rangle=\langle e_i,e_j\rangle=\delta _{ij}

The geodesic equation will be

\displaystyle \Gamma^k_{ij}(\gamma(t))v^iv^j=0

Multiplying {t^2}, we get

\displaystyle \Gamma^k_{ij}(\gamma(t))x^ix^j=0

Since the tangent vector has constant length along {\gamma(t)}, i.e.

\displaystyle g_{ij}(\gamma(t))v^iv^j=g_{ij}(p)v^iv^j=v^iv^i

Multiplying {t^2}, it is equivalent to say

\displaystyle g_{ij}x^ix^j=x^ix^i

Furthermore, recalling the definition of Christoffel symbol, one can get

\displaystyle \frac{1}{2}(\partial_j g_{ik}+\partial_i g_{jk}-\partial_k g_{ij})x^ix^j=0

Or equivalently

\displaystyle \partial_j g_{ik}x^ix^j=\frac{1}{2}\partial_kg_{ij}x^ix^j=\frac{1}{2}\partial_k(g_{ij}x^ix^j)-g_{kj}x^j=x^k-g_{kj}x^j

While on the left hand side

\displaystyle \partial_j g_{ik}x^ix^j=\partial_j(g_{ik}x^i)x^j-g_{ik}x^i

Combining this two facts,

\displaystyle \partial_j(g_{ik}x^i)x^j=x^k

\displaystyle \partial_j(g_{ik}x^i-x^k)x^j=0

This means along {\gamma(t)},

\displaystyle \frac{d}{dt}(g_{ik}x^i-x^k)=0

Since at {p}, {g_{ik}x^i-x^k=0}, one get

\displaystyle g_{ik}x^i=x^k

on any point in the neighborhood. This identity is called Gauss Lemma.

Suppose {g(x)=\text{exp}(h(x))}, {h_{ij}(x)} is a symmetric 2-tensor. Then the above identity means {h_{ij}x^j=0}.

By Sun's World | Also posted in Riemannian Geo | Comments (0)

Torus and Flat Torus

November 24, 2014 – 2:58 pm

Torus

\displaystyle T(u,v)=((R+r\cos u)\cos v,(R+r\cos u)\sin v,r\sin u)

One can calculate the Gauss curvature by the following formula

\displaystyle E=T_u\cdot T_u, F=T_u\cdot T_v=0, G=T_v\cdot T_v

\displaystyle K=-\frac{1}{2\sqrt{EG}}\left(\frac{\partial }{\partial v}\left(\frac{E_v}{\sqrt{EG}}\right)+\frac{\partial }{\partial u}\left(\frac{G_u}{\sqrt{EG}}\right)\right)

In fact

\displaystyle K=\frac{\cos u}{r(R+r\cos u)}

So one can see that some places {K} is positive and some places {K} is negative. There is another torus homeomorphic to this one, which is also {S^1\times S^1} with different metric,

\displaystyle T=(\cos u,\sin u,\cos v,\sin v)

lies in the space {\mathbb{R}^4}. Using the formula above, one can get

\displaystyle E=1,F=0,G=1,\text{ then }K=0

So this torus has a name as flat torus. Flat torus can not be put in {R^3} because the following therem:

Theorem: On every compact surface {M\subset\mathbb{R}^3} there is some point {p} with {K(p)>0}.

Remark:Oprea, John. Differential geometry and its applications. p129

By Sun's World | Also posted in Riemannian Geo | Comments (0)

Invariance under the conformal mapping

July 9, 2014 – 12:34 am

Consider the 2-sphere {\mathbb{S}^2=\{x\in \mathbb{R}^3||x|^2=1\}} with the standard metric {g_0}. All conformal diffeomorphism of {\mathbb{S}^2} are composing a suitable isometries of {\mathbb{S}^2} and some {\pi^{-1}\circ M_t\circ \pi}, where {\pi} is the stereographic projection and {M_t:x\rightarrow tx} on {\mathbb{R}^2}. For any conformal metric {g=e^{2u}g_0}, define

\displaystyle S[u]= \int_{\mathbb{S}^2} |\nabla u|^2+2u\,d\mu_0

If {\phi} is a conformal transformation, then we can find {u_\phi} such that {\phi^*g=e^{2u_\phi}g_0}, where

\displaystyle u_\phi=u\circ\phi+\frac{1}{2}\log \det|d\phi|

here we use the notation {\phi^*g_0=\det|d\phi|g_0}.

Fact: for any conformal map {\phi} of {S^2}, {u=\frac{1}{2}\log\det|d\phi|} satisfies the following identity

\displaystyle \frac 12\Delta \log\det|d\phi|+\det|d\phi|=1

Actually, this identity means {(\mathbb{S}^2,\phi^*g_0)} has Gaussian curvature {1}, the same as {(\mathbb{S}^2, g_0)}.{\hfill\square}

Upon this fact, we have the invariance of {S[u]}.

Proposition: {S[u]=S[u_\phi]}.

Proof:

\displaystyle S[u_\phi]=\int |\nabla (u\circ \phi)+\frac{1}{2}\nabla\log \det|d\phi||^2+ 2u\circ \phi+\log \det|d\phi|

\displaystyle =\int |\nabla(u\circ\phi)|^2+\nabla (u\circ \phi)\cdot\nabla\log \det|d\phi|+2u\circ\phi+S[\frac{1}{2}\log \det|d\phi|]

\displaystyle =\int |\nabla(u\circ\phi)|^2+2u\circ\phi\det|d\phi|+S[\frac{1}{2}\log \det|d\phi|]

from integration by parts of the middle term. Suppose {\nabla u= g^{ij}\frac{\partial u}{\partial x^i}\frac{\partial }{\partial x^j}}, then

\displaystyle \nabla(u\circ\phi)=g^{ik}\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial\phi^\alpha}{\partial x^i}\frac{\partial}{\partial x^k}

So

\displaystyle |\nabla(u\circ\phi)|^2=g^{ik}\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial\phi^\alpha}{\partial x^i}g^{jl}\frac{\partial u}{\partial x^\beta}\circ\phi\cdot\frac{\partial\phi^\beta}{\partial x^j}g_{kl}

\displaystyle =\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial u}{\partial x^\beta}\circ\phi\cdot g^{ij}\frac{\partial\phi^\alpha}{\partial x^i}\frac{\partial\phi^\beta}{\partial x^j}

\displaystyle =\frac{\partial u}{\partial x^\alpha}\circ\phi\cdot\frac{\partial u}{\partial x^\beta}\circ\phi\cdot g^{\alpha\beta}\circ\phi\cdot\det|d\phi|=|(\nabla u)|^2\circ\phi\det|d\phi|

where we have used {\phi^*g_0=\det|d\phi|g_0}. Continuing our simplication of {S[u_\phi]},

\displaystyle S[u_\phi]=\int |(\nabla u)|^2\circ\phi\det|d\phi|+2u\circ\phi\det|d\phi|+S[\frac{1}{2}\log \det|d\phi|]

\displaystyle =S[u]+S[\frac{1}{2}\log \det|d\phi|]

by changing variables. So we only need to prove the last term is {0}, which is

\displaystyle \frac{1}{4}\int |\nabla \log\det|d\phi||^2+\log\det|d\phi|=0

integration by parts, this is equivalent to

\displaystyle \int \log\det|d\phi|=-\int\det|d\phi|\log\det|d\phi| \ \ \ \ \ (1)

From {g_0=\phi^*(\phi^{-1})^*g_0}, we get

\displaystyle \det|d\phi|\circ\phi^{-1}\cdot\det|d\phi^{-1}|=1

Changing variable by {x=\phi^{-1}(y)},

\displaystyle -\int\det|d\phi|(x)\log\det|d\phi|(x)d\mu_0(x)

\displaystyle =-\int\det|d\phi|\circ\phi^{-1}(y)\log\det|d\phi|\circ\phi^{-1}(y)\det|d\phi^{-1}|d\mu_0(y)

\displaystyle =\int \log\det|d\phi^{-1}|(y)d\mu_0(y)

So we only need to justify {\int \log\det|d\phi^{-1}|d\mu_0=\int \log\det|d\phi|d\mu_0}. As mentioned at the begining, up to some isometry, {\phi=\pi^{-1}\circ M_t\circ\pi} for some {t}, where {\pi} is the stereographic projection of north pole. Then {\phi^{-1}=\tilde{\pi}^{-1}\circ M_{1/t}\circ\tilde{\pi}}, where {\tilde{\pi}} is the stereographic projection of south pole. Note that

\displaystyle \det|d\phi|(x)=\det|d\phi^{-1}|(\tilde{x})

where {\tilde{x}=(x_1,x_2,-x_3)} if {x=(x_1,x_2,x_3)}. By changing variables

\displaystyle \int \log\det|d\phi^{-1}|(x)d\mu_0(x)=\int \log\det|d\phi^{-1}|(\tilde{x})d\mu_0(\tilde{x})

\displaystyle =\int \log\det|d\phi^{-1}|(\tilde{x})d\mu_0(x)=\int \log\det|d\phi|(x)d\mu_0(x)

\Box

Remark1: Under sterographic projection, {(\mathbb{S}^2\backslash\{P\},g_0)} is isometric to {(\mathbb{R}^2, \frac{4}{(1+|x|^2)^2}dx^2)}. Then for {\phi:x\rightarrow tx} on {\mathbb{R}^2}

\displaystyle \det|d\phi|=\frac{t^2(1+|x|^2)}{(1+t^2|x|^2)}

Another point of view is thinking {\phi:\mathbb{R}^2\rightarrow \mathbb{R}^2} as a diffeomorphism, then {d\phi} is the transformation of corresponding tangent space. One can also get {\det|d\phi|} is the above expression.

Remark2: Alice Chang, Paul Yang, prescribing curvature on {\mathbb{S}^2}, 1987.

By Sun's World | Also posted in Riemannian Geo | Tagged | Comments (0)
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