Category Archives: Geometry

63411

Basic concepts of Riemanian Geometry

April 28, 2013 – 2:26 am

Defition of Topological manifold
Suppose {M} is a topological space, for every point in {M}, there exists a neighborhood {U} of {x} and {\phi:U\rightarrow \mathbb{R}^n} such that {\psi} is homoemorphism of {U} and an open subset in {\mathbb{R}^n}, then {M} is called {n-}dimensional topological manifold.

In order to have differentiation and integration on manifold, we need {M} have differentiable structure.

Suppose {M} is {n-}dimensional manifold. If a given set of coordinate chars {\mathcal{A}=\{(U_\alpha,\phi_\alpha)|\alpha \text{is some index}\}} satisfies

(1){\bigcup\limits_\alpha U_\alpha} coves {M}

(2) the transition map between two local coordinate charts {(U_\alpha,\phi_\alpha)} and {(U_\beta,\phi_\beta)} are {C^k} smooth. Namely the map

\displaystyle \phi_\alpha\circ \phi_\beta^{-1}: \phi_\beta(U_\alpha\cap U_\beta)\rightarrow \phi_\alpha(U_\alpha\cap U_\beta)

is {C^k} differentiable.

(3) {\mathcal{A}} is maximal. If there exists some {(U,\phi)} such that it is compatible to every {(U_\alpha,\phi_\alpha)}, then {(U,\phi)\subset\mathcal{A} }.

Tangent vector on {M}.

A tangent vector at {p} on {M} is a map {v:C_p^\infty\rightarrow \mathbb{R}^1} satisfies

(1) {v(af+bg)=av(f)+b(g)}, for {f,g\in C^\infty_p} and {a,b} are real numbers;

(2) {v(fg)=v(f)g(p)+f(p)v(g)}.

Suppose {\gamma(t):[0,1]\rightarrow M}, then {\gamma} induces a tangent vector at {\gamma(0)=p} by

\displaystyle v(f)=\frac{d(f\circ\gamma)}{dt}\bigg|_{t=0}

If we have a coordinate {(u^1,u^2,\cdots,u^{n})} near {p}, then

\displaystyle v(f)=\frac{\partial f}{\partial u^i}\frac{\partial u^i(\gamma(t))}{dt}\bigg|_{t=0}

where {\displaystyle \left(\frac{\partial f}{\partial u^i}\right)_p=\left(\frac{\partial f\circ \phi^{-1}}{\partial u^i}\right)_{\phi(p)}}. Then

\displaystyle v=\frac{\partial u^i(\gamma(t))}{dt}\bigg|_{t=0}\frac{\partial}{\partial u^i}

is a tangent vector at {p}.

Tangent space

All tangent vectors form a linear space. For {f\in C^\infty(\mathbb{R}^n)}, there exists {g\in C^\infty(\mathbb{R}^n)} such that

\displaystyle f(u^1,u^2,\cdots,u^n)=f(0,0,\cdots,0)+ u^ig_i(u^1,u^2,\cdots,u^n)

where {g(0,0,\cdots,0)=\frac{\partial f}{\partial u^i}(0,0,\cdots,0)}. Then any tangent vector {v}, we have

\displaystyle v(f)=v(u^ig_i(u^1,u^2,\cdots,u^n))=\left(\frac{\partial f}{\partial u^i}\right)_pv(u^i)=\left(\frac{\partial f}{\partial u^i}\right)_p\frac{\partial}{\partial u^i}

Moreover we can prove that {\left\{\displaystyle \frac{\partial}{\partial u^i}\right\}_{i=0}^n} is linearly independent. So {\left\{\displaystyle \frac{\partial}{\partial u^i}\right\}_{i=0}^n} forms a basis of the tangent space.

Tangent bundle

\displaystyle TM=\bigcup_p T_pM=\{X_p\in T_pM|p\in M\}

We can assign topology and differentiable stucture on tangent bundle.

Suppose {M} is {n-}dimensional manifold. Then

\displaystyle \pi:TM\mapsto M

\displaystyle X_p\rightarrow p

For every {X_p\in M}, there exists a coordinates of {p} on {M}, say {(U,\phi)}, then Let {\tilde{U}=\pi^{-1}(U)=\{x\in T_pM|p\in U \}}, define

\displaystyle \tilde{\phi}:\tilde{U}\mapsto \phi(U)\times \mathbb{R}^n

\displaystyle \phi(X_p)\rightarrow (x^i(p); \xi^i)

where {X_p=\xi^i\frac{\partial}{\partial x^i}}. Obviously, {\tilde{\phi}} is bijective. Define the open set of {TM} is {\{\tilde{\phi}^{-1}(A\times B)|A \subset \phi(U) \text{ is open } B\subset \mathbb{R}^n\text{ is open }\}}. So {TM} is a {2n-}dimensional manifold, and {\pi:TM\rightarrow M} is continuous.

Next suppose {M} has {C^k} differentiable structure. There exists an open cover {(U_\alpha,\phi_\alpha)}, then {TM} has an open cover {(\tilde{U}_\alpha, \tilde{\phi}_\alpha)} and for {\tilde{U}_\alpha\cap \tilde{U}_\beta\neq \emptyset}

\displaystyle \tilde{\phi}_\alpha\circ \tilde{\phi}_\beta^{-1}: \tilde{\phi}_\beta(\tilde{U}_\alpha\cap \tilde{U}_\beta)\mapsto \tilde{\phi}_\alpha(\tilde{U}_\alpha\cap \tilde{U}_\beta)

\displaystyle (x_\beta^i; \xi_\beta^i)\rightarrow (x_\alpha^i; \xi_\alpha^i)

where

\displaystyle x_\alpha^i=(\phi_\alpha\circ\phi_\beta^{-1})^i(x_\beta^1,\cdots,x_\beta^n)

\displaystyle \xi_\alpha^i=\xi^k_\beta\frac{\partial (\phi_\alpha\circ\phi_\beta^{-1})^j}{\partial x_\beta^k}

Hence {\tilde{\phi}_\alpha\circ \tilde{\phi}_\beta^{-1}} is {C^{k-1}} transition function. {TM} has a {C^{k-1}} differentiable structure.

Metric

Similarly {T^*M} is the cotangent bundle which is also a manifold. Riemanian metric is a smooth function

\displaystyle g: M\rightarrow T^*M\otimes T^*M=\bigcup_pT^*_pM\otimes T^*_pM

satisfies

(1) {g_{ij}(p)=g_{ji}(p)}

(2) matrix {(g_{ij}(p))>0}

 

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

Gauss lemma and geodesics

April 28, 2013 – 12:19 am

From the Gauss lemma to the minimality of geodesics

Lemma:(Gauss) Suppose {p\in M}, {v\in T_u(T_pM)}

\displaystyle \langle u,v\rangle=\langle(d\exp_p)_uu,(d\exp_p)_uv\rangle

Here we make natural equivalence of {T_u(T_pM)} and {T_pM}. This lemma means {d\exp} presvers the inner product.

For any {p\in M} there exists a neighborhood {U} of {p} such that the exponential map is diffeomorphism from {B_\epsilon(0)\subset T_p(M)} to {U_p}.

Thm: {p\in M}, for such {U_p}, suppose {B=\{\exp_pv|||v||<\epsilon\}} is any geodesic ball contained in {U_p}. Then the shortest length connecting a point on {\partial B} and {p} is the geodesic ray from {p}.

Proof: Suppose {\gamma(t)=\exp_p\sigma(t):[0,1]\rightarrow M} connects {p} and {q\in \partial B}, where {\sigma(t)} is a curve in {T_pM}. Suppoe {\displaystyle \alpha(t)=\frac{\sigma(t)}{|\sigma(t)|}} is the unit normal along {\sigma(t)}, then {\dot{\sigma}(t)=\langle \dot{\sigma}(t),\alpha(t)\rangle \alpha(t)+\beta(t)} with {\beta(t)} is perpendicular to {\alpha(t)}. Then by Gauss lemma

\displaystyle \dot{\gamma}(t)=(d\exp_p)_{\sigma(t)}\dot{\sigma}(t)=(d\exp_p)_{\sigma(t)}\left[\langle\dot{\sigma}(t),\alpha(t)\rangle \alpha(t)+\beta(t)\right]

\displaystyle =\langle\dot{\sigma}(t),\alpha(t)\rangle(d\exp_p)_{\sigma(t)}\alpha(t)+(d\exp_p)_{\sigma(t)}\beta(t)

\displaystyle l(\gamma)=\int_0^1 |\dot{\gamma}(t)|dt\geq \int_0^1|\langle\dot{\sigma}(t),\alpha(t)\rangle(d\exp_p)_{\sigma(t)}\alpha(t)|dt

\displaystyle =\int_0^1 |\langle\dot{\sigma}(t),\alpha(t)\rangle|dt(\text{ since }||\alpha(t)||=1)

\displaystyle \geq \int_0^1 \langle\dot{\sigma}(t),\alpha(t)\rangle dt=\int_0^1 \langle {\sigma}(t),\alpha(t)\rangle'dt=\langle {\sigma}(1),\alpha(1)\rangle

\displaystyle =||\sigma(1)||=\text{length of }\exp t\sigma(1)= \text{geodesic ray from }p \text{ to } q

We use the fact {\langle{\sigma}(t),\dot{\alpha}(t)\rangle=0}, because {||\alpha(t)||=1} which implies {\langle{\alpha}(t),\dot{\alpha}(t)\rangle=0}.

Remark: I own this proof to Prof. Xiaochun Rong.

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

Metric of sphere and geodesics

April 11, 2013 – 6:15 pm

From the embedding of {i:\mathbb{S}^n(r)\rightarrow\mathbb{R}^{n+1}}, we have the induced metric {i^*g_0} on {\mathbb{S}^n}, {g_0} is the Euclidean metric on {\mathbb{R}^{n+1}} We can write this metric explicitly under stereographic projection.

\displaystyle \phi:\mathbb{S}^n(r)\rightarrow \mathbb{R}^n

\displaystyle (x^1,x^2,\cdots,x^{n+1})\backslash{N}\rightarrow(u^1,u^2,\cdots,u^n)

where {N} is the north pole {(0,0,\cdots,1)}

\displaystyle u^i=\frac{rx^i}{r-x^{n+1}}

and

\displaystyle \psi=\phi^{-1}:(u^1,u^2,\cdots,u^n)\backslash{N}\rightarrow(x^1,x^2,\cdots,x^{n+1})

by

\displaystyle x^i=\frac{2r^2u^i}{r^2+|u|^2},\quad x^{n+1}=\frac{(|u|^2-r^2)r}{r^2+|u|^2}

{g=\psi^*i^*g_0} will a metric on {\mathbb{R}^{n}},

\displaystyle g\left(\frac{\partial}{\partial u^i}, \frac{\partial}{\partial u^j}\right)=g_0\left((i\psi)_*\frac{\partial}{\partial u^i},(i\psi)_*\frac{\partial}{\partial u^j}\right)=\frac{4r^2\delta_{ij}}{(r^2+|u|^2)^2}

so

\displaystyle g=\frac{4r^2du^i\otimes du^i}{(r^2+|u|^2)^2}

From this we define 1-form basis {\displaystyle w^i=\frac{2rdu^i}{r^2+|u|^2}}, By the cartan structure equation

\displaystyle \begin{cases}dw^i=w^j\wedge w_j^i, w^i_j+w^j_i=0\\ dw_j^i=w_j^k\wedge w_k^i+\frac{1}{2}R_{jkl}^i w^k\wedge w^l\end{cases}

we can find that

\displaystyle w^i_j=\frac{u^idu^j-u^jdu^i}{r^2+|u|^2}

\displaystyle R_{jkl}^i=\frac{1}{r^2}(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk})

Also from the metric, we can obtain the christoffel symbols

\displaystyle \Gamma^k_{ij}=\frac{1}{2}g^{kh}\left\{\frac{\partial g_{ih}}{\partial u^j}+\frac{\partial g_{jh}}{\partial u^i}-\frac{\partial g_{ij}}{\partial u^h}\right\}

\displaystyle =\frac{-2}{r^2+|u|^2}\left\{u^j\delta_{ik}+u^i\delta_{jk}-u^k\delta_{ij}\right\}

We want to verify great circles on sphere are geodesics. Under some rotation, assume the great circle is

\displaystyle \gamma(\theta)=(r\sin\theta,0,\cdots,-r\cos\theta)

where {\theta} is the angle of {\gamma(t)} with the {x^{n+1}} axis, thus {\theta} represents the arc length of {\gamma(\theta)}. So after stereographic projection, we can get

\displaystyle \phi\gamma(\theta)=\left(\frac{r\sin\theta}{1+\cos\theta},0,\cdots,0\right)=(r\tan\frac{\theta}{2},0,\cdots)

Readers can verify it satisfies the equation of geodesics

\displaystyle \frac{d^2u^k}{d\theta^2}+\Gamma^k_{ij}\frac{du^i}{d\theta}\frac{du^j}{d\theta}=0

Remark: It is easy to make mistake here. My original intension is using {\alpha(t)=(t,0,\cdots,0)} on {\mathbb{R}^n} and try to verify {\alpha(t)} satisfies the geodesic equation. Since the image of {\alpha(t)} is a line, then {\phi^{-1}\alpha(t)} should be a great circle hence a geodesic.

But one can easily know {\alpha(t)} fails to satisfy the geodesic equation. The subtle error I made is pointed out by Bin Guo, which lies in the parametization of a curve. The geodesics are parametrized by arc length in usual.

To illustrate this, check {\gamma(t)=(t^2,0)} on {\mathbb{R}^2}, the image of {\gamma(t)} is a line, which is a geodesic under the Euclidean metric. But {\gamma(t)} does not satisfy the geodesic equation, in which {\Gamma^k_{ij}=0}.

For the same reason, in {\alpha(t)}, {t} is not such parameter. That is why I use {\theta} to define a curve.

There is an easy way to verify that great arc is geodesic on sphere even if we don’t know what the \Gamma_{ij}^k are.

For an embedded manifold in \mathbb{R}^n, suppose X=(x^1(t),\cdots, x^n(t)) and Y=(y^1(t),\cdots, y^n(t)) are two vector fields. Then

\displaystyle \nabla_XY=\pi\left(\frac{dY}{dt}\right)=\frac{dY}{dt}-\langle\frac{dY}{dt},\nu\rangle\cdot \nu

One can use this method to prove that \nabla_{\gamma'}\gamma'=0, for any \gamma is a part of great circle.

By Sun's World | Also posted in Riemannian Geo | Tagged | Comments (2)

Measure on sphere

March 21, 2013 – 1:12 am

Considering an n-sphere, which is the unit sphere in \mathbb{R}^n, we usually encounter this kind of integration

\displaystyle \int_{\mathbb{S}^n}f(x)dS_x

here dS_x is the sphere measure. For example find the value of \displaystyle \int_{\mathbb{S}^n}x^+_ndS_x.

To calculate this integral, we have to understand dS_x. What is the relation to Lebesgue measure in Euclidean space?

The key point is viewing \mathbb{S}^n as a manifold, then dS_x is the volume element of \mathbb{S}^n, which can be written as dS_x=\sqrt{g}du_1\wedge du_2\wedge \cdots \wedge du_{n-1} if u_1,u_2,\cdots,u_{n-1} is the local coordinates of \mathbb{S}^n.

\displaystyle \psi: B_1(0)\mapsto \mathbb{S}_+^n

\displaystyle (u_1,u_2,\cdots,u_{n-1})\to (u_1,u_2,\cdots,u_{n-1}, \gamma)

where B_1(0)\subset \mathbb{R}^{n-1} and \gamma=\sqrt{1-u^2_1-u^2_2-\cdots-u^2_{n-1}}.

Suppose the metric of \mathbb{R}^n is g_0, then it induces a metric \psi^*g_0 on \mathbb{S}^n.

\displaystyle \psi^*g_0\left(\frac{\partial }{\partial u_i}, \frac{\partial }{\partial u_j}\right)=\delta_{ij}+\frac{\partial \gamma}{\partial u_i}\frac{\partial \gamma}{\partial u_j}.

Since (u_1,u_2,\cdots,u_{n-1}) can serve as a local coordinates of \mathbb{S}_+^n, then

\displaystyle dS_x=\sqrt{\psi^*g_0}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}=\sqrt{1+|D\gamma|^2}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}.

Or we can get from the volume form

\displaystyle \omega = \sum_{j=1}^{n} (-1)^{j-1} x_j \,dx_1 \wedge \cdots \wedge dx_{j-1} \wedge dx_{j+1}\wedge \cdots \wedge dx_{n}

and substitute x_{n} by \gamma.

So

\displaystyle \int_{\mathbb{S}^n}x^+_ndS_x=\int_{B_1(0)}\gamma\sqrt{1+|D\gamma|^2}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}

which can be calculated by the polar coordinates.

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

A note on Huisken’s paper 1984

February 17, 2013 – 12:43 am

\mathbf{Thm(Myers):} If a Riemannian manifold has Ric\geq (n-1)K, then diam(M)\leq \frac{\pi}{K}.

Suppose we know a function h:M\to \mathbb{R}^+, M is compact and the gradient estimate

|\nabla h|\leq \eta^2 h^2_{\max} on M      (1)

and Ric_M\geq (n-1)c^2h^2 . Consider the relation \displaystyle h_{\min}\geq (1-\eta)h_{\max}.

\mathbf{Proof:}

Let x is a point on M, h achieves the maximum. Since (1), then

h(x)>(1-\eta)h_{\max} in B=B(x,\eta^{-1}h^{-1}_{\max}),

By the Myers thm, we get

\displaystyle diam(B)\leq \frac{\pi}{(1-\eta)h_{\max}}

But we know if \eta is small enough, then \displaystyle \eta^{-1}h^{-1}_{\max}>\frac{\pi}{(1-\eta)h_{\max}}. This forces diam(M)<\eta^{-1}h^{-1}_{\max}, then

\displaystyle h_{\min}\geq (1-\eta)h_{\max}.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Gerhard Huisken. Flow by mean curvature of convex surfaces into spheres.1984. p258 lemma 8.6.

The parameter \eta has relation with M and h_{\max}. We can not apply Myers thm directly. Guo Bin told me this method.

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

Second fundamental form of hypersurface

February 10, 2013 – 8:44 pm

\mathbf{Problem:} Suppose F:\Omega\to \mathbb{R}^{n+1} is an embedding map. \Omega\subset\mathbb{R}^nis an open set. Consider the Gauss equation of hypersurface

\displaystyle \overline{\nabla}_XY=\nabla^M_XY+h(X,Y)\nu\quad (1),

\nu is the outer normal vector. The Codazzi equation

\displaystyle \overline{\nabla}_X\nu=-A(X)\quad (2)

Then the second fundamental form of M is B=h_{ij}\partial_iF\otimes\partial_j F, where

\displaystyle h_{ij}=h(\partial_iF,\partial_jF)

There are two ways to express the second fundamental form, one is from (1), the other is from (2). Firstly, let us consider from (2)

\displaystyle h(\partial_iF,\partial_jF)\langle A(\partial_iF),\partial_jF\rangle_g=-\langle \displaystyle \overline{\nabla}_{\partial_iF}\nu,\partial_jF\rangle_g

Extend \displaystyle \partial_iF=F_*(\frac{\partial}{\partial u^i}) and \nu to a vector field X and Y on \mathbb{R}^{n+1}

\displaystyle X^A\big|_{F(p)}=\frac{\partial F^A}{\partial u^i}\bigg|_p, \displaystyle Y^A\big|_{F(p)}=\nu^A\big|_{F(p)}

\displaystyle \overline{\nabla}_{\partial_iF}\nu\big|_{F(p)}=\overline{\nabla}_{X}Y\big|_{F(p)}=X^A\frac{\partial}{\partial x^A}\left(Y^B\right)\frac{\partial}{\partial x^B}\bigg|_{F(p)}=\frac{\partial F^A}{\partial u^i}\bigg|_{p}\frac{\partial}{\partial x^A}\left(\nu^B\right)\bigg|_{F(p)}\frac{\partial}{\partial x^B}\bigg|_{F(p)}

Note the fact that \displaystyle \partial_i\nu^B=\frac{\partial}{\partial u^i}\nu^B=\frac{\partial}{\partial x^A}\left(\nu^B\right)\bigg|_{F(p)}\frac{\partial F^A}{\partial u^i}\bigg|_{p}, we get

\displaystyle \overline{\nabla}_{\partial_iF}\nu\big|_{F(p)}=\partial_i\nu^B\frac{\partial}{\partial x^B}=\partial_i \nu

\displaystyle h_{ij}=-\langle \partial_i\nu,\partial_jF\rangle_g=-\partial_i\nu\cdot \partial_jF

Secondly, let us consider from (1). Using the same extension process, one can prove

\displaystyle \overline{\nabla}_{\partial_iF}\partial_jF=\partial_i\partial_jF,    \displaystyle \nabla^M_{\partial_iF}\partial_jF=\Gamma_{ij}^k\partial_{k}F

So (1) is equivalent to

\partial_i\partial_jF=\Gamma_{ij}^k\partial_{k}F+h_{ij}\nu

\text{Q.E.D}\hfill \square

\mathbf{Remark:}see the notation on https://sunlimingbit.wordpress.com/2013/02/10/tangential-gradient-on-the-hypersurface/

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

Tangential gradient on the hypersurface

February 10, 2013 – 4:55 pm

\mathbf{Problem:}

Let the (u^1,u^2,\cdots,u^n) be the coordinates of \Omega and (x^1,x^2,\cdots,x^n,x^{n+1}) be the coordinates of \mathbb{R}^{n+1}. Denote \displaystyle\frac{\partial }{\partial x^i}=E_i. Suppose F:\Omega\to \mathbb{R}^{n+1} is an embedding map with F(\Omega)=M where \Omega\subset\mathbb{R}^n.

Then \displaystyle F_*\left(\frac{\partial }{\partial u^i}\right)=\frac{\partial F^l}{\partial u^i}\frac{\partial }{\partial x^l}=\partial_iF, 1\leq i\leq n form a basis for T_{F(p)}M at every p\in \Omega. Define the metric on M as

\displaystyle g_{ij}=\partial_iF\cdot \partial_j F

h:M\to \mathbb{R} is a function, we can define the tangential gradient of h as

\displaystyle \nabla^Mh=g^{ij}\partial_jh\partial_iF\quad (1)

Here we view h as a function on \Omega and \displaystyle \partial_jh=\frac{\partial }{\partial u^j}h(u^1,u^2,\cdots,u^n)

Consider the case when F is a graph of some function.

\displaystyle F:\Omega\mapsto \mathbb{R}^{n+1}

\displaystyle (u^1,u^2,\cdots,u^n)\rightarrow (u^1,u^2,\cdots,u^n, f(u_1,u_2,\cdots,u_n))

Then \partial_iF=E_i+p_iE_{n+1}, here \displaystyle p_i=\frac{\partial f}{\partial u^i}.

\displaystyle g_{ij}=\delta_{ij}+p_ip_j and W=\sqrt{1+\sum p_i^2}. \displaystyle g^{ij}=\delta_{ij}-\frac{1}{W^2}p_ip_j

The unit normal vector of M is e_{n+1}=\frac{1}{W}(\sum_ip_iE_i-E_{n+1}).

Naturally, the tangential vector of h should be defined as

\overline{\nabla} h-(\overline{\nabla} h\cdot e_{n+1})e_{n+1}\quad (2)

here \overline{\nabla} is respect to \mathbb{R}^{n+1}. Verify that this definition of tangential gradient is consistent with (1)

\mathbf{Proof:} Let us find the explicit expression of (2).

\displaystyle \overline{\nabla}h =(h_{x^1}, h_{x^2},\cdots,h_{x^{n+1}}) here \displaystyle h_{x^A}=\frac{\partial }{\partial x^A}h(x^1,x^2,\cdots,x^{n+1})

\displaystyle \overline{\nabla}h\cdot e_{n+1}=\frac{1}{W}\sum_ip_ih_{x^i}-\frac{1}{W}h_{x^{n+1}}

\displaystyle (\overline{\nabla}h\cdot e_{n+1})e_{n+1}=\left(\frac{1}{W}\sum_ip_ih_{x^i}-\frac{1}{W}h_{x^{n+1}}\right)\frac{1}{W}\left(\sum_ip_iE_i-E_{n+1}\right)

\displaystyle (3)\quad \overline\nabla h-(\overline \nabla h\cdot e_{n+1})e_{n+1}=\left(h_{x^i}-\frac{1}{W^2}\sum_{j}p_ip_jh_{x^j}+\frac{1}{W^2}p_ih_{x^{n+1}}\right)E_i+\left(h_{n+1}+\frac{1}{W^2}\sum_jp_jh_{x^j}-\frac{1}{W^2}h_{x^{n+1}}\right)E_{n+1}

Next, let us calculate (1)

\partial_ih=h_{x^i}+p_ih_{x^{n+1}}, \partial_j F=E_j+p_jE_{n+1}

\displaystyle (1)=\sum_{ij}\left(\delta_{ij}-\frac{1}{W^2}p_ip_j\right)\left(h_{x^i}+h_{x^{n+1}}p_i\right)\left(E_j+p_jE_{n+1}\right)

Expand this by using the fact W=\sqrt{1+\sum_ip^2_i}, one will see this is equal to (3)

\text{Q.E.D}\hfill \square

But there should have a simple way to understand how (1) is defined. Actually, this can be seen from the identity

\langle\nabla^Mh, X\rangle_g=dh(X)=X(h)\quad (4)

holds for every vector field X on M. The definition of \nabla^Mh should satisfies (4), which can also be seen from chern’s book.

Plugging in  X=\partial_iF in (4), we get

\displaystyle \langle\nabla^Mh, \partial_i F\rangle_g=\partial_iF(h)=F_*(\frac{\partial}{\partial u^i})h=\partial_ih

considering g_{ij}=\partial_iF\cdot\partial_iF=\langle\partial_iF,\partial_jF\rangle_g, we get (1)

\mathbf{Remark:} Klaus Ecker, Regularity theory for mean curvature flow. Appendix A

By Sun's World | Also posted in Mean curvature flow, PDE, Riemannian Geo | Tagged , | Comments (2)

Some geometric facts of Cylinder

January 15, 2013 – 9:22 pm

Let S^1\times \mathbb{R}^1\subset \mathbb{R}^3  be a cylinder with radius 1.

\displaystyle \begin{cases}x=\cos\theta\\ y=\sin\theta\\ z=z\end{cases}

Pull back the metric of \mathbb{R}^3 to cylinder ds^2=d\theta^2+dz^2. Then cylinder is locally flat Riemanian manifold. Its curvature tensor is 0 and thus the sectional curvature is also 0.

\textbf{Thm:} If M has non-positive sectional curvature, then any point x\in M does not have conjugate points.

Suppose p\in M. Then Conj(p)=M. Actually, any two points can be connected by infinitely many geodesics. But the Seg(p) is the cylinder except a line apposite to p.

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

Left and right fundamental differential form on Lie group

January 3, 2013 – 2:07 am

Suppose G is a Lie group. R_a and L_a are right and left translations. Assume e_i is a base of T_eG, the tangent space at the unit element e. X_i denote the right-invariant vector field obtained by the right translation and \tilde{X}_i by left translation. Let w^i and \tilde{w}^i denote the dual of X_i and \tilde{X}_i. Then we have the Maurer-Cartan equation

\displaystyle \begin{cases}dw^i=-\frac{1}{2}\sum\limits_{l,k=1}^n c^i_{lk}w^j\wedge w^k\\ c^i_{lk}+c^i_{kl}=0\end{cases}

Here c^i_{lk} are constants. Also we have the structure equation with respect to \tilde{w}^i. We see that there is a relation between \tilde{c}^i_{lk} and c^i_{lk}, that is

\tilde{c}^i_{lk}=-{c}^i_{lk}

\textbf{Proof: } Choose a local coordinate system (U,x^i) at e. Then

\displaystyle dy^i=\frac{\partial \phi^i(x,y)}{\partial x^j}\bigg|_{x=e}w^j

where \phi(x,y)=x\cdot y. Apply exterior differentiation on both sides

\displaystyle 0=\frac{\partial^2 \phi^i(x,y)}{\partial x^j\partial y^p}\bigg|_{x=e}dy^p\wedge w^j+\frac{\partial \phi^i(x,y)}{\partial x^j}\bigg|_{x=e}dw^j

Letting y=e, we get

\displaystyle dw^i=\frac{\partial^2 \phi^i(x,y)}{\partial x^k\partial y^l}\bigg|_{(e,e)}w^l\wedge w^k=-\frac{1}{2} \left(\frac{\partial^2 \phi^i(x,y)}{\partial x^l\partial y^k}\bigg|_{(e,e)}-\frac{\partial^2 \phi^i(x,y)}{\partial x^k\partial y^l}\bigg|_{(e,e)}\right)w^l\wedge w^k

Then it is easy to see \tilde{c}^i_{lk}=-{c}^i_{lk}.

\textbf{Remark: } see Chern’s lectures on differential geometry. p182

By Sun's World | Tagged | Comments (0)
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