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63411

Mean curvature of sphere cap

May 2, 2014 – 7:11 pm

Suppose one has the disc ${B_1=\{x\in \mathbb{R}^n||x|\leq 1\}}$ , ${n\geq 3}$ , prescribe the ball with metric

$\displaystyle g_{ij}=4u^{\frac{4}{n-2}}\delta_{ij}, \quad u=\left(\frac{\epsilon}{\epsilon^2+|x|^2}\right)^{(n-2)/2}$

What is the mean curvature of the boundary? As we all know that under the Euclidean metric, the boundary of unit ball has mean curvature ${h=1}$ . We want to use the formula of mean curvature under the conformal transmformation. Namely, suppose the ${(M,g_0)}$ has mean curvature ${h_0}$ , then under metric ${g=v^{\frac{4}{n-2}}g_0}$ , the mean curvature of ${(M,g)}$ will be

$\displaystyle h_g=\frac{2}{n-2}v^{-\frac{n}{n-2}}\left(\frac{\partial v}{\partial \eta}+\frac{n-2}{2}h_0 v\right)$

where ${\eta}$ is the normal outer unit vector under ${g_0}$ . Using the above principle, let ${v=2^{\frac{n-2}{2}}u}$ , ${g_0}$ be the Euclidean flat metric, then ${h_0=1}$ .

$\displaystyle \frac{\partial v}{\partial \eta}=(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)$

$\displaystyle \frac{n-2}{2}h_0 v=\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}$

$\displaystyle h_g=\frac{2}{n-2}\frac{(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n}{2}}(2-n)+\frac{n-2}{2}(2\epsilon)^{\frac{n-2}{2}}(\epsilon^2+1)^{-\frac{n-2}{2}}}{(2\epsilon)^{-\frac{n}{2}}(\epsilon^2+1)^{-\frac{n}{2}}}=\frac{\epsilon^2-1}{2\epsilon}$

Remark: Escobar. Conformal Defromation of a Riemannnian metric to a constant scalar curvature metric with constant mean curvature on the boundary. Indiana University Mathematics Journal 1996.

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

Conformally invariant Laplacian

December 15, 2013 – 10:10 pm

On a compact manifold ${(M^n,g)}$ , ${n\geq 3}$ ,

$\displaystyle L_g=-\Delta_g+\frac{n-2}{4(n-1)}R_g$

is called conformal laplacian operator. This follows from the following fact. Suppose ${\tilde{g}=\phi^{\frac{4}{n-2}}g}$ , then

$\displaystyle L_{\tilde{g}}(f)=\phi^{-\frac{n+2}{n-2}}L_g(\phi f), \quad\forall\,f\in C^\infty(M)$

Proof: Firstly suppose ${f>0}$ , define ${\bar{g}=(\phi f)^{\frac{4}{n-2}}g}$ , then

$\displaystyle L_{\bar{g}}(\phi f)=\frac{n-2}{4(n-1)}R_{\bar{g}}(\phi f)^{\frac{n+2}{n-2}}$

on the other hand ${\bar{g}=f^{\frac{4}{n-2}}\tilde{g}}$

$\displaystyle L_{\tilde{g}}(f)=\frac{n-2}{4(n-1)}R_{\bar{g}}(f)^{\frac{n+2}{n-2}}$

So we proved the equality. For general ${f\in C^\infty(M)}$ , ${\exists\, C>0}$ such that ${f+C>0}$ . By the special case

$\displaystyle L_{\tilde{g}}(f+C)=\phi^{-\frac{n+2}{n-2}}L_g(\phi (f+C))$

$\displaystyle L_{\tilde{g}}(C)=\phi^{-\frac{n+2}{n-2}}L_g(\phi (C))$

Thus the general case is also true. $\Box$

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

Cartan-Hadmard Theorem

November 7, 2013 – 3:13 pm

${\textbf{Thm: }}$ Suppose ${(M,g)}$ is a complete manifold

$\displaystyle \text{Sec}_M\leq 0$

Then ${\text{exp}_p:T_pM\rightarrow M}$ is a covering map. Moreover if ${M}$ is simply connected, then ${M}$ is diffeomorphic to ${\mathbb{R}^n}$ .

Remark: ${p}$ is the north pole of ${\mathbb{S}^n}$ , ${\text{exp}_p:\mathbb{R}^n\rightarrow \mathbb{S}^n}$ is the exponential map, then ${\text{exp}_p^*g}$ will not be a metric on ${\mathbb{R}^n}$ and ${\text{exp}_p^{-1}(-p)}$ are concentric circles. Proof: Since ${\text{Sec}_M\leq 0}$ , we have no conjugate points of ${p}$ , ${d\text{exp}_p}$ is nonsingular on ${T_pM}$ . So ${\tilde{g}=\text{exp}_p^*g}$ will be a metric on ${M}$ . In fact ${(T_pM,\tilde{g})}$ is a complete manifold. This is implied from Hopf-Rinow thm and the fact that lines through origin of ${T_pM}$ are geodesics under ${\tilde{g}}$ and they can be extended to infinity.

Since ${\text{exp}_p}$ is a local isometry between ${(T_pM,\tilde{g})}$ and ${(M,g)}$ , then

$\displaystyle \{V_i|\text{exp}_pV_i=q\}$

is a discrete set. Choose ${0<\delta<\text{injrad}(q)}$ . Denote ${\text{exp}_p=\phi}$

$\displaystyle \phi^{-1}(B_\delta(q))=\cup_iB_{\delta}(V_i)$

$\displaystyle \phi: B_{\delta}(V_i)\rightarrow B_\delta(q)\text{ is homeomorphism}$

$\displaystyle B_i\cap B_j=\emptyset,\quad i\neq j$

namely ${\phi}$ will be the covering map.

For any ${p'\in B_\delta(q)}$ , there exists a geodesic ${\gamma}$ connects ${p'}$ and ${q}$ . For any ${W_i\in \text{exp}^{-1}_pp'}$ , there exists a local lifting of ${\gamma}$ , say ${\tilde{\gamma}}$ , such that ${\gamma=\phi\circ\tilde\gamma}$ . Since ${\phi}$ is a local isometry, then it is length preserving. Since ${\tilde{\gamma}}$ ends up at some ${V_{i_0}}$ , then ${W_i}$ must be in ${B_\delta(V_{i_0})}$ .

Next ${\phi:B_{\delta}(V_i)\rightarrow B_\delta(q)}$ is injective. Otherwise suppose ${W_1, W_2\in B_{\delta}(V_i)}$ with ${\phi(W_1)=\phi(W_2)=p'}$ . Suppose ${\tilde\gamma_1}$ and ${\tilde\gamma_2}$ are geodesics connecting ${V_i}$ and ${W_1}$ and ${W_2}$ respectively, then both ${\gamma_1=\phi\circ\tilde\gamma_1}$ and ${\gamma_2=\phi\circ\tilde\gamma_2}$ will connect ${q}$ and ${p'}$ . ${\gamma_1}$ and ${\gamma_2}$ will be two different geodesics because ${\phi}$ is a local isometry. This can not happen because ${p'}$ is inside the injective domain.

${\phi:B_{\delta}(V_i)\rightarrow B_\delta(q)}$ is onto. Suppose ${\text{exp}_qtv}$ be the minimal geodesic from ${q}$ to ${p'}$ . Then there exist ${w\in T_{V_i}T_pM}$ such that ${d\phi (w)=v}$ , then ${\phi(\text{exp}_{V_i}tw)}$ will be a geodesic in ${M}$ with initial speed ${v}$ . Therefore it must concide with ${\text{exp}_ptv}$ . Since ${\phi}$ preserve the length, we have ${p'\in (B_\delta(V_i))}$ . ${\phi}$ is onto.

Suppose ${w\in B_\delta(V_i)\cap B_\delta(V_j)}$ , ${\tilde{\gamma}_i}$ and ${\tilde\gamma_j}$ are two minimal geodesics from ${w}$ to ${V_i}$ and ${V_j}$ . Then ${\tilde\gamma_i}$ and ${\tilde\gamma_j}$ are different near ${w}$ , then ${\gamma_i=\phi\circ\tilde{\gamma}_i}$ and ${\gamma_j=\phi\circ\tilde{\gamma}_j}$ will be two different geodesics connecting ${q}$ and ${\phi(w)}$ . Contracdtion implies ${B_\delta(V_i)}$ and ${B_\delta(v_j)}$ are disjoint. $\Box$

Remark: If we change the restriction to ${\text{Ric}_M\leq 0}$ , the theorem will not be true. Counter example comes from the Calabi-Yau manifold, which is compact, simply connected and has vanishing ricci curvature.

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

Divergence on manifold

October 31, 2013 – 12:38 am

Divergence on manifold Suppose ${(M,g)}$ is a Riemannian manifold. ${X}$ is smooth vector field on it. ${\{e_i\}}$ is a frame at ${p}$ . Suppose ${X=X^ie_i}$ , then define

$\displaystyle \text{div} X=\sum_iX^i_{,i}$

Under a local coordinates near ${p}$ , choose ${\{\frac{\partial}{\partial u^i}\}}$ as the basis of tangent space, then

$\displaystyle \text{div} X=\frac{1}{\sqrt{G}}\sum_i\frac{\partial(\sqrt{G}X^i)}{\partial u^i}$

where ${G=\det(g_{ij})}$ . One way to prove is straightforward verification. Let us see a more intuitive way, from the divergence theorem

$\displaystyle \int_{M}\text{div } XdV=\int_{\partial M}X\cdot \vec{n}\,dS$

Choose ${h\in C_c^\infty(U(p))}$ , where ${U(p)}$ is a neighborhood of ${p}$ , apply the divergence theorem to ${hX}$

$\displaystyle \int_{U}\text{div } (hX)dV=0$

which means

$\displaystyle \int_U h\text{div}\,XdV=-\int_U X\cdot \nabla h \,dV=-\int_U X^i\frac{\partial h}{\partial u^i}\sqrt{G}du^1\wedge\cdots\wedge du^n$

Denote ${\phi:U\rightarrow\mathbb{R}^n}$ is the diffeomorphism and ${\tilde{X}=X\circ\phi^{-1}}$ , ${\tilde{h}=h\circ \phi^{-1}}$ . From Green’s formula

$\displaystyle \int_U X^i\frac{\partial h}{\partial u^i}\sqrt{G}du^1\wedge\cdots\wedge du^n=\int_{\phi(U)}\tilde{X}^i\frac{\partial \tilde{h}}{\partial u^i}\sqrt{G}du^1\cdots du^n$

$\displaystyle =-\int_{\phi(U)}\tilde{h}\frac{\partial (\tilde{X}^i\sqrt{G})}{\partial u^i}du^1\cdots du^n=-\int_{U}\frac{1}{\sqrt{G}}\sum_i\frac{\partial(\sqrt{G}X^i)}{\partial u^i}hdV$

Combining the two equalities, we get that

$\displaystyle \text{div} X=\frac{1}{\sqrt{G}}\sum_i\frac{\partial(\sqrt{G}X^i)}{\partial u^i}$

When you are on a hypersurface, suppose ${F:U\rightarrow \mathbb{R}^{n+1}}$ is the embedding. Metric is ${g_{ij}=\partial_iF\cdot\partial_j F}$

Denote ${e_i=\partial_iF}$ , ${X}$ is any smooth vector field, not necessaritly tangent to ${M}$ , then

$\displaystyle \langle\nabla^M_{e_i}X,e_l\rangle=\langle\bar{\nabla}_{e_i}X,e_l\rangle=\partial_{i}X\cdot e_l$

So ${\nabla^M_{e_i}X=g^{jl}(\partial_iX\cdot e_l)e_j}$

$\displaystyle \text{div}_M\,X=X^i_{,i}=g^{il}\partial_iX\cdot e_l$

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

The thing about Hopf Rinow Theorem

August 16, 2013 – 3:05 am

Lemma 1: ${M}$ is a Riemannian manifold. ${\forall p\in M}$ , ${\exists\,}$ neighborhood ${N}$ of ${p}$ and ${\epsilon>0}$ such that ${\exp_p:B_\epsilon(0)\rightarrow N}$ is a diffeomorphism and any two points in ${N}$ can ve connected by a unique geodesic with length smaller than ${\epsilon}$ .

Lemma 2: ${p\in M}$ , ${r}$ is small enough such that ${\exp_p:B_r(0)\rightarrow N_r(p)=\exp(B_r(0))}$ is a differeomorphism, then for any ${q\not \in N_r(p)}$ , ${\exists \,q'\in \partial N_r(p)}$ such that

$\displaystyle d(p,q)=r+d(q',q)$

Proof: ${d(\cdot,q)}$ is a continuous function on ${\partial N_r(p)}$ . Since ${\partial N_r(p)}$ is compact, then ${\exists\, q'\in \partial N_r(p)}$ such that

$\displaystyle d(q',q)=\inf\{d(\tilde{q},q)|\tilde{q}\in\partial N_r(p)\}$

Suppose ${\gamma_n}$ is a minimizing sequence of ${d(p,q)}$ . For any ${\gamma_n}$ , there exists ${q_n\in \partial N_r(p)\cap \gamma_n}$ , then

$\displaystyle L(\gamma_n)\geq d(p,q_n)+d(q_n,q)\geq r+d(q',q)$

Letting ${n\rightarrow \infty}$ , ${d(p,q)\geq r+d(q',q)}$ . By the triangle inequality,

$\displaystyle d(p,q)= r+d(q',q)$

$\Box$

Thm(Hopf-Rinow) The following statements are equivalent for Riemannian manifold ${M}$ :
(1) ${M}$ is a complete metric space, where the metric is induced by

$\displaystyle d(p,q)=\inf\{L(\gamma)|\gamma\text{ is a curve connects } p,q\}$

(2)The closed and bounded sets of ${M}$ are compact.
(3) ${\exists \,p\in M}$ such that ${\exp_p}$ is defined on all of ${T_pM}$ .
(4) ${\forall\,p\in M}$ , ${\exp_p}$ is defined on ${T_pM}$
Furthermore, each of the statements ${(1-4)}$ implies
(5) Any two points ${p,q\in M}$ can be connected by a geodesic of shortest length. Proof: Let us prove ${(3)\Rightarrow (5)}$ firstly.
By lemma 2, there exists ${r>0}$ and ${q'}$ on ${\partial N_r(p)}$ such that

$\displaystyle d(p,q)=r+d(q',q)$

Suppose ${q'=\exp_ptv}$ for some ${v\in T_pM}$ , ${||v||=1}$ , then ${\gamma(t)=\exp_ptv}$ is defined on ${[0,\infty)}$ by assumption. Consider ${\{t|t\in [r,d(p,q)]\}}$ satisfies

$\displaystyle d(p,q)=t+d(\gamma(t),q),$

denote such points as ${I}$ . ${I}$ is nonempty as ${r\in I}$ and it is closed by the sake of continuity.
Suppose ${t_0=\max_{t\in I}t}$ . If ${t_0=d(p,q)}$ , we are done. Otherwise consider ${q_0=\gamma(t_0)}$ , by lemma 2, ${\exists\,\epsilon>0}$ small enough, ${q''\in\partial N_\epsilon(q_0)}$ such that

$\displaystyle q\not\in N_{\epsilon}(q_0),\quad d(q_0,q)=\epsilon+d(q'',q)$

then

$\displaystyle d(p,q)=t_0+\epsilon+d(q'',q)$

We are going to prove ${q''=\gamma(t_0+\epsilon)}$ . From the triangle inequality

$\displaystyle d(p,q'')\geq d(p,q)-d(q'',q)=t_0+\epsilon$

$\displaystyle d(p,q'')\leq d(p,q_0)+d(q_0,q'')=t_0+\epsilon$

then we must have ${d(p,q'')=t_0+\epsilon}$ . Then the union of ${\gamma([0,t_0])}$ and the geodesic from ${q_0}$ to ${q''}$ constitutes a shortest curve from ${p}$ to ${q''}$ , therefore it must be a geodesic. By lemma 1, such geodesic is unique with given initial values. So it must concider with ${\gamma(t)}$ in the ${N_\epsilon(q_0)}$ , which means ${q''=\gamma(t_0+\epsilon)}$ . Then

$\displaystyle d(p,q)=t_0+\epsilon+d(\gamma(t_0+\epsilon),q)$

This contradicts the fact ${t_0}$ is maximal. So ${t_0}$ must be ${d(p,q)}$ , and the curve ${\gamma([0,t_0])}$ is just the minimal geodesic connects ${p}$ and ${q}$ .

${(4)\Rightarrow (3)}$ Evidently

${(3)\Rightarrow (2)}$ Suppose ${K}$ is a bounded set in ${M}$ . From ${(5)}$ we know, ${K}$ can be contained in ${\exp_p B_r(0)}$ where ${B_r(0)\subset T_pM}$ for some ${r}$ large enough. Since ${B_r(0)}$ is compact and ${\exp_p}$ is a continuous mapping, then ${\exp_p B_r(0)}$ is compact, therefore its closed subset ${K}$ is compact.

${(2)\Rightarrow(1)}$ Any cauchy sequence in ${M}$ is bounded, its closure is compact by ${(2)}$ . Then it must have a converge subsequence and being cauchy, it has to converge itself.

${(1)\Rightarrow (4)}$ For any ${p\in M}$ , we need to prove ${\exp_ptv}$ is defined on ${[0,\infty)}$ for any ${v\in T_pM}$ . Consider ${t_n\nearrow T<\infty}$ and ${\exp_p tv}$ is defined on each ${t_n}$ , denote ${\exp_pt_nv=q_n}$ .

Since ${d(q_n,q_m)\leq |t_n-t_m|}$ , ${\{q_n\}}$ is a cauchy sequence, ${\exists\, q\in M}$ such that ${q_n\rightarrow q}$ . Since there exists a neighborhood of ${q}$ , say ${N_\epsilon(q)}$ , such that ${\forall z\in N_\epsilon(q)}$ , any geodesic starting from ${z}$ can be extended at least up to length ${\rho_0>0}$ .

For sufficiently large ${n}$ , ${q_n\in N_\epsilon(p)}$ and ${d(q_n,q)<\rho_0}$ . ${\exp_ptv|_{[t_n,t_{n+1}]}}$ is a curve starting from ${\gamma(t_n)}$ , then it can be extended at least up to length ${\rho_0}$ . By the uniqueness of geodesic, ${\exp_ptv}$ can be extened beyond ${q}$ , namely ${q=\exp_pTv}$ is well defined. So ${\exp_p}$ is defined on whole ${[0,\infty)}$ .

$\Box$

Remark: Should thank Bin Guo’s picture

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

Non-compact and complete manifold

June 15, 2013 – 6:49 pm

Thm: A normal geodesic ${\gamma:[0,\infty)\rightarrow M}$ is called shortest if and only if ${\gamma|_{[0,t]}}$ is shortest. A complete Riemannian manifold is non-compact if and only if for any point ${p\in M}$ , there is a shortest geodesic ${\gamma:[0,\infty)\rightarrow M}$ such that ${\gamma(0)=p}$

Proof: If there is a shortest geodesic ${\gamma:[0,\infty)\rightarrow M}$ , then ${\{\gamma(n), n=1,2,\cdots\}}$ is a sequence of points with no converging subsequence, because ${|\gamma(i)-\gamma(j)|\geq 1}$ , ${i\neq j}$ . So ${M}$ is non-compact.

If ${M}$ is complete and non-compact, from the Hopf-Rinow theorem, there exists ${p_n}$ , ${n\geq 0}$ such that ${d(p,p_n)\rightarrow \infty}$ as ${\infty}$ . Suppose ${\gamma_n=\exp_p{tv_n}:[0,a_n]\rightarrow M}$ is the shortest normal geodesic connecting ${p}$ and ${p_n}$ , ${v_n\in B_1(0)\subset T_pM}$ . Then ${\exists\, v\in B_1(0)}$ such that ${v_n\rightarrow v}$ , going to subsequence if necessary. Consider ${\exp_ptv:[0,\infty)\rightarrow M}$ , if it is not shortest for some ${t_0}$ , then there exists ${t_0>d(p,\gamma(t_0))+\epsilon_0}$ . Since ${\gamma_n(t)\rightarrow \gamma(t)}$ for any fixed ${t>0}$ ,

$\displaystyle t_0=d(p,\gamma_n(t_0))\leq d(p,\gamma(t_0))+d(\gamma(t_0),\gamma_n(t_0))<t_0-\epsilon_0-d(\gamma(t_0),\gamma_n(t_0))$

when ${n}$ is large enough, we get contradiction. So ${\gamma}$ is shortest. $\Box$

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

Index form and Jacobi field

June 12, 2013 – 7:32 pm

Assume ${\gamma(t):[0,a]\rightarrow M}$ is a normal geodesic, ${\gamma(t,s):[0,a]\times[-\epsilon,\epsilon]\rightarrow M}$ is one parameter smooth variation of curve ${\gamma(t)}$ . Denote ${\gamma_s(t)=\gamma(t,s)}$ .

$\displaystyle V(t)=\frac{\partial \gamma_s}{\partial s}\bigg|_{s=0}$

is called the variation field. As we all know if the variation is geodesic one, then ${V}$ is a Jacobi field. If ${L(s)=L(\gamma_s(t))}$ , then

$\displaystyle L''(0)=\langle\nabla_VV,T\rangle|_0^a+\int_0^a |\dot{V}(t)|^2-R(T,V,T,V)dt$

Let us define $\mathcal{V}=\{V|V\perp \dot{\gamma}\text{ and } V \text{ is piecewise smooth along }\gamma\}$ and ${\mathcal{V}_0=\{V\in\mathcal{V}|V(0)=V(a)=0\}}$ is a subspace of ${\mathcal{V}}$ . Index form

$\displaystyle I(X,Y)=\int_{0}^a\langle\dot{X},\dot{Y}\rangle-R(T,X,T,Y)dt$

for ${X,Y\in\mathcal{V}}$ .

Thm1: If ${\gamma}$ does not have conjugate point, then ${I(X,X)>0}$ for any ${X\in \mathcal{V}_0}$ and ${X\neq 0}$ . This is equivalent to say index form is positive definite.

Thm2(Minimality of Jacobi field): Assume ${\gamma}$ does not have conjugate point. Suppose ${V,X\in \mathcal{V}}$ and ${V(0)=X(0)}$ and ${V(a)=X(a)}$ . If ${V}$ is a Jacobi field on ${\gamma}$ , then

$\displaystyle I(V,V)\leq I(X,X)$

Thm1 and Thm2 are equivalent.
Thm1 ${\Rightarrow}$ Thm2:

${V-X\in \mathcal{V}}$ , then

$\displaystyle 0\leq I(V-X,V-X)=I(V,V)-2I(V,X)+I(X,X)$

$\displaystyle =\langle\dot{V},V\rangle|_0^a-2\langle\dot{V},X\rangle|_0^b+I(X,X)=I(V,V)-I(X,X)$

Thm2 ${\Rightarrow}$ Thm1: Obviously

One way to prove Thm1 is to prove Thm2. However, we will prove Thm1 directly.

Proof: Suppose ${\gamma(t)=\exp_ptv}$ , ${t\in [0,a]}$ . Since ${\gamma}$ does not have conjugate point, then ${d\exp}$ is non-degenerate in ${tv}$ , ${t\in [0,a]}$ . There exists a neighborhood ${U}$ of ${tv}$ such that ${d\exp:U\rightarrow M}$ is an immersion. From the fact that

${\sigma:[0,a]\rightarrow \exp_pU}$ is any piecewise smooth curve connecting ${\gamma(0)}$ and ${\gamma(a)}$ , then ${L(\gamma)\leq L(\sigma)}$ when the equality holds, ${\sigma}$ must be monotone reparametirzation of ${\gamma}$

After proving this, we can construct ${\gamma_s\subset \exp_pU}$ a variation of ${\gamma}$ such that the variation field is ${X}$ , then

$\displaystyle 0\leq L''(0)=I(X,X)$

Suppose ${I(X,X)=0}$ for some ${X\neq 0}$ , then for any ${Y\in \mathcal{V}_0}$

$\displaystyle 0\leq I(X+\epsilon Y,X+\epsilon Y)$

$\displaystyle 0\leq I(X-\epsilon Y,X-\epsilon Y)$

this means ${I(X,Y)=0}$ , ${\forall Y\in \mathcal{V}_0}$ . This means ${X}$ must be a Jacobi field. But ${X\in \mathcal{V}_0}$ and ${\gamma}$ has no conjugate point, ${X}$ must be 0.

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

Smoothness of distance function

June 11, 2013 – 5:53 pm

Suppose ${M}$ is a complete Riemannian manifold. ${p\in M}$ define ${\rho(x)=d(x,p)}$ . Obviously ${\rho(x)}$ is continuous, what can we say about the smoothness of ${\rho}$ .

(1) ${\rho(x)}$ is not ${C^1}$ near ${p}$ ;

(2) If ${M}$ is compact, then ${\rho(x)}$ is not ${C^1}$ in ${M\backslash \{p\}}$

It seems that ${\rho(x)}$ is not so smooth, let us consider ${\rho^2(x)}$

(3) ${\rho^2(x)}$ is smooth at neighborhood ${U}$ of ${p}$ and ${D^2\rho^2}$ is positive definite in ${U}$ .

(4) If ${M}$ is simply connected complete manifold with ${Sec_M\leq 0}$ , then ${\rho^2}$ is ${C^\infty}$ on whole ${M}$ and ${D^2\rho^2}$ is positive definite on ${M}$ .

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

Euclidean Laplacian in spherical polar coordinates

May 29, 2013 – 11:16 pm

We can express laplace of Euclidean space in sphere polar form,

$\displaystyle \Delta f= \frac{\partial^2 f}{\partial r^2}+ \frac{N-1}{r} \frac{\partial f}{\partial r}+ \frac{1}{r^2} \Delta_{\mathbb{S}^{N-1}} f$

Suppose ${r}$ is the distance function to the origin. ${(\theta^1,\cdots,\theta^{n-1})}$ is a local coordinates of ${\mathbb{S}^{N-1}}$ . And ${\phi:(\theta^1,\cdots,\theta^{n-1})\rightarrow \mathbb{S}^{n-1}}$ is the diffeomorphism. Then ${(r,\theta^1,\cdots,\theta^{n-1})}$ is a coordinates of ${\mathbb{R}^N}$ . Metric under this coordiates is

$\displaystyle g=\sum_i(d(r\phi^i))^2=\sum_i(\phi^idr+rd\phi^i)^2=(dr)^2+r^2\sum_i(d\phi^i)^2$

$\displaystyle =(dr)^2+r^2h_{ij}(\theta)d\theta^i d\theta^j$

$\displaystyle \Delta f=tr(Hess(f))=Hess(f)(\frac{\partial }{\partial r},\frac{\partial }{\partial r})+\frac{1}{r^2}h^{ij}Hess(f)(\frac{\partial }{\partial \theta^i},\frac{\partial }{\partial\theta^j})$

One can prove that

$\displaystyle Hess(f)(\frac{\partial }{\partial r},\frac{\partial }{\partial r})=\frac{\partial^2 f}{\partial r^2}+ \frac{N-1}{r} \frac{\partial f}{\partial r}$

Note that ${h_{ij}(\theta)d\theta^i d\theta^j}$ is the stardand metric of ${\mathbb{S}^{N-1}}$ , then

$\displaystyle \Delta_{\mathbb{S}^{N-1}}f=h^{ij}Hess(f)(\frac{\partial }{\partial \theta^i},\frac{\partial }{\partial\theta^j})$

Our formula is proved.

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

Meyer’s Thm

April 29, 2013 – 6:43 pm

Thm:(Meyers) Suppose ${M}$ is a ${n-}$ dimensional complete manifold and

$\displaystyle \text{Ric}_M\geq (n-1)K>0$

Then ${\text{diam}M<\frac{\pi}{\sqrt{K}}}$ hence ${M}$ is compact and ${\pi_1(M)}$ is finite.

This theorem is proved by the following lemma.

Lemma: Suppose ${\text{Ric}_M\geq (n-1)K>0}$ , then every geodesic which is longer than $\frac{\pi}{\sqrt{K}}$ on ${M}$ must have conjugate points hence not a minimal geodesic.

Proof: Suppose ${\gamma(t):[0,l]\rightarrow M}$ is a mininal normal geodesic with length $l>\frac{\pi}{\sqrt{K}}$ , ${t}$ is the arc length parameter. Then there exists $K'=\left(\frac{\pi}{l}\right)^2$ such that ${K'<K}$ .

For ${T_{\gamma(0)}M}$ , there exists an orthonormal basis ${\{\frac{\partial }{\partial t}, V_i, i=1,\cdots,n-1\}}$ . Let ${V_i(t)}$ be the parallel displacement of ${V_i}$ along ${\gamma}$ , ${T}$ is the one of ${\frac{\partial }{\partial t}}$ .

Define ${W_i=V_i(t)\sin \sqrt{K'}t }$ . Then the second variation of ${\gamma}$ along ${W_i}$ is

$\displaystyle I(W_i,W_i)=\int_{0}^l\langle\nabla_TW_i, \nabla_TW_i \rangle+R(T,W_i,T,W_i)dt$

$\displaystyle =\int_0^l-\langle W_i, \nabla_T\nabla_TW_i\rangle+ R(T,W_i,T,W_i)dt$

$\displaystyle =\int_0^l K'\sin^2\sqrt{K'}t+\sin^2\sqrt{K'}tR(T,V_i,T,V_i)dt$

$\displaystyle \sum_1^{n-1} I(W_i,W_i)=\int_0^l[(n-1)K'-\text{Ric}_M(V_i)]\sin^2\sqrt{K'}tdt<0$

So there exists a ${I(W_i,W_i)<0}$ , for such ${i}$ , this means ${\gamma}$ is locally maximal. This is contradiction, because we choose ${\gamma}$ is a minimal geodesic.

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

lim Practice= Perfect

Category Archives: Geometry

Mean curvature of sphere cap

Conformally invariant Laplacian

Cartan-Hadmard Theorem

Divergence on manifold

The thing about Hopf Rinow Theorem

Non-compact and complete manifold

Index form and Jacobi field

Smoothness of distance function

Euclidean Laplacian in spherical polar coordinates

Meyer’s Thm

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