Weighted isoperimetric inequalities in warped product manifolds

1. Introduction

The classical isoperimetric inequality on the plane states that for a simple closed curve on {\mathbb R^2}, we have {L^2\ge 4\pi A}, where {L} is the length of the curve and {A} is the area of the region enclosed by it. The equality holds if and only if the curve is a circle. The classical isoperimetric inequality has been generalized to hypersurfaces in higher dimensional Euclidean space, and to various ambient spaces. For these generalizations, we refer to the beautiful article by Osserman [O] and the references therein. For a more modern account see [Ros]. Apart from two-dimensional manifolds and the standard space forms {\mathbb R^n}, {\mathbb H^n} and {\mathbb S^n}, there are few manifolds for which the isoperimetric surfaces are known. According to [BM], known examples include {\mathbb R\times \mathbb H^n}, {\mathbb RP^3}, {\mathbb S^1\times \mathbb R^2}, {T^2\times \mathbb R}, {\mathbb R\times \mathbb S^n}, {\mathbb S^1\times \mathbb R^n}, {\mathbb S^1\times \mathbb S^2}, {\mathbb S^1\times \mathbb H^2} and the Schwarzschild manifold, most of which are warped product manifolds over an interval or a circle. There are also many applications of the isoperimetric inequalities. For example, isoperimetric surfaces were used to prove the Penrose inequality [B], an inequality concerning the mass of black holes in general relativity, in some important cases.

In this post, we prove both classical and weighted isoperimetric results in warped product manifolds, or more generally, multiply warped product manifolds. We also relate them to inequalities involving the higher order mean-curvature integrals. Some applications to geometric inequalities and eigenvalues are also given.

For the sake of simplicity, let us describe our main results on a warped product manifold. The multiply warped product case is only notationally more complicated and presents no additional conceptual difficulty.

Let {\displaystyle M=[0, l)\times N} ({l\le \infty}) be a product manifold. Equip {M} with the warped product Riemannian metric {\displaystyle g=dr^2+ s(r)^2 g_{N}} for some continuous {s(r)\ge 0}, where {g_{N}} is a Riemannian metric on the {m}-dimensional manifold {N}, which we assume to be compact and oriented. Define {B_R:=\{(r, \theta)\in M: r< R\}}. We define the functions {A(r)} and {v(r)} by

\displaystyle \begin{array}{rl} \displaystyle   A(r):= s(r)^m \; \textrm{ and }\; v(r):= \int_{0}^{r} A(t) dt. \end{array}

Up to multiplicative constants, they are just the area of {\partial B_r} and the volume of {B_r} respectively. For a bounded domain {\Omega} in {M}, we define {\Omega^\#} to be the region {B_R} which has the same volume as {\Omega}, i.e. {\mathrm{Vol} (B_R) =\mathrm{Vol} (\Omega) }. We denote the area of {\partial \Omega} and {\partial \Omega^\#} by {|\partial \Omega|} and {|\partial \Omega^\#|} respectively.

One of our main results is the following isoperimetric theorem, which is a special case of Theorem 11.

Theorem 1 Let {\Omega} be a bounded open set in {(M,g)} with Lipschitz boundary. Assume that

  1. The projection map {\pi: \partial \Omega\subset \mathbb [0, l)\times N\rightarrow N} defined by {(r, \theta)\mapsto \theta} is surjective.
  2. {s(r)} is non-decreasing.
  3. {A\circ v^{-1}} is convex, or equivalently, {s(r)s''(r)-s'(r)^2\ge 0} for {r>0} if {s} is twice differentiable.

Then the isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle |\partial \Omega|\ge|\partial \Omega^\#|.\end{array}

The equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=\mathrm{constant}\}}.

We remark that our notion of convexity does not require the function to be differentiable: {f} is convex on {I} if and only if {f((1-t)x+ty)\le (1-t)f(x)+tf(y)} for any {t\in(0,1)} and {x, y\in I}.

One feature of our result is that except compactness, we do not impose any condition on the fiber manifold {N}. We will see in Section 7 that without further restriction on {N} or {s}, our conditions are optimal in a certain sense.

The expression {ss''-s'^2} comes from the observation that if {s} is twice differentiable, then as {v'=s^m}, the convexity of {A(v^{-1} (u))=s\left( v^{-1}\left( u \right)\right)^m} is equivalent to

\displaystyle \begin{array}{rl} \displaystyle  \frac{d^2}{du^2} A \left(v^{-1} (u)\right) =\frac{m}{ s(r)^{m+2}}\left(s(r)s''(r)-s'(r)^2\right)\ge 0, \ \ \ \ \ (1)\end{array}

where {r=v^{-1} (u)}.

The expression {ss''-s'^2} also has a number of geometric and physical meanings. It is related to the stability of the slice {\Sigma=\{r=r_0\}} as a constant-mean-curvature (CMC) hypersurface (i.e. whether it is a minimizer of the area among nearby hypersurfaces enclosing the same volume). Indeed, it can be shown ([GLW] Proposition 6.2) that {\lambda_1(g_N)\ge m\left(s(r_0)'^2-s(r_0)s''(r_0)\right)} if and only if {\Sigma} is a stable CMC hypersurface, where {\lambda_1(g_N)} is the first Laplacian eigenvalue of {(N, g_N)}. It is also related to the so called “photon spheres” in relativity (see [GLW] Proposition 6.1).

From (1), {A\circ v^{-1}} is convex if and only if {s} is {\log}-convex. So if {f(r)} is non-decreasing and convex, then {s(r)=\exp(f(r))} satisfies the convexity condition. One example of such a function is {s(r)=e^r}.

If {\partial \Omega} is star-shaped in the sense that it is a graph over {N}, i.e. of the form {\partial \Omega=\{(r, \theta): r=\rho(\theta), \theta\in N\}}, we can remove the assumption on the monotonicity of {s(r)}.

Theorem 2 Suppose {\Omega} is a bounded open set in {(M,g)} with Lipschitz boundary. Assume that

  1. {\partial \Omega} is star-shaped.
  2. {A\circ v^{-1}} is convex, or equivalently, {ss''-s'^2\ge 0} if {s} is twice differentiable.

Then the isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle |\partial \Omega|\ge|\partial \Omega^\#|.\end{array}

The equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=\mathrm{constant}\}}.

If the classical isoperimetric inequality already holds on {(M,g)}, we can extend it by the following result, which is a special case of Theorem 8.

Theorem 3 (Weighted isoperimetric inequality) Let {\Omega} be a bounded open set in {(M,g)} with Lipschitz boundary. Assume that

  1. The classical isoperimetric inequality holds on {(M,g)}, i.e. {|\partial \Omega|\ge|\partial \Omega^\#|}.
  2. {a(r)} is a non-negative function such that {\psi(r):=b(r)A(r)} is non-decreasing and non-negative, where {b(r):=a(r)-a(0)}.
  3. The function {\psi\circ v^{-1} } is convex.


\displaystyle \begin{array}{rl} \displaystyle   \int_{\partial \Omega} a(r)dS \ge\int_{\partial \Omega^{\#}} a(r)dS. \end{array}

If {b(r)A(r)>0} for {r>0}, then the equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=\mathrm{constant}\}}.

Using a volume preserving flow, recently Guan, Li and Wang [GLW] (see also [GL2]) proved the following related result, assuming {s} is smooth ({\partial \Omega} is called “graphical” in [GLW]):

Theorem 4 ([GLW][Theorem 1.2]) Suppose {\Omega} is a domain in {(M,g)} with smooth star-shaped boundary. Assume that

  1. The Ricci curvature {\mathrm{Ric}_N} of {g_N} satisfies {\mathrm{Ric}_N\ge (m-1)K g_N}, where {K>0} is constant.
  2. {0\le s'^2-ss''\le K}.

Then the isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle |\partial \Omega|\ge|\partial \Omega^\#|.\end{array}

If { s'^2-ss''<K}, then the equality holds if and only if {\partial \Omega} is a coordinate slice {\{r=r_0\}}.

We note that our assumption {ss''-s'^2\ge 0} in Theorem 2 complements that of [GLW]. This does not contradict the result in [GLW]. In fact, we will show the necessity of this and other conditions in Section 7 (cf. Proposition 28). We also notice that except the obvious case that {M} has constant curvature, the equality holds only when {\partial \Omega} is a coordinate slice. Indeed, combining Theorem 4 with Theorem 3, we can generalize Theorem 4 as follows.

Theorem 5 (Theorem 10) Suppose {\Omega} is a domain in {(M,g)} with smooth star-shaped boundary. Assume that the Ricci curvature {\mathrm{Ric}_N} of {g_N} satisfies {\mathrm{Ric}_N\ge (m-1)K g_N} and {0\le s'^2-ss''\le K}, where {K>0} is constant. Suppose {a(r)} is a positive function such that {b( v^{-1} (u))s( v^{-1} (u))^m} is convex, where {b(r):=a(r)-a(0)}. Then the weighted isoperimetric inequality holds:

\displaystyle \begin{array}{rl} \displaystyle \int_{\partial \Omega}a(r)dS\ge \int_{\partial \Omega^\#}a(r)dS.\end{array}

The equality holds if and only if either

  1. {(M,g)} has constant curvature, {a(r)} is constant on {\partial \Omega}, and {\partial \Omega} is a geodesic hypersphere, or
  2. {\partial \Omega} is a slice {\{r=r_0\}}.

Combining Theorem 4, Theorem 2 and the proof of Proposition 28, we get the following picture for the isoperimetric problem in warped product manifolds:

Theorem 6 Let {M} be the product manifold {[0, l)\times N} equipped with the warped product metric {g=dr^2+s(r)^2 g_N}.

  1. Suppose {s'^2-ss''\le 0}. Then the star-shaped isoperimetric hypersurfaces are precisely the coordinate slices {\{r=r_0\}}.
  2. Suppose {0\le s'^2-ss''\le K} and {\mathrm{Ric}_N\ge (m-1)Kg_N} where {K>0} is constant. Then the star-shaped isoperimetric hypersurfaces are either geodesic hyperspheres if {(M,g)} has constant curvature, or the coordinate slices {\{r=r_0\}}.
  3. Suppose {s'^2-ss''>K} and {\lambda_1(g_N)\le mK} where {K>0} is constant. Then the coordinate slices {\{r=r_0\}} cannot be isoperimetric hypersurfaces.

We also prove isoperimetric type theorems involving the integrals of higher order mean curvatures {(H_k)} in warped product manifolds. For simplicity, let us state the result when the ambient space is {\mathbb R^n} (Corollary 18), which follows from a more general theorem (Theorem 17)

Theorem 7 (Corollary 18) Let {\Sigma} be a closed embedded hypersurface in {\mathbb R^{m+1}} which is star-shaped with respect to {0} and {\Omega} is the region enclosed by it. Assume that {H_k>0} on {\Sigma}. Then for any integer {l\ge 0},

\displaystyle \begin{array}{rl} \displaystyle   n \beta_n ^{-\frac{l-1}{n}}\mathrm{Vol} (\Omega)^{\frac{n-1+l}{n}} \le\int_{\Sigma} H_k r^{l+k}dS, \end{array}

where {\beta_n} is the volume of the unit ball in {\mathbb R^n}. If {l\ge 1}, the equality holds if and only if {\Sigma} is a hypersphere centered at {0}.

Note that when {k=l=0}, this reduces to {n {\beta_n}^{\frac{1}{n}}\mathrm{Vol} (\Omega)^{\frac{n-1}{n}} \le \mathrm{Area} (\Sigma) }, which is the classical isoperimetric inequality. In fact, we prove a stronger result (14) :

\displaystyle \begin{array}{rl} \displaystyle   n \beta_n^{-\frac{l-1}{n}}\mathrm{Vol} (\Omega)^{\frac{n-1+l}{n}} \le \int_{\partial \Omega} H_0 r ^{l} dS \le \int_{\partial \Omega} H_1 r ^{l+1} dS \le \cdots \le \int_{\partial \Omega} H_k r ^{l+k} dS. \end{array}

This can be compared to the following result of Guan-Li [GL][Theorem 2]:

\displaystyle \begin{array}{rl} \displaystyle   \left(\frac{1}{ \beta_n }\mathrm{Vol} (\Omega)\right)^{\frac{1}{n}}\le \left(\frac{1}{n\beta_n}{\int_\Sigma H_0dS}\right)^{\frac{1}{n-1}}\le \cdots \le \left(\frac{1}{n\beta_n} \int_\Sigma H_k dS \right) ^{\frac{1}{n-k-1}} \end{array}

under the same assumption.

Some applications of the weighted isoperimetric inequalities will also be given in Section 4 and Section 6.

The rest of this post is organized as follows. In Section 2, we first prove Theorem 3. In Section 3, we prove the isoperimetric inequality involving a weighted volume (Theorem 11), which implies Theorem 1 and Theorem 2. Although Theorem 3 can also be stated using the weighted volume, we prefer to prove the version involving only the ordinary volume for the sake of clarity, and indicates the changes needed to prove Theorem 11. In Section 4, we illustrate how we can obtain interesting geometric inequalities in space forms by using Theorem 8. In Section 5, we introduce the weighted Hsiung-Minkowski formulas in warped product manifolds, and combine them with the isoperimetric theorem to obtain new isoperimetric results involving the integrals of the higher order mean curvatures. In Section 6, we give further applications of our results to obtain some sharp eigenvalue estimates for some second order differential operators related to the extrinsic geometry of hypersurfaces and an eigenvalue estimate for the Steklov differential operator (also known as Dirichlet-to-Neumann map). A Pólya-Szegö inequality and a Faber-Krahn type theorem are also derived. Finally in Section 7, we show that the conditions of Theorem 1 are necessary by giving counterexamples where the isoperimetric inequality fails if any one of the conditions is violated.

Acknowledgements. We would like to thank Professor Frank Morgan for pointing out the reference [H] to us, and Professor Mu-Tao Wang for useful comments and discussion.
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Faber-Krahn inequality

I record a proof of the Faber-Krahn inequality here, mainly for my own benefit.

Let {M} be one of the standard space forms: the Euclidean space {\mathbb R^n}, the unit sphere {\mathbb S^n}, or the hyperbolic space {\mathbb H^n}. Suppose {\Omega } is a bounded domain in {M} with smooth boundary {\partial \Omega} which is a closed hypersurface. Then we can define {\Omega^\#} to be the geodesic ball in {M} which has the same volume as {\Omega}.

Let {\lambda_1(\Omega)} be the first Laplacian eigenvalue of {\Omega} under the Dirichlet boundary condition. i.e. {\lambda_1(\Omega)>0} is the smallest number {\lambda} such that there exists a non-zero smooth function {f} with

\displaystyle \begin{array}{rl} \displaystyle   \Delta f =-\lambda f\;  \displaystyle \textrm{in }\Omega, \quad f =0 \;\textrm{on }\partial \Omega. \end{array}

Theorem 1 (Faber-Krahn inequality) Suppose {\Omega } is a bounded domain in {M} with smooth boundary {\partial \Omega}. Then

\displaystyle \begin{array}{rl} \displaystyle   \lambda_1(\Omega)\ge \lambda_1(\Omega^\#).  \end{array}

The equality holds if and only if {\Omega} is a geodesic ball.

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Why is a² + b² ≥ 2ab ?

This post can be regarded as a sequel to my previous (and very ancient) post on 1+2+3+…. Though these two posts are not quite logically related, they share the same spirit (I’m asking a dumb question again).

How can one prove the following?

Theorem 1

\displaystyle \begin{array}{rl} \displaystyle a^2+b^2\ge 2ab.\end{array}

This extremely simple inequality (just next to the simplest but arguably the most important inequality {a^2\ge 0}) turns out to be the testing ground for a lot of more advanced inequalities (e.g. Cauchy-Schwarz, or Newton’s inequality).

In this post, I record some proofs of this inequality that I can think of. Some of them are quite similar, or may be considered almost the same depending how you measure the level of similarity. I try to group proofs of roughly the same idea together. In some of these proofs, we have actually used more advanced inequalities (which perhaps are even proved from this simple inequality). For such a simple result, I guess there may be over a hundred proofs.

So what’s the point of doing all these, since only one proof is all that suffices? Because not many people would have the patience to read till the end (or even this paragraph), let me put the summary here:

Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis? (Paul Halmos, “I want to be a mathematician: an automathography”)

So, let’s begin.

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Posted in Calculus, Discrete Mathematics, Geometry, Inequalities, Linear Algebra, Probability | 1 Comment

A remark on the divergence theorem

The divergence theorem states that for a compact domain {D} in {\mathbb R^3} with piecewise smooth boundary {\partial D}, then for a smooth vector field {\textbf{F}} on {D}, we have

\displaystyle \begin{array}{rl} \displaystyle \iint_{\partial D}\textbf{F}\cdot \textbf{n}\,dS=\iiint_D \nabla \cdot\textbf{F}\,dV,\end{array}

where {\textbf{n}} is the unit outward normal and {\nabla \cdot\textbf{F}} is the divergence of {\textbf{F}}.

In most textbooks, the divergence theorem is proved by the following strategy:

  1. Prove that

    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(0, 0, R)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial R}{\partial z}\;dV \ \ \ \ \ (1)\end{array}

    where {D} is a so called type {1} domain, i.e. {D=\{(x, y, z): (x, y)\in \Omega, \underline f(x, y)\le z\le \bar f(x, y)\}} for some region {\Omega\subset \mathbb R^2} and {\bar f, \underline f\in C^1(R)}. i.e. {D} is bounded by the graphs of two functions with variables {x, y}. This can be quite easily done using the fundamental theorem of calculus.

  2. Similarly prove

    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(P, 0, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial P}{\partial x}\;dV \ \ \ \ \ (2)\end{array}


    \displaystyle \begin{array}{rl} \displaystyle  \iint _{\partial D}(0, Q, 0)\cdot \textbf{n} \;dS= \iiint _D\frac{\partial Q}{\partial y}\;dV \ \ \ \ \ (3)\end{array}

    for the so called type {2} domain and type {3} domain respectively.

  3. Argue that for sufficiently nice {D}, it can be cut into finitely many pieces {D_i}, each of which are of both type {1}, {2}, and {3} (the so called type {4} domain). Thus we can compute the divergence integral {\displaystyle \iiint_D \nabla \cdot \textbf{F}\,dV} by adding the integrals of the pieces and apply the previous result (using linearity). Argue that the orientation of the common face of these {D_i} (if any) is opposite to each other and thus a cancellation argument will give the surface integral {\displaystyle \iint _{\partial D}\textbf{F}\cdot \textbf{n}\;dS}.

I am slightly disappointed by this approach because even for the so called type {1} domain, we still cannot prove the divergence theorem very directly (say, by reducing to a double integral on some planar domain {\Omega} and apply Green’s theorem), but have to further cut it into even smaller type {4} domains.

In this note we modify the strategy above by proving (1), (2), (3) all hold on a type {1} domain. In particular, I will show that (2) and (3) can be proved for a type {1} domain by using Green’s theorem.
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The Cauchy-Schwarz inequality and the Lagrange identity

The classical Lagrange identity is the following:

\displaystyle \begin{array}{rl} \displaystyle \left(\sum_{i=1}^n a_ib_i\right)^2+\sum _{1\le i<j\le n} \left(a_ib_j-a_jb_i\right)^2 =\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{j=1}^{n}b_j^2\right).\end{array}

This can be proven by expanding {\displaystyle \sum _{1\le i<j\le n}\left(a_ib_j-a_jb_i\right)^2 } and separating the terms into the cross-terms part and the non cross-terms part.

The Lagrange identity implies the Cauchy-Schwarz inequality in {\mathbb R^n}. And when {n=3}, this can be rephrased as

\displaystyle \begin{array}{rl} \displaystyle |{a}|^2|{b}|^2= \left({a}\cdot {b}\right) ^2+|{a}\times {b}|^2\end{array}

for {{a}, {b}\in \mathbb R^3}. In general, the term {\displaystyle \sum _{1\le i<j\le n} \left(a_ib_j-a_jb_i\right)^2} can be identified as the norm squared of the wedge product {{a}\wedge {b}}.

In this note, we give the less well-known extension of this identity and the corresponding Cauchy-Schwarz type inequality.
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Posted in Algebra, Group theory, Inequalities, Linear Algebra | Leave a comment

On the existence of a metric compatible with a given connection

Question: Suppose we are given a torsion-free (i.e. the torsion tensor vanishes) affine connection {\nabla } on a smooth connected manifold {M}. Does there exist a Riemannian metric {g} such that its Levi-Civita connection is {\nabla }? If so, is it unique if we prescribe its value at a point?

Let {(x^i)} be a local coordinates on {M} and let {\nabla _i\partial _j=\Gamma_{ij}^k\partial _k} (using Einstein’s notation). The torsion-free condition is then equivalent to {\Gamma_{ij}^k=\Gamma_{ji}^k}. We want to solve the linear system of partial differential equations (Note that {\Gamma_{ij}^k} are {C^1} as {\nabla } is a connection.)

\displaystyle \begin{array}{rl} \displaystyle  \partial _k g_{ij}=\Gamma_{ki}^pg_{pj}+\Gamma_{kj}^pg_{pi}. \ \ \ \ \ (1)\end{array}

We impose an initial condition at {x_0} by

\displaystyle \begin{array}{rl} \displaystyle g_{ij}(x_0)= h_{ij}\end{array}

for some fixed symmetric {h_{ij}}.

It is clear from the initial condition and from (1) that {g_{ij}} is symmetric if the solution exists. According to the theory of first order partial differential equations system (cf. e.g. Stoker’s Differential Geometry Appendix B), the system (1) is uniquely solvable if and only if the compatibility conditions hold:

\displaystyle \begin{array}{rl} \displaystyle   & \displaystyle \partial _l\Gamma_{ki}^pg_{pj} +\Gamma_{ki}^p \partial _lg_{pj} +\partial _l\Gamma_{kj}^pg_{ip} +\Gamma_{kj}^p \partial _lg_{ip}\\ =& \displaystyle \partial _k\Gamma_{li}^pg_{pj} +\Gamma_{li}^p \partial _kg_{pj} +\partial _k\Gamma_{lj}^pg_{ip} +\Gamma_{lj}^p \partial _kg_{ip}.  \ \ \ \ \ (2)\end{array}

In view of (1), this is equivalent to

\displaystyle \begin{array}{rl} \displaystyle   & \displaystyle \partial _l\Gamma_{ki}^pg_{pj} +\Gamma_{ki}^p (\Gamma_{lp}^m g_{mj}+\Gamma_{lj}^m g_{pm}) +\partial _l\Gamma_{kj}^pg_{ip} +\Gamma_{kj}^p (\Gamma_{li}^m g_{mp}+\Gamma_{lp}^m g_{im})\\ =& \displaystyle \partial _k\Gamma_{li}^pg_{pj} +\Gamma_{li}^p (\Gamma_{kp}^m g_{mj}+\Gamma_{kj}^m g_{pm}) +\partial _k\Gamma_{lj}^pg_{ip} +\Gamma_{lj}^p (\Gamma_{ki}^m g_{mp}+\Gamma_{kp}^m g_{im}).  \ \ \ \ \ (3)\end{array}

To shed some light on the following computation, let us introduce the curvature tensor associated with {\nabla }.

Definition 1 Given a connection {\nabla } and vector fields {X}, {Y}, {Z}, we define the curvature tensor {R(X, Y)Z} by

\displaystyle \begin{array}{rl} \displaystyle  R (X, Y)Z=\nabla _X\nabla _YZ-\nabla _Y\nabla _XZ-\nabla _{[X,Y]}Z. \ \ \ \ \ (4)\end{array}

Here {[\cdot,\cdot]} is the Lie bracket of vector fields. It is readily checked that {R(\cdot, \cdot)\cdot} is a tensor field regardless of whether {\nabla } is torsion-free or not. In local coordinates, we define {R_{ijk}^l} by

\displaystyle \begin{array}{rl} \displaystyle R(\partial _i, \partial _j)\partial _k=R_{ijk}^l\partial _l.\end{array}

It is not hard to see that if {\nabla } is torsion-free and the {\nabla } is compatible with {g} in the sense of (1), then we have the local formula

\displaystyle \begin{array}{rl} \displaystyle R_{ijk}^l = \partial _i \Gamma_{jk}^l - \partial _j \Gamma_{ik }^l + \Gamma _{jk}^m \Gamma_{im}^l- \Gamma _{ik}^m \Gamma _{jm}^l.\end{array}

We subtract RHS from LHS of (3) and get

\displaystyle \begin{array}{rl} \displaystyle   & \displaystyle \partial _l\Gamma_{ki}^pg_{pj} +\Gamma_{ki}^p (\Gamma_{lp}^m g_{mj}+\Gamma_{lj}^m g_{pm}) +\partial _l\Gamma_{kj}^pg_{ip} +\Gamma_{kj}^p (\Gamma_{li}^m g_{mp}+\Gamma_{lp}^m g_{im})\\ & \displaystyle -\left(\partial _k\Gamma_{li}^pg_{pj} +\Gamma_{li}^p (\Gamma_{kp}^m g_{mj}+\Gamma_{kj}^m g_{pm}) +\partial _k\Gamma_{lj}^pg_{ip} +\Gamma_{lj}^p (\Gamma_{ki}^m g_{mp}+\Gamma_{kp}^m g_{im})\right)\\ =& \displaystyle  \left(\partial _l \Gamma_{ki}^m-\partial _k \Gamma_{li}^m+\Gamma_{ki}^p\Gamma_{lp}^m-\Gamma_{li}^p\Gamma_{kp}^m\right)g_{mj} +\left(\partial _l \Gamma_{kj}^m-\partial _k \Gamma_{lj}^m+\Gamma_{kj}^p\Gamma_{lp}^m-\Gamma_{lj}^p\Gamma_{kp}^m\right)g_{mi}\\ & \displaystyle +\Gamma_{ki}^p\Gamma_{lj}^m g_{pm}+\Gamma_{kj}^p\Gamma_{li}^m g_{mp}-\Gamma_{li}^p \Gamma_{kj}^m g_{pm}-\Gamma_{lj}^p\Gamma_{ki}^m g_{mp}\\ =& \displaystyle  \left(\partial _l \Gamma_{ki}^m-\partial _k \Gamma_{li}^m+\Gamma_{ki}^p\Gamma_{lp}^m-\Gamma_{li}^p\Gamma_{kp}^m\right)g_{mj} +\left(\partial _l \Gamma_{kj}^m-\partial _k \Gamma_{lj}^m+\Gamma_{kj}^p\Gamma_{lp}^m-\Gamma_{lj}^p\Gamma_{kp}^m\right)g_{mi}\\ =& \displaystyle R_{lki}^m g_{mj}+R_{lkj}^m g_{mi}.  \ \ \ \ \ (5)\end{array}

Let us show that (1) implies {R_{lki}^m g_{mj}+R_{lkj}^m g_{mi}=0}, so by (5), we conclude that the compatibility condition (2) holds. To see this, using the compatibility of metric (1) (We do not sum over {k} in the following):

\displaystyle \begin{array}{rl} \displaystyle   \frac{1}{2}\partial _i\partial _j g_{kk} =\partial _i\left(\Gamma_{jk}^l g_{lk}\right) =\partial _i\Gamma_{jk}^l g_{lk}+\Gamma_{jk}^l\Gamma_{il}^m g_{mk} +\Gamma_{jk}^l\Gamma_{ik}^m g_{ml} \end{array}

and similarly

\displaystyle \begin{array}{rl} \displaystyle \frac{1}{2}\partial _j \partial _i g_{kk} =\partial _j\Gamma_{ik}^l g_{lk}+\Gamma_{ik}^l\Gamma_{jl}^m g_{mk} +\Gamma_{ik}^l\Gamma_{jk}^m g_{ml}. \end{array}

Subtracting the two equations,

\displaystyle \begin{array}{rl} \displaystyle   0 =\partial _i\Gamma_{jk}^l g_{lk}-\partial _j\Gamma_{ik}^l g_{lk}+\Gamma_{jk}^l\Gamma_{il}^m g_{mk} -\Gamma_{ik}^l\Gamma_{jl}^m g_{mk} = R_{ijk}^lg_{lk}. \end{array}

This implies that {g(R(X, Y)Z, Z)=0} if {g} is defined by {g:=g_{ij}dx^idx^j}. This is easily seen to be equivalent to {R_{lki}^m g_{mj}+R_{lkj}^m g_{mi}=0}. We conclude that (2) holds, and therefore we have proved the existence and uniqueness of the solution to the system

\displaystyle \begin{array}{rl} \displaystyle   \begin{cases} g_{ij,k}=\Gamma_{ki}^pg_{pj}+\Gamma_{kj}^pg_{ip}\\ g_{ij}(x_0)=h_{ij}. \end{cases} \end{array}

Gometrically, if we prescribe an inner product on {T_{p_0}M} for some {p_0\in M}, then there is a unique Riemannian metric which is compatible with the torsion-free connection {\nabla }. So we have proved

Theorem 2 Suppose {\nabla } is a torsion-free affine connection on a smooth connected manifold {M}. Let {g_0} be an inner product on {T_{p_0}M}, where {p_0\in M}. Then there exists a unique Riemannian metric {g} such that {\nabla } is the Levi-Civita connection of {g} and {g(p_0)=g_0}.

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A curious identity on the median triangle

I just came across a curious identity about the angles of the “median triangle” of a given triangle, while I was reviewing a paper from a team participating in the Hang Lung Mathematics Award. Of course, I am not going to reveal the identity for the obvious reason.

Let me first describe the setting. Let {\Delta ABC} be a triangle. (By abuse of notations, we regard (for example) {A} both as a vertex, a vector (in {\mathbb R^2} or {\mathbb R^3}), and the angle of the triangle {\Delta ABC} at the vertex {A}.) We can then draw the three medians on the triangle which, as is well-known, intersect at the so called centroid of the triangle. Let {D, E}, and {F} be the angles at the centroid as shown:triangle1.png

Theorem 1 We have the identity

\displaystyle \frac{1}{\sin^2 D}+ \frac{1}{\sin^2 E}+ \frac{1}{\sin^2 F} =\frac{1}{\sin^2 A}+ \frac{1}{\sin^2 B}+ \frac{1}{\sin^2 C}.

This result certainly looks very elegant (and is new to me). However, the proof in that paper consists of several pages of computations which to me is not very enlightening. So I set out to write a proof myself, which will be described below. Nevertheless, I have to resort to coordinates to prove the result. It would be desirable to know if there is a more classical proof without using coordinates. (Of course, all the computations using coordinates can theoretically be translated to classical statements, e.g. the cosine law is just the expansion of the inner product { |A-B|^2}. However, I am not quite willing to do such kind of line-by-line translation. ) Continue reading

Posted in Geometry | 4 Comments