The Rigidity of Hypersurfaces in Euclidean Space?

Chunhe LI Yanyan XU

Abstract In the present paper,the rigidity of hypersurfaces in Euclidean space is revisited.The Darboux equation is highlighted and two new proofs of the rigidity are given via energy method and maximal principle,respectively.

Keywords Global rigidity,In finitesimal rigidity,Energy method,Maximal principle

1 Introduction

The isometric embedding problem is one of the fundamental problems in differential geometry.Since Riemannian manifold was formulated by Riemann in 1868,naturally there arose the question of whether an abstract Riemannian manifold is simply a submanifold of some Euclidean space with its induced metric.In other words,it is the question of reality of Riemannian manifold(see more details in an expository note(cf.[9])).

Mathematically,the isometric embedding problem is to solve the following system.For any given Riemannian manifold(M,g),there is a surface~r:M 7→Rn+1such that

where·denotes the Euclidean inner product.In the present paper we assume that~r is a hypersurface,i.e.,M is a manifold of n dimension.

As is known the uniqueness of solution in PDEs is related to the existence,hence it is another important topic.The counterpart of uniqueness in isometric embedding is global rigidity.The rigidity is to characterize isometric deformation of surfaces which is closely related to the global isometric embedding of surfaces.

Definition 1.1 An immersed surface:M→R3is rigid if every immersion:M→R3,with the same induced metric,is congruent to,that is,differs fromby an isometry of R3.

The linearized version of global rigidity is in finitesimal rigidity.We say thatyields a first order isometric deformation ofif the induced metrichas a critical point at t=0,

As is known,the isometry group of Rn+1is orthogonal group O(n+1)and translation(cf.[11]),namely affine group.Hence thegenerated by its Lie algebra is always the solution to homogeneous linearized equation,where A∈o(n+1)is a skew matrix andis a constant vector.Suchis called a trivial solution to(1.2).For n=2,it is equivalent tofor any constantand.

Definition 1.2 The surface is in finitesimally rigid if(1.2)has only trivial solutions.

In the present paper we will revisit several kinds of rigid surfaces and give new proof which is based on the equivalence of isometric embedding equation(1.1),Gauss-Codazzi equations and Darboux equation.

For the case of n=2,Cohn-Voseen[3]and Blaschke[2]proved the following theorems.

Theorem 1.1 Let M be a smooth closed surface with nonnegative curvature and let the vanishing set of the curvature have no interior points.Then M is globally rigid.

Theorem 1.2 Let M be a smooth closed surface with nonnegative curvature and let the vanishing set of the curvature have no interior points.Then M is in finitesimally rigid.

Another rigid surface is Alexandrov’s annuli(cf.[1]).

Definition 1.3 The 2-dimensional multiply-connected Riemannian manifold(M,g)satisfies Alexandrov’s assumption:

The following rigidity theorems are due to Alexandrov[1]and Yau[15],respectively.

Theorem 1.3 Alexandrov’s annuliis globally rigid.

Theorem 1.4 Alexandrov’s annuliis in finitesimally rigid.

Ivan Izmestiev[10]proved the in finitesimal rigidity of convex surface in R3via the second derivative of the Hilbert-Einstein functional.In[14],Lin and Wang proved the in finitesimal rigidity of convex surface in H3.Li and Wang[13]reproved Lin-Wang’s theorem by Beltrami map.Li,Miao and Wang[12]reproved Lin-Wang’s theorem by integral method.For the case of n≥3,Dajczer-Rodriguez[4]proved the following theorem.

Theorem 1.5 If the rank of the matrix(hij)is greater than 2,where h=hijdxidxjis the second fundamental form,then the hypersurface is globally and in finitesimally rigid.

Remark 1.1 Compared with the case of n=2,Dajczer-Rodriguez’s theorem is local without any topological restriction on M.

In[7],Guan and Shen proved a rigidity theorem for hypersurfaces in higher dimensional space forms.In[13],Li and Wang showed that if a spherically symmetric(n+1)-manifold with metric

the sphere of symmetry r=c is not globally rigid and in finitesimally rigid unless g is a space form.

2 Set up and Formulation

Before discussing the rigidity of Alexandrov’s annuli,we need some geometric preliminaries.

We use the geodesic coordinates(s,t)=(x1,x2)based on?M,

where B(s,t)is a sufficiently smooth function and B(s,t)is periodic in s,and kgis geodesic curvature.

Under the geodesic coordinates,Alexandrov proved the following lemma(cf.[1]or[9]).

Lemma 2.1 For Alexandrov’s annuli,the coefficients of the second fundamental form of,L,M and N satisfy:At t=0,

Lemma 2.2 The components of boundary(?M)are some planar curves σk,1 ≤ k ≤ m,which are determined completely by their metric,and lie on the plane πktangential to~r along σk.

At the same time,Dong[5]proved the following lemma.

Lemma 2.3 If there exists sufficiently smooth isometric embedding

then we have

In what follows we will formulate the rigidity.

Let

We have

and

where h=hijdxidxj,are the second fundamental forms,respectively,K is the Gaussian curvature.

Let Wij=and Φ=?ρ.By(2.6)–(2.7)we have

Taking the difference of(2.8)–(2.9)and the two sides of(2.10)yields

Gauss-Codazzi equations say

There exists an orthogonal mapping which sends the frame{r1,r2,n}to.Let the associated matrix be A,if h andcoincide which means A is constant,i.e.,W=Wijdxidxj=0,anddiffer from an isometry and so it’s globally rigid.

and

Note that uidxi=·is a globally well-defined 1-form,and w is a well-defined function.Then we have

Then for

which implies that d~Y is parallel to the tangent plane.Let,k=1,2,where wijdxidxjis a symmetric tensor.=0 means

where h=hijdxidxjis the second fundamental form and(hij)=(hij)?1.

Let

We have

Combining(2.24)–(2.25),we have

If the support functionμ≠0,wij=0 if and only ifis constant sinceare linearly independent,i.e.,.For convex surface,by a translation we can assume the support functionμ>0.Throughout the paperμ>0 if not specified.

3 The Rigidity of Surfaces in R3

In this section we will reprove Theorem 1.1,Theorem 1.3 and Theorem 1.2,Theorem 1.4.The main ideas are from an unpublished note(cf.[12]).

To prove Theorem 1.1 and Theorem 1.3,we introduce the following inner product:For any two(0,2)-symmetric tensors α = αikdxi? dxk,β = βjldxj? dxl,

and the metric induces a metric on the tensor bundle T?M?T?M,

In what follows we will show the tensor W=0 by(W,W)=0,where W=Wijdxidxjis the solution to(2.14)–(2.15),hence prove Theorem 1.1 and Theorem 1.3.

A direct computation shows

For i=1,

If M=S2,in the integral by parts the boundary term vanishes;if M is Alexandrov’s annuli,on the boundary W=0 by Lemma 2.1 hence the boundary term vanishes too.Both of the two terms in(3.4)vanish,(W,W)=0,W≡0.

To prove Theorem 1.2 and Theorem 1.4,we introduce the following inner product:For any two(0,2)-symmetric tensors α = αikdxidxk,β = βjldxjdxl,

In what follows we will show the tensor w=0 by(w,w)=0,where w=wijdxidxjis the solution to(2.21)–(2.22),hence prove Theorem 1.2 and Theorem 1.4.

A direct computation shows

If M=S2,a similar argument in(3.5)yields(w,w)=0,w≡0.

If M is Alexandrov’s annuli,we have

Note the right-hand side of(3.7)is invariant under coordinate change.So we use geodesic coordinates based on?M.Without loss of generality,we merely consider the case that M is a disk,and then ?M is a planar curve denoted by σ.On the boundary,we have h11=h12=0,w11=0 andμis constant.

where in the third equality we use the fact h11=h12=0,w11=0 and in the fourth equality we use hijwij=0.

In what follows we will show

where F=w12μ.

Recall on the boundary σ,h11=h12=0,w11=0 and.By(2.26),we have on the boundary

which is nothing else but an ODE of ?sand ?t.We can rewrite(3.9)in complex form

For convenience,we introduce a new variableand let c1= ?s(0),c2= ?t(0).Then the solution to(3.9)is

Suppose that the boundary lies on the plane z=0.By the motion of moving frame we have on the boundary

It is easy to check

where α is a fixed constant.

where we use(2.3)–(2.4).

Hence

We define a new closed planar curve Γ by parameter equations

A direct computation shows that the curvature of Γ is kgand the area bounded by the curve is

And we introduce two new functions

where

Then we have U′(θ)cotθ=V′(θ)and U(0)=U(π)=0.Therefore

and integral by parts yields

where in the third equality we use(3.13).

Combining(3.14)–(3.17),we have

and then

In what follows we give another proof of Theorem 1.2 and Theorem 1.4.The proof is more geometric than above,correspondingly for Theorem 1.4 we restrict that the component number of boundary of Alexandrov’s positive annuli is 1(disk)or 2(annulus).We need the following lemma.

Lemma 3.1 For any vector valued:M 7→R3satisfying

the 1-form defined on M,

is closed.

Proof It is obvious that ω is a 1-form.Exterior differentiation yields

By(3.20),we have

Hence by(3.22)we get

ω is a closed 1-form.

Case 1 Let M be a disk D called Alexandrov’s positive disk,be the normal along the boundary σ,and~andform an orthogonal basis.Assume=0 on the boundary σ,and>0 at the interior points.We have

For convenience,we write

Hence we have

We will show that ψ is constant hence ω =0,which is one key step to prove Theorem 1.4.

It is worth pointing out that the following idea is borrowed from[8]which proves the rigidity in prescribed curvature problem.

A simple computation shows

Then

By[8,Lemma 4],we have

We conclude that

Similarly,we have

Hopf’s strong maximum principle(cf.[6, §3.2,Theorem 3.5])tells us that ψ is a constant function on the disk since on the boundary ψ is a constant,hence

Let

We have in DS,at least one of the following mixed products is nonzero:

Recall that

Case 2 M is Alexandrov’s positive annulus.Lemma 2.2 says that the boundary consists of two planar curves.We will discuss two different cases,respectively.Subcase 2.1:The two boundary planes are parallel;Subcase 2.2:The two boundary planes are not parallel.

Diff erent from Case 1,we need some extra topology preliminary.

hence there exists some smooth function ψ defined on the M,such that

Proof Integral by parts yields

where we use(2.20).

Note that on M at least one ofis not zero,otherwiseis parallel to some normal on M,but as a convex surface,its Gauss map is one-to-one and any normal on M differs from the normals on?M therefore is not parallel to.Hence the coefficient determinantsays trh(wij)=tr(h?1w)=0,in addition,then h?1w=0 and w=0 because h and w are symmetric,i.e.,=0.

For Subcase 2.2,let the constant normals on σ1,σ2be(σ1),(σ2),and the constant support functions on σ1,σ2be μ(σ1),μ(σ2),respectively.We chooseas

where c1,c2solves

Similar to Subcase 2.1,if at least one ofis not zero,the tensor w=wijdxidxj=0.We will see the set

is of zero measure.Then w=0 everywhere on M by the continuity.

We have that Spis contained in the level set{p∈ M,?M(p)=0}since~n is parallel toon Sp.We will check on the level set,

hence aj=0 and=0.is regular surface and the translationis regular too,then the{p∈M,(p)=0}is finite.The level set{p∈M,?M(p)=0}is zero measured.As a subset of{p∈ M,?M(p)=0},Spis also zero measured.

Remark 3.1 If M=S2,i.e.,the case of closed convex surface,we choose

Similar but simpler argument yields=0.Thus we complete the proof of Theorem 1.2.

As we have seen,the new proofs we give highlight the roles that the function ρ defined in(2.5)and its linearized version ? defined in(2.23)play.In fact we can extract all information from ρ which satisfies Darboux equation in isometric embedding problem as we work on the support function in Minkowski problem.

4 The Rigidity of Hypersurfaces in Rn+1,n≥3

Similarly in the case of higher dimension,for the equation(1.2)we can assume that

Let

and

By

we have

where δ is generalized Kronecker symbol.Obviously eα∧ eβ= ?eβ∧ eα,we set

i.e.,

Define a basis Eγ,1 ≤ γ ≤ n+1 in Gr(n,n+1)～=Gr(1,n+1)by

hence

i.e.,for fixed i,j and γ,

hence

We claim the following lemma.

Lemma 4.1 If 1≤i,j,γ≤n,then

For the left-hand side of(4.9),by(4.2)and(4.8),

And on the other hand,for the right-hand side of(4.9),by(2.2)and(4.8)

so

Hence we can rewrite

At the same time note that for fixed i,j,

We rewrite(4.12)as

In what follows we will compute the covariant derivative of el∧en+1.

At first we notice that

therefore

Since

and for k2

Similarly we have

Thus

Remark 4.1(4.20)shows that wijis Codazzi.In fact,(4.20)–(4.21)is a homogeneous linearized Gauss-Codazzi system.

Similar to the case of n=2,

Hence for hypersurface in Rn+1,we can use maximal principle to get the in finitesimal rigidity.But we can make use of(4.21)to reprove Theorem 1.5.

Proof of Theorem 1.5 We want to show wij=0.In view that wijdxidxjis invariant under variable transformation,we consider the diagonal case,i.e.,hij=0,i ≠j,since at any point on the hypersurface we can diagonalize the matrix(hij)by variable transformation.

If the rank of the matrix(hij)is greater than 2,without loss of generality we can assume h11,h22,h33≠0.By(4.21),

Since hij=0,i ≠j,(4.23)is just a linear system of w11,w22,w33,

The coefficient matrix in(4.24)is invertible,hence w11=w22=w33.For other wij,by(4.21),

since i ≠1,j ≠1 and w11=0,h11wij=0.

As for the part of global rigidity,without loss of generality we assume that the block H3=(hij)3×3is of full rank,then its adjoint matrixis of full rank too.By Gauss equation,every element inis an entry of Riemannian curvature tensor which is totally determined by metric.Thereforeis intrinsic and we can recover H3from.H3is intrinsic too,and as we proceed in the part of in finitesimal rigidity the H=(hij)n×nis intrinsic too.

In the proof of Theorem 1.5,we just deal with the algebraic equations,Gauss equations or its linearized equations,so we can say Theorem 1.5 is algebraic.

AcknowledgementsThe authors wish to thank Professor Pengfei Guan and Professor Zhizhang Wang for their valuable suggestions and comments.Part of the content also comes from Professor Wang’s contribution.The first author wishes to thank China Scholarship Council for its financial support.The first author also would like to thank McGill University for their hospitality.