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Anais da Academia Brasileira de Ciências

versão impressa ISSN 0001-3765versão On-line ISSN 1678-2690

An. Acad. Bras. Ciênc. vol.90 no.3 Rio de Janeiro jul./set. 2018

http://dx.doi.org/10.1590/0001-3765201820170689 

Mathematical Sciences

A new characterization of the Euclidean sphere

ABDÊNAGO A. BARROS1 

CÍCERO P. AQUINO2 

JOSÉ N.V. GOMES3 

1Departamento de Matemática, Universidade Federal do Ceará, Campus do Pici, Bloco 914, 60455-760 Fortaleza, CE, Brazil

2Departamento de Matemática, Universidade Federal do Piauí, Campus Universitário Ministro Petrônio Portella, 64049-550 Teresina, PI, Brazil

3Departamento de Matemática, Universidade Federal do Amazonas, Av. General Rodrigo Octávio, 6200, 69080-900 Manaus, AM, Brazil

Abstract

In this paper, we obtain a new characterization of the Euclidean sphere as a compact Riemannian manifold with constant scalar curvature carrying a nontrivial conformal vector field which is also conformal Ricci vector field.

Key words Conformal vector field; Scalar curvature; Euclidean sphere; Einstein manifold

INTRODUCTION

In the middle of the last century many geometers tried to prove a conjecture concerning the Euclidean sphere as the unique compact orientable Riemannian manifold (Mn,g) admitting a metric of constant scalar curvature S and carrying a nontrivial conformal vector field X . Among them, we cite Bochner and Yano (1953), Goldberg and Kobayashi (1962), Lichnerowicz (1955), Nagano and Yano (1959), Obata (1962), Obata and Yano (1965, 1970) and Tashiro (1965). The attempts to prove this conjecture resulted into the rich literature which has currently been attracting a lot of attention in the mathematical community. We address the reader to the book of Yano (1970) for a summary of those results. Ejiri gave a counter example to this conjecture building metrics of constant scalar curvature on the warped product 𝕊1×fFn1 , where Fn1 is a compact Riemannian manifold of constant scalar curvature, while 𝕊1 stands for the Euclidean circle. In his example the conformal vector field is X=f(d/dt) , where d/dt is a unit vector field on 𝕊1 and f satisfies a certain ordinary differential equation, see Ejiri (1981) for details.

The primary concept involved in the study of this subject is of Lie derivatives. After all, what is the geometric meaning of the Lie derivative of a tensor T (or of a vector field Y ) with respect to a vector field X? This is a method that uses the flow of X to push values of T back to p and then differentiate. The result is called the Lie derivative XT of T with respect to X . A vector field X on Riemannian manifold (Mn,g) is called conformal if Xg is a multiple of g . There are important applications of Lie derivatives in the study of how geometric objects such as Riemannian metrics, volume forms, and symplectic forms behave under flows. For instance, it is well-known that the Lie derivative of a vector field Y with respect to X is zero if and only if Y is invariant under the flow of X . It turns out that the Lie derivative of a tensor has exactly the same interpretation. For more details see the book of Lee (2003).

We highlight that Nagano and Yano (1959) have proved that the aforementioned conjecture is true when (Mn,g) is an Einstein manifold, i.e., the Ricci tensor of metric g satisfies Ric=Sng . In this case, S is constant for dimensions n3 . Thus if X is the conformal vector field with conformal factor ρ, that is, Xg=2ρg, we deduce XRic=2ρRic. With this setting we define a conformal Ricci vector field on a Riemannian manifold (Mn,g) as a vector field X satisfying

XRic=2βRic, (1)

for some smooth function β:Mn. In particular, on Einstein manifolds this concept is equivalent to the classical conformal vector field. With this additional condition the aforementioned conjecture is true. More precisely, we have the following theorem.

Theorem 1. Let (Mn,g),n3, be a compact orientable Riemannian manifold with constant scalar curvature carrying a nontrivial conformal vector field X which is also a conformal Ricci vector field. Then Mn is isometric to a Euclidean sphere 𝕊n(r) . Moreover, up to a rescaling, the conformal factor ρ is given by

ρ=τhvn

and X is the gradient of the Hodge-de Rham function which in this case, up to a constant, is the function 1nhv , where hv is a height function on a unitary sphere 𝕊n and τ is an appropriate constant.

We point out that Obata and Yano (1965) have obtained the same conclusion of the preceding theorem under the hypothesis XRic=αg, for some smooth function α defined in Mn. Moreover, when Mn is an Einstein compact Riemannian manifold the result of Nagano and Yano (1959) is a consequence of Theorem 1. We also observe that the compactness of Mn in our result is an essential hypothesis. In fact, let us consider the hyperbolic space n as a hyperquadric of the Lorentz-Minkowski space 𝕃n+1 and v a nonzero fixed vector in 𝕃n+1 . After a straightforward computation, it is easy to verify that the orthogonal projection v of v onto the tangent bundle Tn provides a nontrivial conformal vector field on n for an appropriate choice of v . Consequently, since n is Einstein, it follows that v is also a nontrivial conformal Ricci vector field on n .

PRELIMINARIES AND AUXILIARY RESULTS

To start with, we consider the Hilbert-Schmidt norm for tensors on a Riemannian manifold (Mn,g) , i.e., the inner product T,S=tr(TS*). It is important to notice that for an orthonormal basis {e1,,en} , we can use the natural identification of (0,2) -tensors with (1,1) -tensors, T(ei,ej)=g(Tei,ej) , to write

T,S=i,jTijSij=ig(Tei,Sei).

We recall that the divergence of a (1,r) -tensor T on Mn is the (0,r) -tensor given by

(divT)(v1,,vr)(p)=tr(w(wT)(v1,,vr)(p)),

where pMn and (v1,,vr)TpM××TpM.

Let X be a smooth vector field on Mn , and let φ be its flow. For any pMn , if t is sufficiently close to zero, φt is a diffeomorphism from a neighborhood of p to a neighborhood of φt(p) , so φt* pulls back tensors at φt(p) to ones at p .

Given a (0,r) -tensor T on Mn , the Lie derivative of T with respect to X is the (0,r) -tensor XT given by

(XT)p=ddt|t=0(φt*T)p=limt01t[φt*(Tφ(t,p))Tp].

Fortunately, there is a simple formula for computing the Lie derivative without explicitly finding the flow. For any (0,r) -tensor T on Mn ,

(XT)(Y1,,Yr)=X(T(Y1,,Yr))T([X,Y1],Y2,,Yr)T(Y1,,Yr1,[X,Yr]),

where Y1,,Yr are any smooth vector fields on Mn , and [X,Yi] stands for the Lie bracket of X and Yi.

In what follows we prove some lemmas and integral formulas which will be required later.

Lemma 1. For any symmetric (0,2) -tensor T on a Riemannian manifold (Mn,g) and X𝔛(M) , holds

X|T|2=2XT,T2T2,Xg.

In particular, if X is a conformal vector field with conformal factor ρ, then we have X|T|2=2XT,T4ρ|T|2.

Proof. Let {e1,,en} be a geodesic orthonormal frame at pMn . Then we have

X|T|2=X(i,jTijTij)=2i,jTijX(Tij)=2i,jTij{(XT)ij+T([X,ei],ej)+T(ei,[X,ej])}=2i,jTij(XT)ij2i,jTij{T(eiX,ej)+T(ei,ejX)}.

Whence, we use that T is symmetric to deduce

X|T|2=2T,XT2i,jTij{g(eiX,Tej)+g(Tei,ejX)}=2T,XTj2g(TejX,Tej)i2g(Tei,TeiX)=2T,XT2i(Xg)(Tei,Tei)=2T,XT2i,jTij2(Xg)ij,

which completes the proof of the lemma.

Corollary 1. Under the assumptions of Lemma 1 we have X|T|2=0, provided that Xg=2ρg and XT=2ρT.

Another useful result is given by the following lemma.

Lemma 2. Let (Mn,g) be a Riemannian manifold endowed with a symmetric (0,2) -tensor T. Then it holds

XT,g=XT,gT,Xg.

In particular, if XT=2βT, then

2βtr(T)=(tr(T)),X+T,Xg.

Proof. Let {e1,,en} be a geodesic orthonormal frame at a point pMn. By the symmetry of T , we have

XT,g=tr(XT)=i(X(Tii)+2T(eiX,ei))=XT,g+2T,X=XT,g+T,Xg,

that finishes the proof of the lemma.

Now we remind the reader that the traceless tensor of a symmetric (0,2) -tensor T on a Riemannian manifold (Mn,g) is given by

T̊=Ttr(T)ng. (2)

With this setting we prove the next corollary.

Collary 2. Let (Mn,g) be a Riemannian manifold endowed with a symmetric (0,2) -tensor T such that XT=2βT , with Xg=2ρg. Then we have:

  1. 2(βρ)tr(T)=(tr(T)),X.

  2. If tr(T) is a non null constant, then β=ρ and XT̊=2ρT̊.

Proof. Since Xg=2ρg we have immediately the first item. Now, if tr(T) is a non zero constant, then we have β=ρ, which implies XT̊=XTtr(T)nXg=2ρT̊, finishing the proof of the corollary.

Next, we apply the previous results to the Ricci tensor. First of all, given a conformal vector field X on a Riemannian manifold Mn such that Xg=2ρg, we have the next well-known formulae, see e.g. Obata and Yano (1970).

XRic=(n2)2ρ(Δρ)g, (2)
XS=2(n1)Δρ2Sρ, (2)

and

XG=(n2)(2ρ1n(Δρ)g), (2)

where G=RicSng=Ric̊.

We claim that if Mn is compact with constant scalar curvature S and ρ is not constant, then equation (4) allows us to infer that S is positive. Indeed, since ρ is not constant Sn1 belongs to the spectrum of the Laplacian of Mn . Therefore, we deduce the next lemma.

Lemma 3. Let (Mn,g) be a compact Riemannian manifold with constant scalar curvature such that XRic=2βRic and Xg=2ρg . Then we have β=ρ and X|G|2=0.

Proof. Since XRic=2βRic and S=tr(Ric) is a positive constant we get by Corollary 2 that β=ρ and XG=2ρG. Therefore, applying Corollary 1 we have X|G|2=0, which completes the proof of the lemma.

Taking into account the second contracted Bianchi identity: div(Ric)=12dS and using the identity div(Sng)=1ndS we obtain the following relation

div(G)=n22ndS.

Therefore, we can write

ρdiv(G)(ρ)=n24nS,ρ2.

The next equation is well-known. For details of a more general case see for example Lemma 1 in Barros and Gomes (2013).

div(ρG(ρ))=ρdiv(G)(ρ)+ρ2ρ,G+G(ρ,ρ). (6)

Since G,g=0 , from (5) we have

XG,G=(n2)2ρ,G.

Applying Lemma 1 to this identity we obtain

2ρ,G=1n2(12X|G|2+2ρ|G|2). (7)

Comparing (6) and (7) we infer

div(ρG(ρ))=n24nS,ρ21n2(ρ2X|G|2+2ρ2|G|2)+G(ρ,ρ). (8)

In what follows we assume that (M,g) is an orientable Riemannian manifold. If M is not orientable, we take the orientable double covering M̃ of M and induce, in the natural manner, the Riemannian metric g̃ on M̃ . Then (M,g) and (M̃,g̃) have the same local geometry.

As a direct consequence from (8) and Stokes’ Theorem we obtain the following lemma.

Lemma 4. Let (Mn,g) be a compact orientable Riemannian manifold endowed with a conformal vector field X of conformal factor ρ , then

MG(ρ,ρ)dM=1n2M(ρ2X|G|2+2ρ2|G|2)dM+n24nMρ2ΔSdM.

Before stating our next result we note that |2ρΔρng|2=|2ρ|21n(Δρ)2 . Accordingly, the Bochner formula becomes

12Δ|ρ|2=G(ρ,ρ)+Sn|ρ|2+|2ρΔρng|2+1n(Δρ)2ρ,(Sρn1+XS2(n1)),

where in the last term we use equation (4). Then,

12Δ|ρ|2=G(ρ,ρ)+|2ρΔρng|2S(|ρ|2+ρΔρ)n(n1)12n(n1)((XS)Δρ+nS,ρ2+nρ,(XS)).

By integration,

M(G(ρ,ρ)+|2ρΔρng|2+12n(ρ2ΔS+(XS)Δρ))dM=0. (9)

In the notation of (2) we have |2̊ρ|2=|2ρΔρng|2 . Comparing Lemma 4 with equation (9) we get the lemma.

Lemma 5. Let (Mn,g) be a compact orientable Riemannian manifold endowed with a conformal vector field X of conformal factor ρ , then

  1. M(ρn2(12X|G|2+2ρ|G|2)+|2̊ρ|2+S2div(ρρ)+XS2nΔρ)dM=0.

  2. M(ρn2XG,G+|2̊ρ|2+S2div(ρρ)+XS2nΔρ)dM=0.

Proof. First assertion is a direct combination of Lemma 4 and equation (9), while the second one follows from Lemma 1 and the first assertion. Indeed, from this latter lemma we have 12X|G|2+2ρ|G|2=XG,G, which completes our proof.

We are in the right position to prove our main result.

PROOF OF THEOREM 1

Firstly, we observe that we are supposing that there exists a vector field X on Mn such that Xg=2ρg and XRic=2βRic , for some smooth functions ρ and β on Mn , where ρ is non-constant. Consequently, from Lemma 3 we obtain β=ρ and X|G|2=0 . Secondly, from item (1) of Lemma 5 and equation (4) we get

M(2n2ρ2|G|2+|2ρ+Sρn(n1)g|2)dM=0.

Taking into account that ρ is non-constant the preceding identity allows us to achieve G=0 and 2ρ=Sn(n1)ρg. Therefore, we can apply a classical result due to Obata (1962), for instance, to conclude that Mn is isometric to a Euclidean sphere 𝕊n(r) . Moreover, 2ρ=Δρng and Δ(Δρ)+Sn1Δρ=0 (see equation (4)). Rescaling the metric we can assume that S=n(n1) . Then we conclude that Δρ is the first eigenvalue of the unitary sphere 𝕊n. Whence, there exists a fixed vector v𝕊n such that Δρ=hv=1nΔhv . Thus we have Δ(ρ+1nhv)=0, which gives ρ=τ1nhv . Setting u=ρ we obtain

ug=22u=22ρ=2ρg=Xg.

It is also true that uRic=2ρRic . Besides, by Hodge-de Rham decomposition theorem we can write X=Y+ , for some vector field Y with divY=0 and is the Hodge-de Rham function. So, Δu=divX=Δ which implies u is constant. Note that this is sufficient to complete our proof.

A MORE GENERAL CASE

We notice that XG=(n2)2̊ρ and XG=XRic1nX(Sg) give

XRic=1nX(Sg)(n2)2̊ρ=2ρ(Sng)+1n(XS)g(n2)2̊ρ=2ρ(RicG)+1n(XS)g(n2)2̊ρ=2ρRic+1n(XS)g+T,

where T=2ρG(n2)2̊ρ. Therefore, we deduce

XRic=2ρRic+1n(XS)g+T,

where tr(T)=0.

Let us suppose that XRic=2βRic+T and Xg=2ρg , where T is a (0,2) -tensor on Mn . By using (3) and (4) we deduce

tr(T)=2(n1)Δρ2βS

and

XS=tr(T)+2(βρ)S.

In particular, if S is a non null constant and ρ0 , we have

tr(T)=0if, and only, ifβ =ρ . (10)

On the other hand, XG=XRic1nX(Sg) gives

XG=2βG+Ttr(T)ng (11)

and

X|G|2=4(βρ)|G|2+2T,G. (12)

Lemma 6. Let (Mn,g),n3 , be a compact orientable Riemannian manifold endowed with a conformal vector field X whose conformal factor is ρ. If XRic=2βRic+T , then the following integral formula holds:

M(1n2(2βρ|G|2+ρT,G)+|2ρΔρng|2+S2div(ρρ)+XS2nΔρ)dM=0.

Proof. First we notice that from (12) we obtain

ρ2X|G|2+2ρ2|G|2=2βρ|G|2+ρT,G. (13)

Therefore, using (13) in the first assertion of Lemma 5, we have

M(1n2(2βρ|G|2+ρT,G)+|2ρΔρng|2+S2div(ρρ)+XS2nΔρ)dM=0

which completes the proof of the lemma.

Proceeding we use this lemma to obtain the following theorem.

Let (Mn,g) , n3 , be a compact orientable Riemannian manifold with constant scalar curvature S . Suppose that there exists a nontrivial conformal vector field X on Mn such that Xg=2ρg and XRic=2βRic+T . If tr(T)=0 and M(2ρ2|G|2+ρG,T)dM0 , then Mn is isometric to a Euclidean sphere.

Moreover, up to a rescaling, the conformal factor ρ is given by ρ=τhvn and X is the gradient of the Hodge-de Rham function which in this case, up to a constant, is the function 1nhv , where hv is a height function on a unitary sphere 𝕊n and τ is an appropriate constant.

Proof. It follows from (10) that β=ρ . Now we may use Lemma 6 to deduce that |2ρΔρng|2=0. But, from (4) we have Δρ=Sn1ρ. Therefore, we deduce 2ρ=Sn(n1)ρg , from which we may apply Obata’s Theorem, consult Obata (1962), to conclude that Mn is isometric to a Euclidean sphere. Moreover, from (11) we have that T is null tensor. So, the latter claim is proved following the same steps given in the proof of Theorem 1.

Notice that if T=λρG , with (2+λ)0 , then the conditions of the previous theorem are verified. In particular, for λ=2 , we have XRic=2ρSng , which allows us to obtain the result due to Obata and Yano (1965).

ACKNOWLEDGMENTS

The authors would like to thank the referees for their careful reading and useful comments which improved the paper. José N.V. Gomes would like to thank the Department of Mathematics at Lehigh University, where part of this work was carried out. He is grateful to Huai-Dong Cao and Mary Ann for their warm hospitality and constant encouragement. The authors are partially supported by Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq), do Ministério da Ciência, Tecnologia e Inovação (MCTI) do Brasil.

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Received: September 06, 2017; Accepted: April 04, 2018

Correspondence to: José Nazareno Vieira Gomes E-mail: jnvgomes@pq.cnpq.br/jov217@lehigh.edu

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