CURVATURE HOMOGENEITY FOR FOUR-DIMENSIONAL MANIFOLDS
KOUEI SEKIGAWA, HIROSHI SUGA AND LIEVEN VANHECKE
1. Introduction and preliminaries
Let (M,g) be an n-dimensional, connected Riemannian manifold with Levi Civita connection V and Riemannian curvature tensor R defined by
Rxy = [Vx, Vy] - Vrx,Yj
for all smooth vector fields X, Y. VR, ... ,VkR, ... denote the succes- sive covariant derivatives and we assume VOR = R.
In [17] I.M. Singer studied infinitesimally homogeneous spaces and introduced the following condition :
P(.e) : for every x, y EM there exists a linear isometry</> : TzM ~
TyM such that
</>*«VkR)y)= (VkR)z for k = 0,1"" ,.e.
A Riemannian manifold such that P(O) holds is said to be curvature homogeneous and if P(i) holds, the manifold is said to be curvature homogeneous up to orderE, Further, for any point x E M, let G; be the Lie group
G: ={a E O(TzM)I(V~R)a=(ViR)z,i =0,1, ... ,s].
Its Lie algebra g; consists of all skew-symmetric endomorphisms A of TzM such that A .(ViR)z = 0 for i = 0, ... ,s. Here A acts as a derivation of the tensor algebra. Clearly, there always exists a first
Received October 23,1993.
1991 Mathematics Subject Classification: 53C20, 53C30.
Key words and phrases: Riemannian manifold, homogeneous spaces, curvature homogeneous spaces.
integer kx such that gk", =gk",+l. Moreover, if P(R.) is satisfied, then gf and gf are conjugated for 0 ~ i ~ R.. Hence, if P(kx +1) holds, kz does not depend on x. In this case we put gf = gi,kM = kz • A Riemannian manifold satisfying the conditionP(kM +1)is said to be infinitesimally homogeneous [17] and Singer's main result in [17] is the following
THEOREM 1. A conuecied, simply connected, complete, infinitesi- mally homogeneousRiemannian manifold is a homogeneous Riemann- ian space.
It is clear that kM + 1 :::; ~n(n - 1) but a better estimate, namely kM+1< ~n, is givenin [3, p. 165].
The following useful lemma also follows from [17].
LEMMA 2. H P(r) is satisfied, tben there existsamaximal principal .snbbundleF:of the orthonormal frame bundleOeM;g) ~.Monwhich ibe componentsRijkt. andRh•... ht,ijkt.,1 ~ hl,··· ,ha,i,j,k,R. ~n,1 :::;
s :::; r, are constants and which contains a given frame b E O(M,g).
Moreover, the connected component of tbe identity ofG;, x E M being arbitrary, is the structure group ofF:.
Here we used the notational convection
Rh•... ht,ijkt. = g«'Vt ...htR)e;ej ek, et.) where {ei'i = 1, ... ,n} is an orthonormal frame.
There is no lack of examples of non-homogeneous curvature homo- geneous manifolds (i.e., (M,g) satisfying P(O». We refer to [6] - [13], [16], [18] - [21] for more details, more references and up-to-date infor- mation, in particular for the three-· and four-dimensional case.
For dim M = 3, Singer's estimate is kM+1 ~ 3 but in [14] the first author proved the following sharper result :
THEOREM 3. Let (M,g) bea three-dimensional, connected, simply connected, complete Riemannian manifold wbicb.is curvature homoge- neousup to order 1. Then (M, g) is homogeneous and moreover, (M, g) is either symmetric or a group space witb a left invariant metric.
We also refer to [5] for a short proof of the homogeneity.
When dimM = 4, Singer's estimate gives kM +1 ~6 and Gromov's estimate is kM +1< 6. Moreover, we proved in [15] :
THEOREM 4. Let (M,g) be a four-dimensional, connected simply connected and complete Riemannian manifold which is curvature ho- mogeneous up to order two. Then (M,g) is homogeneous and more- over, (M,g) is either symmetric or a group space with a left invariant metric.
The second part of this statement is proved in [1], [4].
The main purpose of this note is to prove the following improvement of Theorem 4:
THEOREM 5. Let (M,g) be a four-dimensional, connected, simply connected and complete Riemannian manifold which is curvature ho- mogeneous up to order one. Then (M,g)is homogeneous and moreover, (M,g)is either symmetric or a group space with a left invariant metric.
2. Sketch of the proof of the main result
Because of [1], [4] we have only to prove the homogeneity. So, let u =(el' ...,en) be a smooth local cross section of OeM,g)and put
n
v,», = L::rijkek
k=l
Then the local functions rijk satisfy
, i,j = 1, ... ,n.
i,j,k = 1, ...,n.
For dimM = n = 4 and x E (M,g), we may choose an orthonormal basis [e,,i = 1, ... ,4} ofTxM such that
where Q denotes the Ricci endomorphism and when (M, g) satisfies P(O), all the eigenvalues Ai are constant on M. Then we have to consider the following five cases :
(1) four different Ricci eigenvalues ,
(II) three different Ricci eigenvalues ,
(III) two Ricci eigenvalues with multiplicity two , (IV) three equal Ricci eigenvalues ,
(V) four equal Ricci eigenvalues
and without loss of generality, we may suppose : (I) Al> A2 >A3 > A4,
(II) Al= A2, A3 i-A4 i-Ali-A3, (III) Al=A2 i-A3 =A4'
(IV) Al=A2 = A3i-A4, (V) Al=A2= A3=A4.
We start by considering the cases (I) and (V).
LEMMA A. An(M,g) of type(I) is homogeneous.
Proof. For. such an (M,g) we have tJo={O}Qrequivalently, kM=O.
Then the result follows from Singer's theorem.·
LEMMA B. An (M,g) of type(V) is homogeneous.
Proof. Inthis case (M,g)is a curvature homogeneous Einstein space and hence symmetric as follows from a still unpublished result of A.
Derdziriski [2].
So we are left with the cases (II), (III) and (IV). More specifically we have to consider the following subcases :
(1) (11h :go= {OJ,
(11)z :go= 50(2)EB{OJ= £II, (11h :go = 50(2} EEl{O},gl = {OJ;
(2) (111h :go= {OJ,
(111h :go=50(2)EEl {OJ= £11,
(111h :go= 50(2)EEl{O},gl = {OJ;
(111)4 :go=50(2)EEl50(2)=£11,
(111)s :go= 50(2)EEl50(2) , £II = 50(2)EB{OJ, (111)6: go=50(2) EEl 50(2) , £11= {OJ;
(3) (IVh :go = {O},
(IV)2 :go =50(2)EEl{O}=gt,
(IV)3 :go = 50(2)EEl{O},gl ={O};
(IV)4 : go = 50(3)= gI,
(IV)s :go =50(3), gl = 50(2)EEl {O},
(IV)6 :go = 50(3),gl = {O};
First we note that the cases (IIIh, (IIIh cannot occur. Further, we have
LEMMA C. The theorem holds for the cases (IIh, (II)2, (IIIh, (111)4, (IVh, (IVh, (IV)4.
Proof. As is Lemma A the result follows at once from Singer's result.
For the six remaining cases we note that the method of proof is similar to the one used in [15] but the explicit computations are now considerably longer. For that reason we only give a brief sketch of the proofs.
LEMMA D. The theorem holds for the case (IIh.
Proof. The hypothesis implies that we may choose a global or- thonormal frame field U= (eI,e2,e3,e4) such that Qei = Aiei,1 ::;
i ::; 4, and such that the functions Rabcd(U),Ri,abcd(U),1 ::;i,a ::;4, are constant on M. Then it follows by considering the components of the covariant derivative Vpof the Ricci tensorpof type (0,2) that the functions 1\13,ri14,ri23,ri24,ri34,1 ::;i ::;4, are also constant. More- over the frame field may be chosen such that, up to sign, the non-zero componentsof R are given by
(4) {R1212 =a,R1313 =R2323 =(3,R1414 = R2424 ="R3434 = a,
R 1324 =€,R 1423 = -€,R 1234 = 2€.
Using (4), the components of VR and the Bianchi identities, direct but long computations lead to the consideration of the following cases :
(ii) (1'123+1'213, 1'124+1'214, 1'113 - 1'223, 1'114 - 1'224)=(0,0,0,0) and (1'134,1'234) =J(0,0);
(iii) (1'123+1'213, 1'124+1'214, 1'113 - 1'223,1'114 - 1'224)=(0,0,0,0)
and (1'134,1'234) =(0,0) and ge2 - (a - (3)(6 - (3) = 0,ge2 - (0 - ,)(6 - ,)=J°(respectively ge2- (a - (3)(6 - (3) =J0, ge2-
(0 - ,)(6 - ,)= 0).
Inall these cases one derives that (M,g) is a group space.
LEMMA E. The theorem holds for the cases (111)5 and (111)6.
Proof. It is already proved in [15] that an (M,g) of type (111)5 is a direct product of two surfaces of different constant curvature. So, in this case, (M,g) is a symmetric space.
Hence, it suffices to consider the case (111)6' Again, the hypothe- sis implies that we may choose a global orthonormal frame field u = (el,e2, e3, e4) such that Qei = Aiei,1 ::;i ::;4 and such that the func- tions Rabcd(u), Ri,abcd(u),1 ::; i,a ::; 4, are constant. Then the func- tions 1'i13,1'i23,1'i14, 1'i24 ,1 ::;i ::;4, are constant. Further, the frame field may be chosen such that, up to sign, the non-zero components of
R are given by
(5) {
R1212 =a,R 3434 =6,R 1313=R 2323 =R 1414 =R 2424 =(3,
R 1324 = e,R 1423 = - € ,R1234 = 2e.
Now, we proceed as in Lemma D and consider first the case (a - (3,e)=J (0,0),(6 - (3, e):f:.(0,0). Then we get
and further direct computations lead here to the consideration of the following cases :
(i) (1'123,1'113) =J ±(1'U4,-1'124) and (1'314,1'313)
:f:. ±( -1'323,1'324);
(ii) (1'134,1'313) = (-1'323,1'324) respectively (1'323, -1'324), and
(1'123,1'113) =J ±(1'114,-1'124);
(ii)' (1'314,1'313) =J ±(-1'323,1'324) and (1'123,1'113) = (-1'114,1'124) respectively (1'114, -1'124);
(iii) ( f314,f313) = (-f323,f324) respectively ( f323 , - f324) and
(f123,f113) = ( - f114, f 124)respectively( f114, - f142)together with(f313,f323) -I(0,0);
(iv) the same conditions as in (iii) but now writh (r313,f323) = (0,0).
Inall these cases we find that(M,g)is a group space or a direct product of two surfaces of different constant curvature.
For the case (a - (3, e)-I(0,0),(6 - (3,e) =(0,0) respectively (a- (3,e)=(0,0), (6 - (3,e)-I(0,0), we also deduce the same result.
LEMMA F. The theorem holds for the cases (IV)a,(IV)s,(IVk Proof. For (IV)s and (IV)6 the result follows from [15]. In these cases(M,g)is a direct product of a three-dimensional space of non-zero constant curvature and R, and hence it is a symmetric space.
So, we are left with the case (IV)3. Now we choose again a global orthonormal frame field u = (e1, e2, e3,ea) on M such that Qei =
Aiei,1 ~ i ~ 4, and such that Rabcd(U), Ri,abcd(U),1 ~ i,a ~ 4 are constant. Then it follows that fi14, ri24, fi34,1~ i ::;4, are constant functions. This frame field may be specialized further such that, up to sign, the non-vanishing components ofR are given by
(6) {R1212 = a,R1313 = R 2323= (3,R3434 = 6,R1414 = R 2424 = "
R 1324= e,R 1423 = -e,R 1234 = 2e.
First, let e -I O. Then also fi13,fi23,1 ::; i ::; 4, are constants.
Further direct computations then lead to the following cases :
(i) ( f123+f 213, f 124+f214, f113 - f 223, f114 - f224) =1=(0,0,0,0);
(ii) ( f123+f 213, f124+f 214, f113 - r 223, fU4 - f 224) = (0,0,0,0) and( f134, f 234) -I(0,0).
In both cases (M,g) turns out to be a group space.
Next, let e= O. It follows that I'i13,I'i23,1 ~ i ~ 4, are constants.
Here we find that (M,g) is a group space or a direct product of a three-dimensional space of non-zero curvature and R.
The main result follows nowdirectly from these lemmas.
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Department of Mathematics Niigata University
Niigata 950-21, Japan
C. Itoh Techno-Science Co. Ltd.
Komazawa 1-16-7, Setagaya 154Tokyo, Japan
Department of Mathematics Katholieke Universiteit Leuven Celestijnenlaan 200B
3001 Leuven, Belgium
