SYMMETRIES OF SECTIONAL CURVATURE ON 3 MANIFOLDS. May 27, 1992

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1 SYMMETRIES OF SECTIONAL CURVATURE ON 3 MANIFOLDS Luis A. Cordero 1 Phillip E. Parker,3 Dept. Xeometría e Topoloxía Facultade de Matemáticas Universidade de Santiago Santiago de Compostela Spain cordero@zmat.usc.es pparker@zmat.usc.es May 7, 199 Abstract In dimension three, there are only two signatures of metric tensors: Lorentzian and Riemannian. We find the possible pointwise symmetry groups of Lorentzian sectional curvatures considered as rational functions, and determine which can be realized on naturally reductive homogeneous spaces. We also give some examples. MSC (1991): Primary 53C50; Secondary 53B30, 53C30. 1 Supported by Project XUGA , Xunta de Galicia, Spain. On leave from Math. Dept., Wichita State Univ., Wichita KS 6760, U.S.A., pparker@twsuvm.uc.twsu.edu 3 Partially supported by DGICYT-Spain.

2 1 Introduction Let M be a smooth 3-manifold and g a pseudoriemannian metric tensor on M. Let G (M) denote the Grassmannian bundle with fibers G (T x M), the space of (-dimensional) planes in the tangent space T x M at a point x M. Observe that each G (T x M) may be regarded as a (real) algebraic variety, diffeomorphic to the (real) projective plane P. As in [1], we shall regard the sectional curvature K x at each point x M as a rational mapping of algebraic varieties G (T x M) R, or a rational function for short. The group of all automorphisms of G (T x M) is isomorphic to P GL 3 P GL 3 (R), the group of projective automorphisms of P. We may then ask: what is the largest subgroup of P GL 3 which leaves K x invariant as a rational function? We shall refer to this group as the symmetry group of K at x. Throughout this paper, we shall concentrate on the Lorentzian case, giving complete details. On the other hand, we shall merely summarize the Riemannian case, omitting details and leaving them for the reader to supply. Thus, in Section, we determine the possible symmetry groups for Lorentzian sectional curvature, finding them to be P GL 3, P O, P O 1 1, P HT, Z Z, Z, 1, the first case characterizing constant curvature at a point. For this, we use the analysis of canonical forms for selfadjoint operators on a Minkowskian (Lorentzian) vector space [10, pp.61 6]. For comparison, we list the corresponding results for Riemannian sectional curvature. In Section 3, we study the existence of naturally reductive spaces in which the symmetry group of K is the same at every point. All continuous symmetry groups (see Theorem.) can be realized on compact models, but there are no naturally reductive models of constant negative curvature or with any discrete symmetry group. We do not seek a neat classification here, so questions about effectiveness of actions are of no relevance to us, for example. (Cf. [, p. 58f ] for a list of isometry classes of simply connected Riemannian homogeneous spaces of dimension 3.) Finally, Section 4 contains some particular examples. In most cases, we are able to obtain models which are trivial circle bundles over surfaces. We defer to another paper [3] the exhibition of more explicit forms of metric tensors on these spaces; the techniques used there are not directly related to those used here. We also defer to other papers [3, 4] the exhibition of examples which are not naturally reductive, including all left-invariant Lorentzian metric tensors on 3-dimensional Lie groups. Our Lorentzian metric tensors will have signature +. In some cases, we shall have to distinguish among the possible orderings +, +, +. (To convert to the other signature convention + +, see [10, p.9].) Thus a vector v is timelike if g(v, v) > 0, lightlike or null if g(v, v) = 0, spacelike if g(v, v) < 0, and causal if g(v, v) 0. 1

3 If A is a matrix regarded as a linear transformation R n R n, then the induced mapping A: R n R n is given by the matrix classically called the second compound of A, the matrix whose entries are the determinants of submatrices of A in an appropriate ordering [7, Sec. 7.]. Throughout, we shall regard the Riemann tensor R ijkl as a quadratic form on T M and thus, via the Plücker embedding, on G (M); cf. [1]. Then the sectional curvature appears as a rational function on G (M) in the form of a quotient of two quadratic functions: K = R g. Also recall that the associated tensor R ij kl represents the curvature operator R: T M T M in local coordinates. Note that if R and R are written as matrices with respect to the same local coordinates, then R = ( g) R. We denote the Lorentz group in (n = p + q) dimensions of signature (p, q) by O q p = O q p(r), thus the (usual) orthogonal group by O n = O n (R). Projectivization of any group of linear transformations is indicated by a prefixed P ; for example P GL 3 = GL 3 /{ai ; 0 a R} = SL 3. We thank Graham Hall for pointing out to us the symmetry of horocyclic translations, and for reminding us that eigenvalue degenerations produce symmetries. Most of the results in Section were previously presented by Parker at the A.M.S. Winter Meetings in 1984 and 1985, and at the Bolyai Colloquium on Differential Geometry in Parker thanks Cordero and the Departamento at Santiago for their extraordinary hospitality during his visits. Canonical Forms and Symmetry Groups We begin by finding canonical forms for the Riemann tensor R considered as a quadratic form on T M. We shall do this by first finding canonical forms for the selfadjoint operator R: T M T M. It suffices to work pointwise, so consider R 3 with the Minkowski inner product given by η = diag[1, 1, 1] and let R be a η-selfadjoint operator R 3 R 3. Note, η = diag[ 1, 1, 1]. From O Neill [10, pp. 61 6, ex. 19], we find that there are four canonical forms for selfadjoint operators such as R. Let e 1, e, e 3 denote the usual basis for R 3. We choose e 1 = e 1 e, e 13 = e 1 e 3, e 3 = e e 3 as our associated basis for R 3. Changing bases from O Neill s to ours, we have Lemma.1 The selfadjoint operator R appears, with respect to some η- orthonormal basis of R 3, in precisely one of the following forms on the associated basis of R 3 :

4 B C 0, 0 0 A B λ ± 1 λ ± 1, B A F, 0 F A λ λ 0 0 λ 1 where A, B, C, F, λ are independent real constants. Now let M be a 3-manifold with Lorentzian metric tensor g. We agree that whenever we choose local coordinates at x M which give rise to a g-orthonormal basis (e 1, e, e 3 ) of T x M with g(e 1, e 1 ) = 1, we shall use (e 1, e 13, e 3 ) as the associated basis of T x M. With these choices, the Riemann tensor R x appears as a 3 3 symmetric matrix R 11 R 113 R 13 R 113 R 1313 R 133 R 13 R 133 R 33 which we regard as a quadratic form on T x M as in [1]. Applying Lemma.1 pointwise and using R = ( g) R, we have Theorem. At each point x of a Lorentzian 3-manifold (M, g), there exists a choice of g-orthonormal coordinates with respect to which the Riemann tensor R x on T x M takes on exactly one of these canonical forms:, CF1 diag [B, C, A] ; B 0 0 CF 0 A F, F 0; 0 F A CF3 B λ ± 1 ± 1 ;, CF4 0 ± 1 λ ± 1 λ λ λ We note that these forms can also be characterized in terms of eigenvectors of R: CF1 corresponds to a timelike eigenvector, CF to a spacelike eigenvector, CF3 to a double null eigenvector, and CF4 to a triple null eigenvector; compare [6, 4.3]. We now determine the pointwise symmetry group of each associated sectional curvature K x = R x / g x. Before we begin, let us note that these 3

5 Table 1: Lorentzian sectional curvature symmetry groups Canonical form of R x CF1 : diag[b, C, A] Symmetry group of K x = R x / g x A = B = C P GL 3 B = C A } P O A = B C P O1 A = C 1 B generic Z Z CF Z CF3 B λ Z B = λ P HT CF4 1 symmetry groups may also be regarded as parameterizing the choices of local orthonormal coordinates with respect to which R x appears in its canonical form. Observe that if R x = c g x, then K x = c is a constant and is invariant under all automorphisms of the Grassmannian G (T x M). Therefore, in this case the symmetry group is P GL 3. In terms of our canonical forms, constant sectional curvature at x is characterized by CF1 with A = B = C. Note that in all other cases, the symmetry group will be a subgroup of P O1( +); see Table 1. The procedure for determining the remaining symmetry groups is as follows. First, the invariance group of R x on T x M, expressed with respect to an appropriate orthonormal basis of T x M, is {A GL 3 ; A t R x A = R x }. Note that with respect to the same basis, the invariance group of g x is O1( +) GL 3. The desired symmetry group of K x is then the projectivization of the intersection of these two invariance groups. Now consider CF1. Letting A = [a ij ] with 1 i, j 3, it is easy to see that if B = C, then a 13 = a 3 = a 31 = a 3 = 0 and [ ] a11 a 1 O a 1 a O1( +). Hence a 33 = ±1 and the symmetry group is isomorphic to P O. Similarly, if A = B or A = C, it follows that the symmetry group is isomorphic to P O 1 1. Finally, if none of these hold, direct calculation from A t R x A = R x shows that A is diagonal with all diagonal entries ±1, independently of each other. Projectivized, this group is isomorphic to Z Z. Next, consider CF. Again, one begins with A t R x A = R x and calculates directly to obtain the invariance group, finding that A is diagonal with 4

6 Table : Riemannian sectional curvature symmetry groups Canonical form of R x CF1 : diag[b, C, A] Symmetry group of K x = R x / g x A = B = C P GL 3 A = B C A = C B B = C P O A generic Z Z all three diagonal entries ±1. But this time, one also obtains a = a 33. Projectivizing, we obtain a symmetry group isomorphic to Z. Next, consider CF3. Using the same procedure, we find two cases. When B λ, we obtain the group isomorphic to Z as for CF. When B = λ, we obtain the group HT of horocyclic translations (called null rotations in relativity because there is a fixed null direction). The identity component of this group consists of all the matrices exp 0 t t t 0 0 = t t t t 1 t t t t 1 + t, t R, and appears naturally as the nilpotent group in an Iwasawa decomposition of SO1 +. Under the canonical double covering SL SO1 +, it is the image of the parabolic elements of the form [ ] 1 t. 0 1 See [5] for more about this subgroup. Each component of O 1( +) contains one component of HT. Projectivizing, we obtain P HT. Finally, consider CF4. Direct calculation from A t R x A = R x again yields a diagonal A, now with a 11 = a = a 33 = ±1. Projectivizing, we obtain the trivial group 1. This completes the determination of all possible pointwise symmetry groups for the sectional curvature function K x of a Lorentzian 3-manifold, as listed in Table 1. We conclude this section with the Riemannian case. Here, selfadjoint transformations correspond to symmetric matrices, which are always orthogonally diagonalizable. Thus only CF1 occurs for Riemannian metrics, and we obtain Table. 5

7 3 Existence of Naturally Reductive Models In this section, we determine which CF types can be realized on naturally reductive spaces so that the CF type of K (and thus the symmetry group) is the same at every point. Thus we now consider naturally reductive homogeneous spaces M = G/H of dimension 3 with a left-invariant Lorentzian metric tensor. According to Corollary.11, p.198, Theorem 3.3 (), p.01, and the definition preceding Proposition 3.4, p.0, of [8], naturally reductive Lorentzian homogeneous spaces are complete. From Proposition 33, p.55, and Lemma 8, p.53, of [10], it follows that dim G 6. For connected M, dim G = 6 implies constant curvature. Thus we need to consider only G of dimensions 3, 4 and 5. Let g and h denote the Lie algebras of G and H, respectively. In order that G/H be reductive, we must have g = h m and [h, m] m. (3.1) It then suffices to consider ad H -invariant nondegenerate symmetric bilinear forms B on m such that B(X, [Y, Z]) B([Z, X], Y ) = 0 (3.) for all X, Y m and Z h. The naturally reductive condition is then B(X, [Y, Z] m ) B([Z, X] m, Y ) = 0 (3.3) for all X, Y, Z m. (Cf. [8, pp ].) We begin by showing that all these spaces with dim G = 3 are flat or of constant positive curvature. We shall use the summation convention without further explicit notice. Theorem 3.1 If M = G is an irreducible, naturally reductive, Lorentzian homogeneous space of dimension 3, then either M is flat or of constant positive curvature. In the former case, M is Minkowskian 3-space or one of its quotients by a discrete group of translations. In the latter, M is SO1 + (R) or one of its coverings or quotients by a discrete subgroup. Proof. Let g have basis (e 1, e, e 3 ) chosen so that B(e 1, e 1 ) = B(e, e ) = B(e 3, e 3 ) = 1 and B(e i, e j ) = 0 for i j. Let [e i, e j ] = c k ije k define the structure constants. Now m = g and h = 0, so (3.1) is trivial and (3.) becomes B(X, [Y, Z]) B([Z, X], Y ) = 0 for X, Y, Z g. Considering two and three different arguments in this equation, we obtain the structure equations of g as [e 1, e ] = a e 3, [e, e 3 ] = a e 1, [e 3, e 1 ] = a e. 6

8 If a = 0, then the Lie algebra is abelian and we obtain the flat cases. If a 0, then the Lie algebra is isomorphic to so 1. We compute the curvature of the latter. Using the formula from [8, Thm. 3.3], the Levi-Civitá connection is given by X Y = 1 [X, Y ]. A direct computation then yields R 11 = 1 4 a, R 1313 = 1 4 a, R 33 = 1 4 a, R 113 = R 13 = R 133 = 0, and we find an example of CF1 with group P GL 3. We observe that this is the situation for SL (R) considered by Nomizu [9]. 3.1 dim G = 4 Let g have basis (e 1, e, e 3, e 4 ) with B(e i, e i ) = ε i = ±1 and B(e i, e j ) = 0 for i j. Again, [e i, e j ] = c k ije k defines the structure constants. We take h = e 4 so that m = [[e 1, e, e 3 ]], where denotes the algebra generated by what is enclosed and [[ ]] denotes the linear span of what is enclosed. Thus exactly one ε i = 1 for 1 i 3, but the choice of ε 4 is free. We want G/H to be naturally reductive, (3.3) and (3.), and B to be ad H -invariant (3.1). Working all these out, we find that the structure equations for g become [e 1, e ] = c 3 1 e 3 + c 4 1 e 4, [e, e 3 ] = ε 1 ε 3 c 3 1 e 1 + c 4 3 e 4, [e 3, e 1 ] = ε ε 3 c 3 1 e + c 4 31 e 4, [e 1, e 4 ] = ε 1 ε c 1 4 e + c 3 14 e 3, [e, e 4 ] = c 1 4 e 1 + ε ε 3 c 43 e 3, [e 3, e 4 ] = ε 1 ε 3 c 3 41 e 1 + c 34 e. (3.4) From the Jacobi identity, we obtain ε 1 c 1 4c 4 31 ε 3 c 3 14c 4 1 = 0, ε c 34c 4 1 ε 1 c 1 4c 4 3 = 0, ε 3 c 3 14c 4 3 ε c 34c 4 31 = 0. (3.5) Thus all the connected, naturally reductive, Lorentzian homogeneous spaces with dim G = 4 are obtained from all simultaneous solutions of (3.4) and 7

9 (3.5) and their quotients by discrete subgroups of G. We calculate their Riemann tensors using the formula R(X, Y )Z = 1 4 [X, [Y, Z] m] m 1 4 [Y, [X, Z] m] m 1 [[X, Y ] m, Z] m [[X, Y ] h, Z] m, X, Y, Z m, (3.6) found in the proof of Proposition 3.4 of [8, p. 0], obtaining R 11 = 1 4 ε 3 (c 3 1) + ε 1 c 1 4 c 4 1, R 1313 = 1 4 ε (c 3 1) + ε 3 c 3 14 c 4 31, R 33 = 1 4 ε 1 (c 3 1) + ε c 34 c 4 3, R 113 = ε 3 c 3 41 c 4 1 = ε 1 c 1 4 c 4 13, R 13 = ε 1 c 1 4 c 4 3 = ε c 34 c 4 1, R 133 = ε c 34 c 4 13 = ε 3 c 3 41 c 4 3. (3.7) Examining (3.7) and the canonical forms CF1 CF4, we claim that only those cases of CF1 in which two of A, B, C have the same absolute value and those of CF3 in which λ = B 0 are possible when dim G = 4, and that CF and CF4 do not occur. We proceed to show this. In order to obtain CF1, suppose first that at least two of the c 4 ij in (3.4) are nonzero, whence all c i j4 = 0, or that all c 4 ij vanish, whence all c i j4 are arbitrary. Calculating, we find the curvature matrix 1 ( c 3 4 1) diag[ε 3, ε, ε 1 ] which represents constant nonnegative curvature and a symmetry group of P GL 3. If exactly one c 4 ij 0, then c i j4 is arbitrary and the others vanish. If c 4 1 0, then the curvature matrix is 1 4 ε 3(c 3 1) + ε 1 c 1 4c ε (c 3 1) ε 1 (c 3 1) For c 1 4 = 0 we have a symmetry group of P GL 3, and for c we have a symmetry group of P O1 1 when ε 1 = ε and P O when ε 1 = ε. If c , then we find the curvature matrix 1 4 ε 3(c 3 1) ε (c 3 1) + ε 3 c 3 14c ε 1 (c 3 1) 8

10 For c 3 14 = 0 the symmetry is P GL 3, and for c it is P O1 1 when ε 1 = ε 3 and P O when ε 1 = ε 3. Finally, if c we obtain the curvature matrix 1 4 ε 3(c 3 1) ε (c 3 1) ε 1 (c 3 1) + ε c 34c 4 3. For c 34 = 0 the symmetry is P GL 3, and for c 34 0 it is P O when ε = ε 3 and P O1 1 when ε = ε 3. We observe that in all cases of constant curvature it is nonnegative; it is not possible to obtain constant negative curvature with dim g = 4. Also, in all cases at least two of A, B, C have the same absolute value and the symmetry group Z Z cannot be realized. There are no other constraints on the production of CF1. Note that while the form of CF1 does not depend on the ordering of (e 1, e, e 3 ), all of CF 4 do. Thus we may as well assume from now on that (e 1, e, e 3 ) are ordered so that 1 = ε 1 = ε = ε 3. In order to obtain CF, it follows from the structure equations (3.4) and the canonical form of Theorem. that this can occur if and only if c and ε 1 + ε 0. This contradicts our assumption, so CF cannot be produced with dim g = 4. In order to obtain CF3, we must have R 113 = R 13 = 0 whence c 1 4 = c 4 1 = 0. We must also have R 133 = ±1/ whence c 34c 4 31 = c 3 14c 4 3 = ±1/. Now 1 4 (c3 1) = B so we can produce only those CF3 in which B 0, but this is the only constraint on B. If we regard B as given and substitute using the preceding relations, then we obtain c3 14 c 34 = λ ± 1 B, c 34 c 3 14 = λ ± 1 + B. Using the reciprocity of the left-hand sides, this system has a unique solution and we can produce CF3 only for λ = B 0. Finally, in order to have CF4 we must have R 133 = 0 whence c 34c 4 31 = c 3 14c 4 3 = 0. We must also have R 113 = R 13 = 1/ whence c 3 14c 4 1, c 1 4c 4 31, c 1 4c 4 3, c 34c These conditions are contradictory so CF4 cannot be produced with dim g = dim G = 5 Let g have basis (e 1, e, e 3, e 4, e 5 ) and again assume that it is B-orthonormal. This time we take h = e 4, e 5, so [[e 4, e 5 ]] must be a subalgebra, and m = 9

11 [[e 1, e, e 3 ]] again. We continue with 1 = ε 1 = ε = ε 3, and now both ε 4 and ε 5 are free. Working again from the reductive condition (3.1), the ad H -invariance condition (3.), and the naturally reductive condition (3.3), and requiring that h be a subalgebra, the structure equations become [e 1, e ] = α e 3 + a 4 e 4 + a 5 e 5, [e, e 3 ] = α e 1 + b 4 e 4 + b 5 e 5, [e 3, e 1 ] = α e + c 4 e 4 + c 5 e 5, [e 4, e 5 ] = n 4 e 4 + n 5 e 5, [e 1, e 4 ] = β 1 e + β 3 e 3, [e 1, e 5 ] = γ 1 e + γ 3 e 3, [e, e 4 ] = β 1 e 1 β e 3, [e, e 5 ] = γ 1 e 1 γ e 3, [e 3, e 4 ] = β 3 e 1 + β e, [e 3, e 5 ] = γ 3 e 1 + γ e. (3.8) From the Jacobi identity, we obtain β 1 c 4 + γ 1 c 5 + β 3 a 4 + γ 3 a 5 = 0, β a 4 + γ a 5 + β 1 b 4 + γ 1 b 5 = 0, β 3 b 4 + γ 3 b 5 β c 4 γ c 5 = 0, β c 4 + a 5 n 4 β 3 b 4 = 0, β c 5 + a 5 n 5 β 3 b 5 = 0, γ c 4 a 4 n 4 γ 3 b 4 = 0, γ c 5 a 4 n 5 γ 3 b 5 = 0, β 1 b 4 c 5 n 4 + β a 4 = 0, β 1 b 5 c 5 n 5 + β a 5 = 0, γ 1 b 4 + c 4 n 4 + γ a 4 = 0, γ 1 b 5 + c 4 n 5 + γ a 5 = 0, β 3 a 4 b 5 n 4 + β 1 c 4 = 0, β 3 a 5 b 5 n 5 + β 1 c 5 = 0, γ 3 a 4 + b 4 n 4 + γ 1 c 4 = 0, γ 3 a 5 + b 4 n 5 + γ 1 c 5 = 0, β 3 γ β γ 3 β 1 n 4 γ 1 n 5 = 0, β γ 1 β 1 γ β 3 n 4 γ 3 n 5 = 0, β 1 γ 3 β 3 γ 1 + β n 4 + γ n 5 = 0. (3.9) Thus all the connected, naturally reductive, Lorentzian homogeneous spaces with dim G = 5 are obtained from all simultaneous solutions of (3.8) and 10

12 (3.9) and their quotients by discrete subgroups of G. We calculate their Riemann tensors using (3.6), obtaining R 11 = 1 4 α + β 1 a 4 + γ 1 a 5, R 113 = β 3 a 4 + γ 3 a 5 = β 1 c 4 γ 1 c 5, R 13 = β a 4 γ a 5 = β 1 b 4 + γ 1 b 5, R 1313 = 1 4 α β 3 c 4 γ 3 c 5, R 133 = β c 4 + γ c 5 = β 3 b 4 + γ 3 b 5, (3.10) R 33 = 1 4 α β b 4 γ b 5. As before, not all cases of all canonical forms occur when dim G = 5. Only the continuous symmetry groups from CF1 and CF3 occur and CF and CF4 do not. First, observe that adding and subtracting certain equations from (3.9) yields the additional conditions a 5 n 4 a 4 n 5 = 0, b 5 n 4 b 4 n 5 = 0, c 5 n 4 c 4 n 5 = 0. For example, the first of these is obtained by taking the third equation plus the fourth plus the seventh from (3.9). It follows that unless a 4 b 5 a 5 b 4 = 0, a 4 c 5 a 5 c 4 = 0, b 4 c 5 b 5 c 4 = 0, (3.11) then n 4 = n 5 = 0. Thus we shall initially assume that (3.11) holds. In order to obtain CF1, we must have R 113 = R 13 = R 133 = 0, which via (3.10) is equivalent to β 1 b 4 + γ 1 b 5 = 0, β 1 c 4 + γ 1 c 5 = 0, β a 4 + γ a 5 = 0, β c 4 + γ c 5 = 0, β 3 a 4 + γ 3 a 5 = 0, β 3 b 4 + γ 3 b 5 = 0. Note that our assumption of (3.11) implies that β i and γ i do not necessarily all vanish, so it is possible to obtain nonconstant curvature. If all a i, b i, c i vanish, however, we would also have constant curvature. Thus we also need 11

13 at least one of the following satisfied: β 1 γ β γ 1 = 0 ; (3.1) β 1 γ 3 β 3 γ 1 = 0 ; (3.13) β γ 3 β 3 γ = 0. (3.14) If all three hold, then we are again reduced to constant nonnegative curvature. Thus we shall consider only the cases where exactly one or two hold. If one does not hold and the other two do, then in each case it now follows from two of the last three equations in (3.9) that n 4 = n 5 = 0, whence the one in fact holds. Therefore this case is not possible. If (3.1) holds and the other two do not, then a i = b i = 0 so R 11 = R 33 0 and the symmetry group is P O1 1 when β 3 c 4 + γ 3 c 5 0, P GL 3 when it is. If (3.13) holds and the other two do not, then a i = c i = 0 so R 11 = R and the symmetry group is P O when β b 4 + γ b 5 0, P GL 3 when it is. If (3.14) holds and the other two do not, then b i = c i = 0 so R 1313 = R 33 0 and the symmetry group is P O1 1 when β 1 a 4 + γ 1 a 5 0, P GL 3 when it is. Note that in all these cases of constant curvature, it is also nonnegative. The analysis for CF begins similarly. We must have β 3 c 4 + γ 3 c 5 + β b 4 + γ b 5 = 0, β 3 a 4 + γ 3 a 5 = β 1 c 4 + γ 1 c 5 = 0, β a 4 + γ a 5 = β 1 b 4 + γ 1 b 5 = 0, β c 4 + γ c 5 = β 3 b 4 + γ 3 b 5 0. It follows that (3.11), (3.1), and (3.13) are satisfied. Whether (3.14) holds or not, it turns out that β, γ, β 3, γ 3, b 4, b 5, c 4, c 5 0 and β 1 = γ 1 = a 4 = a 5 = 0. Since b 4 c 5 b 5 c 4 = 0 (else β 3 = γ 3 = 0), b 4 e 4 + b 5 e 5 and c 4 e 4 + c 5 e 5 are linearly dependent. It follows that (3.14) holds, whence β 3 e 1 + β e and γ 3 e 1 + γ e are linearly dependent. But this implies β c 4 + γ c 5 = β 3 b 4 + γ 3 b 5 = 0, contradicting the requirements for CF. Therefore CF cannot be produced with dim G = 5. In order to obtain CF3, we must have β 3 c 4 + γ 3 c 5 + β b 4 + γ b 5 = 1, β 3 a 4 + γ 3 a 5 = β 1 c 4 + γ 1 c 5 = 0, β a 4 + γ a 5 = β 1 b 4 + γ 1 b 5 = 0, β c 4 + γ c 5 = β 3 b 4 + γ 3 b 5 = ± 1. These are sufficiently different from those for CF, and the analysis again proceeds along similar lines to that for CF1. Using the first of the preceding conditions (equivalent to R R 33 = ±1), we obtain λ = B 0 when (3.11) holds; when it fails, CF3 cannot be produced. 1

14 We come at last to CF4. Considering (3.11) both holding and not, it is easy to see that this form cannot be produced with dim G = 5. As noted at the end of Section, only CF1 occurs in the Riemannian case. Theorem 3.1 continues to hold if Lorentzian is changed to Riemannian, Minkowskian to Euclidean, and SO1 + to SO 3. If dim G = 4, we take ε 1 = ε = ε 3 = 1 and now obtain only those CF1 in which two of A, B, C have the same positive value. Thus there is no naturally reductive Riemannian model with dim G = 4 and symmetry group Z Z. When dim G = 5, we also take ε 1 = ε = ε 3 = 1 and still obtain only those symmetry groups from CF1 except Z Z. Note that these actions are not effective. This concludes our summary of the Riemannian case. 4 Naturally Reductive Examples Here we shall present a few selected examples with dim G = 4 and 5. All the cases in which we were able to identify the Lie algebra can be realized as circle bundles over surfaces, but the surfaces may be open. We do not know if this is a general feature; this should be investigated further. The first examples will be those with dim G = 4. Recall the general structure equations (3.4). We also recall that here only those CF1 in which two of A, B, C have the same absolute value and only those CF3 in which λ = B 0 are possible. We begin with CF1. Referring to the discussion following (3.7), the case where all c 4 ij 0 is typical for at least two of them not vanishing. Signature + is also typical here, so we shall write only it down. Then g has structure equations [e 1, e ] = c 3 1 e 3 + c 4 1 e 4, [e, e 3 ] = c 3 1 e 1 + c 4 3 e 4, [e 3, e 1 ] = c 3 1 e + c 4 31 e 4. If c 3 1 0, then g is a central extension of so 1 by e 4 and we obtain compact models with constant positive curvature as circle bundles over closed surfaces Σ g of genus g. If c 3 1 = 0, then g is a 4-dimensional nilpotent Lie algebra with center e 4 and we obtain the compact model of the flat torus T 3. If all c 4 ij = 0, then g has the structure equations [e 1, e ] = c 3 1 e 3, [e, e 3 ] = c 3 1 e 1, [e 3, e 1 ] = c 3 1 e, [e 1, e 4 ] = c 1 4 e + c 3 14 e 3, 13

15 [e, e 4 ] = c 1 4 e 1 c 34 e 3, [e 3, e 4 ] = c 3 14 e 1 + c 34 e. Now g = g 1 θ e 4 is a semidirect product with θ : e 4 Der(g 1 ) given by 0 c 1 4 c 3 14 θ(e 4 ) = c c 34. c 3 14 c 34 0 If c 3 1 = 0, then g 1 is abelian and we obtain the compact model of a flat T 3 if and only if exp(θ) preserves a lattice in R 3. If c 3 1 0, then g 1 = so 1 and we obtain compact models with constant positive curvature as circle bundles over closed surfaces Σ g of genus g if and only if exp(θ) preserves a suitable lattice in SO1 + = P SL (R). For exactly one c 4 ij 0, the case c is typical. The structure equations of g are [e 1, e ] = c 3 1 e 3 + c 4 1 e 4, [e, e 3 ] = ε 1 ε 3 c 3 1 e 1, [e 3, e 1 ] = ε ε 3 c 3 1 e, [e 1, e 4 ] = ε 1 ε c 1 4 e, [e, e 4 ] = c 1 4 e 1, [e 3, e 4 ] = 0. If c 1 4 = 0 and c 3 1 0, then we have a central extension of so 1 again and constant positive curvature. If c and c 3 1 = 0, then the result depends on the signature. If ε 1 = ε, then g = so 1 e 3 and the symmetry group is P O 1 1. If ε 1 = ε, then g = g 1 e 3 with g 1 = so3 if c 4 1 c 1 4 > 0 and g 1 = so 1 if c 4 1 c 1 4 < 0, with the symmetry group of P O in both cases. In all these cases e 4 is not the compact generator of so 1, so we can obtain a compact model S 1 S only with so 3. Finally, if c 1 4, c 3 1 0, then g is unidentified and the symmetry group is P O 1 1 or P O according to the signature. For CF3, consider the special case c 4 3 = c and c 3 1 = 0 so B = λ = 0. Replacing e 1 by e 1 + e and leaving e, e 3, e 4 as is, the structure equations for g can be written as [e 1, e ] = 0, [e, e 3 ] = a e 4, [e 3, e 1 ] = 0, [e 1, e 4 ] = 0, [e, e 4 ] = b e 3, [e 3, e 4 ] = b e 1, 14

16 where a = c 4 3 and ab = 1/. We obtain a semidirect product g = g 1 θ e with g 1 = e 3, e 4, e 1 isomorphic to the Heisenberg algebra and θ(e ) = 0 b 0 a We can get compact models if and only if exp(θ) preserves a suitable lattice in G. Next, some examples with dim G = 5. When β i = γ i = 0, we have constant nonnegative curvature with a central extension of so 1 by the abelian e 4, e 5. We omit the details. As a typical example of the other CF1 cases, suppose that β i = γ 1 = γ 3 = a 5 = c 5 = 0 and γ, b 5 0. Then (3.9) implies that a 4 = c 4 = n 4 = n 5 = 0. The structure equations of g are [e 1, e ] = α e 3, [e, e 3 ] = α e 1 + b 4 e 4 + b 5 e 5, [e 3, e 1 ] = α e, [e, e 5 ] = γ e 3, [e 3, e 5 ] = γ e. If α = 0, then g = e 1 g 1 where g 1 is a central extension (a product if b 4 = 0) of so 1 by e 4 and we obtain models with symmetry group P O. Since e 5 is not the compact generator of so 1, none of them are compact. If α 0, then g is a central extension (a product if b 4 = 0) by e 4 of an unidentified 4-dimensional Lie algebra. All of the other Lie algebras remain unidentified. We hope to return to these later. References [1] J.K. Beem and P.E. Parker, Values of pseudoriemannian sectional curvature, Comment. Math. Helv. 59 (1984) [] L. Bérard-Bergery, Les espaces homogènes riemanniens de dimension 4, in Géométrie Riemannienne en Dimension 4, ed. L. Bérard-Bergery et al. Paris: Cedic, pp [3] L. A. Cordero and P. E. Parker, Examples of sectional curvature with prescribed symmetry on 3-manifolds, Czech. Math. J. 45 (1995) 7 0. [4] L. A. Cordero and P. E. Parker, Left-invariant Lorentzian metrics on 3-dimensional Lie groups, U. Santiago preprint,

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