COMPACT FLAT SPACETIMES

Mohammed Guediri

Abstract

We give a precise description of compact flat spacetimes and use this to draw conclusions about the existence of closed timelike and null geodesics. Other related results and comments are also given.

1. Introduction

Recall that a spacetime (M, g) is a time-oriented Lorentz manifold with signature −, +, …, +. A tangent vector X to M is said to be timelike, spacelike or null according to whether g(X, X) < 0, > 0 or = 0, respectively. Similarly, a smooth curve in M is said to be timelike, spacelike or null if its tangent vector field is always timelike, spacelike or null, respectively. We are interested in compact flat spacetimes which are, according to [5], quotients of the form ℝ_{1}^{n}/Γ, where ℝ_{1}^{n} denotes the n-dimensional Minkowski space and Γ is a discrete cocompact subgroup of isometries which, by [13], must be virtually polycyclic (equivalently, contains a solvable subgroup of finite index).

As mentioned in [16], it was our great wish to complete our line of investigation of compact flat spacetimes by classifying them, but after having almost had finished the classification we read in [6] that this had been done already in [14], where it was shown that a compact flat n-dimensional spacetime is, up to a finite cover, either a k-step nilmanifold with k ≤ 3, or a solvmanifold T_{A}^{n} = ℝ × T^{n−1}/ (t, z) ∼ (t + 1, Az) , where A ∈ Sl_{n−1} (ℤ) is a Lorentz matrix, that is, A is diagonalizable over ℂ and its eigenvalues are 1, λ, λ^{−1}, a_{1}, …, a_{n−3}, where λ is positive real and each a_{i} satisfies |a_{i}| = 1.

2000 Mathematics Subject Classification. 53C50, 53C22, 22E25.

Key words. Transitive actions of Lie groups, compact flat spacetimes, closed geodesics.

The aim of this paper is to give a precise classification of compact flat spacetimes using different techniques, and then use this to study the existence of the closed timelike and null geodesics. Our approach is purely geometric, given that the methods we employed can be regarded as an adaptation to the n-dimensional case of the ideas that had been used in [8] for the case n = 4. This relies on an exact description by means of exponential coordinates of the Lie groups which may act isometrically and simply transitively on Minkowski space, and on an extension of the classical Bieberbach’s theorem to discrete subgroups of the affine group Aff (ℝ^{n}) (cf. [6]). The paper is organized as follows. In Section 2, we classify the unipotent Lie groups which can act isometrically and simply transitively on Minkowski space. We show that they are very special k-step nilpotent Lie groups U_{(a,b)} that we parametrize by pairs (a, b) . In this case we see that the nilpotency class k is ≤ 3, and that U_{(a,b)} is abelian (i.e., k = 1) if and only if a = b = 0; it is 2-step nilpotent if and only if a = 0 and b ≠ 0; and it is 3-step nilpotent otherwise (i.e., a ≠ 0). Since a geodesically complete and flat left invariant Lorentz metric on a simply connected Lie group N is the same as an isometric simply transitive action of N on Minkowski space ℝ_{1}^{n}, we deduce from this classification that a nilpotent Lie group can admit a flat left invariant Lorentz metric if and only if it is isomorphic to some U_{(a,b)}.

In Section 3, we classify the nonunipotent Lie groups that can act isometrically and simply transitively on Minkowski space. We show that they fall to two categories depending on whether the linear part of the unipotent radical of the algebraic hull of the group contains a hyperbolic automorphism. The fundamental result of this section is Theorem 3.2. It asserts that if G is a nonunipotent Lie group that arises as the crystallographic hull of some discrete cocompact subgroup, then G is isomorphic to a 2-step sovable Lie group which may be obtained by extending (in some precise sense) the three-dimensional 2-step solvable Lie group G_{3} having X, Y, Z with [X, Y] = Y, [X, Z] = −Z, [Y, Z] = 0 as a basis for its Lie algebra. This result is obtained by using a result in [6] generalizing the first theorem of Bieberbach on crystallographic groups to discrete subgroups of Aff (ℝ^{n}). Theorem 3.2 will be afterwards a crucial tool for the study of the existence of closed timelike and null geodesics in compact flat spacetimes.

In Section 4, we classify compact flat spacetimes in light of the results obtained in the previous sections. We prove in Theorem 4.1 that, up to finite covers, one such spacetime is either a k-step nilmanifold with k ≤ 3, or a 2-step solvmanifold that is diffeomorphic to the product T^{m} × T_{A}^{2k+3}, where n = m + 2k + 3 and A is some Lorentz matrix in Sl_{2k+3} (ℤ). This contains the results of [2] for n = 3 and results of [8] for n = 4 as special cases. We should also mention the paper [7] from that a more precise classification for the case n = 3 might be deduced.

In Section 5 we prove, using Theorem 4.1, a special case of an old conjecture of Auslander [3] stating that every nilpotent Lie group N which acts affinely and simply transitively on ℝ^{n} has a nontrivial central translation. Namely, we prove that if N acts affinely and simply transitively on ℝ^{n} and the action preserves a flat Lorentz metric on ℝ^{n}, then N contains a nontrivial translation in its center. More precisely, such a translation lies in the derived subgroup [N, N] of N.

In Section 6, we discuss the problem of existence of closed nonspacelike geodesics in compact flat spacetimes. We prove that a compact flat spacetime contains no closed timelike geodesic if and only if it is a nilmanifold U_{(a,b)}/Γ with b ≠ 0. We also prove that any compact flat spacetime contains a closed null geodesic, and we deduce an alternate proof for a previous result [16] stating that a compact flat spacetime M contains a closed timelike geodesic if and only if the fundamental group π_{1} (M) contains a nontrivial timelike translation.

2. Unipotent Lie groups acting isometrically and simply transitively on ℝ_{1}^{n}

The following result classifies unipotent Lie groups which can act isometrically and simply transitively on Minkowski space.

Theorem 2.1. Let U be a unipotent Lie group acting isometrically and simply transitively on Minkowski space (ℝ_{1}^{n}, g) . Then, with respect to suitable linear coordinates (x, y, z) on ℝ^{n} ≅ ℝ × ℝ^{n−2}×ℝ with y = (y_{1}, …, y_{n−2}) , the metric g has the form 2dxdz + dy_{2}^{2} + … + dy_{n−2}^{2} and U is conjugate to the unipotent group U_{(a,b)} which consists of all pairs

(( \begin{matrix}1 & −w & −\frac{1}{2}ww^{t} \\ 0 & Id_{n−2} & w^{t} \\ 0 & 0 & 1 \end{matrix} ) , [ \begin{matrix}x − \frac{1}{2}wy^{t} − \frac{1}{6}ww^{t} \\ y^{t} + \frac{1}{2}zw^{t} \\ z \end{matrix} ]) ,

with w = (a_{2}y_{2}, b_{2}z, a_{4}y_{2}, b_{4}z, …, a_{n−2}y_{n−2}, b_{n−2}z) if n is even, and w = (b_{1}z, a_{2}y_{2}, b_{3}z, a_{4}y_{4}, …, a_{n−3}y_{n−3}, b_{n−2}z) if n is odd; here a = (a_{2}, …, a_{n−2}) , b = (b_{2}, …, b_{n−2}) if n is even, and a = (a_{2}, …, a_{n−3}), b = (b_{1}, …, b_{n−2}) if n is odd, and the conjugacy class of U_{(a,b)} is uniquely determined by (a, b) up to rescaling (a, b) → (λa, λ^{2}b) , with λ > 0.

Proof. Recall that Isom (ℝ_{1}^{n}) = O (n − 1, 1) ⋉ℝ^{n}, where the factor ℝ^{n} acts on ℝ_{1}^{n} by translations. Let π : Isom (ℝ_{1}^{n}) → ℝ^{n} be the natural projection which associates each element of Isom (ℝ_{1}^{n}) to its translation part. Since U acts simply transitively, we see that π is bijective from U onto ℝ^{n}. This allows us to define a map ϕ : ℝ^{n} → O (n − 1, 1) so that U = Graph ϕ. In this respect, write (ϕ (v) , v) for elements of U. Since U is unipotent, it is algebraic and therefore coincides with its unipotent radical. Let L : Isom (ℝ_{1}^{n}) → O (n − 1, 1) denote the morphism which associates each element of Isom (ℝ_{1}^{n}) = O (n − 1, 1) ⋉ℝ^{n} to its linear part. Since U is unipotent, L (U) is conjugate to a subgroup of the group U (n, ℝ) of upper triangular matrices with diagonal entries 1. On the other hand, we can easily see that L (U) preserves a null vector (compare [8], p. 388). So, we choose linear coordinates x_{1}, …, x_{n} in ℝ^{n} so that L (U) fixes the vector e_{1} = (1, 0, …, 0)^{t} , where t denotes the transpose. Denote by J the matrix of the metric g relative to the pseudo-orthonormal basis {e_{1}, …, e_{n}} associated to the above system of coordinates, with g (e_{1}, e_{1}) = g (e_{n}, e_{n}) = 0 and g (e_{1}, e_{n}) = 1. Since B = ϕ (v) satisfies B^{t}JB = J and Be_{1} = e_{1} we get B^{t} (Je_{1}) = Je_{1}, and since Je_{1} is null and B^{t} (e_{n}) = e_{n} we may assume that Je_{1} = e_{n}, that is, g has the form 2dx_{1}dx_{n} + dx_{2}^{2} + … + dx_{n−1}^{2}. Relative to the above basis, an element (A (v) , v) of the Lie algebra U of U takes the form

A (v) = ( \begin{matrix}0 & α (v) & β (v) \\ 0 & K (v) & γ (v) \\ 0 & 0 & 0 \end{matrix} ) ,

with α (v) = (α_{1} (v) , …, α_{n−2} (v)), β (v) ∈ ℝ, γ (v) = (γ_{1} (v) , …, γ_{n−2} (v))^{t} and K (v) is a an upper triangular (n − 1)-squared matrix with diagonal entries 0.

¿From the identity A^{t}J+JA = 0 it follows that K^{t} = −K, β = 0, and γ^{t} = −α. That is, K ≡ 0, β = 0, and γ_{2} = −α_{1}, …, γ_{n−1} = α_{n−1}. Set v = (x, y, z)^{t} with y = (y_{1}, …, y_{n−2}) , and write α_{i} ((x, y, z)^{t}) = c_{i}x+∑_{j=1}^{n−2} α_{i,j}y_{j}+b_{i}z. For v, w ∈ ℝ^{n}, we have [(A (v) , v) , (A (w) , w)] = (A (v) A (w) − A (w) A (v) , A (v) w − A (w) v) .

We check easily that A (v) A (w) − A (w) A (v) = 0 and that (x̃, ỹ, z̃)^{t} = A (v) w − A (w) v = (−α (v) w_{2} + α (w) v_{2}, α^{t} (v) w_{3} − α^{t} (w) v_{3}, 0)^{t} , where here v = (v_{1}, v_{2}, v_{3})^{t} and w = (w_{1}, w_{2}, w_{3})^{t} . Since [(A (v) , v) , (A (w) , w)] ∈ U, we get α ((x̃, ỹ, z̃)^{t}) = 0 for arbitrary v, w ∈ U. On the one hand, for v = (1, 0, 0)^{t} and w = (0, k_{i}, 0)^{t} , with k_{i} = (0, …, 0, 1, 0, …, 0) where the 1 is at the ith rank, we see that α (v) = (c_{1}, …, c_{n−2}) and α (w) = (α_{1,i}, …, α_{n−2,i}) . Thus (x̃, ỹ, z̃)^{t} = (−c_{i}, 0, 0)^{t} , and hence α ((x̃, ỹ, z̃)^{t}) = −c_{i} (c_{1}, …, c_{n−2}) = 0. Since i is arbitrary, this yields c_{1} = … = c_{n−2} = 0. On the other hand, for v = (0, 0, 1)^{t} and w = (0, k_{i}, 0)^{t} we see that α (v) = (b_{1}, …, b_{n−2})^{t} and α (w) = (α_{1,i}, …, α_{n−2,i}) . Thus (x̃, ỹ, z̃)^{t} = (b_{i}, (α_{1,i}, …, α_{n−2,i}) , 0)^{t} , and hence

α ((x̃, ỹ, z̃)^{t}) = ( ∑_{j=1}^{n−2} α_{1,j}α_{j,i}, …, ∑_{j=1}^{n−2} α_{n−2,j}α_{j,i})^{t} = 0  (2.1)

Setting [α] = (α_{i,j})_{1≤i,j≤n−2} it follows from (2.1) by letting i vary that [α]^{2} = 0.

We distinguish two cases. Assume first that n is even and remark that, in this case, there exists a basis relative to which [α] is a product of Jordan blocks of the form

( \begin{matrix}0 & a \\ 0 & 0 \end{matrix} ) , a ∈ ℝ.

Notice that since [α] acts on a spacelike subspace, we may choose Euclidean orthonormal coordinates, say y_{1}, …, y_{n−2}, relative to which [α] is a product of Jordan blocks of the above form. In that coordinates, we see that the α_{i} have the form α_{2k−1} ((x, y, z)^{t}) = a_{2k}y_{2k} + b_{2k−1}z and α_{2k} ((x, y, z)^{t}) = b_{2k}z, 1 ≤ k ≤ \frac{n−2}{2}. Without loss of generality, by assuming b_{n−2} ≠ 0 and a_{2k} ≠ 0, 1 ≤ k ≤ \frac{n−2}{2}, we may change the coordinates y_{2}, y_{4}, …, y_{n−2} so that b_{2k−1} = 0 for 1 ≤ k ≤ \frac{n−2}{2}. In matricial words, we may suppose that (α_{1}, α_{3}, …, α_{n−3})^{t} = diag {a_{2}, a_{4,} …, a_{n−2}} (y_{2}, y_{4}, …, y_{n−2})^{t}. This finally allows us to choose an orthonormal basis relative to which we have

(A (v) , v) = (( \begin{matrix}0 & −w & 0 \\ 0 & 0 & w^{t} \\ 0 & 0 & 0 \end{matrix} ) , [ \begin{matrix}x \\ y^{t} \\ z \end{matrix} ]) ,

with w = (a_{2}y_{2}, b_{2}z, a_{4}y_{4}, b_{4}z, …, a_{n−2}y_{n−2}, b_{n−2}z) and y = (y_{1}, …, y_{n−2}) . By computing

exp ( \begin{matrix}A (v) & v \\ 0 & 0 \end{matrix} ) ,

we see that this is precisely an element of the unipotent group U_{(a,b)} described by Theorem 2.1 when n is even. Now, the second case where n is odd can be handled similarly. This completes the proof of Theorem 2.1.

In light of Theorem 2.1, we easily see that the nilpotency class k of U_{(a,b)} is ≤ 3. Precisely, U_{(a,b)} is abelian (i.e., k = 1) if and only if a = b = 0; it is 2-step nilpotent if and only if a = 0 and b ≠ 0; and it is 3-step nilpotent otherwise (i.e., a ≠ 0). In the case U_{(a,b)} is 2-step nilpotent, we see that U_{(a,b)} is a trivial central extension of the three-dimensional Heisenberg group H_{3}. More generally, we proved in [17] that a 2-step nilpotent Lie group N can admit a flat left invariant Lorentz metric if and only if N is a trivial central extension of H_{3}.

Theorem 2.2. A nilpotent Lie group N can admit a flat left invariant Lorentz metric if and only if it is isomorphic to some unipotent group U_{(a,b)}. In particular, the nilpotency class of N is ≤ 3.

Proof. It is clear that a geodesically complete and flat left invariant Lorentz metric on a simply connected Lie group N is the same as a simply transitive action by isometries of N on Minkowski space ℝ_{1}^{n}. On the other hand, the following result is contained in [1] : a flat left invariant pseudo-Riemannian metric on a Lie group G is geodesically complete if and only if G is unimodular (that is, the linear transformation Ad_{g} has determinant ±1 for every g ∈ G). Since any nilpotent Lie group is unimodular, Theorem 2.2 follows now as a consequence of Theorem 2.1 and the result above.

Corollary 2.3. A k-step nilpotent Lie group with k > 3 cannot admit a flat left invariant Lorentz metric.

3. Nonunipotent Lie groups acting isometrically and simply transitively on ℝ_{1}^{n}

Nonunipotent Lie groups which can act isometrically and simply transitively on Minkowski space are classified by the following result.

Theorem 3.1. Let G be a nonunipotent Lie group that acts isometrically and simply transitively on Minkowski space (ℝ_{1}^{n}, g) , n ≥ 3. Then, with respect to suitable linear coordinates (x, y, z) on ℝ^{n} ≅ ℝ × ℝ^{n−2}×ℝ with y = (y_{1}, …, y_{n−2}) , the metric g has the form 2dxdz + dy_{2}^{2} + … + dy_{n−2}^{2} and, with the notation \overline{y} = (y_{2k+1}, …, y_{n−2}) and n = m + 2 (k + 1), the group G is conjugate to either one of the following solvable Lie groups

(i) G_{1,n} = \{( \begin{matrix}e^{μ(\overline{y})} & & & & & & & 0 \\ & e^{J_{1}(\overline{y})} \\ & & . & . & . \\ & & & & & e^{J_{k}(\overline{y})} \\ & & & & & & Id_{m} \\ 0 & & & & & & & e^{−μ(\overline{y})} \end{matrix} ) , [ x ; y_{1} ; ⋮ ; y_{2k} ; \overline{y} ; z ]\} ,

where μ and the θ_{i} are linear functions depending only in \overline{y}, and

e^{J_{i}(\overline{y})} = ( \begin{matrix}cos θ_{i} (\overline{y}) & sin θ_{i} (\overline{y}) \\ − sin θ_{i} (\overline{y}) & cos θ_{i} (\overline{y}) \end{matrix} ) , 1 ≤ i ≤ k.

(ii) G_{2,n} = \{( \begin{matrix}1 & & & & & & & 0 \\ & e^{J_{1}(x,\overline{y},z)} \\ & & . & . & . \\ & & & & & e^{J_{k}(x,\overline{y},z)} \\ & & & & & & Id_{m} \\ 0 & & & & & & & 1 \end{matrix} ) , [ x ; y_{1} ; ⋮ ; y_{2k} ; \overline{y} ; z ]\} ,

where the θ_{i} are linear functions depending in x, \overline{y}, z, and

e^{J_{i}(x,\overline{y},z)} = ( \begin{matrix}cos θ_{i} (x, \overline{y}, z) & sin θ_{i} (x, \overline{y}, z) \\ − sin θ_{i} (x, \overline{y}, z) & cos θ_{i} (x, \overline{y}, z) \end{matrix} ) , 1 ≤ i ≤ k.

Proof. By [13], G is solvable. Since G is nonunipotent, it follows (see [4]) that the algebraic hull A (G) (that is, the Zariski closure) of G decomposes as a semidirect product T ⋉U, where U denotes the unipotent radical of A (G) and T is a nontrivial reductive abelian subgroup of O (n − 1, 1). Let L : Isom (ℝ_{1}^{n}) → O (n − 1, 1) denote the morphism which associates each element of Isom (ℝ_{1}^{n}) = O (n − 1, 1) ⋉ℝ^{n} to its linear part. Since U is unipotent, L (U) is conjugate to a subgroup of the group U (n, ℝ) of upper triangular matrices with diagonal entries 1. On the one hand, it is easy to see that the normalizer N (U) of U in Isom (ℝ_{1}^{n}) decomposes as N (U) = R⋉U, where R is a product of a maximal torus of SO (n − 1) with a reductive one-parameter subgroup of O (n − 1, 1) acting by hyperbolic automorphisms. Notice that since U is normal in A (G) , we have A (G) ⊆ N (U) by definition of N (U) . On the other hand, as we can check easily, L (U) preserves a null vector (compare [8], p. 388). We choose linear coordinates x_{1}, …, x_{n} in ℝ^{n} so that L (U) fixes the vector e_{1} = (1, 0, …, 0)^{t} , where t denotes the transpose. Denote by J the matrix of the metric g relative to the peudo-orthonormal basis {e_{1}, …, e_{n}} associated to the above system of coordinates, with g (e_{1}, e_{1}) = g (e_{n}, e_{n}) = 0 and g (e_{1}, e_{n}) = 1. Let ϕ = (ϕ (v) , v) be an element of G, and write as usual ϕ (v) = ϕ_{u} (v) .ϕ_{s} (v) = ϕ_{s} (v) .ϕ_{u} (v) for unique semi-simple ϕ_{s} (v) and unipotent ϕ_{u} (v) (notice that (ϕ_{u} (v) , v) ∈ U and ϕ_{s} (v) ∈ R ). Since A = ϕ_{u} (v) satisfies A^{t}JA = J and Ae_{1} = e_{1}, we get A^{t} (Je_{1}) = Je_{1}. Since Je_{1} is null and A^{t} (e_{n}) = e_{n}, we may assume that Je_{1} = e_{n}, that is, g has the form 2dx_{1}dx_{n} + dx_{2}^{2} + … + dx_{n−1}^{2}. Relative to the above basis, ϕ takes the form

ϕ = (ϕ (v) , v) = (( \begin{matrix}e^{μ(v)} & a_{1,2} & & ⋯ & & a_{1,n} \\ 0 & e^{J_{1}(v)} & a_{2,4} & ⋯ \\ & & ... & ... & & ⋮ \\ ⋮ & & ... & e^{J_{k}(v)} & ... \\ & & & & Id_{m} & a_{n−1,n} \\ 0 & & ⋯ & & 0 & e^{−μ(v)} \end{matrix} ) , [ v_{1} ; ⋮ ; ⋮ ; v_{2k+1} ; ⋮ ; v_{n} ]) ,

where μ, the θ_{i,} and the a_{i,j} are functions in v, and

J_{i} (v) = ( \begin{matrix}0 & θ_{i} (v) \\ −θ_{i} (v) & 0 \end{matrix} ) , 1 ≤ i ≤ k.

In this case, ϕ_{s} (v) = diag (e^{μ(v)}, e^{J_{1}(v)}, …, e^{J_{k}(v)}, 1, …, 1, e^{−μ(v)}) and ϕ_{u} (v) equals the matrix obtained by taking μ (v) = θ_{1} (v) = ⋯ = θ_{k} (v) = 0 in the matrix representing ϕ (v) . Since ϕ_{u} (v) .ϕ_{s} (v) = ϕ_{s} (v) .ϕ_{u} (v) we see that a_{1,2} = … = a_{1,n} = a_{2,n} = … = a_{n−1,n} = 0. Moreover, we deduce from the equation g = ϕ (v)^{∗} g that all the remaining a_{i,j} are identically equal to zero.

Now we turn to the Lie algebra G of G. Clearly, G consists of elements of the form (A (v) , v) = (diag (μ (v) , J_{1} (v) , …, J_{k} (v) , 0, …, 0, −μ (v)) , v) . Note that μ and the θ_{i} are linear functions in v_{1}, …, v_{n}. In order to simplify computations, change the notation v = (v_{1}, …, v_{n}) with v = (r, v_{1}, w_{1}, …, v_{k}, w_{k}, u_{1}, …, u_{m}, s) , where n = 2 (k + 1) + m. With this notation, we see that for two elements v and v′ we get [(A (v) , v) , (A (v′) , v′)] = (0, v′′), with v′′ having r′′ = μ (v) r′ − μ (v′) r, v′′_{i} = θ_{i} (v) w′_{i} − θ_{i} (v′) w_{i}, w′′_{i} = −θ_{i} (v) v′_{i} + θ_{i} (v′) v_{i}, and s′′ = −μ (v) s′ + μ (v′) s as components. Since G is a Lie algebra, it follows that

μ (v′′) = θ_{1} (v′′) = ⋯ = θ_{k} (v′′) = 0.  (3.1)

We distinguish two cases according to whether μ is identically zero or not. If μ ≡ 0, set θ_{i} (v) = a_{i}r + b_{i}s + ∑_{j=1}^{k} (a_{ij}v_{j} + b_{ij}w_{j}) + ∑_{j=1}^{m} δ_{ij}u_{j}. In this case, we have for v′′ that r′′ = s′′ = u′′_{j} = 0, 1 ≤ j ≤ m. Thus, the equation θ_{i} (v′′) = 0 reduces to

∑_{j=1}^{k} {(a_{ij}w′_{j} − b_{ij}v′_{j}) θ_{i} (v) − (a_{ij}w_{j} − b_{ij}v_{j}) θ_{i} (v′)} = 0.

Letting v and v′ such that r = r′ = s = s′ = u_{j} = u′_{j} = 0, 1 ≤ j ≤ m, we deduce from the above equation that a_{ii}^{2} + b_{ii}^{2} = 0, that is, a_{ii} = b_{ii} = 0. Now, if there exists j_{0} such that a_{ij_{0}} ≠ 0 or b_{ij_{0}} ≠ 0, letting v with v_{j} = w_{j} = u_{l} = 0, 1 ≤ j ≤ k, 1 ≤ l ≤ m, and v′ with r′ = s′ = 0, we deduce that θ_{j_{0}} ≡ 0. It follows that (A (v) , v) = diag (0, J_{1} (r, u, s) , …, J_{k} (r, u, s) , 0, …, 0) , where J_{i} (r, u, s) =

( \begin{matrix}0 & θ_{i} (r, u, s) \\ −θ_{i} (r, u, s) & 0 \end{matrix} ) with θ_{i} ≢ 0, 1 ≤ i ≤ k, (the notation θ_{i} (r, u, s) means

that θ_{i} depends only in r, u, s with u = (u_{1}, …, u_{m})). This covers case (ii).

Assume now that μ ≢ 0. In this case, we deduce from equations (3.1) that μ does not depend in r, s and that each θ_{i} depends only in u = (u_{1}, …, u_{m}) . Setting μ (v) = ∑_{i=1}^{k} (α_{i}v_{i} + β_{i}w_{i}) + ∑_{i=1}^{m} δ_{i}u_{i} and θ_{i} (v) = ∑_{j=1}^{m} δ_{ij}u_{j}, we see that the equation μ (v′′) = 0 yields

∑_{i=1}^{k} {(α_{i}w′_{i} − β_{i}v′_{i}) ∑_{j=1}^{m} δ_{ij}u_{j} + (β_{i}v_{i} − α_{i}w_{i}) ∑_{j=1}^{m} δ_{ij}u′_{j}} = 0.

Note that we deduce nothing from the equations θ_{1} (v′′) = ⋯ = θ_{k} (v′′) = 0, because u′′ = 0 and each θ_{i} depends only in u.

If there exists i_{0} such that α_{i_{0}} ≠ 0 or β_{i_{0}} ≠ 0, say α_{i_{0}} ≠ 0, then by fixing j_{0} and letting u_{j_{0}} = 1, u_{j} = 0 for any j ≠ j_{0}, and u′ = (u′_{1}, …, u′_{m}) = (0, …, 0) , we see that ∑_{i=1}^{k} (α_{i}w′_{i} − β_{i}v′_{i}) δ_{ij_{0}}. Letting in addition w_{i_{0}} = 1, w_{i} = 0 for i ≠ i_{0}, and v′ = (v′_{1}, …, v′_{k}) = (0, …, 0) , we get finally that α_{i_{0}}δ_{i_{0}j_{0}} = 0, that is, δ_{i_{0}j_{0}} = 0. This implies that θ_{i_{0}} ≡ 0, given that j_{0} has been chosen arbitrarily. It follows that either (A (v) , v) = (diag (μ (v, w, u) , 0, …, 0, −μ (v, w, u)) , v) or (A (v) , v) = (diag (μ (u) , J_{1} (u) , …, J_{k} (u) , 0, …, 0, −μ (u)) , v) , where J_{i} (u) =

( \begin{matrix}0 & θ_{i} (u) \\ −θ_{i} (u) & 0 \end{matrix} ) with θ_{i} ≢ 0, 1 ≤ i ≤ k, (the notation μ (v, w, u) (resp. μ (u)

and θ_{i} (u)) means that μ (resp. μ and θ_{i}) depends in v_{1}, w_{1}, …, v_{k}, w_{k}, u_{1}, …, u_{m} (resp. u = (u_{1}, …, u_{m}))). This covers case (i), and therefore achieves the proof of Theorem 3.1.

We refine Theorem 3.1 in the case that G is a nonunipotent Lie group arising as the crystallographic hull of some discrete cocompact subgroup.

Theorem 3.2. Let Γ be a nonunipotent group that acts discontinuously and isometrically on Minkowski space (ℝ_{1}^{n}, g) with compact quotient ℝ_{1}^{n}/Γ, n ≥ 3. Then, with respect to suitable linear coordinates (x, t, y, z) on ℝ^{n} ≅ ℝ × ℝ × ℝ^{n−3}×ℝ with y = (y_{1}, …, y_{n−3}) , the metric g has the form 2dxdz + dt^{2} + dy_{1}^{2} + … + dy_{n−3}^{2} and Γ has a subgroup of finite index which is conjugate to a discrete subgroup of the 2-step solvable Lie group

G_{c,k}^{n} = \{( \begin{matrix}e^{t} & & & & & & & & 0 \\ & 1 \\ & & e^{J_{1}(t)} \\ & & & . & . & . \\ & & & & & & e^{J_{k}(t)} \\ & & & & & & & Id_{m} \\ 0 & & & & & & & & e^{−t} \end{matrix} ) , [ x ; t ; y_{1} ; ⋮ ; ⋮ ; y_{n−3} ; z ]\} ,

with c = (c_{1}, …, c_{k}) ∈ ℝ^{k}, k and m satisfying n = m + 2k + 3, and

e^{J_{i}(t)} = ( \begin{matrix}cos (c_{i}t) & sin (c_{i}t) \\ − sin (c_{i}t) & cos (c_{i}t) \end{matrix} ) , 1 ≤ i ≤ k.

Proof. By [13] Γ is virtually solvable, that is, it contains a solvable subgroup of finite index. Assume without loss of generality that Γ is itself contained in a solvable Lie group G. Being nonunipotent, G is conjugate in Isom (ℝ_{1}^{n}) to either G_{1,n} or G_{2,n} by Theorem 3.1. Let L be the morphism which associates each element of Isom (ℝ_{1}^{n}) = O (n − 1, 1) ⋉ℝ^{n} to its linear part, and let \overline{L (Γ)}^{0} be the identity component of the topological closure of L (Γ) . Setting L (Γ)_{nd} = L (Γ) ∩ \overline{L (Γ)}^{0} and Γ_{nd} = Γ ∩ L^{−1} (L (Γ)_{nd}) , and recalling that Γ is cocompact we deduce by Theorem 1.2.1 of [6] that Γ_{nd} is unipotent. This implies that G is not conjugate to G_{2,n}, and thus G is conjugate to G_{1,n}. The same argument of Theorem 1.2.1 in [6] applied to Γ considered as a subgroup of G_{1,n} shows that if μ (\overline{y}) = 0 then θ_{i} (\overline{y}) = 0 for all i. In other words, G is conjugate to G_{1,n} with the additional

condition that ker μ ⊆ ⋂_{i=1}^{k} ker θ_{i}.

Let us now see to what extent can this last inclusion be simplified. First, with the notation of Theorem 3.1, note that by setting μ (\overline{y}) = ∑_{j=1}^{m} δ_{j}y_{2k+j} and θ_{i} (\overline{y}) = ∑_{j=1}^{m} α_{ij}y_{2k+j}, where \overline{y} = (y_{2k+1}, …, y_{n−2}) , it follows from the above inclusion that there exists some j_{0} such that δ_{j_{0}} ≠ 0 and α_{ij_{0}} ≠ 0 for all i, 1 ≤ i ≤ k. Thus the implication

[μ (\overline{y}) = 0] ⇒ [θ_{i} (\overline{y}) = 0 for all i]

yields the implication

[y_{2k+j_{0}} = −\frac{1}{δ_{j_{0}}} ∑_{j≠j_{0}} δ_{j}y_{2k+j}] ⇒ [∑_{j≠j_{0}} (α_{ij} − \frac{α_{ij_{0}}}{δ_{j_{0}}} δ_{j}) y_{2k+j} = 0 for all i] .

Since \overline{y} is arbitrary in the vector space ker μ, we deduce that α_{ij} = \frac{α_{ij_{0}}}{δ_{j_{0}}} δ_{j} for all i, j. Thus, by setting v′ = (x, y_{1}, …, y_{2k}, y′_{2k+1}, …, y′_{n−2}, z) with y′_{2k+j_{0}} = ∑_{j=1}^{m} δ_{j}y_{2k+j} and y′_{2k+j} = y_{2k+j} for j ≠ j_{0}, 1 ≤ j ≤ m, we see that μ (v′) = y′_{2k+j_{0}} and θ_{i} (v′) = \frac{α_{ij_{0}}}{δ_{j_{0}}} y′_{2k+j_{0}} for all i.

By setting c_{i} = \frac{α_{ij_{0}}}{δ_{j_{0}}} , 1 ≤ i ≤ k, and choosing the new linear coordinates x′_{1}, …, x′_{n} in ℝ^{n} such that x′_{l_{0}} = ∑_{j=1}^{m} δ_{j}x_{j} and x′_{l} = x_{l} for l ≠ l_{0} where l_{0} = n − m + j_{0} − 1, we see that

(A (v) , v) = (diag (y_{2k+j_{0}}, J_{1} (y_{2k+j_{0}}) , …, J_{k} (y_{2k+j_{0}}) , 0, …, 0, −y_{2k+j_{0}}) , v) ,

with J_{i} (y_{2k+j_{0}}) = ( \begin{matrix}0 & c_{i}y_{2k+j_{0}} \\ −c_{i}y_{2k+j_{0}} & 0 \end{matrix} ) , 1 ≤ i ≤ k. The proof of Theorem 3.2

is complete.

Remark 1. To see that G_{c,k}^{n} is 2-step solvable, consider its Lie algebra G_{c,k}^{n}. This consists of elements of the form

(A (v) , v) = (diag (t, 0, J_{1} (t) , …, J_{k} (t) , 0, …, 0, −t) , v) ,

where v = (x, t, y_{1}, …, y_{n−3}, z) and J_{i} (t) = ( \begin{matrix}0 & c_{i}t \\ −c_{i}t & 0 \end{matrix} ) , 1 ≤ i ≤ k.

Let {e_{1}, …, e_{n}} be the canonical basis of ℝ^{n}, that is, e_{i} is the vector having all components equal to 0 except the i-th one which is equal to 1. Recalling that n = m + 2k + 3, and setting X = (A (e_{2}) , e_{2}) , Y = (A (e_{1}) , e_{1}) , Z = (A (e_{n}) , e_{n}) , V_{i} = (A (e_{2(i+1)−1}) , e_{2(i+1)−1}) , W_{i} = (A (e_{2(i+1)}) , e_{2(i+1)}) with 1 ≤ i ≤ k, and U_{j} = (A (e_{2(k+1)+j}) , e_{2(k+1)+j}) with 1 ≤ j ≤ m, we get a basis for G_{c,k}^{n} with [X, Y] = Y, [X, Z] = −Z, [X, V_{i}] = −c_{i}W_{i}, [X, W_{i}] = c_{i}V_{i}, 1 ≤ i ≤ k, and all the other brackets are zeros. It follows that [[G_{c,k}^{n}, G_{c,k}^{n}] , [G_{c,k}^{n}, G_{c,k}^{n}]] = 0, that is, G_{c,k}^{n} is a 2-step solvable Lie algebra.

Remark 2. It should be noticed that the group G_{c,k}^{n} is not algebraic and its unipotent radical U (G_{c,k}^{n}) equals the n-dimensional group T_{n} of translations of ℝ_{1}^{n}. To see this, set v = (0, t, 0, …, 0)^{t} and consider the element

(ϕ (v) , v) = (diag (e^{t}, 1, e^{J_{1}(t)}, …, e^{J_{k}(t)}, 1, …, 1, e^{−t}) , v)

of G_{c,k}^{n}. This splits as a product (ϕ (v) , v) = gh = hg with g = (ϕ_{u} (v) = Id, v) and h = (ϕ_{s} (t) , 0) , where ϕ_{s} (t) = diag (e^{t}, 1, e^{J_{1}(t)}, …, e^{J_{k}(t)}, 1, …, 1, e^{−t}) . Since U (G_{c,k}^{n}) = {(ϕ_{u} (v) , v) : (ϕ (v) , v) ∈ G_{c,k}^{n}} it follows that U (G_{c,k}^{n}) = T_{n}. This agrees with Theorem III.1 in [3] that if a connected, simply connected solvable Lie group G acts transitively by affine transformations on ℝ^{n}, then the unipotent radical U of the algebraic hull A (G) also acts simply transitively by affine transformations on ℝ^{n}.

4. Classification of compact flat spacetimes

Compact flat spacetimes are classified by the following result.

Theorem 4.1. Let M be an n-dimensional compact flat spacetime, n ≥ 3. Then, up to a finite cover, M is congruent to either

(i) a k-step nilmanifold ℝ_{1}^{n}/Γ with k ≤ 3, where Γ is a lattice in some unipotent Lie group U_{(a,b)} that is described by Theorem 2.1, or

(ii) a 2-step solvmanifold ℝ_{1}^{n}/Γ, where Γ is a lattice in some nonunipotent Lie group G_{c,k}^{n} that is described by Theorem 3.2.

Proof. By [5] the universal covering of M is Minkowski space (ℝ_{1}^{n}, g) . Therefore, without loss of generality M ≅ ℝ_{1}^{n}/Γ where , by [13], Γ ≅ π_{1} (M) is a lattice in a connected and simply connected n-dimensional solvable Lie group G that acts isometrically and simply transitively on ℝ_{1}^{n}. If G is unipotent, then it is isomorphic to some group U_{(a,b)} that is described by Theorem 2.1. In this case, since Γ ⊂ G is cocompact and G is nilpotent, it follows that G must be rational [20]. This covers case (i). If G is nonunipotent, then by Theorem 3.2 G is conjugate to some 2-step solvable Lie group G_{c,k}^{n}. In this case, it is easy to see that Γ coincides with the subgroup of G_{c,k}^{n} consisting of elements (r, t, q, s) ∈ ℤ^{n} ≅ ℤ × ℤ × ℤ^{n−3}×ℤ, with q = (q_{1}, …, q_{n−3}). This covers case (ii), and therefore completes the proof of Theorem 4.1.

We shall now describe and classify the lattices of the 2-step solvable Lie groups G_{c,k}^{n}. Recall first that a matrix A ∈ GL_{n} (ℤ) is called of Lorentz type in [14] if it is diagonalizable over ℂ and its eigenvalues are 1, λ, λ^{−1}, a_{1}, …, a_{n−3}, where λ is positive real and each a_{i} satisfies |a_{i}| = 1. It is then clear that given a Lorentz matrix A ∈ SL_{n−1} (ℤ) there exist t ∈ ℝ, an integer k ≥ 0, and c = (c_{1}, …, c_{k}) ∈ ℝ^{k} such that A is conjugate in SL_{n−1} (ℝ) to the matrix A_{c,k}^{t} = diag (e^{t}, e^{J_{1}(t)}, …, e^{J_{k}(t)}, 1, …, 1, e^{−t}) , where J_{i} (t) stands for the 2 × 2-matrix appearing in the notation of Theorem 3.2. This defines for A a mapping

ℝ → SL_{n−1} (ℝ)
t ↦ A_{c,k}^{t}.

Let G_{A} be the Lie group diffeomorphic to ℝ^{n} = ℝ × ℝ^{n−1} with the group law

(t, w) (t′, w′) = (t + t′, A_{c,k}^{t}w + w′) , t, t′ ∈ ℝ and w, w′ ∈ ℝ^{n−1}.

Clearly G_{A} is isomorphic to G_{c,k}^{n}. Let Γ_{A} ≅ ℤ⋉ℤ^{n−1} be the discrete subgroup of G_{A} consisting of elements with integral coordinates. It is then easy to see that Γ_{A} is a lattice in G_{A} ≅ G_{c,k}^{n}.

With this notation, we deduce from Theorem 3.2 the following result which generalizes in higher dimensions results in [7] and [8]. This also generalizes Proposition II.2.2 in [12].

Theorem 4.2. Every n-dimensional compact flat spacetime of type G_{c,k}^{n}/Γ_{A} is finitely covered by a T^{n−1}-bundle over the circle. More precisely, up to a finite cover, G_{c,k}^{n}/Γ_{A} is diffeomorphic to T^{m} × T_{A}^{2k+3} with :

T_{A}^{2k+3} = ℝ × T^{2(k+1)}/ (t, w) ∼ (t + 1, Ã(w)) ,

where n = m + 2k + 3, and Ã denotes the automorphism of the torus T^{2(k+1)} induced by A.

Remark 3. It is interesting to note that G_{A} depends in general on the matrix A. However, if A is conjugate to a matrix of type A_{c,k}^{t} with c = (c_{1}, …, c_{k}) = (0, …, 0) , that is, A is now purely a hyperbolic automorphism of the torus, then G_{A} becomes isomorphic to the solvable group E (1, 1) = O (1, 1) ⋉ℝ^{2} of rigid motions of Minkowski plane ℝ_{1}^{2}. In this particular case, if Γ_{A} is a lattice in G_{A} then the 2-step solvmanifold M = ℝ_{1}^{n}/Γ_{A} becomes congruent to a product of a Euclidean torus T^{n−3} with the three-dimensional Lorentz 2-step solvmanifold ℝ_{1}^{3}/Γ, where Γ is a lattice in the 2-step solvable Lie group G_{3} = ℝ⋉ℝ^{2} with the factor ℝ acting on ℝ_{1}^{2} by hyperbolic automorphisms and the normal factor ℝ^{2} consisting purely of translations. Notice that lattices in G_{3} exist (see [10], p. 511, or [12], p. 291-292).

In [12], Proposition II.2.3, it is shown that if Γ is lattice in the 2-step solvable Lie group G_{3} described in the remark above then there exists a matrix A in SL_{2} (ℤ) with trace tr (A) > 2 such that Γ is conjugate in G_{3} to Γ_{A}. The following proposition which appears as a simple consequence of Theorem 3.2 generalizes this result. This shows that the spacetimes G_{c,k}^{n}/Γ_{A} described above are the only ones to consider.

Proposition 4.3. Let Γ be a lattice in some 2-step solvable Lie group G_{c,k}^{n}, n ≥ 3. Then, there exists a Lorentz matrix A in SL_{n−1} (ℤ) such that Γ is conjugate in G_{c,k}^{n} to Γ_{A}.

5. Around a conjecture of L. Auslander

In [3], Auslander has been conjectured that every nilpotent Lie group N which acts affinely and simply transitively on ℝ^{n} has a nontrivial central translation. This is true if n ≤ 3 [7] or if N has a degree of nilpotence k = 1, that is, N is abelian [9]. This has also been proved in the case of filiform Lie groups (i.e., the index of the Lie algebra is maximal) with odd dimension [19]. However, this conjecture does not hold in general (see [9] for a counter-example in dimension 4 which is 3-step nilpotent). We believe that Auslander’s conjecture holds for 2-step nilpotent Lie groups, and we are actively working toward this result. In light of Theorem 2.1, we prove here that Auslander’s conjecture holds if N is a nilpotent subgroup of the group E (1, n − 1) of affine Lorentz motions (see also [1] for an alternate proof of this fact).

Theorem 5.1. Let N be a nilpotent Lie group acting affinely and simply transitively on ℝ^{n}. If the action preserves a flat Lorentz metric on ℝ^{n} then N contains a nontrivial translation in its center. More precisely, such a translation lies in the derived subgroup [N, N] of N.

Proof. Being a nilpotent group and acting simply transitively by affine transformations, N is unipotent [21]. By Theorem 2.1, N is conjugate to U_{(a,b)} for some pair (a, b) . Now, the element τ_{x} of U_{(a,b)} defined by w = 0, y = 0, and z = 0 is clearly a nontrivial translation which lies in the derived subgroup of U_{(a,b)}.

6. Closed geodesics in compact flat spacetimes

Establishing the existence of closed timelike geodesics in compact spacetimes is one of the important problems in Lorentzian geometry. Based on an old result in [22], it is natural to conjecture that any compact spacetime which admits a globally hyperbolic regular covering contains a closed timelike geodesic. In a recent work [15] we provided, among other results, examples of compact flat Lorentz space forms without closed timelike geodesics. Since each of such spacetimes admits Minkowski space as its universal covering, this provides a negative answer to the above conjecture. We proved, however, that any compact flat spacetime contains a closed timelike or null geodesic. In a second paper [17], we proved that in N is a simply connected 2-step nilpotent Lie group with a flat left invariant Lorentz metric g, then for any lattice Γ in N the compact flat Lorentz 2-step nilmanifold (N/Γ, g) contains no closed timelike geodesic. A natural question that arises in this context is : Does any compact flat Lorentz nilmanifold contain no closed timelike geodesic ?

The following result which characterizes completely the closed timelike and null geodesics in compact flat spacetimes also answers negatively the above question.

Theorem 6.1. Let M be a compact, flat spacetime of dimension n ≥ 3. Then M necessarily contains a closed null geodesic, and it contains a closed timelike geodesic if and only if it is congruent to either :

(i) a nilmanifold ℝ_{1}^{n}/Γ, where Γ is a lattice in some U_{(a,b)} with b = 0. More precisely, M is a torus if a = 0, or a 3-step nilmanifold if a ≠ 0.

(ii) a 2-step solvmanifold ℝ_{1}^{n}/Γ, where Γ is a lattice in some G_{c,k}^{n}.

Proof. Recall that the universal covering of M is Minkowski space ℝ_{1}^{n}, and that the geodesics of M are projections under the natural projection π : ℝ_{1}^{n} → M of geodesics (i.e., straight lines) of ℝ_{1}^{n}. In light of Theorem 4.1, we distinguish two cases.

Case 1. M is a nilmanifold ℝ_{1}^{n}/Γ, where Γ is a lattice in some U_{(a,b)}. In this case, Γ is precisely the subgroup of U_{(a,b)} consisting of elements with integral entries. Let c(t) = (αt + x_{0}, β_{1}t + y_{1,0}, …, β_{n−2}t + y_{n−2,0}, δt + z_{0}) be a geodesic in ℝ_{1}^{n} that is translated an amount ω by an element ϕ ∈ Γ, that is, ϕ · c (t) = c (t + ω) for all t ∈ ℝ. According to the form can ϕ take (Theorem 2.1), this is equivalent to the following system of equations

{−w(βt + y_{0}) − 1/2ww^{t}(δt + z_{0}) + x − 1/2wy^{t} − 1/6ww^{t} = αω ; (δt + z_{0})w^{t} + y^{t} + 1/2zwt = ωβ^{t} ; z = δω,  (6.1)

where β = (β_{1}, …, β_{n−2})^{t} , y_{0} = (y_{1,0}, …, y_{n−2,0})^{t} , and w is as in Theorem 2.1. Since t is arbitrary, we necessarily get wβ + 1/2δww^{t} = 0 and δw^{t} = 0. If b ≠ 0, there is some b_{i_{0}} ≠ 0, and so it follows from δw^{t} = 0 that δz = 0. Substituting this into the third equation of (6.1), we deduce that δ = 0. Thus g(ċ, ċ) = 2αδ + β^{2} = β^{2} ≥ 0, and hence c cannot be timelike. This however shows that M contains closed null geodesics, as it suffices to choose β = 0 in c (t).

If b = 0, letting ϕ (x, y, z) be an element of Γ with y = 0 we see that the geodesic c(t) = (αt + x_{0}, 0, …, 0, δt + z_{0}) in ℝ_{1}^{n} with x = αω, and z = δω may be translated an amount ω by ϕ (x, 0, z) . Since x, z in ϕ (x, 0, z) may be chosen so that xz < 0 or = 0, we deduce that M contains both closed timelike and null geodesics.

Case 2. M is a 2-step solvmanifold ℝ_{1}^{3}/Γ, where Γ is a lattice in some G_{c,k}^{n}. Let c(t) = (αt + x_{0}, ηt + t_{0}, β_{1}t + y_{1,0}, …, β_{n−3}t + y_{n−3,0}, δt + z_{0}) be a geodesic in ℝ_{1}^{n} that is translated an amount ω by an element ϕ (x, s, y, z) ∈ Γ, where y = (y_{1}, …, y_{n−3}) . In this case, with the notation of Theorem 3.2, the equation ϕ · c (t) = c (t + ω) is equivalent to the following system of equations

{e^{s}(αt + x_{0}) + x = α(t + ω) + x_{0} ; s = ηω ; e^{J_{i}(s)}(β_{i}t + y_{i,0}) + y_{i} = β_{i}(t + ω) + y_{i,0}, 1 ≤ i ≤ n − 3 ; e^{−s}(δt + z_{0}) + z = δ(t + ω) + z_{0},

Since t is arbitrary, we get from the first and the last equations that s = 0 (that is, ϕ is a pure translation) or α = δ = 0 (that is, c is spacelike). Now, by letting for instance ϕ (a, 0, 0, b) be an element of Γ with either ab = −1, or else a = 0 and b ≠ 0, we see that the geodesic c (t) = (at, 0, …, 0, bt) of ℝ_{1}^{n} is translated an amount 1 by ϕ (a, 0, 0, b) . This shows that M contains closed timelike and null geodesics. The proof of Theorem 6.1 is complete.

Theorem 6.1 and its proof can also provide an alternate proof to the following result established already in [16].

Theorem 6.2. A compact flat spacetime M contains a closed timelike geodesic if and only if the fundamental group π_{1} (M) contains a nontrivial timelike translation.

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Department of Mathematics College of Science King Saud University P.O. Box 2455, Riyadh 11451 Saudi Arabia E-mail : mguediri@ksu.edu.sa