On Completeness of Sliced Spaces under the Alexandrov Topology

Abstract

We show that in a sliced spacetime $(V,g)$, global hyperbolicity in V is equivalent to $T_A$-completeness of a slice, if and only if the product topology $T_P$, on V, is equivalent to $T_A$, where $T_A$ denotes the usual spacetime Alexandrov “interval” topology.

1. Preliminaries

Sliced spaces have attracted the attention of several authors in studies related to systems of Einstein equations (see [1]), completeness (see [2]), global hyperbolicity (see [3,4]), as well as in problems of a more geometric nature on quantum cosmology (see [5,6]).

Definition 1.

Let $V=M\times ℝ$, where M is an n-dimensional smooth manifold, such that V is equipped with an $n+1$-dimensional Lorentz metric g, which splits in the following way:

$$g=-N^2{\left({\theta }^0\right)}^2+g_{ij}{\theta }^i{\theta }^j,$$

where ${\theta }^0=dt$, ${\theta }^i=d{x}^i+{\beta }^{i}dt$, $N=N(t,x^i)$ is called lapse function, ${\beta }^i(t,x^j)$ is called shift function and $M_t=M\times \left\{t\right\}$, called spatial slices of V, are spacelike submanifolds equipped with the time-dependent spatial metric $g_t=g_{ij}d{x}^{i}d{x}^j$. Such a product space V is called a sliced space.

Let $(V,g)$ be a sliced space. A base for the product topology $T_P$, on V, consists of all sets of the form $A\times B$, where $A\in T_M$ and $B\in T_{ℝ}$. Here $T_M$ denotes the natural topology of the manifold M where, for an appropriate Riemann metric h, it has a base consisting of open balls $B_{ϵ}^h\left(x\right)$ and $T_{ℝ}$ is the usual topology on the real line.

The Alexandrov topology (or “interval topology”) $T_A$ on a spacetime V has a base consisting of open sets of the form $<x,y>=I^{+}\left(x\right)\cap I^{-}\left(y\right)$, where $I^{+}\left(x\right)=\{z\in V:x\ll z\}$ and $I^{-}\left(y\right)=\{z\in V:z\ll y\}$, where ≪ is the chronological order defined as $x\ll y$ iff there exists a future oriented timelike curve, joining x with y. By $J^{+}\left(x\right)$ one denotes the topological closure of $I^{+}\left(x\right)$ and by $J^{-}\left(y\right)$ that one of $I^{-}\left(y\right)$ (see [7]).

A spacetime V is strongly causal, if and only if it is strongly causal at every point, that is, for every point $p\in V$, p has arbitrarily small causally convex neighbourhoods. We say that V is globally hyperbolic, if and only if V is strongly causal and every set $J^{+}\left(x\right)\cap J^{-}\left(y\right)$ (called a “closed diamond”) is compact. Global hyperbolicity is considered the strongest causality condition in the causal hierarchy of spacetimes (see [8]) and is equivalent to the existence of a Cauchy hypersurface S for V (see Section 5, in [7]); this supplies us with the benefit to construct on V well-defined initial-value problems (see [9,10], Theorem 10.2.2). One can also view global hyperbolicity as a property on a spacetime which guarantees the absence of naked singularities in V (for its role in the strong cosmic censorship, see [11]).

In the next section we will show that global hyperbolicity in a sliced spacetime $(V,g)$ is equivalent to completeness with respect to the Alexandrov topology of a slice $(M_t,g_t)$. Although completeness of the Alexandrov topology $T_A$, by itself, is not a criterion of nonsingularity (in the Schwartzschild space and the Friedmann-Robertson-Walker cosmologies, for example, $T_A$ is complete, but these spaces are singular; see [12]), it is interesting that in the particular case of sliced spacetimes that are equipped with their natural product topology, completeness of a slice with respect to $T_A$ can be considered as a criterion of global hyperbolicity for the entire space.

Thoughout our text, for topological terms like Hausdorff space and completeness, we refer to the seminal book of Engelking, [13].

2. A Topological Condition for the Completeness of a Sliced Space

In [3], sliced spaces are being considered to have uniformly bounded lapse, shift and spatial metric, in order to achieve the equivalence of global hyperbolicity of $(V,g)$ with the completeness of the slice $(M_0,\gamma )$ (Theorem 2.1). Being motivated by this result, in the Theorem that follows, we consider global topological conditions, for showing the equivalence of global hyperbolicity of $(V,g)$ with a slice $(M_t,g_t)$ being $T_A$-complete. Our Theorem 1, below, differs from Theorem 2.1 of [3] in that the slices in [3] are complete Riemannian manifolds (with uniformly bounded spatial metric, lapse and shift functions) while in our case the slices are $T_A$-complete. We discuss this further in Section 3.

Theorem 1.

Let $(V,g)$ be a sliced space, with respect to its natural product topology $T_P$, where $V=M\times ℝ$, M is an n-dimensional manifold and g the $n+1$-Lorentz “metric” on V. Let also $T_A$ be the Alexandrov topology on V. Then, the following statements are equivalent:

1. 

$(V,g)$ is globally hyperbolic.

2. 

$T_P\equiv T_A$.

3. 

$(M_t,g_t)$ is complete with respect to $T_A$.

Proof. 

1. has been shown to be equivalent to 2. in [4].

To show that 2. implies 3., we first notice that since $(V,g)$ is globally hyperbolic, it is also strongly causal. Since, also, $T_P\equiv T_A$, we have that for every $t\in ℝ$, $M_t$ is a subset of a spacetime V, with nondegenerate spacetime metric, with subspace topology $T_A$ inherited from V, such that $M_t$ is strongly causal. Hence, $T_A$, on $M_t$, is complete (see [12], Theorem 2).

For proving that 3. implies 1., for each $t\in ℝ$, we let $(M_t,g_t)$ to be complete with respect to $T_A$, where each $M_t$ is a spacelike submanifold with time dependent spatial metric $g_t\equiv g_{ij}d{x}^{i}d{x}^j$. But since each $M_t$ is complete, the Alexandrov topology $T_A$, on $M_t$, is strongly causal (see, again, [12]). So, each point of $M_t$ is strongly causal, which means that for every point $P\in M_t$ there exists an arbitrarily small convex neighbourhood. But, $V={\bigcup }_{t\in ℝ}M\times \left\{t\right\}$, so $P\in V$ if and only if there exists $M_t=M\times \left\{t\right\}$, such that $P\in M_t$, and hence V is strongly causal with respect to $T_A$. That the closed $T_A$-diamonds, in V, are compact, has been shown in Theorem 3, from [4]. Thus, $(V,g)$ is globally hyperbolic. □

3. Discussion

Question 1: Can our Theorem 1 hold, if one substituted in 3. “$(M_t,g_t)$ is complete with respect to $T_A$” with the statement “$(M_t,g_t)$ is a complete Riemannian manifold”? The answer is negative, since in a spacetime manifold, $T_A$ is usually a coarser topology than the spacetime topology, and it is equivalent to the manifold topology only if it is Hausdorff (see [7], Theorem 4.24). So, in order for this question to have a positive answer, one would have to add in Theorem 1 the extra condition that $T_A$, on V, is Hausdorff. As a continuation of this question, we ask whether the spacetimes considered in [3] may well have their Alexandrov topology $T_A$ not being Hausdorff. In such a case, strong causality will fail due to this (see, for example, Remark 4.25 of [7]). Spacetimes where $T_A$ fails to be Hausdorff, according to Penrose, admit a null geodesic along which strong causality fails and this is one aspect of a general result concerning the region of strong causality failure in a spacetime [7]. Given the above argument, we conjecture that for a physically reasonable spacetime, the statements of Theorem 2.1 of [3] and of Theorem 1 here should be equivalent. A rigorous proof showing the equivalence of a uniformly bounded spatial metric, lapse and shift functions with the condition of the topologies $T_P$ and $T_A$ to be equivalent, will be of a great interest.

In [3], there are conditions introduced, so that global hyperbolicity to be equivalent to geodesic completeness. In particular, in Theorem 3.1, the term trivially sliced space is introduced, so that a slice is a complete Riemannian manifold, if and only if the space $(V,g)$ is geodesically complete. The “disadvantage” of this condition is that the spatial metric $g_{ij}$ is time-independent.

Question 2: Can one relate slice-completeness and geodesic completeness of $(V,g)$ with a time-dependent spatial metric $g_{ij}$?

Question 2 does not seem to have a trivial answer. In a possible variation of Theorem 1, towards an answer to this question, one could make use of the classical Hopf-Rinow Theorem (see [14]), which gives that metric completeness, in a spacetime, is equivalent to geodesic completeness. Again, $T_A$ should be Hausdorff.