# What is the Lebesgue covering dimension of this topological space?

Take the 4 dimensional time-oriented spacetime $$(M,g)$$ such that it's not strongly causal.

Take the induced topology defined by the Lorentzian metric called Alexandrov topology.

This topology matches the manifold topology if and only if the spacetime is strongly causal.

It can be shown for our spacetime which is not strongly causal, this topology is strictly coarser than that of the manifold topology:

$$\tau_{\text{Alexandrov}} \subsetneq \tau_{\text{Manifold}}$$

Moreover, it is shown that such topology cannot lead to any topological manifold, since it cannot be locally Euclidean and strictly coarser than the manifold topology at the same time.

My question is:

What is the Lebesgue covering dimension of this topological space? What about inductive dimensions?

EDIT:

I would be rather interested in spaces where the metric cannot be globally decomposed into a direct sum of two subspace metrics of lower dimensions with signatures $$(1+1)⊕2$$, although a $$(1+2) \oplus 1$$ signature for a possible global decomposition is acceptable.

EDIT: any such spacetime better be a solution of Einstein's vacuum field equations.

• In the case the Alexandrov topology is indiscrete, the covering dimension is 0. OTOH, many of the standard examples of non strongly causal 2 dimensional manifolds have infinite Lebesgue covering dimension (in fact, this applies to all of the examples I checked so far). Commented Feb 21 at 16:15
• Good, but first of all, dimension two is too restrictive for interesting objects to appear. Second, I would be rather interested in spaces where the metric cannot be globally decomposed into direct sum of two subspaces of the form $(1+1) \oplus 2$@WillieWong Commented Feb 21 at 23:09

Concerning the Lebesgue covering dimension, absolutely nothing can be said, if you work with manifolds of total space-time dimension 3 or higher.

### Preamble

We will consider spacetimes with closed timelike curves. Let $$(M,g)$$ be Lorentzian. A point $$p\in M$$ has a closed timelike curve through it if and only if $$p \in I^+(p)$$; this is further equivalent to $$p\in I(p,p)$$ where $$I(p,q) = I^+(p)\cap I^-(q)$$. Note that if $$q\in I(p,p)$$, then $$q$$ both precedes and succeeds $$p$$, and hence there is a closed timelike curve through $$q$$. Furthermore, in this case $$I^\pm(p) = I^\pm(q)$$. And hence we have that

For each $$p,q\in M$$, either $$I(p,p)\cap I(q,q) = \emptyset$$ or $$I(p,p) = I(q,q)$$.

We will let $$M_c := \cup_{p\in M} I(p,p)$$, the set of all points through which a closed timelike curve passes.

### Trivial case

If $$M = M_c$$, then the Alexandrov topology is indiscrete, and the covering dimension is 0.

### Finite positive case

#### A basic construction

Let $$M = S \times (-1,1) \times \mathbb{S}^1$$ be a topological cylinder where $$S$$ is any Riemannian manifold; parametrize it with $$(x,z,t)$$. The Riemannian metric on $$S$$ is denoted by $$h$$.

Consider the metric on $$M$$ given by $$g = h + ( dt + \sin^2(z) dz) \otimes (-\sin^2(z) dt + dz) + (-\sin^2(z) dt + dz) \otimes (dt + \sin^2(z) dz) \tag{*}$$ or $$g = h - 2 \sin^2(z) dt^2 + 2\sin^2(z) dz^2 + (1-\sin^4(z)) (dt \otimes dz + dz\otimes dt)$$ One sees that $$g$$ is non-degenerate of signature $$(-,+,+,\dots)$$. The vector field $$\partial_t - \partial_z$$ is timelike everywhere, and so $$g$$ is time-orientable. And finally:

When $$z \neq 0$$, the vector field $$\partial_t$$ is time-like, but when $$z = 0$$ the vector field $$\partial_t$$ is null.

In particular, we have that:

• When $$p$$ is such that $$z \leq 0$$, then $$I^+(p) = \{z < 0\}$$; similarly when $$p$$ is such that $$z \geq 0$$, then $$I^-(p) = \{z > 0\}$$.
• When $$p$$ is such that $$z > 0$$, then $$I^+(p) = M$$; when $$p$$ is such that $$z < 0$$, then $$I^-(p) = M$$.

So the Alexandrov toplogy of $$M$$ has $$\tau = \{ \emptyset, \{z < 0\}, \{z > 0\}, M\}$$

#### The construction refined

Let $$n \geq 2$$ and $$N$$ any natural number. Take $$N+1$$ copies of $$\mathbb{S}^n$$ and label them $$S_0, \ldots, S_N$$. Place $$S_0$$ in the center, and arrange the remaining spheres to surround it. Connect each $$S_i$$ (when $$i > 0$$ to $$S_0$$ with a cylinder that is topologically $$\mathbb{S}^{n-1}\times (-1,1)$$. So in the case $$n = 2$$ and $$N = 4$$ it looks a bit like the chemistry picture below.

Call the resulting manifold (which is still topologically a sphere) $$\Sigma$$. Let $$M_N = \Sigma \times \mathbb{S}^1$$. We will place a metric on $$M_N$$ such that on each of the $$S_i$$ it is basically just the product metric $$h - dt^2$$ where $$h$$ is the round sphere metric. Near the center of the connecting rods the metric will look like (*), with the negative z side near the center and the positive z side near the outside. This can be smoothly glued together.

Let $$S'_i$$ for $$i > 0$$ denote the $$S_i$$ together with the portion of the neck corresponding to $$z >0$$. (Roughly each of the white regions in the picture above). And let $$S'_0$$ denote $$S_0$$ together with all the neck portions corresponding to $$z < 0$$. (Roughly the magenta portion.) And let $$\Gamma_i$$ be the boundaries.

To help you think about this: the $$\Gamma_i$$ (when crossed with $$\mathbb{S}^1$$) are null hypersurfaces in $$M_N$$. They function as one-way gates: a future-directed time-like curve can cross from $$S_i'$$ through $$\Gamma_i$$ into $$S_0'$$ in the forward time direction, but cannot go back.

Using a similar analysis as before, given a point $$p$$, we have that (for this argument, we need that $$S_0$$ with finitely many disjoint closed discs removed is path connected. This is where we need $$n \geq 2$$ (and hence spacetime dimension at least 3))

• If $$p\in S'_0$$, then $$I^+(p) = S'_0$$ and $$I^-(p) = M_N$$.
• If $$p\in S'_i$$, $$i > 0$$, then $$I^+(p) = S'_0 \cup \Gamma_i \cup S'_i$$, and $$I^-(p) = S'_i$$.
• If $$p\in \Gamma_i$$, then $$I^+(p) = S'_0$$ and $$I^-(p) = S'_i$$.

In particular, given $$i$$, let $$p\in S'_i$$ and $$q\in S'_0$$, then $$I(p,q) = S'_0 \cup \Gamma_i \cup S'_i$$ is open in the Alexandrov topology. In addition, it is the smallest open set in the Alexandrov topology that intersects $$\Gamma_i$$. Call this set $$O_i$$.

It is clear now that $$\{O_1, \ldots, O_N\}$$ is an open cover of $$M_N$$ in the Alexandrov topology. As $$O_i$$ are as small as possible, any refinement of the cover will necessarily still contain $$\{O_1, \ldots, O_N\}$$, and any cover of $$M_N$$ may be refined to $$\{O_1, \ldots, O_N\}$$. But now any $$p\in S_0$$ will be covered $$N$$ times, this shows that any natural number may arise as the covering dimension of a Lorentzian manifold under the Alexandrov topology.

### The infinite case

I conclude with the case where the covering dimension is infinite. This can be done using a 2 dimensional example, but it can be easily generalized to higher dimensions.

Let $$M' = \mathbb{R}\times \mathbb{S}^1 \ni (x,t)$$ with the product metric (we identify $$\mathbb{S}^1 = \mathbb{R} / \mathbb{Z}$$ so has period 1). Consider $$M = M' \setminus (-\infty,0] \times \{0\}$$. Then $$M_c = \{(x,t) \in \mathbb{R}\times [-1/2,1/2) : x > - |t|\}$$. In particular, if $$p\in \partial M_c$$, then any open set in the Alexandrov topology that contains $$p$$ must contain $$M_c$$.

For $$\epsilon > 0$$, the set $$O_\epsilon = M_c \cup \{(x,t)\in \mathbb{R}\times [-1,0): t < \epsilon \}$$ is open. And altogether they form an open cover of $$M$$. Key observation is that if $$\mathcal{V}$$ is a collection of finitely many sets, each of which is a refinement of some $$O_\epsilon$$, then $$\mathcal{V}$$ cannot cover $$\partial M_c$$, never mind $$M$$. Hence any refinement that covers $$M$$ needs to contain infinitely many open subsets of the $$O_\epsilon$$s, each of which intersecting $$\partial M_c$$; but as every such open set contains $$M_c$$ we see that the covering dimension must be infinite.

### Remarks

A modification of the procedure in the finite positive case provides a 2 dimensional example with covering dimension 1. I do not know whether in 2 dimensions, other finite covering dimensions are possible.

• Since you also asked about inductive dimensions: consider the case of $M_N$. (In the Alexandrov topology) You can check that each $S'_i$ has no proper open subset, and that $S'_i$ is not closed (its closure is $S'_i \cup \Gamma_I$). Hence the small inductive dimension of $M_N$ is $\infty$. // Similarly the large inductive dimension of $M_N$ is $\infty$ when $N > 1$. (These you can check yourself, as the topology of $M_N$ has finitely many sets and for small $N$ you can actually do a case-by-case analysis.) Commented Feb 22 at 7:06
• First thing that makes me skeptical about the construction is that the spatial part $\Sigma$ is compact while this is not realistic from a physical stand point. Second is the decomposablility of the of the topological manifold into the cartesian product of 3 topological spaces $S, I, \mathbb{S^1}$ where $I=(-1,1)$. It kills so much topological twists that can bear significant consequences for the final Alexandrov topology. Third is that I am curious what the final remark results in if one works on a $3+1$ and $2+1$ generalisation of the construct with the modification claimed by the writer. Commented Feb 25 at 9:30
• And I'd be happy to know about the modificatation in $1+1$ dimensions too. Thanks @WillieWong Commented Feb 25 at 10:43
• Maybe I better edit the question even further to be more detailed. @WillieWong Commented Feb 25 at 13:39
• @BastamTajik: the compactness is entirely inessential in the argument; if you'd looked at it carefully you'd see that the sphere factors $S_1, \ldots, S_N$ can easily be replaced by copies of any manifold you want, which allows you to make the spatial topology anything you want. The point is that once you have space-time dimension 3 or above, topological data from the manifold topology says absolutely nothing about the covering dimension of the Alexandrov topology, as every extended natural number is possible. Commented Feb 26 at 2:34