# What physically determines the point-set topology of a spacetime manifold?

Like any manifold, the pseudo-Riemannian manifold of spacetime in special or general relativity is a topological space, so there is a notion of open sets (or equivalently, neighborhoods) that allows us to talk about continuity, connectedness, etc. We implicitly use this structure whenever we frame the equivalence principle as saying that any spacetime "locally looks like Minkowski space" - the "locally" really means "in very small neighborhoods within the manifold". This point-set-topological structure is in a sense even more fundamental than anything relating to the metric, because any manifold has such a structure, whether or not is is pseudo-Riemannian (or even differentiable).

But what physically defines these open sets? For a Riemannian manifold (or more generally any metric space), in practice we always use the topology induced by the metric. But this doesn't work for a pseudo-Riemannian manifold, because the indefinite metric signature prevents it from being a metric space (in the mathematical sense). For example, if I emit a photon which "later" gets absorbed in the Andromeda Galaxy, then there is clearly a physical sense in which the endpoints of the null photon world line are "not infinitesimally close together", even though the spacetime interval separating them is zero (e.g. we could certainly imagine a physical field whose value varies significantly over the null trajectory). Is there a physical, coordinate- and Lorentz-invariant way to define the open sets of the spacetime?

(Note that I'm not talking about the global/algebraic topology of the spacetime, which is a completely separate issue.)

## 4 Answers

There's no need to define the topology of the manifold from the metric. While a nice feature, the topology of the manifold is defined primarily by its atlas, which, from a physical perspective, correspond to the coordinates. A spacetime with a set of coordinates $\{ x^i \}$ will have a topology defined by the mapping of open sets from $\mathbb{R}^n$ to the manifold via the chart $\phi$.

If you wish, though, there are some things in general relativity that do define the spacetime topology.

A common basis of the spacetime topology is the Alexandrov topology. If your spacetime is strongly causal, the Alexandrov topology is equivalent to the manifold topology. Its basis is defined by the set of causal diamonds :

$$\{ C | \forall p, q \in M, C = I^+(p) \cap I^-(q) \}$$

It's easy to find counterexamples (the Alexandrov topology is just $\varnothing$ and $M$ for the Gödel spacetime), but if it is strongly causal, that will give you back the manifold topology.

There are lots of different possible ways of defining a manifold, some of which are not quite equivalent but all of which are equivalent for physics purposes. E.g., you can define a manifold in terms of a triangulation.

You could just start with the manifold, say defined using a triangulation. Then it has a definite topology, and only after that do you need to worry about putting a metric on it.

If you use the definition of a manifold in terms of a chart with smooth transition maps, then you get a topology for free from the charts. I think this is essentially what enumaris is saying.

But we should also be able to talk about these things in a coordinate-independent way. A metric can just exist on a manifold, regardless of whether the manifold was ever defined in terms of any coordinate charts. Then I think you still get a topology induced by the metric. This is because the metric defines geodesics, and it also defines affine parameters along those geodesics. So in your example of sending a photon to the Andromeda galaxy, the photon travels along a geodesic, we can define an affine parameter, and we can tell that the emission and reception of the photon do not lie in an arbitrarily small neighborhood of one another, because they lie at a finite affine distance.

• Yes, I was thinking the same thing about using the affine parameter as a "distance" measure along null geodesics, but I couldn't find any references to topologies induced by pseudo-Riemannian "metrics". Do you know of any? – tparker May 2 '18 at 19:06

I don't know about "physically" what defines open sets since open sets are a (afaik) purely mathematical construction, but what defines the open sets on the spacetime manifold is simply the open sets in $\mathbb{R}^4$. Open sets in $\mathbb{R}^4$ gets mapped to open sets in the manifold by definition. The topology of manifolds is induced naturally this way.

• So for Minkowski space, the topology is generated by the balls $({\bf x} - {\bf x}_0)^2 + (t - t_0)^2 < r$, even though these sets aren't Lorentz invariant, since that's the Euclidean topology of the domain of the coordinate chart? – tparker May 2 '18 at 18:09
• Yes, by definition, the induced topology on an manifold comes from the underlying Euclidean space to which the manifold locally maps. Maybe another way to put it is the maps which define the Atlas must be homeomorphisms. – enumaris May 2 '18 at 18:24
• Perhaps this thread can shed some more light on your question: mathoverflow.net/q/266903 It appears that if we restrict ourselves to strongly causal spacetimes then the topology induced by the metric will be equal to the topology induced by the charts. Good question though, I had not come across this subtle detail before. – enumaris May 2 '18 at 19:51

The indefinite metric of a pseudo-Riemannian manifold does prevent it from being a metric space, and hence use this route to define a topological space.

However, we can still relax the axioms of a metric space and still be able to define a topological space. In this case, we have the definition of a pseudo-metric and then the construction of the topology goes through as with the usual case.

Mathematically, a more important aporia (a missing but important property) is that manifolds do not have the exponential property:

If $M$ and $N$ are manifolds. Then $M+N$ and $MN$ are manifolds (the former is the disjoint union and the latter the Cartesian product). However, whilst the exponential $M^N$ exists both at the point set and topological level, it does not as manifold. There are many attempts to get around this, but a method that seems to be finding increasing favour and which was first put forward by Souriou and later named diffeology uses techniques inspired by sheaf theory.

• Could you clarify what you mean by "the construction of the topology goes through as with the usual case"? In the usual case, the topology is generated by the balls $|{\bf x} - {\bf x}_0| < r$. In the Lorentzian case, these "balls" always include the entire future and past light cones and their interiors. Surely open sets are allowed to be compact in the time direction. – tparker May 2 '18 at 23:38
• @tparker: well, the same definition works; it's simplest to show that the pseudo-metric defines a topology as in this article (see under the sub-heading topology), and then show that a Lorentz metric is actually a pseudo-metric. Open sets are never compact. – Mozibur Ullah May 2 '18 at 23:56
• I can't find any sub-heading "topology" in that article. – tparker May 2 '18 at 23:58
• @tparker: Sorry, I linked to the wrong article; it's this one, on pseudo-metrics. – Mozibur Ullah May 3 '18 at 0:10
• The metric tensor does not actually induce a pseudometric structure on a pseudo-Riemannian manifold, because a pseudometric is required to be nonnegative and the spacetime interval on a pseudo-Riemannian manifold can be negative. (It's unfortunate that same prefix "pseudo-" is used in incompatible ways in "pseudo-Riemannian manifold" and in "pseudometric".) I take your point that we can still use the invariant interval to define a topology on an arbitrary pseudo-Riemannian manifold, but it seems so much vastly coarser than the usual topology as to be pretty useless for any applications. – tparker May 3 '18 at 0:18