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In physics, we often say that spacetime is a collection (set) of all events (idealized occurrences of zero extension in space-time, the "here and now"s). Moreover, spacetime is said to be a continuum. By continuum, at least in the Euclidean case $(\mathbb{R}^3)$, a naive intuition is given which reads: if there are two points, no matter how close to each other they are there will always be more points between them. Somehow, this intuition is carried over to the case of spacetime as well, and I do not entirely understand how.

I would like to know what the most rigorous definition and minimalist construction of the spacetime continuum is. From some expositions on differential geometry introduced in general relativity courses, I naively guess the following:

  1. The notion of continuity is studied in topology. Thus one models the spacetime as a $4$-dimensional topological manifold (locally isomorphic to $\mathbb{R}^4$).
  2. In the previous intuitive definition ("if there are two points ... more points between them") we considered at least two points (events). Therefore we must be able to distinguish two points on the given manifold. As far as I understand, to achieve this one would require some separability axiom. As physicists like their spacetime well behaved, generally, it is assumed that spacetime manifold has Hausdorff property.
  3. Now we must understand the closeness of the points. To me, it sounds like we need a metric space to have a notion of distance. So we must consider a metric on the manifold. From this point, I do not understand how to go about this. Because spacetime comes with a metric of Lorentzian signature (signature $2$, pseudo-Riemannian geometry). That is, a manifold with metric $(\mathscr{M},\mathbf{g})$ is locally isomorphic to $(\mathbb{R}^4,\eta)$, where the Lorentzian metric $\eta$ is ${\rm diag}(-1,1,1,1)$. Therefore, my intuition regarding standard metric topology on $\mathbb{R}^4$, using which one could have defined open $\epsilon$-Balls, breaks down. For the distances defined with $\eta$ metric is not positive definite.

Here is the question(s):

How does one mathematically define the notion of a spacetime continuum? Is such a definition possible without the metric and only at the level of some primitive topological constructs or do we need a metric to define a continuum? If we do need a metric then how do we deal with a non-positive-definite metric as one encounters in pseudo-Reimannian geometry?

To summarize

What are the necessary and sufficient mathematical notions to construct a spacetime "continuum"?

The definition of spacetime given by Hawking and Ellis (one of the most mathematically rigorous books on the subject) may be helpful in this context:

The mathematical model we shall use for space-time, i.e. the collection of all events, is a pair $(\mathscr{M}, \mathbf{g})$ where $\mathscr{M}$ is a connected four-dimensional Hausdorff $C^\infty$ manifold and $\mathbf{g}$ is a Lorentz metric (i.e. a metric of signature + 2) on $\mathscr{M}$.

(P.S.: I am a physics student and have very little experience with abstract mathematics. Brief physical/intuitive explanations of the mathematical concepts used in the answer would be most helpful and much appreciated!)

VacuuM
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    "if there are two points, no matter how close to each other they are there will always be more points between them." This isn't a great description of $\mathbb{R}^n,\ $ for example $\mathbb{R}$ or $\mathbb{R}^3,\ $ because the description also describes $\mathbb{Q}^n,$ for example $\mathbb{Q}$ or $\mathbb{Q}^3.\ $ A better description of $\mathbb{R}\ $ is that it is, up to isomorphism, the unique complete ordered field. And then $\mathbb{R}^3 = \mathbb{R} \times \mathbb{R} \times \mathbb{R}. $ – Adam Rubinson Jun 16 '23 at 07:43
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    See What is the "full" topological requirements for a (classical) spacetime? to get you started. Now, your question has something extra stuff not discussed there, but just get acquainted with that first. – peek-a-boo Jun 16 '23 at 07:48

3 Answers3

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First, let me remark that there is no such thing as a “necessary and sufficient condition” for modelling spacetime. It’s a model, so it can’t be an exact description (and Physics doesn’t answer the question of “what exactly is happening” either).

The intuitive notion of continuum is not just something like “between any two points there are more points”, because, as mentioned in the comments, this would allow the rationals $\Bbb{Q}$ or $\Bbb{Q}^n$. What we want is a form of completeness, i.e “no holes”. So, we really should work with $\Bbb{R}^n$ as our model.

Now, we want to generalize so we consider topological manifolds $M$. With this, we still retain locally being homeomorphic to $\Bbb{R}^n$; this is good because as a classical observer, space/time around you has “no holes”, “no start/end” etc, i.e things look nice and as expected. Next, within the definition of a topological manifold, are the Hausdorff condition, and second-countability.

  • For Hausdorffness, one of the nice consequences is that limits are unique; without this property all our undergraduate analysis thoery can be thrown out (or at the very least, being very generous, requires serious modifications).
  • Second-countability is a little harder to motivate, but it’s a condition that roughly says “the space isn’t too big, while taking the topology into account”. It has some very nice technical consequences, when coupled with the other properties, such as the existence of partitions of unity. This allows one (atleast when talking about smooth manifolds) to define integrals, and obviously integration is a very important subject.
  • Some authors require manifolds to be connected, while others don’t. For physical reasons, we should require $M$ connectedness because Physics deals ultimately with experiments (hence measurements), which means we have to interact with our surroundings, and there is no way to interact (connect, via a path for example) with a different component. Having said this, sometimes, it’s nice for mathematical convenience to allow a disconnected manifold.

So, our notions of space and time as a continuum are captured, preliminarily, by the notion of a topological manifold. What I mean by this is that topological manifolds capture the idea of “having a nice blob of stuff no matter how close you look”. At the level of a topological manifold, we have not made any statements about the Physics of space and time, we have simply said what the “background arena” ought to look like.

Next, we come to the idea that a whole bunch of Physics is concerned with understanding changes in things; changes of things in “position”, changes in “time”, so we now upgrade our hypothesis to having a smooth manifold $M$. With a smooth manifold, we can now elevate all our familiar differential and integral calculus from $\Bbb{R}^n$ to the manifold.

Ok, so thus far we have agreed that in order to mathematically model a Physical theory of space and time, one should work with a smooth manifold $M$. Now, we come to the Physical input. Of course the VIP is the specification of a Lorentzian metric $g$; the $(-,+,\cdots,+)$ signature encodes our understanding of time and space (along with the assumption $\dim M=4$ of course, but we are free to study other dimensions:). Also, you’re right that in Lorentzian geometry, unlike Riemannian geometry, you cannot talk about distances between points, but you should not think of this as saying that we have no way of talking about “closeness” of points. One of the things a topology gives us is a notion of “closeness”, because otherwise, one could not even talk about limits.

Next, I’ll mention that the fact that we use a Lorentzian metric, and hence are unable to define distances generally, is very much on purpose! We do not want to be able to make absolute statements like “Alice and Bob are 10 meters apart” or “everyone experiences time in the same way”; this was one of the insights gained in SR alone, not even GR. Everyone is unique, meaning that it is only for a given smooth curve that we can talk about the “length” of that curve, using the formula $L(\gamma):=\int_a^b\sqrt{|g(\dot{\gamma}(\lambda),\dot{\gamma}(\lambda))|}\,d\lambda$. Of course I put the word length in quotation marks because we interpret this as (proper) time for a timelike curve. So, if you now have two different points in spacetime, you can connect try to connect them by a timelike curve, and then you can meaningfully make a statement like “in order to travel from point $A$ to point $B$ along this specific trajectory, ___ amount of (proper) time will elapse”. This number we speak of depends on the path used to join the two points, so if you take a different path between two points, you can expect a different answer (once we recognize this, the twin phenomenon, and so many other ‘paradoxes’ are so much easier to digest).

Ok, so we now have a smooth Lorentzian manifold $(M,g)$, and for physical reasons, $\dim M=4$ and $M$ be connected. Because of topology, we have a notion of “closeness”. I should mention though, that even though we can’t quantify “closeness” using the Lorentzian metric $g$, the manifold is still metrizable, so I can impose some distance metric $d$. The only drawback (and this is a major one) is that it has no direct relationship with the Lorentzian metric $g$ whatsoever, so for all intents and purposes, it is useless. Also, for physical reasons, one also wants a time-orientation, so that we know what “future” is.

Finally, let me make the remark that if in GR, unlike in geometry, we are not given the manifold $M$ a-priori, nor are we given the Lorentzian metric $g$. What we have to do is consider maximal globally hyperbolic developments of initial data sets. This is where the dynamical nature of GR appears (read Hawking and Ellis for more), meaning you “build your manifold and the Lorentzian metric” by “solving” (this term requires a precise definition) Einstein’s equations.

peek-a-boo
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  • +1 Very nice answer! However, where I am having a problem is in understanding statements like: "topological manifolds capture the idea of “having a nice blob of stuff no matter how close you look”", "Because of topology, we have a notion of “closeness”" etc. probably because I don't yet have a good intuition about topology. I thought in std. topology on $\mathbb{R}^n$ one creates $\epsilon$-ball neighborhoods to talk about closeness. But surely there must be other notions of closeness. Can you please elaborate on how in topology we understand closeness without a metric? – VacuuM Jun 17 '23 at 01:01
  • @VacuuM “having a nice blob of stuff” is my attempt at saying in a hopefully descriptive manner “a topological manifold is locally isomorphic to $\Bbb{R}^n$”. Next, topology gives us the concept of an open set/open neighborhood. As the very (appropriately chosen) word “neighborhood” suggests, it is talking in some sense of closeness. If you’re not convinced, look at the definition of “$f$ is continuous at $a$”: “for every open neighborhood $U$ of $f(a)$, there is an open neighborhood $V$ of $a$ such that $f(V)\subset U$”. The notion of open sets (i.e topology) tells us what closeness is. – peek-a-boo Jun 17 '23 at 01:14
  • The fact that the standard topology of $\Bbb{R}^n$ is most intuitively described (not just intuitively, but it’s also what we’ve been taught all our lives… we grew up learning Euclidean geometry) using a distance function, does not negate the fact that topology is meant to generalize our notions of closeness. Topological spaces generalizes a metric space. So, if you don’t find it obvious/intuitive, then good, it means you’re thinking :) Continue doing so. This is also partly an experience issue: the more you study topology, the more intuitive it becomes (as with any subject). – peek-a-boo Jun 17 '23 at 01:18
  • I mean there’s a reason topology proper gained its proper definitions only in the early parts of the last century, whereas for over 2000 years, we’ve been happily working with the Euclidean picture. It’s a new concept, or at the very least, difficult to distill into its bare bones essentials, hence it took so much time. So it’s perfectly normal to have doubts about it – peek-a-boo Jun 17 '23 at 01:23
  • Thanks for the comments. Is the following way of thinking correct?: Say there's a topological manifold $(M,\mathcal{O})$, with $\mathcal{O}$ being some topology. Then a chart on $M$, which is open by definition, maps to an open set in $\mathbb{R}^n$ irrespective of what $\mathcal{O}$ has been chosen. But now for an open set in $\mathbb{R}^n$ we can have a notion of closeness either by $\epsilon$-balls or otherwise. Thus, because of the bijection between the chart on $M$ and open set $\mathbb{R}^n$, we can transfer the notion of closeness onto $M$ from $\mathbb{R}^n$. – VacuuM Jun 17 '23 at 01:33
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    @VacuuM no it’s more like you already have a notion of closeness on $M$ because you have a topology $\mathcal{O}$. But the fact that the topology is locally-cartesian, means given your chart $(U,x)$, it maps $U\subset M$ homeomorphically to the open set $x[U]$ in $\Bbb{R}^n$, meaning that the notion of closeness at the manifold level (a notion which exists purely by virtue of having a topology) is “essentially the same” as the notion of closeness that we have in $\Bbb{R}^n$, i.e it’s like a “compatibility of the notion of closeness”. – peek-a-boo Jun 17 '23 at 01:40
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I think it is nice to think about the different notions of manifold.

  • A topological manifold is an object that locally resembles $\mathbb R^n$ in terms of topology (that is, in terms of information that you can derive from knowing who are the open sets; basically, the topology tells you the “shape” of the space, or, “how points of the space are glued together”). For instance, you can tell whether a function defined on the manifold is continuous or not.
  • A differentiable manifold is a topological manifold with, in addition, a smooth structure, which basically gives you a way to define derivatives of functions whose domain or codomain is the space itself. You can define the tangent space to a point of the manifold; you can tell whether a function defined on the manifold is smooth or not; you can uniquely solve first order ODEs… but you do not have a definition of distance between points of the manifold. This notion, as the previous one, makes no distinction between “space only” and “spacetime” objects: for that you need the next step.
  • A Riemannian manifold is a differentiable manifold with, in addition, a metric. This allows you to define the distance between two points, unlike with the previous two cases. You can actually do much more, for instance construct geodesics, … Of course, in the case of physical spacetime, one uses the notion of Pseudoriemannian manifold instead.

So, if you want to define a distance, you need the third structure. But if you only care about your spacetime being “a continuum”, I would say that the first notion is enough, that is, you simply need to ensure that your space is locally homeomorphic to $\mathbb R^4$ (that is, isomorphic in the topological sense). Morally, you can simply assume that your spacetime “locally looks like $\mathbb R^4$”, even if in this case your manifold does not necessarily have a canonical notion of distance between points. (Topology is way less information that distance, but still allows you to say a lot about the space).

You also want to assume that your spacetime is Hausdorff (which is not always included in the definition of topological manifold, it depends on the author). So essentially, the first two points in your list are enough to define a spacetime continuum, and you don’t need a metric (this is the cool fact of topological manifolds and also of differentiable manifolds).

I is worth pointing out that the fact that spacetime is a continuum is something that does not really make it different from a “space only” object: you can have one Riemannian manifold and one pseudo-Riemannian manifold that are isomorphic to each other as topological manifolds (e.g., the standard Minkowski spacetime $\mathbb R\times \mathbb R^3$ (equipped with the usual pseudometric with signature $(-,+,+,+)$) and $\mathbb R^4$ (equipped with the Eucledian metric) have the exact same topology and are isomorphic as topological manifolds, even though they don’t have the same metric structure). The fact of being “a coutinuum” does not really care about the signature of the metric: the only thing that characterizes the local properties of a topological manifold is the number of dimentions (4 in the case of a physical spacetime).

Just one (very mathematical) side note: in the case of a Riemannian manifold, the Riemannian metric is enough to univocally determine the topology, because as you noticed you have a well behaved distance. For the pseudo-riemannian metric this fact is less obvious, but it is still true (under further assumptions), even though the metric is not positive definite. So, the fact that the distance can be zero, is not a problem per se under suitable assumptions. If you don’t see why don’t worry, it is not that important, but just know that even if the metric is not positive definite, there is still some hope to recover the topology from the metric.

Conclusion. If you want to know how you mathematically think about the fact that the space is “a continuum”, I would say this corresponds to saying that your spacetime is a (Hausdorff) topological manifold with dimention 4. Of course, depending on what you want to do, you might want to consider broader or narrower definitions (see the comment and answer of @peek-a-boo). Your spacetime being separable, 2nd countable, with finitely (or countably) many connected components… are all desirable properties that you want to assume most of the times (a manifold which is not 2nd countable is extremely degenerate; see the long line if you are curious), but they are all “global properties”, i.e., they talk about the global structure of your spacetime (e.g., given two points A and B, is it always possible to connect them with a curve which lies on the manifold?). On the other hand, “being a continuum” is (I guess…) a local property, and is essentially captured by the notion of Hausdorff topological manifold. You can ask yourself whether you care or not about global properties of the spacetime when you think about the idea of “continuum” to decide whether or not you want to assume those additional properties.

Lorenzo Pompili
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With respect to space-time: QM says, it does not exist, since it has no measurable attributes.

Classical mechanics of many particle systems says, that momentum space-energy-time is intimately represented by local Maxwell fields, that are differential forms cotangent to an idealized space-time. Pointlike fields don't exist.

While the cotangent fields interplay nicely with the phase space of position and momentum of charged massive point particles, mankind, by its optical observation equipment, takes the space with its topological and metric properties, bought from momentum space, as the primary physical arena. But thats an illusion, as Platon already mentions.

During the last three decades, even the enegineers branch reacted: Astronomical time is replaced by the spin-flip frequency in atoms, metric distance in space is replaced by the quantum Hall resistance.

For subatomar "distances" all ideas of a continuum of space-time cease to produce any substantial output besides the classical philosophical discussions in endless word associations.

The idea of global metric space time manifold of the universe lacks direct evidence simply by the short time of 50 years of observation. All far-away measurements are again model dependent reconstructions from spectral data of radio- and optical data. Reconstruction of the closed big bang model of space time is an area of popular physics text since Hawkings book, but is by no means something "real".

Roland F
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