In all the references/textbooks that I have looked at, the precise definition of spacetime is never really clear. By gathering the hypothesis that we need to make, I get the following definition: $$\text{spacetime is a smooth connected and orientable Lorentzian manifold with or without boundary}.$$ But it seems that we never really specify the set on which we put the manifold structure. Is this okay? For example, let's take the Schwarzschild metric, on which manifold is this metric defined? $\mathbb{R}^4$ with the usual/trivial smooth structure?


I'm familiar with the basics notions of differential and Riemannian geometry, my question is, perhaps better formulated: what is the set $M$ on wich we define a smooth atlas $A$ to make the smooth manifold $(M,A)$?

  • 3
    $\begingroup$ What is the definition in the most rigurous 3 texts on GR? SACHS & WU, WALD and Hawking & Ellis. Look them up, please. $\endgroup$
    – DanielC
    Oct 21, 2020 at 14:49
  • $\begingroup$ What kind of properties are you interested in? $\endgroup$ Oct 21, 2020 at 14:52
  • $\begingroup$ I'm interested in knowing on wich sets we put the smooth structure. For Schwarzschild for example. $\endgroup$
    – xpsf
    Oct 21, 2020 at 14:56
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    $\begingroup$ It can be any set that admits a four-dimensional manifold structure (which is just any set with the cardinality of the contiuum). The definition doesn't ask for a specific one. $\endgroup$
    – Javier
    Oct 21, 2020 at 15:49
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    $\begingroup$ @DanielC It is not generally helpful to tell people to look in textbooks they don't necessarily have (or want to get - these things do cost money). $\endgroup$ Oct 21, 2020 at 22:40

4 Answers 4


A manifold is a set - you don't need to put the manifold structure onto anything. Take a look at the first line of the wikipedia page for a manifold: a manifold is defined as a topological space which satisfies certain properties (and a topological space is a set of points).

Intuitively: a manifold is a set which looks flat if you zoom in close enough on any of its points. This is where your notion of $\mathbb{R}^4$ comes in - since any spacetime in GR is a manifold, this means it looks like flat Minkowski space ($\mathbb{R}^4$) if you zoom in close enough on any of its points.

If you zoom out, the space may be curved and not resemble $\mathbb{R}^4$ at all (as is the case for Schwarzschild). The thing you use $\mathbb{R}^4$ for when describing Schwarzschild space is to describe points on the manifold with coordinates (see coordinate chart on wikipedia).

Coordinate charts always can map/describe a region $U \subseteq \mathcal{M}$ of a manifold $\mathcal{M}$, but sometimes they fail to describe the entire manifold (ie. sometimes $U \neq \mathcal{M}$ for a particular coordinate chart). Or the coordinates might have singularities at certain points on the manifold as well (as is the case in Schwarzschild space: ordinary Schwarzschild coordinates famously break down at the horizon).

  • $\begingroup$ I'm familiar with Differential Geometry and the basics of Riemaniann Geometry but what I don't understand is, perhaps better formulated, "What is the set $M$ on wich we define a smooth atlas ?". $\endgroup$
    – xpsf
    Oct 21, 2020 at 15:04
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    $\begingroup$ I'm wondering if you're expecting Schwarzschild space to be embedded in some other space. To illustrate what I mean, consider the 2-sphere $\mathcal{S}^2 := \{ \mathbf{x} \in \mathbb{R}^3 : |\mathbf{x}| = 1 \}$ which is a manifold. Furthermore, it is an embedded manifold, or submanifold of $\mathbb{R}^3$, which essentially means that it is defined as a surface living inside of some larger space (in this case $\mathbb{R}^3$). $\endgroup$ Oct 21, 2020 at 15:11
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    $\begingroup$ Continued: Note that a manifold generally does not need to be embedded in some larger space / a submanifold of some other manifold. You can have manifolds that just exist in their own right (and are not simple things like $\mathbb{R}^n$). $\endgroup$ Oct 21, 2020 at 15:13
  • $\begingroup$ In GR, it's almost always the case that the manifolds are not understood as embeddings/submanifolds in some larger spaces, and are just manifolds in their own right. This is true for Schwarzschild space as far as I know. (Note however: there are some spacetimes which can be abstractly defined in some higher dimensional space as a way to understand the geometry - for example, De Sitter space can be understood as a surface in a 5-dimensional Minkowski space, but as far as I know that is a special case [also, the 5D Minkowski space there should not be taken as anything physically meaningful]) $\endgroup$ Oct 21, 2020 at 15:15
  • $\begingroup$ You might be right but I don't think that I suppose that. From what I understand, to define a smooth manifold $(M,A)$, we need a set $M$ (without any specific structure) and a smooth atlas $A$ on $M$ such that the induced topology satisfies some properties. The coordinate maps are of the form $\phi:U\subset M\to\phi(U)\subset\mathbb{R}^n$. So, for me, we need to define $M$ in order to define $\phi$. $\endgroup$
    – xpsf
    Oct 21, 2020 at 15:20

The set is not predetermined but arises from physical/mathematical requirements of the given solution.

GR is local theory and sufficinetly small region of spacetime is assumed to be isomorphic to open region of $\mathbb{R}^4.$ Globally the set is given by "gluing" these regions together until you arive at global solution you are satisfied with. GR does not strictly speaking enforce this. However, it is reasonable to demand some properties like smoothness of the metric, maximal extension and so on. In the case of Schwarzschild spacetime, these requirements are strong enough to guarantee uniqueness.


Mathematical conditions such as Hausdorf or paracompactness apply to mathematical models of reality. They are introduced to prove theorems that apply to these models. Do not confuse mathematical models of reality with reality itself. Whether the universe is everywhere Hausdorf or paracompact is something to be decided by experiment. No amount of studying the continuum hypothesis or alternative axiom systems for the real line can tell us anything about the space in which we live.


General Relativity is a mathematical framework within which we can construct Lorentzian manifold models of reality. In general, structures (e.g. the spacetime manifold) of a given model taken to represent some aspect of observable reality need not be physically real aspects of nature, whatever that might mean. Indeed, they're almost certainly not-- what are the odds that the formalism we decided was most natural to us perfectly captures the nature of reality? In practice, all we can ask of a model is that it provides a means of unequivocally (within discernible error) predicting some observations. We like (accurate) models that are broad in scope, offering a means of predicting many different types of observations, and that sit well philosophically within the larger network of other successful models, but even these cannot be presumed to present "actual" reality.

All that to say: the only crucially significant feature of a model is the collection of predictions it makes. Details of a model's structure that don't impact its predictions are ultimately of little import, especially if they also don't impact the model's philosophical interpretation. The particular set of cardinality $2^{\aleph_0}$ one chooses to think of as underlying the Lorentzian manifold of a GR model is probably one of the least impactful (both observationally and philosophically) features I can imagine, and for this reason, GR makes no such choice outright-- any one at all will do.

A comment worth making is that the set in question is essentially universally given an interpretation as the set of spacetime events, pairs of "space" and "time" instances that characterize where and when something can occur, but this is only a heuristic intuition, not a rigorous definition that nails down the set theoretic object under consideration. Since all candidate sets are bijective, such an interpretation on one of them induces the same on all others.


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