Is the topology of Minkowski space the same as that of $\mathbb{R}^4$? My thoughts would be no, because of the very different inner products define very different metrics, and because the metric determines the open balls, it determines the topology.

  • 10
    $\begingroup$ In the theory of manifolds, the topology of your space is set before you put a metric tensor on the space. So the topology of Minkowski space is the usual topology on $\mathbb{R}^4$. $\endgroup$
    – gj255
    Apr 13, 2017 at 15:41
  • 1
    $\begingroup$ See also: math.stackexchange.com/q/1607335 and physics.stackexchange.com/q/228669 $\endgroup$
    – gj255
    Apr 13, 2017 at 15:47
  • $\begingroup$ @gj255 You should make those comments an answer. $\endgroup$
    – ACuriousMind
    Apr 13, 2017 at 17:18
  • 5
    $\begingroup$ To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. $\endgroup$
    – balu
    Apr 13, 2017 at 18:24
  • $\begingroup$ Now on Meta: physics.meta.stackexchange.com/q/9820 $\endgroup$
    – Kyle Kanos
    Apr 20, 2017 at 10:12

3 Answers 3


The topology is $M^{1,3}~=~\mathbb R\times\mathbb R^3$ which is the product of the number line for time and the three space. The Lorentz group is a system of transformations between $\mathbb R$ and any direction within $\mathbb R^3$. Since light cones or null rays are invariant these transformations can't change the topology of the spacetime. Hence the topology of $M^{1,3}$ is not changed by Lorentz transformations.

This carries over to general relativity as well. The diffeomorphsims of spacetime are such that the topology of the spacetime is not changed. This has a parallel with quantum mechanics. If you could change the topology of the spatial part $\mathbb R^3$ so it is multiply connected then a Lorentz boost can convert this into a time loop. This means one could clone quantum states. This suggests that $no~topology~change~=~no~cloning$. That general relativity prevents topology change seems to prevent an attack of the quantum clones.

  • 2
    $\begingroup$ Isn't the standard topology of $\mathbb{R} \times \mathbb{R}^3$ the same as that of $\mathbb{R}^4$? $\endgroup$ Apr 13, 2017 at 16:16
  • 4
    $\begingroup$ You're not really answering the question, that is, you're just stating what the topology is and oddly phrasing it as $\mathbb{R}\times\mathbb{R}^3$ instead of the equivalent $\mathbb{R}^4$ the question mentions, but you're not explaining why. A better explanation is given by gj255 in the comments and its linked questions. $\endgroup$
    – ACuriousMind
    Apr 13, 2017 at 17:19
  • 3
    $\begingroup$ Everyone knows that $\mathbb R^4~=~\mathbb R\times\mathbb R^3$. I just wrote it this way to illustrate a bit the argument the Lorentz group does not change topology. $\endgroup$ Apr 13, 2017 at 21:23

The topologies coincide since Minkowski spacetime is strongly causal.


gj255, ACuriousMind, Balu, Lawrence B. Crowell, Jackson Burzynski

$\mathbb{R}^4$ is not the metric of space-time. It is just a space we begin with to indicate a coordinate system, in terms of which the metric or pseudo-metric may be described. The true topology (or pseudo-topology!) of space-time is to be found from the metric, or more precisely, pseudo-metric, defined by the invariant interval. It is not necessary to set up a topology to define a metric. One can use a metric to define a topology.

Similar examples: (1) If we begin with a rectangular strip defined by $0<\phi<2\pi$ and $0<\theta<\pi$, and use the metric $ds^2=d\theta^2+sin^2\theta d\phi^2$, we get the metric for the surface of a sphere, which implies the topology of that surface.

(2) If we begin with a rectangular strip defined by $0<\phi<2\pi$ and $0<\psi<2\pi$, and use the metric $ds^2= d\phi^2+d\psi^2$, we get the metric for the surface of a torus, which implies the topology of that surface.

Use of variables conventionally employed for angles is a way of implying that they are assumed to be periodic with period $2\pi$. For the sphere, values of $\theta>\pi$ will result in a second (alternative) designation of points on the sphere, so they are usually omitted.

  • $\begingroup$ I don't see why this got a -2. It looks like an excellent explanation to me. $\endgroup$ May 7, 2017 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.