# Topology of Minkowski Space

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.

• 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 $\mathbb{R}^4$. Apr 13 '17 at 15:41
• Apr 13 '17 at 15:47
• @gj255 You should make those comments an answer. Apr 13 '17 at 17:18
• 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.
– balu
Apr 13 '17 at 18:24
• Now on Meta: physics.meta.stackexchange.com/q/9820 Apr 20 '17 at 10:12

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.

• Isn't the standard topology of $\mathbb{R} \times \mathbb{R}^3$ the same as that of $\mathbb{R}^4$? Apr 13 '17 at 16:16
• 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. Apr 13 '17 at 17:19
• 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. Apr 13 '17 at 21:23

The topologies coincide since Minkowski spacetime is strongly causal.
https://en.wikipedia.org/wiki/Spacetime_topology

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.

• I don't see why this got a -2. It looks like an excellent explanation to me. May 7 '17 at 20:49