# Does the general topology of Minkowski space-time change under a Lorentz transformation?

Does the general topology of Minkowski space-time change under a Lorentz transformation? Open balls in $\mathbb{R}^{4}$ (with the standard topology) are not invariant under Lorentz transformations. Does this mean for example that observers in one reference frame would have different notions of convergence, continuity etc?

-
See a related one: physics.stackexchange.com/q/83596 – Idear Jun 4 '14 at 23:46

No, because a Lorentz transformation is continuous with a continuous inverse. While an open ball is not mapped to itself, it is mapped to some other open set, in an invertible way. (That a Lorentz transformation is continuous of course follows from that it is linear.)

-
Thanks! So I could phrase it this way: Lorentz transformations are homeomorphisms, so even though they open sets not invariant, all topological notions are still preserved? – Ryan Lafferty Jun 10 '14 at 20:27
Yes, that is correct. – Robin Ekman Jun 10 '14 at 21:46
@RobinEkman My answer is not meant to be a criticism of yours, it's just that in my experience the fact that linear is not needfully continuous in an infinite dimensional vector space comes as a bit of a surprise to some people, so I wanted to make sure that the OP understands this for when he looks at QM more deeply. – WetSavannaAnimal aka Rod Vance Jul 1 '14 at 10:20

Homeomorphism is indeed the key concept here, and I wish to add a very slight nitpick with Robin's answer so that there is no risk of your being confused in contexts other than SR/GR (for example, in quantum mechanics): you also need the information that Minkowski spacetime is finite dimensional to infer continuity from linearity and homeomorphism from linear and invertible (which of course is a given in Minkowski spacetime). In infinite dimensions, not all linear maps are continuous: witness the Dirac delta on $\mathcal{L}^2(\mathbb{R})$ for example. The difference between the concepts of general linear and the strictly more specialised "linear continuous" in, say the standard, countably infinite dimensional Hilbert space $\mathcal{L}^2(\mathbb{R})$ is actually the very reason for being of distributions and the framework of rigged Hilbert space for talking about them; see my answer here for more information.