# 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?

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. – Selene Routley 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.