Let $(M,g)$ be a time-oriented smooth Lorentzian manifold, with Lorentzian metric $g$. In the following thread:


<https://physics.stackexchange.com/questions/228669/why-pseudo-riemannian-metric-cannot-define-a-topology/228676?noredirect=1#comment730891_228676>


physicist @ValterMoretti makes the following claim:


"As a matter of fact, a (connected) Lorentzian smooth metric $g$ over the time-oriented smooth manifold $M$ does define a topology, the same already present on $M$."


I have not been able to find a proof of the statement above, unless some relatively strong assumptions are made on $(M,g)$ (for example imposing that $(M,g)$ is strongly causal). Can the statement be proved in general without further extra-assumptions? I was under the impression that in fact one of the key points of departure between Riemannian and Lorentzian geometry is the failure for a Lorentzian metric to define in general a topology equivalent to the pre-existing manifold topology, in contrast to what happens in Riemannian geometry.


Thanks.



On the topology induced by a Lorentzian metric
In general, this topology is coarser than the original topology of the manifold, and, without further assumptions, strictly coarser. It coincides with the original one iff the Lorentz manifold is strongly causal, see e.g. Prop 3.11 in Beem-Ehrlich-Easley. To get counter-example consider the cylinder $\mathbb{S}^1 \times \mathbb{R}$ with time direction being $\mathbb{S}^1$, i.e. periodic, and the usual flat metric. Then you can connect any two points by a timelike curve, thus the only non-empty open diamond is the whole spacetime. In this case the induced topology is the in-discrete one.


