I assume that for a Lorentzian manifold (i.e. with Minkowski signature), the analog of an open ball is the interior of a light cone. My question is motivated by the observation that whereas any point on the boundary of an open ball on a Rimannian manifold (i.e. with Euclidean signature) can be considered to be simultaneously interior to an infinite number of other open balls (and exterior to an infinite number of others) the boundary of a light cone is associated with a metric interval that is distinct from the timelike and spacelike intervals. For that reason, I wonder whether this introduces additional subtleties/restrictions on constructing a spacetime topology. Related to this question is under what circumstances (if any) can individual points be associated with certain kinds of intervals (e.g. Spacelike, timelike, null).

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    $\begingroup$ The subtleties of determining a topology of the Minkowski-signature manifolds are different from those of the Euclidean-signature manifolds - in some sense, more subtle. Replacing balls by light cones will surely not do the job. One surely has to learn things like the Penrose causal diagrams, geodesic completeness, surprising coordinate singularities that are nevertheless OK in other coordinate systems, and so on. It's a whole subject and it's not like saying three sentences to completely exhaust it. $\endgroup$ Commented Jun 6, 2013 at 9:36
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    $\begingroup$ Topology non-metrical. Light cones are metrical. A manifold has a topology even when it isn't equipped with a metric at all. Since topology is more basic than the metrical stuff, it doesn't really make sense to redo the topology based on the metric. $\endgroup$
    – user4552
    Commented Jun 6, 2013 at 13:14

1 Answer 1


As Luboš says, there is no point in trying to define the topology (I assume that you are here trying to construct the topology, in its most rigrorous sense, from a base -- the set of open balls) via lightcones. The reason is that for a reasonable topology you should have arbitrary small(in some intuitive sense) open sets, so that the statements like "there exists an open set such that.." indeed mean what we want (i.e. "there exists a small enough open ball such that"). The lightcones are not in any sense small, and your topology will not see the continuity we are used to.

Well, one may say, forget about intuition, lets see where these open cones will lead us.

  • We cannot say that the interior of full light cones, defined as all space-time points in a timelike separation from a given point, form a base of the topology. The reason is that an intersection of two open lightcones can not be represented as a union of an infinite ammount of open light cones.
  • We can, however, require not just a timelike separation, but a timelike separation greater than something. Or allow only positive half-lightcones. This will work as a base of a topology (at least the second option), but in the resulting topology any open set will contain together with a point the whole positive lightcone stemming from that point. This topology is not a one you are used to. It is not Hausdorff, it 'glues together' casually-connected pairs of points.
  • Another option is to claim that the lightcones form a prebase of the topology. Then the intersection of two lightcones one atop the other gives you something very similar to a unit ball. I guess that in this case you will get the topology of the underlying manifold. Em, yes, you have to have an underlying manifold in order to define a Lorenzian metric and the lightcones thenselves..

Now, I think that another question should be asked in this context: while a paracompact manifold always admits a positive-definite Riemannian metric, there is no such theorem for pseudo-Riemannian metrics. The reason is that you have to glue timelike curves from different coordinate charts in a sensible manner. What are the conditions for a manifold to admit a pseudo-Riemannian metric? I have found some usefull info here, but I do not have a complete answer.

  • $\begingroup$ That last paragraph would make for a good question itself, either here or on math.SE. $\endgroup$
    – user10851
    Commented Jun 6, 2013 at 19:54

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