How are the pseudo-Riemannian metric tensor properties restricted by the manifold topology in pseudo-Riemannian manifolds?

Question

My understanding is that a pseudo-Riemannian metric tensor induces a topology that is not compatible with the manifold topology, and obviously the manifold topology prevails if we are to have a manifold like in this case.

But then it is hard not to wonder how a pseudo-Riemannian manifold, generalization of Riemannian manifolds, departs from the latter. I mean they are formally distinguished by the different metric tensor structure (indefinite vs definite positive) on top of the smooth manifold, but if the pseudo-Riemannian metric tensor is restricted by the base manifold topology to the space metric properties of a Riemannian metric I can only see left as difference between Riemannian and pseudo-Riemannian manifolds those referred locally to a manifold point's tangent space. But even there the manifold topology makes some restrictions AFAICS. Timelike or spacelike vectors at a point are allowed but no nonzero null vectors, and therefore no light-cone structure, if we are to go strictly by the $\Bbb R^4$ topology. I have read the wikipedia article on spacetime topology that refers to Zeeman and Hawking topologies, but those are not manifold topologies (they are not locally compact, nor metrizable), the other topologies mentioned there are equivalent to the manifold topology. Am I missing something here or is the above basically correct?

[Edit: The next two paragraphs centered on Minkowski space are easily answered just by considering it an affine space rather than a smooth manifold]

To be more specific when Zeeman writes in his paper of 1967 in Topology Vol. 6, 161-170 'The topology of Minkowski space': "LET M denote Minkowski space, the real 4-dimensional space-time continuum of special relativity. It is customary to think of M as having the topology of real 4-dimensional Euclidean space, although there are reasons why this is wrong. In particular:The 4-dimensional Euclidean topology is locally homogeneous, whereas M is not; every point has associated with it a light cone separating space vectors from time vectors."

Does this mean that the Euclidean topology, wich happens to be the same as the manifold topology, is incompatible with light cones, and if so how is Minkowski 4 dimensional manifold different from Euclidean 4-space, other than for the curvature invariant of the immersed hyperboloid preserved by the local isometry that gives rise to the indefinite form in 4 dimensions to begin with?

Answer 1

It seems hard to me to give an "answer" to this quesiton, but let me try to share some thoughts. On a Riemannian manifold, the Riemannian metric can be used to define a distance function (sometimes also referred to as a metric) which in turn defines a topology on the manifold. As you note, it is a nice feature, that this topology coincides with the manifold topology. But this also implies that the topology is completely independent of the Riemannian metric in question. For example, you can put a Riemannian metric on $\mathbb R^4$ which does not admit a single isometry. But still, this will induce the usual topology (which of course is locally homogeneous).

In the pseudo-Riemannian case, this goes wrong in several respects. If you try to define a notion of "distance" parallel to the Riemannian case, you can get positive and negative distances and (even worse) different points may have distance zero (if they lie on a lightlike geodesic). This certainly does not fit into the setting of metric spaces, and while one could try to use it to define a topology, this topology would certainly be very badly behaved. (Any point which can be reached from $x$ by a lightlike geodesic would be considered as being arbitrarily close to $x$.)

The basic point in my opinion is that in mathematics you usually look at the same object from different perspectives. This is usually expressed by looking at different classes of "(iso-)morphisms". Any Riemannian or pseudo-Riemannain manifold has an underlying smooth manifold, which in turns has an underlying topological space (and if you want to push things further you can look at the underlying measure spaces or even the underlying set). The difference between these pictures is whether you look at isometries or at diffeomorphisms (respectively smooth maps), homeomorphisms (respectively continous maps) or measurable maps and isomorphisms. Now any smooth manifold is homogeneous under its group of diffeomorphisms (and much more than that is true), whereas (pseudo-)Riemannian manifolds which are homogeneous under their isometry group are rather rare. This of course implies that the topologies of smooth manifolds are always homogeneous. Even worse, any two compact manifolds (regardless of their dimension) are isomorphic as measure spaces.

So the "answer" from my point of view would be that Minkowski space is the smooth manifold $\mathbb R^4$ endowed with a flat Lorentzian metric. There are many properties of this metric, which are by no means reflected in the topology of $\mathbb R^4$, but this is also true for Riemannian metrics on $\mathbb R^4$. There certainly are some structures resembling topologies, which can be used to encode interesting properties of Minkowski space (I don't know the Zeeman and Hawking topologies you refer to, but I would guess that they are such structures, and causal structures are another example). But I don't think that they should be used as a replacement for the vector space topology on $\mathbb R^4$ ...

Let me finally remark that Minkowski space is homogeneous as a pseudo-Riemannian manifold (since translations are isometries). They are not homogeneous on an infinitesimal level, since there are different directions emanating from a point. (This is another bit of confusion. Minkowski space should not be considered as a vector space endowed with an inner product. Otherwise there would be a distinguished point - the origin. Mathematically speaking it is an affine space togehter with the inner product on each tangent space.)

Answer 2

There is a basic problem with the premise of the question as stated in the first paragraph. A pseudo-Riemannian(Lorentzian) metric tensor cannot produce a metric in the mathematical sense of a distance function(in semi-Riemannian manifolds the concept of length does not have a meaning), so it doesn't induce any topology, compatible or not, with the manifold topology.

The causal structure derived from the Lorentzian metric tensor can be used to define many different topologies according to the causality conditions of the specific Lorentzian manifold but unless they coincide with the manifold topology they can't replace the manifold topology(as explained by Andreas Cap above). See for example:"On the Topology of Lorentzian manifolds" by Renee Hoekzema.

The problems enters in relativity and physics in general when trying to relate measurable quantities or lengths, that are essential physically, with the causal structure topology in the context of a Lorentzian manifold.

Apparently the only topology coincident with a manifold topology is the Alexandrov topology in a strongly causal manifold. But this topology that is graphically depicted by two light cones united in the form of an infinitely elongated rhombus, paradoxically loses the causal light cone at that strongly causal limit, wich would render it equivalent to $E^4$ and erasing the distinction with a Riemannian manifold.

This is actually realized in Minkowski space, that is strongly causal, and that is the only motive I see for authors trying to avoid treating Minkowski space as a manifold as it would stress this fact.

In the case of curved semi-Riemannian manifolds that are strongly causal the Hawking-Penrose singularity theorems apply.