I just looked on the proof in Wikipedia Existence of Riemannian metric on smooth manifold, but I don’t see why this proof failed for pseudo-Riemannian metric, could anyone point out why this is not applicable for pseudo-Riemannian metric?
Why the proof of existence of Riemannian metric does not apply to pseudo Riemannian metric
metric-spacesriemannian-geometrysemi-riemannian-geometrysmooth-manifolds
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As an American president once said "it all depends on what the meaning of the word is is". If you look at the definition, it is clear that Riemannian metrics, as defined, are not distance functions. When you say that a (connected) Riemannian manifold $(M,g)$ is a metric space, you are actually saying that there exists a certain functor from the category of Riemannian manifolds to the category of metric spaces: $$ \Phi: (M,g)\mapsto (M, d_g) $$ where $d_g$ is the Riemannian distance function. (If you do not know what functors and categories are, just think of functors as maps and categories as sets, even though this is, strictly speaking, false.) This functor is like a conversion procedure from, say, kilograms to pounds (except, it is not invertible). The functor $\Phi$ is defined by first assigning a quantity, which we call length $L_g(p)$ to each (rectifiable) path $p$ in $M$ and then minimizing.
The step $g\mapsto L_g$ makes sense even for pseudo-Riemannian manifolds, just you get some paths of negative length. It is the 2nd step which fails: If you try to minimize, you will (frequently) get $-\infty$, which is not particularly useful.
Remark. What I said above about isometries: Both categories of Riemannian manifolds and metric spaces have their own notions of isometric maps and these two notions do not correspond to each other under the functor $\Phi$. For instance, a Riemannian geometer would think of the circle of length $2\pi$ isometrically embedded into $R^2$ (with the image of the embedding being the standard unit circle). But this embedding does not preserve distances between points! This is another reason I do not like to use the word is for the relation between Riemannian manifolds and metric spaces. However, assuming that you know what functors are, otherwise, just ignore this: $\Phi$ does become a functor if we use 1-Lipschitz maps between spaces as morphisms for both Riemannian manifolds and metric spaces.
Now, to your question of why do we call pseudo-Riemannian metrics metrics, it is all matter of habit and tradition. You can think of three different worlds:
Metric geometry
Riemannian geometry.
Pseudo-Riemannian geometry.
They all have their notions of metrics (and isometries), but these notions have different meanings. It is as if people who speak different languages can occasionally use the same word, but it has different meaning in these languages. My favorite example is the word application, which has different meaning in English and in French.
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.)
Best Answer
It fails even at the pointwise level: in the Riemannian case, $$\tag{1} \sum_\beta \tau_\beta g_\beta$$ is always positive definite since $g_\beta >0$, $\tau _\beta \ge 0$ and $\tau_{\beta_0} >0$ for some $\beta_0$. But if each $g_\beta$ is only non-degenerate, then (1) might give you a degenerate symmetric two tensor.
Related question about the existence of Puesdo Riemannian metric: here