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.)
Let me try to answer your question to some extent; given the vague nature of the question, there will be no canonical answer.
You should think in terms of three different areas of mathematics:
These areas have different agendas, different tools, different "favorite toys." These areas, however, share some common toys, such as projective spaces of various dimensions and their projective subspaces.
For instance, the real-projective space ${\mathbb R}P^n$ (and its complex analogue ${\mathbb C}P^n$) appears as one of these common toys. In order to become one, some further choices have to be made in RG and AG. From RG viewpoint, one needs to choose a Riemannian metric, the most important one is the Fubini-Study metric. From the AG viewpoint, one needs to make the projective space $P$ into an algebraic variety. The standard choice is to identify $P$ with ${\mathbb R}P^n$ (resp. ${\mathbb C}P^n$). But other important choices are Veronese embeddings. They are all isomorphic to the "standard" projective space but the choice of an embedding is important.
Moreover, from the AG viewpoint, it is also important to work with other fields and rings, e.g. say with the projective spaces over $p$-adic numbers, over finite fields, over rings of polynomials, etc. These will have increasingly less and less in common with PG and RG.
As a projective geometer, you can think of all projective lines in the projective plane (or space) as congruent, but images of the Veronese embeddings of the projective line in the projective planes are genuinely different ones. From the AG viewpoint, projective lines in the projective plane are the degree 1 rational curves. Veronese curves/surfaces have higher degree.
From PG viewpoint, "everything takes place inside of a fixed projective space." To some extent, this is also true from the AG viewpoint, except you may have to work with different projective embeddings and some (or many, depending on what you do) of the objects of AG are not projective varieties and do not embed as subsets in a projective space. But from the RG viewpoint, one usually considers Riemannian manifolds abstractly and not as embedded isometrically in a projective space. (Caveat: There are many exceptions to this rule, for instance, much of the theory of minimal surfaces deals with surfaces in the Euclidean 3-space.)
I did not even get started on Kahler geometry that links (complex) AG and RG, but, again, AG and RG agendas are different here.
As for some of your more specific questions:
i. A question in a comment
"Why do the projective straight lines (which are primitive notions in the axiomatic treatment of projective geometry) just happen to agree with the geodesics in the Euclidean plane?"
the answer is that this is not quite true, as Tabes says. For one thing, you have to work "over the real numbers" (i.e. with real projective spaces). Secondly, you have to remove a point from the projective line to make it an affine line. Or, if you like, you remove the line at infinity from the projective plane, which will make it into an affine plane and then affine lines become complete Euclidean geodesics (after you choose a metric!). With these caveats, the fact that you have mention is true but is an accident of geometry of a space of constant curvature. "Most" Riemannian manifolds do not have constant sectional curvature and do not admit even a local a "model" where their geodesics are (pieces of) Euclidean straight lines. (OK, I am lying a bit here, this is not a complete accident and spherical, Euclidean and hyperbolic geometries were and continue to be a source of inspiration of RG.)
ii.
"Do smooth projective varieties (at least over the real numbers) have a natural projective connection that somehow agrees with the variety structure?"
No, in general they do not. Once you fix an embedding in a projective space then, yes, you can induce the ambient connection from the FS metric. But there are many different projective embeddings of the same variety, so you do not have a canonical choice. As for "somehow agrees with the variety structure," I do not even know how to interpret this. Connections, curvature, Chern classes, etc, play important role in AG, but one should definitely not limit oneself to the tangent bundle here. Frequently, other bundles and their connections are more informative. For instance, sometimes you work with the canonical or anticanonical line bundle, sometimes you work with all line bundles simultaneously, or work with all stable vector bundles, etc.
iii.
"Which concepts from projective geometry and from Riemannian geometry have natural analogs in the intrinsic geometry of projective varieties?"
I would say that the curvature sign (more precisely, the sign of one of your favorite curvatures or even positivity of the curvature operator) and the notion of "positivity" in algebraic geometry which manifests itself in different ways. Assuming that a particular curvature is positive/negative, semipositive, etc., usually makes it possible to prove interesting theorems in RG. Ditto in AG:
Lazarsfeld, Robert, Positivity in algebraic geometry. I. Classical setting: line bundles and linear series, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge 48. Berlin: Springer (ISBN 3-540-22533-1/hbk). xviii, 387 p. (2004). ZBL1093.14501.
Lazarsfeld, Robert, Positivity in algebraic geometry. II. Positivity for vector bundles, and multiplier ideals., Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge 49. Berlin: Springer (ISBN 3-540-22534-X/hbk). xviii, 385 p. (2004). ZBL1093.14500.
Also the key dichotomy: Positive curvature-negative curvature (RG), rational variety-general type variety (AG).
iv.
"As raised above, what about straight lines? Do these have a generalization to projective surfaces/varieties?"
As far as I am concerned, this is too vague to be answerable. One can say that these are rational curves. (Which play an important role in huge chunk of AG.) But in many cases one works with varieties that simply do not contain rational curves. Then you consider curves of higher genus: Thinking of all these curves at once is frequently quite useful. Or, you become a complex-analyst and work with the Kobayashi metric; the Kobayashi-extremal disks can be regarded as analogues of straight lines.
I declare myself done here....
The bottom line is: Treat the three geometries as separate areas of math. Occasionally, they meet in one place and share their toys, tools and theorems.
Best Answer
This answer was written by @M.G. on MO:
I believe the following is meant:
Every smooth (real) manifold $M$ has a (unique) real-analytic structure compatible with the smooth structure. So, cover $M$ with real-analytic charts, i.e. whose transition functions are real-analytic diffeomorphisms $$ \phi_{ij}:=\phi_j^{-1}\circ\phi_i: U_{ij}:=\phi_i^{-1}(\phi_i(U_i)\cap\phi_j(U_j))\to U_{ji} $$ One can find open subsets $U_i^{\mathbb{C}}\subseteq\mathbb{C}^n$ with $U_i^{\mathbb{C}}\cap\mathbb{R}^n=U_i$ and $U_{ij}^{\mathbb{C}}\cap\mathbb{R}^n=U_{ij}$ such that the (real-analytic) $\phi_{ij}$ extend to biholomorphisms $\phi_{ij}^{\mathbb{C}}:U_{ij}^{\mathbb{C}}\to U_{ji}^{\mathbb{C}}$ satisfying the usual cocycle conditions. Then the complexification $M^{\mathbb{C}}$ is defined as a quotient space of the disjoint union, $\left(\coprod_i U_i^{\mathbb{C}}\right)/\sim$, where $z_i\sim z_j$ iff $z_i\in U_{ij}^{\mathbb{C}}$ and $z_j = \phi_{ij}^{\mathbb{C}}(z_i)$ (this works because of the cocycle conditions). The maps $U_i^{\mathbb{C}}\hookrightarrow\coprod U_i^{\mathbb{C}}$ induce coordinate charts $U_i^{\mathbb{C}}\to M^{\mathbb{C}}$ with biholomorphic transition functions.
This and the details around it are part (of the proof of) Bruhat-Whitney's theorem* on the existence of $M^{\mathbb{C}}$. Moreover, complexification is functorial in the obvious way. By Grauert, $M^{\mathbb{C}}$ is in fact a Stein manifold.
*F. Bruhat and H. Whitney, Quelques propriétés fondamentales des ensembles analytiques-réels, Comment. Math. Helv. 33, 132-160 (1959).