> I am confused. Let me write down the definition I know, and let's error-correct to a point of agreement?

Fixing the definitions is a good idea because they may vary slightly in the literature.

> - A Lorentzian manifold is a pseudo-Riemannian manifold with a metric tensor that can be positive or negative definite, but not zero.

A bit unconventional definition as it places some of the characteristics of a pseudo-Riemannian manifold into Lorentzian territory. Usually the Lorentzian manifold restricts the signature of the metric to (1, n-1) or (n-1, 1). That is to say, restricts the pseudo-Riemannian manifold to one with only 1 time dimension.

> - A neighbourhood of a Lorentzian manifold is flat if one can assign a coordinate system where we have the minowski metric ds^2 = dx1^2 dx2^2 + ... + dx(n-1)^2 - dxn^2. So it's hyperbolic space of dimension n, Hn. This space has constant negative curvature; You can check by computing the curvature tensor.

If the space is flat (Riemann curvature tensor = 0) how can it have constant negative curvature? Hope you see with this question that something is off. Here is a more detailed explanation:

There is a tight link between Hyperbolic space and the Lorentz transformations. The proper (i.e. no space inversion) ortochronous (i.e. no time inversion) Lorentz group of transformations SO^+(1,n) is an isometry group of H^n. To be noted here the dimensionality difference, this means that a hyperbolic space H^n can be embedded in R^(1,n) and that the action of Lorentz transformations leaves H^n invariant. You can see a neat visualization in the wiki: https://en.wikipedia.org/wiki/Lorentz_group#orthochronous, the Lorentz transformations leave invariant the hyperboloids shown in the picture. The action of a Lorentz transformation in SO^+(1,n) on any vector in one those hyperboloids H^n produces another vector in the same hyperboloid. But H^n is not Minkowski space, it's just one "shell" embedded in a Minkowski space of one additional dimension that is left invariant by SO^+(1,n). These shells have constant negative curvature but the space they are embedded into (Minkowski) is flat.

This is equivalent to the relation between Euclidean space and spheres. The group of rotations in n-dimensional Euclidean space E^n leaves invariant the spheres S^(n-1). The spheres have positive constant curvature and are embedded in a higher dimensional flat space.

> - Thanks for that particular model of Schwarzschild's solution! I wasn't aware of it.

> - Do you have more examples of a similar flavour? I don't care if the solution obeys the einstein field equations. I'm mostly just interested in examples of Lorentzian manifolds with curvature.

Not that I can think of, to visualize a curved space you always have to embed it in a higher dimensional flat space. To "see" one curved dimension, you need (at least) 2 flat ones in which to embed it. This is what Flamm's paraboloid does, it shows the curved radial dimension of Schwarzschild solution embedded in 2D so we can see how distances change along the way. The problem is that we run very quickly out of "real estate" to show the curvature. If you are a highly visual person I could recommend you to get a copy of Gravitation by Misner, Thorne, Wheeler. It is a masterpiece and comes with huge amounts of visual intuition.