Riemannian surfaces as one dimensional complex manifolds

TrickyDicky

Ok, so we have this object that is non-compact, and that is an embedding of an infinite plane in H^3. It is not the Real projective plane though, because horospherical surfaces, unlike the real projective plane, are oriented (holler if you disagree with this), and also because the real projective plane can't be embedded in 3-space (it intersects with itself), only immersed as the Boy's surface.
The real projective plane can however be embedded as a closed surface in E^4, so I wondered if the horosphere could be embedded as closed surface in a Lorentzian 4-manifold. I would think so, but I wouldn't advice anyone to take my word for it.
what do you think?

homeomorphic

It doesn't matter where you embed it. The things we talked about were instrinsic. The horosphere is always missing a point, by definition. If you want to close it up, it's not going to have a metric on that extra point.

TrickyDicky

Yeah, the real projective plane is compact to begin with.
I'm having a hard time visualizing the horosphere as an object (as an independent entity) with a missing point, by definition all of its points are at infinite distance from the center and yet it only misses one. Does the fact that it misses a point mean that it is a fundamentally incomplete object? Or is it just a mathematical definitions type of issue?

homeomorphic

The horosphere is basically just a Euclidean plane. R^2. Makes sense to call that a "sphere" of infinite radius. If you add a point at infinity, you get a topological sphere.

It touches that boundary at infinity at a point, but remember the boundary at infinity in H^3 is infinitely far away.