Riemannian surfaces as one dimensional complex manifolds

[homeomorphic]

I see, acouple of doubts though,
Isn't the integral of zero a constant that depends on the boundary conditions? And are there not situations where integrating over infinity is not meaningless (like Ben did when integrating the boundary term)?

No, this is a definite integral, so no constant.

As I said, the problem is that you would have to make sense out of the curvature being a delta function.

It must be that people have thought about things like that, but I don't know how to do it. Black holes have singularities in them, so I imagine it's something like that. But then, that singularity can't be part of the Lorentzian manifold, so it's just like the situation here. Actually, with black holes, I guess the curvature will not exactly be a delta funtion, but I could imagine maybe you would want some kind of Green's function or something like that that solves the Einstein equation when the source is a delta function.

Here's a paper where it sounds like they do something like that, but I just read the abstract.

http://arxiv.org/abs/gr-qc/0411038

Ok, let's say it is not compact, then a problem arise with the gauss-bonnet formula if we say there is no boundary term, how do we get an Euler charachteristic of one (that of a plane)?

Gauss-Bonnet is for CLOSED (compact with no boundary) surfaces, so it doesn't apply to the plane.

 

[Ben Niehoff]

As I said, the problem is that you would have to make sense out of the curvature being a delta function.

The Ricci scalar measures local angular deficit density, so a delta function in the scalar curvature indicates a conical singularity: the space looks locally like a cone.

This does step outside strict Riemannian geometry, but it is not hard to make physical sense of it, so it is useful to us physicists (and turns up a lot in string theory).

Gauss-Bonnet is for CLOSED (compact with no boundary) surfaces, so it doesn't apply to the plane.

Yes, but it seems to give the correct result for the infinite plane considered as a limit of disks of increasing size (or alternatively, one may simply use the fact that the infinite plane is homeomorphic to the open disk and be done with it).


Also, I see this thread has gone absolutely nowhere while I was gone. I'm not going to argue with Tricky anymore. I might jump in if anyone says anything interesting.

 

[TrickyDicky]

It must be that people have thought about things like that, but I don't know how to do it. Black holes have singularities in them, so I imagine it's something like that. But then, that singularity can't be part of the Lorentzian manifold, so it's just like the situation here. Actually, with black holes, I guess the curvature will not exactly be a delta funtion, but I could imagine maybe you would want some kind of Green's function or something like that that solves the Einstein equation when the source is a delta function.

Here's a paper where it sounds like they do something like that, but I just read the abstract.

http://arxiv.org/abs/gr-qc/0411038

Ok, thanks Homeomorphic, you've been of great help (and patient, I can't help being stubborn until I'm convinced of something, Ben knows it from other encounters and yet I still seem to manage to exasperate him).
I guess I need to get used to the mathematical definitions and terminology not making intuitive sense to me.
I found the comparison you made with singularities really illuminating.

 

[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]

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.

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.