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Riemannian surfaces as one dimensional complex manifolds 
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#145
Nov2511, 06:02 PM

P: 1,197




#146
Nov2611, 05:53 AM

P: 3,012

Also see here http://eom.springer.de/s/s086640.htm where they show that horospheres can be defined as a set of points in S^2. 


#147
Nov2611, 12:52 PM

P: 3,012

Correction of post #136: It was quite obvious that the gaussian curvature had to be zero but if even the "experts" blunder I guess it's no big deal that I do (and quite a few times). My only excuse is my math ignorance and that I have been misled to some extent (about the core of the matter, certainly not about the Gaussian curvature)
The important thing is that the underlying theme of the thread which I stated in my second post (#4), that a topological sphere can have a flat metric in hyperbolic space and that a horosphere is a topological sphere in H^3 is still alive. Now to the correction of #136, it should have said: So since the horosphere is closed it has no boundary term (it is compact without boundary): We only need the integral of the gaussian curvature to obtain the Euler characteristic, and since it has infinite volume: [tex]\begin{align} 2 \pi \, \chi(M) &= \int_Ʃ K \, dA = \\ &= \lim_{R \to \infty} \int_Ʃ \frac{1}{R^2} \, dA=4\pi \\ \chi(M) &= 2 \end{align} [/tex] Let me know if there's any problem with this. 


#148
Nov2611, 07:33 PM

P: 1,197

So, maybe it is semantics, but your terminology is extremely nonstandard, plus I'm not entirely sure if this curvature as a delta function thing can be made precise. But, it's true that the horosphere would be kind of a limiting case of spheres with positive curvature, with the curvature getting more and more concentrated near infinity and becoming flat everywhere else. Even if you insist on using your bad and misleading phrasing, you must admit that it is not the entire sphere that has Euclidean geometry, but the sphere minus a point that has Euclidean geometry. I offered this solution to you before, but you didn't accept it. That was to say that if you insist that the horosphere has to be a sphere, you can include that extra point (with the cost that the metric is not defined there), but that's not standard. Jeopardy music plays... 


#149
Nov2611, 07:41 PM

P: 1,197

Here's an example of people who are clearly implying that horospheres are missing a point, in some kind of peerreviewed paper:
http://www.intlpress.com/JDG/archive/1977/124481.pdf 


#150
Nov2711, 06:55 AM

P: 3,012

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)? 


#151
Nov2711, 11:17 AM

P: 1,197

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/grqc/0411038 


#152
Nov2711, 01:06 PM

Sci Advisor
P: 1,588

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). 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. 


#153
Nov2711, 03:58 PM

P: 3,012

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. 


#154
Nov2811, 08:02 AM

P: 3,012

Ok, so we have this object that is noncompact, 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 3space (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 4manifold. I would think so, but I wouldn't advice anyone to take my word for it. what do you think? 


#155
Nov2811, 10:36 AM

P: 1,197

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.



#156
Nov2811, 01:31 PM

P: 3,012

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? 


#157
Nov2811, 02:28 PM

P: 1,197

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. 


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