Riemannian surfaces as one dimensional complex manifolds


by TrickyDicky
Tags: complex, dimensional, manifolds, riemannian, surfaces
homeomorphic
homeomorphic is offline
#145
Nov25-11, 06:02 PM
P: 1,028
Euclidean geometry implies 0 gaussian curvature:

http://en.wikipedia.org/wiki/Flat_manifold
TrickyDicky
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#146
Nov26-11, 05:53 AM
P: 2,889
Quote Quote by homeomorphic View Post
Euclidean geometry implies 0 gaussian curvature:

http://en.wikipedia.org/wiki/Flat_manifold
That's fine, but remember that the relevant quantity for the gauss-bonnet theorem is the total curvature and what happens when integrating over infinity.
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.
TrickyDicky
TrickyDicky is offline
#147
Nov26-11, 12:52 PM
P: 2,889
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.
homeomorphic
homeomorphic is offline
#148
Nov26-11, 07:33 PM
P: 1,028
That's fine, but remember that the relevant quantity for the gauss-bonnet theorem is the total curvature and what happens when integrating over infinity.
Now, I see what you are trying to say, I think. But the integral is zero, since the curvature is zero. The idea of integrating over infinity is mathematically meaningless, as it stands. I suppose you might try to say the curvature is a delta function, so that all the curvature is concentrated at that point at infinity, but a Riemannian metric is a smoothly varying thing, so it can't really have a delta function as the curvature of its connection and remain within the realm of Riemannian geometry.

So, maybe it is semantics, but your terminology is extremely non-standard, 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.


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.
That is a terribly written article. I think they mean it is within S^2, as in, the interior of the closed ball bounded by S^2. To say that it is a subset of S^2 is nonsense, and, if you look at their awful equation that they give you with no explanation, it doesn't make sense that it would be in S^2 because it has an x, y, and z as variables, and the equation wouldn't make any sense if you put in points on the unit sphere.


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)
Perhaps, you need to fill in some of the gaps in your background in order to not make these mistakes, rather than quoting other people's statements that you don't completely understand.


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.
It is not alive. You are saying the curvature is zero, when you say it's flat. What happens when you integrate a function that is identically equal to 0 over a manifold? What do you get?

Jeopardy music plays...



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):
It is NOT closed. It is true that it has no boundary, hence no boundary term, but it is not compact, since it is missing a point.


We only need the integral of the gaussian curvature to obtain the Euler characteristic, and since it has infinite volume:

2πχ(M)χ(M)=∫ƩKdA==limR→∞∫Ʃ1R2dA=4π=2


Let me know if there's any problem with this.
Yes, there's a problem. The result of that limiting process is not a Riemannian manifold, since it has no metric at one point.
homeomorphic
homeomorphic is offline
#149
Nov26-11, 07:41 PM
P: 1,028
Here's an example of people who are clearly implying that horospheres are missing a point, in some kind of peer-reviewed paper:

http://www.intlpress.com/JDG/archive/1977/12-4-481.pdf
TrickyDicky
TrickyDicky is offline
#150
Nov27-11, 06:55 AM
P: 2,889
Quote Quote by homeomorphic View Post
Now, I see what you are trying to say, I think. But the integral is zero, since the curvature is zero. The idea of integrating over infinity is mathematically meaningless, as it stands.
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)?
Quote Quote by homeomorphic View Post
So, maybe it is semantics, but your terminology is extremely non-standard, 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.
This is how I picture it.

Quote Quote by homeomorphic View Post
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
This interests me.



Quote Quote by homeomorphic View Post
Perhaps, you need to fill in some of the gaps in your background in order to not make these mistakes, rather than quoting other people's statements that you don't completely understand.
Point taken.



Quote Quote by homeomorphic View Post
It is not alive. You are saying the curvature is zero, when you say it's flat. What happens when you integrate a function that is identically equal to 0 over a manifold? What do you get?

Jeopardy music plays...
See above



Quote Quote by homeomorphic View Post
It is NOT closed. It is true that it has no boundary, hence no boundary term, but it is not compact, since it is missing a point.
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)?


Quote Quote by homeomorphic View Post
Yes, there's a problem. The result of that limiting process is not a Riemannian manifold, since it has no metric at one point.
Ok, certainly the horosphere is a very special surface, I should have picked an easier one to debate.
homeomorphic
homeomorphic is offline
#151
Nov27-11, 11:17 AM
P: 1,028
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
Ben Niehoff is offline
#152
Nov27-11, 01:06 PM
Sci Advisor
P: 1,562
Quote Quote by homeomorphic View Post
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
TrickyDicky is offline
#153
Nov27-11, 03:58 PM
P: 2,889
Quote Quote by homeomorphic View Post
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
TrickyDicky is offline
#154
Nov28-11, 08:02 AM
P: 2,889
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
homeomorphic is offline
#155
Nov28-11, 10:36 AM
P: 1,028
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
TrickyDicky is offline
#156
Nov28-11, 01:31 PM
P: 2,889
Quote Quote by homeomorphic View Post
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
homeomorphic is offline
#157
Nov28-11, 02:28 PM
P: 1,028
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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