Riemannian surfaces as one dimensional complex manifolds

[homeomorphic]

So you claim that the formula for umbilical points in a surface: (mean curvature)^2=Gausian curvature holds only for R^3, right? is that really what you are saying? Hadn't we agreed that Gaussian curvature is intrinsic?

Yes. The Gaussian curvature is intrinsic, but the way you compute it is not. You can't compute it as the product of principal curvatures in H^3. Only in R^3 can you compute it that way.

It is only true in R^3 that the Gaussian curvature is the determinant of the shape operator. Why would you think it would be valid anywhere else? It is only the end result that is invariant.

It is not my definition, it is mathematics definition. The Riemannian metric extends to that point because the horosphere belongs to H^3. This fact is in every non-euclidean geometry book.

No. It may be in books that the horosphere belongs to H^3. But it either does NOT include that point or the metric doesn't extend (but that would be a non-standard definition, I think).

 

[TrickyDicky]

Yes. The Gaussian curvature is intrinsic, but the way you compute it is not. You can't compute it as the product of principal curvatures in H^3. Only in R^3 can you compute it that way

I'm not talking about the way to compute it, I'm saying it shouldn't depend on the embedding.
I haven't mentioned anything about principle curvatures products, my formula is H^2 > or equal to K, the equality only for umbilical points.
With H being the mean curvature

 

[TrickyDicky]

it either does NOT include that point or the metric doesn't extend

Oh, sure, that is a trivial fact for curved manifolds, one needs more than one chart to cover all the surface, in the case of the sphere at least two. So I guess you are actually admitting the horosphere has positive gaussian curvature.

 

[homeomorphic]

I'm not talking about the way to compute it, I'm saying it shouldn't depend on the embedding.
I haven't mentioned anything about principle curvatures products, my formula is H^2 > or equal to K, the equality only for umbilical points.
With H being the mean curvature

Your formula COMES from the fact that the Gaussian curvature is equal to the product of principal curvatures, so you may not have mentioned it, but it's implicit.

If not, then how are you going to prove your formula?

Oh, sure, that is a trivial fact for curved manifolds, one needs more than one chart to cover all the surface, in the case of the sphere at least two.

That has nothing to do with it.

So I guess you are actually admitting the horosphere has positive gaussian curvature.

Absolutely not. If the geometry is Euclidean, the curvature is zero! End of story.

You keep insisting that a surface with positive Gaussian curvature can have INTRINSIC Euclidean geometry. If the Gaussian curvature is INTRINSIC, then how can this be? You must give up your claim that the instrinsic geometry is Euclidean if you are insisting on positive curvature (but the right thing to do is give up on the positive curvature because it's wrong).

 

[homeomorphic]

Euclidean geometry implies 0 gaussian curvature:

http://en.wikipedia.org/wiki/Flat_manifold

 

[TrickyDicky]

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]

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:

$$\begin{align} 2 \pi \, \chi(M) &= \int_Ʃ K \, dA = \\ &= \lim_{R \to \infty} \int_Ʃ \frac{1}{R^2} \, dA=4\pi \\ \chi(M) &= 2 \end{align}$$

Let me know if there's any problem with this.

 

[homeomorphic]

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=2Let 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]

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]

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)?

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.

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.

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.

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

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)?

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]

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.