## Riemannian surfaces as one dimensional complex manifolds

 The fact that you are not even addressing the simple differential geometry facts I mentioned in post #129 could be misunderstood as a weakness, I myself wonder why you guys avoid it.
If you prove something wrong, it doesn't matter what other facts you bring to the table. Gauss-Bonnet and the fact that the geometry of the horospheres is Euclidean means that they could not possibly be topological spheres. You are claiming that they have positive curvature and are Euclidean. That is impossible. I guess maybe you are trying to say extrinsic curvature with respect to hyperbolic space, but we tried to confront you about that earlier and you would not explain what you meant. But, it doesn't matter. Gaussian curvature is 0, you integrate it over the horosphere, and you get 0. So, it's not a topological sphere, end of discussion. It doesn't matter if it has positive curvature in some other bizarre sense. It has 0 curvature in the intrinsic sense, so Gauss-Bonnet does apply.

But let's ignore that for the moment.

 It says in WP:"In Riemannian geometry, a Bryant surface is a 2-dimensional surface embedded in 3-dimensional hyperbolic space with constant mean curvature equal to 1."
I assume this means mean curvature with respect to hyperbolic space.

 Horospheres are the only Bryant surfaces in wich all of the surface points are umbilical points.
I'm not familiar with that fact, but it could be right.

 And it is a well known fact of differential geometry that only for umbilical points the gaussian curvature equals the square of the mean curvature at that point,
Nope. That is for surfaces in R^3. An umbilic point is where the principal curvatures are equal. So the mean curvature equals the principal curvatures. So, with respect to an embedding in Euclidean space, the mean curvature equals Gaussian curvature. But the Gaussian curvature can't be computed in the same way with respect to an embedding in hyperbolic space--the Gaussian curvature is the product of principal curvatures with respect to an embedding in R^3. That is your mistake.

 as a conclusion horospheres have positive gaussian curvature and therefore gauss-bonnet theorem is fine with them.
If it were so, then you would have derived a contradiction and destroyed the whole of mathematics.

 Do you accept that horospheres are objects in H^3?
Yes, and therefore, they are missing a point because the whole sphere has a point in the boundary of the closed ball, which is not in H^3.

 Do you agree that a horosphere has a extrinsic positive curvature in H^3?
What do you mean?

 Do you recognize the formula that relates extrinsic mean curvature and gaussian curvature for submanifolds that are totally umbilical?
Yes, but for surfaces in R^3.

 Do you admit horospheres are totally umbilical surfaces?
I think so.

 Are you able to leave for a moment the Euclidean ambient mindframe?
Yes, but that is irrelevant. If you prove something using the Euclidean mindframe, it is still correct regardless of whether you switch to a different mindframe.

 Do you understand that manifolds and submanifolds are different objects with respect to their Riemannian metric?
Unclear statement. Usually, you use the same metric, but you restrict it to the submanifold. But the geometry can be different, yes.

 Do you see for example that a torus has 0 total curvature and yet when embedded in Euclidean space doesn't have euclidean metric (the external sides have round metric and the internal hyperbolic metric)
Yes, I was the one trying to tell YOU that earlier.

 or that a Clifford torus embedded in S^3 has round metric and that doesn't make it a topological sphere?
A torus can't have a "round" metric, by which I think you mean positive curvature (it may have positive curvature in some places).

 Why can't you understand that a topological sphere like the horosphere embedded in hyperbolic sphere can have a Euclidean metric due to its ambient space?
Because that contradicts Gauss-Bonnet. When you take the point out, yes, it can have a Euclidean metric. But you keep insisting it's not Euclidean (in contradiction to yourself) since it has positive curvature.

 Do you see that a globally Euclidean metric means 0 gaussian curvature for manifolds and for submanifolds embedded in Euclidean space?
Not just Euclidean space. The Gaussian curvature is intrinsic, so it does not depend on the embedding:

"Gauss's Theorema Egregium (Latin: "remarkable theorem") states that Gaussian curvature of a surface can be determined from the measurements of length on the surface itself. In fact, it can be found given the full knowledge of the first fundamental form and expressed via the first fundamental form and its partial derivatives of first and second order. Equivalently, the determinant of the second fundamental form of a surface in R3 can be so expressed. The "remarkable", and surprising, feature of this theorem is that although the definition of the Gaussian curvature of a surface S in R3 certainly depends on the way in which the surface is located in space, the end result, the Gaussian curvature itself, is determined by the inner metric of the surface without any further reference to the ambient space: it is an intrinsic invariant. In particular, the Gaussian curvature is invariant under isometric deformations of the surface."

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

 Do you admit the gauss-bonnet theorem?
What an odd thing to ask when you are the one denying it.

 Can you calculate the Euler characteristic of the horosphere taking into account that it has positive gaussian curvature?
It does not have positive curvature.

 Did you understand that this gaussian positive curvature is derived from the fact that a horosphere is a submanifold with positive mean curvature and totally umbilical?
I understand that you said it, and I understand that you were incorrect

 "Let γ(∞) denote the endpoint of γ on the sphere at infinity ∂H3 ≃ S2, a closed horosphere centered at γ(∞) is a closed surface Ʃ ⊂ H3 which is orthogonal to all geodesic lines in H3 with endpoint γ(∞)."
They are saying that they consider that missing point as part of the horosphere. Problem is, the metric won't extend to that point. So you can say a horosphere is a topological sphere if that is your definition, but then, you have to live with the fact that the metric is undefined at a point.

 Quote by homeomorphic Nope. That is for surfaces in R^3. An umbilic point is where the principal curvatures are equal. So the mean curvature equals the principal curvatures. So, with respect to an embedding in Euclidean space, the mean curvature equals Gaussian curvature. But the Gaussian curvature can't be computed in the same way with respect to an embedding in hyperbolic space--the Gaussian curvature is the product of principal curvatures with respect to an embedding in R^3. That is your mistake.
Ok, if you think you have identified here where the mistake is let's center on this point.

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?

 Quote by homeomorphic They are saying that they consider that missing point as part of the horosphere. Problem is, the metric won't extend to that point. So you can say a horosphere is a topological sphere if that is your definition, but then, you have to live with the fact that the metric is undefined at a point.
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.

 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).

 Quote by homeomorphic 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

 Quote by homeomorphic 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.

 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).

 Euclidean geometry implies 0 gaussian curvature: http://en.wikipedia.org/wiki/Flat_manifold

 Quote by homeomorphic 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.

 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.

 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.

 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

 Quote by homeomorphic 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 by homeomorphic 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 by homeomorphic 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 by homeomorphic 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 by homeomorphic 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 by homeomorphic 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 by homeomorphic 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.

 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.

Recognitions:
 Quote by homeomorphic 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.

 Quote by homeomorphic 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.