Riemannian surfaces as one dimensional complex manifolds

[TrickyDicky]

But is there a hyperbolic manifold that has as interior the H^3 space manifold and as boundary at infinity the extended complex plane? I mean, is there a way to have a noncompact infinite manifold as interior of a compact boundary at infinity? This would imply of course the boundary should have infinite radius.

 

[homeomorphic]

But is there a hyperbolic manifold that has as interior the H^3 space manifold and as boundary at infinity the extended complex plane? I mean, is there a way to have a noncompact infinite manifold as interior of a compact boundary at infinity? This would imply of course the boundary should have infinite radius.

Yes, as a manifold, but not as an hyperbolic manifold.

It is a smooth manifold only. The metric is only defined in the interior, so only the interior is a Riemannian manifold (i.e. has a Riemannian metric). The hyperbolic metric does not extend to the whole closed ball. That's what I have been arguing for.

To clarify, my definition of hyberbolic manifold is a Riemannian manifold (a smooth manifold equipped with a Riemannian metric) with constant negative curvature.

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

(this defines it as curvature -1, but it's essentially the same as my definition)

 

[TrickyDicky]

Hi, homeomorphic
I'm having some problems reconciling what you say (and also what Ben Niehoff said) with what I read about hyperbolic manifolds. Are you familiar with the Ending laminations and Tame endings theorems(they can be found in WP)? They seem to imply there are hyperbolic 3-manifolds (not just manifolds) with infinite volume and Riemann sphere as boundary at infinity.
An interesting discussion of this can be found in google books:
http://books.google.com/books?id=e0...&resnum=1&ved=0CDIQ6AEwAA#v=onepage&q&f=false
pages 15-26

 

[TrickyDicky]

The metric is only defined in the interior, so only the interior is a Riemannian manifold (i.e. has a Riemannian metric). The hyperbolic metric does not extend to the whole closed ball. That's what I have been arguing for.

But the hyperbolic geometry of the manifold is only defined from the interior of the ball if it is to be a hyperbolic manifold right? I mean inhabitants of the hyperbolic manifold don't have access to the exterior part of the sphere boundary.

 

[homeomorphic]

Hi, homeomorphic
I'm having some problems reconciling what you say (and also what Ben Niehoff said) with what I read about hyperbolic manifolds. Are you familiar with the Ending laminations and Tame endings theorems(they can be found in WP)? They seem to imply there are hyperbolic 3-manifolds (not just manifolds) with infinite volume and Riemann sphere as boundary at infinity.

Semantics. That doesn't contradict what we said. Riemann sphere as BOUNDARY AT INFINITY. That means it's not part of the hyperbolic manifold itself, hence not contrary to what we've been saying.

But the hyperbolic geometry of the manifold is only defined from the interior of the ball if it is to be a hyperbolic manifold right? I mean inhabitants of the hyperbolic manifold don't have access to the exterior part of the sphere boundary.

Yes. The hyperbolic manifold is just the interior. The boundary at infinity is not part of the hyperbolic manifold because there is no metric there, which is part of the definition of hyperbolic manifold. It is the boundary at infinity of it. Not part of it.

 

[TrickyDicky]

Maybe it is just semantics, because I don't think I'm arguing that the boundary is part of the interior.
Consider this example, the topological boundary of a closed disk viewed as a topological space is empty, while its boundary in the sense of manifolds is the circle surrounding the disk. Wich is itself a different manifold with one less dimension.
Is the circle part of the closed disk? That dependes on how you are considering it.

I guess I'm considering as a whole both the interior hyperbolic space and its Riemann sphere boundary, in the sense that you can't have one without the other, hyperbolic manifolds are always quotient spaces of H^3 by a Kleinian group (even if it is the trivial group).

 

[homeomorphic]

Maybe it is just semantics, because I don't think I'm arguing that the boundary is part of the interior.

You could say it's semantics, but if you go with the standard definitions, what we are saying is the technically correct way of saying it.

Consider this example, the topological boundary of a closed disk viewed as a topological space is empty, while its boundary in the sense of manifolds is the circle surrounding the disk. Wich is itself a different manifold with one less dimension.
Is the circle part of the closed disk? That dependes on how you are considering it.

It's kind of like that, but we are not saying that the boundary of a manifold with boundary is not part of the manifold (except in this case H^3 is DEFINED to be the interior). We are saying that the boundary at infinity is not and cannot be part of the manifold when considered as a Riemannian manifold.

 

[lavinia]

BTW: I think ones can show that no non-orientable closed surface without boundary can be embedded in 3 space using a Mayer-Vietoris sequence argument or Van kampen's Theorem. if one assume that the surface can be embedded in 3 space then a tube around the surface has an orientable boundary and thus the boundary has no torsion in its fundamental group.

But a non-orientable surface always has a torsion curve. Such a curve is trapped in the interior of the tube and can not be slid out to the boundary. But then it can not be shrunk to the point at infinity in the 3 sphere (the one point compactification of R^3). But the 3 sphere is simply connected so this is an impossible situation.

Correct this if it is wrong.

 

[lavinia]

Maybe it is just semantics, because I don't think I'm arguing that the boundary is part of the interior.
Consider this example, the topological boundary of a closed disk viewed as a topological space is empty, while its boundary in the sense of manifolds is the circle surrounding the disk. Wich is itself a different manifold with one less dimension.
Is the circle part of the closed disk? That dependes on how you are considering it.

The circle is part of the closed disk - by definition, the circle is a subset of it. It does not depend on how you look at it.

I guess I'm considering as a whole both the interior hyperbolic space and its Riemann sphere boundary, in the sense that you can't have one without the other, hyperbolic manifolds are always quotient spaces of H^3 by a Kleinian group (even if it is the trivial group).

The riemann sphere boundary is not a boundary in the manifold sense nor is it part of hyperbolic space. It is not contained in hyperbolic space at all.

 

[homeomorphic]

The riemann sphere boundary is not a boundary in the manifold sense nor is it part of hyperbolic space. It is not contained in hyperbolic space at all.

It is a boundary in the manifold sense, but it is the boundary of the closed ball containing H^3, not a boundary of H^3 in the manifold sense.

 

[TrickyDicky]

It is a boundary in the manifold sense, but it is the boundary of the closed ball containing H^3, not a boundary of H^3 in the manifold sense.

Yes, and this ball is a 3-manifold, right? and considering that this ball has as interior H^3 and as boundary the Riemann sphere manifold,and being of infinite volume, I wonder just what type of 3-manifold this ball might be: I woul have thought it is a hyperbolic manifold but now I'm not so sure.

 

[TrickyDicky]

The circle is part of the closed disk - by definition, the circle is a subset of it. It does not depend on how you look at it.

So I guess you would say the sphere is part of the closed ball,by definition, no?

 

[homeomorphic]

Yes, and this ball is a 3-manifold, right? and considering that this ball has as interior H^3 and as boundary the Riemann sphere manifold,and being of infinite volume,

It has an undefined volume because it doesn't have a metric or a volume form specified on it. IT DOES NOT HAVE INFINITE VOLUME.

I wonder just what type of 3-manifold this ball might be: I woul have thought it is a hyperbolic manifold but now I'm not so sure.

As I have been saying, it's just a closed ball as a smooth manifold with boundary--THERE IS NO RIEMANNIAN METRIC ON THE WHOLE THING (unless you put one on it, but it will not agree with the hyperbolic metric in the interior). A Riemannian metric is a prerequisite to being a hyperbolic manifold.

So, actually, you could make it into a hyperbolic manifold that is isometric to a proper subset of H^3, but it's not inherently hyperbolic. It is just a closed ball. Nothing that complicated or mysterious about it.

 

[lavinia]

So I guess you would say the sphere is part of the closed ball,by definition, no?

yes.

An the Poincare ball is an open ball not a closed ball. The sphere is not in it - also by definition

 

[TrickyDicky]

In post #23 I said this:

Horospheres have the particular property of being tangent to infinity. So here we have a geometrical 2D-object that having euclidean geometry can be thought of as containing the complex plane numbers and that also contains a point at infinity.

It turns out this is not so, it has been pointed out to me by someone that horospheres are not actually tangent to the boundary at infinity (even if in the two dimensional Poincare disk representation they seem to be which was what got me confused), so no point of the horospherical surfaces lies on the boundary at infinity. They have their center at infinity though (infinite radius or more accurately the radius tends to infinity ). With all the planes cutting the horosphere orthogonally giving circles, except for the one tangent to the horosphere that gives horocycles.
I think it is easy to see now that horospheres are embedded in H^3 and are basically spheres of infinite radius, so they have euclidean geometry on them and at the same time I would say they are homeomorphic to the topological sphere (and therefore to the Riemann sphere).
Would you guys agree with this?

 

[homeomorphic]

It turns out this is not so, it has been pointed out to me by someone that horospheres are not actually tangent to the boundary at infinity (even if in the two dimensional Poincare disk representation they seem to be which was what got me confused), so no point of the horospherical surfaces lies on the boundary at infinity.

I think the convention is that that point you want to say is at the boundary at infinity is not included in the horosphere, by definition, since the horosphere is supposed to live inside hyperbolic space (so it's not actually a sphere, but a plane, topologically). But the horosphere plus that point is tangent to the boundary at infinity.

They have their center at infinity though (infinite radius or more accurately the radius tends to infinity ). With all the planes cutting the horosphere orthogonally giving circles, except for the one tangent to the horosphere that gives horocycles.

That is a very unclear statement.

I think it is easy to see now that horospheres are embedded in H^3 and are basically spheres of infinite radius, so they have euclidean geometry on them

True, with the inherited metric from hyperbolic space, they are actually Euclidean. I don't know if I would call them spheres of infinite radius, since they are not spheres, topologically.

and at the same time I would say they are homeomorphic to the topological sphere (and therefore to the Riemann sphere).

No, they are homeomorphic to R^2, since we don't include that point at infinity (since it is not in hyperbolic space). Also, it is kind of superfluous to say that something is homeomorphic to the Riemann sphere because the thing that makes it a Riemann sphere is its complex structure and the homeomorphism doesn't respect that, so you may as well just call it a sphere, not a Riemann sphere.

 

[lavinia]

manifolds fall into different types that are determined by the way coordinate charts overlap. In a topological manifold, charts overlap by homeomorphims. In smooth manifolds that overlap with smooth diffeomorphisms. With complex manifolds they overlap with analytic maps. These are distinct types. The sphere can be a topological manifold in which case its charts overlap continuously but not necessarily smoothly or analytically. It is a smooth manifold when viewed as a submanifold of Euclidean space. It is a Riemann surface when the charts overlap in conformal maps for instance when one uses the two charts z and 1/z.

 

[TrickyDicky]

I think the convention is that that point you want to say is at the boundary at infinity is not included in the horosphere, by definition, since the horosphere is supposed to live inside hyperbolic space (so it's not actually a sphere, but a plane, topologically). But the horosphere plus that point is tangent to the boundary at infinity.

Forget about points at infinity, why do you say now a topological sphere needs to have a point at infinity?EDIT:I erased a part because I think is not correct.

True, with the inherited metric from hyperbolic space, they are actually Euclidean. I don't know if I would call them spheres of infinite radius, since they are not spheres, topologically.

But they are.

"A horosphere has a critical amount of (isotropic) curvature." according to wikipedia.
And as I wrote in post #90 and has not been refuted yet:
The theorem of Micallef-Moore says that a compact simply connected manifold of positive isotropic curvature is homeomorphic to the sphere.

 

[homeomorphic]

Forget about points at infinity, why do you say now a topological sphere needs to have a point at infinity?

Horospheres are not topological spheres. They are spheres missing a point because that point is not in H^3.

A unit sphere in R^3 can be cut by parallel planes from the equator,(plane sections) that will draw circles progressively smaller at the intersection, the last plane will be tangent to the sphere and will have a point. In the case of the horosphere the only change is that the plane tangent to the horosphere instead of a point intersects a horocycle.

I don't know what you are saying. Tangent planes typically intersect things at a point. If you take a plane THROUGH (not tangent to) the point at infinity, you get a horocycle.

But they are.

No, they are spheres minus a point, by convention, since we want them to live in hyperbolic space.

"A horosphere has a critical amount of (isotropic) curvature." according to wikipedia.

Didn't we just say they were Euclidean (i.e. flat)?

And as I wrote in post #90 and has not been refuted yet:
The theorem of Micallef-Moore says that a compact simply connected manifold of positive isotropic curvature is homeomorphic to the sphere.

They are not compact. If you insist on including that extra point that makes them compact, you get the same problem we have been talking about. The metric won't extend to that point. It's Euclidean. There's no such thing as a Euclidean sphere, by Gauss-Bonnet. Therefore, by your own assertions, we must leave out that point at infinity and define the horosphere to be a sphere minus that point, which is then, non-compact and homeomorphic to R^2.

 

[lavinia]

But they are.

"A horosphere has a critical amount of (isotropic) curvature." according to wikipedia.
And as I wrote in post #90 and has not been refuted yet:
The theorem of Micallef-Moore says that a compact simply connected manifold of positive isotropic curvature is homeomorphic to the sphere.

The horosphere is not compact so the theorem does not apply.

 

[TrickyDicky]

Horospheres are not topological spheres. They are spheres missing a point because that point is not in H^3.





No, they are spheres minus a point, by convention, since we want them to live in hyperbolic space.




Didn't we just say they were Euclidean (i.e. flat)?



They are not compact. If you insist on including that extra point that makes them compact, you get the same problem we have been talking about. The metric won't extend to that point. It's Euclidean. There's no such thing as a Euclidean sphere, by Gauss-Bonnet. Therefore, by your own assertions, we must leave out that point at infinity and define the horosphere to be a sphere minus that point, which is then, non-compact and homeomorphic to R^2.

I corrected my previous assertions severl posts above, the horosphere is completely embedded in H^3, in contrast to what I thought, it is not missing any point, it doesn't have any point at the boundary, how could it have the same problem in both situations, looks as if you weren't acknowledging the change.

 

[TrickyDicky]

There's no such thing as a Euclidean sphere

From "Three-dimensional geometry and topology, Volumen 1" page 61 by Thurston:

"A horizontal Euclidean plane is not a plane in hyperbolic geometry, it lies entirely on one side of a true hyperbolic plane tangent to it, which is a Euclidean sphere... These horizontal surfaces are examples of horospheres"

 

[homeomorphic]

From "Three-dimensional geometry and topology, Volumen 1" page 61 by Thurston:

"A horizontal Euclidean plane is not a plane in hyperbolic geometry, it lies entirely on one side of a true hyperbolic plane tangent to it, which is a Euclidean sphere... These horizontal surfaces are examples of horospheres"

Yes, he means sphere minus a point. You know Gauss-Bonnet, so I don't see what your objection is.

 

[homeomorphic]

I corrected my previous assertions severl posts above, the horosphere is completely embedded in H^3, in contrast to what I thought, it is not missing any point, it doesn't have any point at the boundary, how could it have the same problem in both situations, looks as if you weren't acknowledging the change.

Yes, it IS missing a point. It doesn't have a point at the boundary, but that is because it is approaching that point at the boundary.

 

[lavinia]

From "Three-dimensional geometry and topology, Volumen 1" page 61 by Thurston:

"A horizontal Euclidean plane is not a plane in hyperbolic geometry, it lies entirely on one side of a true hyperbolic plane tangent to it, which is a Euclidean sphere... These horizontal surfaces are examples of horospheres"

ok. so can you say in your own words what Thurston means by a Euclidean sphere? Tell us.

 

[lavinia]

In post #23 I said this:

I think it is easy to see now that horospheres are embedded in H^3 and are basically spheres of infinite radius, so they have euclidean geometry on them and at the same time I would say they are homeomorphic to the topological sphere (and therefore to the Riemann sphere).
Would you guys agree with this?

horospheres are not homeomorphic to the topological sphere. But... you have insisted that they are despite repeated explanations that they are not. So there must be a homeomorphism that you have in mind. Could you tell us what you are thinking of? What is the homeomorphism?

 

[TrickyDicky]

ok. so can you say in your own words what Thurston means by a Euclidean sphere? Tell us.

I certainly don't think he meant a sphere with a point missing or else he would have stated it so.
He seems to refer to a sphere the way we are used to in euclidean embedding. I think he means horospheres are homeomorphic to the topological sphere, that they have positive isotropic curvature, thus his insistence that, in hyperbolic space, the euclidean plane is different from euclidean space.

 

[lavinia]

I certainly don't think he meant a sphere with a point missing or else he would have stated it so.
He seems to refer to a sphere the way we are used to in euclidean embedding. I think he means horospheres are homeomorphic to the topological sphere, that they have positive isotropic curvature, thus his insistence that, in hyperbolic space, the euclidean plane is different from euclidean space.

so you are not sure

 

[TrickyDicky]

horospheres are not homeomorphic to the topological sphere. But... you have insisted that they are despite repeated explanations that they are not. So there must be a homeomorphism that you have in mind. Could you tell us what you are thinking of? What is the homeomorphism?

Ever heard about a Bryant surface, the horosphere is one very important example of a Bryant surface.
http://en.wikipedia.org/wiki/Bryant_surface
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."
Horospheres are the only Bryant surfaces in which all of the surface points are umbilical points. 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, as a conclusion horospheres have positive gaussian curvature and therefore gauss-bonnet theorem is fine with them. Ben should correct his calculation of the Euler characteristic in a previous post.

Some References:
R. Aiyama, K. Akutagawa, Kenmotsu-Bryant type representation formula for
constant mean curvature spacelike surfaces in H3, Differential Geom.
Appl. 9 (1998), 251–272.
L. Bianchi, Lezioni di Geometria Differenziale, Ast´erisque, 154-155 (1987),
321–347.
R.L. Bryant, Surfaces of mean curvature one in hyperbolic space, terza Edizione,
Bologna (1987).
P. Collin, L. Hauswirth, H. Rosenberg, The geometry of finite topology Bryant
surfaces, Ann. of Math. 153 (2001), 623–659.
J.A. G´alvez, A. Mart´ınez, F. Mil´an, Flat surfaces in the hiperbolic 3-space,
Math. Ann., 316 (2000), 419–435.

 

[TrickyDicky]

Besides what is said above:

WP:"In topology, an n-sphere is defined as a space homeomorphic to the boundary of an (n+1)-ball; thus, it is homeomorphic to the Euclidean n-sphere, but perhaps lacking its metric."
A horosphere is the boundary of a horoball, which is the limit of a sequence of increasing balls in hyperbolic space, a ball in hyperbolic space is homeomorphic to a ball in Euclidean space (since hyperbolic space is homeomorphic to Euclidean space), so this should prove a horosphere is homeomorphic to a sphere in Hyperbolic space (therefore also to the Riemann sphere logically) even if it has Euclidean metric. I know is counterintuitive, and hard to swallow but I don't know how else explain it.

 

[lavinia]

Besides what is said above:

WP:"In topology, an n-sphere is defined as a space homeomorphic to the boundary of an (n+1)-ball; thus, it is homeomorphic to the Euclidean n-sphere, but perhaps lacking its metric."
A horosphere is the boundary of a horoball, which is the limit of a sequence of increasing balls in hyperbolic space, a ball in hyperbolic space is homeomorphic to a ball in Euclidean space (since hyperbolic space is homeomorphic to Euclidean space), so this should prove a horosphere is homeomorphic to a sphere in Hyperbolic space (therefore also to the Riemann sphere logically) even if it has Euclidean metric. I know is counterintuitive, and hard to swallow but I don't know how else explain it.

I am tired of your answering with quotes from Wikipedia. what's the matter? Can't you explain things in your own words? Tell me - without some language that you find in a book or on line - what is meant by a topological sphere? How is a horosphere homeomorphic to a topological sphere? What is the homeomorphism? write it down.

 

[TrickyDicky]

I am tired of your answering with quotes from Wikipedia. what's the matter? Can't you explain things in your own words? Tell me - without some language that you find in a book or on line - what is meant by a topological sphere? How is a horosphere homeomorphic to a topological sphere. What is the homeomorphism? write it down.

I do so to avoid errors, I'm not a mathgematician nor have any formal training in math, so I can only use my intuition and try to search for places where it comes explained in a better way than I can do it. Did you read the previous post?
Why the ad hominem attacks though? Reply only the content of the posts please.

 

[lavinia]

This thread has made me read a little about hyperbolic geometry. I write this post to be helpful to you,TD, since I know that the others here already know this stuff much better that I do.

I will explain what I think horospheres are from the perspective of two models, the Poincare ball model and the upper half space model, and then show the connection between the two models.

First the Poincare ball model.

The point set is the open unit ball in Euclidean n-space. It does no include the bounding sphere.

The metric is radially symmetric and makes rays from the origin infinitely long.

A horosphere as a point set is the subset of the unit ball that is the intersection of the open ball with the point set of a Euclidean sphere that is tangent to the bounding sphere. As a point set it is this sphere minus the point of tangency.
It's topology is the subset topology that it inherits from the open unit ball and is homeomorphic to Euclidean n-1 space.(Tricky: Why not try to write down the homeomorphism. It would be a good exercise for you.)

Metrically these horospheres have the geometry of Euclidean space.

Choose a conformal transformation of Euclidean n space that maps the open ball to the upper half plane and maps the bounding unit sphere to the hyperplane at its boundary. Such a map must send one of the points on the unit sphere to infinity. All of the horospheres that are tangent to the sphere at this point get mapped to parallel Euclidean planes. the other horospheres, those which come from spheres that are tangent at other points, get mapped to points sets that come from spheres that are tangent to the the bounding hyperplane. Like the horospheres in the unit ball these point sets do not include the point at the boundary and are homeomorphic to Euclidean space not to spheres. Their geometry is still Euclidean as well.

 

[homeomorphic]

Tricky, you are not even reading Thurston carefully. I have that book.

He clearly says that the GEOMETRY of the horospheres is Euclidean. That means it has 0 curvature. He just expects you to realize that the geometry of the whole sphere can't be Euclidean.


And you are not reading wikipedia carefully, either.

"A horosphere has a critical amount of (isotropic) curvature: if the curvature were any greater, the surface would be able to close, yielding a sphere,"

So, according to wikipedia, it is not a sphere, if you read carefully. The very thing you were trying to bring to help you disagrees with you in the very same line you tried to quote. Let's try to be a little more careful. It has zero curvature. If it had greater curvature (i.e. positive curvature), THEN you could get a sphere. But that would not be a horosphere because those have 0 curvature.

 

[TrickyDicky]

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.
A few easy questions about which we need to reach a consensus to have any meaningful discussion:

Do you accept that horospheres are objects in H^3?
Do you agree that a horosphere has a extrinsic positive curvature in H^3?
Do you recognize the formula that relates extrinsic mean curvature and gaussian curvature for submanifolds that are totally umbilical? Do you admit horospheres are totally umbilical surfaces?
Are you able to leave for a moment the Euclidean ambient mindframe?
Do you understand that manifolds and submanifolds are different objects with respect to their Riemannian metric?
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) or that a Clifford torus embedded in S^3 has round metric and that doesn't make it a topological sphere?
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?
Do you see that a globally Euclidean metric means 0 gaussian curvature for manifolds and for submanifolds embedded in Euclidean space?
Do you admit the gauss-bonnet theorem? Can you calculate the Euler characteristic of the horosphere taking into account that it has positive gaussian 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?

Unless you guys address this simple differential geometry questions I don't think we can have a productive discussion without getting lost in semantics.

 

[TrickyDicky]

Since Ben works with strings he might find interesting this quote from an article on strings on page 10:
"Triangulated Surfaces in Twistor Space:A Kinematical Set up for
Open/Closed String Duality" http://arxiv.org/abs/hep-th/0607146

"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(∞)."


So since the horosphere is closed it has no boundary term (it is compact without boundary) and your calculation was incorrect on that too: We only need the integral of the gaussian curvature to obtain the Euler characteristic, and since it has positive gaussian curvature due to its mean curvature +1 and being totally umbilical:

\begin{align}<br /> 2 \pi \, \chi(M) &amp;= \int_Ʃ K \, dV = 4 \pi \\<br /> \chi(M) &amp;= 2<br /> \end{align}<br />

 

[homeomorphic]

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 which 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

 

[homeomorphic]

"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.

 

[TrickyDicky]

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?

 

[TrickyDicky]

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.

 

[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}<br /> 2 \pi \, \chi(M) &amp;= \int_Ʃ K \, dA = \\<br /> &amp;= \lim_{R \to \infty} \int_Ʃ \frac{1}{R^2} \, dA=4\pi \\<br /> \chi(M) &amp;= 2<br /> \end{align}<br />

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.