Riemannian surfaces as one dimensional complex manifolds

lavinia

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

TrickyDicky

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

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, wich is a Euclidean sphere... These horizontal surfaces are examples of horospheres"

homeomorphic

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

homeomorphic

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

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

lavinia

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

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

so you are not sure

TrickyDicky

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 wich 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, wich 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

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 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:

[tex]\begin{align}
2 \pi \, \chi(M) &= \int_Ʃ K \, dV = 4 \pi \\
\chi(M) &= 2
\end{align}
[/tex]