Riemannian surfaces as one dimensional complex manifolds

[Ben Niehoff]

Accordingly I would expect a horospherical surface in a hyperbolic closed (compactified by the Mobius group ) 3-manifold

You forgot some words here. What you mean is quotiented (not compactified) by a discrete subgroup of the Mobius group.

would be homeomorphic to the topology of a surface with positive total curvature (surface integral of the gaussian curvature) and could also have a flat Riemannian metric on it.

I don't think so. If we take some quotient of H^3, any given horosphere will appear as an open disk (polygon-shaped) in the fundamental domain. This open disk must remain flat after taking the quotient, but it also must have some of its polygonal edges identified*. I think the only possible results are a torus or a Klein bottle.

* Actually, the situation is more complicated. For example, if the fundamental domain is a dodecahedron (as in a Seifert-Weber space), then opposite faces must be identified with a twist. Therefore a horosphere section will be twisted so that its edges do not line up. Several horosphere sections in the fundamental domain will end up getting glued together into some kind of surface knot with self-intersections. In any case, since it is flat, this surface knot must be homeomorphic to either the torus or the Klein bottle.

To be clear I'm referring to hyperbolic Riemannian manifolds that include the point at infinity so the horospheres are compact and closed.

Is this more correct?

No, because such things do not exist. Asking for a "hyperbolic Riemannian manifold that includes the sphere at infinity" is like asking for a 10-foot pole that fits in your pocket. The two requirements are contradictory.

- H^3 does not include the sphere at infinity.

- The conformal compactification of H^3 can include the sphere at infinity, but this space is the closed ball in E^3. It has a flat metric, and hence is not hyperbolic.*

- The quotient of H^3 by a discrete group is hyperbolic, compact, closed, and also has finite volume. Hence it does not include any part of the sphere at infinity.


* One can also embed the closed 3-ball into H^3, of course. This will be a hyperbolic manifold with boundary. However, the boundary will not be the sphere at infinity; it will be the sphere at radius 2 (for example). And in this embedding of B^3, flat surfaces will intersect the boundary rather than lie tangent to it.


That's all for now, I'm going on vacation for a week. Lavinia and Homeomorphic know what they're talking about, though.

 

[TrickyDicky]

Ok, have a nice holiday.

 

[TrickyDicky]

- H^3 does not include the sphere at infinity.

This should be clarified:
The Euclidean model that represents hyperbolic space doesn't include it, there is a reason hyperbolic space can't be embedded in euclidean space.

But hyperbolic space as a manifold has a boundary at infinity, namely the Riemann sphere complex manifold (conformal infinity). And its points "belong" to the infinite hyperbolic manifold just like the point at infinity of the extended complex plane is part of it even if it lies at infinite distance.

 

[homeomorphic]

This should be clarified:
The Euclidean model that represents hyperbolic space doesn't include it, there is a reason hyperbolic space can't be embedded in euclidean space.

What do you mean by embedding? If you mean as a Riemannian manifold, then it can't be embedded because the curvature is different. If you mean in a differential topology sense, then, it can be embedded very easily, but then it's an only an open ball, as a smooth manifold with no metric, not H^3.

But hyperbolic space as a manifold has a boundary at infinity, namely the Riemann sphere complex manifold (conformal infinity). And its points "belong" to the infinite hyperbolic manifold just like the point at infinity of the extended complex plane is part of it even if it lies at infinite distance.

H^3, by definition, does not include the points at infinity. The metric on H^3 does not extend to the points at infinity. Proof: the metric defines a distance. If the metric could be extended to the whole ball, then this distance function would be defined on a compact set. Therefore, the distance function would be bounded. This is a contradiction because we know that there are points that are arbitrarily far apart in H^3. Thus, the boundary at infinity cannot be considered part of the Riemannian manifold H^3.

 

[TrickyDicky]

What do you mean by embedding? If you mean as a Riemannian manifold, then it can't be embedded because the curvature is different.

I mean the difficulty lies precisely in representing the boundary at infinity, we have perfect modelds of spherical geometry(manifolds without boudary) that can be embedded in Euclidean spaces even if the curvature is different.

H^3, by definition, does not include the points at infinity. The metric on H^3 does not extend to the points at infinity. Proof: the metric defines a distance. If the metric could be extended to the whole ball, then this distance function would be defined on a compact set. Therefore, the distance function would be bounded. This is a contradiction because we know that there are points that are arbitrarily far apart in H^3. Thus, the boundary at infinity cannot be considered part of the Riemannian manifold H^3.

Maybe this is a bit of a semantic issue or reflects some subtle differences between the language used in topology and in manifolds, I am considering H^3 as the interior of a manifold that has as boundary at infinity the extended complex plane.
From WP:
"The terminology is somewhat confusing: every topological manifold is a topological manifold with boundary, but not vice-versa.

Let M be a manifold with boundary. The interior of M, denoted Int M, is the set of points in M which have neighborhoods homeomorphic to an open subset of Rn. The boundary of M, denoted ∂M, is the complement of Int M in M. The boundary points can be characterized as those points which land on the boundary hyperplane (xn = 0) of Rn+ under some coordinate chart.

If M is a manifold with boundary of dimension n, then Int M is a manifold (without boundary) of dimension n and ∂M is a manifold (without boundary) of dimension n − 1."

The first sentence of this quote that refers to confusing termiology is itself quite confusing by the way.
So I would consider Int M to be H^3 and ∂M to be the conformal boundary at infinity.
So I'm not sure what would be the correct terminology for the topological manifold with boundary, M, that I would consider to include the points in ∂M. I'm not even sure if such M can be metrizable ( I guess it would be locally metrizable only).

 

[homeomorphic]

What do you mean by embedding? If you mean as a Riemannian manifold, then it can't be embedded because the curvature is different.

I mean the difficulty lies precisely in representing the boundary at infinity, we have perfect modelds of spherical geometry(manifolds without boudary) that can be embedded in Euclidean spaces even if the curvature is different.

You missed the point. Do you have an example of a manifold being embedded in a manifold of the same dimension with different curvature? The imbedded sphere is 2-dimensional and you are embedding it in R^3.

Maybe this is a bit of a semantic issue or reflects some subtle differences between the language used in topology and in manifolds, I am considering H^3 as the interior of a manifold that has as boundary at infinity the extended complex plane.

No, topologists like me actually study manifolds (though we typically care less about metrics). That's what we do. The terminology is the same. The division lies WITHIN topology. We use the same word in different ways. However, that is not the relevant issue here.

From WP:
"The terminology is somewhat confusing: every topological manifold is a topological manifold with boundary, but not vice-versa.

This is different from boundary at infinity. This does apply to our ball. It IS a manifold with boundary. However, it is not part of the Riemannian manifold H^3. You have to forget the metric before you can consider the boundary as part of the ball as a manifold with boundary. Or you can consider it as a topological manifold containing a Riemannian manifold. But the Riemannian structure does not extend to the whole thing.

Also, H^3, by definition, and this is a convention, does not include the boundary at infinity.

So I would consider Int M to be H^3 and ∂M to be the conformal boundary at infinity.

Yes, H^3 is the interior. It's not the whole thing.

So I'm not sure what would be the correct terminology for the topological manifold with boundary, M, that I would consider to include the points in ∂M.

Topological manifold with boundary (or smooth manifold with boundary).

I'm not even sure if such M can be metrizable ( I guess it would be locally metrizable only).

Of course, it's metrizable. It's a closed ball in R^3. Subspaces of metric spaces are metric spaces. You can put a Riemannian metric on it, too, but it won't restrict to that of H^3 in the interior.

 

[TrickyDicky]

Thanks Homeomorphic.

You can put a Riemannian metric on it, too, but it won't restrict to that of H^3 in the interior.

I have some difficulty to fully understand this part. Specially in relation with the bit you mentioned about forgetting the metric.
and what do you mean by a Riemannian metric that won't restrict to that of H^3? what kind of Riemannian metric would it be? Wouldn't the topological manifold with boundary (M) be a hyperbolic manifold then? And would it have finite or infinite volume?

 

[homeomorphic]

I have some difficulty to fully understand this part. Specially in relation with the bit you mentioned about forgetting the metric.
and what do you mean by a Riemannian metric that won't restrict to that of H^3?

For example, you could take the metric inherited from R^3, since the ball is sitting in R^3. That is defined on the whole ball, but it is obviously not the same metric as the hyperbolic metric.

what kind of Riemannian metric would it be?

Any metric that a closed ball can have.

Wouldn't the topological manifold with boundary (M) be a hyperbolic manifold then?

No.

You could actually put a metric of constant negative curvature on it, but it wouldn't be the same as H^3. Just imagine shrinking the ball and then giving the shrunken ball the metric inherited from H^3. But that would be a proper subset of H^3. It cannot contain a copy of H^3 because it is compact, as I argued earlier.

And would it have finite or infinite volume?

Finite volume. Compact manifolds have to have finite volume. This is because the volume is locally finite--you can choose a neighborhood of each point that has finite volume. That gives you an open cover, which has to have a finite subcover. Thus, the volume is finite.

 

[TrickyDicky]

To be sure , is there such thing in math as a hyperbolic manifold with boundary of infinite volume?

 

[homeomorphic]

To be sure , is there such thing in math as a hyperbolic manifold with boundary of infinite volume?

Yes. Take half of hyperbolic space.

The problem is not having boundary. The problem is compactness.

 

[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.