Riemannian surfaces as one dimensional complex manifolds

[homeomorphic]

is it your impression that a conformal surface is always orientable? if the definition is a surface given an equivalence class of riemannian metrics at each point under scaling, it seems any riemannian surface determines a conformal surface. that would seem to include non orientable ones, another argument that the statement in WP that their geometry is precisely that of riemann surfaces is false. i.e. it would seem the klein bottle has a conformal structure.

I don't think so. Why not consider a conformal structure on the Klein bottle? It's probably false, but all you need to do to fix it is say that the geometry of Riemann surfaces is the oriented case. I guess you can't get a complex structure if you're non-orientable (since a complex structure determines an orientation), so you don't have the same connection with complex analysis anymore, but I don't see anything wrong with considering metrics up to scale, which was their definition.

All manifolds are compact, oriented, connected...whatever you want them to be. The definition sometimes shifts depending on the writer or speaker's willingness to keep saying all the words that they need.

 

[mathwonk]

i agree it may be true to say that oriented conformal geometry agrees with the geometry of riemann surfaces, but that is not what they said - they said all conformal geometry is the same as the geometry of riemann surfaces. i agree that is an easy fix, but we are very picky in math, if there i an easy fix then make that fix before speaking. there is true and there is false, this is not horseshoes. or else do as i do, and admit one is not an expert.

 

[homeomorphic]

I was being sarcastic.

It's okay to fudge things a little if everyone involved understands what's going on (a big if) or if you are clear on the fact that you are fudging. Which isn't really the case, here. I agree it should be fixed. I was just pointing out that the error was a small one. Wikipedia works pretty well most of the time, but I guess we found some cracks.

Personally, it's hard for me to say I'm an expert on anything, except maybe a couple very very specialized topics.

 

[TrickyDicky]

This is a bit OT but I'll give it a shot.

is it your impression that a conformal surface is always orientable? if the definition is a surface given an equivalence class of riemannian metrics at each point under scaling, it seems any riemannian surface determines a conformal surface. that would seem to include non orientable ones, another argument that the statement in WP that their geometry is precisely that of riemann surfaces is false. i.e. it would seem the klein bottle has a conformal structure.

Can you give any example of a non orientable conformal surface? I don't think the klein bottle has a conformal structure.
Wikipedia says:
"A complex structure gives rise to a conformal structure by choosing the standard Euclidean metric given on the complex plane and transporting it to X by means of the charts. Showing that a conformal structure determines a complex structure is more difficult"

"Complex manifolds are canonically oriented, not just orientable".

Taking all this into account, I think the Wikipedia claim is basically correct if it is restricted, as it is, to conformal geometry in two real dimensions if we agree that riemann surfaces are complex manifolds.
But if you were able to provide counterexamples like showing how the klein bottle has a conformal structure that would be interesting.

Going back to topic for a minute, I'd like to say again that the topology of the extended complex plane (positive curvature) is preserved by a flat metric if we think about it in terms of relative curvature wrt the ambient space so it keeps being homeomorphic to the topology of the sphere, this means IMO there can be a globally flat metric on the extended plane that induces the usual sphere topology if the ambient space negative curvature allows the euclidean metric to be considered positively curved wrt to it.

 

[lavinia]

i agree it may be true to say that oriented conformal geometry agrees with the geometry of riemann surfaces, but that is not what they said - they said all conformal geometry is the same as the geometry of riemann surfaces. i agree that is an easy fix, but we are very picky in math, if there i an easy fix then make that fix before speaking. there is true and there is false, this is not horseshoes. or else do as i do, and admit one is not an expert.

Since I am still reading this thread I would like to rejoin if that is OK. My apologies to Tricky and everyone else for getting frustrated.

There are many conformal structures on the Klein bottle (by which I mean a smooth surface that is homeomorphic to a topological Klein bottle).

One may obtain different conformal flat Klein bottles from different conformal flat tori. Given a flat torus mod out by an isometry that is a fixed point free involution to get a flat Klein bottle. For example, if the flat torus is obtained from the standard lattice in the plane then add the transformation (x,y) -> (x+1/2,-y). Note that the square of this transformation is a lattice point so it projects to an involution of the flat torus. In fact it projects to a fixed point free isometry of the flat torus so the quotient is a flat Klein bottle. For other lattices the procedure is the same.

Conversely, starting with a flat Klein bottle there is a flat torus that covers it by a local isometry. So conformal classes of flat Klein bottles correspond to conformal classes of flat tori.

 

[lavinia]

here are some more comments that confused me recently. what does it mean to have an isometric embedding of a manifold into euclidean space?

consider the sphere in euclidean space. it has two natural metrics (not riemannian metrics, but metric space metrics), both induced by the embedding, namely we can measure the distance between two points by using the distance between them as points of R^3, or we can use the distance measured along a great circle on the sphere.

since the second one is induced by the notion of lengths of curves on the sphere, it is the metric on the sphere, induced by the riemannian metric on the sphere. now suppose we think an isometric embedding is an embedding that preserves the metric, rather than the riemannian metric. if this is the case, and we ask for such an embedding which takes the second metric of the sphere to the intrinsic metric of euclidean space, then there is no such embedding.

i.e. there is no isometric embedding of the usual sphere in R^3, if by that you mean the embedding takes the metric measured along great circles, to the restriction of the euclidean metric. that is because that would require geodesics in the riemannian metric to map to geodesics in the embedded metric, which would mean all great circles would map to straight lines.

although that does happen for stereographic projection, that projection does not preserve lengths. i.e. under projection, finite great circles go to infinite straight lines.

so apparently an isometric embedding of riemannian manifolds means one such that the riemannian metric is preserved, but not the metric induced by the riemannian metric. i am a little over the line from dinner so could be wrong even here. but it is confusing hey?>

For a smooth Riemannian manifold, an isometric embedding means that the inner product on the tangent space is preserved. Another way of saying this is that curves on the manifold have the same length after the embedding. Isometry means that the distance measure on the manifolds is preserved.

- If a manifold is already embedded in a Riemannian manifold of higher dimension then it inherits a Riemannian metric on its tangent space from the higher dimensional manifold. With this metric it is embedded isometrically in the bigger manifold. I am used to the terminology which says that this inherited metric is "induced".

If the map is merely an immersion, then the manifold also inherits a metric because the differential of an immersion is injective on each tangent plane.

- This idea of induced metrics generalizes to Riemannian vector bundles. A vector bundle,E, over a manifold is Riemannian if it has an inner product on each fiber that varies smoothly across the manifold.

An arbitrary vector bundle with a Riemannian metric still has a curvature tensor and still has an idea of parallel translation. But there is in general no way to translate this into a distance measure on the manifold. Therefore, a Riemannian metric on a vector bundle over a manifold in general does not induce a topology. But the topology already exists - otherwise you could not define the vector bundle in the first place.

In the case of the tangent bundle one does get a distance measure - the distance measure determined by geodesics. One can prove that it determines the same topology that the manifold already has. This is a basic theorem of Riemannian geometry.

So any Riemannian metric on a manifold determines the same topology.- When one says that a manifold is embedded in another manifold one is already assuming that the manifold has a topology. Embedding literally means that the map is smooth and injective, and that its image is surrounded by a tubular neighborhood. This means – prove it for yourself – that the image set of the embedding in the subspace topology is diffeomorphic to the original manifold. Its topology therefore will be the same as the subspace topology it has with respect to any metric that determines the topology of the ambient manifold.. The point is that embedding is an idea of differential topology. There is absolutely no need for any metric.

 

[lavinia]

This is a bit OT but I'll give it a shot.

Going back to topic for a minute, I'd like to say again that the topology of the extended complex plane (positive curvature) is preserved by a flat metric if we think about it in terms of relative curvature wrt the ambient space so it keeps being homeomorphic to the topology of the sphere, this means IMO there can be a globally flat metric on the extended plane that induces the usual sphere topology if the ambient space negative curvature allows the euclidean metric to be considered positively curved wrt to it.

There can not be a globally flat metric on the extended plane if by metric you mean a Riemannian metric on its tangent space and by flat you mean that the Gauss curvature is zero.

In hyperbolic 3 space, as Ben demonstrated by direct calculation, the horosphere has a flat Riemannian metric. Therefore this metric can not be extended to the extended complex plane.

 

[homeomorphic]

Can you give any example of a non orientable conformal surface? I don't think the klein bottle has a conformal structure.

The definition of conformal structure that they give is a metric up to scale. That means, yes, the Klein bottle has a conformal structure on it, since the Klein bottle has a metric on it (as does any manifold, by a partition of unity argument or by embedding it in R^n). And this makes perfect sense. Conformal mappings are just about angles. If you have a metric up to scale, you have angles, so you can talk about conformal mappings.

Wikipedia says:
"A complex structure gives rise to a conformal structure by choosing the standard Euclidean metric given on the complex plane and transporting it to X by means of the charts. Showing that a conformal structure determines a complex structure is more difficult"

There's something fishy about that because a conformal structure is supposed to be a globally defined metric, up to scale, but they are trying to define it using a local, coordinate-dependent procedure. A conformal structure does not determine a complex structure if the surface is not orientable, and, as I have argued above, by their definition, the Klein bottle most certainly has a conformal structure. It may be that a conformal structure determines a complex structure if the surface is oriented.

"Complex manifolds are canonically oriented, not just orientable".

Yes, as I have mentioned.

Taking all this into account, I think the Wikipedia claim is basically correct if it is restricted, as it is, to conformal geometry in two real dimensions if we agree that riemann surfaces are complex manifolds.
But if you were able to provide counterexamples like showing how the klein bottle has a conformal structure that would be interesting.

I just did. Embed it in R^5 by Whitney's embedding theorem. It then inherits a metric from R^5. Consider it up to scale. Done.

Going back to topic for a minute, I'd like to say again that the topology of the extended complex plane (positive curvature)

You're using the word topology in a very problematic way. Of course, you're doing so in reference to Gauss-Bonnet. You shouldn't say that the topology has positive curvature. What could that possibly mean? Curvature is defined in terms of structures other than the topology. Plus, the sphere can have different metrics, and they don't have to be positively curved. It's just that the integral of the curvature over the sphere is positive. You could embed a sphere so it has lots of saddle points and negative curvature, locally. But, of course, the metrics induced by these embeddings aren't the standard metric on the sphere. The usual one is the one inherited from R^3 when you consider it as the unit sphere. That one has constant positive curvature. There's only one topological 2-sphere. But there are many different Riemannian 2-spheres.

is preserved by a flat metric

How can topology be preserved by a metric?

if we think about it in terms of relative curvature wrt the ambient space so it keeps being homeomorphic to the topology of the sphere,

Any metric will induce the same topology, the standard one on the sphere. Maybe that's what you are trying to say?

this means IMO there can be a globally flat metric on the extended plane that induces the usual sphere topology

As we've been trying to say, the topology is already there on the sphere. Any metric on the sphere induces the topology that it already has. The topology comes PRIOR to the metric. So, what you are saying is a tautology (the globally flat part doesn't sound right, though, but I think we don't know what you mean).

if the ambient space negative curvature allows the euclidean metric to be considered positively curved wrt to it.

 

[TrickyDicky]

Since I am still reading this thread I would like to rejoin if that is OK. My apologies to Tricky and everyone else for getting frustrated.

Welcome back Lavinia.

There can not be a globally flat metric on the extended plane if by metric you mean a Riemannian metric on its tangent space and by flat you mean that the Gauss curvature is zero.

Ok, this is understood. Let's change flat by euclidean and let's think about a situation where a euclidean surface could have a extrinsic positive curvature, for instance a negatively curved ambient space, what would be wrong in this picture in your opinion?

The definition of conformal structure that they give is a metric up to scale. That means, yes, the Klein bottle has a conformal structure on it, since the Klein bottle has a metric on it (as does any manifold, by a partition of unity argument or by embedding it in R^n). And this makes perfect sense. Conformal mappings are just about angles. If you have a metric up to scale, you have angles, so you can talk about conformal mappings.

Ok, let me try to ask again: Can you give any example of a non orientable conformal surface in R^3? I'm not trying to cheat by adding this but I think the WP page is referring to euclidean ambient space because it defines orientability as:" a property of surfaces in Euclidean space". Not trying to defend wikipedia here, just pointing out that in this case there seems to be a contextual ambiguity.

There's something fishy about that because a conformal structure is supposed to be a globally defined metric, up to scale, but they are trying to define it using a local, coordinate-dependent procedure. A conformal structure does not determine a complex structure if the surface is not orientable, and, as I have argued above, by their definition, the Klein bottle most certainly has a conformal structure. It may be that a conformal structure determines a complex structure if the surface is oriented.

Ok,I think this is related with what I write above on the context of conformal surfaces in Euclidean space and the klein bottle impossibility to be embedded in euclidean space (R^3). It seems in R^3 all surfaces conformal structures are orientable and therefore determine complex structures.

I just did. Embed it in R^5 by Whitney's embedding theorem. It then inherits a metric from R^5. Consider it up to scale. Done.

See above.

You're using the word topology in a very problematic way. Of course, you're doing so in reference to Gauss-Bonnet. You shouldn't say that the topology has positive curvature. What could that possibly mean? Curvature is defined in terms of structures other than the topology. Plus, the sphere can have different metrics, and they don't have to be positively curved. It's just that the integral of the curvature over the sphere is positive. You could embed a sphere so it has lots of saddle points and negative curvature, locally. But, of course, the metrics induced by these embeddings aren't the standard metric on the sphere. The usual one is the one inherited from R^3 when you consider it as the unit sphere. That one has constant positive curvature. There's only one topological 2-sphere. But there are many different Riemannian 2-spheres.

Yes, I'm using this problematic way of dealing with topology in reference to Gauss-Bonnet, rather than to the "Riemannian metric" use of curvature. I see you caught my drift here.

How can topology be preserved by a metric?

Never mind this, bad choice of words again. What I meant was the topology is of course not changed by embedding in a hyperbolic manifold, but the metric induced on that topology by the Negatively curved ambient Riemannian metric could change and admit a euclidean metric on the topology.

As we've been trying to say, the topology is already there on the sphere. Any metric on the sphere induces the topology that it already has. The topology comes PRIOR to the metric. So, what you are saying is a tautology (the globally flat part doesn't sound right, though, but I think we don't know what you mean).

I see this, that's why I keep talking about the "one-point compactification of the complex plane" topology, but of course my expressions lack mathematical rigor.

I know the globally flat part doesn't sound right, I'm not sure about it either but all the time I've had the feeling that I'm not being able to get across what I mean, I don't mind being wrong but it bothers me not to be able to express what i mean in a sound mathematical way, so I guess I have to study this harder.

 

[lavinia]

The Mobius band is a non-orientable surface in 3 space. It naturally has a flat metric. There are no closed surfaces without boundary in 3 space that are unorientable. For instance the Klein bottle can not be embedded in 3 space.

 

[Ben Niehoff]

I will try to help if you're willing to work with me.

Ok, this is understood. Let's change flat by euclidean and let's think about a situation where a euclidean surface could have a extrinsic positive curvature, for instance a negatively curved ambient space, what would be wrong in this picture in your opinion?

I haven't the faintest idea how to answer this question, because those words do not mean anything when put together in that order.

There are many extrinsic curvatures a surface can have, depending on the dimension of the ambient manifold. For example, I can pinch the north and south poles of a sphere, pull on them, and twist them by any arbitrary amount. All of these are measured by various extrinsic curvatures.

There is one special extrinsic curvature, the Gauss curvature, which Gauss proved is actually intrinsic; that is, independent of the ambient manifold into which the surface is embedded. From the extrinsic point of view, the Gauss curvature is the product of the principal curvatures (which are curvatures of curves on the surface, measured with respect to the ambient space). From the intrinsic point of view, the Gauss curvature is one-half the Ricci scalar intrinsic to the surface. But the two are always equal to each other. This is why it is quite strange to me (and likely everyone else) that you keep harping on embeddings in H^3 being somehow different from embeddings in R^3. They are not.

Ok, let me try to ask again: Can you give any example of a non orientable conformal surface in R^3? I'm not trying to cheat by adding this but I think the WP page is referring to euclidean ambient space because it defines orientability as:" a property of surfaces in Euclidean space". Not trying to defend wikipedia here, just pointing out that in this case there seems to be a contextual ambiguity.

I think the simple explanation here is that Wikipedia is wrong (or in this case, incomplete). Orientability is a property that can be defined without reference to any ambient space. An n-dimensional manifold is orientable if and only if it admits a nowhere-vanishing n-form.

In two dimensions, it turns out that a nowhere-vanishing 2-form can also be interpreted as a complex structure. Hence all orientable 2-surfaces are complex manifolds (in fact, they are Kaehler).

Homeomorphic said:

You're using the word topology in a very problematic way. Of course, you're doing so in reference to Gauss-Bonnet. You shouldn't say that the topology has positive curvature. What could that possibly mean? Curvature is defined in terms of structures other than the topology. Plus, the sphere can have different metrics, and they don't have to be positively curved. It's just that the integral of the curvature over the sphere is positive. You could embed a sphere so it has lots of saddle points and negative curvature, locally. But, of course, the metrics induced by these embeddings aren't the standard metric on the sphere. The usual one is the one inherited from R^3 when you consider it as the unit sphere. That one has constant positive curvature. There's only one topological 2-sphere. But there are many different Riemannian 2-spheres.

--------------------

Yes, I'm using this problematic way of dealing with topology in reference to Gauss-Bonnet, rather than to the "Riemannian metric" use of curvature. I see you caught my drift here.

I think you've missed Homeomorphic's point. A sphere can have different metrics, yes; but the total curvature of any metric on the sphere must be positive! You can imagine pulling on the ends of a sphere and pinching it in the middle to give it regions of negative curvature, but the regions of positive curvature at the ends must win out when you integrate over the whole sphere. This is a basic fact of topology, a direct consequence of the Gauss-Bonnet theorem; it does not depend on the embedding and hence you cannot negate this fact by embedding the sphere in H^3 or anywhere else.

The total curvature of any topological sphere must be positive.

Never mind this, bad choice of words again. What I meant was the topology is of course not changed by embedding in a hyperbolic manifold, but the metric induced on that topology by the Negatively curved ambient Riemannian metric could change and admit a euclidean metric on the topology.

This is where you've gone wrong. These things called "horospheres" are not, in fact, embeddings of the sphere into H^3. Look closely at the definition of "embedding". The embedded manifold must map entirely into the ambient space, and there must be a tubular neighborhood around it in the ambient space.

Using the Poincare ball model, H^3 is the open ball on the interior. It does not include the boundary. In fact, H^3 is homeomorphic to R^3; it has no boundary! (More on that later).

Looking at a horosphere in the Poincare ball model, the horosphere is represented by what looks like a round sphere touching the boundary. However, one point of this round sphere lies on the boundary of the ball model, and therefore does not properly lie within H^3. Therefore, a horosphere is not actually an embedding of the sphere into H^3, because there is one point of the sphere which is not inside H^3.

More properly speaking, the horosphere is the embedding in H^3 of a "sphere with one point removed". Or in other words, it is a 1-point de-compactification of the sphere; i.e., a horosphere is actually an embedding of the infinite plane! As I have pointed out before, this is why it is possible for the horosphere to be globally flat.



OK, so then what is the boundary of the Poincare ball model? This is easiest to explain after we see what the Poincare ball model is.

Start with a more natural model of hyperbolic space: the hyperboloid model. The hyperboloid model is easiest to visualize from by embedding it isometrically in Minkowski space. Starting with the metric

ds_4^2 = dx^2 + dy^2 + dz^2 - dw^2
consider one sheet of the hyperboloid

w^2 - x^2 - y^2 - z^2 = 1
This is a 3-hyperboloid that lies entirely within the future light cone (and asymptotically approaches it at infinity). If we compute the induced metric on this hyperboloid, we get

ds_3^2 = d\rho^2 + \sinh^2 \rho \, d\theta^2 + \sinh^2 \rho \, \sin^2 \theta \, d\phi^2
The reason I call this a more natural model of H^3 is because now it is clear that H^3 is homogenous, isotropic, and infinite in extent in all directions. Also, it embeds isometrically within the future lightcone, so you can imagine what it looks like without distorting distances (whereas, the Poincare ball model must distort distances in order to fit all of H^3 within a ball).

To map between this model and the Poincare ball model, one actually uses stereographic projection! Unfortunately I can't draw a picture here. But imagine lines emanating from the origin of Minkowski space and intersecting the H^3 hyperboloid. Then map each point of H^3 to the unique point where each of these lines intersects the hyperplane w = 1. It is easy to show that the result is the Poincare ball model, with the usual metric.

We can also easily see that the "boundary sphere" of the Poincare ball model is the image of the lightcone itself under this stereographic projection. Since stereographic projection is a conformal transformation, we can call this boundary a "conformal boundary". Under stereographic projection, H^3 itself maps to an open ball in E^3, whereas the lightcone maps to the boundary of this ball. To obtain a closed ball, we must add a whole 2-sphere's worth of points to H^3.

It is important to realize that H^3 itself is not a closed ball. To make it into one, we must conformally compactify H^3 and add a sphere's worth of points. The result turns H^3 into a closed ball in Euclidean space. Therefore if we want to consider spheres in the Poincare ball model that actually touch the boundary, then we can't use the hyperbolic metric to discuss them! After the conformal compactification required to give us the boundary, we will have a Euclidean metric inside the ball, and the spheres touching the boundary are simply ordinary, round spheres.

Alternatively, we can de-compactify, which turns the open interior of the ball into an honest H^3, and turns the horospheres into honest R^2's.

 

[mathwonk]

this is becoming more and more fun! thanks for the question, tricky. this is the way real math conversations go, when everyone is feeling her/his way, (with all respect).

 

[lavinia]

this is becoming more and more fun! thanks for the question, tricky. this is the way real math conversations go, when everyone is feeling her/his way, (with all respect).

I thought about the metric on the sphere obtained from the usual Euclidean distance that you mentioned. I think that it can not be a metric that is derived from a Riemannian matric on the tangent bundle. The reason I think is that the distance between near by points along a geodesic must be additive. The distance from a to c must equal the distance from a to b plus the distance from b to c. But it seems with this metric and any curve the sum of the distances is always greater. What do you think?

 

[lavinia]

Sometimes an embedding does determine a Riemannain metric that is different than the one that the manifold inherits by restricting the ambient inner product to its tangent planes. Take the case of a surface embedded in Euclidean 3-space that has positive Gauss curvature. Then the eigen values of the second fundamental form are strictly positive so they determine a new Riemannian metric. I do not know if the manifold with this new Riemannian metric can be embedded in 3 space. I guess if the change is small it will still have positive curvature.

 

[homeomorphic]

I see this, that's why I keep talking about the "one-point compactification of the complex plane" topology, but of course my expressions lack mathematical rigor.

The one-point compactification gives you the same topology as usual. Makes no difference topologically whether it's the one point compactification of ℂ or the unit sphere as a subspace of ℝ^3. The two are homeomorphic, via stereographic projection.

In two dimensions, it turns out that a nowhere-vanishing 2-form can also be interpreted as a complex structure.

I think you mean in the presence of a suitable Riemannian metric. The 2-form by itself is just a symplectic structure. With a compatible metric, you can then define an almost complex structure, which will turn out to be integrable, so you get a complex structure (actually, a Kahler structure). An orientable surface only has 2 possible orientations, but it has lots and lots of complex structures--actually, a whole moduli space of them.

 

[Ben Niehoff]

I think you mean in the presence of a suitable Riemannian metric. The 2-form by itself is just a symplectic structure. With a compatible metric, you can then define an almost complex structure, which will turn out to be integrable, so you get a complex structure (actually, a Kahler structure). An orientable surface only has 2 possible orientations, but it has lots and lots of complex structures--actually, a whole moduli space of them.

Yes, I agree.

 

[TrickyDicky]

I will try to help if you're willing to work with me.

Sure I'm willing, I'm clarifying lots of stuff here, I admit that includes being frustrating at times, learning is hard and more so in a forum, but as mathwonk says it can be fun too.

There is one special extrinsic curvature, the Gauss curvature, which Gauss proved is actually intrinsic; that is, independent of the ambient manifold into which the surface is embedded. From the extrinsic point of view, the Gauss curvature is the product of the principal curvatures (which are curvatures of curves on the surface, measured with respect to the ambient space). From the intrinsic point of view, the Gauss curvature is one-half the Ricci scalar intrinsic to the surface. But the two are always equal to each other. This is why it is quite strange to me (and likely everyone else) that you keep harping on embeddings in H^3 being somehow different from embeddings in R^3. They are not

I really agree with this. I'll try asking specific questions, so I can get closer to what I mean, but certainly I am aware that the topology is not affected by the embedding, my doubt is about metrics, as to whether a euclidean 2-dimensional metric can be thought of as having positive curvature wrt the ambient space or not. Can this be directly addressed by anyone?

I think the simple explanation here is that Wikipedia is wrong (or in this case, incomplete). Orientability is a property that can be defined without reference to any ambient space. An n-dimensional manifold is orientable if and only if it admits a nowhere-vanishing n-form.

I can see now that mathwonk was completely right to complain about this.

A sphere can have different metrics, yes; but the total curvature of any metric on the sphere must be positive! You can imagine pulling on the ends of a sphere and pinching it in the middle to give it regions of negative curvature, but the regions of positive curvature at the ends must win out when you integrate over the whole sphere. This is a basic fact of topology, a direct consequence of the Gauss-Bonnet theorem; it does not depend on the embedding and hence you cannot negate this fact by embedding the sphere in H^3 or anywhere else.

The total curvature of any topological sphere must be positive.

Fully agree, see above

This is where you've gone wrong. These things called "horospheres" are not, in fact, embeddings of the sphere into H^3.

Ok, I see this. It is also true I've made an effort to differentiate between the H^3 space and the H^3 manifold (quotient space) that includes the boundary. Can then be said according to you that a horosphere is not embedded in the H^3 manifold (not just the H^3 space)? And I would like for you or someone to comment on the fact that the Riemann sphere is the conformal boundary 2-manifold of the hyperbolic 3-manifold. And correct me if I'm wrong here but the R^3 representation of this boundary is a sphere (see Poincare ball model of H^3). Are horospheres not homeomorphic to this conformal boundary?

 

[TrickyDicky]

Using the Poincare ball model, H^3 is the open ball on the interior. It does not include the boundary. In fact, H^3 is homeomorphic to R^3; it has no boundary! (More on that later).

Looking at a horosphere in the Poincare ball model, the horosphere is represented by what looks like a round sphere touching the boundary. However, one point of this round sphere lies on the boundary of the ball model, and therefore does not properly lie within H^3. Therefore, a horosphere is not actually an embedding of the sphere into H^3, because there is one point of the sphere which is not inside H^3.

More properly speaking, the horosphere is the embedding in H^3 of a "sphere with one point removed". Or in other words, it is a 1-point de-compactification of the sphere; i.e., a horosphere is actually an embedding of the infinite plane! As I have pointed out before, this is why it is possible for the horosphere to be globally flat.
OK, so then what is the boundary of the Poincare ball model? This is easiest to explain after we see what the Poincare ball model is.

Start with a more natural model of hyperbolic space: the hyperboloid model. The hyperboloid model is easiest to visualize from by embedding it isometrically in Minkowski space. Starting with the metric

ds_4^2 = dx^2 + dy^2 + dz^2 - dw^2
consider one sheet of the hyperboloid

w^2 - x^2 - y^2 - z^2 = 1
This is a 3-hyperboloid that lies entirely within the future light cone (and asymptotically approaches it at infinity). If we compute the induced metric on this hyperboloid, we get

ds_3^2 = d\rho^2 + \sinh^2 \rho \, d\theta^2 + \sinh^2 \rho \, \sin^2 \theta \, d\phi^2
The reason I call this a more natural model of H^3 is because now it is clear that H^3 is homogenous, isotropic, and infinite in extent in all directions. Also, it embeds isometrically within the future lightcone, so you can imagine what it looks like without distorting distances (whereas, the Poincare ball model must distort distances in order to fit all of H^3 within a ball).

To map between this model and the Poincare ball model, one actually uses stereographic projection! Unfortunately I can't draw a picture here. But imagine lines emanating from the origin of Minkowski space and intersecting the H^3 hyperboloid. Then map each point of H^3 to the unique point where each of these lines intersects the hyperplane w = 1. It is easy to show that the result is the Poincare ball model, with the usual metric.

We can also easily see that the "boundary sphere" of the Poincare ball model is the image of the lightcone itself under this stereographic projection. Since stereographic projection is a conformal transformation, we can call this boundary a "conformal boundary". Under stereographic projection, H^3 itself maps to an open ball in E^3, whereas the lightcone maps to the boundary of this ball. To obtain a closed ball, we must add a whole 2-sphere's worth of points to H^3.

Nicely explained, thanks, it agrees with my previous knowledge about hyperbolic geometry.

It is important to realize that H^3 itself is not a closed ball. To make it into one, we must conformally compactify H^3 and add a sphere's worth of points. The result turns H^3 into a closed ball in Euclidean space. Therefore if we want to consider spheres in the Poincare ball model that actually touch the boundary, then we can't use the hyperbolic metric to discuss them! After the conformal compactification required to give us the boundary, we will have a Euclidean metric inside the ball, and the spheres touching the boundary are simply ordinary, round spheres.

Alternatively, we can de-compactify, which turns the open interior of the ball into an honest H^3, and turns the horospheres into honest R^2's.

You are describing precisely the back and forth between a round metric for the stereographic projection of the extended complex plane in E^3 and a flat metric in H^3 that I've been talking about all the time.
Just one point, as I said in a hyperbolic 3-manifold I would say we'd have a compact conformal boundary that will not require the space in the ball to be euclidean, as it is evident the ambient space in a hyperbolic manifold has a hyperbolic metric.

Maybe it is worth recalling here that there are some differences about some notions in topology and in manifold theory, for instance the concept of boundary in topology differs slightly from that in manifolds, or the concepts of closed and open in certain contexts.

 

[Ben Niehoff]

I'll try asking specific questions, so I can get closer to what I mean, but certainly I am aware that the topology is not affected by the embedding, my doubt is about metrics, as to whether a euclidean 2-dimensional metric can be thought of as having positive curvature wrt the ambient space or not. Can this be directly addressed by anyone?

No, the curvature of a Riemannian metric (assuming Levi-Civita connection) is an intrinsic property; it doesn't care about the ambient space. If you choose some other connection, it is still an intrinsic property, because the connection is an intrinsic object as well. If the surface is isometrically embedded, then its extrinsic Gauss curvature will agree with its intrinsic Ricci scalar.

Ok, I see this. It is also true I've made an effort to differentiate between the H^3 space and the H^3 manifold (quotient space) that includes the boundary. Can then be said according to you that a horosphere is not embedded in the H^3 manifold (not just the H^3 space)? And I would like for you or someone to comment on the fact that the Riemann sphere is the conformal boundary 2-manifold of the hyperbolic 3-manifold. And correct me if I'm wrong here but the R^3 representation of this boundary is a sphere (see Poincare ball model of H^3). Are horospheres not homeomorphic to this conformal boundary?

In this context, "space" and "manifold" mean the same thing, so I don't know what you're trying to say. Also, there's no quotient space anywhere in sight...to form a quotient space, you would quotient out by some family of submanifolds, which isn't being discussed here. (However, for example, the double torus and higher-genus Riemann surfaces can be thought of as quotients of H^2 by a discrete symmetry group).

H^3 as a manifold does not include the conformal boundary. It can't include the conformal boundary, because you can't just add points that are infinitely far away (any interval of infinite length must be open). You can add these points to the conformal compactification of H^3; but the conformal compactification of H^3 is honestly an open ball in E^3. This is analogous to the fact that the conformal compactification of R^2 is honestly an open region of a round sphere. The conformal transformation that compactifies the space actually changes the curvature. It makes H^3 flat, and it makes R^2 curved.

In fact, the coordinates in the Poincare ball model can be thought of as exactly the map from H^3 to E^3 that accomplishes this conformal compactification.

It's easy to see that conformal compactifications change curvature: just stick a conformal factor in front of the metric and recompute the curvature, you'll find it changes in a simple way:

http://en.wikipedia.org/wiki/Ricci_curvature#Behavior_under_conformal_rescaling

 

[TrickyDicky]

No, the curvature of a Riemannian metric (assuming Levi-Civita connection) is an intrinsic property; it doesn't care about the ambient space. If you choose some other connection, it is still an intrinsic property, because the connection is an intrinsic object as well.

Maybe you'll understand better my question if I remind you that I was not referring in this case to a Riemanninan metric but to the conformal geometry class of metrics since I'm considering the extended complex plane as a conformal manifold rather than a Riemannian manifold. So these facts apply according to WP:"Conformal geometry has a number of features which distinguish it from (pseudo-)Riemannian geometry. The first is that although in (pseudo-)Riemannian geometry one has a well-defined metric at each point, in conformal geometry one only has a class of metrics. Thus the length of a tangent vector cannot be defined, but the angle between two vectors still can. Another feature is that there is no Levi-Civita connection because if g and λ2g are two representatives of the conformal structure, then the Christoffel symbols of g and λ2g would not agree. Those associated with λ2g would involve derivatives of the function λ whereas those associated with g would not."

If the surface is isometrically embedded, then its extrinsic Gauss curvature will agree with its intrinsic Ricci scalar.

Right.

In this context, "space" and "manifold" mean the same thing, so I don't know what you're trying to say. Also, there's no quotient space anywhere in sight...to form a quotient space, you would quotient out by some family of submanifolds, which isn't being discussed here.

In the context that I'm presenting it is of great importance to distinguish a space like H^3 from an hyperbolic 3-manifold. My question from the beginning is about manifolds, complex manifolds as submanifolds of hyperbolic manifolds. What do you mean there is no quotient space in sight?: WP:
"A hyperbolic 3-manifold is a 3-manifold equipped with a complete Riemannian metric of constant sectional curvature -1. In other words, it is the quotient of three-dimensional hyperbolic space by a subgroup of hyperbolic isometries acting freely and properly discontinuously." This subgroup happens to be isomorphic to the the automorphism group of the Riemann sphere.

H^3 as a manifold does not include the conformal boundary. It can't include the conformal boundary, because you can't just add points that are infinitely far away (any interval of infinite length must be open). You can add these points to the conformal compactification of H^3; but the conformal compactification of H^3 is honestly an open ball in E^3. This is analogous to the fact that the conformal compactification of R^2 is honestly an open region of a round sphere. The conformal transformation that compactifies the space actually changes the curvature. It makes H^3 flat, and it makes R^2 curved.

Do you mean that a hyperbolic 3-manifold can't have a 2-manifold compact boundary? This boundary manifold being what you call the conformal compactification of H^3.

It's easy to see that conformal compactifications change curvature: just stick a conformal factor in front of the metric and recompute the curvature, you'll find it changes in a simple way:

http://en.wikipedia.org/wiki/Ricci_curvature#Behavior_under_conformal_rescaling

I looked at the link and it says:"For two dimensional manifolds, the above formula shows that if f is a harmonic function, then the conformal scaling g ↦ e2ƒg does not change the Ricci curvature."

 

[Ben Niehoff]

My previous response assumed you were talking about H^3. In your above post, you have changed all the definitions, quite literally changing the rules of the game.

You must be clear what you're talking about in the first place.

 

[Ben Niehoff]

I do know a bit about conformal geometry of surfaces, since this is important to string theory. I can try to respond to your new questions later.

For now, it might help for you to go back and read some of Mathwonk's and Homeomorphic's earlier comments concerning local vs. global conformal transformations. I think the discussion on Wikipedia is incomplete again. Wikipedia seems to focus on conformal classes of local metrics, but these kinds of conformal transformations do not preserve global topology.

For now, I can respond to this:

In the context that I'm presenting it is of great importance to distinguish a space like H^3 from an hyperbolic 3-manifold. My question from the beginning is about manifolds, complex manifolds as submanifolds of hyperbolic manifolds. What do you mean there is no quotient space in sight?: WP:
"A hyperbolic 3-manifold is a 3-manifold equipped with a complete Riemannian metric of constant sectional curvature -1. In other words, it is the quotient of three-dimensional hyperbolic space by a subgroup of hyperbolic isometries acting freely and properly discontinuously." This subgroup happens to be isomorphic to the the automorphism group of the Riemann sphere.

You need to make up your mind. The Poincare ball model is a model of H^3, which is the quotient of H^3 by the trivial group. That's why I say there are no quotient manifolds in play.

If you do quotient H^3 by a discrete group, you get a closed manifold without boundary (and with constant negative curvature). This is analogous to the quotient of R^2 by a discrete group (giving you, e.g., the torus), or the quotient of H^2 by a discrete group (giving you the double torus and higher-genus Riemann surfaces).

You can imagine a quotient of H^3 by a discrete group as a tesselation of H^3 by regular polyhedra, where the quotient manifold is one such polyhedron with various faces identified.

 

[lavinia]

For a surface embedded in a 3 dimensional Riemannian manifold calculation shows that the Riemann curvature tensors of the 3 dimensional manifold and the surface are related by a formula.

R_{ambient}(x,y,x,y) = R_{surface}(x,y,x,y) + A(x,y)^{2}-A(x,x)A(y,y)

where A(x,y) is the normal component of the covariant derivative of y with respect to x.

The Gauss curvature of the surface thus satisfies the equation,

G = R_{ambient}(x,y,x,y)/|x^y|^{2} - (A(x,y)^{2}-A(x,x)A(y,y))/|x^y|^{2}

If the sectional curvature of the ambient manifold is constant,K, as in Eudlidean space or some other flat 3 manifold or in hyperbolic 3 space then this equation becomes

G = K - (A(x,y)^{2}-A(x,x)A(y,y))/|x^y|^{2}

If k = 0 this is the usual equation for the Gauss curvature with respect to the determinant of the second fundamental form.

For the horosphere, G = 0 so the normalized determinant of the second fundamental form is equal to 1.

It is important to realize though that this equation in no way says that if the surface is a sphere that G can be identically 0. G is still the intrinsic curvature no matter how it is embedded in another manifold.

 

[TrickyDicky]

For a surface embedded in a 3 dimensional Riemannian manifold calculation shows that the Riemann curvature tensors of the 3 dimensional manifold and the surface are related by a formula.

R_{ambient}(x,y,x,y) = R_{surface}(x,y,x,y) + A(x,y)^{2}-A(x,x)A(y,y)

where A(x,y) is the normal component of the covariant derivative of y with respect to x.

The Gauss curvature of the surface thus satisfies the equation,

G = R_{ambient}(x,y,x,y)/|x^y|^{2} - (A(x,y)^{2}-A(x,x)A(y,y))/|x^y|^{2}

If the sectional curvature of the ambient manifold is constant,K, as in Eudlidean space or some other flat 3 manifold or in hyperbolic 3 space then this equation becomes

G = K - (A(x,y)^{2}-A(x,x)A(y,y))/|x^y|^{2}

If k = 0 this is the usual equation for the Gauss curvature with respect to the determinant of the second fundamental form.

For the horosphere, G = 0 so the normalized determinant of the second fundamental form is equal to 1.

It is important to realize though that this equation in no way says that if the surface is a sphere that G can be identically 0. G is still the intrinsic curvature no matter how it is embedded in another manifold.

Thanks for the formula, it is actually a tensorial form to express the formula I brought from the Cartan book in post #10.

 

[lavinia]

Thanks for the formula, it is actually a tensorial form to express the formula I brought from the Cartan book in post #10.

Another example of this is the flat torus embedded in the 3 sphere. Here G = 0 so the normalized determinant of the second fundamental form is constantly -1.

I wonder if this means that an observer in the three sphere looking at a flat torus would see it as negatively curved - assuming light followed geodesics in the 3 sphere.

 

[TrickyDicky]

Another example of this is the flat torus embedded in the 3 sphere. Here G = 0 so the normalized determinant of the second fundamental form is constantly -1.

I wonder if this means that an observer in the three sphere looking at a flat torus would see it as negatively curved - assuming light followed geodesics in the 3 sphere.

I think there is some sign problem with your formulas, can you give some reference of where you got them.
I would have expected the torus to look negatively curved for an observer in the hyperbolic manifold case rather than in the 3-sphere where I would imagine it would look extrinsically spherical. Are you sure you meant embedded in the 3-sphere?

In R^3 the Riemannian curvature of the surface (what you call G??) is the determinant of the second fundamental form(shape tensor).
This is generalized to arbitrary ambient curvature embeddings by adding the constant ambient curvature K to the determinant of the second fundamental form.

 

[TrickyDicky]

Accordingly I would expect a horospherical surface in a hyperbolic closed (compactified by the Mobius group ) 3-manifold 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.
It seems that my not using consistently the term manifold (using H^3 space instead) arose confusion.
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?

 

[lavinia]

I think there is some sign problem with your formulas, can you give some reference of where you got them.

i derived the formulas. The calculation is simple. try it. There is no sign error that I see.

I would have expected the torus to look negatively curved for an observer in the hyperbolic manifold case rather than in the 3-sphere where I would imagine it would look extrinsically spherical. Are you sure you meant embedded in the 3-sphere?

I am sure. I just meant that the normalized determinant of the second fundamental for is -1 since the sectional curvature of the 3 sphere is +1. I am not sure how someone in the 3 sphere would view the flat torus but this can be figured out pretty easily.

In R^3 the Riemannian curvature of the surface (what you call G??) is the determinant of the second fundamental form(shape tensor).
This is generalized to arbitrary ambient curvature embeddings by adding the constant ambient curvature K to the determinant of the second fundamental form.

The generalization only works for three manifolds of constant sectional curvature. In general there is no constant and there is no single normal component to the covariant derivative.

For instance if a surface is embedded in R^4, there is a plane of normal directions and the normal component of the covariant derivative in general lies in this plane not on any single normal direction.

 

[lavinia]

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

This language confuses me.

The horosphere is homeomorphic to R^2. It can be given a metric of zero or constant positive or constant negative Gauss curvature. As a subset of H^3 it has zero Gauss curvature.

It seems that my not using consistently the term manifold (using H^3 space instead) arose confusion.
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?

The horosphere is not compact. I think Ben has explained this in detail.

 

[TrickyDicky]

The horosphere is not compact. I think Ben has explained this in detail.

I think he was under the impression I was referring exclusively to the Poincare ball model which is a representation of the H^3 space (and in that case he was right the horospheres are not compact), instead of the hyperbolic Riemannian 3-manifold that is a quotient of the H^3 space by a Kleinian subgroup so it it includes the points at infinity in contrast to just the H^3 space.
Wikipedia entry on the horoball says:"In hyperbolic geometry, a horoball is an object in hyperbolic n-space: the limit of a sequence of increasing balls sharing (on one side) a tangent hyperplane and its point of tangency. Its boundary is called a horosphere.
A horosphere has a critical amount of (isotropic) curvature."
The theorem of Micallef-Moore says that a compact simply connected manifold of positive isotropic curvature is homeomorphic to the sphere.