Riemannian surfaces as one dimensional complex manifolds

[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.

 

[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.