# Riemannian surfaces as one dimensional complex manifolds

If we consider a Riemannian surface as a one-dimensional complex manifold, what does that tell us about its intrinsic curvature? I mean for one-dimensional curves we know they only have extrinsic curvature so it depends on the embedding space, this doesn't seem to be the case for one-dimensional complex manifolds, but I'd like to understand this better.

On the other hand, thinking of Riemannian surfaces as 2-dimensional real manifolds they inmediately have an intrinsic curvature that determines to a certain extent their topology (number of handles:g=0 positive curvature,g=1 flat, g>1 negative). Or said differently, in these manifolds the topology can be determined by their metric, I think this is a property of manifolds up to 3 real dimensions at least for the constant curvature cases but I'm not completely sure. Maybe someone could clarify. Certainly in the GR 4-manifold the metric does not give us the general topology.

Ben Niehoff
Gold Member
If we consider a Riemannian surface as a one-dimensional complex manifold, what does that tell us about its intrinsic curvature?

Having a complex structure means that the structure group of the manifold is reduced from $GL(2n, \mathbb{R})$ to $GL(n, \mathbb{C})$. This means that the curvature 2-form is only allowed to take values within this subgroup.

On a Riemann surface, $SO(2) \simeq U(1)$, so the structure group reduction is tautologous. This is just another way to state the obvious fact that all Riemann surfaces have a complex structure.

However, since this is tautologous, the fact that there is a complex structure tells you nothing about the curvature. There is one independent curvature in 2 dimensions --- the Ricci scalar --- and it can be anything whatsoever.

I mean for one-dimensional curves we know they only have extrinsic curvature so it depends on the embedding space, this doesn't seem to be the case for one-dimensional complex manifolds, but I'd like to understand this better.

1-complex-dimensional curves have 2 real dimensions, hence they can have intrinsic curvature given by the Ricci scalar.

On the other hand, thinking of Riemannian surfaces as 2-dimensional real manifolds they inmediately have an intrinsic curvature that determines to a certain extent their topology (number of handles:g=0 positive curvature,g=1 flat, g>1 negative). Or said differently, in these manifolds the topology can be determined by their metric, I think this is a property of manifolds up to 3 real dimensions at least for the constant curvature cases but I'm not completely sure. Maybe someone could clarify. Certainly in the GR 4-manifold the metric does not give us the general topology.

The metric never tells you everything about the global topology. For example, a flat metric might correspond to an infinite plane, an infinite cylinder, or a torus. Each of these has different first homotopy group.

However, on a closed 2-manifold, you can calculate the Euler characteristic by integrating the Ricci scalar, yes. In higher dimensions, it is a very non-trivial task to find useful topological invariants that can be expressed as integrals of the curvature.

There are some topological invariants that cannot be expressed this way. For example, the property of having a complex structure in even dimensions > 2 is highly-nontrivial. It is still an open question whether some manifolds can be given complex structures, for example the 6-sphere.

lavinia
Gold Member
- For oriented compact surfaces, the Euler characteristic tells you the topology of the surface. The Euler characteristic can be calculated from any Riemannian metric by integrating its Gauss curvature with respect to the volume form given by the metric.

In higher dimensions Euler characteristic does not determine the topology of the manifold even though it can still be calculated from the curvature. For instance the Euler characteristic of any 3 dimensional manifold is zero.

In higher dimensions one has a curvature tensor, rather than just the Gauss curvature. Knowing this tensor still does not tell you the topology. For instance in three dimensions there are 30 something different compact oriented Riemannian manifolds whose curvature tensor is identically zero.

- In this part I am not completely sure if this is right but .....While requiring that coordinate transformations be analytic is the actual definition of a complex manifold, on a surface, a complex manifold may be thought of as a real manifold with a multiplication by i on each tangent plane. Such a multiplication is a linear map on each tangent plane whose square is multiplication by -1.

A surface together with a complex structure is a Riemann surface. This idea does not require the idea of a metric. It is really a different idea. A given topological surface has many inequivalent complex structures. So it can be many different complex manifolds.

However, a Riemannian metric does determine a complex structure on a surface. To multiply a tangent vector by i, just rotate it so that the ordered pair v,iv is positively oriented. One can show that the surface has a coordinate system in which the metric looks like multiplication by a scalar - so called isothermal coordinates - and that the coordinate transformations between isothermal coordinates are analytic. Thus the metric determines a conformal structure on the surface. However, many metrics determine the same conformal structure.

- The Gauss curvature does not tell you the metric. On a torus for instance there are many conformal structures whose metric has Gauss curvature zero.

One can see from this that there is more to Riemann surfaces than geometry and topology. There is also its possible conformal(complex) structures. Study of conformal structures is a whole field of mathematics.

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Hi, Ben. Your answer gives a more rigorous explanation of what I meant to express.
Just a couple of points to further clear up what I was trying to get at.

However, since this is tautologous, the fact that there is a complex structure tells you nothing about the curvature. There is one independent curvature in 2 dimensions --- the Ricci scalar --- and it can be anything whatsoever.

1-complex-dimensional curves have 2 real dimensions, hence they can have intrinsic curvature given by the Ricci scalar.

..on a closed 2-manifold, you can calculate the Euler characteristic by integrating the Ricci scalar
This is correct, and close to my point.
Let me use a specific Riemannian surface as an example: the Riemann sphere (extended complex plane), and borrow the words of the WP page on the Riemann sphere (Metric section):
QUOTE
"A Riemann surface does not come equipped with any particular Riemannian metric. However, the complex structure of the Riemann surface does uniquely determine a metric up to conformal equivalence. (Two metrics are said to be conformally equivalent if they differ by multiplication by a positive smooth function.) Conversely, any metric on an oriented surface uniquely determines a complex structure, which depends on the metric only up to conformal equivalence. Complex structures on an oriented surface are therefore in one-to-one correspondence with conformal classes of metrics on that surface.
Within a given conformal class, one can use conformal symmetry to find a representative metric with convenient properties. In particular, there is always a complete metric with constant curvature in any given conformal class.
In the case of the Riemann sphere, the Gauss-Bonnet theorem implies that a constant-curvature metric must have positive curvature K. It follows that the metric must be isometric to the sphere of radius 1/sqroot of k in R^3 via stereographic projection.
Conversely, let S denote the sphere (as an abstract smooth or topological manifold). By the uniformization theorem there exists a unique complex structure on S. It follows that any metric on S is conformally equivalent to the round metric. All such metrics determine the same conformal geometry. The round metric is therefore not intrinsic to the Riemann sphere, since "roundness" is not an invariant of conformal geometry. The Riemann sphere is only a conformal manifold, not a Riemannian manifold. However, if one needs to do Riemannian geometry on the Riemann sphere, the round metric is a natural choice." End Quote.

So the positive curvature of the Riemann sphere metric is just a "natural" choice, when the metric must be isometric to a sphere in R^3 . But in certain circumstances other choices could be more natural. I'm thinking for instance about hyperbolic space where bijective conformal maps of the Riemann sphere to itself are isomorphic to the group of orientation-preserving isometries of H^3. In this case the natural choice of metric for the extended complex plane would be a flat metric, and in fact the conformal boundary of hyperbolic 3-manifold is usually described in terms of euclidean geometry.

Thanks Lavinia, your post helps me see it clearer.

In higher dimensions Euler characteristic does not determine the topology of the manifold even though it can still be calculated from the curvature. For instance the Euler characteristic of any 3 dimensional manifold is zero.
I have always found intriguing the fact that all compact odd-manifolds have Euler characteristic=0.

lavinia
Gold Member
QUOTE

So the positive curvature of the Riemann sphere metric is just a "natural" choice, when the metric must be isometric to a sphere in R^3 . But in certain circumstances other choices could be more natural. I'm thinking for instance about hyperbolic space where bijective conformal maps of the Riemann sphere to itself are isomorphic to the group of orientation-preserving isometries of H^3. In this case the natural choice of metric for the extended complex plane would be a flat metric, and in fact the conformal boundary of hyperbolic 3-manifold is usually described in terms of euclidean geometry.

Could you explain this? I just am not sure what you are getting at.

I suspect that the amazing fact is that no matter what the conformal structure, the Riemann surface admits a metric of constant curvature. A priori, one might suspect that there are conformal structures that do not admit metrics of constant curvature.

For the torus this implies that every 2 dimensional torus admits a metric of curvature zero. So you can have differomorphic manifolds that are everywhere locally isometric yet are not globally isometric.

Also: The extended complex plane does not admits a flat metric - if by flat you mean zero Gauss curvature.

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The extended complex plane does not admits a flat metric - if by flat you mean zero Gauss curvature.

Did you see the quote from wikipedia. When it says: "The round metric is therefore not intrinsic to the Riemann sphere." I understand that an abstract conformal Riemann sphere (wich is the extended complex plane) admits not only the positive curvature metric but other constant curvature metrics (up to conformal equivalence) like the euclidean metric.

lavinia
Gold Member
Did you see the quote from wikipedia. When it says: "The round metric is therefore not intrinsic to the Riemann sphere." I understand that an abstract conformal Riemann sphere (wich is the extended complex plane) admits not only the positive curvature metric but other constant curvature metrics (up to conformal equivalence) like the euclidean metric.

I think this is the quote from the article that you mean.

All such metrics determine the same conformal geometry. The round metric is therefore not intrinsic to the Riemann sphere, since "roundness" is not an invariant of conformal geometry.

It is saying that all metrics on the sphere determine the same conformal structure - since there is only one. Thus the round metric isn't special since it does not determine a special conformal structure.

BTW: Up to a scalar there is only one metric of constant curvature on the sphere. But it has many metrics of non-constant positive curvature. I wonder if all of these can be realized by embeddings of the sphere into 3 dimensions.

lavinia
Gold Member
Thanks Lavinia, your post helps me see it clearer.

I have always found intriguing the fact that all compact odd-manifolds have Euler characteristic=0.

All odd dimensional compact manifolds without boundary have Euler characteristic zero. 3 - manifolds are not special. But there is a stronger fact for three manifolds which is that they are all parallelizable. I don't know how to prove this.

Thus the round metric isn't special.
Fine but how does this contradict my example:in a hyperbolic embedding, the conformal structure of the complex manifold we are dealing with admits as natural choice a euclidean metric.
Maybe this can be understood in simple geometric terms:
According to Cartan's "Geometry of Riemannian spaces" ,1946 page 190 the gaussian curvature of a closed surface in R^3 can be related to the gaussian curvature in other embeddings of different constant curvature by the following simple formula, in wich K is gaussian curvature in the new embedding and k is the relative gaussian curvature of a point in the surface( k=1/r1r2), we have K=k+curvature of the embedding space wich in the example I chose is -1, thus for a sphere with relative gaussian curvature of 1 K=1-1=0 , that is, a euclidean metric surface.

Ben Niehoff
Gold Member
The possible metrics that can be put on Riemann surfaces are constrained by the Gauss-Bonnet formula (i.e., that the integral of the Ricci scalar is the Euler characteristic).

A sphere cannot have a globally flat metric. You can, however, push all the curvature to a single point. Using stereographic projection to the infinite plane, you can then use the standard flat metric on the infinite plane. However, this coordinate patch covers every point on the sphere except one: the north pole. If you compute carefully, you will find that all the curvature has been pushed to this single point, and the Ricci scalar is a delta function located on the north pole. That is, you have made a conical singularity.

There are more familiar ways to do this. Take a cube for example. A cube is just a sphere where all the curvature has been pushed to 8 conical singularities at the corners (if you think carefully, the edges are not singularities).

Similarly, Riemann surfaces with $\chi < 0$ can be given metrics that push all the curvature to a finite number of points. But you must count those delta functions; you can't ignore them. The only Riemann surface that admits a globally flat metric is the torus, because in this case, $\chi = 0$

If you want to find metrics of globally constant curvature, then the sphere will have to be positively curved, and the surfaces with $\chi < 0$ will have to be negatively curved.

I emphasize the word "global" because you have to remember that a metric is a local object. It only applies to a single coordinate patch. Surfaces of nontrivial topology cannot be covered by a single coordinate patch. So while you are free to "sweep the curvature away" and put a flat metric on any given open region U, that is not the same as saying the curvature is actually gone; it has merely moved outside U.

Also be careful when people talk about "embeddings". Any old embedding is not necessarily an isometric embedding, nor even necessarily a smooth embedding! For example, the cube is a non-smooth embedding of the sphere into R^3.

The possible metrics that can be put on Riemann surfaces are constrained by the Gauss-Bonnet formula (i.e., that the integral of the Ricci scalar is the Euler characteristic).

A sphere cannot have a globally flat metric. You can, however, push all the curvature to a single point. Using stereographic projection to the infinite plane, you can then use the standard flat metric on the infinite plane. However, this coordinate patch covers every point on the sphere except one: the north pole. If you compute carefully, you will find that all the curvature has been pushed to this single point, and the Ricci scalar is a delta function located on the north pole. That is, you have made a conical singularity.

There are more familiar ways to do this. Take a cube for example. A cube is just a sphere where all the curvature has been pushed to 8 conical singularities at the corners (if you think carefully, the edges are not singularities).

Similarly, Riemann surfaces with $\chi < 0$ can be given metrics that push all the curvature to a finite number of points. But you must count those delta functions; you can't ignore them. The only Riemann surface that admits a globally flat metric is the torus, because in this case, $\chi = 0$

If you want to find metrics of globally constant curvature, then the sphere will have to be positively curved, and the surfaces with $\chi < 0$ will have to be negatively curved.

I emphasize the word "global" because you have to remember that a metric is a local object. It only applies to a single coordinate patch. Surfaces of nontrivial topology cannot be covered by a single coordinate patch. So while you are free to "sweep the curvature away" and put a flat metric on any given open region U, that is not the same as saying the curvature is actually gone; it has merely moved outside U.

Also be careful when people talk about "embeddings". Any old embedding is not necessarily an isometric embedding, nor even necessarily a smooth embedding! For example, the cube is a non-smooth embedding of the sphere into R^3.

This is all right but I fail to see its relation with what is being said in my post.
I would only add that I think the gauss-bonnet formula constraint you mention is referring to euclidean space and does not generalize to arbitrary ambient spaces, and also that it is the metric that induces the topology and not the other way around.

lavinia
Gold Member
The possible metrics that can be put on Riemann surfaces are constrained by the Gauss-Bonnet formula (i.e., that the integral of the Ricci scalar is the Euler characteristic).

A sphere cannot have a globally flat metric. You can, however, push all the curvature to a single point. Using stereographic projection to the infinite plane, you can then use the standard flat metric on the infinite plane. However, this coordinate patch covers every point on the sphere except one: the north pole. If you compute carefully, you will find that all the curvature has been pushed to this single point, and the Ricci scalar is a delta function located on the north pole. That is, you have made a conical singularity.

There are more familiar ways to do this. Take a cube for example. A cube is just a sphere where all the curvature has been pushed to 8 conical singularities at the corners (if you think carefully, the edges are not singularities).

Similarly, Riemann surfaces with $\chi < 0$ can be given metrics that push all the curvature to a finite number of points. But you must count those delta functions; you can't ignore them. The only Riemann surface that admits a globally flat metric is the torus, because in this case, $\chi = 0$

If you want to find metrics of globally constant curvature, then the sphere will have to be positively curved, and the surfaces with $\chi < 0$ will have to be negatively curved.

Ben can ou explain why the edges of the cube are not singularities/ Is it because the curvature is zero there, so the delta function is zero?

What about taking the exponential map on any Riemannian manifold. Can one push all the curvature out to the cut locus where if will be a delta function and leave the rest of the manifold flat?

Ben Niehoff
Gold Member
Tricky:

The Gauss-Bonnet formula has nothing to do with ambient spaces. The Ricci scalar is an intrinsic curvature and the Euler characteristic is an intrinsic topological invariant. Therefore the Gauss-Bonnet formula is intrinsically true regardless of the ambient space.

The relevance to your previous business about embedding spheres in hyperbolic spaces in order to somehow put flat metrics on them was to explain that you are wrong. You cannot put a flat metric on a sphere no matter what the ambient space is, because the Gauss-Bonnet formula is a statement about facts intrinsic to the sphere itself. The best you can hope to do is make the sphere locally flat everywhere except at a finite number of points.

Also, the topology is more fundamental than the metric. This is an important thing for people coming from differential geometry to realize. A manifold is a topological object. It has topological features (such as handles, etc.) that exist independently of any local definition of "distance".

When we say a "sphere", we mean a closed 2-dimensional manifold of Euler characteristic 2. There is no need to make any reference to metrics. Then one can ask, what sorts of metrics can be put on a sphere? The Gauss-Bonnet formula gives the only constraint: any metric whose total Ricci curvature is $8 \pi$ (the Ricci scalar is exactly twice the Gauss curvature). Some parts of the manifold might have R = 0 or even R < 0, so long as the integral over the whole manifold is $8 \pi$.

Ben Niehoff
Gold Member
Ben can ou explain why the edges of the cube are not singularities/ Is it because the curvature is zero there, so the delta function is zero?

The edges can be smoothed out. Or in other words, "yes".

This is easy to see by unfolding the cube and tracing the paths of geodesics. There is no bending of geodesics at the edges of the cube.

What about taking the exponential map on any Riemannian manifold. Can one push all the curvature out to the cut locus where if will be a delta function and leave the rest of the manifold flat?

I'm not sure. Are you talking in arbitrary dimensions, or are we still in 2 dimensions?

lavinia
Gold Member
The edges can be smoothed out. Or in other words, "yes".

This is easy to see by unfolding the cube and tracing the paths of geodesics. There is no bending of geodesics at the edges of the cube.

I'm not sure. Are you talking in arbitrary dimensions, or are we still in 2 dimensions?

Arbitrary dimensions.

Tricky:

The Gauss-Bonnet formula has nothing to do with ambient spaces. The Ricci scalar is an intrinsic curvature and the Euler characteristic is an intrinsic topological invariant. Therefore the Gauss-Bonnet formula is intrinsically true regardless of the ambient space.
This is apparently true, but if you think more carefully about it it could be a moot point. The Gauss-Bonnet formula has several formulations. And some of them are conditioned by the ambient space.
In the Wolfram mathworld page http://mathworld.wolfram.com/Gauss-BonnetFormula.html that I think is considered a reliable source in general, they mention several of these formulations: according to them "the simplest one expresses the total Gaussian curvature of an embedded triangle...".

The second one they mention is probably the most known and I think the one you are referring to, is explained like this:the ".. most common formulation of the Gauss-Bonnet formula is that for any compact, boundaryless two-dimensional Riemannian manifold, the integral of the Gaussian curvature over the entire manifold with respect to area is 2pi times the Euler characteristic of the manifold".

The third formulation is introduced with the sentence "Another way of looking at the Gauss-Bonnet theorem for surfaces in three-space" wich leaves room to think that the previous formulation is also for surfaces in in ambient three-space, and closes the paragraph about this third formulation with these words: "Singer and Thorpe (1996) give a 'Gauss's theorema egregium-inspired' proof which is entirely intrinsic, without any reference to the ambient Euclidean space." which also leaves room to think that up to this point the formulas were referring to extrinsic versions of the theorem and this is not contradicted in any way by the fact that the formula contains intrinsic quantities like the Gaussian curvature and the Euler topological invariant.

Anyway I should have made clear from the beguinning that my claims referred to compact immersed submanifolds and the extrinsic version of the Gauss-Bonnet theorem, my fault for not being precise about this . This for instance is taken from a math paper:
"For compact immersed submanifolds M in euclidean spaces, the well known
extrinsic version of the Gauss-Bonnet theorem states that the total Lipschitz-Killing curvature of M is equal to the Euler characteristic χ(M) of M.
For compact immersed submanifolds in hyperbolic spaces, the picture completely
changes: the total Lipschitz-Killing curvature of M is not equal to the Euler
characteristic of M."

The relevance to your previous business about embedding spheres in hyperbolic spaces in order to somehow put flat metrics on them was to explain that you are wrong. You cannot put a flat metric on a sphere no matter what the ambient space is, because the Gauss-Bonnet formula is a statement about facts intrinsic to the sphere itself. The best you can hope to do is make the sphere locally flat everywhere except at a finite number of points.
See above. Also I was centering on the conformal structure of the Riemann sphere, before we can say it is a sphere so to speak.

Also, the topology is more fundamental than the metric.
True. But would you agree that in Riemann surfaces (specifically this one, the simplest of the Riemann surfaces) the complex structure is even more fundamental than the metric induced topology?

When we say a "sphere", we mean a closed 2-dimensional manifold of Euler characteristic 2. There is no need to make any reference to metrics. Then one can ask, what sorts of metrics can be put on a sphere? The Gauss-Bonnet formula gives the only constraint: any metric whose total Ricci curvature is $8 \pi$ (the Ricci scalar is exactly twice the Gauss curvature). Some parts of the manifold might have R = 0 or even R < 0, so long as the integral over the whole manifold is $8 \pi$.
I agree on this, just remember that I'm addressing a step previous to calling it a sphere. That's why I insist on using the term "extended complex plane" for the Riemann "sphere"

lavinia
Gold Member
Fine but how does this contradict my example:in a hyperbolic embedding, the conformal structure of the complex manifold we are dealing with admits as natural choice a euclidean metric.

This is wrong. You are confusing intrinsic with extrinsic curvature I think.

Maybe this can be understood in simple geometric terms:
According to Cartan's "Geometry of Riemannian spaces" ,1946 page 190 the gaussian curvature of a closed surface in R^3 can be related to the gaussian curvature in other embeddings of different constant curvature by the following simple formula, in wich K is gaussian curvature in the new embedding and k is the relative gaussian curvature of a point in the surface( k=1/r1r2), we have K=k+curvature of the embedding space wich in the example I chose is -1, thus for a sphere with relative gaussian curvature of 1 K=1-1=0 , that is, a euclidean metric surface.

The embedding space is not the sphere. It is Euclidean space.

This is wrong. You are confusing intrinsic with extrinsic curvature I think.
Huh? How so? Did you read my previous post?

The embedding space is not the sphere. It is Euclidean space.
I don't know what you are talking about, in my specific example the embedding space is hyperbolic with curvature -1, and the sphere intrinsic curvature (1/r1r2) =1

If you read the reference I gave of the book by Cartan (it is in google books pages 189 and 190), this is explained there, hope you don't think Cartan confuses intrinsic with extrinsic curvature.:

"The intrinsic Riemannian curvature of a surface at one of its points M (I called this K) is equal to the Riemannian curvature of the ambient space at M in the direction of the plane element tangent to the surface, augmented by the total curvature (the product of the principal curvatures) of the surface at M. (I called it k)"

lavinia
Gold Member
Huh? How so? Did you read my previous post?

I don't know what you are talking about, in my specific example the embedding space is hyperbolic with curvature -1, and the sphere intrinsic curvature (1/r1r2) =1

The sphere can never ever have a flat metric no matter how it is embedded. its extrinsic curvature can be zero.

If you read the reference I gave of the book by Cartan (it is in google books pages 189 and 190), this is explained there, hope you don't think Cartan confuses intrinsic with extrinsic curvature.:

No but I think you do. I read the post. Sadly you are being vague in your posts so I have to guess what you are really talking about. What I and Ben have said are true.

Perhaps you are confusing the fact that for embeddings you can deduce the Gauss curvature from the sphere map at least if the surface is embedded in dimension 3. In dimension 4 there is no sphere map but there is a generalization. This does not change the fact that Gauss curvature is intrinsic - which means that it can be deduced from the metric on the tangent space.

"The intrinsic Riemannian curvature of a surface at one of its points M (I called this K) is equal to the Riemannian curvature of the ambient space at M in the direction of the plane element tangent to the surface, augmented by the total curvature (the product of the principal curvatures) of the surface at M. (I called it k)"

This merely says that the intrinsic curvature can be deduced from the shape operator and the curvature of the ambient manifold. It does not in any way mean that the Gauss curvature of the sphere can be flat.

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The sphere can never ever have a flat metric no matter how it is embedded. its extrinsic curvature can be zero.
Once again I'm not talking about the sphere, only about the conformal structure of the extended complex plane.

No but I think you do. I read the post. Sadly you are being vague in your posts so I have to guess what you are really talking about.

But this only shows you are either guessing wrong or trying to use a straw man. Or both.
This does not change the fact that Gauss curvature is intrinsic - which means that it can be deduced from the metric on the tangent space.
But where have I said Gaussian curvature is not intrinsic?

This merely says that the intrinsic curvature can be deduced from the shape operator and the curvature of the ambient manifold. It does not in any way mean that the Gauss curvature of the sphere can be flat.
Once again, you read things in my posts I have not stated. I never said that Gaussian curvature of the sphere can be flat. Gaussian curvature is intrinsic, that's precisely the reason that in a different embedding with different geometric relations a surface curvature that remains invariant can admit a different metric in the new embedding. The conformal structure of the extended complex plane allows this.

lavinia
Gold Member
Once again I'm not talking about the sphere, only about the conformal structure of the extended complex plane.
The extended complex plane is a sphere. Whatever metric you give it, it will not be flat.
But this only shows you are either guessing wrong or trying to use a straw man. Or both.

Whatever

Once again, you read things in my posts I have not stated. I never said that Gaussian curvature of the sphere can be flat. Gaussian curvature is intrinsic, that's precisely the reason that in a different embedding with different geometric relations a surface curvature that remains invariant can admit a different metric in the new embedding. The conformal structure of the extended complex plane allows this.

Have no idea what this means.

Ok, Lavinia, I'm sure you know a lot more than I do about geometry, I just feel you are not making any effort to see my point. But I might be wrong, I'm just trying to learn.

Let's use an example: there is a well known object (a surface) in hyperbolic geometry called the horosphere, it is also a well known geometric fact that the geometry on these horospheric surfaces is euclidean. Horospheres have the particular property of being tangent to infinity. So here we have a geometrical 2D-object that having euclidean geometry can be thought of as containing the complex plane numbers and that also contains a point at infinity. This is quite similar to the complex extended plane, which is the ℂ $\cup$ ∞ if we think of ℂ as ℝ^2
This is purely heuristic so there might be very obvious reasons why the extended complex plane in hyperbolic ambient space has nothing to do with a horospheric surface and can't share metric, but I'm interested in having someone pointing it out to me.

Ben Niehoff
Gold Member
Ok, Lavinia, I'm sure you know a lot more than I do about geometry, I just feel you are not making any effort to see my point. But I might be wrong, I'm just trying to learn.

Let's use an example: there is a well known object (a surface) in hyperbolic geometry called the horosphere, it is also a well known geometric fact that the geometry on these horospheric surfaces is euclidean. Horospheres have the particular property of being tangent to infinity. So here we have a geometrical 2D-object that having euclidean geometry can be thought of as containing the complex plane numbers and that also contains a point at infinity. This is quite similar to the complex extended plane, which is the ℂ $\cup$ ∞ if we think of ℂ as ℝ^2
This is purely heuristic so there might be very obvious reasons why the extended complex plane in hyperbolic ambient space has nothing to do with a horospheric surface and can't share metric, but I'm interested in having someone pointing it out to me.

So then tell me, what's the Ricci curvature of such a horosphere at the point at infinity itself?

Two important points (using the Poincare disk model for clarity):

1. The points on the boundary of the disk are not part of hyperbolic space. Hyperbolic space is the open region within the disk

2. In the hyperbolic metric, the boundary of the disk is infinitely far away from any given point in the interior.

So these "horospheres" are in fact infinite planes. They do not close. Alternatively, if you add the point at infinity, I guarantee you will find all the curvature piled up there in a delta function.

So then tell me, what's the Ricci curvature of such a horosphere at the point at infinity itself?
I think it is the same curvature as in the rest of the surface, just like the point at infinity in the riemann sphere is no special point, just the north pole.
But anyway my point is that a submanifold surface metric is determined by the embedding 3-manifold so it makes little sense to calculate the Ricci tensor of the submanifold, the Ricci curvature in this case is referred to the 3D manifold.

Two important points (using the Poincare disk model for clarity):

1. The points on the boundary of the disk are not part of hyperbolic space. Hyperbolic space is the open region within the disk
This is correct, it is also probably correct that if we consider it in terms of a hyperbolic manifold the boundary can be considered to belong to the hyperbollic manifold in certain cases (there are loads of hyperbolic 3-manifolds with very complicated and different properties.

2. In the hyperbolic metric, the boundary of the disk is infinitely far away from any given point in the interior.
Exactly.

So these "horospheres" are in fact infinite planes. They do not close.
In the non-euclidean literature they are usually considered "limit spheres with origin at infinity".

Alternatively, if you add the point at infinity, I guarantee you will find all the curvature piled up there in a delta function.
Not exactly but it is a way to see it. The curvature is recovered if the ambient space is Euclidean back again. I think the key point here is that the submanifold Riemann sphere conformal structure admits any metric up to conformal equivalence, the curvature is in the change from ambient space with curvature 0 to the ambient space with curvature -1, that to preserve angles and geometric relations changes the surface curvature from 1 to 0.
The important thing to remember here is that we are talking about submanifolds of either euclidean space or hyperbolic space, not about the proper Riemannian sphere wich is intrinsically defined to have positive curvature in the absence of any higher dimension embedding space or in the euclidean ambient space that having 0 curvature doesn't alter the geometric relations between manifold and submanifold.

lavinia
Gold Member
Not exactly but it is a way to see it.

Why isn't it exact?

The curvature is recovered if the ambient space is Euclidean back again.

What does this mean? Explain and give an example.

I think the key point here is that the submanifold Riemann sphere conformal structure admits any metric up to conformal equivalence,

By the submanifold Riemann sphere conformal structure I am guessing that you mean the conformal structure that the sphere inherits from the manifold that it is embedded in, though I have no idea why you would call an embedded sphere the Riemann sphere.

But why do you say that the conformal structure admits any metric up to conformal equivalence? What does that mean? Are you saying that any two metrics on the sphere are conformally equivalent? That certainly must be false.

the curvature is in the change from ambient space with curvature 0 to the ambient space with curvature -1, that to preserve angles and geometric relations changes the surface curvature from 1 to 0.

I don't understand this. What do you mean by surface curvature?

The important thing to remember here is that we are talking about submanifolds of either euclidean space or hyperbolic space, not about the proper Riemannian sphere wich is intrinsically defined to have positive curvature in the absence of any higher dimension embedding space or in the euclidean ambient space that having 0 curvature doesn't alter the geometric relations between manifold and submanifold.

The sphere does not have positive curvature intrinsically. It can have regions of negative curvature and regions where it is flat. Intrinsic means that the curvature can be derived from the metric on the tangent space.

What does this mean? Explain and give an example.
I already did in previous posts and gave a reference from Cartan on how to use a formula to obtain the submanifold curvatures determined by the ambient space.

By the submanifold Riemann sphere conformal structure I am guessing that you mean the conformal structure that the sphere inherits from the manifold that it is embedded in, though I have no idea why you would call an embedded sphere the Riemann sphere.
Yes, I mean that.
Why would you not call it a Riemann sphere? Most of the explanations I have read about why the extended complex plane manifold is called a Riemann sphere rely on embedding the manifold in Euclidean space to see how such a sphere (located on the plane) points correspond with a complex plane by stereographic projection.

Taken from Wikipedia #Riemann sphere article

Why would you not call this a Riemann sphere?

But why do you say that the conformal structure admits any metric up to conformal equivalence?
Because I read in WP that : "the complex structure of the Riemann surface does uniquely determine a metric up to conformal equivalence and two metrics are said to be conformally equivalent if they differ by multiplication by a positive smooth function. It follows that any metric on S is conformally equivalent to the round metric. All such metrics determine the same conformal geometry. The round metric is therefore not intrinsic to the Riemann sphere, since "roundness" is not an invariant of conformal geometry. The Riemann sphere is only a conformal manifold, not a Riemannian manifold"
But if the way I say it confuses you, just refer to the quoted WP paragraph.

What does that mean? Are you saying that any two metrics on the sphere are conformally equivalent? That certainly must be false.
Certainly not any two metrics, only those with constan curvature as explained in the WP paragraph.

I don't understand this. What do you mean by surface curvature?
In this case it can be understood in the terms you used of extrinsic curvature of the embedded one-dimensional complex manifold.

The sphere does not have positive curvature intrinsically.
The 2-sphere as a surface has according to Gauss and many after him an intrinsic positive gaussian curvature. This is too basic for anyone to deny it.

The possible metrics that can be put on Riemann surfaces are constrained by the Gauss-Bonnet formula (i.e., that the integral of the Ricci scalar is the Euler characteristic).
To lead this thread to a sensible end let me recover this point for a moment.
This quote as stated is true. But some considerations can be added for the case we are dealing with. As shown in a previous post extrinsic and intrinsic versions of the Gauss-Bonnet formula can be distinguished depending on whether we are talking about an embedded submanifold (extrinsic version) or a surface without any reference to higher dimensions (intrinsic version, the one commonly used and the one Ben is referring to in the quoted paragraph).
And for a Riemann surface embedded in hyperbolic 3-space (in this case the extended complex plane) the extrinsic version of Gauss-Bonnet constrains the possible metrics that can be assigned to the Riemann sphere differently than it is the case if it was embedded in Euclidean space (where the only metric it admits the positive curvature one).

Can we all agree on this at least?

Ben Niehoff
Gold Member
Can we all agree on this at least?

No, this is complete nonsense.

As shown in a previous post extrinsic and intrinsic versions of the Gauss-Bonnet formula can be distinguished depending on whether we are talking about an embedded submanifold (extrinsic version) or a surface without any reference to higher dimensions (intrinsic version, the one commonly used and the one Ben is referring to in the quoted paragraph).

The Gauss-Bonnet formula does not come in "extrinsic and intrinsic versions". I don't know how this particular confusion of yours arose. The Gauss-Bonnet formula is purely an intrinsic statement.

The entire point of Gauss's Theorema Egregium is that the Gauss curvature, while expressed as a product of extrinsic curvatures $k_1, k_2$, is in fact an intrinsic property. Essentially, Gauss discovered the Ricci scalar on surfaces.

Since every manifold is locally like R^n and extrinsic curvatures are a local property, it doesn't matter whether you embed a surface in R^3 or in H^3: The Gauss curvature, though it may be expressed as the product of the principle extrinsic curvatures, is still an intrinsic property of the surface.

And for a Riemann surface embedded in hyperbolic 3-space (in this case the extended complex plane) the extrinsic version of Gauss-Bonnet constrains the possible metrics that can be assigned to the Riemann sphere differently than it is the case if it was embedded in Euclidean space (where the only metric it admits the positive curvature one).

Absolutely not.

A horosphere in hyperbolic space is NOT a topological sphere. As I mentioned earlier, the boundary sphere (in the Poincare ball model) is not part of hyperbolic space; hyperbolic space is the open region within the boundary sphere. So in particular, the point of tangency between the horosphere and the boundary sphere is not actually in hyperbolic space. Therefore, speaking of the horosphere strictly as a submanifold of hyperbolic space, it is actually a sphere with one point removed; not a full sphere. A sphere with one point removed is homeomorphic to the infinite plane. Hence a "horosphere" is really just an isometric embedding of R^2 into H^3.

You mentioned the fact that there are other hyperbolic 3-manifolds that are closed. One example is the Seifert-Weber space. But these spaces do not contain entire horospheres: they contain only finite sections of them; i.e., plane segments.

No, this is complete nonsense.
The Gauss-Bonnet formula does not come in "extrinsic and intrinsic versions". I don't know how this particular confusion of yours arose.
It arose in several places like math journals written by people that like writing complete nonsense. For instance:
http://www.igt.uni-stuttgart.de/LstDiffgeo/Kuehnel/preprints/totalcurv.pdf
jus take a look at the introduction.
Or: http://www.crm.es/Publications/08/Pr805.pdf
page 18
just to mention a couple that are freely available.
A horosphere in hyperbolic space is NOT a topological sphere. As I mentioned earlier, the boundary sphere (in the Poincare ball model) is not part of hyperbolic space; hyperbolic space is the open region within the boundary sphere. So in particular, the point of tangency between the horosphere and the boundary sphere is not actually in hyperbolic space. Therefore, speaking of the horosphere strictly as a submanifold of hyperbolic space, it is actually a sphere with one point removed; not a full sphere. A sphere with one point removed is homeomorphic to the infinite plane. Hence a "horosphere" is really just an isometric embedding of R^2 into H^3.

You mentioned the fact that there are other hyperbolic 3-manifolds that are closed. One example is the Seifert-Weber space. But these spaces do not contain entire horospheres: they contain only finite sections of them; i.e., plane segments.
This is nonsense (clearly hyperbolic geometry is not your field of expertise) and unrelated to what I wrote.

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Ben Niehoff
Gold Member
It arose in several places like math journals written by people that like writing complete nonsense. For instance:
http://www.igt.uni-stuttgart.de/LstDiffgeo/Kuehnel/preprints/totalcurv.pdf
jus take a look at the introduction.
Or: http://www.crm.es/Publications/08/Pr805.pdf
page 18
just to mention a couple that are freely available.

OK, I see that the Gauss-Bonnet formula can be re-written in terms of various extrinsic curvatures, that's kinda neat! But the point remains that what it is calculating (i.e. the Euler characteristic) is an invariant, intrinsic property (a fact restated many times in those very papers!). This really shouldn't be any surprise; as I've already pointed out, the Gauss curvature itself can be written in terms of extrinsic quantities, and yet it is intrinsic.

In any case, the Euler characteristic of a 2-surface M can always be calculated via

$$2 \pi \, \chi(M) = \int_M K \, dV + \int_{\partial M} k_g \, ds$$
and every quantity appearing here is intrinsic to M; i.e., does not depend on whatever space M might be embedded in.

If a horosphere is a topological sphere then it must have $\chi = 2$, so obviously this will come by paying careful attention to the boundary term (since the Gauss curvature of a horosphere is zero).

This is nonsense (clearly hyperbolic geometry is not your field of expertise) and unrelated to what I wrote.

It is unfortunate that you don't see the relevance, but without pointing out any specific issue, I can't help you much.

You have a habit of misusing mathematical terminology and neglecting to give examples of exactly what you're talking about. Perhaps this is why you get responses that you think are irrelevant.

Earlier you said

But anyway my point is that a submanifold surface metric is determined by the embedding 3-manifold so it makes little sense to calculate the Ricci tensor of the submanifold, the Ricci curvature in this case is referred to the 3D manifold.

which leads me to believe you don't quite understand what's going on. The Euler characteristic is an intrinsic property of the surface, so why do you keep harping on about ambient 3-manifolds? If you have some embedding into an ambient 3-manifold, then it serves one purpose: it induces a metric on the 2-surface. Then this metric can be used to compute the Ricci scalar on the 2-surface, which can be integrated to give the Euler characteristic.

These papers you're reading show that you can also calculate the Euler characteristic via some other routes. The answer will still be the same, though. In fact I would say that's the whole point of those papers.

Ben Niehoff
Gold Member
So, because I'm bored, let's actually calculate the Euler characteristic of a horosphere in H^3. Using the Poincare ball model and spherical coordinates, the metric of hyperbolic space is

$$ds^2 = \frac{4}{(1-r^2)^2} (dr^2 + r^2 \, d\theta^2 + r^2 \, \sin^2 \theta \, d\phi^2)$$
Now, choose a horosphere tangent to the boundary sphere at $(x,y,z) = (0,0,1)$ having the equation

$$r = \cos \theta$$
Hence $dr^2 = \sin^2 \theta \, d\theta^2$, and the induced metric can be written

$$ds_{\mathrm{ind}}^2 = \frac{4}{\sin^4 \theta} (d\theta^2 + \cos^2 \theta \, \sin^2 \theta \, d\phi^2)$$
Now make the change of coordinates $\rho = 2 \cot \theta$, $d\rho^2 = (4/\sin^4 \theta) \, d\theta^2$ to get

$$ds_{\mathrm{ind}}^2 = d\rho^2 + \rho^2 \, d\phi^2$$
The range of $\theta$ was $(0, \pi/2]$, and so the range of $\rho$ is $[0, \infty)$.

This is the standard flat metric on R^2. Therefore the Gauss curvature is zero, and the Euler characteristic can be calculated from the boundary term. We can assume a circular boundary that grows to infinite radius:

\begin{align} 2 \pi \, \chi(M) &= \int_M K \, dV + \int_{\partial M} k_g \, ds \\ &= 0 + \lim_{\rho \to \infty} \int_0^{2\pi} \frac{1}{\rho} \, \rho \, d\theta \\ &= 2 \pi \\ \chi(M) &= 1 \end{align}
And so, as a submanifold of hyperbolic space, a horosphere is NOT a topological sphere, but rather an infinite R^2, just as I said earlier.

The reason for this is precisely the reason I gave earlier: the "point at infinity" that would have closed the sphere is not a point belonging to hyperbolic space. Hence no submanifold of hyperbolic space can contain that point!

The catch here is that the boundary sphere of the Poincare ball model is not actually a boundary of H^3. After all, H^3 is infinite in extent and has no boundary. Rather, the boundary sphere of the Poincare ball model is what's called a "conformal boundary" of H^3. This means the boundary only shows up after using a conformal transformation to map H^3 into a region of finite volume. (The Poincare ball coordinates can be considered such a map.)

The overall lesson is don't be fooled by appearances. A horosphere "looks like" a sphere in Poincare ball coordinates, but in fact it is an infinite flat plane, and is neither compact nor closed!

Hi, Ben

OK, I see that the Gauss-Bonnet formula can be re-written in terms of various extrinsic curvatures, that's kinda neat!
Glad you finally see this. Remember you previously deemed it as "total nonsense".

But the point remains that what it is calculating (i.e. the Euler characteristic) is an invariant, intrinsic property (a fact restated many times in those very papers!). This really shouldn't be any surprise; as I've already pointed out, the Gauss curvature itself can be written in terms of extrinsic quantities, and yet it is intrinsic.

In any case, the Euler characteristic of a 2-surface M can always be calculated via

$$2 \pi \, \chi(M) = \int_M K \, dV + \int_{\partial M} k_g \, ds$$
and every quantity appearing here is intrinsic to M; i.e., does not depend on whatever space M might be embedded in.
You are saying this as if you thought I believe otherwise, this leads me to think that I haven't been able yet to get my point thru to you.
Please remember that quantity K depends on the metric and the metric when we are talking of submanifold structures that admit more than one metric can give rise to K=1 but also K=0 or K=-1.

If a horosphere is a topological sphere then it must have $\chi = 2$, so obviously this will come by paying careful attention to the boundary term (since the Gauss curvature of a horosphere is zero).
This confirms you haven't understood what my point is, I never implied a horosphere is a topological sphere, on the contrary my claim was that it had euclidean metric.

You have a habit of misusing mathematical terminology and neglecting to give examples of exactly what you're talking about.
This I admit freely, I'll try to improve on it.

The Euler characteristic is an intrinsic property of the surface, so why do you keep harping on about ambient 3-manifolds?
I have insisted several times that I think the metric induces the topology on the conformal complex structure, so if we are talking about complex structures immersed in a 3-space that imposes a metric on the complex structure up to conformal equivalence (that is, not any metric can be imposed in complex structure of ℂ$\cup$∞, only constant curvature metrics).

These papers you're reading show that you can also calculate the Euler characteristic via some other routes. The answer will still be the same, though. In fact I would say that's the whole point of those papers.
Those papers also claim that makes a difference to calculate it in euclidean or hyperbolic ambient space, but it would help me to know exactly how you interpret what is written in page 18 of the second reference I linked.

So, because I'm bored, let's actually calculate the Euler characteristic of a horosphere in H^3...
This is the standard flat metric on R^2. Therefore the Gauss curvature is zero, and the Euler characteristic can be calculated from the boundary term. We can assume a circular boundary that grows to infinite radius:

And so, as a submanifold of hyperbolic space, a horosphere is NOT a topological sphere, but rather an infinite R^2, just as I said earlier.

The overall lesson is don't be fooled by appearances. A horosphere "looks like" a sphere in Poincare ball coordinates, but in fact it is an infinite flat plane, and is neither compact nor closed!
First, I'd like to say that I used the horosphere as an example only based on intuition, and I said that obviously I'm not sure if it is a valid example of the extended complex plane in hyperbolic space.
But the key point is that I didn't use it as an example of a topological sphere as you keep saying. Your calculation confirms what I said about horospheres having euclidean metric.
I'd like to call on mathematicians that might be lurking to help me on this and say something about whether an abstract extended complex plane in hyperbolic space can be identified with a horosphere. (Not with a topological sphere)

The reason for this is precisely the reason I gave earlier: the "point at infinity" that would have closed the sphere is not a point belonging to hyperbolic space. Hence no submanifold of hyperbolic space can contain that point!
You are right it doesn't belong to the hyperbolic space, but it can belong to the hyperbolic 3-manifold wich is a quotient space of H^3/$\Gamma$, that is hyperbolic 3-space over the conformal transformations of the Riemann sphere (Kleinian group), and it is equipped with a complete Riemannian metric, being complete is analogous to a closed set:contains all its limit point including the point at infinity.
Thinking about this perhaps the horosphere is not the object that is equivalent to the extended complex plane but the very boundary of the hyperbolic 3-manifold would be a good candidate. Certainly it wouldn't belong to hyperbolic space but to the hyperbolic manifold.

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lavinia
Gold Member
Please remember that quantity K depends on the metric and the metric when we are talking of submanifold structures that admit more than one metric can give rise to K=1 but also K=0 or K=-1.

By K do you mean the Gauss curvature of the surface? If the sumanifold is homeomorphic to an open disk as I guess the horosphere is, then it can have metrics of constant positive, negative of zero Gauss curvature.

I have insisted several times that I think the metric induces the topology on the conformal complex structure, so if we are talking about complex structures immersed in a 3-space that imposes a metric on the complex structure up to conformal equivalence (that is, not any metric can be imposed in complex structure of ℂ$\cup$∞, only constant curvature metrics).

- a metric does not induce a topology
- on a surface a conformal structure is the same as a complex structure
- the extended complex plane can have metrics of non-constant Gauss curvature that it inherits from a manifold that it is embedded in.
- if the sphere is immersed rather than embedded I would like to see a proof that it can inherit a metric of constant Gauss curvature.
[/QUOTE]

But the key point is that I didn't use it as an example of a topological sphere as you keep saying. Your calculation confirms what I said about horospheres having euclidean metric.
I'd like to call on mathematicians that might be lurking to help me on this and say something about whether an abstract extended complex plane in hyperbolic space can be identified with a horosphere. (Not with a topological sphere)

The extended complex plane - as a topological space - is a topological sphere.
Thinking about this perhaps the horosphere is not the object that is equivalent to the extended complex plane but the very boundary of the hyperbolic 3-manifold would be a good candidate. Certainly it wouldn't belong to hyperbolic space but to the hyperbolic manifold.

I don't know anything about these manifolds but if the boundary is covered by the bounding plane of H^3 then it can not be a sphere and therefore can not be the extended complex plane unless you mean something different than what is usually meant when you say extended complex plane.

BTW: You sai that all metrics on the sphere are conformally equivalent. Without a reference, can you give a proof of this?

Maybe it is not so hard. Choose any conformal diffeomorphism of the 2 spheres. Then maybe in isothermal coordinates the diffeomorphism is actually a conformal mapping of the metrics.

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I notice that when talking about submanifolds I've been using the term embedded when I meant immersed, several times and that might lead to confusion, sorry about that.
(Though, I think when referring to Riemannian geometry and Riemannian manifolds like in this case an isometric embedding is an immersion between Riemannian manifolds which preserves the Riemannian metrics)

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