# Riemannian surfaces as one dimensional complex manifolds

1. Oct 31, 2011

### TrickyDicky

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.

2. Oct 31, 2011

### Ben Niehoff

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.

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

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.

3. Nov 1, 2011

### lavinia

- 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.

Last edited: Nov 1, 2011
4. Nov 1, 2011

### TrickyDicky

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.

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.

5. Nov 1, 2011

### TrickyDicky

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.

6. Nov 1, 2011

### lavinia

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.

Last edited: Nov 1, 2011
7. Nov 1, 2011

### TrickyDicky

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.

8. Nov 1, 2011

### lavinia

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

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.

9. Nov 1, 2011

### lavinia

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.

10. Nov 1, 2011

### TrickyDicky

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.

11. Nov 1, 2011

### Ben Niehoff

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.

12. Nov 1, 2011

### TrickyDicky

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.

13. Nov 1, 2011

### lavinia

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?

14. Nov 1, 2011

### Ben Niehoff

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 $8 \pi$.

15. Nov 1, 2011

### Ben Niehoff

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?

16. Nov 1, 2011

### lavinia

Arbitrary dimensions.

17. Nov 2, 2011

### TrickyDicky

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."

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.

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?

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"

18. Nov 2, 2011

### lavinia

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

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

19. Nov 2, 2011

### TrickyDicky

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

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)"

20. Nov 2, 2011

### lavinia

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

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.

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.

Last edited: Nov 2, 2011