# No globally flat geometry on S²

1. Sep 5, 2010

### tom.stoer

Quick question: from which topological consideration can one derive the fact that a sphere S² does not allow for a globally flat geometry?

2. Sep 5, 2010

### quasar987

Because it would mean that there is a diffeomorphism S^2 -->R^2. But continuous maps preserve compactness, so no such map exists.

3. Sep 5, 2010

Well, an infinite cyllinder has a globally flat geometry, but there is no diffeomorphism from the cylinder onto R^2.

But you may check this ion Wikipedia: "Vector fields on spheres". Global flatness of S^2 would imply existence of two nowhere vanishing vector fields.

Last edited: Sep 5, 2010
4. Sep 5, 2010

### quasar987

Ok, I thought globally flat meant there exists a global isometry from S^2 + some metric to R^2 + usual metric.

5. Sep 5, 2010

### lavinia

A flat manifold must have Euler characteristic zero.

This is the content of the Gauss Bonnet theorem.

One can prove that every compact flat Riemannian manifold is covered by flat euclidean space by the action of a group of isometries. Such groups are known as Bieberbach groups and are extensions of lattices by finite groups.

The sphere on the other hand can not be covered by anything but itself because it is simply connected. So it can not be flat.

The only compact flat surfaces without boundary are the flat torus and the flat Klein bottle. In each dimension there are finitely many compact flat Riemannian manifolds. I know that all of the 3 manifolds have been determined and maybe all of the 4 manifolds as well - not sure. Generally, it is daunting to try to calculate all of the flat manifolds in a high dimension.

More generally whenever you have a flat bundle (in this case the tangent bundle of the flat manifold) whose connection is consistent with a Riemannian metric, the Euler class of the bundle is zero. On the other hand there are examples of flat bundles - even over surfaces - whose connections do no derive from a metric and whose Euler class is not zero.

A simple realization of the flat torus in the unit 3 sphere is the set of points (1/2^.5)(sin x, cos x, sin y, cos y).
So the flat torus can be embedded in 4 dimensions. It would be interesting to know if the same is true of the flat Klein bottle - maybe not.

Last edited: Sep 5, 2010
6. Sep 5, 2010

### lavinia

Globally flat means that the curvature tensor is identically zero.

A global isometry would be a diffeomorphism. The flat torus is not diffeomorphic to R^2.

Last edited: Sep 5, 2010
7. Sep 6, 2010

### tom.stoer

@lavinia: thanks for reminding me to these facts

Let me see if I understand. The Euler characteristic of S² is 2. This can be seen by using a triangulation on S² or by calculating

$$\chi_{S^2} = 2 - 2g_{S^2}$$

and using

$$g_{S^2} = 0$$

But by Gauss-Bonnet we know that

$$\chi_{S^2} = \frac{1}{2\pi}\int_{S^2}K$$

Global flatness would imply $$K = 0$$ globally which results in $$\chi_{S^2} = 0$$ which is a contradiction.

Last edited: Sep 6, 2010
8. Sep 6, 2010

### lavinia

right. the Gauss Bonnet theorem works in all dimensions. There is a differential form which is a polynomial in the curvature 2 form that equals the Euler class of the tangent bundle.

9. Sep 6, 2010

### lavinia

generally global flatness only implies the existence of one nowhere vanishing vector field. On an oriented surface, on can rotate one vector field to get another but generally there is no way to do this. On an unorientable surface for instance, this does not work. The Klein bottle has only one independent non zero vector field and yet it admits a flat metric.

Last edited: Sep 6, 2010
10. Sep 6, 2010

### lavinia

On flat surfaces, there is not only a non-zero vector field but the field can be chosen to always be tangent to geodesics - the projections of lines in the plane into the surface. I wonder if this always true of a flat manifold. It is certainly true if the first Betti number of the manifold is not zero.

11. Sep 6, 2010

S^2 is two-dimensional, connected and simply connected. You choose two orthonormal tangent vectors at one point and transport them using parallel transport to every other point. This transport is path independent (zero curvature plus topology). This way you would get two orthonormal vector fields. While by Adams' theorem even one does not exist.

12. Sep 6, 2010

### tom.stoer

Thanks again.

13. Sep 6, 2010

### lavinia

this is wrong. There are no independent vector fields on the 2 sphere. Further the sphere does not have zero curvature.
There are no simply connected compact flat Riemannian manifolds.

14. Sep 6, 2010

### tom.stoer

By inspection of 2-2g = 0 only g=1 allowes for globally flat geometries.

Another related question: The Gauss-Bonnet theorem restricts the existence of globally flat geometries. What about the other way round? Suppose I have a manifold for which I can proof that the Euler characteristic vanishes. Does this automatically guarantuee that a globally flat geometry must exist? I guess not, but what are the other restrictions?

15. Sep 6, 2010

1) That is what I was saying, assuming hypothetically global flatness - as in the original question. Proof by contradiction. There is not even one.

2) Whether a given manifold is globally flat or not depends on the Riemannian metric. R^2 can be made globally flat and can be made non-flat even locally. Up to you.

16. Sep 6, 2010

### lavinia

Absolutely not. There are many manifolds of zero Euler characteristic.

Take any manifold and take its Cartesian product with a circle.

If a compact manifold is flat then all of its Stiefel Whitney numbers must be zero. But there are many non-flat manifolds of zero euler charateristic and all zero Stiefel Whitney numbers so this does not suffice.

the mod 2 reduction of the Euler characteristic is the top Stiefel Whitney number but there are many other Stiefel Whitney numbers as well.

Last edited: Sep 6, 2010
17. Sep 6, 2010

### lavinia

OK. Didn't understand what you were getting at.

A simply connected globally flat manifold can not be compact. The only example is Euclidean space.

18. Sep 6, 2010

### lavinia

I doubt that there are conditions on Characteristic classes that guarantee that the manifold has a flat a metric. For instance the sphere cartesian product with a circle has zero Euler characteristic and all zero Stiefel Whitney numbers.

Flat manifolds have finite holonomy group - so the structure group of the tangent bundle is finite - this is an extremely strict condition.