# Sufficient condition for global flatness

1. Oct 28, 2012

### tom.stoer

given a compact, orientable, n-dim. Riemann manifold, what is a sufficient condition for globally vanishing curvature i.e. global flatness?

I can get necessary conditions from the generalized Gauss-Bonnet theorem, but not sufficient ones

Last edited: Oct 28, 2012
2. Oct 28, 2012

### lavinia

I think that a Levi-Civita connection is flat if and only if paralllel translation around null homotopic loops is the identity map. I think one can show that this means that the global holonomy group is finite. In any case, it is true that if the Riemannian connection is flat then the holonomy group is finite.

For instance the flat Klein bottle has holonomy group equal to Z/2Z and the flat torus has trivial holonomy group. One can easily generalize the Klein bottle to make an orientable three manifold whose holonomy group is Z/2Z.

I also think that you should be able to show that the exponential map is a covering of the manifold by Euclidean space and is a local isometry. It you find a proof let me know.

The necessary condition that the Gauss Bonnet form is zero is only part of the story. All of the Pontryagin forms are also zero. However, the Stiefel-Whitney classes may not be zero as in the case of the Klein bottle. I will try to construct a flat 4 manifold that has non-zero second Stiefel_Whitney class.

BTW: If the connection is not compatible with any Riemannain metric then it may be flat and still have non-zero Euler class. In such a case, the Euler class can not be expressed in term of the curvature 2 form,

Last edited: Oct 28, 2012
3. Oct 29, 2012

### Vargo

If your space is geodesically complete, then it must be a space form:
See http://en.wikipedia.org/wiki/Space_form

In other words, the space must be Euclidean modulo a crystallographic group of isometries.

Spaces that are not geodesically complete are open subsets of geodesically complete spaces?? I think this is true, but I can't recall the proof. See the Hopf Rinow theorem.

4. Oct 29, 2012

### lavinia

The fundamental group must be an extension of a full lattice by a finite group. It must be torsion free and the action of the finite group on the lattice(under conjugation) is a faithful Z-representation of the finite group. The finite group is isomorphic to the holonomy group of the manifold.

Another way of saying this is that the manifold is the quotient of a flat torus by a properly discontinuous action of a finite group of isometries.

This means that the manifold is a K(π,1), that is all of its homotopy groups are zero except its fundamental group.

5. Oct 29, 2012

### tom.stoer

This is great, thanks; I have to invest some time to understand the reasoning and proof behind the space form, but the result is clear and simple

6. Oct 30, 2012

### lavinia

- Zero holonomy for null homotopic loops should be the easiest.

- It is clear that the exponential map can have no singularities since there are no conjugate points. But this would also be true for manifolds of negative curvature.

- A properly discontinuous finite group of isometries on a flat torus clearly makes a flat manifold and the fundamental group is an extension the the lattice of covering transformations of the torus. If any of the isometries is a pure translation I suppose you can just mod out by them first to get another flat torus and then you are left with a finite group that acts non-trivially on the torus's tangent plane. The group of isometries of Euclidean space that covers this manifold is clearly torsion free.

Last edited: Oct 30, 2012