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## Sufficient condition for global flatness

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

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 Quote by 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 thanks in advance

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,
 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.

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## Sufficient condition for global flatness

 Quote by 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.
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.
 Recognitions: Science Advisor 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

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