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Recall that in a Riemannian manifold (or pseudo Riemannian) there is always the unique Levi-Civita connexion that annuls the torsion.

There are also manifolds (not needfully Riemannian) which are curvature free, thus the deviation from Euclideanhood is encoded wholly in the torsion tensor.

Question 1: Are there known sufficient, or necessary, or both necessary and sufficient conditions conditions for a curvature-free connexion (weaken assumptions to e.g. a Finsler manifold if need be) to be defined?

Question 2: Are there known sufficient, or necessary, or both necessary and sufficient conditions conditions that rule out a curvature-free connexion?

Question 3: Now, one thing that is bending my mind is: what happens to the holonomy group if we can have a curvature-free connexion? Obviously the holonomy group is trivial for a curvature free manifold. This seems to imply to me that somehow all connexions for that manifold must be curvature free, because there is no way to "continuously deform" a Lie group from the trivial group to a nontrivial one. Is this right? If not, how does the holonmy group "jump" from being trivial to something nontrivial?

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I think this pretty much answers your questions 1 and 2.

As for question 3, I am not sure I understand what's troubling you. First of all, note that you can't necessarily always find a "continuous deformation" between two given connections. More to the point, sure, if you start with a flat connection and you deform it so that it's no longer flat, the holonomy group will "instantaneously" become non trivial. Said differently, the flat connections are special points in the space of connections where the holonomy group collapses.

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  • $\begingroup$ It seems from the MO discussion that this is not a trivial question. WRT to question 3, granted: there isn't always a cts deformation but can you give an example of where there is a cts deformation and where the holonomy group "collapses"? It's not altogether clear that it does, but not unreasonable either. $\endgroup$ – WetSavannaAnimal Dec 16 '13 at 3:33
  • $\begingroup$ Yes, your questions 1 and 2 seem difficult (much easier if we were talking about Riemannian connections). For question 3, I'll try to think of an "explicit" example but computing the holonomy is difficult in general. But let me say this: take the Euclidean metric on the plane for instance, then under a generic deformation of the metric, the holonomy will be $SO(2)$. $\endgroup$ – Seub Dec 16 '13 at 3:40
  • $\begingroup$ wonderful, clear example, Seub, thanks muchly. Getting an actual example might be a bit harder: I guess I've piqued your curiousity now. Let me know if you come up with anything and I'll ask a question you can write up an answer to. Interesting about Milnor proving "flat" not possible for surfaces of Euler characteristic $\chi\neq 0$: not unexpected, but really nice that the abstract definition of curvature corresponds to intuition. $\endgroup$ – WetSavannaAnimal Dec 16 '13 at 3:54
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It is false that a flat Levi-Civita connection on a manifold necessarily has zero holonomy group. The theorem of Bieberbach - I think - is that the holonomy group must be finite.

For instance,the flat Klein bottle has holonomy Z2.

Here is a 3 dimensional flat orientable Riemannian manifold with holonomy group,Z4.

Take the quotient of R^3 with the standard flat metric by the group of isometries generated by the standard lattice( vectors all of whose coordinates are integers) together with the isometry (x,y,z) -> (x +1/4, z,-y)

It is probably true that any finite group can be the holonomy group of a flat Riemannian manifold.

The classifying theorem is that any compact flat Riemannian manifold without boundary is the quotient of R^n by a torsion free group that satisfies an exact sequence

o -> L -> G ->G/L -> 1

where L is a full lattice and G/L is a finite group. G/L is isomorphic to the holonomy group of the quotient flat manifold. The converse is also true. Any such group is the fundamental group of a flat Riemannian manifold.

There are flat Riemannian manifolds in every dimension as can be seen by taking the Cartesian product of the Klein bottle with flat tori.

I believe that there is no known method for identifying all of the flat manifolds. There may be classifications in low dimensions such as dimension 3 or 4. In dimension 2 the only ones are the flat torus and the flat Klein bottle.

The overwhelming difficulty in classification is that there is no way to classify the inequivalent integral representations of finite groups. Even very small groups - with only several exceptions such as Z2- have infinitely many inequivalent Z representations. This comes up because the quotient group G/L acts on the lattice,L, by conjugation (since the lattice is a normal subgroup of G) and this action is an integral representation. I say this just to illustrate that the condition of a flat Levi - Civita connection is extremely complex.

Even though the Pontryagin classes and Euler class of a flat Riemannian manifold must be zero - by Chern-Weil theory - the Stiefel- Whitney classes do not.For instance the first Stiefel-Whitney class of the Klein bottle is not zero. I would be surprised if there is a theorem restricting the possible Stiefel Whitney classes except that the top class must be zero.

Finally I would point out that the Moebius band made out of a piece of paper is a flat manifold with boundary and its holonomy group is Z2.

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  • $\begingroup$ Great answer: I wasn't aware that a flat Levi-Civita connexion isn't altogether "anholonomic", but the finite holonomy group is an interesting "almost anholonomic" idea that I haven't yet thought about. Wonderful examples, many thanks. And your comment re difficulty in classification is pretty clear: you're talking the Novikov-Boone theorem or something like this, right? $\endgroup$ – WetSavannaAnimal Dec 16 '13 at 4:03
  • $\begingroup$ The proof that there are infinitely many inequivalent Z representations of almost all finite groups I think is due to someone named Jones. I have no idea how the proof goes. $\endgroup$ – lavinia Dec 16 '13 at 4:09
  • $\begingroup$ Also even if you have a Z representation in hand it still may be difficult to find all of the torsion free group extensions of the lattice by the finite group. This is a problem in group cohomology theory. $\endgroup$ – lavinia Dec 16 '13 at 4:14

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