Are there Riemanniann manifolds with zero curvature other than open subsets of $\mathbb{R}^n \times \mathbb{T}^m$, where $\mathbb{T}^m$ is an $m$ dimensional torus and $m,n\geq 0$ ?

Does taking quotients of opens in $\mathbb{R}^n \times \mathbb{T}^m$ by the action of some -possibly discrete- Lie group provide new examples (i.e. that are not themselves isometric to opens in a Torus x Euclideanspace)? [assuming we're in a case where the quotient is a manifold, and we can induce a metric on it]

Does taking (universal or not) coverings enlarge the class of the above examples?

More generally: is there a classification of flat not necessarily complete flat Riemannian manifolds?

If we added "complete", would we obtain only the $\mathbb{R}^n \times \mathbb{T}^m$ 's ?


If you take a quotient of an annulus by a finite-order rotation, you get an example that is diffeomorphic (but not isometric) to a planar annulus. You can do the same in higher dimension with finite subgroups of $\mathrm{O}(n)$. Also, the universal cover of an annulus is diffeomorphic to the plane, but not isometric to a subset of it: it does not contains any infinite line, but contains infinite simple curves of constant curvature (the lift of any homotopically non-trivial circle).

Any complete flat manifold is a quotient of $\mathbb{R}^n$: its universal cover is indeed flat (thus CAT(0)) and simply connected, therefore a ball. They can be classified, and are called Bieberbach manifolds. Some of them are not diffeomorphic to product $\mathbb{R}^k\times\mathbb{T}^l$. The simplest example would be a flat Klein bottle. If you want an orientable manifold, go one dimension higher: a cube with two pairs of opposite sides glued through a translation, and the third pair identified by translation composed with a rotation of angle $\pi$ does the job.

| cite | improve this answer | |
  • $\begingroup$ As for your first example, now I realize that probably a cone with a closed neighbourhood of the vertex chopped off would be an example of a non-planar thing obtained "as an open in a quotient". $\endgroup$ – Qfwfq Apr 15 '10 at 19:21
  • $\begingroup$ As for your second example, what do you mean by "infinite line" and "simple curves"? Thank you. $\endgroup$ – Qfwfq Apr 15 '10 at 19:22
  • $\begingroup$ In your paragraph "Any complete flat [...]", do you also mean that such manifolds can be obtained as quotients by a discrete group? $\endgroup$ – Qfwfq Apr 15 '10 at 19:25
  • $\begingroup$ Also, could I ask you a question: which is the meaning of "CAT(0)"? $\endgroup$ – Qfwfq Apr 15 '10 at 19:26
  • $\begingroup$ Well, an infinite line is simply a infinite geodesic, or straight line (flat metrics being locally isomorphc to $\mathbb{R}^n$, a flat manifolds inherits an affine structure so affine line make sense). Similarly, the usual notion of curvature is defined locally in the plane (up to an orientation), so that this notion makes sense in a flat manifold. This should clarify the notion of constant curvature curve; "simple" means that the curve does not cross itself. $\endgroup$ – Benoît Kloeckner Apr 15 '10 at 20:17

Concerning the non-complete case, there is no hope for a classification.

Any local diffeomorphism $f:M\to\mathbb R^n$ (non-injective and so on) induces a flat metric on $M$, this gives a lot of examples that are not subsets of $\mathbb R^n$ or anything.

But there are good news: every simply connected flat manifold can be obtained this way (i.e. admits a locally isometric map to $\mathbb R^n$).

There are more examples that are not simply connected. Consider any 2-dimensional polyhedral surface (e.g. the boundary of a convex 3-polytope). Remove the vertices and you get a flat Riemannian metric on a 2-manifold.

| cite | improve this answer | |
  • $\begingroup$ Concerning the polyhedral example: if you remove only the vertices from a surface, aren't you left with the "singularities" given by the edges? $\endgroup$ – Qfwfq Apr 15 '10 at 20:15
  • $\begingroup$ No, the intrinsic metric is locally Euclidean, only the embedding into $\mathbb R^3$ is non-smooth. Consider folding a plane along a line - it is still a plane intrinsically. $\endgroup$ – Sergei Ivanov Apr 15 '10 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.