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 ?
Best Answer
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.